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AMC12_740 | Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \le x \le \frac{\pi}{2}$ and $0 \le y \le \frac{\pi}{2}$ . What is the area of the subset of $S$ for which \[\sin^2x-\sin x \sin y + \sin^2y \le \frac34?\]
$\mathrm{(A)}\ \dfrac{\pi^2}{9} \qquad \mathrm{(B)}\ \dfrac{\pi^2}{8} \qquad \mathrm{(C)}\ \dfrac{\pi^2}{6} \qquad \mathrm{(D)}\ \dfrac{3\pi^2}{16} \qquad \mathrm{(E)}\ \dfrac{2\pi^2}{9}$
| We start out by solving the equality first.
\begin{align*} \sin^2x - \sin x \sin y + \sin^2y &= \frac34 \\ \sin x &= \frac{\sin y \pm \sqrt{\sin^2 y - 4 ( \sin^2y - \frac34 ) }}{2} \\ &= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\ &= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\ &= \frac12 \sin y \pm \frac{\sqrt3}{2} \cos y \\ \sin x &= \sin (y \pm \frac{\pi}{3}) \end{align*}
We end up with three lines that matter: $x = y + \frac\pi3$ , $x = y - \frac\pi3$ , and $x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y$ . We plot these lines below.
[asy] size(5cm); D((0,0)--(3,0)--(3,3)--(0,3)--cycle); D((1,-0.1)--(1,0.1)); D((2,-0.1)--(2,0.1)); D((-0.1,1)--(0.1,1)); D((-0.1,2)--(0.1,2)); D((2,0)--(3,1)--(1,3)--(0,2)); MP("\frac{\pi}{6}", (1,0), plain.S); MP("\frac{\pi}{3}", (2,0), plain.S); MP("\frac{\pi}{2}", (3,0), plain.S); MP("\frac{\pi}{6}", (0,1), plain.W); MP("\frac{\pi}{3}", (0,2), plain.W); MP("\frac{\pi}{2}", (0,3), plain.W); [/asy]
Note that by testing the point $(\pi/6,\pi/6)$ , we can see that we want the area of the pentagon . We can calculate that by calculating the area of the square and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.)
\begin{align*} A &= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac12 \cdot \left(\frac{\pi}{6}\right)^2 - \frac12 \cdot \left(\frac{\pi}{3}\right)^2 \\ &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\ &= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\text{(C)}\ \frac{\pi^2}{6}} \end{align*}
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_24 | C |
AMC12_746 | Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$ . What is $\cos(\angle CMD)$ ?
$\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}$
| Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt3.$
Let $O$ be the center of $\triangle ABD,$ so $\overline{CO}\perp\overline{MOD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$
In right $\triangle CMO,$ we have \[\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.\]
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_12 | 1/3 |
AMC12_747 | Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$ . The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?
$\textbf{(A) }\dfrac{\sqrt3}4\qquad \textbf{(B) }\dfrac{\sqrt3}3\qquad \textbf{(C) }\dfrac23\qquad \textbf{(D) }\dfrac{\sqrt2}2\qquad \textbf{(E) }\dfrac{\sqrt3}2$
| Reflect each of the triangles over its respective side. Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle. Hence the desired answer is just its circumradius, or $\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}$ .
(Solution by djmathman)
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_10 | $\frac{\sqrt{3}}{3}$ |
AMC12_749 | The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$
| Setting the first two equations equal to each other, $\log_3 x = \log_x 3$ .
Solving this, we get $\left(3, 1\right)$ and $\left(\frac{1}{3}, -1\right)$ .
Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$ .
Now, by setting the first and third equations equal to each other, we get $\left(1, 0\right)$ .
Pairing the first and fourth or second and third equations won't work because then $\log x \leq 0$ .
Pairing the second and fourth equations will yield $x = 1$ , but since you can't divide by $\log 1 = 0$ , it doesn't work.
After trying all pairs, we have a total of $5$ solutions $\rightarrow \boxed{\textbf{(D)} 5}$
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_16 | 5 |
AMC12_750 | What is $\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)$ ?
$\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020$
| $20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_1 | 40 |
AMC12_752 | The number of $x$ -intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to
$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$
| The function $f(x) = \sin x$ has roots in the form of $\pi n$ for all integers $n$ . Therefore, we want $\frac{1}{x} = \pi n$ on $\frac{1}{10000} \le x \le \frac{1}{1000}$ , so $1000 \le \frac 1x = \pi n \le 10000$ . There are $\frac{10000-1000}{\pi} \approx \boxed{2900} \Rightarrow \mathrm{(A)}$ solutions for $n$ on this interval.
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_23 | 2900 |
AMC12_754 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$
| First, remove all the 90s, since they make no impact. So, we have numbers from $1$ to $10$ . Then, $5$ is the 7th number. Let the sum of the first $6$ numbers be $k$ . Then, $k\equiv 0 \mod 6$ and $k\equiv 3 \mod 7$ . We easily solve this as $k \equiv 24 \mod 42$ . Clearly, the sum of the first $6$ numbers must be $570$ . In order for the first $5$ numbers to sum to a multiple of $5$ , the sixth number must also be a multiple of $5$ , since $30$ is a multiple of $5$ . Thus, the only option is the sixth number is $10$ , which gives $10+90= \boxed{\textbf{(E) }100}$
~ firebolt360
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | 100 |
AMC12_756 | At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?
$(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15$
| Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\frac{1}{2} \cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 \Rightarrow \mathrm{(A)}$ .
Because there are five days, or four transformations between days (day 1 $\rightarrow$ day 2 $\rightarrow$ day 3 $\rightarrow$ day 4 $\rightarrow$ day 5), she makes $48 \cdot \frac{1}{2^4} = \boxed{\mathrm{(A)}\ 3}$
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_1 | 3 |
AMC12_761 | The area of a pizza with radius $4$ is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$ ?
$\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$
| The area of the larger pizza is $16\pi$ , while the area of the smaller pizza is $9\pi$ . Therefore, the larger pizza is $\frac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\frac{7\pi}{9\pi} \cdot 100\% = 77.777....$ , which is closest to $\boxed{\textbf{(E) }78}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_1 | 78 |
AMC12_763 | A circle has a chord of length $10$ , and the distance from the center of the circle to the chord is $5$ . What is the area of the circle?
$\textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad \textbf{(C) }75\pi \qquad \textbf{(D) }100\pi \qquad \textbf{(E) }125\pi \qquad$
| Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ as shown below.
[asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, M; O = (0,0); A = (-5,5); B = (5,5); M = midpoint(A--B); draw(Circle(O,5sqrt(2))); dot("$O$", O, 1.5*S, linewidth(4.5)); dot("$A$", A, 1.5*NW, linewidth(4.5)); dot("$B$", B, 1.5*NE, linewidth(4.5)); dot("$M$", M, 1.5*N, linewidth(4.5)); draw(A--B^^M--O^^A--O^^M--O^^B--O); label("$5$", midpoint(A--M), 1.5*N); label("$5$", midpoint(B--M), 1.5*N); label("$5$", midpoint(O--M), 1.5*E); label("$r$", midpoint(O--A), 1.5*SW); label("$r$", midpoint(O--B), 1.5*SE); [/asy]
Note that $\overline{OM}\perp\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\triangle OMA$ and $\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt2,$ so the area of $\odot O$ is $\pi r^2=\boxed{\textbf{(B) }50\pi}$ .
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_4 | 50π |
AMC12_764 | A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?
$\textbf{(A)}\ 8\sqrt{3}\qquad\textbf{(B)}\ 10\sqrt{2}\qquad\textbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2}$
| Let the side lengths of the rectangular box be $x, y$ and $z$ .
From the information we get
\[4(x+y+z) = 48 \Rightarrow x+y+z = 12\]
\[2(xy+yz+xz) = 94\]
The sum of all the lengths of the box's interior diagonals is
\[4 \sqrt{x^2+y^2+z^2}\]
Squaring the first expression, we get:
\begin{align*}
144 =(x+y+z)^2 &= x^2+y^2+z^2 + 2(xy+yz+xz) \\
&= x^2+y^2+z^2 + 94.
\end{align*}
Hence \[x^2+y^2+z^2 = 50\]
\[4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2}\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_14 | 20√2 |
AMC12_765 | Suppose that $4^{x_1}=5$ , $5^{x_2}=6$ , $6^{x_3}=7$ , ... , $127^{x_{124}}=128$ . What is $x_1x_2...x_{124}$ ?
$\mathrm{(A)}\ {{{2}}} \qquad \mathrm{(B)}\ {{{\frac{5}{2}}}} \qquad \mathrm{(C)}\ {{{3}}} \qquad \mathrm{(D)}\ {{{\frac{7}{2}}}} \qquad \mathrm{(E)}\ {{{4}}}$
| We see that we can re-write $4^{x_1}=5$ , $5^{x_2}=6$ , $6^{x_3}=7$ , ... , $127^{x_{124}}=128$ as $\left(...\left(\left(\left(4^{x_1}\right)^{x_2}\right)^{x_3}\right)...\right)^{x_{124}}=128$ by using substitution. By using the properties of exponents, we know that $4^{x_1x_2...x_{124}}=128$ .
$4^{x_1x_2...x_{124}}=128\\2^{2x_1x_2...x_{124}}=2^7\\2x_1x_2...x_{124}=7\\x_1x_2...x_{124}=\dfrac{7}{2}$
Therefore, the answer is $\boxed{\mathrm{(D)}\,\dfrac{7}{2}}$
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_13 | 7/2 |
AMC12_768 | In the addition shown below $A$ , $B$ , $C$ , and $D$ are distinct digits. How many different values are possible for $D$ ?
\[\begin{tabular}{cccccc}&A&B&B&C&B\\ +&B&C&A&D&A\\ \hline &D&B&D&D&D\end{tabular}\]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$
| From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$ . We know that $A+B = D$ . Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding two distinct natural numbers. Letting $A=1$ , and letting $B=2,3,4,5,6,7,8$ , we have
\[D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_8 | 7 |
AMC12_772 | A graph is $\textit{symmetric}$ about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers $(a,b,c,d)$ , where $|a|,|b|,|c|,|d|\le5$ and $c$ and $d$ are not both $0$ , is the graph of \[y=\frac{ax+b}{cx+d}\] symmetric about the line $y=x$ ?
$\textbf{(A) }1282\qquad\textbf{(B) }1292\qquad\textbf{(C) }1310\qquad\textbf{(D) }1320\qquad\textbf{(E) }1330$
| Symmetric about the line $y=x$ implies that the inverse function $y^{-1}=y$ . Then we split the question into several cases to find the final answer.
Case 1: $c=0$
Then $y=\frac{a}{d}x+\frac{b}{d}$ and $y^{-1}=\frac{d}{a}x-\frac{b}{a}$ .
Giving us $\frac{a}{d}=\frac{d}{a}$ and $\frac{b}{d}=-\frac{b}{a}$
Therefore, we obtain 2 subcases: $b\neq 0, a+d=0$ and $b=0, a^2=d^2$
Case 2: $c\neq 0$
Then $y^{-1}=\frac{b-dx}{cx-a}=\frac{(cx-a)(-\frac{d}{c})+b-\frac{ad}{c}}{cx-a}=-\frac{d}{c}+\frac{b-\frac{ad}{c}}{cx-a}$
And $y=\frac{(cx+d)(\frac{a}{c})+b-\frac{ad}{c}}{cx+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cx+d}$
So $\frac{a}{c}=-\frac{d}{c}$ , or $a=-d$ ( $c\neq 0$ ), and substitute that into $\frac{b-\frac{ad}{c}}{cx-a}=\frac{b-\frac{ad}{c}}{cx+d}$ gives us:
$bc-ad\neq 0$ (Otherwise $y=\frac{a}{c}$ , $y^{-1}=-\frac{d}{c}=\frac{a}{c}$ , and is not symmetric about $y=x$ )
Therefore we get three cases:
Case 1.1: $c= 0, b\neq 0, d\neq 0, a+d=0$
We have 10 choice of $b$ , 10 choice of $d$ and each choice of $d$ has one corresponding choice of $a$ . In total $10\times 10=100$ ways.
Case 1.2: $c= 0, b = 0, d\neq 0, a^2=d^2$
We have 10 choice for $d$ ( $d\neq 0$ ), each choice of $d$ has 2 corresponding choice of $a$ , thus $10\times 2=20$ ways.
Case 2: $c\neq 0, bc-ad\neq 0, a=-d$
$a=0$ : $10\times 10=100$ ways.
$a=\pm 1$ : $(11\times 10-2)\times 2=216$ ways.
$a=\pm 2$ : $(11\times 10-2)\times 2=216$ ways.
$a=\pm 3$ : $(11\times 10-2)\times 2=216$ ways.
$a=\pm 4$ : $(11\times 10-6)\times 2=208$ ways.
$a=\pm 5$ : $(11\times 10-2)\times 2=216$ ways.
In total $100+208+216\times 4= 1172$ ways.
So the answer is $100+20+1172= \boxed{\textbf{(B) }1292}$
~ERiccc
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_25 | 1292 |
AMC12_773 | What is the graph of $y^4+1=x^4+2y^2$ in the coordinate plane?
