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AMC12_1163 | Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72$
| Letting $x$ be the third side, then by the triangle inequality, $20-15 < x < 20+15$ , or $5 < x < 35$ . Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so $\boxed{\textbf{(E)} \, 72}$ is our answer.
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_2 | 72 |
AMC12_1164 | Let $S$ be the set of points $(x,y)$ in the coordinate plane such that two of the three quantities $3$ , $x+2$ , and $y-4$ are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of $S$ ?
$\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point}$
| If the two equal values are $3$ and $x+2$ , then $x=1$ . Also, $y-4\leqslant 3$ because $3$ is the common value. Solving for $y$ , we get $y\leqslant 7$ . Therefore the portion of the line $x=1$ where $y\leqslant 7$ is part of $S$ . This is a ray with an endpoint of $(1, 7)$ .
Similar to the process above, we assume that the two equal values are $3$ and $y-4$ . Solving the equation $3=y-4$ then $y=7$ . Also, $x+2\leqslant 3$ because $3$ is the common value. Solving for $x$ , we get $x\leqslant 1$ . Therefore the portion of the line $y=7$ where $x\leqslant 1$ is also part of $S$ . This is another ray with the same endpoint as the above ray: $(1, 7)$ . (Note that the only answer choice which has rays in it is answer choice $E$ .)
If $x+2$ and $y-4$ are the two equal values, then $x+2=y-4$ . Solving the equation for $y$ , we get $y=x+6$ . Also $3\leqslant y-4$ because $y-4$ is one way to express the common value (using $x-2$ as the common value works as well). Solving for $y$ , we get $y\geqslant 7$ . Therefore the portion of the line $y=x+6$ where $y\geqslant 7$ is part of $S$ like the other two rays. The lowest possible value that can be achieved is also $(1, 7)$ .
Since $S$ is made up of three rays with common endpoint $(1, 7)$ , the answer is $\boxed{E}$ .
Solution by TheMathematicsTiger7
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_9 | E |
AMC12_1166 | How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy \[1-3\sin\theta+5\cos3\theta = 0?\]
$\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$
| We rearrange to get \[5\cos3\theta = 3\sin\theta-1.\]
We can graph two functions in this case: $y=5\cos{3x}$ and $y=3\sin{x} -1$ .
Using transformation of functions, we know that $5\cos{3x}$ is just a cosine function with amplitude $5$ and period $\frac{2\pi}{3}$ . Similarly, $3\sin{x} -1$ is just a sine function with amplitude $3$ and shifted $1$ unit downward:
[asy] import graph; size(400,200,IgnoreAspect); real Sin(real t) {return 3*sin(t) - 1;} real Cos(real t) {return 5*cos(3*t);} draw(graph(Sin,0, 2pi),red,"$3\sin{x} -1 $"); draw(graph(Cos,0, 2pi),blue,"$5\cos{3x}$"); xaxis("$x$",BottomTop,LeftTicks); yaxis("$y$",LeftRight,RightTicks(trailingzero)); add(legend(),point(E),20E,UnFill); [/asy]
So, we have $\boxed{\textbf{(D) }6}$ solutions.
~Jamess2022 (burntTacos)
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_13 | 6 |
AMC12_1167 | All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$ ?
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$
| Let the three points be at $A = (x_1, x_1^2)$ , $B = (x_2, x_2^2)$ , and $C = (x_3, x_3^2)$ , such that the slope between the first two is $2$ , and $A$ is the point with the least $y$ -coordinate.
Then, we have $\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$ . Similarly, the slope of $BC$ is $x_2 + x_3$ , and the slope of $AB$ is $x_1 + x_2 = 2$ . The desired sum is $x_1 + x_2 + x_3$ , which is equal to the sum of the slopes divided by $2$ .
To find the slope of $AC$ , we note that it comes at a $60^{\circ}$ angle with $AB$ . Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1 + 2i$ and $1 + \sqrt{3}i$ . What this does is generate the complex number that is at a $60^{\circ}$ angle with the complex number $1 + 2i$ . Then, we can find the slope of the line between this new complex number and the origin:
\[(1+2i)(1+\sqrt{3}i)\]
\[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\]
\[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\]
\[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\]
\[= \frac{8 + 5\sqrt{3}}{-11}\]
\[= \frac{-8 - 5\sqrt{3}}{11}.\]
The slope $BC$ can also be solved similarly, noting that it makes a $120^{\circ}$ angle with $AB$ :
\[(1+2i)(-1+\sqrt{3}i)\]
\[= -1 - 2\sqrt{3} - 2i + \sqrt{3}i\]
\[\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}\]
\[= \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\]
At this point, we start to notice a pattern: This expression is equal to $\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}$ , except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out.
Our sum is simply $2 - 2\cdot\frac{8}{11} = \frac{6}{11}$ , and thus we can divide by $2$ to obtain $\frac{3}{11}$ , which gives the answer $\boxed{\mathrm{(A)}\ 14}$ .
~mathboy100
| https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_24 | 14 |
AMC12_1172 | Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$
| Solution 1The repeating decimal $0.\overline{ab}$ is equal to
\[\frac{10a+b}{100} + \frac{10a+b}{10000} + \cdots = (10a+b)\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right) = (10a+b) \cdot \frac 1{99} = \frac{10a+b}{99}\]
When expressed in the lowest terms, the denominator of this fraction will always be a divisor of the number $99 = 3\cdot 3\cdot 11$ . This gives us the possibilities $\{1,3,9,11,33,99\}$ . As $a$ and $b$ are not both nine and not both zero, the denominator $1$ can not be achieved, leaving us with $\boxed{\mathrm{(C) }5}$ possible denominators.
(The other ones are achieved e.g. for $ab$ equal to $33$ , $11$ , $9$ , $3$ , and $1$ , respectively.)
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_20 | 5 |
AMC12_1173 | A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$ ?
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$
| The point $(a,b)$ is the foot of the perpendicular from $(12,10)$ to the line $y=-5x+18$ . The perpendicular has slope $\frac{1}{5}$ , so its equation is $y=10+\frac{1}{5}(x-12)=\frac{1}{5}x+\frac{38}{5}$ . The $x$ -coordinate at the foot of the perpendicular satisfies the equation $\frac{1}{5}x+\frac{38}{5}=-5x+18$ , so $x=2$ and $y=-5\cdot2+18=8$ . Thus $(a,b) = (2,8)$ , and $a+b = \boxed{10}$ .
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_13 | 10 |
AMC12_1174 | In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$
$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$
| Call the midpoint of $\overline{AB}$ point $N.$ Draw in $\overline{NM}$ and $\overline{NP}.$ Note that $\angle{NPM}=90^{\circ}$ due to Thales's Theorem.
[asy] /* Made by QIDb602; edited by MRENTHUSIASM */ size(180); pair A, B, C, D, M, N, O, P; O = origin; A = (-1/2,-1/2); B = (-1/2,1/2); C = (1/2,1/2); D = (1/2,-1/2); M = midpoint(C--D); N = midpoint(A--B); path p; p = Circle(O,1/2); P = intersectionpoints(A--M,p)[0]; fill(N--A--M--cycle,yellow); dot("$\omega$",O,1.5*(0,1),linewidth(4)); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*SE,linewidth(4)); dot("$M$",M,1.5*E,linewidth(4)); dot("$N$",N,1.5*W,linewidth(4)); dot("$P$",P,1.5*dir(60),linewidth(4)); markscalefactor=0.00625; draw(rightanglemark(A,N,M),red); draw(rightanglemark(N,P,A),red); draw(A--B--C--D--cycle^^A--M^^P--N--M^^p); [/asy]
Using the Pythagorean theorem, $AM=\frac{\sqrt{5}}{2}.$ Now we just need to find $AP$ using similar triangles $\triangle APN\sim\triangle ANM:$
\begin{align*} \frac{AP}{AN}&=\frac{AN}{AM} \\ \frac{AP}{1/2}&=\frac{1/2}{\sqrt5/2} \\ AP&=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}. \end{align*}
~QIDb602
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_10 | $\frac{\sqrt{5}}{10}$ |
AMC12_1177 | Let $f(x)=|2\{x\}-1|$ where $\{x\}$ denotes the fractional part of $x$ . The number $n$ is the smallest positive integer such that the equation \[nf(xf(x))=x\] has at least $2012$ real solutions. What is $n$ ? Note: the fractional part of $x$ is a real number $y=\{x\}$ such that $0\le y<1$ and $x-y$ is an integer.
