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AMC12_958 | Let $ABCD$ be a square. Let $E, F, G$ and $H$ be the centers, respectively, of equilateral triangles with bases $\overline{AB}, \overline{BC}, \overline{CD},$ and $\overline{DA},$ each exterior to the square. What is the ratio of the area of square $EFGH$ to the area of square $ABCD$ ?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}$
| The center of an equilateral triangle is its centroid, where the three medians meet.
The distance along the median from the centroid to the base is one third the length of the median.
Let the side length of the square be $1$ . The height of $\triangle E$ is $\frac{\sqrt{3}}{2},$ so the distance from $E$ to the midpoint of $\overline{AB}$ is $\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}$
$EG = 2 \cdot \frac{\sqrt{3}}{6}$ (from above) $+ 1$ (side length of the square).
Since $EG$ is the diagonal of square $EFGH$ , $\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}$
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_17 | (B) 2 + √3/3 |
AMC12_961 | For every real number $x$ , let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$ , and let \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\] The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?
$\textbf{(A) }1\qquad \textbf{(B) }\dfrac{\log 2015}{\log 2014}\qquad \textbf{(C) }\dfrac{\log 2014}{\log 2013}\qquad \textbf{(D) }\dfrac{2014}{2013}\qquad \textbf{(E) }2014^{\frac1{2014}}\qquad$
| Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$ . Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$ . In order for this to be less than or equal to $1$ , we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$ . Combining this with the fact that $\lfloor x\rfloor =k$ gives that $x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$ , and so the length of the interval is $\log_{2014}\left(\dfrac{k+1}k\right)$ . We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from $k=1$ to $k=2013$ to get that the desired sum is \[\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1\textbf{ (A)}}.\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_21 | 1 |
AMC12_965 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$
| Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields $6$ possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields $6$ more possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Thus, our answer is $6 + 6 = \boxed{\bold{(D)}\, 12}$
Solution by akaashp11
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_13 | 12 |
AMC12_967 | A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$ , $(-2, 2)$ , $(-2, -2)$ , $(2, -2)$ . A particle starts at $(0,0)$ . Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $1/8$ that the particle will move from $(x, y)$ to each of $(x, y + 1)$ , $(x + 1, y + 1)$ , $(x + 1, y)$ , $(x + 1, y - 1)$ , $(x, y - 1)$ , $(x - 1, y - 1)$ , $(x - 1, y)$ , or $(x - 1, y + 1)$ . The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ?
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 39$
| We let $c, e,$ and $m$ be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. $(1, 1)$ ), an "inside edge" (e.g. $(1, 0)$ ), and the middle respectively.
Starting in the middle, there is a $\frac{4}{8}$ chance of moving to an inside edge and a $\frac{4}{8}$ chance of moving to an inside corner, so
\[m = \frac{1}{2}e + \frac{1}{2}c.\]
Starting at an inside edge, there is a $\frac{2}{8}$ chance of moving to another inside edge, a $\frac{2}{8}$ chance of moving to an inside corner, a $\frac{1}{8}$ chance of moving into the middle, and a $\frac{3}{8}$ chance of reaching an outside edge and stopping. Therefore,
\[e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}\cdot 0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.\]
Starting at an inside corner, there is a $\frac{2}{8}$ chance of moving to an inside edge, a $\frac{1}{8}$ chance of moving into the middle, a $\frac{4}{8}$ chance of moving to an outside edge and stopping, and finally a $\frac{1}{8}$ chance of reaching that elusive outside corner. This gives
\[c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}\cdot 1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.\]
Solving this system of equations gives
\[m = \frac{4}{35},\]
\[e = \frac{1}{14},\]
\[c = \frac{11}{70}.\]
Since the particle starts at $(0, 0),$ it is $m$ we are looking for, so the final answer is
\[4 + 35 = \boxed{\textbf{(E) }39}.\]
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_22 | 39 |
AMC12_968 | For all positive integers $n$ , let $f(n)=\log_{2002} n^2$ . Let $N=f(11)+f(13)+f(14)$ . Which of the following relations is true?
$\text{(A) }N<1 \qquad \text{(B) }N=1 \qquad \text{(C) }1<N<2 \qquad \text{(D) }N=2 \qquad \text{(E) }N>2$
| First, note that $2002 = 11 \cdot 13 \cdot 14$ .
Using the fact that for any base we have $\log a + \log b = \log ab$ , we get that $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{(D) N=2}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_14 | 2 |
AMC12_973 | A circle of radius 2 is centered at $A$ . An equilateral triangle with side 4 has a vertex at $A$ . What is the difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?
$\textbf{(A)}\; 8-\pi \qquad\textbf{(B)}\; \pi+2 \qquad\textbf{(C)}\; 2\pi-\dfrac{\sqrt{2}}{2} \qquad\textbf{(D)}\; 4(\pi-\sqrt{3}) \qquad\textbf{(E)}\; 2\pi-\dfrac{\sqrt{3}}{2}$
| The area of the circle is $\pi \cdot 2^2 = 4\pi$ , and the area of the triangle is $\frac{4^2 \cdot\sqrt{3}}{4} = 4\sqrt{3}$ . The difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle is the same as the difference between the area of the circle and the area of the triangle (because from both pieces we are subtracting the area of the two shapes' intersection), so the answer is $4\pi-4\sqrt{3} = \boxed{\textbf{(D)}\; 4(\pi-\sqrt{3})}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_14 | 4(π-√3) |
AMC12_975 | What number is one third of the way from $\frac14$ to $\frac34$ ?
$\textbf{(A)}\ \frac {1}{3} \qquad \textbf{(B)}\ \frac {5}{12} \qquad \textbf{(C)}\ \frac {1}{2} \qquad \textbf{(D)}\ \frac {7}{12} \qquad \textbf{(E)}\ \frac {2}{3}$
| Solution 1We can rewrite the two given fractions as $\frac 3{12}$ and $\frac 9{12}$ . (We multiplied all numerators and denominators by $3$ .)
Now it is obvious that the interval between them is divided into three parts by the fractions $\boxed{\frac 5{12}}$ and $\frac 7{12}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_3 | 5/12 |
AMC12_982 | A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$ . Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$ . What is the side length of this square?
$\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5}$
| [asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real a=1; real b=2; pair O=(0,0); pair A=(-(sqrt(19)-2)/5,1); pair B=((sqrt(19)-2)/5,1); pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair E=(-(sqrt(19)-2)/5,0); path inner=Circle(O,a); path outer=Circle(O,b); draw(outer); draw(inner); draw(A--B--C--D--cycle); draw(O--D--E--cycle); label("$A$",D,NW); label("$E$",E,SW); label("$O$",O,SE); label("$s+1$",(D--E),W); label("$\frac{s}{2}$",(E--O),S); pair[] ps={A,B,C,D,E,O}; dot(ps); [/asy]The circles have radii of $1$ and $2$ . Draw the triangle shown in the figure above and write expressions in terms of $s$ (length of the side of the square) for the sides of the triangle. Because $AO$ is the radius of the larger circle, which is equal to $2$ , we can write the Pythagorean Theorem.
\begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\ \frac14 s^2 + s^2 + 2s + 1 &= 4\\ \frac54 s^2 +2s - 3 &= 0\\ 5s^2 + 8s - 12 &=0 \end{align*}
Use the quadratic formula.
\[s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}\]
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_12 | $\frac{2\sqrt{19}-4}{5}$ |
AMC12_983 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$ ?
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$
| $x+\tfrac{2}{x}= y+\tfrac{2}{y}$
Since $x\not=y$ , we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$ .
Cross multiply in either equation, giving us $xy=2$ .
$\boxed{\textbf{(D) }{2}}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | 2 |
AMC12_986 | A pyramid has a square base $ABCD$ and vertex $E$ . The area of square $ABCD$ is $196$ , and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91$ , respectively. What is the volume of the pyramid?
$\textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784$
| Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$ . The side length of the base is $14$ . The heights of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$ , respectively. Consider a side view of the pyramid from $\triangle BCE$ . We have a systems of equations through the Pythagorean Theorem:
$13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$
Setting them equal to each other and simplifying gives
$-27+28a=225 \implies a=9$ .
Therefore, $h=\sqrt{15^2-9^2}=12$ , and the volume of the pyramid is $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_18 | 784 |
AMC12_987 | If circular arcs $AC$ and $BC$ have centers at $B$ and $A$ , respectively, then there exists a circle tangent to both $\overarc {AC}$ and $\overarc{BC}$ , and to $\overline{AB}$ . If the length of $\overarc{BC}$ is $12$ , then the circumference of the circle is
[asy] label("A", (0,0), W); label("B", (64,0), E); label("C", (32, 32*sqrt(3)), N); draw(arc((0,0),64,0,60)); draw(arc((64,0),64,120,180)); draw((0,0)--(64,0)); draw(circle((32, 24), 24)); [/asy]
$\textbf {(A)}\ 24 \qquad \textbf {(B)}\ 25 \qquad \textbf {(C)}\ 26 \qquad \textbf {(D)}\ 27 \qquad \textbf {(E)}\ 28$
| Solution (Pythagorean Theorem)First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$ . $\frac {1}{6}({2}{\pi})AB=12 \implies AB=\frac{36}{\pi}$ . Next, connect the center of the circle to side $AB$ , and call this length $r$ , and call the foot $M$ . Since $ABC$ is equilateral, it follows that $MB=\frac{18}{\pi}$ , and $OA$ (where $O$ is the center of the circle) is $\frac{36}{\pi}-r$ . By the Pythagorean Theorem, you get $r^2+\left(\frac{18}{\pi}\right)^2=\left(\frac{36}{\pi}-r\right)^2 \implies r=\frac{27}{2\pi}$ . Finally, we see that the circumference is $2{\pi}\cdot \frac{27}{2\pi}=\boxed{(D)27}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_24 | 27 |
AMC12_988 | Let $f$ be the unique function defined on the positive integers such that \[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\] for all positive integers $n$ . What is $f(2023)$ ?
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$
| First, we note that $f(1) = 1$ , since the only divisor of $1$ is itself.
Then, let's look at $f(p)$ for $p$ a prime. We see that
\[\sum_{d \mid p} d \cdot f\left(\frac{p}{d}\right) = 1\]
\[1 \cdot f(p) + p \cdot f(1) = 1\]
\[f(p) = 1 - p \cdot f(1)\]
\[f(p) = 1-p\]
Nice.
Now consider $f(p^k)$ , for $k \in \mathbb{N}$ .
\[\sum_{d \mid p^k} d \cdot f\left(\frac{p^k}{d}\right) = 1\]
\[1 \cdot f(p^k) + p \cdot f(p^{k-1}) + p^2 \cdot f(p^{k-2}) + \dotsc + p^k f(1) = 1\] .
It can be (strongly) inductively shown that $f(p^k) = f(p) = 1-p$ . Here's how.
We already showed $k=1$ works. Suppose it holds for $k = n$ , then
\[1 \cdot f(p^n) + p \cdot f(p^{n-1}) + p^2 \cdot f(p^{n-2}) + \dotsc + p^n f(1) = 1 \implies f(p^m) = 1-p \; \forall \; m \leqslant n\]
For $k = n+1$ , we have
\[1 \cdot f(p^{n+1}) + p \cdot f(p^{n}) + p^2 \cdot f(p^{n-1}) + \dotsc + p^{n+1} f(1) = 1\] , then using $f(p^m) = 1-p \; \forall \; m \leqslant n$ , we simplify to
\[1 \cdot f(p^{n+1}) + p \cdot (1-p) + p^2 \cdot (1-p) + \dotsc + p^n \cdot (1-p) + p^{n+1} f(1) = 1\]
\[f(p^{n+1}) + \sum_{i=1}^n p^i (1-p) + p^{n+1} = 1\]
\[f(p^{n+1}) + p(1 - p^n) + p^{n+1} = 1\]
\[f(p^{n+1}) + p = 1 \implies f(p^{n+1}) = 1-p\] .
Very nice! Now, we need to show that this function is multiplicative, i.e. $f(pq) = f(p) \cdot f(q)$ for $\textbf{distinct}$ $p,q$ prime.
It's pretty standard, let's go through it quickly.
\[\sum_{d \mid pq} d \cdot f\left(\frac{pq}{d}\right) = 1\]
\[1 \cdot f(pq) + p \cdot f(q) + q \cdot f(p) + pq \cdot f(1) = 1\]
Using our formulas from earlier, we have
\[f(pq) + p(1-q) + q(1-p) + pq = 1 \implies f(pq) = 1 - p(1-q) - q(1-p) - pq = (1-p)(1-q) = f(p) \cdot f(q)\]
Great! We're almost done now.