$\textbf{(A) } \text{two intersecting parabolas} \qquad \textbf{(B) } \text{two nonintersecting parabolas} \qquad \textbf{(C) } \text{two intersecting circles} \qquad \\\\ \textbf{(D) } \text{a circle and a hyperbola} \qquad \textbf{(E) } \text{a circle and two parabolas}$
| Since the equation has even powers of $x$ and $y$ , let $y'=y^2$ and $x' = x^2$ . Then $y'^2 + 1 = x'^2 + 2y'$ . Rearranging gives $y'^2 - 2y' + 1 = x'^2$ , or $(y'-1)^2=x'^2$ . There are two cases: $y' \leq 1$ or $y' > 1$ .
If $y' \leq 1$ , taking the square root of both sides gives $1 - y' = x'$ , and rearranging gives $x' + y' = 1$ . Substituting back in $x'=x^2$ and $y'=y^2$ gives us $x^2+y^2=1$ , the equation for a circle.
Similarly, if $y' > 1$ , we take the square root of both sides to get $y' - 1 = x'$ , or $y' - x' = 1$ , which is equivalent to $y^2 - x^2 = 1$ , a hyperbola.
Hence, our answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$ .
~ Bxiao31415
(Solutions 1 and 2 are in essence the same; Solution 1 lets $(x',y')=\left(x^2,y^2\right)$ for convenience, but the two solutions are otherwise identical.)
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_8 | a circle and a hyperbola |
AMC12_774 | The graph of $y=f(x)$ , where $f(x)$ is a polynomial of degree $3$ , contains points $A(2,4)$ , $B(3,9)$ , and $C(4,16)$ . Lines $AB$ , $AC$ , and $BC$ intersect the graph again at points $D$ , $E$ , and $F$ , respectively, and the sum of the $x$ -coordinates of $D$ , $E$ , and $F$ is 24. What is $f(0)$ ?
$\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8$
| Note that $f(x) - x^2$ has roots $2, 3$ , and $4$ . Therefore, we may write $f(x) = a(x-2)(x-3)(x-4) +x^2$ . Now we find that lines $AB$ , $AC$ , and $BC$ are defined by the equations $y = 5x - 6$ , $y= 6x-8$ , and $y=7x-12$ respectively.
Since we want to find the $x$ -coordinates of the intersections of these lines and $f(x)$ , we set each of them to $f(x)$ and synthetically divide by the solutions we already know exist.
In the case of line $AB$ , we may write $a(x-2)(x-3)(x-4)+x^2-5x+6 = a(x-2)(x-3)(x-r_1)$ for some real number $r_1$ . Dividing both sides by $(x-2)(x-3)$ gives $a(x-4)+1 = a(x-r_1)$ or $r_1 = \frac {4a-1}{a}$ .
For line $AC$ , we have $a(x-2)(x-3)(x-4)+x^2-6x+8 = a(x-2)(x-4)(x-r_2)$ for some real number $r_2$ , which gives $a(x-3)+1 = a(x-r_2)$ or $r_2 = \frac {3a-1}{a}$ .
For line $BC$ , we have $a(x-2)(x-3)(x-4)+x^2-7x+12 = a(x-3)(x-4)(x-r_3)$ for some real number $r_3$ , which gives $a(x-2)+1 = a(x-r_3)$ or $r_3 = \frac {2a-1}{a}$ .
Since $r_1 + r_2 + r_3 = 24$ , we have $\frac {4a-1}{a} + \frac {3a-1}{a} + \frac {2a-1}{a} = 24$ or $\frac {9a-3}{a} = 24$ . Solving for $a$ gives $a = - \frac{1}{5}$ .
Substituting this back into the original equation, we get $f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2$ , and $f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}$
Solution by vedadehhc
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_23 | 24/5 |
AMC12_777 | Define a function on the positive integers recursively by $f(1) = 2$ , $f(n) = f(n-1) + 1$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$ . What is $f(2017)$ ?
$\textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036$
| This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$ . We also know that when $n$ is odd, $f(n)=f(n-2)+2$ . Thus we know that $f(2017)=f(2015)+2$ . Thus we know that n will always be odd in the recursion of $f(2017)$ , and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$ , which is answer
$\boxed{\textbf{(B)}}$ .
Note that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$ , so $f(2017)=2018$
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_7 | B |
AMC12_781 | How many positive two-digit integers are factors of $2^{24}-1$ ?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$
~ pi_is_3.14
| Repeating difference of squares :
$2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)$
$2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7$
$2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7$
The sum of cubes formula gives us:
$2^{12}+1=(2^4+1)(2^8-2^4+1)$
$2^{12}+1 = 17\cdot241$
A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2\cdot5\cdot7\cdot13\cdot17$ , since multiplying by $241$ will make any factor too large.
Multiplying $17$ by $3$ or $5$ will give a two-digit factor; $17$ itself will also work. The next smallest factor, $7$ , gives a three-digit number. Thus, there are $3$ factors that are multiples of $17$ .
Multiplying $13$ by $3$ , $5$ , or $7$ will also give a two-digit factor, as well as $13$ itself. Higher numbers will not work, giving $4$ additional factors.
Multiply $7$ by $3$ , $5$ , or $3^2$ for a two-digit factor. There are no more factors to check, as all factors which include $13$ are already counted. Thus, there are an additional $3$ factors.
Multiply $5$ by $3$ or $3^2$ for a two-digit factor. All higher factors have been counted already, so there are $2$ more factors.
Thus, the total number of factors is $3+4+3+2=\boxed{\textbf{(D) }12}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_15 | 12 |
AMC12_782 | Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$ ?
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$
| Converting both summands to exponential form,
\begin{align*} -1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\ -1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}. \end{align*}
Notice that the two terms in the problem are two of the third roots of unity (that is, both of them equal $1$ when raised to the power of $3$ ).
When we replace the summands with their exponential form, we get \[f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n.\]
When we substitute $n = 2022$ , we get \[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}.\]
We can rewrite $2022$ as $3 \cdot 674$ , how does that help?
\[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} = \left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} = 1^{674} + 1^{674} = \boxed{\textbf{(E)} \ 2}.\]
Since any third root of unity must cube to $1$ .
~ zoomanTV
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11 | 2 |
AMC12_784 | Let $m>1$ and $n>1$ be integers. Suppose that the product of the solutions for $x$ of the equation
\[8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0\]
is the smallest possible integer. What is $m+n$ ?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272$
| Rearranging logs, the original equation becomes
\[\frac{8}{\log n \log m}(\log x)^2 - \left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x - 2013 = 0\]
By Vieta's Theorem, the sum of the possible values of $\log x$ is $\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n \log m}} = \frac{7\log m + 6 \log n}{8} = \log \sqrt[8]{m^7n^6}$ . But the sum of the possible values of $\log x$ is the logarithm of the product of the possible values of $x$ . Thus the product of the possible values of $x$ is equal to $\sqrt[8]{m^7n^6}$ .
It remains to minimize the integer value of $\sqrt[8]{m^7n^6}$ . Since $m, n>1$ , we can check that $m = 2^2$ and $n = 2^3$ work. Thus the answer is $4+8 = \boxed{\textbf{(A)}\ 12}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_22 | 12 |
AMC12_785 | What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?
$\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10$
| We factor $x^2-3x+2$ into $(x-1)(x-2)$ . Thus, either $1$ or $2$ is a root of $x^2-5x+k$ . If $1$ is a root, then $1^2-5\cdot1+k=0$ , so $k=4$ . If $2$ is a root, then $2^2-5\cdot2+k=0$ , so $k=6$ . The sum of all possible values of $k$ is $\boxed{\textbf{(E) }10}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_5 | 10 |
AMC12_786 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$
| First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+d)(c+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4) = 9^3$ or $\boxed{\textbf{(D)}\ 729 }$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | 729 |
AMC12_789 | A $\textit{disphenoid}$ is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?
$\textbf{(A) }\sqrt{3}\qquad\textbf{(B) }3\sqrt{15}\qquad\textbf{(C) }15\qquad\textbf{(D) }15\sqrt{7}\qquad\textbf{(E) }24\sqrt{6}$
| Notice that any scalene $\textit{acute}$ triangle can be the faces of a $\textit{disphenoid}$ . (See proof in Solution 2.)
As a result, we simply have to find the smallest area a scalene acute triangle with integer side lengths can take on. This occurs with a $4,5,6$ triangle (notice that if you decrease the value of any of the sides the resulting triangle will either be isosceles, degenerate, or non-acute). For this triangle, the semiperimeter is $\frac{15}{2}$ , so by Heron’s Formula:
\begin{align*}
A&=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}\\
&=\sqrt{\frac{15^2\cdot7}{16}}\\
&=\frac{15}{4}\sqrt{7}
\end{align*}
The surface area is simply four times the area of one of the triangles, or $\boxed{\textbf{(D) }15\sqrt{7}}$ .
~eevee9406
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_24 | 15√7 |
AMC12_790 | Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ , $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$ ?
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$
| Observe that $\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ since $m\angle MEI = m\angle MAI = 45^\circ$ ), so $MI=2,MC=\frac{3}{\sqrt{2}}$ . With $\angle{MCI}=45^\circ$ , we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$ . The same calculations hold for $BE$ also, and since $CI<BE$ , we deduce that $CI$ is the smaller root, giving the answer of $\boxed{\textbf{(D) }12}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_20 | 12 |
AMC12_792 | Let points $A = (0,0) , \ B = (1,2), \ C = (3,3),$ and $D = (4,0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left (\frac{p}{q}, \frac{r}{s} \right )$ , where these fractions are in lowest terms. What is $p + q + r + s$ ?
$\textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75$
| If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.
Pick's Theorem states that
$A$ = $I$ $+$ $\frac{B}{2}$ - $1$ , where $I$ is the number of lattice points in the interior of the polygon, and $B$ is the number of lattice points on the boundary of the polygon.
In this case,
$A$ = $5$ $+$ $\frac{7}{2}$ - $1$ = $7.5$
so
$\frac{A}{2}$ = $3.75$
The bottom half of the quadrilateral makes a triangle with base $4$ and half the total area, so we can deduce that the height of the triangle must be $\frac{15}{8}$ in order for its area to be $3.75$ . This height is the y coordinate of our desired intersection point.
Note that segment CD lies on the line $y = -3x + 12$ . Substituting in $\frac{15}{8}$ for y, we can find that the x coordinate of our intersection point is $\frac{27}{8}$ .
Therefore the point of intersection is ( $\frac{27}{8}$ , $\frac{15}{8}$ ), and our desired result is $27+8+15+8= \boxed{ \text{B}) \ 58}$ .
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_13 | 58 |
AMC12_793 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\qquad\textbf{(E) }9$
| This problem asks to find largest $x$ that cannot be written as
\[6 a + 10 b + 15 c = x, \hspace{1cm} (1)\]
where $a, b, c \in \Bbb Z_+$ .
Denote by $r \in \left\{ 0, 1 \right\}$ the remainder of $x$ divided by 2.
Modulo 2 on Equation (1), we get
By using modulus $m \in \left\{ 2, 3, 5 \right\}$ on the equation above, we get
$c \equiv r \pmod{2}$ .
Following from Chicken McNugget's Theorem , we have that any number that is no less than $(3-1)(5-1) = 8$ can be expressed in the form of $3a + 5b$ with $a, b \in \Bbb Z_+$ .
Therefore, all even numbers that are at least equal to $2 \cdot 8 + 15 \cdot 0 = 16$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .
All odd numbers that are at least equal to $2 \cdot 8 + 15 \cdot 1 = 31$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .
The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .
Next, we need to prove that 29 cannot be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .
Because 29 is odd, we must have $c \equiv 1 \pmod{2}$ .
Because $a, b, c \in \Bbb Z_+$ , we must have $c = 1$ .
Plugging this into Equation (1), we get $3 a + 5 b = 7$ .
However, this equation does not have non-negative integer solutions.
All analysis above jointly imply that the largest $x$ that has no non-negative integer solution to Equation (1) is 29.
Therefore, the answer is $2 + 9 = \boxed{\textbf{(D) 11}}$ .
~Steven Chen, www.professorchenedu.com
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | 11 |
AMC12_795 | What is the value in simplest form of the following expression? \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]
$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$
| We have \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}.\]
Note: This comes from the fact that the sum of the first $n$ odds is $n^2$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_1 | 10 |
AMC12_796 | Two real numbers are selected independently and at random from the interval $[-20,10]$ . What is the probability that the product of those numbers is greater than zero?
$\textbf{(A)}\ \frac{1}{9} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{4}{9} \qquad \textbf{(D)}\ \frac{5}{9} \qquad \textbf{(E)}\ \frac{2}{3}$
| For the product to be greater than zero, we must have either both numbers negative or both positive.
Both numbers are negative with a $\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$ chance.
Both numbers are positive with a $\frac{1}{3}*\frac{1}{3}=\frac{1}{9}$ chance.
Therefore, the total probability is $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$ and we are done. $\boxed{D}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_9 | D |
AMC12_797 | Let $a$ , $b$ , $c$ , $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ , $b+c$ , $c+d$ and $d+e$ . What is the smallest possible value of $M$ ?
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$
| We want to try make $a+b$ , $b+c$ , $c+d$ , and $d+e$ as close as possible so that $M$ , the maximum of these, is smallest.
Notice that $2010=670+670+670$ . In order to express $2010$ as a sum of $5$ numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): $2010=670+1+670+1+668$ or $2010=670+1+669+1+669$ . We see that in both cases, the value of $M$ is $671$ , so the answer is $671 \Rightarrow \boxed{B}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14 | B |
AMC12_802 | Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$ ?