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\ 64$
| Our goal is to determine how many times the graph of $nf(xf(x))=x$ intersects the graph of $y=x$ . (Conversely, we can also divide the equation by $n$ to get $f(xf(x))=\frac{x}{n}$ and look at the graph $y=\frac{x}{n}$ )
We begin by analyzing the behavior of $\{x\}$ . It increases linearly with a slope of one, then when it reaches the next integer, it repeats itself. We can deduce that the function is like a sawtooth wave, with a period of one. We then analyze the function $f(x)=|2\{x\}-1|$ . The slope of the teeth is multiplied by 2 to get 2, and the function is moved one unit downward. The function can then be described as starting at -1, moving upward with a slope of 2 to get to 1, and then repeating itself, still with a period of 1. The absolute value of the function is then taken. This results in all the negative segments becoming flipped in the Y direction. The positive slope starting at -1 of the function ranging from $u$ to $u.5$ , where u is any arbitrary integer, is now a negative slope starting at positive 1. The function now looks like the letter V repeated within every square in the first row.
It is now that we address the goal of this, which is to determine how many times the function intersects the line $y=x$ . Since there are two line segments per box, the function has two chances to intersect the line $y=x$ for every integer. If the height of the function is higher than $y=x$ for every integer on an interval, then every chance within that interval intersects the line.
Returning to analyzing the function, we note that it is multiplied by $x$ , and then fed into $f(x)$ . Since $f(x)$ is a periodic function, we can model it as multiplying the function's frequency by $x$ . This gives us $2x$ chances for every integer, which is then multiplied by 2 once more to get $4x$ chances for every integer. The amplitude of this function is initially 1, and then it is multiplied by $n$ , to give an amplitude of $n$ . The function intersects the line $y=x$ for every chance in the interval of $0\leq x \leq n$ , since the function is n units high. The function ceases to intersect $y=x$ when $n < x$ , since the height of the function is lower than $y=x$ .
The number of times the function intersects $y=x$ is then therefore equal to $4+8+12...+4n$ . We want this sum to be greater than 2012 which occurs when $n=32 \Rightarrow \boxed{(C)}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_25 | C |
AMC12_1178 | In quadrilateral $ABCD$ , $m\angle B = m \angle C = 120^{\circ}, AB=3, BC=4,$ and $CD=5.$ Find the area of $ABCD.$
$\text{(A) }15 \qquad \text{(B) }9 \sqrt{3} \qquad \text{(C) }\frac{45 \sqrt{3}}{4} \qquad \text{(D) }\frac{47 \sqrt{3}}{4} \qquad \text{(E) }15 \sqrt{3}$
| Solution 1Draw $AE$ parallel to $BC$ and draw $BF$ and $CG \perp AE$ , where $F$ and $G$ are on $AE$ .
It is clear that triangles $AFB$ and $EGC$ are congruent $30-60-90$ triangles. Therefore, $AF = EG = \frac{3}{2}$ and $BF = CG = \frac{3\sqrt{3}}{2}$ .
Therefore, $AE = AF + FG + GC = 4 + (2)(\frac{3}{2}) = 7$ and the area of trapezoid $ABCE$ is $(\frac{1}{2})(4+7)(\frac{3\sqrt{3}}{2}) = \frac{33\sqrt{3}}{4}$ .
It remains to find the area of triangle $AED$ , which is $(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}$ .
Therefore, the total area of quadrilateral $ABCD$ is $\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \boxed{\textbf{(D) }\frac{47\sqrt{3}}{4}}$ .
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_19 | $\frac{47\sqrt{3}}{4}$ |
AMC12_1180 | Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 16 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 40$
| Alice may sit in the center chair, in an end chair, or in a next-to-end chair. Suppose she sits in the center chair. The 2nd and 4th chairs (next to her) must be occupied by Derek and Eric, in either order, leaving the end chairs for Bob and Carla in either order; this yields $2! * 2! = 4$ ways to seat the group.
Next, suppose Alice sits in one of the end chairs. Then the chair beside her will be occupied by either Derek or Eric. The center chair must be occupied by Bob or Carla, leaving the last two people to fill the last two chairs in either order. $2$ ways to seat Alice times $2$ ways to fill the next chair times $2$ ways to fill the center chair times $2$ ways to fill the last two chairs yields $2 * 2 * 2 * 2 = 16$ ways to fill the chairs.
Finally, suppose Alice sits in the second or fourth chair. Then the chairs next to her must be occupied by Derek and Eric in either order, and the other two chairs must be occupied by Bob and Carla in either order. This yields $2 * 2 * 2 = 8$ ways to fill the chairs.
In total, there are $4 + 8 + 16$ ways to fill the chairs, so the answer is $\boxed{\bold{(C)}\ 28}$ .
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_14 | 28 |
AMC12_1185 | Which of the following rigid transformations (isometries) maps the line segment $\overline{AB}$ onto the line segment $\overline{A'B'}$ so that the image of $A(-2, 1)$ is $A'(2, -1)$ and the image of $B(-1, 4)$ is $B'(1, -4)$ ?
$\textbf{(A) }$ reflection in the $y$ -axis
$\textbf{(B) }$ counterclockwise rotation around the origin by $90^{\circ}$
$\textbf{(C) }$ translation by 3 units to the right and 5 units down
$\textbf{(D) }$ reflection in the $x$ -axis
$\textbf{(E) }$ clockwise rotation about the origin by $180^{\circ}$
| We can simply graph the points, or use coordinate geometry, to realize that both $A'$ and $B'$ are, respectively, obtained by rotating $A$ and $B$ by $180^{\circ}$ about the origin. Hence the rotation by $180^{\circ}$ about the origin maps the line segment $\overline{AB}$ to the line segment $\overline{A'B'}$ , so the answer is $\boxed{(\text{E})}$ .
~Dodgers66
| https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_3 | E |
AMC12_1186 | What is $\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}?$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ \frac{5}{36} \qquad \textbf{(C)}\ \frac{7}{12} \qquad \textbf{(D)}\ \frac{147}{60} \qquad \textbf{(E)}\ \frac{43}{3}$
| Solution 1Add up the numbers in each fraction to get $\frac{12}{9} - \frac{9}{12}$ , which equals $\frac{4}{3} - \frac{3}{4}$ . Doing the subtraction yields $\boxed{\frac{7}{12}\ \textbf{(C)}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_1 | 7/12 |
AMC12_1188 | In the equation below, $A$ and $B$ are consecutive positive integers, and $A$ , $B$ , and $A+B$ represent number bases: \[132_A+43_B=69_{A+B}.\]
What is $A+B$ ?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17$
| Change the equation to base 10: \[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\] \[A^2 - 3A - 2B - 4=0\]
Either $B = A + 1$ or $B = A - 1$ , so either $A^2 - 5A - 6, B = A + 1$ or $A^2 - 5A - 2, B = A - 1$ . The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1) = 0, B = A + 1$ . Since A must be positive, $A = 6, B = 7$ and $A+B = \boxed{\textbf{(C)}\ 13}$
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_11 | 13 |
AMC12_1190 | Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$ , where $0 < \theta < 60^{\circ}$ , to form $\triangle DEF$ . See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$ . What is $\tan\theta$ ?