Let's actually plug in $2023 = 7 \cdot 17^2$ into the original formula.
\[\sum_{d \mid 2023} d \cdot f\left(\frac{2023}{d}\right) = 1\]
\[1 \cdot f(2023) + 7 \cdot f(17^2) + 17 \cdot f(7 \cdot 17) + 7 \cdot 17 \cdot f(17) + 17^2 \cdot f(7) + 7 \cdot 17^2 \cdot f(1) = 1\]
Let's use our formulas! We know
\[f(7) = 1-7 = -6\]
\[f(17) = 1-17 = -16\]
\[f(7 \cdot 17) = f(7) \cdot f(17) = (-6) \cdot (-16) = 96\]
\[f(17^2) = f(17) = -16\]
So plugging ALL that in, we have
\[f(2023) = 1 - \left(7 \cdot (-16) + 17 \cdot (-6) \cdot (-16) + 7 \cdot 17 \cdot (-16) + 17^2 \cdot (-6) + 7 \cdot 17^2\right)\]
which, be my guest simplifying, is $\boxed{\textbf{(B)} \ 96}$
~ $\color{magenta} zoomanTV$
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22 | 96 |
AMC12_990 | The acronym AMC is shown in the rectangular grid below with grid lines spaced $1$ unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC $?$
[asy] import olympiad; unitsize(25); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 9; ++j) { pair A = (j,i); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 9; ++j) { if (j != 8) { draw((j,i)--(j+1,i), dashed); } if (i != 2) { draw((j,i)--(j,i+1), dashed); } } } draw((0,0)--(2,2),linewidth(2)); draw((2,0)--(2,2),linewidth(2)); draw((1,1)--(2,1),linewidth(2)); draw((3,0)--(3,2),linewidth(2)); draw((5,0)--(5,2),linewidth(2)); draw((4,1)--(3,2),linewidth(2)); draw((4,1)--(5,2),linewidth(2)); draw((6,0)--(8,0),linewidth(2)); draw((6,2)--(8,2),linewidth(2)); draw((6,0)--(6,2),linewidth(2)); [/asy]
$\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21$
| Each of the straight line segments have length $1$ and each of the slanted line segments have length $\sqrt{2}$ (this can be deducted using $45-45-90$ , pythag, trig, or just sense)
There area a total of $13$ straight lines segments and $4$ slanted line segments. The sum is $\boxed{\textbf{(C) } 13 + 4\sqrt{2}}$
~quacker88
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_2 | 13 + 4√2 |
AMC12_991 | For certain real numbers $a$ , $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$ ?
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$
| Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\ r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$ .
Now applying Vieta's formulas on the constant term of $g(x)$ , the linear term of $g(x)$ , and the linear term of $f(x)$ , we obtain:
\begin{align*} r_1r_2r_3 & = -10\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\ r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100\\ \end{align*}
Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain:
\[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100\]
It follows that $r_4=-90$ . But $r_4=a-1$ so $a=-89$
Now we can factor $f(x)$ in terms of $g(x)$ as
\[f(x)=(x-r_4)g(x)=(x+90)g(x)\]
Then $f(1)=91g(1)$ and
\[g(1)=1^3-89\cdot 1^2+1+10=-77\]
Hence $f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | -7007 |
AMC12_992 | Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\]
has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\]
where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$ ?
$\textbf{(A) } {-}4 \qquad \textbf{(B) } {-}1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4$
| We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$
Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)$ , which can be derived using sine angle addition with $\sin{(2x + x)}$ , we have \[a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)\]
Since $\sin{x} \ne 0$ as it is on the open interval $(0, \pi)$ , we can divide out $\sin{x}$ from both sides, leaving us with \[a\cdot(1+2\cos{x})=4\cos^2{x}-1\]
Now, distributing $a$ and rearranging, we achieve the equation \[4\cos^2{x} - 2a\cos{x} - (1+a) = 0\] which is a quadratic in $\cos{x}$ .
Applying the quadratic formula to solve for $\cos{x}$ , we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}\] and expanding the terms under the radical, we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}\]
Factoring, since $4a^2+16a+16 = (2a+4)^2$ , we can simplify our expression even further to \[\cos{x} =\frac{a\pm(a+2)}{4}\]
Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$ .
Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \frac{2\pi}{3}$ , we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.
As $x\in (0, \pi)$ , $\cos{x}\in (-1, 1)$ , and therefore $\frac{a+1}{2}\in (-1, 1)$ , and $a\in(-3,1)$ .
There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$ , as this would lead to a double root for $\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \ne -2$ .
Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$ , so the answer is $-3-2+1 = \boxed{\textbf{(A) } {-}4}$ .
- DavidHovey
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_17 | -4 |
AMC12_993 | How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$ ?
$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$
| The ranges of $\frac{\pi}2 \sin x$ and $\frac{\pi}2 \cos x$ are both $\left[-\frac{\pi}2, \frac{\pi}2 \right],$ which is included in the range of $\arcsin,$ so we can use it with no issues.
\begin{align*} \frac{\pi}2 \cos x &= \arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \arcsin \left( \sin \left( \frac{\pi}2 - \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x \\ \cos x &= 1 - \sin x \\ \cos x + \sin x &= 1. \end{align*}
This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi],$ because one of $\sin$ and $\cos$ must be $1$ and the other $0.$ Therefore, the answer is $\boxed{\textbf{(C) }2}.$
~Tucker
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_19 | 2 |
AMC12_995 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ , $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$ ?
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$
| Solution 1Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$ , so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ ( $\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$ , so we're good there).
$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}$
These are in the ratio $3:2:4$ , so our minimal triangle has side lengths $2$ , $3$ , and $4$ . $\boxed{\textbf{(A) } 9}$ is our answer.
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | 9 |
AMC12_998 | Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?
$\mathrm{(A)}\ \dfrac{1}{36} \qquad\mathrm{(B)}\ \dfrac{1}{24} \qquad\mathrm{(C)}\ \dfrac{1}{18} \qquad\mathrm{(D)}\ \dfrac{1}{12} \qquad\mathrm{(E)}\ \dfrac{1}{9}$
| The first point is placed anywhere on the circle, because it doesn't matter where it is chosen.
The next point must lie within $60$ degrees of arc on either side, a total of $120$ degrees possible, giving a total $\frac{1}{3}$ chance. The last point must lie within $60$ degrees of both points.
The minimum area of freedom we have to place the third point is a $60$ degrees arc(if the first two are $60$ degrees apart), with a $\frac{1}{6}$ probability.
The maximum amount of freedom we have to place the third point is a $120$ degree arc(if the first two are the same point), with a $\frac{1}{3}$ probability.
As the second point moves farther away from the first point, up to a maximum of $60$ degrees, the probability changes linearly (every degree it moves, adds one degree to where the third could be).
Therefore, we can average probabilities at each end to find $\frac{1}{4}$ , the average probability we can place the third point based on a varying second point.
Therefore the total probability is $1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$ or $\boxed{\text{(D)}}$
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_25 | D |
AMC12_1000 | There are lily pads in a row numbered $0$ to $11$ , in that order. There are predators on lily pads $3$ and $6$ , and a morsel of food on lily pad $10$ . Fiona the frog starts on pad $0$ , and from any given lily pad, has a $\frac{1}{2}$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6$ ?
$\textbf{(A) } \frac{15}{256} \qquad \textbf{(B) } \frac{1}{16} \qquad \textbf{(C) } \frac{15}{128}\qquad \textbf{(D) } \frac{1}{8} \qquad \textbf{(E) } \frac14$
| Firstly, notice that if Fiona jumps over the predator on pad $3$ , she must land on pad $4$ . Similarly, she must land on $7$ if she makes it past $6$ . Thus, we can split the problem into $3$ smaller sub-problems, separately finding the probability Fiona skips $3$ , the probability she skips $6$ (starting at $4$ ) and the probability she doesn't skip $10$ (starting at $7$ ). Notice that by symmetry, the last of these three sub-problems is the complement of the first sub-problem, so the probability will be $1 - \text{the probability obtained in the first sub-problem}$ .
In the analysis below, we call the larger jump a $2$ -jump, and the smaller a $1$ -jump.
For the first sub-problem, consider Fiona's options. She can either go $1$ -jump, $1$ -jump, $2$ -jump, with probability $\frac{1}{8}$ , or she can go $2$ -jump, $2$ -jump, with probability $\frac{1}{4}$ . These are the only two options, so they together make the answer $\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$ . We now also know the answer to the last sub-problem is $1-\frac{3}{8}=\frac{5}{8}$ .
For the second sub-problem, Fiona must go $1$ -jump, $2$ -jump, with probability $\frac{1}{4}$ , since any other option would result in her death to a predator.
Thus, since the three sub-problems are independent, the final answer is $\frac{3}{8} \cdot \frac{1}{4} \cdot \frac{5}{8} = \boxed{\textbf{(A) }\frac{15}{256}}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_16 | 15/256 |
AMC12_1003 | A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$
| Using point slope form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$ . Simplifying, we get $6x-y=210$ and $2x-y=50$ . Letting $y=0$ in both equations and solving for $x$ gives the $x$ -intercepts: $x=35$ and $x=25$ , respectively. Thus the distance between them is $35-25=\boxed{\textbf{(B) } 10}$ .
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_3 | 10 |
AMC12_1004 | A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$ -axis or $y$ -axis. Let $A = (-3, 2)$ and $B = (3, -2)$ . Consider all possible paths of the bug from $A$ to $B$ of length at most $20$ . How many points with integer coordinates lie on at least one of these paths?
$\textbf{(A)}\ 161 \qquad \textbf{(B)}\ 185 \qquad \textbf{(C)}\ 195 \qquad \textbf{(D)}\ 227 \qquad \textbf{(E)}\ 255$
| We declare a point $(x, y)$ to make up for the extra steps that the bug has to move. If the point $(x, y)$ satisfies the property that $|x - 3| + |y + 2| + |x + 3| + |y - 2| \le 20$ , then it is in the desirable range because $|x - 3| + |y + 2|$ is the length of the shortest path from $(x,y)$ to $(3, -2)$ and $|x + 3| + |y - 2|$ is the length of the shortest path from $(x,y)$ to $(-3, 2)$ .
If $-3\le x \le 3$ , then $-7\le y \le 7$ satisfy the property. there are $15 \times 7 = 105$ lattice points here.
else let $3< x \le 8$ (and for $-8 \le x < -3$ because it is symmetrical) We set 8 as the upper bound for x because the shortest distance from $(-3, 2)$ to $(x, y)$ added to the shortest distance from $(x, y)$ to $(3, -2)$ is $|x - 3| + |y + 2| + |x + 3| + |y - 2|$ . Since the minimum value for the difference between the y-coordinates is at $y = 0$ , we get $2x + 4 = 20$ or $-2x + 4 = 20$ . Thus, the upper and lower bounds for $x$ are $8$ and $-8$ , respectively.
Now we test each value for x satisfying $3< x \le 8$ and double the result because of symmetry.
For $x = 4$ , the possibles values of y are such that $|2y| \le 12$ for a total of $13$ lattice points,
for $x = 5$ , the possibles values of y are such that $|2y| \le 10$ for a total of $11$ lattice points,
for $x = 6$ , the possibles values of y are such that $|2y| \le 8$ for a total of $9$ lattice points,
for $x = 7$ , the possibles values of y are such that $|2y| \le 6$ for a total of $7$ lattice points,
for $x = 8$ , the possibles values of y are such that $|2y| \le 4$ for a total of $5$ lattice points,
Hence, there are a total of $105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{(C) 195}$ lattice points.
One may also obtain the result by using Pick's Theorem(how?).
$i = a - \frac{b}{2} - 1$ (Suggestion)
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_23 | 195 |
AMC12_1007 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$ ?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
| The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.
Derive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule,
\begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\ f’’(x) &= 6x-2a \end{align*}
So $x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).
The function with the minimum $a$ :
\[f(x)=\left(x-\frac{a}{3}\right)^3\]
\[x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}\]
Since this is equal to the original equation $x^3-ax^2+bx-a$ , equating the coefficients, we get that
\[\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}\]
\[b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}\]
The actual function:
$f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}$
$f(x)=0\rightarrow x=\sqrt{3}$ triple root. "Complete the cube."
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | 9 |
AMC12_1009 | A circle has center $(-10, -4)$ and has radius $13$ . Another circle has center $(3, 9)$ and radius $\sqrt{65}$ . The line passing through the two points of intersection of the two circles has equation $x+y=c$ . What is $c$ ?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}$
| The equations of the two circles are $(x+10)^2+(y+4)^2=169$ and $(x-3)^2+(y-9)^2=65$ . Rearrange them to $(x+10)^2+(y+4)^2-169=0$ and $(x-3)^2+(y-9)^2-65=0$ , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation $(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65$ . We can simplify this like the following. $(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65 \rightarrow (x^2+20x+100)+(y^2+8y+16)-(x^2-6x+9)-(y^2-18y+81)=104 \rightarrow 26x+26y+26=104 \rightarrow 26x+26y=78 \rightarrow x+y=3$ . Thus, $c = \boxed{\textbf{(A)}\ 3}$ .
Solution by TheUltimate123
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_9 | 3 |
AMC12_1011 | Suppose that $a$ , $b$ , $c$ and $d$ are positive integers satisfying all of the following relations.
\[abcd=2^6\cdot 3^9\cdot 5^7\]
\[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\]
\[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\]
\[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\]
\[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\]
\[\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2\]
\[\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2\]
What is $\text{gcd}(a,b,c,d)$ ?
$\textbf{(A)}~30\qquad\textbf{(B)}~45\qquad\textbf{(C)}~3\qquad\textbf{(D)}~15\qquad\textbf{(E)}~6$
| Denote by $\nu_p (x)$ the number of prime factor $p$ in number $x$ .
We index Equations given in this problem from (1) to (7).
First, we compute $\nu_2 (x)$ for $x \in \left\{ a, b, c, d \right\}$ .
Equation (5) implies $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$ .
Equation (2) implies $\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3$ .
Equation (6) implies $\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2$ .
Equation (1) implies $\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6$ .
Therefore, all above jointly imply $\nu_2 (a) = 3$ , $\nu_2 (d) = 2$ , and $\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)$ or $\left( 1, 0 \right)$ .