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$
| By: dragonfly
First of all, lets plug in all of the $x$ 's into the equation.
$\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$
Then we simplify to get
$\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$
which simplifies into
$\bigg|$ $2016$ $\bigg|$ $+2016$
and finally we get $\boxed{\textbf{(D)}\ 4032}$
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3 | 4032 |
AMC12_805 | Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial
\[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$
| For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$
Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$
Since $\gcd(m,4)=1,$ we conclude that \[\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}\] is the complete residue system modulo $m$ for all integers $k.$
Given any ordered quadruple $(a'_1, a'_2, a'_3, a'_4)$ in modulo $m,$ it follows that exactly one of these $m$ ordered quadruples has sum $0$ modulo $m:$
\[\begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] \end{array}\]
We conclude that $q(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ so \[q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.\]
By Vieta's Formulas, we get $c_1=1\cdot2+1\cdot3+2\cdot3=\boxed{\textbf{(E)}\ 11}.$
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25 | 11 |
AMC12_810 | The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$
| Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus,
$\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4$ .
$\boxed{ \textbf{C}}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_2 | C |
AMC12_812 | A circle of radius $r$ is concentric with and outside a regular hexagon of side length $2$ . The probability that three entire sides of hexagon are visible from a randomly chosen point on the circle is $1/2$ . What is $r$ ?
$\mathrm{(A) \ } 2\sqrt{2}+2\sqrt{3}\qquad \mathrm{(B) \ } 3\sqrt{3}+\sqrt{2}\qquad \mathrm{(C) \ } 2\sqrt{6}+\sqrt{3}\qquad \rm{(D) \ } 3\sqrt{2}+\sqrt{6}\qquad \mathrm{(E) \ } 6\sqrt{2}-\sqrt{3}$
| Project any two non-adjacent and non-opposite sides of the hexagon to the circle ; the arc between the two points formed is the location where all three sides of the hexagon can be fully viewed. Since there are six such pairs of sides, there are six arcs. The probability of choosing a point is $1 / 2$ , or if the total arc degree measures add up to $\frac{1}{2} \cdot 360^{\circ} = 180^{\circ}$ . Each arc must equal $\frac{180^{\circ}}{6} = 30^\mathrm{\circ}$ .
Call the center $O$ , and the two endpoints of the arc $A$ and $B$ , so $\angle AOB = 30^{\circ}$ . Let $P$ be the intersections of the projections of the sides of the hexagon corresponding to $\overline{AB}$ . Notice that $\triangle APO$ is an isosceles triangle : $\angle AOP = 15^{\circ}$ and $\angle OAP = OAB - 60^{\circ} = \frac{180^{\circ}-30^{\circ}}{2} - 60^{\circ} = 15^{\circ}$ . Since $OA$ is a radius and $OP$ can be found in terms of a side of the hexagon, we are almost done.
If we draw the altitude of $APO$ from $P$ , then we get a right triangle . Using simple trigonometry, $\cos 15^{\circ} = \frac{\frac{r}{2}}{2\sqrt{3}} = \frac{r}{4\sqrt{3}}$ .
Since $\cos 15^{\circ} = \cos (45^{\circ} - 30^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}$ , we get $r = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \cdot 4\sqrt{3} = 3\sqrt{2} + \sqrt{6} \Rightarrow \boxed{D}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_22 | D |
AMC12_813 | The measures of the smallest angles of three different right triangles sum to $90^\circ$ . All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$ . What is the perimeter of the third triangle?
$\textbf{(A) } 40 \qquad\textbf{(B) } 126 \qquad\textbf{(C) } 154 \qquad\textbf{(D) } 176 \qquad\textbf{(E) } 208$
| Let $\alpha$ and $\beta$ be the smallest angles of the $3-4-5$ and $5-12-13$ triangles respectively. We have
\[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\]
Then
\[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\]
Let $\theta$ be the smallest angle of the third triangle. Consider
\[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\]
In order for this to be undefined, we need
\[1-\frac{56}{33}\cdot\tan{\theta}=0\]
so
\[\tan{\theta}=\frac{33}{56}\]
Hence the base side lengths of the third triangle are $33$ and $56$ . By the Pythagorean Theorem, the hypotenuse of the third triangle is $65$ , so the perimeter is $33+56+65=\boxed{\textbf{(C) }154}$ .
~ kafuu_chino
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_21 | 154 |
AMC12_814 | In $\triangle BAC$ , $\angle BAC=40^\circ$ , $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$ ?
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$
| Let $C_1$ be the reflection of $C$ across $\overline{AB}$ , and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$ . Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$ . (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As $A$ lies on both $AB$ and $AC$ , we have $C_2A=C_1A=CA=6$ . Furthermore, $\angle CAC_1=2\angle CAB=80^\circ$ by the nature of the reflection, so $\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$ . Therefore by the Law of Cosines \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{\textbf{(D) }14}.\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | 14 |
AMC12_818 | Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$ .
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$
| Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$ . Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$ , while the magnitude of $a-bi$ is $s$ . We get that $s^{2002}=s$ , hence either $s=0$ or $s=1$ .
For $s=0$ we get a single solution $(a,b)=(0,0)$ .
Let's now assume that $s=1$ . Multiply both sides by $a+bi$ . The left hand side becomes $(a+bi)^{2003}$ , the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$ . Hence the solutions for this case are precisely all the $2003$ rd complex roots of unity, and there are $2003$ of those.
The total number of solutions is therefore $1+2003 = \boxed{2004}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_24 | 2004 |
AMC12_823 | Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$
| Rewrite $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ . Now let $a = \cos{2x}$ , and let $b = \cos{\left( \frac{2014\pi^2}{x} \right) }$ . We have:
\[2a(a - b) = 2a^2 - 2\]
Therefore, \[ab = 1\] .
Notice that either $a = 1$ and $b = 1$ or $a = -1$ and $b = -1$ . For the first case, $a = 1$ only when $x = k\pi$ and $k$ is an integer. $b = 1$ when $\frac{2014\pi^2}{k\pi}$ is an even multiple of $\pi$ , and since $2014 = 2*19*53$ , $b =1$ only when $k$ is an odd divisor of $2014$ . This gives us these possible values for $x$ :
\[x= \pi, 19\pi, 53\pi, 1007\pi\]
For the case where $a = -1$ , $\cos{2x} = -1$ , so $x = \frac{m\pi}{2}$ , where m is odd. $\frac{2014\pi^2}{\frac{m\pi}{2}}$ must also be an odd multiple of $\pi$ in order for $b$ to equal $-1$ , so $\frac{4028}{m}$ must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for $m$ , and therefore no cases where $a = -1$ and $b = -1$ . Therefore, the sum of all our possible values for $x$ is
\[\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_25 | 1080π |
AMC12_825 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.
$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.
$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.
$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.
$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.
| Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$
Thus, $Y_1 = Z_1 + 3$ , $Y_2 = Y_1 + 18$ , $Y_2 = Z_2 + 3$ .
Denote by $A_{12}$ the average of person with initial $A$ in the full year.
Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number in $\left( Z_1 , Z_2 \right)$ .
Therefore, the impossible solution is
$\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | Yolanda's quiz average for the academic year was 22 points higher than Zelda's. |
AMC12_827 | Triangle $ABC$ has $AB=13,BC=14$ and $AC=15$ . Let $P$ be the point on $\overline{AC}$ such that $PC=10$ . There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$
$\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18$
| Toss on the Cartesian plane with $A=(5, 12), B=(0, 0),$ and $C=(14, 0)$ . Then $\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}$ by the trapezoid condition, where $D, E\in\overline{BP}$ . Since $PC=10$ , point $P$ is $\tfrac{10}{15}=\tfrac{2}{3}$ of the way from $C$ to $A$ and is located at $(8, 8)$ . Thus line $BP$ has equation $y=x$ . Since $\overline{AD}\parallel\overline{BC}$ and $\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$ -coordinate as $A$ , except it'll also lie on the line $y=x$ . Therefore, $D=(12, 12). \, \blacksquare$
To find the location of point $E$ , we need to find the intersection of $y=x$ with a line parallel to $\overline{AB}$ passing through $C$ . The slope of this line is the same as the slope of $\overline{AB}$ , or $\tfrac{12}{5}$ , and has equation $y=\tfrac{12}{5}x-\tfrac{168}{5}$ . The intersection of this line with $y=x$ is $(24, 24)$ . Therefore point $E$ is located at $(24, 24). \, \blacksquare$
The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$ , which is $\boxed{\textbf{(D) }12\sqrt2}$ .
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_11 | 12√2 |
AMC12_828 | Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove $2$ or $4$ coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove $1$ or $3$ coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with $2013$ coins and when the game starts with $2014$ coins?
$\textbf{(A)}$ Barbara will win with $2013$ coins and Jenna will win with $2014$ coins.
$\textbf{(B)}$ Jenna will win with $2013$ coins, and whoever goes first will win with $2014$ coins.
$\textbf{(C)}$ Barbara will win with $2013$ coins, and whoever goes second will win with $2014$ coins.
$\textbf{(D)}$ Jenna will win with $2013$ coins, and Barbara will win with $2014$ coins.
$\textbf{(E)}$ Whoever goes first will win with $2013$ coins, and whoever goes second will win with $2014$ coins.
| We split into 2 cases: 2013 coins, and 2014 coins.
$\textbf{2013 coins:}$
Notice that when there are $5$ coins left, whoever moves first loses, as they must leave an amount of coins the other person can take. If Jenna goes first, she can take $3$ coins. Then, whenever Barbara takes coins, Jenna will take the amount that makes the total coins taken in that round $5$ . (For instance, if Barbara takes $4$ coins, Jenna will take $1$ ). Eventually, since $2010=0 (\text{mod }5)$ it will be Barbara's move with $5$ coins remaining, so she will lose. If Barbara goes first, on each round, Jenna will take the amount of coins that makes the total coins taken on that round $5$ . Since $2013=3 (\text{mod }5)$ , it will be Barbara's move with $3$ coins remaining, so she will have to take $2$ coins, allowing Jenna to take the last coin. Therefore, Jenna will win with $2013$ coins.
$\textbf{2014 coins:}$
If Jenna moves first, she will take $1$ coin, leaving $2013$ coins, and she wins as shown above. If Barbara moves first, she can take $4$ coins, leaving $2010$ . After every move by Jenna, Barbara will then take the number of coins that makes the total taken in that round $5$ . Since $2010=0\text{(mod }5)$ , it will be Jenna's turn with $5$ coins left, so
Barbara will win. In this case, whoever moves first wins.
Based on this, the answer is $\boxed{\textbf{(B)}}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_18 | B |
AMC12_829 | Susie pays for $4$ muffins and $3$ bananas. Calvin spends twice as much paying for $2$ muffins and $16$ bananas. A muffin is how many times as expensive as a banana?
$\textbf{(A)}\ \frac{3}{2}\qquad\textbf{(B)}\ \frac{5}{3}\qquad\textbf{(C)}\ \frac{7}{4}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{13}{4}$
| Let $m$ stand for the cost of a muffin, and let $b$ stand for the value of a banana. We need to find $\frac{m}{b}$ , the ratio of the price of the muffins to that of the bananas. We have
\[2(4m + 3b) = 2m + 16b\]
\[6m = 10b\]
\[\frac{m}{b} = \boxed{\textbf{(B)}\ \frac{5}{3}}\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_4 | 5/3 |
AMC12_831 | Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ . What is $r_1 + r_2$ ?
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$
| Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$ .
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$ .
As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$ .
Using Vieta's formulas we get that the sum of these two roots is $\boxed{1}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_17 | 1 |
AMC12_833 | What is the value of \[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?\] $\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\frac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5$
| \[\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}\]
Note that $\log_{40}{2}=\frac{1}{\log_{2}{40}}$ , and similarly $\log_{20}{2}=\frac{1}{\log_{2}{20}}$
\[= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\]
\[=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}\]
\[=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}\]
Expanding, \[2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{20})^2-3\log_{2}{20}-(\log_{2}{20})^2\]
All the log terms cancel, so the answer is $2\implies\boxed{\text{(D)}}$ .
~ SoySoy4444
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_9 | 2 |
AMC12_836 | A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n$ , $9\leq n\leq 2017$ , can be the number of participants?
$\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562$
| Let there be $T$ teams. For each team, there are ${n-5\choose 4}$ different subsets of $9$ players that includes a given full team, so the total number of team-(group of 9) pairs is
\[T{n-5\choose 4}.\]
Thus, the expected value of the number of full teams in a random set of $9$ players is
\[\frac{T{n-5\choose 4}}{{n\choose 9}}.\]
Similarly, the expected value of the number of full teams in a random set of $8$ players is
\[\frac{T{n-5\choose 3}}{{n\choose 8}}.\]
The condition is thus equivalent to the existence of a positive integer $T$ such that
\[\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.\]
\[T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}\]
\[T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}\]
Note that this is always less than ${n\choose 5}$ , so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to
\[2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).\]
It is obvious that $5$ divides the RHS, and that $7$ does if $n\equiv 0,1,2,3,4\mod 7$ . Also, $3^2$ divides it if $n\not\equiv 5,8\mod 9$ . One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\mod 32$ .
Note that $2016 = 7*9*32$ so by using all numbers from $2$ to $2017$ , inclusive, it is clear that each possible residue $\mod 7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\cdot 7\cdot 16 = 560$ . However, we must subtract the number of "working" $2\leq n\leq 8$ , which is $3$ . Thus, the answer is $\boxed{\textbf{(D) } 557}$ .