[asy] // Credit to shihan for this diagram. defaultpen(fontsize(13)); size(200); pair O=(0,0),A=dir(225),B=dir(-15),C=dir(105),D=rotate(38.21,O)*A,E=rotate(38.21,O)*B,F=rotate(38.21,O)*C; draw(A--B--C--A,gray+0.4);draw(D--E--F--D,gray+0.4); draw(A--D--B--E--C--F--A,black+0.9); dot(O); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy]
$\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\frac{4}{5}\qquad\textbf{(D)}~\frac{11}{13}\qquad\textbf{(E)}~\frac{7\sqrt{3}}{13}$
| Let O be circumcenter of the equilateral triangle
Easily get $OF = \frac{14\sqrt{3}}{3}$
$2 \cdot \triangle(OFC) + \triangle(OCE) = OF^2 \cdot \sin(\theta) + OF^2 \cdot \sin(120 - \theta)$
\[= \frac{14^2 \cdot 3}{9} ( \sin(\theta) + \sin(120 - \theta) )\]
\[= \frac{196}{3} ( \sin(\theta) + \sin(120 - \theta) )\]
\[= 2 \cdot {\frac{1}{3} } \cdot(ABCDEF) = 2\cdot \frac{91\sqrt{3}}{3}\]
\[\sin(\theta) + \sin(120 - \theta) = \frac{13\sqrt{3}}{14}\]
\[\sin(\theta) + \frac{ \sqrt{3}}{2}\cos( \theta) +\frac{ \sqrt{1}}{2}\sin( \theta) = \frac{13\sqrt{3}}{14}\]
\[\sqrt{3} \sin( \theta) + \cos( \theta) = \frac{13 }{7}\]
\[\cos( \theta) = \frac{13 }{7} - \sqrt{3} \sin( \theta)\]
\[\frac{169 }{49} - \frac{26\sqrt{3} }{7} \sin( \theta) + 4 \sin( \theta)^2 = 1\]
\[\sin( \theta) = \frac{5\sqrt{3} }{14} or \frac{4\sqrt{3} }{7}\]
$\frac{4\sqrt{3} }{7}$ is invalid given $\theta \leq 60^\circ$ , $\sin(\theta ) < \sin( 60^\circ ) = \frac{\sqrt{3} }{2} = \frac{\sqrt{3} \cdot 3.5}{7}$
\[\cos( \theta) = \frac{11 }{14}\]
\[\tan( \theta) = \frac{5\sqrt{3} }{11} \boxed{B }\]
~ luckuso
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_19 | B |
AMC12_1195 | Suppose that $a_1 = 2$ and the sequence $(a_n)$ satisfies the recurrence relation \[\frac{a_n -1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}\] for all $n \ge 2.$ What is the greatest integer less than or equal to \[\sum^{100}_{n=1} a_n^2?\]
$\textbf{(A) } 338{,}550 \qquad \textbf{(B) } 338{,}551 \qquad \textbf{(C) } 338{,}552 \qquad \textbf{(D) } 338{,}553 \qquad \textbf{(E) } 338{,}554$
| Multiply both sides of the recurrence to find that $n(a_n-1)=(n-1)(a_{n-1}+1)=(n-1)(a_{n-1}-1)+(n-1)(2)$ .
Let $b_n=n(a_n-1)$ . Then the previous relation becomes
\[b_n=b_{n-1}+2(n-1)\]
We can rewrite this relation for values of $n$ until $1$ and use telescoping to derive an explicit formula:
\[b_n=b_{n-1}+2(n-1)\]
\[b_{n-1}=b_{n-2}+2(n-2)\]
\[b_{n-2}=b_{n-3}+2(n-3)\]
\[\cdot\]
\[\cdot\]
\[\cdot\]
\[b_2=b_1+2(1)\]
Summing the equations yields:
\[b_n+b_{n-1}+\cdots+b_2=b_{n-1}+b_{n-2}+\cdots+b_1+2((n-1)+(n-2)+\cdots+1)\]
\[b_n-b_1=2\cdot\frac{n(n-1)}{2}\]
\[b_n-1=n(n-1)\]
\[b_n=n(n-1)+1\]
Now we can substitute $a_n$ back into our equation:
\[n(a_n-1)=n(n-1)+1\]
\[a_n-1=n-1+\frac{1}{n}\]
\[a_n=n+\frac{1}{n}\]
\[a_n^2=n^2+\frac{1}{n^2}+2\]
Thus the sum becomes
\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} n^2+ \sum^{100}_{n=1} \frac{1}{n^2}+ \sum^{100}_{n=1} 2\]
We know that $\sum^{100}_{n=1} n^2=\frac{100\cdot101\cdot201}{6}=338350$ , and we also know that $\sum^{100}_{n=1} 2=200$ , so the requested sum is equivalent to $\sum^{100}_{n=1} \frac{1}{n^2}+338550$ . All that remains is to calculate $\sum^{100}_{n=1} \frac{1}{n^2}$ , and we know that this value lies between $1$ and $2$ (see the note below for a proof). Thus,
\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550<2+338550=338552\]
\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550>1+338550=338551\]
so
\[\sum^{100}_{n=1} a_n^2\in(338551,338552)\]
and thus the answer is $\boxed{\textbf{(B) }338551}$ .
~eevee9406
Note: $\sum^{100}_{n=1} \frac{1}{n^2}< \sum^{\infty}_{n=1} \frac{1}{n^2}=\frac{\pi^2}{6}<2$ . It is obvious that the sum is greater than 1 (since it contains $\frac{1}{1^2}$ as one of its terms).
If you forget this and have to derive this on the exam, here is how:
\[\sum^{100}_{n=1} \frac{1}{n^2}<\sum^{\infty}_{n=1} \frac{1}{n^2}=\frac{1}{1^1}+\left(\frac{1}{2^2}+\frac{1}{3^2}\right)+\left(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)+\cdots\]
\[<1+\left(\frac{1}{2^2}+\frac{1}{2^2}\right)+\left(\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}\right)+\left(\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}\right)+\cdots\]
\[=1+\frac{2}{2^2}+\frac{4}{4^2}+\frac{8}{8^2}+\cdots\]
\[=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=2\]
and it is clear that $\sum^{100}_{n=1} \frac{1}{n^2}<2$ . ~eevee9406
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_21 | 338551 |
AMC12_1197 | For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 4$
| Denote three roots as $r_1 < r_2 < r_3$ .
Following from Vieta's formula, $r_1r_2r_3 = -6$ .
Case 1: All roots are negative.
We have the following solution: $\left( -3, -2, -1 \right)$ .
Case 2: One root is negative and two roots are positive.
We have the following solutions: $\left( -3, 1, 2 \right)$ , $\left( -2, 1, 3 \right)$ , $\left( -1, 2, 3 \right)$ , $\left( -1, 1, 6 \right)$ .
Putting all cases together, the total number of solutions is
$\boxed{\textbf{(A) 5}}$ .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_14 | 5 |
AMC12_1198 | Let $A, M,$ and $C$ be nonnegative integers such that $A + M + C = 12$ . What is the maximum value of $A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C$ ?