Second, we compute $\nu_3 (x)$ for $x \in \left\{ a, b, c, d \right\}$ .
Equation (2) implies $\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2$ .
Equation (3) implies $\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3$ .
Equation (4) implies $\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3$ .
Equation (1) implies $\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9$ .
Therefore, all above jointly imply $\nu_3 (c) = 3$ , $\nu_3 (d) = 3$ , and $\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)$ or $\left( 2, 1 \right)$ .
Third, we compute $\nu_5 (x)$ for $x \in \left\{ a, b, c, d \right\}$ .
Equation (5) implies $\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2$ .
Equation (2) implies $\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3$ .
Thus, $\nu_5 (a) = 3$ .
From Equations (5)-(7), we have either $\nu_5 (b) \leq 1$ and $\nu_5 (c) = \nu_5 (d) = 2$ , or $\nu_5 (b) = 2$ and $\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2$ .
Equation (1) implies $\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7$ .
Thus, for $\nu_5 (b)$ , $\nu_5 (c)$ , $\nu_5 (d)$ , there must be two 2s and one 0.
Therefore,
\begin{align*} {\rm gcd} (a,b,c,d) & = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\ & = 2^0 \cdot 3^1 \cdot 5^0 \\ & = \boxed{\textbf{(C) 3}}. \end{align*}
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_24 | 3 |
AMC12_1013 | For every $m$ and $k$ integers with $k$ odd, denote by $\left[\frac{m}{k}\right]$ the integer closest to $\frac{m}{k}$ . For every odd integer $k$ , let $P(k)$ be the probability that
\[\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]\]
for an integer $n$ randomly chosen from the interval $1 \leq n \leq 99!$ . What is the minimum possible value of $P(k)$ over the odd integers $k$ in the interval $1 \leq k \leq 99$ ?
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{44}{87} \qquad \textbf{(D)}\ \frac{34}{67} \qquad \textbf{(E)}\ \frac{7}{13}$
| Solution 1Answer: $(D) \frac{34}{67}$
First of all, you have to realize that
if $\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$
then $\left[\frac{n - k}{k}\right] + \left[\frac{100 - (n - k)}{k}\right] = \left[\frac{100}{k}\right]$
So, we can consider what happen in $1\le n \le k$ and it will repeat. Also since range of $n$ is $1$ to $99!$ , it is always a multiple of $k$ . So we can just consider $P(k)$ for $1\le n \le k$ .
Let $\text{fpart}(x)$ be the fractional part function
This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider $k = 99$ , $87$ , $67$ , $13$ . $1\le n \le k$
For $k > \frac{200}{3}$ , $\left[\frac{100}{k}\right] = 1$ . 3 of the $k$ that we should consider land in here.
For $n < \frac{k}{2}$ , $\left[\frac{n}{k}\right] = 0$ , then we need $\left[\frac{100 - n}{k}\right] = 1$
else for $\frac{k}{2}< n < k$ , $\left[\frac{n}{k}\right] = 1$ , then we need $\left[\frac{100 - n}{k}\right] = 0$
For $n < \frac{k}{2}$ , $\left[\frac{100 - n}{k}\right] = \left[\frac{100}{k} - \frac{n}{k}\right]= 1$
So, for the condition to be true, $100 - n > \frac{k}{2}$ . ( $k > \frac{200}{3}$ , no worry for the rounding to be $> 1$ )
$100 > k > \frac{k}{2} + n$ , so this is always true.
For $\frac{k}{2}< n < k$ , $\left[\frac{100 - n}{k}\right] = 0$ , so we want $100 - n < \frac{k}{2}$ , or $100 < \frac{k}{2} + n$
$100 <\frac{k}{2} + n < \frac{3k}{2}$
For k = 67, $67 > n > 100 - \frac{67}{2} = 66.5$
For k = 69, $69 > n > 100 - \frac{69}{2} = 67.5$
etc.
We can clearly see that for this case, $k = 67$ has the minimum $P(k)$ , which is $\frac{34}{67}$ . Also, $\frac{7}{13} > \frac{34}{67}$ .
So for AMC purpose, answer is $\boxed{\textbf{(D) }\frac{34}{67}}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_25 | 34/67 |
AMC12_1015 | A unit cube has vertices $P_1,P_2,P_3,P_4,P_1',P_2',P_3',$ and $P_4'$ . Vertices $P_2$ , $P_3$ , and $P_4$ are adjacent to $P_1$ , and for $1\le i\le 4,$ vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2$ , $P_1P_3$ , $P_1P_4$ , $P_1'P_2'$ , $P_1'P_3'$ , and $P_1'P_4'$ . What is the octahedron's side length?
[asy] import three; size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye); triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]}; // draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype("4 4")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype("4 4")); // draw cube for(int i = 0; i < 4; ++i){ draw(P[1]--P[i]); draw(Pp[1]--Pp[i]); for(int j = 0; j < 4; ++j){ if(i == 1 || j == 1 || i == j) continue; draw(P[i]--Pp[j]); draw(Pp[i]--P[j]); } dot(P[i]); dot(Pp[i]); dot(pt(i)); dot(ptp(i)); } label("$P_1$", P[1], dir(P[1])); label("$P_2$", P[2], dir(P[2])); label("$P_3$", P[3], dir(-45)); label("$P_4$", P[4], dir(P[4])); label("$P'_1$", Pp[1], dir(Pp[1])); label("$P'_2$", Pp[2], dir(Pp[2])); label("$P'_3$", Pp[3], dir(-100)); label("$P'_4$", Pp[4], dir(Pp[4])); [/asy]
$\textbf{(A)}\ \frac{3\sqrt{2}}{4}\qquad\textbf{(B)}\ \frac{7\sqrt{6}}{16}\qquad\textbf{(C)}\ \frac{\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \frac{\sqrt{6}}{2}$
|
Observe the diagram above. Each dot represents one of the six vertices of the regular octahedron. Three dots have been placed exactly x units from the $(0,0,0)$ corner of the unit cube. The other three dots have been placed exactly x units from the $(1,1,1)$ corner of the unit cube. A red square has been drawn connecting four of the dots to provide perspective regarding the shape of the octahedron. Observe that the three dots that are near $(0,0,0)$ are each $(x)(\sqrt{2}$ ) from each other. The same is true for the three dots that are near $(1,1,1).$ There is a unique $x$ for which the rectangle drawn in red becomes a square. This will occur when the distance from $(x,0,0)$ to $(1,1-x, 1)$ is $(x)(\sqrt{2}$ ).
Using the distance formula we find the distance between the two points to be: $\sqrt{{(1-x)^2} + {(1-x)^2} + 1}$ = $\sqrt{2x^2 - 4x +3}$ . Equating this to $(x)(\sqrt{2}$ ) and squaring both sides, we have the equation:
$2{x^2} - 4x + 3$ = $2{x^2}$
$-4x + 3 = 0$
$x$ = $\frac{3} {4}$ .
Since the length of each side is $(x)(\sqrt{2}$ ), we have a final result of $\frac{3 \sqrt{2}}{4}$ . Thus, Answer choice $\boxed{\text{A}}$ is correct.
(If someone can draw a better diagram with the points labeled P1,P2, etc., I would appreciate it).
-- Jm314 14:55, 26 February 2012 (EST)
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_19 | A |
AMC12_1018 | In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
$\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$
| If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: \[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5\] Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Then, the first side has length $3*7=21$ , the second side has length $7$ , the third side has length $15$ , and so the perimeter is $21+7+15=43 \Rightarrow \boxed{\text {(A)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_10 | A |
AMC12_1019 | Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which one of these statements necessarily follows logically?
$\textbf{(A)}\ \text{ If Lewis did not receive an A, then he got all of the multiple choice questions wrong.} \\ \qquad\textbf{(B)}\ \text{ If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.} \\ \qquad\textbf{(C)}\ \text{ If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A.} \\ \qquad\textbf{(D)}\ \text{ If Lewis received an A, then he got all of the multiple choice questions right.} \\ \qquad\textbf{(E)}\ \text{ If Lewis received an A, then he got at least one of the multiple choice questions right.}$
| Taking the contrapositive of the statement "if he got all of them right, he got an A" yields "if he didn't get an A, he didn't get all of them right", yielding the answer $\boxed{\textbf{(B)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_3 | B |
AMC12_1020 | In $\triangle ABC$ , $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$
| Draw a good diagram! Now, let's call $BD=t$ , so $DC=2t$ . Given the rather nice angles of $\angle ABD = 45^\circ$ and $\angle ADC = 60^\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$ ; call this point $H$ . We realize that there is no specific factor of $t$ we can call this just yet, so let $AH=kt$ . Notice that in $\triangle{ABH}$ we get $BH=kt$ . Using the 60-degree angle in $\triangle{ADH}$ , we obtain $DH=\frac{\sqrt{3}}{3}kt$ . The comparable ratio is that $BH-DH=t$ . If we involve our $k$ , we get:
$kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$ . Eliminating $t$ and removing radicals from the denominator, we get $k=\frac{3+\sqrt{3}}{2}$ . From there, one can easily obtain $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$ . Now we finally have a desired ratio. Since $\tan\angle ACH = 2+\sqrt{3}$ upon calculation, we know that $\angle ACH$ can be simplified. Indeed, if you know that $\tan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and cosine using $\sin(x)^2+\cos(x)^2=1$ , and perhaps the half- or double-angle formulas, you get $\boxed{75^\circ}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | 75 |
AMC12_1024 | What is the product of all solutions to the equation
\[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A) } (\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B) } \log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C) } 1 \\ \qquad \textbf{(D) } \log_{7}2023\cdot \log_{289}2023\qquad \textbf{(E) } (\log_7 2023\cdot\log_{289} 2023)^2$
| For $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$ , transform it into $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$ . Replace $\ln x$ with $y$ . Because we want to find the product of all solutions of $x$ , it is equivalent to finding the exponential of the sum of all solutions of $y$ . Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled. By using Vieta, we see that since there does not exist a $by$ term, $\sum y=0$ and $\prod x=e^0=\boxed{\textbf{(C)} 1}$ .
~plasta
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_19 | 1 |
AMC12_1025 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl's garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$
| To start, use algebra to determine the number of posts on each side. You have (the long sides count for $2$ because there are twice as many) $6x = 20 + 4$ (each corner is double counted so you must add $4$ ) Making the shorter end have $4$ , and the longer end have $8$ . $((8-1) \cdot 4) \cdot ((4-1) \cdot 4) = 28 \cdot 12 = 336$ . Therefore, the answer is $\boxed{\textbf{(B)}\ 336}$
~Albert471
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_9 | 336 |
AMC12_1027 | Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?
$\textbf{(A)}\; 8 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 14 \qquad\textbf{(D)}\; 15 \qquad\textbf{(E)}\; 18$
| Let $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$ . Plugging in $a = 28$ doesn't yield an integer for $b$ , so it must be that $b = 28$ , and we get $2a + 84 = 100$ . Solving for $a$ , we obtain $a = \boxed{\textbf{(A)}\; 8}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_3 | 8 |
AMC12_1031 | Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$ . What is $x$ ?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26$
| Use SFFT to get $(x+1)(y+1)=81$ . The terms $(x+1)$ and $(y+1)$ must be factors of $81$ , which include $1, 3, 9, 27, 81$ . Because $x > y$ , $x+1$ is equal to $27$ or $81$ . But if $x+1=81$ , then $y=0$ and so $x=\boxed{\textbf{(E)}\ 26}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_10 | 26 |
AMC12_1032 | LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid A dollars and Bernardo had paid B dollars, where $A < B.$ How many dollars must LeRoy give to Bernardo so that they share the costs equally?
$\textbf{(A)}\ \frac{A+B}{2} \qquad \textbf{(B)}\ \frac{A-B}{2} \qquad \textbf{(C)}\ \frac{B-A}{2} \qquad \textbf{(D)}\ B-A \qquad \textbf{(E)}\ A+B$
| The total amount of money that was spent during the trip was
\[A + B\]
So each person should pay
\[\frac{A+B}{2}\]
if they were to share the costs equally. Because LeRoy has already paid $A$ dollars of his part, he still has to pay
\[\frac{A+B}{2} - A =\]
\[= \boxed{\frac{B-A}{2} \textbf{(C)}}\]
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_3 | B-A/2 |
AMC12_1033 | For positive integers $m$ and $n$ such that $m+10<n+1$ , both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$ . What is $m+n$ ?
$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$
| The mean and median are \[\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,\] so $3m+17=2n$ and $m+11=n$ . Solving this gives $\left(m,n\right)=\left(5,16\right)$ for $m+n=\boxed{\textbf{(B)}~21}$ . (trumpeter)
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_6 | 21 |
AMC12_1035 | Set $u_0 = \frac{1}{4}$ , and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence \[u_{k+1} = 2u_k - 2u_k^2.\]
This sequence tends to a limit; call it $L$ . What is the least value of $k$ such that \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$
| Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac12\right)$ and are strictly increasing.
Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$
The given equation becomes \[L=2L-2L^2,\] from which $L=\frac12.$
The given inequality becomes \[\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},\] and we only need to consider $\frac12-\frac{1}{2^{1000}} \leq u_k.$
We have
\begin{alignat*}{8} u_0 &= \phantom{1}\frac14 &&= \frac{2^1-1}{2^2}, \\ u_1 &= \phantom{1}\frac38 &&= \frac{2^2-1}{2^3}, \\ u_2 &= \ \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\ u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\ & \phantom{1111} \vdots \end{alignat*}
By induction, it can be proven that \[u_k=\frac{2^{2^k}-1}{2^{2^k+1}}=\frac12-\frac{1}{2^{2^k+1}}.\]
We substitute this into the inequality, then solve for $k:$
\begin{align*} \frac12-\frac{1}{2^{1000}} &\leq \frac12-\frac{1}{2^{2^k+1}} \\ -\frac{1}{2^{1000}} &\leq -\frac{1}{2^{2^k+1}} \\ 2^{1000} &\leq 2^{2^k+1} \\ 1000 &\leq 2^k+1. \end{align*}
Since $2^9+1 \leq 1000 \leq 2^{10}+1,$ the least such value of $k$ is $\boxed{\textbf{(A)}\: 10}.$
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_18 | 10 |
AMC12_1036 | Let $ABCD$ be a parallelogram with area $15$ . Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$
| Let $X$ denote the intersection point of the diagonals $AC$ and $BD$ . Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$ , so $XP = XQ = 3$ and $XR = XS = 4$ . Now note that since $\angle APB = \angle ARB = 90^\circ$ , quadrilateral $ARPB$ is cyclic, and so
\[XR\cdot XA = XP\cdot XB,\] which implies $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$ .
Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$ . Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$ , and so \[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15\] Solving this for $x^2$ yields $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$ , and so \[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\] The requested answer is $32 + 8 + 41 = \boxed{\textbf{(A)} ~81}$ .
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_24 | 81 |
AMC12_1038 | In $\triangle ABC$ , $\angle C = 90^\circ$ and $AB = 12$ . Squares $ABXY$ and $CBWZ$ are constructed outside of the triangle. The points $X$ , $Y$ , $Z$ , and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32$
| [asy] pair A,B,C,M,E,W,Z,X,Y; A=(2,0); B=(0,2); C=(0,0); M=(A+B)/2; W=(-2,2); Z=(-2,-0); X=(2,4); Y=(4,2); E=(W+Z)/2; draw(A--B--C--cycle); draw(W--B--C--Z--cycle); draw(A--B--X--Y--cycle); dot(M); dot(E); label("W",W,NW); label("Z",Z,SW); label("C",C,S); label("A",A,S); label("B",B,N); label("X",X,NE); label("Y",Y,SE); label("E",E,1.5*plain.W); label("M",M,NE); draw(circle(M,sqrt(10))); [/asy]
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of $WZ$ and $XY$ and finding their intersection point. This point happens to be the midpoint of $AB$ , the hypotenuse. Let this point be $M$ . To find the radius, determine $MY$ , where $MY^{2} = MA^2 + AY^2$ , $MA = \frac{12}{2} = 6$ , and $AY = AB = 12$ . Thus, the radius $=r =MY = 6\sqrt5$ .
Next we let $AC = b$ and $BC = a$ . Consider the right triangle $ACB$ first. Using the Pythagorean theorem, we find that $a^2 + b^2 = 12^2 = 144$ .
[asy] pair A,B,C,M,E,W,Z,X,Y; A=(2,0); B=(0,2); C=(0,0); M=(A+B)/2; W=(-2,2); Z=(-2,-0); X=(2,4); Y=(4,2); E=(W+Z)/2; draw(A--B--C--cycle); draw(W--B--C--Z--cycle); draw(A--B--X--Y--cycle); dot(M); dot(E); label("W",W,NW); label("Z",Z,SW); label("C",C,S); label("A",A,S); label("B",B,N); label("X",X,NE); label("Y",Y,SE); label("E",E,1.5*plain.W); label("M",M,NE); draw(circle(M,sqrt(10))); draw(E--Z--M--cycle,dashed); [/asy]
Now, we let $E$ be the midpoint of $WZ$ , and we consider right triangle $ZEM$ . By the Pythagorean theorem, we have that $\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180$ . Expanding this equation, we get that
\[\frac{1}{4}(a^2+b^2) + a^2 + ab = 180\]
\[\frac{144}{4} + a^2 + ab = 180\]
\[a^2 + ab = 144 = a^2 + b^2\]
\[ab = b^2\]
\[b = a\]
This means that $ABC$ is a 45-45-90 triangle, so $a = b = \frac{12}{\sqrt2} = 6\sqrt2$ . Thus the perimeter is $a + b + AB = 12\sqrt2 + 12$ which is answer $\boxed{\textbf{(C)}\; 12 + 12\sqrt2}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_19 | 12 + 12√2 |
AMC12_1039 | Let $f(x) = \sin{x} + 2\cos{x} + 3\tan{x}$ , using radian measure for the variable $x$ . In what interval does the smallest positive value of $x$ for which $f(x) = 0$ lie?
$\textbf{(A)}\ (0,1) \qquad \textbf{(B)}\ (1, 2) \qquad\textbf{(C)}\ (2, 3) \qquad\textbf{(D)}\ (3, 4) \qquad\textbf{(E)}\ (4,5)$
| We must first get an idea of what $f(x)$ looks like:
Between $0$ and $1$ , $f(x)$ starts at $2$ and increases; clearly there is no zero here.
Between $1$ and $\frac{\pi}{2}$ , $f(x)$ starts at a positive number and increases to $\infty$ ; there is no zero here either.
Between $\frac{\pi}{2}$ and 3, $f(x)$ starts at $-\infty$ and increases to some negative number; there is no zero here either.
Between $3$ and $\pi$ , $f(x)$ starts at some negative number and increases to -2; there is no zero here either.
Between $\pi$ and $\pi+\frac{\pi}{4} < 4$ , $f(x)$ starts at -2 and increases to $-\frac{\sqrt2}{2} + 2\left(-\frac{\sqrt2}{2}\right) + 3\left(1\right)=3\left(1-\frac{\sqrt2}{2}\right)>0$ . There is a zero here by the Intermediate Value Theorem. Therefore, the answer is $\boxed{\textbf{(D)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_15 | D |
AMC12_1041 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$ )
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$
| For $6$ to $8$ heads, we are guaranteed to hit $4$ , so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37$ .
For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$ .
For $5$ heads, we either start off with $4$ heads, which gives us $_4\text{C}_1=4$ ways to arrange the other flips, or we start off with five heads and one tail, which has $6$ ways minus the $2$ overlapping cases, $\text{HHHHHTTT}$ and $\text{HHHHTHTT}$ . Total ways: $8$ .
Then we sum to get $46$ . There are a total of $2^8=256$ possible sequences of $8$ coin flips, so the probability is $\frac{46}{256}=\frac{23}{128}$ . Summing, we get $23+128=\boxed{\textbf{(B) }151}$ .
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | 151 |
AMC12_1042 | A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("$1$", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("$4$", B -- (B+(4,0)), N ); label("$A$",A,E); label("$B$",B,W); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]
$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$
| Solution 1[asy] unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$S$",S,N); filldraw( circle(S,4/9), lightgray, black ); dot(S); draw( rightanglemark(A,C,B) ); draw( S -- A ); draw( S -- B ); [/asy]
In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$ , thus by the Pythagorean theorem we have $AC=4$ .
Let $r$ be the radius of the small circle, and let $s$ be the perpendicular distance from $S$ to $\overline{AC}$ . Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$ .
We have $SA = \sqrt{s^2 + (1-r)^2}$ and $SB=\sqrt{(4-s)^2 + (4-r)^2}$ . Hence we get the following two equations:
\begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*}
Simplifying both, we get
\begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*}
As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\left( \frac{4-s}s \right)^2 = 4$ .
Now there are two possibilities: either $\frac{4-s}s=-2$ , or $\frac{4-s}s=2$ .
In the first case clearly $s<0$ , which puts the center on the wrong side of $A$ , so this is not the correct case.
(Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the $4:1$ ratio between radii of $A$ and $B$ , this circle turns out to have the same radius as circle $B$ , with center directly left of center $B$ , and tangent to $B$ directly above center $A$ .)
The second case solves to $s=\frac 43$ . We then have $4r = s^2 = \frac {16}9$ , hence $r = \boxed{\frac 49}$ .
More generally, for two large circles of radius $a$ and $b$ , the radius $c$ of the small circle is $c = \frac{ab}{\left(\sqrt{a}+\sqrt{b}\right)^2} = \frac{1}{\left(1/\sqrt{a}+1/\sqrt{b}\right)^2}$ .
Equivalently, we have that $1/\sqrt{c} = 1/\sqrt{a} + 1/\sqrt{b}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_18 | 4/9 |
AMC12_1043 | How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ , $b$ , $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$ )
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$
| Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$ . We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{2\pi i / 3}$ . That is because of Euler's Formula : $e^{ix} = \cos(x) + i \cdot \sin(x)$ . $\frac{-1+i\sqrt{3}}{2}$ = $-\frac{1}{2} + i \cdot \frac {\sqrt{3}}{2}$ = $\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}$ .
In order $r$ to be a root of $P$ , $re^{2\pi i / 3}$ must also be a root of P, meaning that 3 of the roots of $P$ must be $r$ , $re^{i\frac{2\pi}{3}}$ , $re^{i\frac{4\pi}{3}}$ . However, since $P$ is degree 5, there must be two additional roots. Let one of these roots be $w$ , if $w$ is a root, then $we^{2\pi i / 3}$ and $we^{4\pi i / 3}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$ , $re^{2\pi i / 3}$ , or $re^{4\pi i / 3}$ . Thus we know that the polynomial $P$ can be written in the form $(x-r)^m(x-re^{2\pi i / 3})^n(x-re^{4\pi i / 3})^p$ . Moreover, by Vieta's, we know that there is only one possible value for the magnitude of $r$ as $||r||^5 = 2020$ , meaning that the amount of possible polynomials $P$ is equivalent to the possible sets $(m,n,p)$ . In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{2\pi i / 3}$ and $re^{4 \pi i / 3}$ being conjugates and since $m+n+p = 5$ , (as the polynomial is 5th degree) we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{\textbf{(C) } 2}$ .
~Murtagh
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_17 | 2 |
AMC12_1046 | Suppose $x$ and $y$ are positive real numbers such that
\[x^y=2^{64}\text{ and }(\log_2{x})^{\log_2{y}}=2^{7}.\]
What is the greatest possible value of $\log_2{y}$ ?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }3+\sqrt{2} \qquad \textbf{(D) }4+\sqrt{3} \qquad \textbf{(E) }7$
| Take the base-two logarithm of both equations to get
\[y\log_2 x = 64\quad\text{and}\quad (\log_2 y)(\log_2\log_2 x) = 7.\]
Now taking the base-two logarithm of the first equation again yields
\[\log_2 y + \log_2\log_2 x = 6.\]
It follows that the real numbers $r:=\log_2 y$ and $s:=\log_2\log_2 x$ satisfy $r+s=6$ and $rs = 7$ . Solving this system yields
\[\{\log_2 y,\log_2\log_2 x\}\in\{3-\sqrt 2, 3 + \sqrt 2\}.\]
Thus the largest possible value of $\log_2 y$ is $3+\sqrt 2 \implies \boxed{\textbf {(C)}}$ .
cr. djmathman
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_16 | C |
AMC12_1049 | For all integers $n \geq 9,$ the value of
\[\frac{(n+2)!-(n+1)!}{n!}\] is always which of the following?
$\textbf{(A) } \text{a multiple of 4} \qquad \textbf{(B) } \text{a multiple of 10} \qquad \textbf{(C) } \text{a prime number} \qquad \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}$
| We first expand the expression:
\[\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}.\]
We can now divide out a common factor of $n!$ from each term of the numerator:
\[(n+2)(n+1)-(n+1).\]
Factoring out $(n+1),$ we get \[[(n+2)-1](n+1) = (n+1)^2,\]
which proves that the answer is $\boxed{\textbf{(D) } \text{a perfect square}}.$
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_6 | a perfect square |
AMC12_1050 | What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$ ?
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$
| If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$ , where $m$ is any integer from $1$ to $119$ , inclusive: $|mx-1|=0\implies mx=1\implies x=\frac{1}{m}$ .
The minimum value of $f(x)$ occurs where the absolute value of the sum of the slopes is at a minimum $\ge 0$ , since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some $\frac{1}{m}$ .
The sum of the slopes at $x = \frac{1}{m}$ is
\begin{align*}&\sum_{i=m+1}^{119}i - \sum_{i=1}^{m}i\\ &=\sum_{i=1}^{119}i - 2\sum_{i=1}^{m}i\\ &=-m^2-m+7140\end{align*}
Now we want to minimize $-m^2-m+7140$ . The zeros occur at $-85$ and $84$ , which means the slope is $0$ where $m = 84, 85$ .
We can now verify that both $x=\frac{1}{84}$ and $x=\frac{1}{85}$ yield $\boxed{49\ \textbf{(A)}}$ .
You can also think of the slopes playing 'tug of war', where the slope of each absolute function upon passing its $x$ -intercept is negated, positively tugging on the remaining negative slopes.
The sum of the slopes is $1+2+3+4\dots 119=\sum_{m=1}^{119}m=\frac{119\cdot 120}{2}=60\cdot 119=7140$
So we need to find the least integer $a$ such that $1+2+3+\dots a=\sum_{n=1}^an=\frac{a(a+1)}{2}\ge \frac{7140}{2}=3570:$
\[a(a+1)\ge 7140\implies a^2+a-7140\ge 0\rightarrow a=84\text{ exactly!}\]
This "exactly" means that the slope is ZERO between the whole interval $x\in\left(\frac{1}{85},\frac{1}{84}\right)$ . We can explicitly evaluate both to check that they are both equal to the desired minimum value of $f(x)$ :
\[\frac{84+83+\dots+2+1+1+2+\dots+33+34}{85}=\frac{84(85)/2+34(35)/2}{85}=\frac{85(14+84)/2}{85}=49\]
\[\frac{83+82+\dots+2+1+1+2+\dots+34+35}{84}=\frac{83(84)/2+35(36)/2}{84}=\frac{84(15+83)/2}{84}=49\]
Thus the minimum value of $f(x)$ is $49$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_22 | 49 |
AMC12_1051 | Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ , $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$ , and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$ . What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?