Alternatively, it is enough to approximate by finding the floor of $2017 \cdot \frac57 \cdot \frac79 \cdot \frac12 - 3$ to get $\boxed{\textbf{(D) } 557}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_25 | 557 |
AMC12_837 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$ . What is the sum of the possible values of $w$ ?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$
| Assume that $y-z=a, x-y=b, w-x=c.$
$w-z$ results in the greatest pairwise difference, and thus it is $9$ .
This means $a+b+c=9$ . $a,b,c$ must be in the set ${1,3,4,5,6}$ .
The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$ .
$a+b$ , and $b+c$ then must be the remaining two numbers which are $4$ and $6$ .
The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$ .
\begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*}
\begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*}
The sum of the two w's is $15+16=31$ $\boxed{B}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_13 | 31 |
AMC12_838 | For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$
| Solution 1To find the answer it was enough to play around with $f$ . One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$ , and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_19 | 802 |
AMC12_839 | Two circles of radius 1 are to be constructed as follows. The center of circle $A$ is chosen uniformly and at random from the line segment joining $(0,0)$ and $(2,0)$ . The center of circle $B$ is chosen uniformly and at random, and independently of the first choice, from the line segment joining $(0,1)$ to $(2,1)$ . What is the probability that circles $A$ and $B$ intersect?
$\textbf{(A)} \; \frac {2 + \sqrt {2}}{4} \qquad \textbf{(B)} \; \frac {3\sqrt {3} + 2}{8} \qquad \textbf{(C)} \; \frac {2 \sqrt {2} - 1}{2} \qquad \textbf{(D)} \; \frac {2 + \sqrt {3}}{4} \qquad \textbf{(E)} \; \frac {4 \sqrt {3} - 3}{4}$
| Circles centered at $A$ and $B$ will overlap if $A$ and $B$ are closer to each other than if the circles were tangent. The circles are tangent when the distance between their centers is equal to the sum of their radii. Thus, the distance from $A$ to $B$ will be $2$ . Since $A$ and $B$ are separated by $1$ vertically, they must be separated by $\sqrt{3}$ horizontally. Thus, if $|A_x-B_x|<\sqrt{3}$ , the circles intersect.
Now, plot the two random variables $A_x$ and $B_x$ on the coordinate plane. Each variable ranges from $0$ to $2$ . The circles intersect if the variables are within $\sqrt{3}$ of each other. Thus, the area in which the circles don't intersect is equal to the total area of two small triangles on opposite corners, each of area $\frac{(2-\sqrt{3})^2}{2}$ . So, the total area of the 2 triangles sums to $(2-\sqrt{3})^2$ . Since the total $2 \times 2$ square has an area of $4$ , the probability of the circles not intersecting is $\frac{(2-\sqrt{3})^2}{4}$ . But remember, we want the probability that they do intersect. We conclude the probability the circles intersect is: \[1-\frac{(2-\sqrt{3})^2}{4}=\boxed{\textbf{(E)}\frac{4\sqrt{3}-3}{4}}.\]
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_21 | $\frac{4\sqrt{3}-3}{4}$ |
AMC12_841 | Circles with centers $O$ and $P$ have radii 2 and 4, respectively, and are externally tangent. Points $A$ and $B$ are on the circle centered at $O$ , and points $C$ and $D$ are on the circle centered at $P$ , such that $\overline{AD}$ and $\overline{BC}$ are common external tangents to the circles. What is the area of hexagon $AOBCPD$ ?
[asy] // from amc10 problem series unitsize(0.4 cm); defaultpen(linewidth(0.7) + fontsize(11)); pair A, B, C, D; pair[] O; O[1] = (6,0); O[2] = (12,0); A = (32/6,8*sqrt(2)/6); B = (32/6,-8*sqrt(2)/6); C = 2*B; D = 2*A; draw(Circle(O[1],2)); draw(Circle(O[2],4)); draw((0.7*A)--(1.2*D)); draw((0.7*B)--(1.2*C)); draw(O[1]--O[2]); draw(A--O[1]); draw(B--O[1]); draw(C--O[2]); draw(D--O[2]); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SW); label("$D$", D, NW); dot("$O$", O[1], SE); dot("$P$", O[2], SE); label("$2$", (A + O[1])/2, E); label("$4$", (D + O[2])/2, E);[/asy]
$\textbf{(A) } 18\sqrt {3} \qquad \textbf{(B) } 24\sqrt {2} \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 24\sqrt {3} \qquad \textbf{(E) } 32\sqrt {2}$
| Draw the altitude from $O$ onto $DP$ and call the point $H$ . Because $\angle OAD$ and $\angle ADP$ are right angles due to being tangent to the circles, and the altitude creates $\angle OHD$ as a right angle. $ADHO$ is a rectangle with $OH$ bisecting $DP$ . The length $OP$ is $4+2$ and $HP$ has a length of $2$ , so by pythagorean's, $OH$ is $\sqrt{32}$ .
$2 \cdot \sqrt{32} + \frac{1}{2}\cdot2\cdot \sqrt{32} = 3\sqrt{32} = 12\sqrt{2}$ , which is half the area of the hexagon, so the area of the entire hexagon is $2\cdot12\sqrt{2} = \boxed{(B)24\sqrt{2}}$
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_15 | 24√2 |
AMC12_842 | Right triangle $ACD$ with right angle at $C$ is constructed outwards on the hypotenuse $\overline{AC}$ of isosceles right triangle $ABC$ with leg length $1$ , as shown, so that the two triangles have equal perimeters. What is $\sin(2\angle BAD)$ ?
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.016233639805293cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */ draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); /* draw figures */ draw((-2.,3.)--(-2.,-1.), linewidth(2.)); draw((-2.,-1.)--(2.,-1.), linewidth(2.)); draw((2.,-1.)--(-2.,3.), linewidth(2.)); draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); /* dots and labels */ dot((-2.,3.),linewidth(4.pt) + dotstyle); dot((-2.,-1.),linewidth(4.pt) + dotstyle); dot((2.,-1.),linewidth(4.pt) + dotstyle); dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
$\textbf{(A) } \dfrac{1}{3} \qquad\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad\textbf{(C) } \dfrac{3}{4} \qquad\textbf{(D) } \dfrac{7}{9} \qquad\textbf{(E) } \dfrac{\sqrt{3}}{2}$
| Solution 1Firstly, note by the Pythagorean Theorem in $\triangle ABC$ that $AC = \sqrt{2}$ . Now, the equal perimeter condition means that $BC + BA = 2 = CD + DA$ , since side $AC$ is common to both triangles and thus can be discounted. This relationship, in combination with the Pythagorean Theorem in $\triangle ACD$ , gives $AC^2+CD^2=\left(\sqrt{2}\right)^2+\left(2-DA\right)^2=DA^2$ . Hence $2 + 4 - 4DA + DA^2 = DA^2$ , so $DA = \frac{3}{2}$ , and thus $CD = \frac{1}{2}$ .
Next, since $\angle BAC = 45^{\circ}$ , $\sin{\left(\angle BAC\right)} = \cos{\left(\angle BAC\right)} = \frac{1}{\sqrt{2}}$ . Using the lengths found above, $\sin{\left(\angle CAD\right)} = \frac{\left(\frac{1}{2}\right)}{\left(\frac{3}{2}\right)} = \frac{1}{3}$ , and $\cos{\left(\angle CAD\right)} = \frac{\sqrt{2}}{\left(\frac{3}{2}\right)} = \frac{2 \sqrt{2}}{3}$ .
Thus, by the addition formulae for $\sin$ and $\cos$ , we have
\[\begin{split}\sin{\left(\angle BAD\right)}&=\sin{\left(\angle BAC + \angle CAD\right)}\\&=\sin{\left(\angle BAC\right)}\cos{\left(\angle CAD\right)}+\cos{\left(\angle BAC\right)}\sin{\left(\angle CAD\right)}\\&=\frac{1}{\sqrt{2}}\cdot\frac{2 \sqrt{2}}{3} + \frac{1}{\sqrt{2}}\cdot\frac{1}{3} \\ &= \frac{2 \sqrt{2} + 1}{3 \sqrt{2}}\end{split}\]
and
\[\begin{split}\cos{\left(\angle BAD\right)}&=\cos{\left(\angle BAC + \angle CAD\right)}\\&=\cos{\left(\angle BAC\right)}\cos{\left(\angle CAD\right)}-\sin{\left(\angle BAC\right)}\sin{\left(\angle CAD\right)}\\&=\frac{1}{\sqrt{2}}\cdot\frac{2 \sqrt{2}}{3} - \frac{1}{\sqrt{2}}\cdot\frac{1}{3} \\ &= \frac{2 \sqrt{2} - 1}{3 \sqrt{2}}\end{split}\]
Hence, by the double angle formula for $\sin$ , $\sin{\left(2\angle BAD\right)} = 2\sin{\left(\angle BAD\right)}\cos{\left(\angle BAD\right)} = \frac{2(8-1)}{18} = \boxed{\textbf{(D) } \frac{7}{9}}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_12 | 7/9 |
AMC12_845 | Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her
watch first reads 10:00 PM?
$\text {(A) 10:22 PM and 24 seconds} \qquad \text {(B) 10:24 PM} \qquad \text {(C) 10:25 PM} \qquad \text {(D) 10:27 PM} \qquad \text {(E) 10:30 PM}$
| For every $60$ minutes that pass by in actual time, $57+\frac{36}{60}=57.6$ minutes pass by on Cassandra's watch. When her watch first reads, 10:00 pm, $10(60)=600$ minutes have passed by on her watch. Setting up a proportion,
\[\frac{57.6}{60}=\frac{600}{x}\]
where $x$ is the number of minutes that have passed by in actual time. Solve for $x$ to get $625$ minutes, or $10$ hours and $25$ minutes $\Rightarrow \boxed{\textbf{(C)}\ \text{10:25 PM}}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_11 | 10:25 PM |
AMC12_846 | Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$ , and $\overline{CN}$ is the angle bisector of $\angle{ACB}$ with $N$ on $\overline{AB}$ . Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$ . In addition $\triangle BXN$ is equilateral with $AC=2$ . What is $BX^2$ ?
$\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}$
| Let $BN=x$ and $NA=y$ . From the conditions, let's deduct some convenient conditions that seem sufficient to solve the problem.
is the midpoint of side .
This implies that $[ABX]=[CBX]$ . Given that angle $ABX$ is $60$ degrees and angle $BXC$ is $120$ degrees, we can use the area formula to get
\[\frac{1}{2}(x+y)x \frac{\sqrt{3}}{2} = \frac{1}{2} x \cdot CX \frac{\sqrt{3}}{2}\]
So, $x+y=CX$ .....(1)
is angle bisector.
In the triangle $ABC$ , one has $BC/AC=x/y$ , therefore $BC=2x/y$ .....(2)
Furthermore, triangle $BCN$ is similar to triangle $MCX$ , so $BC/CM=CN/CX$ , therefore $BC = (CX+x)/CX = (2x+y)/(x+y)$ ....(3)
By (2) and (3) and the subtraction law of ratios, we get
\[BC=2x/y = (2x+y)/(y+x) = y/x\]
Therefore $2x^2=y^2$ , or $y=\sqrt{2}x$ . So $BC = 2x/(\sqrt{2}x) = \sqrt{2}$ .
Finally, using the law of cosine for triangle $BCN$ , we get
\[2 = BC^2 = x^2 + (2x+y)^2 - x(2x+y) = 3x^2 + 3xy + y^2 = \left(5+3\sqrt{2}\right)x^2\]
\[x^2 = \frac{2}{5+3\sqrt{2}} = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}}.\]
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_24 | 10-6√2/7 |
AMC12_847 | Rational numbers $a$ and $b$ are chosen at random among all rational numbers in the interval $[0,2)$ that can be written as fractions $\frac{n}{d}$ where $n$ and $d$ are integers with $1 \le d \le 5$ . What is the probability that
\[(\text{cos}(a\pi)+i\text{sin}(b\pi))^4\]
is a real number?
$\textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}$
| Let $\cos(a\pi) = x$ and $\sin(b\pi) = y$ . Consider the binomial expansion of the expression:
\[x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.\]
We notice that the only terms with $i$ are the second and the fourth terms. Thus for the expression to be a real number, either $\cos(a\pi)$ or $\sin(b\pi)$ must be $0$ , or the second term and the fourth term cancel each other out (because in the fourth term, you have $i^2 = -1$ ).
$\text{Case~1:}$ Either $\cos(a\pi)$ or $\sin(b\pi)$ is $0$ .
The two $\text{a's}$ satisfying this are $\tfrac{1}{2}$ and $\tfrac{3}{2}$ , and the two $\text{b's}$ satisfying this are $0$ and $1$ . Because $a$ and $b$ can both be expressed as fractions with a denominator less than or equal to $5$ , there are a total of $20$ possible values for $a$ and $b$ :
\[0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},\]
\[\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},\]
\[\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},\]
\[\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.\]
Calculating the total number of sets of $(a,b)$ results in $20 \cdot 20 = 400$ sets.
Calculating the total number of invalid sets (sets where $a$ doesn't equal $\tfrac{1}{2}$ or $\tfrac{3}{2}$ and $b$ doesn't equal $0$ or $1$ ), resulting in $(20-2) \cdot (20-2) = 324$ .
Thus the number of valid sets is $76$ .