$\textbf{(A) \ } 62 \qquad \textbf{(B) \ } 72 \qquad \textbf{(C) \ } 92 \qquad \textbf{(D) \ } 102 \qquad \textbf{(E) \ } 112$
| It is not hard to see that
\[(A+1)(M+1)(C+1)=\]
\[AMC+AM+AC+MC+A+M+C+1\]
Since $A+M+C=12$ , we can rewrite this as
\[(A+1)(M+1)(C+1)=\]
\[AMC+AM+AC+MC+13\]
So we wish to maximize
\[(A+1)(M+1)(C+1)-13\]
Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$ , we set $A=M=C=4$
Which gives us
\[(4+1)(4+1)(4+1)-13=112\]
so the answer is $\boxed{\textbf{(E) }112}.$
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_12 | 112 |
AMC12_1201 | Define $\blacktriangledown(a) = \sqrt{a - 1}$ and $\blacktriangle(a) = \sqrt{a + 1}$ for all real numbers $a$ . What is the value of \[\frac{\blacktriangledown(20 + \blacktriangle(2024))}{\blacktriangledown(\blacktriangle(24))}~?\]
$\textbf{(A)}~ 1 \qquad \textbf{(B)}~ 2 \qquad \textbf{(C)}~ 4 \qquad \textbf{(D)}~ 8 \qquad \textbf{(E)}~ 16$
| The value of the expression is \[\frac{\sqrt{20+\sqrt{2024+1}-1}}{\sqrt{\sqrt{24+1}-1}}=\frac{\sqrt{20+\sqrt{2025}-1}}{\sqrt{\sqrt{25}-1}}=\frac{\sqrt{20+45-1}}{\sqrt{5-1}}=\frac{\sqrt{64}}{\sqrt{4}}=\frac{8}{2}=\boxed{\textbf{(C)}~4}.\]
| https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_2 | 4 |
AMC12_1204 | The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD$ . Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD$ ?
$\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250$
| Since the angles of Quadrilateral $ABCD$ form an arithmetic sequence, we can assign each angle with the value $a$ , $a+d$ , $a+2d$ , and $a+3d$ . Also, since these angles form an arithmetic progression, we can reason out that $(a)+(a+3d)=(a+d)+(a+2d)=180$ .
For the sake of simplicity, lets rename the angles of each similar triangle. Let $\angle ADB = \angle CBD = \alpha$ , $\angle DBA = \angle DCB = \beta$ , $\angle CDB = \angle BAD = \gamma$ .
Now the four angles of $ABCD$ are $\beta$ , $\alpha + \beta$ , $\gamma$ , and $\alpha + \gamma$ .
As for the similar triangles, we can name their angles $y$ , $y+b$ , and $y+2b$ . Therefore $y+y+b+y+2b=180$ and $y+b=60$ . Because these 3 angles are each equal to one of $\alpha, \beta, \gamma$ , we know that one of these three angles is equal to 60 degrees.
Now we we use trial and error. Let $\alpha = 60^{\circ}$ . Then the angles of ABCD are $\beta$ , $60^{\circ} + \beta$ , $\gamma$ , and $60^{\circ} + \gamma$ . Since these four angles add up to 360, then $\beta + \gamma= 120$ . If we list them in increasing value, we get $\beta$ , $\gamma$ , $60^{\circ} + \beta$ , $60^{\circ} + \gamma$ . Note that this is the only sequence that works because the common difference is less than 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, $\alpha, \beta, \gamma$ also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.
If we apply the same reasoning to $\beta$ and $\gamma$ , we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, $\boxed{\textbf{(D) }240}$ is the correct answer.
| https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_13 | 240 |
AMC12_1206 | Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?
$\text {(A) } 0 \qquad \text {(B) } 13 \qquad \text {(C) } 37 \qquad \text {(D) } 64 \qquad \text {(E) } 83$
| Where $a,b,c$ is the number of nickels, dimes, and quarters, respectively, we can set up two equations:
\[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\]
Eliminate $a$ by subtracting $(2)$ from $(1)/5$ to get $b+4c=67$ . Of the integer solutions $(b,c)$ to this equation, the number of dimes $b$ is least in $(3,16)$ and greatest in $(67,0)$ , yielding a difference of $67-3=\boxed{\textbf{(D)}\ 64}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_7 | 64 |
AMC12_1207 | A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?
$\text{(A) }8! \qquad \text{(B) }2^8 \cdot 8! \qquad \text{(C) }(8!)^2 \qquad \text{(D) }\frac {16!}{2^8} \qquad \text{(E) }16!$
| Solution 1Let the spider try to put on all $16$ things in a random order. Each of the $16!$ permutations is equally probable. For any fixed leg, the probability that he will first put on the sock and only then the shoe is clearly $\frac{1}{2}$ . Then the probability that he will correctly put things on all legs is $\frac{1}{2^{8}}$ . Therefore the number of correct permutations must be $\boxed{\frac {16!}{2^8}}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_16 | 16!/2^8 |
AMC12_1209 | A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$ . When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$ . How many cards were in the deck originally?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$
| If the probability of choosing a red card is $\frac{1}{3}$ , the red and black cards are in ratio $1:2$ . This means at the beginning there are $x$ red cards and $2x$ black cards.
After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the ratio of red to black cards is $1:3$ . This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.
So, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.
So, the answer is $8+4 = 12 = \boxed{\textbf{(C) }12}$ .
--abhinavg0627
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_6 | 12 |
AMC12_1212 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$ ?
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$
| Factor the quadratic into
\[\left(5x + \frac{12}{n}\right)\left(x + n\right) = 0\]
where $-n$ is our integer solution. Then,
\[k = \frac{12}{n} + 5n,\]
which takes rational values between $-200$ and $200$ when $|n| \leq 39$ , excluding $n = 0$ . This leads to an answer of $2 \cdot 39 = \boxed{\textbf{(E) } 78}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | 78 |
AMC12_1215 | Suppose $a$ , $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$ ?
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 501$
| The power of $10$ for any factorial is given by the well-known algorithm
\[\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots\]
It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher powers of $5$ . As we list out the powers of 5, it is clear that $5^{4}=625$ is less than 2006 and $5^{5}=3125$ is greater. Therefore, set $a$ and $b$ to be 624. Thus, c is $2006-(624\cdot 2)=758$ . Applying the algorithm, we see that our answer is $152+152+188= \boxed{492}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_22 | 492 |
AMC12_1216 | At Rachelle's school, an A counts $4$ points, a B $3$ points, a C $2$ points, and a D $1$ point. Her GPA in the four classes she is taking is computed as the total sum of points divided by 4. She is certain that she will get A's in both Mathematics and Science and at least a C in each of English and History. She thinks she has a $\frac{1}{6}$ chance of getting an A in English, and a $\tfrac{1}{4}$ chance of getting a B. In History, she has a $\frac{1}{4}$ chance of getting an A, and a $\frac{1}{3}$ chance of getting a B, independently of what she gets in English. What is the probability that Rachelle will get a GPA of at least $3.5$ ?
$\textbf{(A)}\; \frac{11}{72} \qquad\textbf{(B)}\; \frac{1}{6} \qquad\textbf{(C)}\; \frac{3}{16} \qquad\textbf{(D)}\; \frac{11}{24} \qquad\textbf{(E)}\; \frac{1}{2}$
| The probability that Rachelle gets a C in English is $1-\frac{1}{6}-\frac{1}{4} = \frac{7}{12}$ .
The probability that she gets a C in History is $1-\frac{1}{4}-\frac{1}{3} = \frac{5}{12}$ .
We see that the sum of Rachelle's "point" scores must be at least 14 since $4*3.5 = 14$ . We know that in Mathematics and Science we have a total point score of 8 (since she will get As in both), so we only need a sum of 6 in English and History. This can be achieved by getting two As, one A and one B, one A and one C, or two Bs. We evaluate these cases.