$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$
| For all nonnegative integers $n$ , let $\angle C_nA_nB_n=x_n$ , $\angle A_nB_nC_n=y_n$ , and $\angle B_nC_nA_n=z_n$ .
Note that quadrilateral $A_0B_0A_1B_1$ is cyclic since $\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ$ ; thus, $\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0$ . By a similar argument, $\angle A_0A_1C_1=\angle A_0C_0C_1=90^\circ-x_0$ . Thus, $x_1=\angle A_0A_1B_1+\angle A_0A_1C_1=180^\circ-2x_0$ . By a similar argument, $y_1=180^\circ-2y_0$ and $z_1=180^\circ-2z_0$ .
Therefore, for any positive integer $n$ , we have $x_n=180^\circ-2x_{n-1}$ (identical recurrence relations can be derived for $y_n$ and $z_n$ ). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to $n$ (and the coefficient of $x_0$ is $(-2)^n$ ). Hence, we let $x_n=pq^n+r+(-2)^nx_0$ . We will solve for $p$ , $q$ , and $r$ by iterating the recurrence to obtain $x_1=180^\circ-2x_0$ , $x_2=4x_0-180^\circ$ , and $x_3=540-8x_0$ . Letting $n=1,2,3$ respectively, we have \begin{align} pq+r&=180 \\ pq^2+r&=-180 \\ pq^3+r&=540 \end{align}
Subtracting $(1)$ from $(3)$ , we have $pq(q^2-1)=360$ , and subtracting $(1)$ from $(2)$ gives $pq(q-1)=-360$ . Dividing these two equations gives $q+1=-1$ , so $q=-2$ . Substituting back, we get $p=-60$ and $r=60$ .
We will now prove that for all positive integers $n$ , $x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60$ via induction. Clearly the base case of $n=1$ holds, so it is left to prove that $x_{n+1}=(-2)^{n+1}(x_0-60)+60$ assuming our inductive hypothesis holds for $n$ . Using the recurrence relation, we have \begin{align*} x_{n+1}&=180-2x_n \\ &=180-2((-2)^n(x_0-60)+60) \\ &=(-2)^{n+1}(x_0-60)+60 \end{align*}
Our induction is complete, so for all positive integers $n$ , $x_n=(-2)^n(x_0-60)+60$ . Identical equalities hold for $y_n$ and $z_n$ .
The problem asks for the smallest $n$ such that either $x_n$ , $y_n$ , or $z_n$ is greater than $90^\circ$ . WLOG, let $x_0=60^\circ$ , $y_0=59.999^\circ$ , and $z_0=60.001^\circ$ . Thus, $x_n=60^\circ$ for all $n$ , $y_n=-(-2)^n(0.001)+60$ , and $z_n=(-2)^n(0.001)+60$ . Solving for the smallest possible value of $n$ in each sequence, we find that $n=15$ gives $y_n>90^\circ$ . Therefore, the answer is $\boxed{\textbf{(E) } 15}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_25 | 15 |
AMC12_1054 | If $a,b,c,d$ are positive real numbers such that $a,b,c,d$ form an increasing arithmetic sequence and $a,b,d$ form a geometric sequence, then $\frac ad$ is
$\mathrm{(A)}\ \frac 1{12} \qquad\mathrm{(B)}\ \frac 16 \qquad\mathrm{(C)}\ \frac 14 \qquad\mathrm{(D)}\ \frac 13 \qquad\mathrm{(E)}\ \frac 12$
| Solution 1We can let $a=1$ , $b=2$ , $c=3$ , and $d=4$ . $\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_9 | 1/4 |
AMC12_1055 | Four positive integers $a$ , $b$ , $c$ , and $d$ have a product of $8!$ and satisfy:
\[\begin{array}{rl} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{array}\]
What is $a-d$ ?
$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$
| Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:
\begin{align*} (a+1)(b+1) & = 525 \\ (b+1)(c+1) & = 147 \\ (c+1)(d+1) & = 105 \end{align*}
Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$ . We get:
\begin{align*} ef & = 3\cdot 5\cdot 5\cdot 7 \\ fg & = 3\cdot 7\cdot 7 \\ gh & = 3\cdot 5\cdot 7 \end{align*}
Clearly $7^2$ divides $fg$ . On the other hand, $7^2$ can not divide $f$ , as it then would divide $ef$ . Similarly, $7^2$ can not divide $g$ . Hence $7$ divides both $f$ and $g$ . This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$ .
The first case solves to $(e,f,g,h)=(75,7,21,5)$ , which gives us $(a,b,c,d)=(74,6,20,4)$ , but then $abcd \not= 8!$ . We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$ . (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.)
The second case solves to $(e,f,g,h)=(25,21,7,15)$ , which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$ , and we have $a-d=24-14 =\boxed{10}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_21 | 10 |
AMC12_1058 | The square
| All the unknown entries can be expressed in terms of $b$ .
Since $100e = beh = ceg = def$ , it follows that $h = \frac{100}{b}, g = \frac{100}{c}$ ,
and $f = \frac{100}{d}$ . Comparing rows $1$ and $3$ then gives
$50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}$ ,
from which $c = \frac{20}{b}$ .
Comparing columns $1$ and $3$ gives
$50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}$ ,
from which $d = \frac{c}{5} = \frac{4}{b}$ .
Finally, $f = 25b, g = 5b$ , and $e = 10$ . All the entries are positive integers
if and only if $b = 1, 2,$ or $4$ . The corresponding values for $g$ are $5, 10,$ and
$20$ , and their sum is $\boxed{\mathbf{(C)}35}$ .
Credit to Solution B goes to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf , a page with a play-by-play explanation of the solutions to this test's problems.
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_22 | 35 |
AMC12_1063 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$ ?
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$
| For each digit $a=1,2,\ldots,9$ there are $10\cdot10$ (ways of choosing $b$ and $c$ ) palindromes. So the $a$ s contribute $(1+2+\cdots+9)(100)(10^4+1)$ to the sum.
For each digit $b=0,1,2,\ldots,9$ there are $9\cdot10$ (since $a \neq 0$ ) palindromes. So the $b$ s contribute $(0+1+2+\cdots+9)(90)(10^3+10)$ to the sum.
Similarly, for each $c=0,1,2,\ldots,9$ there are $9\cdot10$ palindromes, so the $c$ contributes $(0+1+2+\cdots+9)(90)(10^2)$ to the sum.
It just so happens that \[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\] so the sum of the digits of the sum is $\boxed{\textbf{(B)}\; 18}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | 18 |
AMC12_1065 | Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$
| The ratio $\frac{400 \text{ euros}}{500 \text{ dollars}}$ can be simplified using conversion factors: \[\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04\] which means the money is greater by $\boxed{ \textbf{(B)} \ 4 }$ percent.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_4 | 4 |
AMC12_1067 | Let $ABCD$ be an isosceles trapezoid with $\overline{BC}\parallel \overline{AD}$ and $AB=CD$ . Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$ , as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ$ , $AX = 3$ , $XY = 1$ , and $YC = 2$ . What is the area of $ABCD?$
[asy] size(10cm); usepackage("mathptmx"); import geometry; void perp(picture pic=currentpicture, pair O, pair M, pair B, real size=5, pen p=currentpen, filltype filltype = NoFill){ perpendicularmark(pic, M,unit(unit(O-M)+unit(B-M)),size,p,filltype); } pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),Y=(2,0),X=(3,0),A=(6,0),B=(2,sqrt(5.6)),D=(3,-sqrt(12.6)); draw(A--B--C--D--cycle,p); draw(A--C,p); draw(B--Y,p); draw(D--X,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(X,q); dot(Y,q); label("2",C--Y,S); label("1",Y--X,S); label("3",X--A,S); label("$A$",A,2*E); label("$B$",B,2*N); label("$C$",C,2*W); label("$D$",D,2*S); label("$Y$",Y,2*sqrt(2)*NE); label("$X$",X,2*N); perp(B,Y,C,8,p); perp(A,X,D,8,p); [/asy]
$\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}$
| First realize that $\triangle BCY \sim \triangle DAX.$ Thus, because $CY: XA = 2:3,$ we can say that $BY = 2s$ and $DX = 3s.$ From the Pythagorean Theorem, we have $AB^2 =(2s)^2 + 4^2 = 4s^2 + 16$ and $CD^2 = (3s)^2 + 3^2 = 9s^2 + 9.$ Because $AB = CD,$ from the problem statement, we have that \[4s^2 + 16 = 9s^2 + 9.\] Solving, gives $s = \frac{\sqrt{7}}{\sqrt{5}}.$ To find the area of the trapezoid, we can compute the area of $\triangle ABC$ and add it to the area of $\triangle ACD.$ Thus, the area of the trapezoid is $\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \left(\frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 \right) = \frac{15\sqrt{7}}{\sqrt{5}} = 3\sqrt{35}.$ Thus, the answer is $\boxed{\textbf{(C)} \: 3\sqrt{35}}.$
~NH14
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_21 | 3√35 |
AMC12_1068 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$ ?
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$
| $(625^{\log_5 2015})^\frac{1}{4}=((5^4)^{\log_5 2015})^\frac{1}{4}=(5^{4 \cdot \log_5 2015})^\frac{1}{4}=(5^{\log_5 2015 \cdot 4})^\frac{1}{4}=((5^{\log_5 2015})^4)^\frac{1}{4}=(2015^4)^\frac{1}{4}=\boxed{\textbf{(D)}\; 2015}$
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | 2015 |
AMC12_1069 | Two parabolas have equations $y= x^2 + ax +b$ and $y= x^2 + cx +d$ , where $a, b, c,$ and $d$ are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have at least one point in common?
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{25}{36}\qquad\textbf{(C)}\ \frac{5}{6}\qquad\textbf{(D)}\ \frac{31}{36}\qquad\textbf{(E)}\ 1$
| Solution 1Set the two equations equal to each other: $x^2 + ax + b = x^2 + cx + d$ . Now remove the x squared and get $x$ 's on one side: $ax-cx=d-b$ . Now factor $x$ : $x(a-c)=d-b$ . If $a$ cannot equal $c$ , then there is always a solution, but if $a=c$ , a $1$ in $6$ chance, leaving a $1080$ out $1296$ , always having at least one point in common. And if $a=c$ , then the only way for that to work, is if $d=b$ , a $1$ in $36$ chance, however, this can occur $6$ ways, so a $1$ in $6$ chance of this happening. So adding one thirty sixth to $\frac{1080}{1296}$ , we get the simplified fraction of $\frac{31}{36}$ ; answer $\boxed{(D)}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_13 | 31/36 |
AMC12_1071 | For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ , $9$ , and $10$ . For example, $R(15) = 1+0+3+0+3+1+7+6+5=26$ . How many two-digit positive integers $n$ satisfy $R(n) = R(n+1)\,?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$
| Note that we can add $9$ to $R(n)$ to get $R(n+1)$ , but must subtract $k$ for all $k|n+1$ . Hence, we see that there are four ways to do that because $9=7+2=6+3=5+4=4+3+2$ . Note that only $7+2$ is a plausible option, since $4+3+2$ indicates $n+1$ is divisible by $6$ , $5+4$ indicates that $n+1$ is divisible by $2$ , $6+3$ indicates $n+1$ is divisible by $2$ , and $9$ itself indicates divisibility by $3$ , too. So, $14|n+1$ and $n+1$ is not divisible by any positive integers from $2$ to $10$ , inclusive, except $2$ and $7$ . We check and get that only $n+1=14 \cdot 1$ and $n+1=14 \cdot 7$ give possible solutions so our answer is $\boxed{\textbf{(C) }2}$ .
- kevinmathz
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_25 | 2 |
AMC12_1075 | A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n$ ?
$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280$
| The first box has volume $2\times3\times5=30\text{ cm}^3$ , and the second has volume $(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3$ . The second has a volume that is $6$ times greater, so it holds $6\times40=\boxed{\textbf{(D)}\ 240}$ grams.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_3 | 240 |
AMC12_1076 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$ ?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$
| The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.
The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$
As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$ , for some nonnegative integer $k$ .
$a^2-8a=k^2$
$a(a-8)=k^2$
$((a-4)+4)((a-4)-4)=k^2$
$(a-4)^2-4^2=k^2$
$(a-4)^2=k^2+4^2$
From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$ .
In the case $k=0$ , the equation simplifies to $|a-4|=4$ . From this equation, we have $a=0,8$ . For both $a=0$ and $a=8$ , $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")
If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$ . Hence, $a=-1,9$ . Again, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$ , so these two values also satisfy the original constraints.
There are a total of four possible values for $a$ : $-1,0,8,$ and $9$ . Hence, the sum of all of the possible values of $a$ is $\boxed{\textbf{(C) }16}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | 16 |
AMC12_1077 | Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?
$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}$
| Each can of soda costs $\frac QS$ quarters, or $\frac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\frac{D}{\left(\tfrac{Q}{4S}\right)}=\boxed{\textbf{(B) } \frac{4DS}{Q}}$ cans of soda.