$\text{Case~2}$ : The two terms cancel.
We then have:
\[\cos^3(a\pi) \cdot \sin(b\pi) = \cos(a\pi) \cdot \sin^3(b\pi).\]
So:
\[\cos^2(a\pi) = \sin^2(b\pi),\]
which means for a given value of $\cos(a\pi)$ or $\sin(b\pi)$ , there are $4$ valid values(one in each quadrant).
When either $\cos(a\pi)$ or $\sin(b\pi)$ are equal to $1$ , however, there are only two corresponding values. We don't count the sets where either $\cos(a\pi)$ or $\sin(b\pi)$ equals $0$ , for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if $a$ is $\tfrac{1}{5}$ , then $b$ must be $\tfrac{3}{10}$ , which we don't have). Thus the total number of sets for this case is $4 \cdot 4 + 2 \cdot 2 = 20$ .
Thus, our final answer is $\frac{(20 + 76)}{400} = \frac{6}{25}$ , which is $\boxed{\text{(D)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_24 | D |
AMC12_848 | Let $c = \frac{2\pi}{11}.$ What is the value of
\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{10}{11} \qquad\textbf{(E)}\ 1$
| Plugging in $c$ , we get
\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\]
Since $\sin(x+2\pi)=\sin(x),$ $\sin(2\pi-x)=\sin(-x),$ and $\sin(-x)=-\sin(x),$ we get
\[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{\textbf{(E)}\ 1}.\]
~kingofpineapplz
~Ziyao7294 (minor edit)
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_13 | 1 |
AMC12_849 | Suppose that $\left|x+y\right|+\left|x-y\right|=2$ . What is the maximum possible value of $x^2-6x+y^2$ ?
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$
| Plugging in some values, we see that the graph of the equation $|x+y|+|x-y| = 2$ is a square bounded by $x= \pm 1$ and $y = \pm 1$ .
Notice that $x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9$ means the square of the distance from a point $(x,y)$ to point $(3,0)$ minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point $(3,0)$ , which is $(-1, \pm 1)$ . Either one, when substituting into the function, yields $\boxed{\textbf{(D)}\ 8}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_18 | 8 |
AMC12_852 | Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$ ?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$
| In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).
In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.
Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.
Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
1/2 (216°-144°) = 1/2 (72°) $=\boxed{36 \textbf{(C)}}.$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6 | 36 |
AMC12_854 | An object moves $8$ cm in a straight line from $A$ to $B$ , turns at an angle $\alpha$ , measured in radians and chosen at random from the interval $(0,\pi)$ , and moves $5$ cm in a straight line to $C$ . What is the probability that $AC < 7$ ?
$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$
| By the Law of Cosines ,
\begin{align*} AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\ \cos \alpha &> \frac 12\\ \end{align*}
It follows that $0 < \alpha < \frac {\pi}3$ , and the probability is $\frac{\pi/3}{\pi} = \boxed{\textbf{(D) } \frac13 }$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_21 | 1/3 |
AMC12_855 | The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 504 \qquad \textbf{(D)}\ 1008 \qquad \textbf{(E)}\ 2015$
| Since the harmonic mean is $2$ times their product divided by their sum, we get the equation
$\frac{2\times1\times2016}{1+2016}$
which is then
$\frac{4032}{2017}$
which is finally closest to $\boxed{\textbf{(A)}\ 2}$ .
-dragonfly
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_2 | 2 |
AMC12_864 | The first three terms of a geometric progression are $\sqrt 3$ , $\sqrt[3]3$ , and $\sqrt[6]3$ . What is the fourth term?
$\textbf{(A) }1\qquad\textbf{(B) }\sqrt[7]3\qquad\textbf{(C) }\sqrt[8]3\qquad\textbf{(D) }\sqrt[9]3\qquad\textbf{(E) }\sqrt[10]3\qquad$
| The terms are $\sqrt 3$ , $\sqrt[3]3$ , and $\sqrt[6]3$ , which are equivalent to $3^{\frac{3}{6}}$ , $3^{\frac{2}{6}}$ , and $3^{\frac{1}{6}}$ . So the next term will be $3^{\frac{0}{6}}=1$ , so the answer is $\boxed{\textbf{(A)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_7 | A |
AMC12_865 | In $\triangle{ABC}$ with side lengths $AB = 13$ , $AC = 12$ , and $BC = 5$ , let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$ . What is the area of $\triangle{MOI}$ ?
$\textbf{(A)}\ \frac52\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{13}{4}\qquad\textbf{(E)}\ \frac72$
| In this solution, let the brackets denote areas.
We place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$
Since $\triangle ABC$ is a right triangle with $\angle ACB=90^\circ,$ its circumcenter is the midpoint of $\overline{AB},$ from which $O=\left(6,\frac52\right).$ Note that the circumradius of $\triangle ABC$ is $\frac{13}{2}.$
Let $s$ denote the semiperimeter of $\triangle ABC.$ The inradius of $\triangle ABC$ is $\frac{[ABC]}{s}=\frac{30}{15}=2,$ from which $I=(2,2).$
Since $\odot M$ is also tangent to both coordinate axes, its center is at $M=(a,a)$ and its radius is $a$ for some positive number $a.$ Let $P$ be the point of tangency of $\odot O$ and $\odot M.$ As $\overline{OP}$ and $\overline{MP}$ are both perpendicular to the common tangent line at $P,$ we conclude that $O,M,$ and $P$ are collinear. It follows that $OM=OP-MP,$ or \[\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.\]
Solving this equation, we have $a=4,$ from which $M=(4,4).$
Finally, we apply the Shoelace Theorem to $\triangle MOI:$ \[[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.\]
Remark
Alternatively, we can use $\overline{MI}$ as the base and the distance from $O$ to $\overleftrightarrow{MI}$ as the height for $\triangle MOI:$
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_21 | 7/2 |
AMC12_868 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$
| Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$ . Let $CX=a$ and $EX=b$ . This implies that $DE = a - b$ . Since $CE = CX + EX = a + b$ , we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$ . Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50$ . Thus, our answer is $2\times50=\boxed{\textbf{(E)}\ 100}$ .
~JHawk0224
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 100 |
AMC12_870 | Let
$S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}$
and
$S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}$ .
What is the ratio of the area of $S_2$ to the area of $S_1$ ?
$\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102$
| Looking at the constraints of $S_1$ :
$x+y > 0$
$\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10} 10 +\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10}(10x+10y)$
$1+x^2+y^2 \le 10x+10y$
$x^2-10x+y^2-10y \le -1$
$x^2-10x+25+y^2-10y+25 \le 49$
$(x-5)^2 + (y-5)^2 \le (7)^2$
$S_1$ is a circle with a radius of $7$ . So, the area of $S_1$ is $49\pi$ .
Looking at the constraints of $S_2$ :
$x+y > 0$
$\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)$
$\log_{10}(2+x^2+y^2)\le \log_{10} 100 +\log_{10}(x+y)$
$\log_{10}(2+x^2+y^2)\le \log_{10}(100x+100y)$
$2+x^2+y^2 \le 100x+100y$
$x^2-100x+y^2-100y \le -2$
$x^2-100x+2500+y^2-100y+2500 \le 4998$
$(x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2$
$S_2$ is a circle with a radius of $7\sqrt{102}$ . So, the area of $S_2$ is $4998\pi$ .
So the desired ratio is $\frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{E}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_21 | 102 |
AMC12_872 | A triangle has area $30$ , one side of length $10$ , and the median to that side of length $9$ . Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$ ?
$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10}$
| [asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0); draw(A--B--C--cycle); draw(D--C); draw(E--C); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,SW); label("$E$",E,S); label("$9$",(D--C),NW); label("$6$",(E--C)); label("$10$",(A--B),SE); [/asy]$AB$ is the side of length $10$ , and $CD$ is the median of length $9$ . The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$ . To find $\sin{\theta}$ , just use opposite over hypotenuse with the right triangle $\triangle DCE$ . This is equal to $\frac69=\boxed{\textbf{(D)}\ \frac23}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_10 | 2/3 |
AMC12_873 | Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$ ?
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$
| First note that by log properties $a\diamondsuit b = 7^{(\log_7a)(\log_7b)}$ and $a \heartsuit b = 7^{\frac{\log_7a}{\log_7b}} = 7^{\log_ba}$ .
Now, define $b_n = \log_7(a_n)$ . Thus $b_3 = \log_7(3\heartsuit 2) = \log_7(7^{\log_23}) = \log_23$ .
Taking logs of both sides of the recursion and using the definition of $\diamondsuit$ gives $\log_7(a_n) = \log_7(7^{\log_7n\heartsuit (n-1)\log_7a_{n-1}})$ .
The logs and the exponent cancel to $\log_7((n\heartsuit (n-1))^{\log_7(a_{n-1})})$ , and by the definition of $\heartsuit$ , ths is $\log_7(7^{(\log_{n-1}n)\log_7(a_{n-1})})$ , which quickly simplifies to $\log_7(a_{n-1})\log_{n-1}n$ $= b_{n-1}\log_{n-1}n$ .
Thus $b_n = b_{n-1}\log_{n-1}n$ . From this, we have $b_4 = b_3\log_34 = \log_23\log_34 = \log_24$ , $b_5 = \log_45\log_24 = \log_25$ , and in general, $b_n = \log_2n$ .
Finally, $\log_7(a_{2019}) = b_{2019}= \log_22019$ .
Since $2^{11} = 2048$ and $2019$ is slightly less than $2048$ , $\log_22019 \approx \boxed{\text{(D) }11}$ .
- NamelyOrange
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_23 | 11 |
AMC12_876 | Circles with centers $(2,4)$ and $(14,9)$ have radii $4$ and $9$ , respectively. The equation of a common external tangent to the circles can be written in the form $y=mx+b$ with $m>0$ . What is $b$ ?
[asy] size(150); defaultpen(linewidth(0.7)+fontsize(8)); draw(circle((2,4),4));draw(circle((14,9),9)); draw((0,-2)--(0,20));draw((-6,0)--(25,0)); draw((2,4)--(2,4)+4*expi(pi*4.5/11)); draw((14,9)--(14,9)+9*expi(pi*6/7)); label("4",(2,4)+2*expi(pi*4.5/11),(-1,0)); label("9",(14,9)+4.5*expi(pi*6/7),(1,1)); label("(2,4)",(2,4),(0.5,-1.5));label("(14,9)",(14,9),(1,-1)); draw((-4,120*-4/119+912/119)--(11,120*11/119+912/119)); dot((2,4)^^(14,9)); [/asy]
$\mathrm{(A) \ } \frac{908}{119}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ } \frac{912}{119}$
| Let $L_1$ be the line that goes through $(2,4)$ and $(14,9)$ , and let $L_2$ be the line $y=mx+b$ . If we let $\theta$ be the measure of the acute angle formed by $L_1$ and the x-axis, then $\tan\theta=\frac{5}{12}$ . $L_1$ clearly bisects the angle formed by $L_2$ and the x-axis, so $m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}$ . We also know that $L_1$ and $L_2$ intersect at a point on the x-axis. The equation of $L_1$ is $y=\frac{5}{12}x+\frac{19}{6}$ , so the coordinate of this point is $\left(-\frac{38}{5},0\right)$ . Hence the equation of $L_2$ is $y=\frac{120}{119}x+\frac{912}{119}$ , so $b=\frac{912}{119}$ , and our answer choice is $\boxed{\mathrm{E}}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_19 | E |
AMC12_877 | Suppose that on a parabola with vertex $V$ and a focus $F$ there exists a point $A$ such that $AF=20$ and $AV=21$ . What is the sum of all possible values of the length $FV?$
$\textbf{(A) }13 \qquad \textbf{(B) }\frac{40}3 \qquad \textbf{(C) }\frac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\frac{43}3$
| Let $\ell$ be the directrix of $\mathcal P$ ; recall that $\mathcal P$ is the set of points $T$ such that the distance from $T$ to $\ell$ is equal to $TF$ . Let $P$ and $Q$ be the orthogonal projections of $F$ and $A$ onto $\ell$ , and further let $X$ and $Y$ be the orthogonal projections of $F$ and $V$ onto $AQ$ . Because $AF < AV$ , there are two possible configurations which may arise, and they are shown below.
[asy] import olympiad; size(230); defaultpen(linewidth(0.8)+fontsize(11pt)); real d = 1.1, edge = 2.5, Ax = 1.6; real f(real x) { return 1/(4*d) * x * x; } pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax)); pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q); draw(P--F--A--V--Y^^F--X--Q^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(graph(f,-2.5,2.5)); draw(la -- lb); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red); [/asy]
[asy] import olympiad; size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); real d = 0.7, edge = 2.5, Ax = 1.9; real f(real x) { return 1/(4*d) * x * x; } pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax)); pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q); draw(Q--A--F--P^^F--X^^A--V--Y^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(la -- lb); draw(graph(f,-2.5,2.5)); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red); [/asy]
Set $d = FV$ , which by the definition of a parabola also equals $VP$ . Then as $AQ = AF = 20$ , we have $AY = 20 - d$ and $AX = |20 - 2d|$ . Since $FXYV$ is a rectangle, $FX = VY$ , so by Pythagorean Theorem on triangles $AFX$ and $AVY$ ,
\begin{align*} 21^2 - (20 - d)^2 &= AV^2 - AY^2 = VY^2\\ &= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2 \end{align*}
This equation simplifies to $3d^2 - 40d + 41 = 0$ , which has solutions $d = \tfrac{20\pm\sqrt{277}}3$ . Both values of $d$ work - the smaller solution with the bottom configuration and the larger solution with the top configuration - and so the requested answer is $\boxed{\textbf{(B)}\ \tfrac{40}{3}}$ .