The probability that she gets two As is $\frac{1}{6}\cdot\frac{1}{4} = \frac{1}{24}$ .
The probability that she gets one A and one B is $\frac{1}{6}\cdot\frac{1}{3} + \frac{1}{4}\cdot\frac{1}{4} = \frac{1}{18}+\frac{1}{16} = \frac{8}{144}+\frac{9}{144} = \frac{17}{144}$ .
The probability that she gets one A and one C is $\frac{1}{6}\cdot\frac{5}{12} + \frac{1}{4}\cdot\frac{7}{12} = \frac{5}{72}+\frac{7}{48} = \frac{31}{144}$ .
The probability that she gets two Bs is $\frac{1}{4}\cdot\frac{1}{3} = \frac{1}{12}$ .
Adding these, we get $\frac{1}{24} + \frac{17}{144} + \frac{31}{144} + \frac{1}{12} = \frac{66}{144} = \boxed{\mathbf{(D)}\; \frac{11}{24}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_15 | 11/24 |
AMC12_1218 | What is the value of \[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\] where $\log$ denotes the base-ten logarithm?
$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$
| Let $\text{log } 2 = x$ . The expression then becomes \[(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.\]
-bluelinfish
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_14 | 2 |
AMC12_1221 | How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$ ?
$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$
| We want to find the number of perfect square factors in the product of all the factorials of numbers from $1 - 9$ . We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$ . To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$ . To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{\mathrm{(B)}}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_23 | 672 |
AMC12_1222 | If $f(c)=\frac{3}{2c-3}$ , find $\frac{kn^2}{lm}$ when $f^{-1}(c)\times c \times f(c)$ equals the simplified fraction $\frac{kc+l}{mc+n}$ , where $k,l,m,\text{ and }n$ are integers.
| Since $f(f^{-1}(x))=x$ , it follows that $a(bx+a)+b=x$ , which implies $abx + a^2 +b = x$ . This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$ .
Then $b = -a^2$ . Substituting into the equation $ab = 1$ , we get $-a^3 = 1$ . Then $a = -1$ , so $b = -1$ , and \[f(x)=-x-1.\] Likewise \[f^{-1}(x)=-x-1.\] These are inverses to one another since \[f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.\] \[f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.\] Therefore $a+b=\boxed{-2}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_13 | -2 |
AMC12_1224 | Consider sequences of positive real numbers of the form $x, 2000, y, \dots$ in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of $x$ does the term $2001$ appear somewhere in the sequence?
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) more than }4$
| It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that $\forall$ (for all) $n>1:~ a_n = a_{n-1}a_{n+1} - 1$ . This can be rewritten as $a_{n+1} = \frac{a_n +1}{a_{n-1}}$ . We have $a_1=x$ and $a_2=2000$ , and we compute:
\begin{align*} a_3 & = \frac{a_2+1}{a_1} = \frac{2001}x \\ a_4 & = \frac{a_3+1}{a_2} = \frac{ \dfrac{2001}x + 1 }{ 2000 } = \frac{2001 + x}{2000x} \\ a_5 & = \frac{a_4+1}{a_3} = \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } = \frac{ \frac{2001 + 2001x}{2000x} }{ \frac{2001}x } = \frac{1+x}{2000} \\ a_6 & = \frac{a_5+1}{a_4} = \frac{ \frac{1+x}{2000} + 1 }{ \frac{2001 + x}{2000x} } = \frac{ \frac{2001+x}{2000} }{ \frac{2001 + x}{2000x} } = x \\ a_7 & = \frac{a_6+1}{a_5} = \frac{ x+1 }{ \frac{1+x}{2000} } = 2000 \end{align*}
At this point we see that the sequence will become periodic: we have $a_6=a_1$ , $a_7=a_2$ , and each subsequent term is uniquely determined by the previous two.
Hence if $2001$ appears, it has to be one of $a_1$ to $a_5$ . As $a_2=2000$ , we only have four possibilities left. Clearly $a_1=2001$ for $x=2001$ , and $a_3=2001$ for $x=1$ . The equation $a_4=2001$ solves to $x = \frac{2001}{2000\cdot 2001 - 1}$ , and the equation $a_5=2001$ to $x=2000\cdot 2001 - 1$ .
No two values of $x$ we just computed are equal, and therefore there are $\boxed{4}$ different values of $x$ for which the sequence contains the value $2001$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_25 | 4 |
AMC12_1225 | Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that
$\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 \qquad \mathrm{(D)}\ \frac 15 \qquad \mathrm{(E)}\ \frac 14$
| Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$ ?
The equation $\lfloor\log_{10}x\rfloor = k$ can be rewritten as $10^k \leq x < 10^{k+1}$ . The second one gives us $10^k \leq 4x < 10^{k+1}$ . Combining these, we get that both hold at the same time if and only if $10^k \leq x < \frac{10^{k+1}}4$ .
Hence for each integer $k$ we get an interval of values for which $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ . These intervals are obviously pairwise disjoint.
For any $k\geq 0$ the corresponding interval is disjoint with $(0,1)$ , so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$ . Hence our answer is the sum of the lengths of the intervals for $k<0$ .
For a fixed $k$ the length of the interval $\left[ 10^k, \frac{10^{k+1}}4 \right)$ is $\frac 32\cdot 10^k$ .
This means that our result is $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_20 | 1/6 |
AMC12_1226 | Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$ , how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$ ?
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$
| Each of the $\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles — one facing towards the center, and one outwards — for a total of 72 triangles. However, the three triangles $A_1A_4A_7$ , $A_2A_5A_8$ , and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\boxed{66}$ distinct equilateral triangles.
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_14 | 66 |
AMC12_1227 | How many $15$ -letter arrangements of $5$ A's, $5$ B's, and $5$ C's have no A's in the first $5$ letters, no B's in the next $5$ letters, and no C's in the last $5$ letters?
$\textrm{(A)}\ \sum_{k=0}^{5}\binom{5}{k}^{3}\qquad\textrm{(B)}\ 3^{5}\cdot 2^{5}\qquad\textrm{(C)}\ 2^{15}\qquad\textrm{(D)}\ \frac{15!}{(5!)^{3}}\qquad\textrm{(E)}\ 3^{15}$
| The answer is $\boxed{\textrm{(A)}}$ .
Note that the first five letters must be B's or C's, the next five letters must be C's or A's, and the last five letters must be A's or B's. If there are $k$ B's in the first five letters, then there must be $5-k$ C's in the first five letters, so there must be $k$ C's and $5-k$ A's in the next five letters, and $k$ A's and $5-k$ B's in the last five letters. Therefore the number of each letter in each group of five is determined completely by the number of B's in the first 5 letters, and the number of ways to arrange these 15 letters with this restriction is $\binom{5}{k}^3$ (since there are $\binom{5}{k}$ ways to arrange $k$ B's and $5-k$ C's). Therefore the answer is $\sum_{k=0}^{5}\binom{5}{k}^{3}$ .
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_20 | A |
AMC12_1230 | Corners are sliced off a unit cube so that the six faces each become regular octagons. What is the total volume of the removed tetrahedra?
$\mathrm{(A)}\ \frac{5\sqrt{2}-7}{3}\qquad \mathrm{(B)}\ \frac{10-7\sqrt{2}}{3}\qquad \mathrm{(C)}\ \frac{3-2\sqrt{2}}{3}\qquad \mathrm{(D)}\ \frac{8\sqrt{2}-11}{3}\qquad \mathrm{(E)}\ \frac{6-4\sqrt{2}}{3}$
|
Since the sides of a regular polygon are equal in length, we can call each side $x$ . Examine one edge of the unit cube: each contains two slanted diagonal edges of an octagon and one straight edge. The diagonal edges form $45-45-90 \triangle$ right triangles , making the distance on the edge of the cube $\frac{x}{\sqrt{2}}$ . Thus, $2 \cdot \frac{x}{\sqrt{2}} + x = 1$ , and $x = \frac{1}{\sqrt{2} + 1} \cdot \left(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\right) = \sqrt{2} - 1$ .