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_6 | 4DS/Q |
AMC12_1078 | Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7$ . What is the sum of the digits of $N$ ?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$
| $N$ must be divisible by every positive integer less than $7$ , or $1, 2, 3, 4, 5,$ and $6$ . Each number that is divisible by each of these is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$ , so each number divisible by these is a multiple of $60$ . The smallest multiple of $60$ is $60$ , so the second smallest multiple of $60$ is $2\times60=120$ . Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3\ \textbf{(A)}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_5 | 3 |
AMC12_1079 | Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$
| First, $\textbf{(B)}$ is $2\text{cis}(150)$ , $\textbf{(C)}$ is $2\text{cis}(135)$ , $\textbf{(D)}$ is $2\text{cis}(120)$ .
Taking the real part of the $5$ th power of each we have:
$\textbf{(A): }(-2)^5=-32$
$\textbf{(B): }32\cos(750)=32\cos(30)=16\sqrt{3}$
$\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$
$\textbf{(D): }32\cos(600)=32\cos(240)<0$
$\textbf{(E): }(2i)^5=32i$ , whose real part is $0$
Thus, the answer is $\boxed{\textbf{(B) }{-}\sqrt3+i}$ .
~JHawk0224
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_13 | B |
AMC12_1081 | Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$ ?
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$
| Let $\log_2{x} = \log_y{16}=k$ , so that $2^k=x$ and $y^k=16 \implies y=2^{\frac{4}{k}}$ . Then we have $(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6$ .
We therefore have $k+\frac{4}{k}=6$ , and deduce $k^2-6k+4=0$ . The solutions to this are $k = 3 \pm \sqrt{5}$ .
To solve the problem, we now find
\begin{align*} (\log_2\tfrac{x}{y})^2&=(\log_2 x - \log_2 y)^2\\ &=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 \\ &= (3 \pm \sqrt{5} - [3 \mp \sqrt{5}])^2\\ &= (3 \pm \sqrt{5} - 3 \pm \sqrt{5})^2\\ &=(\pm 2\sqrt{5})^2 \\ &= \boxed{\textbf{(B) } 20}. \\ \end{align*}
~Edits by BakedPotato66
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_12 | 20 |
AMC12_1082 | The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two digit number and its reverse?
$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$
| Let the two digits be $a$ and $b$ . Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$ , or $2a = 7b$ . This yields $a = 7$ and $b = 2$ because $a, b < 10$ . Then, $72 + 27 = \boxed{\textbf{(D) }99}.$
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_6 | 99 |
AMC12_1083 | Let $n$ be the smallest positive integer such that $n$ is divisible by $20$ , $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$ ?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
| We know that $n^2 = k^3$ and $n^3 = m^2$ . Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$ . Let $a = n^6$ . Now we know $a$ must be a perfect $36$ -th power because $lcm(9,4) = 36$ , which means that $n$ must be a perfect $6$ -th power. The smallest number whose sixth power is a multiple of $20$ is $10$ , because the only prime factors of $20$ are $2$ and $5$ , and $10 = 2 \times 5$ . Therefore our is equal to number $10^6 = 1000000$ , with $7$ digits $\Rightarrow \boxed {E}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_9 | E |
AMC12_1084 | Points $A$ and $B$ are on the parabola $y=4x^2+7x-1$ , and the origin is the midpoint of $AB$ . What is the length of $AB$ ?
$\mathrm{(A)}\ 2\sqrt5 \qquad \mathrm{(B)}\ 5+\frac{\sqrt2}{2} \qquad \mathrm{(C)}\ 5+\sqrt2 \qquad \mathrm{(D)}\ 7 \qquad \mathrm{(E)}\ 5\sqrt2$
| Let the coordinates of $A$ be $(x_A,y_A)$ . As $A$ lies on the parabola, we have $y_A=4x_A^2+7x_A-1$ .
As the origin is the midpoint of $AB$ , the coordinates of $B$ are $(-x_A,-y_A)$ .
We need to choose $x_A$ so that $B$ will lie on the parabola as well. In other words, we need $-y_A = 4(-x_A)^2 + 7(-x_A) - 1$ .
Substituting for $y_A$ , we get: $-4x_A^2 - 7x_A + 1 = 4(-x_A)^2 + 7(-x_A) - 1$ .
This simplifies to $8x_A^2 - 2 = 0$ , which solves to $x_A = \pm 1/2$ . Both roots lead to the same pair of points: $(1/2,7/2)$ and $(-1/2,-7/2)$ . Their distance is $\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_18 | 5√2 |
AMC12_1085 | A $4\times 4\times h$ rectangular box contains a sphere of radius $2$ and eight smaller spheres of radius $1$ . The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is $h$ ?
| Let $A$ be the point in the same plane as the centers of the top spheres equidistant from said centers. Let $B$ be the analogous point for the bottom spheres, and let $C$ be the midpoint of $\overline{AB}$ and the center of the large sphere. Let $D$ and $E$ be the points at which line $AB$ intersects the top of the box and the bottom, respectively.
Let $O$ be the center of any of the top spheres (you choose!). We have $AO=1\cdot\sqrt{2}$ , and $CO=3$ , so $AC=\sqrt{3^2-\left(\sqrt2\right)^2}=\sqrt{7}$ . Similarly, $BC=\sqrt{7}$ . $\overline{AD}$ and $\overline{BE}$ are clearly equal to the radius of the small spheres, $1$ . Thus the total height is $AD+AC+BC+BE=2+2\sqrt7$ , or $\boxed{\textbf{(A)}}$ .
~AwesomeToad
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_17 | 2+2√7 |
AMC12_1087 | Which of these describes the graph of $x^2(x+y+1)=y^2(x+y+1)$ ?
$\textbf{(A)}\ \text{two parallel lines}\\ \textbf{(B)}\ \text{two intersecting lines}\\ \textbf{(C)}\ \text{three lines that all pass through a common point}\\ \textbf{(D)}\ \text{three lines that do not all pass through a common point}\\ \textbf{(E)}\ \text{a line and a parabola}$
| The equation $x^2(x+y+1)=y^2(x+y+1)$ tells us $x^2=y^2$ or $x+y+1=0$ . $x^2=y^2$ generates two lines $y=x$ and $y=-x$ . $x+y+1=0$ is another straight line. The only intersection of $y=x$ and $y=-x$ is $(0,0)$ , which is not on $x+y+1=0$ . Therefore, the graph is three lines that do not have a common intersection, or $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$
| https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_7 | three lines that do not all pass through a common point |
AMC12_1090 | Kymbrea's comic book collection currently has $30$ comic books in it, and she is adding to her collection at the rate of $2$ comic books per month. LaShawn's collection currently has $10$ comic books in it, and he is adding to his collection at the rate of $6$ comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25$
| Kymbrea has $30$ comic books initially and every month, she adds two. This can be represented as $30 + 2x$ where x is the number of months elapsed. LaShawn's collection, similarly, is $10 + 6x$ . To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation $2(2x + 30) = 6x + 10$ and get $x = \boxed{\textbf{(E) } 25}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_1 | 25 |
AMC12_1091 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$ ?
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$
| Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle.
Applying the Extended Law of Sines to $\triangle PQR,$ we find the radius of $\odot X:$ \[XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.\]
Alternatively, by the Inscribed Angle Theorem, $\triangle QRX$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle with base $QR=3\sqrt3.$ Dividing $\triangle QRX$ into two congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, we get that the radius of $\odot X$ is $XQ=XR=3$ by the side-length ratios.
Let $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\overline{OM}\perp\overline{QR}$ and $\overline{XM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.
By the SAS Congruence, we have $\triangle QXM\cong\triangle RXM,$ both of which are $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles. By the side-length ratios, we obtain $RM=\frac{3\sqrt3}{2}, RX=3,$ and $XM=\frac{3}{2}.$ By the Pythagorean Theorem on right $\triangle ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $\triangle OXP,$ we get $OP=4.$
Let $C$ be the foot of the perpendicular from $P$ to $\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\overline{PC},$ as shown below:
[asy] /* Made by MRENTHUSIASM */ size(300); pair O, X, A, B, P, Q, R, M, C, D; O = (0,0); X = (4,3); A = (-7,0); B = (7,0); P = (4,0); Q = intersectionpoints(Circle(O,7),Circle(X,3))[0]; R = intersectionpoints(Circle(O,7),Circle(X,3))[1]; M = midpoint(Q--R); C = foot(P,Q,R); D = foot(X,P,C); fill(P--Q--R--cycle,yellow); dot("$O$",O,S); dot("$X$",X,N); dot("$A$",A,SW); dot("$B$",B,SE); dot("$P$",P,S); dot("$Q$",Q,E); dot("$R$",R,N); dot("$M$",M,dir(M)); dot("$C$",C,NE); dot("$D$",D,SE); markscalefactor=0.0375; draw(rightanglemark(O,M,R),red); draw(rightanglemark(P,C,M),red); draw(rightanglemark(P,D,X),red); draw(rightanglemark(O,P,X),red); draw(P--Q--R--cycle); draw(arc(O, 7, 0, 180)^^A--B^^Circle(X,3)); draw(O--M^^X--P); draw(P--C^^X--D,dashed); [/asy]
Clearly, quadrilateral $XDCM$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with the ratio of similitude $\frac{XP}{OX}=\frac 35.$ Therefore, we get $PD=\frac 95$ and $PC=PD+DC=PD+XM=\frac 95 + \frac 32 = \frac{33}{10}.$
The area of $\triangle PQR$ is \[\frac12\cdot QR\cdot PC=\frac12\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\] from which the answer is $99+3+20=\boxed{\textbf{(D) } 122}.$
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | 122 |
AMC12_1092 | A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?
$\textbf{(A) }1 \qquad \textbf{(B) }\frac{1+\sqrt2}{2} \qquad \textbf{(C) }\sqrt2 \qquad \textbf{(D) }\frac32 \qquad \textbf{(E) }\frac{2+\sqrt2}{3} \qquad$
| To find the height of the pyramid, we need the length from the center of the octagon (denote as $I$ ) to its vertices and the length of AV.
From symmetry, we know that $\overline{AV} = \overline{DV}$ , therefore $\triangle{AVD}$ is a 45-45-90 triangle. Denote $\overline{AV}$ as $x$ so that $\overline{AD} = x\sqrt{2}$ . Doing some geometry on the isosceles trapezoid $ABCD$ (we know this from the fact that it is a regular octagon) reveals that $\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}$ and $\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2$ .
To find the length $\overline{IA}$ , we cut the octagon into 8 triangles, each with a smallest angle of 45 degrees. Using the law of cosines on $\triangle{AIB}$ we find that ${\overline{IA}}^2=(2+\sqrt{2})/2$ .
Finally, using the pythagorean theorem, we can find that ${\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}$ which is answer choice $\boxed{B}$ .
～username2333
~hashbrown2009
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_23 | B |
AMC12_1093 | If $x,y,$ and $z$ are positive numbers satisfying
\[x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}\]
Then what is the value of $xyz$ ?
$\text {(A)}\ \frac{2}{3} \qquad \text {(B)}\ 1 \qquad \text {(C)}\ \frac{4}{3} \qquad \text {(D)}\ 2 \qquad \text {(E)}\ \frac{7}{3}$
| We multiply all given expressions to get:
\[(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}\]
Adding all the given expressions gives that
\[(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}\]
We subtract $(2)$ from $(1)$ to get that $xyz + \frac{1}{xyz} = 2$ . Hence, by inspection, $\boxed{xyz = 1 \rightarrow B}$ .
\[\]
~AopsUser101
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_20 | 1 |
AMC12_1097 | What is the reciprocal of $\frac{1}{2}+\frac{2}{3}$ ?
$\mathrm{(A)}\ \frac{6}{7}\qquad\mathrm{(B)}\ \frac{7}{6}\qquad\mathrm{(C)}\ \frac{5}{3}\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{7}{2}$
| Here's a cheapshot:
Obviously, $\frac{1}{2}+\frac{2}{3}$ is greater than $1$ . Therefore, its reciprocal is less than $1$ , and the answer must be $\boxed{\frac{6}{7}}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_2 | 6/7 |
AMC12_1101 | Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is 50% more than the regular. After both consume $\frac{3}{4}$ of their drinks, Ann gives Ed a third of what she has left, and 2 additional ounces. When they finish their lemonades they realize that they both drank the same amount. How many ounces of lemonade did they drink together?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$
| Let the size of Ed's drink equal $x$ ounces, and let the size of Ann's drink equal $\frac{3}{2}x$ ounces. After both consume $\frac{3}{4}$ of their drinks, Ed and Ann have $\frac{x}{4}$ and $\frac{3x}{8}$ ounces of their drinks remaining. Ann gives away $\frac{x}{8} + 2$ ounces to Ed.
In the end, Ed drank everything in his original lemonade plus what Ann gave him, and Ann drank everything in her original lemonade minus what she gave Ed. Thus we have
\[x + \frac{x}{8} + 2 = \frac{3x}{2} - \frac{x}{8} - 2\]
\[x = 16\]
The total amount the two of them drank is simply
\[x + \frac{3}{2}x = 16 + 24 = \boxed{\textbf{(D)}\ 40}\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_6 | 40 |
AMC12_1105 | The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$
| Let the coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$ . We have by shoelace's theorem, that the area is
\begin{align*} &\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2} \\ &=\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} \\ &= \ln \left( \sqrt{\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \right) \\ &= \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) \\ &= \ln \left( \frac{91}{90} \right). \end{align*}
We know that the numerator must have a factor of $13$ , so given the answer choices, $n$ is either $12$ or $11$ . If $n=11$ , the expression $\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\frac{91}{90}$ , but if $n=12$ , the expression evaluates to $\frac{91}{90}$ . Hence, our answer is $\boxed{12}$ .