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_20 | 40/3 |
AMC12_883 | Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r < m < s$ . What is $r + s$ ?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$
| Let $y = m(x - 20) + 14$ . Equating them:
$x^2 = mx - 20m + 14$
$x^2 - mx + 20m - 14 = 0$
For there to be no solutions, the discriminant must be less than zero:
$m^2 - 4(20m - 14) < 0$
$m^2 - 80m + 56 < 0$ .
So $m < 0$ for $r < m < s$ where $r$ and $s$ are the roots of $m^2 - 80m + 56 = 0$ and their sum by Vieta's formulas is $\boxed{\textbf{(E)}\ 80}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_17 | 80 |
AMC12_886 | How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array
\[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\]
satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$
| Note that the arrays and the sum configurations have one-to-one correspondence. Furthermore, the row sum configuration and the column sum configuration are independent of each other. Therefore, the answer is $(4!)^2=\boxed{\textbf{(D) }576}.$
Remark
For any given sum configuration, we can uniquely reconstruct the array it represents. Conversely, for any array, it is clear that we can determine the unique sum configuration associated with it. Therefore, this establishes a one-to-one correspondence between the arrays and the sum configurations.
~bad_at_mathcounts ~MRENTHUSIASM ~tsun26
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17 | 576 |
AMC12_888 | For each $x$ in $[0,1]$ , define
\[\begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\ f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array}\]
Let $f^{[2]}(x) = f(f(x))$ , and $f^{[n + 1]}(x) = f^{[n]}(f(x))$ for each integer $n \geq 2$ . For how many values of $x$ in $[0,1]$ is $f^{[2005]}(x) = \frac {1}{2}$ ?
\[(\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}\]
| For the two functions $f(x)=2x,0\le x\le \frac{1}{2}$ and $f(x)=2-2x,\frac{1}{2}\le x\le 1$ ,as long as $f(x)$ is between $0$ and $1$ , $x$ will be in the right domain, so we don't need to worry about the domain of $x$ .
Also, every time we change $f(x)$ , the expression for the final answer in terms of $x$ will be in a different form(although they'll all satisfy the final equation), so we get a different starting value of $x$ . Every time we have two choices for $f(x$ ) and altogether we have to choose $2005$ times. Thus, $2^{2005}\Rightarrow\boxed{E}$ .
Note: the values of x that satisfy $f^{[n]}(x) = \frac {1}{2}$ are $\frac{1}{2^{n+1}}$ , $\frac{3}{2^{n+1}}$ , $\frac{5}{2^{n+1}}$ , $\cdots$ , $\frac{2^{n+1}-1}{2^{n+1}}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_20 | E |
AMC12_890 | At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$ . Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$ ?
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textbf{(D)}\ 406 \qquad \textbf{(E)}\ 1024$
| We start with $2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19$ . After rearranging, we get $\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)$ .
Since $\lfloor\log_{2}(N-1)\rfloor$ is a positive integer, $\frac{N+19}{2}$ must be in the form of $2^{m}$ for some positive integer $m$ . From this fact, we get $N=2^{m+1}-19$ .
If we now check integer values of N that satisfy this condition, starting from $N=19$ , we quickly see that the first values that work for $N$ are $45$ and $109$ , that is, $2^6-19$ and $2^7 -19$ , giving values of $5$ and $6$ for $m$ , respectively. Adding up these two values for $N$ , we get $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_19 | C |
AMC12_892 | On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $R$ be the region formed by the union of the square and all the triangles, and $S$ be the smallest convex polygon that contains $R$ . What is the area of the region that is inside $S$ but outside $R$ ?
$\textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$
| [asy] real a = 1/2, b = sqrt(3)/2; draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0)); draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0)); filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9)); filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9)); filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9)); filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9)); [/asy]
The equilateral triangles form trapezoids with side lengths $1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $1$ and an angle of $30^{\circ}$ in between them, so the total area of these triangles (which is the area of $S - R$ ) is, $4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1$ which makes the answer $\boxed{C}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_15 | C |
AMC12_893 | Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?
$\text {(A) } 129 \qquad \text {(B) } 137 \qquad \text {(C) } 174 \qquad \text {(D) } 223 \qquad \text {(E) } 411$
| Francesca makes a total of $100+100+400=600$ grams of lemonade, and in those $600$ grams, there are $25$ calories from the lemon juice and $386$ calories from the sugar, for a total of $25+386=411$ calories per $600$ grams. We want to know how many calories there are in $200=600/3$ grams, so we just divide $411$ by $3$ to get $137\implies\boxed{(\text{B})}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_6 | 137 |
AMC12_897 | Joy has $30$ thin rods, one each of every integer length from $1 \text{ cm}$ through $30 \text{ cm}$ . She places the rods with lengths $3 \text{ cm}$ , $7 \text{ cm}$ , and $15 \text{cm}$ on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 19 \qquad\textbf{(E)}\ 20$
| The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7 = 25$ . This means Joy can use the $19$ possible integer rod lengths that fall into $[6, 24]$ . However, she has already used the rods of length $7$ cm and $15$ cm so the answer is $19 - 2 = 17$ $\boxed{\textbf{(B)}}$
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_6 | 17 |
AMC12_901 | Sreshtha needs to estimate the quantity $\frac{a}{b} - c$ , where $a, b,$ and $c$ are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of $\frac{a}{b} - c$ ?
$\textbf{(A)}\ \text{She rounds all three numbers up.}\\ \qquad\textbf{(B)}\ \text{She rounds } a \text{ and } b \text{ up, and she rounds } c \text{down.}\\ \qquad\textbf{(C)}\ \text{She rounds } a \text{ and } c \text{ up, and she rounds } b \text{down.} \\ \qquad\textbf{(D)}\ \text{She rounds } a \text{ up, and she rounds } b \text{ and } c \text{down.}\\ \qquad\textbf{(E)}\ \text{She rounds } c \text{ up, and she rounds } a \text{ and } b \text{down.}$
| To maximize our estimate, we want to maximize $\frac{a}{b}$ and minimize $c$ , because both terms are positive values. Therefore we round $c$ down. To maximize $\frac{a}{b}$ , round $a$ up and $b$ down. $\Rightarrow \boxed{\textbf{(D)}}$
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_5 | D |
AMC12_902 | The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$ . What is $a+b$ ?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3$
| Solution 1Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).$ Then $0 = 4a - 2 \rightarrow a = 0.5$ , and $0 = 4 - 4b \rightarrow b = 1.$ Then $a + b = \boxed{\textbf{(B)}\ 1.5}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_12 | 1.5 |
AMC12_903 | A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$
| Substituting $a = b$ we get
\[f(2a) + f(0) = 2f(a)^2\]
Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\]
This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\]
Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$ , so answer choice $\boxed{\textbf{(E) -2}}$ is impossible.
~AtharvNaphade
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22 | -2 |
AMC12_904 | Suppose that the roots of the polynomial $P(x)=x^3+ax^2+bx+c$ are $\cos \frac{2\pi}7,\cos \frac{4\pi}7,$ and $\cos \frac{6\pi}7$ , where angles are in radians. What is $abc$ ?
$\textbf{(A) }{-}\frac{3}{49} \qquad \textbf{(B) }{-}\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt[3]7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}$
| Let $z=e^{\frac{2\pi i}{7}}.$ Since $z$ is a $7$ th root of unity, we have $z^7=1.$ For all integers $k,$ note that $\cos\frac{2k\pi}{7}=\operatorname{Re}\left(z^k\right)=\operatorname{Re}\left(z^{-k}\right)$ and $\sin\frac{2k\pi}{7}=\operatorname{Im}\left(z^k\right)=-\operatorname{Im}\left(z^{-k}\right).$ It follows that
\begin{alignat*}{4} \cos\frac{2\pi}{7} &= \frac{z+z^{-1}}{2} &&= \frac{z+z^6}{2}, \\ \cos\frac{4\pi}{7} &= \frac{z^2+z^{-2}}{2} &&= \frac{z^2+z^5}{2}, \\ \cos\frac{6\pi}{7} &= \frac{z^3+z^{-3}}{2} &&= \frac{z^3+z^4}{2}. \end{alignat*}
By geometric series, we conclude that \[\sum_{k=0}^{6}z^k=\frac{1-1}{1-z}=0.\]
Alternatively, recall that the $7$ th roots of unity satisfy the equation $z^7-1=0.$ By Vieta's Formulas, the sum of these seven roots is $0.$
As a result, we get \[\sum_{k=1}^{6}z^k=-1.\]
Let $(r,s,t)=\left(\cos{\frac{2\pi}{7}},\cos{\frac{4\pi}{7}},\cos{\frac{6\pi}{7}}\right).$ By Vieta's Formulas, the answer is
\begin{align*} abc&=[-(r+s+t)](rs+st+tr)(-rst) \\ &=(r+s+t)(rs+st+tr)(rst) \\ &=\left(\frac{\sum_{k=1}^{6}z^k}{2}\right)\left(\frac{2\sum_{k=1}^{6}z^k}{4}\right)\left(\frac{1+\sum_{k=0}^{6}z^k}{8}\right) \\ &=\frac{1}{32}\left(\sum_{k=1}^{6}z^k\right)\left(\sum_{k=1}^{6}z^k\right)\left(1+\sum_{k=0}^{6}z^k\right) \\ &=\frac{1}{32}(-1)(-1)(1) \\ &=\boxed{\textbf{(D) }\frac{1}{32}}. \end{align*}
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_22 | \frac{1}{32} |
AMC12_907 | The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6$ . The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?
$\textbf{(A)}\ 3\sqrt{2} \qquad \textbf{(B)}\ \frac{13}{3} \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ \frac{13}{2}$
| Let $ABCDE$ be the pyramid with $ABCD$ as the square base. Let $O$ and $M$ be the center of square $ABCD$ and the midpoint of side $AB$ respectively. Lastly, let the hemisphere be tangent to the triangular face $ABE$ at $P$ .
Notice that $\triangle EOM$ has a right angle at $O$ . Since the hemisphere is tangent to the triangular face $ABE$ at $P$ , $\angle EPO$ is also $90^{\circ}$ . Hence $\triangle EOM$ is similar to $\triangle EPO$ .
$\frac{OM}{2} = \frac{6}{EP}$
$OM = \frac{6}{EP} \times 2$
$OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2}$
The length of the square base is thus $2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_15 | A |
AMC12_909 | A data set containing $20$ numbers, some of which are $6$ , has mean $45$ . When all the 6s are removed, the data set has mean $66$ . How many 6s were in the original data set?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$
Solution
| Because the set has $20$ numbers and mean $45$ , the sum of the terms in the set is $45\cdot 20=900$ .
Let there be $s$ sixes in the set.
Then, the mean of the set without the sixes is $\frac{900-6s}{20-s}$ . Equating this expression to $66$ and solving yields $s=7$ , so we choose answer choice $\boxed{\textbf{(D) }7}$ .
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_5 | 7 |
AMC12_911 | A geometric sequence $(a_n)$ has $a_1=\sin x$ , $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$ ?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
| By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$ . Since $\tan x=\frac{\sin x}{\cos x}$ , we can rewrite this as $\cos^3x=\sin^2x$ .
The common ratio of the sequence is $\frac{\cos x}{\sin x}$ , so we can write
\[a_1= \sin x\]
\[a_2= \cos x\]
\[a_3= \frac{\cos^2x}{\sin x}\]
\[a_4=\frac{\cos^3x}{\sin^2x}=1\]
\[a_5=\frac{\cos x}{\sin x}\]
\[a_6=\frac{\cos^2x}{\sin^2x}\]
\[a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}\]
\[a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}\]
Since $\cos^3x=\sin^2x=1-\cos^2x$ , we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$ , which is $a_8$ , making our answer $8 \Rightarrow \boxed{E}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_20 | E |
AMC12_914 | A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?
$\text{(A) }\frac {1 + i \sqrt {11}}{2} \qquad \text{(B) }\frac {1 + i}{2} \qquad \text{(C) }\frac {1}{2} + i \qquad \text{(D) }1 + \frac {i}{2} \qquad \text{(E) }\frac {1 + i \sqrt {13}}{2}$
| Let the polynomial be $P$ and let the two integer zeros be $z_1$ and $z_2$ . We can then write $P(x)=(x-z_1)(x-z_2)(x^2+ax+b)$ for some integers $a$ and $b$ .
If a complex number $p+qi$ with $q\not=0$ is a root of $P$ , it must be the root of $x^2+ax+b$ , and the other root of $x^2+ax+b$ must be $p-qi$ .
We can then write $x^2+ax+b = (x-p-qi)(x-p+qi) = (x-p)^2 - (qi)^2 = x^2 - 2px + p^2 + q^2$ .
We can now examine each of the five given complex numbers, and find the one for which the values $-2p$ and $p^2+q^2$ are integers. This is $\boxed{\frac {1 + i \sqrt {11}}{2}}$ , for which we have $-2p = -2\cdot\frac 12 = -1$ and $p^2+q^2 = \left( \frac 12 \right)^2 + \left( \frac {\sqrt{11}}2 \right)^2 = \frac 14 + \frac {11}4 = \frac {12}4 = 3$ .