Each of the cut off corners is a pyramid , whose volume can be calculated by $V = \frac 13 Bh$ . Use the base as one of the three congruent isosceles triangles , with the height being one of the edges of the pyramid that sits on the edges of the cube. The height is $\frac x{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$ . The base is a $45-45-90 \triangle$ with leg of length $1 - \frac{1}{\sqrt{2}}$ , making its area $\frac 12\left(1 - \frac 1{\sqrt{2}}\right)^2 = \frac{3 - 2\sqrt{2}}4$ . Plugging this in, we get that the volume of one of the tetrahedra is $\frac 13 \left(1 - \frac{1}{\sqrt{2}}\right)\left(\frac{3 - 2\sqrt{2}}4\right) = \frac{10 - 7\sqrt{2}}{24}$ . Since there are 8 removed corners, we get an answer of $\frac{10 - 7\sqrt{2}}{3} \Longrightarrow \boxed{\mathrm{B}}$
| https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_20 | B |
AMC12_1231 | Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?
$\mathrm{(A)} \ 0.10 \qquad \mathrm{(B)} \ 0.15 \qquad \mathrm{(C)} \ 0.20 \qquad \mathrm{(D)} \ 0.25 \qquad \mathrm{(E)} \ 0.30 \qquad$
| If $A$ and $B$ meet, their paths connect $(0,0)$ and $(5,7).$ There are $\binom{12}{5}=792$ such paths. Since the path is $12$ units long, they must meet after each travels $6$ units, so the probability is $\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}$ .
Note: The number of paths, $\binom{12}{5}$ comes from the fact that there must be 5 ups/downs and 7 lefts/rights in one path. WLOG, for Object A, the number of paths would be the amount of combinations of the sequence of letters with 5 "U"s 7 "R"s (i.e. UUUUURRRRRRR). This is $\frac{12!}{5!7!}$ , which is equivalent to $\binom{12}{5}$ .
~bearjere
| https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_22 | C |
AMC12_1232 | If $x$ and $y$ are positive integers for which $2^x3^y=1296$ , what is the value of $x+y$ ?
$(\mathrm {A})\ 8 \qquad (\mathrm {B})\ 9 \qquad (\mathrm {C})\ 10 \qquad (\mathrm {D})\ 11 \qquad (\mathrm {E})\ 12$
| $1296 = 2^4 3^4$ and $4+4=\boxed{8} \Longrightarrow \mathrm{(A)}$ .
| https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_3 | 8 |
AMC12_1233 | Find $i + 2i^2 +3i^3 + ... + 2002i^{2002}.$
$\text{(A) }-999 + 1002i \qquad \text{(B) }-1002 + 999i \qquad \text{(C) }-1001 + 1000i \qquad \text{(D) }-1002 + 1001i \qquad \text{(E) }i$
| Note that $i^4 = 1$ , so $i^n = i^{4m+n}$ for all integers $m$ and $n$ . In particular, $i = 1$ , $i^2 = -1$ , and $i^3 = -i$ .
We group the positive and negative real terms together and group the positive and negative imaginary parts together.
The positive real terms have exponents on $i$ that are multiples of 4. Therefore, the positive real part evaluates to
\[4 + 8 + ... + 2000\]
The negative real terms have exponents on $i$ that are of the form $4k + 2$ for integers $k$ . Therefore, the negative real part evaluates to \[-(2 + 6 + ... + 2002)\]
The positive imaginary terms have exponents on $i$ that are of the form $4k + 1$ for integers $k$ . Therefore, the negative real part evaluates to \[(1 + 5 + ... + 2001)i\]
The negative imaginary terms have exponents on $i$ that are of the form $4k + 3$ for integers $k$ . Therefore, the negative real part evaluates to \[-(3 + 7 + ... + 1999)i\]
Putting everything together, we have \[i + 2i^2 + ... + 2002i^{2002} = (-2 + 4 - 6 + 8 - ... + 2000 - 2002) + (1 - 3 + 5 - 7 + ... - 1999 + 2001)i\]
Group every 2 consecutive terms as shown below \[((-2 + 4)+(-6 + 8) + ... + (-1998 + 2000)-2002) + ((1 - 3)+(5 - 7) + ... + (1997 - 1999) + 2001)i\]
Now we evaluate each small bracket with 2 terms. We get $500(2) = 1000$ in the real part and $500(-2) = -1000$ in the imaginary part. Therefore, the sum becomes $(1000 - 2002) + (-1000 + 2001)i = \boxed {\text{(D) }-1002 + 1001i}$ .
Note: This problem is similar to 2009 AMC 12A Problem 15
| https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_14 | -1002 + 1001i |
AMC12_1239 | Lian, Marzuq, Rafsan, Arabi, Nabeel, and Rahul were in a 12-person race with 6 other people. Nabeel finished 6 places ahead of Marzuq. Arabi finished 1 place behind Rafsan. Lian finished 2 places behind Marzuq. Rafsan finished 2 places behind Rahul. Rahul finished 1 place behind Nabeel. Arabi finished in 6th place. Who finished in 8th place?
$\textbf{(A)}\; \text{Lian} \qquad\textbf{(B)}\; \text{Marzuq} \qquad\textbf{(C)}\; \text{Rafsan} \qquad\textbf{(D)}\; \text{Nabeel} \qquad\textbf{(E)}\; \text{Rahul}$
| Let --- denote any of the 6 racers not named. Then the correct order looks like this:
\[-, \text{Nabeel}, \text{Rahul}, -, \text{Rafsan}, \text{Arabi}, -, \text{Marzuq}, -, \text{Lian}, -, -\]
Thus the 8th place runner is $\boxed{\textbf{(B)}\; \text{Marzuq}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_4 | Marzuq |
AMC12_1242 | What is
\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100{,}100 \qquad \textbf{(B) }500{,}500\qquad \textbf{(C) }505{,}000 \qquad \textbf{(D) }1{,}001{,}000 \qquad \textbf{(E) }1{,}010{,}000 \qquad$
| Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that
\begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\ &= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\ &= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\ &= 100\cdot5050 + 100\cdot5050 \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*}
~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9 | 1,010,000 |
AMC12_1243 | Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)
$\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$
| Solution 1The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\overrightarrow{AB}$ and $\overrightarrow{AD}$ .
In our setting where $A=(0,0)$ , $B=(s,s)$ , and $D=(t,kt)$ this is simply $s\cdot kt - s\cdot t = (k-1)st$ .
In other words, we need to count the triples of integers $(k,s,t)$ where $k>1$ , $s,t>0$ and $(k-1)st = 1,\!000,\!000 = 2^6 5^6$ .
These can be counted as follows: We have $6$ identical red balls (representing powers of $2$ ), $6$ blue balls (representing powers of $5$ ), and three labeled urns (representing the factors $k-1$ , $s$ , and $t$ ). The red balls can be distributed in ${8\choose 2} = 28$ ways, and for each of these ways, the blue balls can then also be distributed in $28$ ways. (See Distinguishability for a more detailed explanation.)
Thus there are exactly $28^2 = 784$ ways how to break $1,\!000,\!000$ into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is $784 \longrightarrow \boxed{C}$ .
| https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_22 | C |
AMC12_1246 | A frog located at $(x,y)$ , with both $x$ and $y$ integers, makes successive jumps of length $5$ and always lands on points with integer coordinates. Suppose that the frog starts at $(0,0)$ and ends at $(1,0)$ . What is the smallest possible number of jumps the frog makes?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$
| Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (3-4-5 right triangle).