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_17 | 12 |
AMC12_1111 | Convex quadrilateral $ABCD$ has $AB=3$ , $BC=4$ , $CD=13$ , $AD=12$ , and $\angle ABC=90^{\circ}$ , as shown. What is the area of the quadrilateral?
[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A); label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,E); draw(rightanglemark(A,B,C,25)); [/asy]
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5$
| Note that by the pythagorean theorem, $AC=5$ . Also note that $\angle CAD$ is a right angle because $\triangle CAD$ is a right triangle. The area of the quadrilateral is the sum of the areas of $\triangle ABC$ and $\triangle CAD$ which is equal to
\[\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}\]
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_9 | 36 |
AMC12_1112 | The numbers $\log(a^3b^7)$ , $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$ ?
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$
| Solution 1Let $A = \log(a)$ and $B = \log(b)$ .
The first three terms of the arithmetic sequence are $3A + 7B$ , $5A + 12B$ , and $8A + 15B$ , and the $12^\text{th}$ term is $nB$ .
Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$ .
Since the first three terms in the sequence are $13B$ , $22B$ , and $31B$ , the $k$ th term is $(9k + 4)B$ .
Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}$ .
| https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_16 | D |
AMC12_1113 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850$
| Solution 1The first pirate takes $\frac{1}{12}$ of the $x$ coins, leaving $\frac{11}{12} x$ .
The second pirate takes $\frac{2}{12}$ of the remaining coins, leaving $\frac{10}{12}\cdot \frac{11}{12}*x$ .
Note that
$12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}$
$11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$
All the $2$ s and $3$ s cancel out of $11!$ , leaving
$11 \cdot 5 \cdot 7 \cdot 5 = 1925$
in the numerator.
We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, $x$ is the denominator, leaving $\boxed{\textbf{(D) }1925}$ coins for the twelfth pirate.
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | 1925 |
AMC12_1114 | Let $a$ and $b$ be real numbers such that $\sin{a} + \sin{b} = \frac{\sqrt{2}}{2}$ and $\cos {a} + \cos {b} = \frac{\sqrt{6}}{2}.$ Find $\sin{(a+b)}.$
$\text{(A) }\frac{1}{2} \qquad \text{(B) }\frac{\sqrt{2}}{2} \qquad \text{(C) }\frac{\sqrt{3}}{2} \qquad \text{(D) }\frac{\sqrt{6}}{2} \qquad \text{(E) }1$
| Sum to product gives us
\[2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{2}}{2}\]
\[2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{6}}{2}\]
Dividing these equations tells us that $\tan(\frac{a+b}{2}) = \frac{1}{\sqrt{3}}$ , so $\frac{a+b}{2} = \frac{\pi}{6} + \pi n$ for an integer $n$ , so $\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}$ . The answer is $\boxed{\text{(C)}}$ .
~alexanderruan
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_25 | C |
AMC12_1116 | For how many values of $a$ is it true that the line $y = x + a$ passes through the
vertex of the parabola $y = x^2 + a^2$ ?
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 10 \qquad \mathrm{(E)}\ \text{infinitely many}$
| We see that the vertex of the quadratic function $y = x^2 + a^2$ is $(0,\,a^2)$ . The y-intercept of the line $y = x + a$ is $(0,\,a)$ . We want to find the values (if any) such that $a=a^2$ . Solving for $a$ , the only values that satisfy this are $0$ and $1$ , so the answer is $\boxed{\mathrm{(C)}\ 2}$
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_8 | 2 |
AMC12_1122 | What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501$
| By the Binomial Theorem, each term in the expansion is of the form \[\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},\] where $k\in\{0,1,2,\ldots,1000\}.$
This problem is equivalent to counting the values of $k$ such that both $\frac k3$ and $\frac{1000-k}{2}$ are integers. Note that $k$ must be a multiple of $3$ and a multiple of $2,$ so $k$ must be a multiple of $6.$ There are $\boxed{\textbf{(C)}\ 167}$ such values of $k:$ \[6(0), 6(1), 6(2), \ldots, 6(166).\]
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_12 | 167 |
AMC12_1123 | Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$ . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?
$\textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2$
| Set up an isosceles triangle between the center of the 8th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the 7th and 8th spheres, and the points of tangency between the 7th sphere and 2 of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of r (the radius of sphere 8) and h (the height of the first triangle). You can then use Pythagorean Theorem to set up two equations for the two triangles, and find the values of h and r.
$(1+r)^2=2^2+h^2$
$(3\sqrt{2})^2=3^2+(h+r)^2$
$r = \boxed{\textbf{(B) }\frac{3}{2}}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_18 | 3/2 |
AMC12_1124 | A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n$ ?
$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$
| We may let $n = 7^k \cdot m$ , where $m$ is not divisible by 7. Using the fact that the number of divisors function $d(n)$ is multiplicative, we have $d(n) = d(7^k)d(m) = (k+1)d(m) = 60$ . Also, $d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80$ . These numbers are in the ratio 3:4, so $\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{\mathrm{C}}$ .
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_21 | C |
AMC12_1125 | Let $S$ be the set of ordered triples $(x,y,z)$ of real numbers for which
\[\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.\]
There are real numbers $a$ and $b$ such that for all ordered triples $(x,y,z)$ in $S$ we have $x^{3}+y^{3}=a \cdot 10^{3z} + b \cdot 10^{2z}.$ What is the value of $a+b?$
$\textbf{(A)}\ \frac {15}{2} \qquad \textbf{(B)}\ \frac {29}{2} \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ \frac {39}{2} \qquad \textbf{(E)}\ 24$
| Let $x + y = s$ and $x^2 + y^2 = t$ .
Then, $\log(s)=z$ implies $\log(10s) = z+1= \log(t)$ ,so $t=10s$ .
Therefore, $x^3 + y^3 = s\cdot\dfrac{3t-s^2}{2} = s(15s-\dfrac{s^2}{2})$ .
Since $s = 10^z$ , we find that $x^3 + y^3 = 15\cdot10^{2z} - (1/2)\cdot10^{3z}$ .
Thus, $a+b = \frac{29}{2}$ $\Rightarrow$ $\boxed{B}$
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_23 | B |
AMC12_1127 | How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is $ax^2+bx+c,a\neq 0,$ and the roots are $r$ and $s,$ then the requirement is that $\{a,b,c\}=\{r,s\}$ .)
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}$
| Firstly, if $r=s$ , then $a=b=c$ , so the equation becomes $ax^2 + ax + a = 0 \Rightarrow x^2 + x + 1=0$ , which has no real roots.
Hence there are three cases we need to consider:
Case 1 : $a=b=r$ and $c=s \neq r$ :
The equation becomes $ax^2+ax+c=0$ , and by Vieta's Formulas, we have $a+c=-1$ and $ac = \frac{c}{a}$ . This second equation becomes $(a^2-1)c=0$ . Hence one possibility is $c=0$ , in which case $a=-1$ , giving the equation $-x^2 - x = 0$ , which has roots $0$ and ${-}1$ . This gives one valid solution. On the other hand, if $c\neq 0$ , then $a^2-1=0$ , so $a = \pm 1$ . If $a=1$ , we have $c=-2$ , and the equation is $x^2 + x - 2 = 0$ , which clearly works, giving a second valid solution. If $a=-1$ , then we have $c=0$ , which has already been considered, so this possibility gives no further valid solutions.
Case 2 : $a=c=r$ , $b=s \neq r$ :
The equation becomes $ax^2 + bx + a = 0$ , so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = 1$ . These equations reduce to $a^3 + a + 1 = 0$ . By sketching a graph of this function, we see that there is exactly one real root. (Alternatively, note that as it is of odd degree, there is at least one real root, and by differentiation, it has no stationary points, so there is at most one real root. Combining these gives exactly one real root.) This gives a third valid solution.
Case 3 : $a=r$ , $b=c=s \neq r$ :
The equation becomes $ax^2 + bx+b=0$ , so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = \frac{b}{a}$ .
Observe that $b \neq 0$ , as if it were $0$ , the equation would just have one real root, $0$ , so this would not give a valid solution. Thus, taking the second equation and dividing both sides by $b$ , we deduce have $a=\frac{1}{a}$ , so $a=\pm 1$ . If $a=1$ , we have $1+b=-b$ , giving $b=-\frac{1}{2}$ , so the equation is $x^2 - \frac{1}{2}x - \frac{1}{2} = 0$ , which is a fourth valid solution. If $a=-1$ , we have $1+b=b$ , which is a contradiction, so this case gives no further valid solutions.
Hence the total number of valid solutions is $\boxed{\textbf{(B) } 4}$ .
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_21 | 4 |
AMC12_1129 | A point $P$ is selected at random from the interior of the pentagon with vertices $A = (0,2)$ , $B = (4,0)$ , $C = (2 \pi + 1, 0)$ , $D = (2 \pi + 1,4)$ , and $E=(0,4)$ . What is the probability that $\angle APB$ is obtuse?
$\text{(A) }\frac {1}{5} \qquad \text{(B) }\frac {1}{4} \qquad \text{(C) }\frac {5}{16} \qquad \text{(D) }\frac {3}{8} \qquad \text{(E) }\frac {1}{2}$
| The angle $APB$ is obtuse if and only if $P$ lies inside the circle with diameter $AB$ . (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.)
[asy] defaultpen(0.8); real pi=3.14159265359; pair A=(0,2), B=(4,0), C=(2*pi+1, 0), D=(2*pi+1,4), E=(0,4), F=(0,0); draw(A--B--C--D--E--cycle); draw(circle((A+B)/2,length(B-A)/2)); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,SW); draw(A--F--B,dashed); [/asy]
The area of $AFB$ is $[AFB] = \frac {AF\cdot FB}2 = 4$ , and the area of $ABCDE$ is $CD\cdot DE - [AFB] = 4\cdot (2\pi+1) - 4 = 8\pi$ .
From the Pythagorean theorem the length of $AB$ is $\sqrt{2^2 + 4^2} = 2\sqrt{5}$ , thus the radius of the circle is $\sqrt{5}$ , and the area of the half-circle that is inside $ABCDE$ is $\frac{ 5\pi }2$ .
Therefore the probability that $APB$ is obtuse is $\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\text{(C) } \frac 5{16}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_17 | 5/16 |
AMC12_1132 | Triangle $ABC$ has $AB = 13$ and $AC = 15$ , and the altitude to $\overline{BC}$ has length $12$ . What is the sum of the two possible values of $BC$ ?
$\mathrm{(A)}\ 15\qquad \mathrm{(B)}\ 16\qquad \mathrm{(C)}\ 17\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 19$
| Let $D$ be the foot of the altitude to $\overline BC$ . Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$ . Thus $BC = BD + BC = 5 + 9 = 14$ or assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$ . The sum of the two possible values is $14 + 4 = \boxed{18}$ . The answer is $\mathrm{(D)}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_13 | 18 |
AMC12_1133 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35$
| We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$ . In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$ . If we let the list be $1,2,3,8,8,9,10,11,12,13,j$ , then $j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$ .
Next, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$ . Following the same process as above, we get $j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$ . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\boxed{\textbf{(E) }35}$
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_11 | 35 |
AMC12_1138 | Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$
respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$ ?
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$
| First examine the formula $(x-10)^2+y^2=36$ , for the circle $C_1$ . Its center, $D_1$ , is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$ , at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram:
[asy] unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9); draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0)); pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0)); label("$C_1$",(10,0) + 6*dir(-45), SE ); label("$C_2$",(-15,0) + 9*dir(225), SW ); label("$D_1$",(10,0), SE ); label("$D_2$",(-15,0), SW ); label("$Q$", p2[0], NE ); label("$P$", p1[1], SW ); label("$O$", (0,0), SW ); [/asy]
Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles $D_2QO$ and $D_1PO$ similar by AA, with a scale factor of $6:9$ , or $2:3$ . Next, we must subdivide the line $D_2D_1$ in a 2:3 ratio to get the length of the segments $D_2O$ and $D_1O$ . The total length is $10 - (-15)$ , or $25$ , so applying the ratio, $D_2O$ = 15 and $D_1O$ = 10 . These are the hypotenuses of the triangles. We already know the length of $D_2Q$ and $D_1P$ , 9 and 6 (they're radii). So in order to find $PQ$ , we must find the length of the longer legs of the two triangles and add them.
$15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$
$\sqrt{144} = 12$
$10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$
$\sqrt{64} = 8$
Finally, the length of PQ is $12+8=\boxed{20}$ , or (C) .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_18 | 20 |
AMC12_1143 | Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1$ , and with $|z| = 2$ . In the complex plane, the four points $0$ , $z$ , $z^{2}$ , and $z^{3}$ are the vertices of a quadrilateral with area $15$ . What is the imaginary part of $z$ ?
$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }1\qquad\textbf{(C) }\frac{4}{3}\qquad\textbf{(D) }\frac{3}{2}\qquad\textbf{(E) }\frac{5}{3}$
| By making a rough estimate of where $z$ , $z^2$ , and $z^3$ are on the complex plane, we can draw a pretty accurate diagram (like above.)
Here, points $Z_1$ , $Z_2$ , and $Z_3$ lie at the coordinates of $z$ , $z^2$ , and $z^3$ respectively, and $O$ is the origin.