(As an example, the polynomial $x^4 - 2x^3 + 4x^2 - 3x$ has zeroes $0$ , $1$ , and $\frac {1 \pm i \sqrt {11}}{2}$ .)
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_23 | (1 + i√11)/2 |
AMC12_916 | For each real number $a$ with $0 \leq a \leq 1$ , let numbers $x$ and $y$ be chosen independently at random from the intervals $[0, a]$ and $[0, 1]$ , respectively, and let $P(a)$ be the probability that
\[\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1\]
What is the maximum value of $P(a)?$
$\textbf{(A)}\ \frac{7}{12} \qquad\textbf{(B)}\ 2 - \sqrt{2} \qquad\textbf{(C)}\ \frac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \frac{5}{8}$
| Let's start first by manipulating the given inequality.
\[\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1\]
\[\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}\]
Let's consider the boundary cases: $\sin{(\pi x)}=\pm \cos{(\pi y)}$ .
\[\sin{(\pi x)}=\pm \cos{(\pi y)}=\sin{(\tfrac 12 {\pi}\pm \pi y)}\]
Solving the first case gives us \[y=\tfrac{1}{2}-x \quad \textrm{and} \quad y=x-\tfrac{1}{2}.\] Solving the second case gives us \[y=x+\tfrac{1}{2}\quad \textrm{and} \quad y=\tfrac{3}{2}-x.\] If we graph these equations in $[0,1]\times[0,1]$ , we get a rhombus shape.
[asy] defaultpen(fontsize(9)+0.8); size(200); pen p=fontsize(10); pair A,B,C,D,A1,A2,B1,B2,C1,C2,D1,D2,I,L; A=MP("(0,0)",origin,down+left,p); B=MP("(1,0)",right,down+right,p); C=MP("(1,1)",right+up,up+right,p); D=MP("(0,1)",up,up+left,p); A1=MP("",extension(A,B,(0.5,0),(0,0.5)),2*down,p); dot(A1); A2=MP("",extension(A,D,(0.5,0),(0,0.5)),2*left,p); dot(A2); B1=MP("",extension(B,C,(0.5,0),(0,-0.5)),2*right,p); dot(B1); B2=MP("",extension(C,D,(0.5,1),(0,0.5)),2*up,p); dot(B2); real a=0.7; draw(A1--B1--B2--A2--cycle, gray+0.6); draw(a*right--a*right+up, royalblue); draw(A1--B2, royalblue+dashed); draw(A--B--C--D--A, black+1.2); dot("$(a,0)$",(a,0),down); dot("$(a,1)$",(a,1),up); [/asy]
Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement.
From the region graph, notice that in order to maximize $P(a)$ , $a\geq\tfrac{1}{2}$ . We can solve the rest with geometric probability.
Instead of maximizing $P(a)$ , we minimize $Q(a)=1-P(a)$ . $Q(a)$ consists of two squares (each broken into two triangles), one of area $\tfrac{1}{4}$ and another of area $(a-\tfrac 12)^2$ . To calculate $Q(a)$ , we divide this area by $a$ , so \[Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= a+\frac 1{2a}-1.\]
By AM-GM, $a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}$ , which we can achieve by setting $a=\frac{\sqrt{2}}{2}$ .
Therefore, the maximum value of $P(a)$ is $1-\min(Q(a))$ , which is $1-(\sqrt{2}-1) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_25 | 2 - √2 |
AMC12_918 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ , $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$ ?
[asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); [/asy]
$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$
| [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(1.5)); draw((3-sqrt(3),3-sqrt(3)) -- (3-sqrt(3),sqrt(3)-2) -- (sqrt(3)-2,sqrt(3)-2) -- (sqrt(3)-2,3-sqrt(3)) -- cycle,linewidth(1.5)); label("$O (0, 0)$",(0.5,0.5),S); dot((0.5,0.5)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (sqrt(3) - 2, 3-sqrt(3)), NW); label("$M$", (0, sqrt(3)), NW); label("$N$", (1, sqrt(3)), NE); [/asy]
Refer to the diagram above.
Let the origin be at the center of the square, $A$ be the intersection of the top and right hexagons, $B$ be the intersection of the top and left hexagons, and $M$ and $N$ be the top points in the diagram.
By symmetry, $A$ lies on the line $y = x$ . The equation of line $AN$ is $y = -x\sqrt{3} + \frac{3}{2}\sqrt{3} - \frac{1}{2}$ (due to it being one of the sides of the top hexagon). Thus, we can solve for the coordinates of $A$ by finding the intersection of the two lines:
\[x = -x\sqrt{3} + \frac{3\sqrt{3} - 1}{2}\]
\[x(\sqrt{3} + 1) = \frac{3\sqrt{3} - 1}{2}\]
\[x = \frac{3\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3} + 1}\]
\[= \frac{3\sqrt{3}-1}{2(\sqrt{3} + 1)} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\]
\[= \frac{10 - 4\sqrt{3}}{4}\]
\[= \frac{5}{2} - \sqrt{3}\]
\[\therefore A = \left(\frac{5}{2} - \sqrt{3}, \frac{5}{2} - \sqrt{3}\right).\]
This means that we can find the length $AB$ , which is equal to $2(\frac{5}{2} - \sqrt{3}) = (5 - 2\sqrt{3}$ . We will next find the area of trapezoid $ABMN$ . The lengths of the bases are $1$ and $5 - 2\sqrt{3}$ , and the height is equal to the $y$ -coordinate of $M$ minus the $y$ -coordinate of $A$ . The height of the hexagon is $\sqrt{3}$ and the bottom of the hexagon lies on the line $y = \frac{1}{2}$ . Thus, the $y$ -coordinate of $M$ is $\sqrt{3} - \frac{1}{2}$ , and the height is $2\sqrt{3} - 3$ . We can now find the area of the trapezoid:
\[[ABMN] = (2\sqrt{3} - 3)\left(\frac{1 + 5 - 2\sqrt{3}}{2}\right)\]
\[= (2\sqrt{3} - 3)(3 - \sqrt{3})\]
\[= 6\sqrt{3} + 3\sqrt{3} - 9 - 6\]
\[= 9\sqrt{3} - 15.\]
The total area of the figure is the area of a square with side length $AB$ plus four times the area of this trapezoid:
\[\textrm{Area} = (5 - 2\sqrt{3})^2 + 4(9\sqrt{3} - 15)\]
\[= 37 - 20\sqrt{3} + 36\sqrt{3} - 60\]
\[= 16\sqrt{3} - 23.\]
Our answer is $16 + 3 - 23 = \boxed{\textbf{(B) }-4}$ .
~mathboy100
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25 | -4 |
AMC12_919 | Let $\alpha$ be the radian measure of the smallest angle in a $3{-}4{-}5$ right triangle. Let $\beta$ be the radian measure of the smallest angle in a $7{-}24{-}25$ right triangle. In terms of $\alpha$ , what is $\beta$ ?
$\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad$
| From the question,
\[\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}\]
\[\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\]
\[\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\]
\[\tan(\alpha+\beta)=\frac{4}{3}\]
\[\alpha+\beta=\tan^{-1}\left(\frac{4}{3}\right)\]
\[\alpha+\beta=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right)\]
\[\alpha+\beta=\frac{\pi}{2}-\alpha\]
\[\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}\]
~lptoggled
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_10 | $\frac{\pi}{2}-2\alpha$ |
AMC12_920 | Let $S$ be the square one of whose diagonals has endpoints $(1/10,7/10)$ and $(-1/10,-7/10)$ . A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0 \le x \le 2012$ and $0\le y\le 2012$ . Let $T(v)$ be a translated copy of $S$ centered at $v$ . What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coefficients in its interior?
$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25}$
| [asy] pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1); draw (A--B--C--D--A); draw(A--C); draw(B--D); draw(W--X); draw(Y--Z); label("\((0.1,0.7)\)",A,NE); label("\((-0.1,-0.7)\)",C,SW); label("\(x\)",X,NW); label("\(y\)",Y,NE); [/asy]The unit square's diagonal has a length of $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$ . Because $S$ square is not parallel to the axis, the two points must be adjacent.
Now consider the unit square $U$ with vertices $(0,0), (1,0), (1,1)$ and $(0,1)$ . Let us first consider only two vertices, $(0,0)$ and $(1,0)$ . We want to find the area of the region within $U$ that the point $v=(x,y)$ will create the translation of $S$ , $T(v)$ such that it covers both $(0,0)$ and $(1,0)$ . By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.
For $T(v)$ to contain the point $(0,0)$ , $v$ must be inside square $S$ . Similarly, for $T(v)$ to contain the point $(1,0)$ , $v$ must be inside a translated square $S$ with center at $(1,0)$ , which we will call $S'$ . Therefore, the area we seek is Area $(U \cap S \cap S')$ .
To calculate the area, we notice that Area $(U \cap S \cap S') = \frac{1}{2} \cdot$ Area $(S \cap S')$ by symmetry. Let $S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)$ . Let $M = (0.7, 0.4)$ be the midpoint of $S'_1S'_2$ , and $N = (0.7, 0.7)$ along the line $S_1S'_1$ . Let $I$ be the intersection of $S$ and $S'$ within $U$ , and $J$ be the intersection of $S$ and $S'$ outside $U$ . Therefore, the area we seek is $\frac{1}{2} \cdot$ Area $(S \cap S') = \frac{1}{2} [IS'_2JS_2]$ . Because $S_2, M, N$ all have $x$ coordinate $0.7$ , they are collinear. Noting that the side length of $S$ and $S'$ is $1$ (as shown above), we also see that $S_2M = MS'_1 = 0.5$ , so $\triangle{S'_1NM} \cong \triangle{S_2IM}$ . If follows that $IS_2 = NS'_1 = 1.1 - 0.7 = 0.4$ and $IS'_2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2$ . Therefore, the area is $\frac{1}{2} \cdot$ Area $(S \cap S') = \frac{1}{2} [IS'_2JS_2] = \frac{1}{2} \cdot 0.2 \cdot 0.4 = 0.04$ .
Because there are three other regions in the unit square $U$ that we need to count, the total area of $v$ within $U$ such that $T(v)$ contains two adjacent lattice points is $0.04 \cdot 4 = 0.16$ .
By periodicity, this probability is the same for all $0 \le x \le 2012$ and $0 \le y \le 2012$ . Therefore, the answer is $0.16 = \boxed{\frac{4}{25} \textbf{(C)} }$
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_23 | 4/25 |
AMC12_921 | The number $21!=51,090,942,171,709,440,000$ has over $60,000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
$\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}$
| We can consider a factor of $21!$ to be odd if it does not contain a $2$ ; hence, finding the exponent of $2$ in the prime factorization of $21!$ will help us find our answer. We can start off with all multiples of $2$ up to $21$ , which is $10$ . Then, we find multiples of $4$ , which is $5$ . Next, we look at multiples of $8$ , of which there are $2$ . Finally, we know there is only one multiple of $16$ in the set of positive integers up to $21$ . Now, we can add all of these to get $10+5+2+1=18$ . We know that, in the prime factorization of $21!$ , we have $2^{18}$ , and the only way to have an odd number is if there is not a $2$ in that number's prime factorization. This only happens with $2^{0}$ , which is only one of the 19 different exponents of 2 we could have (of which having each exponent is equally likely). Hence, we have $\boxed{\text{(B)} \dfrac{1}{19}}.$
Solution by: armang32324
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_16 | 1/19 |
AMC12_924 | What value of $x$ satisfies
\[\frac{\log_2x \cdot \log_3x}{\log_2x+\log_3x}=2?\]
$\textbf{(A) } 25 \qquad\textbf{(B) } 32 \qquad\textbf{(C) } 36 \qquad\textbf{(D) } 42 \qquad\textbf{(E) } 48$
| We have
\begin{align*}
\log_2x\cdot\log_3x&=2(\log_2x+\log_3x) \\
1&=\frac{2(\log_2x+\log_3x)}{\log_2x\cdot\log_3x} \\
1&=2(\frac{1}{\log_3x}+\frac{1}{\log_2x}) \\
1&=2(\log_x3+\log_x2) \\
\log_x6&=\frac{1}{2} \\
x^{\frac{1}{2}}&=6 \\
x&=36
\end{align*}
so $\boxed{\textbf{(C) }36}$
~kafuu_chino
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_8 | 36 |
AMC12_926 | Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is
$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$
Solution
| By the Pythagorean identity we can rewrite the given expression as follows.
\[\sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}} = \sqrt{\sin^4{x} + 4(1 - \sin^2{x})} - \sqrt{\cos^4{x} + 4(1 - \cos^2{x})}\]
Expanding each bracket gives \[\sqrt{\sin^4{x} - 4\sin^2{x} + 4} - \sqrt{\cos^4{x} - 4\cos^2{x} + 4}\]
The expressions under the square roots can be factored to get \[\sqrt{(\sin^2{x} - 2)^2} - \sqrt{(\cos^2{x} - 2)^2}\]
Since $\sin^2{x} < 2$ and $\cos^2{x} < 2$ for all real $x$ , the expression must evaluate to $(2 - \sin^2{x}) - (2 - \cos^2{x})$ , which simplifies to $\cos^2{x} - \sin^2{x} = \boxed {\text{(E) }\cos{2x}}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_17 | cos(2x) |
AMC12_927 | How many distinct four-tuples $(a,b,c,d)$ of rational numbers are there with
\[a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?\]
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 17 \qquad \mathrm{(D)}\ 2004 \qquad \mathrm{(E)}\ \text{infinitely many}$
| Using the laws of logarithms , the given equation becomes
\[\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005\]
\[\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005\]
\[\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\]
As $a,b,c,d$ must all be rational, and there are no powers of $3$ or $7$ in $10^{2005}$ , $b=d=0$ . Then $2^{a}\cdot 5^{c}=2^{2005}\cdot 5^{2005} \Rightarrow a=c=2005$ .