Because either $1$ , $5$ , or $7$ is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it can't go from $(0,0)$ to $(1,0)$ in an even number of moves. Therefore, the frog cannot reach $(1,0)$ in two moves.
However, a path is possible in 3 moves: from $(0,0)$ to $(3,4)$ to $(6,0)$ to $(1,0)$ .
Thus, the answer is $= \boxed{3 \textbf{}}$ .
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_11 | 3 |
AMC12_1247 | There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$ ?
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$
| $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ can be simplified to $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.$
The equation is then $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$ which implies that $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
$a$ has to be $2$ since $\frac{25}{36}>\frac{7}{12}$ . $\frac{7}{12}$ is the result when $a, b,$ and $c$ are $3, 2,$ and $2$
$b$ being $3$ will make the fraction $\frac{2}{3}$ which is close to $\frac{25}{36}$ .
Finally, with $c$ being $6$ , the fraction becomes $\frac{25}{36}$ . In this case $a, b,$ and $c$ work, which means that $b$ must equal $\boxed{\textbf{(B) } 3.}$ ~lopkiloinm
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_13 | 3 |
AMC12_1248 | A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?
[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2)); draw(A--B--C--D--E--F--G--H--cycle); draw(A--D); draw(B--G); draw(C--F); draw(E--H);[/asy]
$\textbf{(A)}\ \frac{\sqrt{2} - 1}{2} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{2 - \sqrt{2}}{2} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{4} \qquad \textbf{(E)}\ 2 - \sqrt{2}$
| Let's assume that the side length of the octagon is $x$ . The area of the center square is just $x^2$ . The triangles are all $45-45-90$ triangles, with a side length ratio of $1:1:\sqrt{2}$ . The area of each of the $4$ identical triangles is $\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}$ , so the total area of all of the triangles is also $x^2$ . Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is $x$ and the other side length is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$ , so the area of all of the rectangles is $2x^2\sqrt{2}$ . The ratio of the area of the square to the area of the octagon is $\dfrac{x^2}{2x^2+2x^2\sqrt{2}}$ . Cancelling $x^2$ from the fraction, the ratio becomes $\dfrac{1}{2\sqrt2+2}$ . Multiplying the numerator and the denominator each by $2\sqrt{2}-2$ will cancel out the radical, so the fraction is now $\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_12 | (A) (√2 - 1)/2 |
AMC12_1249 | Jesse cuts a circular disk of radius 12, along 2 radii to form 2 sectors, one with a central angle of 120. He makes two circular cones using each sector to form the lateral surface of each cone. What is the ratio of the volume of the smaller cone to the larger cone?
$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{\sqrt{10}}{10}\qquad\textbf{(D)}\ \frac{\sqrt{5}}{6}\qquad\textbf{(E)}\ \frac{\sqrt{5}}{5}$
| If the original radius is $12$ , then the circumference is $24\pi$ ; since arcs are defined by the central angles, the smaller arc, a $120$ degree angle, is half the size of the larger sector. so the smaller arc is $8\pi$ , and the larger is $16\pi$ . Those two arc lengths become the two circumferences of the new cones; so the radius of the smaller cone is $4$ and the larger cone is $8$ . Using the Pythagorean theorem, the height of the larger cone is $4\cdot\sqrt{5}$ and the smaller cone is $8\cdot\sqrt{2}$ , and now for volume just square the radii and multiply by $\tfrac{1}{3}$ of the height to get the volume of each cone: $128\cdot\sqrt{2}$ and $256\cdot\sqrt{5}$ [both multiplied by three as ratio come out the same. now divide the volumes by each other to get the final ratio of $\boxed{\textbf{(C) } \frac{\sqrt{10}}{10}}$
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_15 | $\frac{\sqrt{10}}{10}$ |
AMC12_1250 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ , $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$
| Each path must go through either the second or the fourth quadrant.
Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$ , $(-3,3)$ , and $(-2,2)$ .
There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$ , $(3,-3)$ , and $(2,-2)$ .
Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Hence the total number of paths is $2(1+64+784) = \boxed{1698}$ .
| https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_18 | 1698 |
AMC12_1257 | What is the value of the following expression?
\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80}$
| Using difference of squares to factor the left term, we get
\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.\]
Cancelling all the terms, we get $\boxed{\textbf{(A) } 1}$ as the answer.
| https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_2 | 1 |
AMC12_1261 | Two circles intersect at points $A$ and $B$ . The minor arcs $AB$ measure $30^\circ$ on one circle and $60^\circ$ on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?
$\textbf{(A) }2\qquad \textbf{(B) }1+\sqrt3\qquad \textbf{(C) }3\qquad \textbf{(D) }2+\sqrt3\qquad \textbf{(E) }4\qquad$
| Let the radius of the larger and smaller circles be $x$ and $y$ , respectively. Also, let their centers be $O_1$ and $O_2$ , respectively. Then the ratio we need to find is
\[\dfrac{\pi x^2}{\pi y^2} = \dfrac{x^2}{y^2}\]
Draw the radii from the centers of the circles to $A$ and $B$ . We can easily conclude that the $30^{\circ}$ belongs to the larger circle, and the $60$ degree arc belongs to the smaller circle. Therefore, $m\angle AO_1B = 30^{\circ}$ and $m\angle AO_2B = 60^{\circ}$ . Note that $\Delta AO_2B$ is equilateral, so when chord AB is drawn, it has length $y$ . Now, applying the Law of Cosines on $\Delta AO_1B$ :
\[y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2\]
\[\dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}\]
(Solution by brandbest1)
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_12 | 2 + √3 |
AMC12_1262 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$ ?
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
| Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\}$ and $C\in\{1,5,7,9\},$ there are $8$ possibilities for $A$ and $4$ possibilities for $C.$ Note that each ordered pair $(A,C)$ determines the value of $B$ modulo $3,$ so $B$ can be any element in one of the sets $\{0,6,9\},\{1,4,7\},$ or $\{2,5,8\}.$ We conclude that there are always $3$ possibilities for $B.$
By the Multiplication Principle, the answer is $8\cdot4\cdot3=\boxed{\textbf{(A) } 96}.$
~Plasma_Vortex ~MRENTHUSIASM
| https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 96 |
AMC12_1263 | Three balls are randomly and independantly tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,2,3,....$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\frac pq,$ where $p$ and $q$ are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins $3,17,$ and $10.$ ) What is $p+q?$
$\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59$
| "Evenly spaced" just means the bins form an arithmetic sequence.
Suppose the middle bin in the sequence is $x$ . There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\cdot 2^{-3x} = 6\cdot \frac{1}{8^x}$ , so the probability $x$ is the middle bin is $6\cdot\frac{x-1}{8^x}$ . Then, we want the sum
\begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}+\cdots \right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4} + \cdots \right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3} + \cdots \right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*}
The answer is $6+49=\boxed{\textbf{(A) }55}.$
| https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_23 | 55 |
AMC12_1265 | The line $12x+5y=60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?