We're given $|z|=2$ , so $|z^2|=|z|^2=4$ and $|z^3|=|z|^3 = 8$ . This gives us $OZ_1=2$ , $OZ_2=4$ , and $OZ_3=8$ .
Additionally, we know that $\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}$ (since every power of $z$ rotates around the origin by the same angle.) We set these angles equal to $\theta$ .
We have that
\begin{align*}
[OZ_1Z_2Z_3]&=[OZ_1Z_2]+[OZ_2Z_3] \\
&=\frac{1}{2}\cdot2\cdot4 \sin\theta+\frac{1}{2}\cdot4\cdot8 \sin\theta \\
&=4\sin\theta+16\sin\theta \\
&=20 \sin\theta
\end{align*}
Since this is equal to $15$ , we have $20\sin\theta=15$ , so $\sin\theta=\frac{3}{4}$ .
Thus, $\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}$ .
~nm1728, ShortPeopleFartalot
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_12 | 3/2 |
AMC12_1145 | Small lights are hung on a string $6$ inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of $2$ red lights followed by $3$ green lights. How many feet separate the 3rd red light and the 21st red light?
Note: $1$ foot is equal to $12$ inches.
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5$
| We know the repeating section is made of $2$ red lights and $3$ green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the 11th section. There would then be a total of $44$ lights in between the 3rd and 21st red light, translating to $45$ $6$ -inch gaps. Since the question asks for the answer in feet, the answer is $\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_7 | 22.5 |
AMC12_1146 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ , $BP=\sqrt{3}$ , and $CP=2$ . What is $s$ ?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$
| Solution 1(a)We begin by rotating $\triangle{ APB}$ counterclockwise by $60^{\circ}$ about $A$ , such that $P\mapsto Q$ and $B\mapsto C$ . We see that $\triangle{ APQ}$ is equilateral with side length $1$ , meaning that $\angle APQ = 60^{\circ}$ . We also see that $\triangle{CPQ}$ is a $30$ - $60$ - $90$ right triangle, meaning that $\angle CPQ= 60^{\circ}$ . Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$ .
[asy] size(200); pen p = fontsize(10pt)+gray+0.5; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); B=origin; A=s*dir(60); C=s*right; P=IP(CR(A,1),CR(C,2)); Q=rotate(60,A)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C^^A--Q--C, p); draw(P--Q, p+dashed); //draw(A--A+C--C, p); label("$A$", A, up, q); label("$B$", B, 0.5*(B-P), q); label("$C$", C, 0.5*(C-P), q); label("$P$", P, dir(180), q); label("$Q$", Q, 0.25*(Q-B), q); label("$\sqrt{3}$",B--P, right, p); label("$2$",C--P, 2*left, p); label("$1$",A--P, 1.5**dir(-10), p); label("$1$", A--Q, dir(250), p); label("$1$",P--Q, down, p); label("$\sqrt{3}$",C--Q, right, p); [/asy]
We can now use the law of cosines as following:
\begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2\cdot AP\cdot CP\cdot \cos{\angle{APC}} \\ &= 1 + 4 - 2\cdot 1\cdot 2\cdot \cos{120^{\circ}} \\ &= 5 - 4\left(-\frac{1}{2}\right) \\ &= 7, \end{align*}
giving us that $s = \boxed{ \sqrt{7} \ \textbf{(B)}}$ .
~ciceronii
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | √7 |
AMC12_1147 | How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$ ?
$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$
| Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set $\{-3, -2, -1, 0, 1, 2, 3\}$ . It is easy to see that we can get any integer between $-6$ and $6$ inclusive as the sum of three elements from this set, for the total of $\boxed{\mathrm{(A) } 13}$ integers.
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_10 | 13 |
AMC12_1150 | Cyrus the frog jumps $2$ units in a direction, then $2$ more in another direction. What is the probability that he lands less than $1$ unit away from his starting position?
$\textbf{(A)}~\frac{1}{6}\qquad\textbf{(B)}~\frac{1}{5}\qquad\textbf{(C)}~\frac{\sqrt{3}}{8}\qquad\textbf{(D)}~\frac{\arctan \frac{1}{2}}{\pi}\qquad\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}$
| Let Cyrus's starting position be $S$ . WLOG, let the place Cyrus lands at for his first jump be $O$ . From $O$ , Cyrus can reach all the points on $\odot O$ . The probability that Cyrus will land less than $1$ unit away from $S$ is $\frac{4 \alpha }{ 2 \pi}$ .
\[\sin \alpha = \frac{ \frac12 }{2} = \frac14, \quad \alpha = \arcsin \frac14\]
Therefore, the answer is
\[\frac{4 \arcsin \frac14 }{ 2 \pi} = \boxed{\textbf{(E) } \frac{2 \arcsin \frac{1}{4}}{\pi}}\]
~ isabelchen
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_20 | $\frac{2 \arcsin \frac{1}{4}}{\pi}$ |
AMC12_1151 | A sequence of numbers is defined by $D_0=0,D_1=0,D_2=1$ and $D_n=D_{n-1}+D_{n-3}$ for $n\ge 3$ . What are the parities (evenness or oddness) of the triple of numbers $(D_{2021},D_{2022},D_{2023})$ , where $E$ denotes even and $O$ denotes odd?
$\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)$
| We construct the following table:
\[\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c} &&&&&&&&&&& \\ [-2.5ex] \textbf{Term} &\boldsymbol{D_0}&\boldsymbol{D_1}&\boldsymbol{D_2}&\boldsymbol{D_3}&\boldsymbol{D_4}&\boldsymbol{D_5}&\boldsymbol{D_6}&\boldsymbol{D_7}&\boldsymbol{D_8}&\boldsymbol{D_9}&\boldsymbol{\cdots} \\ \hline \hline &&&&&&&&&&& \\ [-2.25ex] \textbf{Value} & 0&0&1&1&1&2&3&4&6&9&\cdots \\ \hline &&&&&&&&&&& \\ [-2.25ex] \textbf{Parity} & E&E&O&O&O&E&O&E&E&O&\cdots \end{array}\]
Note that $(D_7,D_8,D_9)$ and $(D_0,D_1,D_2)$ have the same parities, so the parity is periodic with period $7.$ Since the remainders of $(2021\div7,2022\div7,2023\div7)$ are $(5,6,7),$ we conclude that $(D_{2021},D_{2022},D_{2023})$ and $(D_5,D_6,D_7)$ have the same parities, namely $\boxed{\textbf{(C) }(E,O,E)}.$
~JHawk0224 ~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_8 | (E,O,E) |
AMC12_1152 | Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}\ 39\qquad\textbf{(E)}\ 41$
| She has $p$ pennies and $n$ nickels, where $n + p = 13$ . If she had $n+1$ nickels then $n+1 = p$ , so $2n+ 1 = 13$ and $n=6$ . So she has 6 nickels and 7 pennies, which clearly have a value of $\boxed{\textbf{(C)}\ 37}$ cents.
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_1 | 37 |
AMC12_1154 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$
| Let the integer have digits $a$ , $b$ , and $c$ , read left to right. Because $1 \leq a<b<c$ , none of the digits can be zero and $c$ cannot be 2. If $c=4$ , then $a$ and $b$ must each be chosen from the digits 1, 2, and 3. Therefore there are $\binom{3}{2}=3$ choices for $a$ and $b$ , and for each choice there is one acceptable order. Similarly, for $c=6$ and $c=8$ there are, respectively, $\binom{5}{2}=10$ and $\binom{7}{2}=21$ choices for $a$ and $b$ . Thus there are altogether $3+10+21=\boxed{34}$ such integers.
(Edited by HMSSONI82)
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | 34 |
AMC12_1156 | The sequence $S_1, S_2, S_3, \cdots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, \[S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.\] Suppose that $S_9 = 110$ and $S_7 = 42$ . What is $S_4$ ?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad$
| $S_9 = 110$ , $S_7 = 42$
$S_8 = S_9 - S_ 7 = 110 - 42 = 68$
$S_6 = S_8 - S_7 = 68 - 42 = 26$
$S_5 = S_7 - S_6 = 42 - 26 = 16$
$S_4 = S_6 - S_5 = 26 - 16 = 10$
Therefore, the answer is $\boxed{\textbf{(C) }{10}}$
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_7 | 10 |
AMC12_1159 | Let $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$ ?
$\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$
| Notice that $\omega=e^{\frac{2i\pi}{3}}$ , which is one of the cube roots of unity. We wish to find the span of $(a+b\omega+c\omega^2)$ for reals $0\le a,b,c\le 1$ .
Observe also that if $a,b,c>0$ , then replacing $a$ , $b$ , and $c$ by $a-\min(a,b,c), b-\min(a,b,c),$ and $c-\min(a,b,c)$ leaves the value of $a+b\omega+c\omega^2$ unchanged. Therefore, assume that at least one of $a,b,c$ is equal to $0$ . If exactly one of them is $0$ , we can form an equilateral triangle of side length $1$ using the remaining terms. A similar argument works if exactly two of them are $0$ . In total, we get $3+{3 \choose 2} = 6$ equilateral triangles, whose total area is $6 \cdot \frac{\sqrt{3}}{4} = \boxed{\textbf{(C) } \frac{3}{2}\sqrt3}$ .
Note : A diagram of the six equilateral triangles is shown below.
[asy] size(200,200); draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((0,0)--(1/2,sqrt(3)/2)--(-1/2,sqrt(3)/2)--cycle); draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--cycle); draw((0,0)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle); draw((0,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--cycle); draw((0,0)--(1/2,-sqrt(3)/2)--(1,0)--cycle); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); [/asy]
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_24 | $\frac{3}{2}\sqrt3$ |
AMC12_1160 | How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$ , inclusive, satisfy the equation $(\log_b a)^{2017}=\log_b(a^{2017})?$
$\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597$
| By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$ . If $x\neq 0$ , we may divide by it and get $x^{2016}=2017$ , which implies $x=\pm \root{2016}\of{2017}$ . Hence, we have $3$ possible values $x$ , namely
\[x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad x=-2017^{\frac1{2016}}.\]
Since $\log_b a=x$ is equivalent to $a=b^x$ , each possible value $x$ yields exactly $199$ solutions $(b,a)$ , as we can assign $a=b^x$ to each $b=2,3,\dots,200$ . In total, we have $3\cdot 199=\boxed{\textbf{(E) } 597}$ solutions.
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_20 | 597 |
AMC12_1161 | Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ , $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c$ ?
| [asy] defaultpen(fontsize(8)+0.8); size(100); pair A=(0,0); pair B=(1.732,3); pair C=(3,3); pair D=(3,1.732); draw(A--(0,3)--C--(3,0)--A, lightgray+dashed); draw(A--B--C--A); draw(A--D--C, gray); label("$A$",A,W); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,E); label("$E$",(0,3),NW); label("$F$",(3,0),E); [/asy]
The $24$ -sided polygon is made out of $24$ shapes like $\triangle ABC$ . Then $\angle BAC=360^\circ/24=15^\circ$ , and $\angle EAC = 45^\circ$ , so $\angle{EAB} = 30^{\circ}$ . Then $EB=AE\tan 30^\circ = \sqrt{3}$ ; therefore $BC=EC-EB=3-\sqrt{3}$ . Thus
\[[ABC] = \frac{BC}{EC}\cdot [ACE] = \frac{3-\sqrt{3}}{3}\cdot \frac 92\] and the required area is $24\cdot[ABC] =108-36\sqrt{3}$ . Finally $108+36+3=\boxed{(\textbf{E})\ 147}$ .
~lopkiloinm
Note: Drop an altitude from $A$ to $\overline{BC}$ to construct point $E$ . This creates right triangles. ~erringbubble
| https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_15 | 147 |
AMC12_1162 | Let $f_{1}(x)=\sqrt{1-x}$ , and for integers $n \geq 2$ , let $f_{n}(x)=f_{n-1}(\sqrt{n^2 - x})$ . If $N$ is the largest value of $n$ for which the domain of $f_{n}$ is nonempty, the domain of $f_{N}$ is $\{c\}$ . What is $N+c$ ?
$\textbf{(A)}\ -226 \qquad \textbf{(B)}\ -144 \qquad \textbf{(C)}\ -20 \qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ 144$
| The domain of $f_{1}(x)=\sqrt{1-x}$ is defined when $x\leq1$ .
\[f_{2}(x)=f_{1}\left(\sqrt{4-x}\right)=\sqrt{1-\sqrt{4-x}}\]
Applying the domain of $f_{1}(x)$ and the fact that square roots must be positive, we get $0\leq\sqrt{4-x}\leq1$ . Simplifying, the domain of $f_{2}(x)$ becomes $3\leq x\leq4$ .
Repeat this process for $f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}$ to get a domain of $-7\leq x\leq0$ .
For $f_{4}(x)$ , since square roots must be nonnegative, we can see that the negative values of the previous domain will not work, so $\sqrt{16-x}=0$ . Thus we now arrive at $16$ being the only number in the of domain of $f_4 x$ that defines $x$ . However, since we are looking for the largest value for $n$ for which the domain of $f_{n}$ is nonempty, we must continue checking until we arrive at a domain that is empty.
We continue with $f_{5}(x)$ to get a domain of $\sqrt{25-x}=16 \implies x=-231$ . Since square roots cannot be negative, this is the last nonempty domain. We add to get $5-231=\boxed{\textbf{(A)}\ -226}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_21 | -226 |
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