Only the four-tuple $(2005,0,2005,0)$ satisfies the equation, so the answer is $\boxed{1} \Rightarrow \mathrm{(B)}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_17 | 1 |
AMC12_930 | Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$ . What is the length of the shorter diagonal of $ABCD$ ?
$\textbf{(A) }\frac{31}7 \qquad \textbf{(B) }\frac{33}7 \qquad \textbf{(C) }5 \qquad \textbf{(D) }\frac{39}7 \qquad \textbf{(E) }\frac{41}7 \qquad$
| [asy] import geometry; size(200); pair A = (-1.66, 0.33); pair B = (-9.61277, 1.19799); pair C = (-7.83974, 3.61798); pair D = (-4.88713, 4.14911); draw(circumcircle(A, B, C)); draw(A--C); draw(A--D); draw(C--D); draw(B--C); draw(A--B); label("$A$", A, E); label("$B$", B, W); label("$C$", C, NW); label("$D$", D, N); label("$7$", midpoint(A--C), SW); label("$5$", midpoint(A--D), NE); label("$3$", midpoint(C--D)+ dir(135)*0.3, N); label("$3$", midpoint(B--C)+dir(180)*0.3, NW); label("$8$", midpoint(A--B), S); markangle(Label("$60^\circ$", Relative(0.5)), A, B, C, radius=10); markangle(Label("$120^\circ$", Relative(0.5)), C, D, A, radius=10); [/asy]
~diagram by erics118
First, $\angle CBA=60 ^\circ$ by properties of cyclic quadrilaterals.
Let $AC=u$ . Apply the Law of Cosines on $\triangle ACD$ :
\[u^2=3^2+5^2-2(3)(5)\cos120^\circ\]
\[u=7\]
Let $AB=v$ . Apply the Law of Cosines on $\triangle ABC$ :
\[7^2=3^2+v^2-2(3)(v)\cos60^\circ\]
\[v=\frac{3\pm13}{2}\]
\[v=8\]
By Ptolemy’s Theorem ,
\[AB \cdot CD+AD \cdot BC=AC \cdot BD\]
\[8 \cdot 3+5 \cdot 3=7BD\]
\[BD=\frac{39}{7}\]
Since $\frac{39}{7}<7$ ,
The answer is $\boxed{\textbf{(D) }\frac{39}{7}}$ .
~lptoggled,eevee9406, meh494
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_19 | 39/7 |
AMC12_934 | Consider the polynomial
\[P(x)=\prod_{k=0}^{10}(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)\]
The coefficient of $x^{2012}$ is equal to $2^a$ . What is $a$ ?
\[\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24\]
| Every term in the expansion of the product is formed by taking one term from each factor and multiplying them all together. Therefore, we pick a power of $x$ or a power of $2$ from each factor.
Every number, including $2012$ , has a unique representation by the sum of powers of two, and that representation can be found by converting a number to its binary form. $2012 = 11111011100_2$ , meaning $2012 = 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4$ .
Thus, the $x^{2012}$ term was made by multiplying $x^{1024}$ from the $(x^{1024} + 1024)$ factor, $x^{512}$ from the $(x^{512} + 512)$ factor, and so on. The only numbers not used are $32$ , $2$ , and $1$ .
Thus, from the $(x^{32} + 32), (x^2+2), (x+1)$ factors, $32$ , $2$ , and $1$ were chosen as opposed to $x^{32}, x^2$ , and $x$ .
Thus, the coefficient of the $x^{2012}$ term is $32 \times 2 \times 1 = 64 = 2^6$ . So the answer is $6 \rightarrow \boxed{B}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_20 | B |
AMC12_937 | The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$
| Let the lengths of the two congruent sides of the triangle be $x$ , then the product desired is $x^2$ .
Notice that the product of the base and twice the height is $4$ times the area of the triangle.
Set the vertex angle to be $a$ , we derive the equation:
$x^2=4\left(\frac{1}{2}x^2\sin(a)\right)$
$\sin(a)=\frac{1}{2}$
As the triangle is obtuse, $a=150^\circ$ only. We get $\boxed{\textbf{(D)} \ 150}.$
~Wilhelm Z
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_8 | 150 |
AMC12_939 | In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$
| The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((0,3)--(0,4)--(1,4)--(1,3)--cycle, gray(0.9)); filldraw((3,3)--(3,4)--(4,4)--(4,3)--cycle, gray(0.9)); filldraw((4,3)--(4,4)--(5,4)--(5,3)--cycle, gray(0.9)); filldraw((2,2)--(2,3)--(3,3)--(3,2)--cycle, gray(0.9)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9)); filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
where the light gray boxes are the ones we have filled. Counting these, we get $\boxed{\textbf{(D) } 7}$ total boxes.
~ciceronii
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_6 | 7 |
AMC12_940 | How many unordered pairs of edges of a given cube determine a plane?
$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$
| Without loss of generality, choose one of the $12$ edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.
For two lines in space to determine a common plane, they must either intersect or be parallel (in other words, they cannot be skew lines). If all $12$ line segments are extended to lines, the first (arbitrarily chosen) edge's line intersects $4$ lines and is parallel to another $3$ . Thus $4+3=7$ of the $12-1=11$ remaining line segments (which could be chosen for the second edge) give a pair of lines determining a common plane. To see this, observe that in the diagram below, the red edge is parallel to the $3$ green edges and intersects with the $4$ blue edges.
[asy] import three; import three; unitsize(1cm); size(200); currentprojection=perspective(-6/5,-8/5,7/8); draw((0,1,0)--(0,0,0)--(1,0,0), blue); draw((1,0,0)--(1,1,0)--(0,1,0)); draw((0,0,0)--(0,0,1), red); draw((0,1,0)--(0,1,1), green); draw((1,1,0)--(1,1,1), green); draw((1,0,0)--(1,0,1), green); draw((0,1,1)--(0,0,1)--(1,0,1), blue); draw((1,0,1)--(1,1,1)--(0,1,1)); [/asy]
This means that the probability that a randomly-selected pair of edges determine a plane is $\frac{7}{11}$ , and we calculate that there are ${12 \choose 2} = 66$ total pairs of edges that could be chosen (without the restriction). Thus the answer is $\frac{7}{11} \cdot 66 =\boxed{\textbf{(D) }42}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_11 | 42 |
AMC12_941 | For every integer $n\ge2$ , let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$ . For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$ . What is the largest integer $m$ such that $2010^m$ divides
| Because 67 is the largest prime factor of 2010, it means that in the prime factorization of $\prod_{n=2}^{5300}\text{pow}(n)$ , there'll be $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ where $x$ is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times $67$ is incorporated into the giant product.
All numbers $n=67 \cdot x$ , given $x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}$ such that for any integer $x$ between $1$ and $m$ , prime $p_x$ must be less than $67$ , contributes a 67 to the product. Considering $67 \cdot 79 < 5300 < 67 \cdot 80$ , the possible values of x are $1,2,...,70,72,74,...78$ , since $x=71,73,79$ are primes that are greater than 67. However, $\text{pow}\left(67^2\right)$ contributes two $67$ s to the product, so we must count it twice. Therefore, the answer is $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_25 | 77 |
AMC12_944 | A circle with center $C$ is tangent to the positive $x$ and $y$ -axes and externally tangent to the circle centered at $(3,0)$ with radius $1$ . What is the sum of all possible radii of the circle with center $C$ ?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$
| Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$ . For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$ , the distance between $C$ and $(3,0)$ must be exactly $r+1$ .
By the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\sqrt{ (r-3)^2 + r^2 }$ , hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$ .
Simplifying, we obtain $r^2 - 8r + 8 = 0$ . By Vieta's formulas the sum of the two roots of this equation is $\boxed{8}$ .
(We should actually solve for $r$ to verify that there are two distinct positive roots. In this case we get $r=4\pm 2\sqrt 2$ . This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.)
[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_16 | 8 |
AMC12_946 | A square and an equilateral triangle have the same perimeter. Let $A$ be the area of the circle circumscribed about the square and $B$ the area of the circle circumscribed around the triangle. Find $A/B$ .
$\mathrm{(A) \ } \frac{9}{16}\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } \frac{27}{32}\qquad \mathrm{(D) \ } \frac{3\sqrt{6}}{8}\qquad \mathrm{(E) \ } 1$
| Suppose that the common perimeter is $P$ .
Then, the side lengths of the square and triangle, respectively, are $\frac{P}{4}$ and $\frac{P}{3}$
The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is $\frac{P\sqrt{2}}{4}$
Therefore, the radius is $\frac{P\sqrt{2}}{8}$ and the area of the circle is
$\pi \cdot \left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64}=\frac{P^2 \pi}{32}=A$
Now consider the circle circumscribed around the equilateral triangle. Due to symmetry, the circle must share a center with the equilateral triangle. The radius of the circle is simply the distance from the center of the triangle to a vertex.
This distance is $\frac{2}{3}$ of an altitude. By $30-60-90$ right triangle properties, the altitude is $\frac{\sqrt{3}}{2} \cdot s$ where s is the side.
So, the radius is $\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \cdot \frac{P}{3} = \frac{P\sqrt{3}}{9}$
The area of the circle is $\pi \cdot \left(\frac{P\sqrt{3}}{9}\right)^2=\pi \cdot \frac{3P^2}{81}=\frac{P^2\pi}{27}=B$
So, $\frac{A}{B}=\frac{\frac{P^2 \pi}{32}}{\frac{P^2 \pi}{27}}=\frac{P^2 \pi}{32} \cdot \frac{27}{P^2\pi}=\boxed{\frac{27}{32} \implies \mathrm{(C) \ } \frac{27}{32}}$
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_11 | 27/32 |
AMC12_949 | A circle having center $(0,k)$ , with $k>6$ , is tangent to the lines $y=x$ , $y=-x$ and $y=6$ . What is the radius of this circle?
$\mathrm{(A)}\ 6\sqrt{2}-6 \qquad \mathrm{(B)}\ 6 \qquad \mathrm{(C)}\ 6\sqrt{2} \qquad \mathrm{(D)}\ 12 \qquad \mathrm{(E)}\ 6+6\sqrt{2}$
| Let $R$ be the radius of the circle. Draw the two radii that meet the points of tangency to the lines $y = \pm x$ . We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are $R$ and the diagonal is $k = R+6$ . The diagonal of a square is $\sqrt{2}$ times the side length. Therefore, $R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow \boxed{\mathrm{E}}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_14 | E |
AMC12_952 | Points $A$ and $B$ lie on the graph of $y=\log_{2}x$ . The midpoint of $\overline{AB}$ is $(6, 2)$ . What is the positive difference between the $x$ -coordinates of $A$ and $B$ ?
$\textbf{(A)}~2\sqrt{11}\qquad\textbf{(B)}~4\sqrt{3}\qquad\textbf{(C)}~8\qquad\textbf{(D)}~4\sqrt{5}\qquad\textbf{(E)}~9$
| Let $A(6+m,2+n)$ and $B(6-m,2-n)$ , since $(6,2)$ is their midpoint. Thus, we must find $2m$ . We find two equations due to $A,B$ both lying on the function $y=\log_{2}x$ . The two equations are then $\log_{2}(6+m)=2+n$ and $\log_{2}(6-m)=2-n$ . Now add these two equations to obtain $\log_{2}(6+m)+\log_{2}(6-m)=4$ . By logarithm rules, we get $\log_{2}((6+m)(6-m))=4$ . By raising 2 to the power of both sides, we obtain $(6+m)(6-m)=16$ . We then get \[36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}\] . Since we're looking for $2m$ , we obtain $(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}$
~amcrunner (yay, my first AMC solution)
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_6 | 4√5 |
AMC12_953 | Let $K$ be the number of sequences $A_1$ , $A_2$ , $\dots$ , $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ , $\{5, 7\}$ , $\{2, 5, 7\}$ , $\{2, 5, 7\}$ , $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$ .What is the remainder when $K$ is divided by $10$ ?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$
| Consider any sequence with $n$ terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the $n$ th spot, which means every number has $(n+1)$ choices to show up in the sequence. Consequently, for each sequence with length $n$ , there are $(n+1)^{10}$ possible ways.
Thus, the desired value is $\sum_{i=1}^{10}(i+1)^{10}\equiv \boxed{\textbf{(C) } 5}\pmod{10}$
~bluesoul
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 5 |
AMC12_956 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$
| The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$ . There are 3 ways to paint each, giving us $\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\underline{3}$ ways to paint $2$ , but without loss of generality, let it be painted red. $4$ cannot be the same color as $2$ or $8$ , so there are $\underline{2}$ ways to paint $4$ , which automatically determines the color for $8$ . $6$ cannot be painted red, so there are $\underline{2}$ ways to paint $6$ , but WLOG, let it be painted blue. There are $\underline{2}$ choices for the color for $3$ , which is either red or green in this case. Lastly, there are $\underline{2}$ ways to choose the color for $9$ .
$9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = \boxed{\textbf{(E) }432}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | 432 |
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