$\textbf{(A)}\; 20 \qquad\textbf{(B)}\; \dfrac{360}{17} \qquad\textbf{(C)}\; \dfrac{107}{5} \qquad\textbf{(D)}\; \dfrac{43}{2} \qquad\textbf{(E)}\; \dfrac{281}{13}$
| Clearly the line and the coordinate axes form a right triangle. Since the x-intercept and y-intercept are 5 and 12 respectively, 5 and 12 are two sides of the triangle that are not the hypotenuse, and are thus two of the three heights. In order to find the third height, we can use different equations of the area of the triangle. Using the lengths we know, the area of the triangle is $\tfrac{1}{2} \times 5 \times 12= 30$ . We can use the hypotenuse as another base to find the third height. Using the distance formula, the length of the hypotenuse is $\sqrt{5^2+12^2}=13$ . Then $\frac{1}{2} \times 13 \times h=30$ , and so $h = \frac{60}{13}$ . Therefore the sum of all the heights is $5+12+\frac{60}{13}=\boxed{\textbf{(E)}\; \frac{281}{13}}$ .
| https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_11 | 281/13 |
AMC12_1266 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$ , and $SR$ , respectively. What is the sum of the coordinates of the center of the square $PQRS$ ?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5$
| [asy] size(14cm); pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8); dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H); draw(A--SS--D--cycle); draw(P--Q--R^^B--Q--C); draw(EE--M--F^^G--B^^C--H,dotted); label("A",A,SW); label("B",B,S); label("C",C,S); label("D",D,SE); label("E",EE,S); label("F",F,S); label("P",P,W); label("Q",Q,NW); label("R",R,NE); label("S",SS,N); label("M",M,S); label("G",G,W); label("H",H,NE);[/asy]
Construct the midpoints $E=(4,0)$ and $F=(10,0)$ and triangle $\triangle EMF$ as in the diagram, where $M$ is the center of square $PQRS$ . Also construct points $G$ and $H$ as in the diagram so that $BG\parallel PQ$ and $CH\parallel QR$ .
Observe that $\triangle AGB\sim\triangle CHD$ while $PQRS$ being a square implies that $GB=CH$ . Furthermore, $CD=6=3\cdot AB$ , so $\triangle CHD$ is 3 times bigger than $\triangle AGB$ . Therefore, $HD=3\cdot GB=3\cdot HC$ . In other words, the longer leg is 3 times the shorter leg in any triangle similar to $\triangle AGB$ .
Let $K$ be the foot of the perpendicular from $M$ to $EF$ , and let $x=EK$ . Triangles $\triangle EKM$ and $\triangle MKF$ , being similar to $\triangle AGB$ , also have legs in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$ , so $10x=EF=6$ . It follows that $EK=0.6$ and $MK=1.8$ , so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $4.6+1.8 = 6.4 =$ $\boxed{\mathbf{(C)}\ 32/5}$ .
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_17 | 32/5 |
AMC12_1267 | How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$
| Solution 1There are $\binom{20}{2}$ selections; however, we count these twice, therefore
$2\cdot\binom{20}{2} = \boxed{\textbf{(D)}\ 380}$ . The wording of the question implies D, not E.
However, MAA decided to accept both D and E.
| https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_12 | 380 |
AMC12_1271 | Let $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$ , and $h_n(x) = h_1(h_{n-1}(x))$ for integers $n \geq 2$ . What is the sum of the digits of $h_{2011}(1)$ ?
$\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099$
| $g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$
$h_{1}(x)=g(f(x))\text{ = }g(10^{10x})=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$
Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ :
For $n=1$ , $h_{1}(x)=10x - 1$
Assume $h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ is true for n:
\begin{align*} h_{n+1}(x)&= h_{1}(h_{n}(x))\\ &=10 h_{n}(x) - 1\\ &=10 (10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\\ &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\\ &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) \end{align*}
Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n.
$h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$ , which is the 2011-digit number 8888...8889
The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089\textbf{(B)}}$
| https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_17 | 16089 |
AMC12_1276 | The circle having $(0,0)$ and $(8,6)$ as the endpoints of a diameter intersects the $x$ -axis at a second point. What is the $x$ -coordinate of this point?
$\textbf{(A) } 4\sqrt{2} \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 5\sqrt{2} \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$
| Because the two points are on a diameter, the center must be halfway between them at the point $(4,3)$ . The distance from $(0,0)$ to $(4,3)$ is 5 so the circle has radius 5. Thus, the equation of the circle is $(x-4)^2+(y-3)^2=25$ .
To find the x-intercept, y must be 0, so $(x-4)^2+(0-3)^2=25$ , so $(x-4)^2=16$ , $x-4=4$ , $x=8, \boxed{D}$ .
Written by: SilverLion
| https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_6 | 8 |
AMC12_1278 | How many positive integers $b$ have the property that $\log_{b} 729$ is a positive integer?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }$
| If $\log_{b} 729 = n$ , then $b^n = 729$ . Since $729 = 3^6$ , $b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $b \Longrightarrow \boxed{\mathrm{E}}$ .
| https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_7 | E |
AMC12_1279 | A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
| Define $T$ to be the set of all integral triples $(a, b, c)$ such that $a \ge b \ge c$ , $b+c > a$ , and $a, b, c < 5$ . Now we enumerate the elements of $T$ :
$(4, 4, 4)$
$(4, 4, 3)$
$(4, 4, 2)$
$(4, 4, 1)$
$(4, 3, 3)$
$(4, 3, 2)$
$(3, 3, 3)$
$(3, 3, 2)$
$(3, 3, 1)$
$(3, 2, 2)$
$(2, 2, 2)$
$(2, 2, 1)$
$(1, 1, 1)$
It should be clear that $|S|$ is simply $|T|$ minus the larger "duplicates" (e.g. $(2, 2, 2)$ is a larger duplicate of $(1, 1, 1)$ ). Since $|T|$ is $13$ and the number of higher duplicates is $4$ , the answer is $13 - 4$ or $\boxed{\textbf{(B)}\ 9}$ .
| https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_12 | 9 |
AMC12_1280 | Five rectangles, $A$ , $B$ , $C$ , $D$ , and $E$ , are arranged in a square as shown below. These rectangles have dimensions $1\times6$ , $2\times4$ , $5\times6$ , $2\times7$ , and $2\times3$ , respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle?
[asy] size(150); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); [/asy]
$\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E$
| The area of this square is equal to $6 + 8 + 30 + 14 + 6 = 64$ , and thus its side lengths are $8$ . The sum of the dimensions of the rectangles are $2 + 7 + 5 + 6 + 2 + 3 + 1 + 6 + 2 + 4 = 38$ . Thus, because the perimeter of the rectangle is $32$ , the rectangle on the inside must have a perimeter of $6 \cdot 2 = 12$ . The only rectangle that works is $\boxed{\textbf{(B) }B}$ .
~mathboy100
| https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_3 | B |
AMC12_1282 | Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits
of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a
two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$ ?
$\text{(A)}\ 2\qquad \text{(B)}\ 3\qquad \text{(C)}\ 6\qquad \text{(D)}\ 8\qquad \text{(E)}\ 9$
| Denote $a$ and $b$ as the tens and units digit of $N$ , respectively. Then $N = 10a+b$ . It follows that $10a+b=ab+a+b$ , which implies that $9a=ab$ . Since $a\neq0$ , $b=9$ . So the units digit of $N$ is $\boxed{\textbf{(E) }9}$ .
| https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_2 | 9 |
AMC12_1284 | Isosceles $\triangle ABC$ has a right angle at $C$ . Point $P$ is inside $\triangle ABC$ , such that $PA=11$ , $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers. What is $a+b$ ?
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]
$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$
| [asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); MP("\alpha",C,5*dir(80),f); MP("90^\circ-\alpha",C,3*dir(30),f); MP("s",(A+C)/2,plain.S,f); MP("s",(B+C)/2,plain.W,f); [/asy]
Using the Law of Cosines on $\triangle PBC$ , we have:
\begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*}
Using the Law of Cosines on $\triangle PAC$ , we have:
\begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*}
Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$ .
\begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 \end{align*}
Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$ . Thus, $a+b=85+42=\boxed{127}$ .
| https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_23 | 127 |
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