link stringlengths 75 84 | letter stringclasses 5
values | answer float64 0 2,935,363,332B | problem stringlengths 14 5.33k | solution listlengths 1 13 |
|---|---|---|---|---|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14 | null | 592 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ ... | [
"In this solution, all angle measures are in degrees.\nLet $M$ be the midpoint of $\\overline{BC}$ so that $\\overline{OM}\\perp\\overline{BC}$ and $A,G,M$ are collinear. Let $\\angle ABC=13k,\\angle BCA=2k$ and $\\angle XOY=17k.$\nNote that:\nTogether, we conclude that $\\triangle OAM \\sim \\triangle OXY$ by AA, ... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_4 | C | 4 | Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ$ . What is the smallest possible degree measure for $\angle CBD$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12$ | [
"$\\angle ABD$ and $\\angle ABC$ share ray $AB$ . In order to minimize the value of $\\angle CBD$ $D$ should be located between $A$ and $C$\n$\\angle ABC = \\angle ABD + \\angle CBD$ , so $\\angle CBD = 4$ . The answer is $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_12 | null | 21 | Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$ . A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$ , side $\overline{CD}$ lies on $\overline{QR}$ , and one of the remaining vertices lies on $\ov... | [
"First, find that $\\angle R = 45^\\circ$ .\nDraw $ABCDEF$ . Now draw $\\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$ . The height of $ABCDEF$ is $\\sqrt{3}$ , so the length of base $QR$ is $2+\\sqrt{3}$ . Let the equation of $\\overline{RP}$ be $y = x$ . Then, the equation of $\\over... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_13 | E | 10 | Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x$ . For example, $\clubsuit(8)=8$ and $\clubsuit(123)=1+2+3=6$ . For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x))=3$
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10$ | [
"Let $y=\\clubsuit (x)$ . Since $x \\leq 99$ , we have $y \\leq 18$ . Thus if $\\clubsuit (y)=3$ , then $y=3$ or $y=12$ . The 3 values of $x$ for which $\\clubsuit (x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\\clubsuit (x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\\boxed{10}$ values in a... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_8 | E | 10 | Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x$ . For example, $\clubsuit(8)=8$ and $\clubsuit(123)=1+2+3=6$ . For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x))=3$
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10$ | [
"Let $y=\\clubsuit (x)$ . Since $x \\leq 99$ , we have $y \\leq 18$ . Thus if $\\clubsuit (y)=3$ , then $y=3$ or $y=12$ . The 3 values of $x$ for which $\\clubsuit (x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\\clubsuit (x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\\boxed{10}$ values in a... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | null | 244 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn fo... | [
"We can use recursion to solve this problem:\n1. Fix 7 points on $\\ell_A$ , then put one point $B_1$ on $\\ell_B$ . Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\\ell_B$ . For example, $f(1) = 6$ because there are 6 regions.\n2. Now, put the sec... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_30 | E | 4 | Let $\left\lfloor x\right\rfloor$ be the greatest integer less than or equal to $x$ . Then the number of real solutions to $4x^2-40\left\lfloor x\right\rfloor+51 = 0$ is
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4$ | [
"We rearrange the equation as $4x^2 = 40\\left\\lfloor x\\right\\rfloor-51$ , where the right-hand side is now clearly an integer, meaning that $4x^2 = n$ for some non-negative integer $n$ . Therefore, in the case where $x \\geq 0$ , substituting $x = \\frac{\\sqrt{n}}{2}$ gives \\[40\\left\\lfloor\\frac{\\sqrt{n}}... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | C | 199 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | [
"This rewrites itself to $x^2=10,000\\{x\\}$ where $\\lfloor x \\rfloor + \\{x\\} = x$\nGraphing $y=10,000\\{x\\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$ , then $1$ to $2$ with a hole at $x=2$ etc.\nHere is a graph of $y=x^2$ and $y=16\\... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_25 | null | 199 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | [
"Subtracting $10000\\lfloor x\\rfloor$ from both sides gives $x^2=10000(x-\\lfloor x\\rfloor)=10000\\{x\\}$ . Dividing both sides by $10000$ gives $\\left(\\frac{x}{100}\\right)^2=\\{x\\}<1$ $\\left(\\frac{x}{100}\\right)^2<1$ when $-100<x<100$ so the answer is $\\boxed{199}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | C | 199 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | [
"This rewrites itself to $x^2=10,000\\{x\\}$ where $\\lfloor x \\rfloor + \\{x\\} = x$\nGraphing $y=10,000\\{x\\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$ , then $1$ to $2$ with a hole at $x=2$ etc.\nHere is a graph of $y=x^2$ and $y=16\\... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | null | 199 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | [
"Subtracting $10000\\lfloor x\\rfloor$ from both sides gives $x^2=10000(x-\\lfloor x\\rfloor)=10000\\{x\\}$ . Dividing both sides by $10000$ gives $\\left(\\frac{x}{100}\\right)^2=\\{x\\}<1$ $\\left(\\frac{x}{100}\\right)^2<1$ when $-100<x<100$ so the answer is $\\boxed{199}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_USAJMO_Problems/Problem_1 | null | 1 | Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\] | [
"The answer is $\\boxed{1}$ , which works.\nTo show it is necessary, we first get $f(1)=f(1)^2$ , so $f(1)=1$ .\nThen, we get $f(2)=f(1^2 + 1^2)=f(1)^2 =1$"
] |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_15 | null | 721 | Let $\mathcal{B}$ be the set of rectangular boxes with surface area $54$ and volume $23$ . Let $r$ be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of $\mathcal{B}$ . The value of $r^2$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive... | [
"Observe that the \"worst\" possible box is one of the maximum possible length. \nBy symmetry, the height and the width are the same in this antioptimal box. (If the height and width weren't the same, the extra difference between them could be used to make the length longer.) Thus, let the width and height be of le... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_11 | null | 365 | Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$ . When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$ , the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$ , where $m$ $n$ , and $p$ ... | [
"The inequalities are equivalent to $y \\ge x/3 + 5, y \\le 10 - |x - 8|$ . We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \\Longrightarrow |x - 8| = 5 - x/3$ . This implies that one of $x - 8, 8 - x = 5 - x/3$ , from which we find that $(x,y) = \\left(\\frac 92, \\frac {13}2\... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_13 | A | 13 | Let $\mathcal{R}$ be the region in the complex plane consisting of all complex numbers $z$ that can be written as the sum of complex numbers $z_1$ and $z_2$ , where $z_1$ lies on the segment with endpoints $3$ and $4i$ , and $z_2$ has magnitude at most $1$ . What integer is closest to the area of $\mathcal{R}$
$\textbf... | [
"\nIf $z$ is a complex number and $z = a + bi$ , then the magnitude (length) of $z$ is $\\sqrt{a^2 + b^2}$ . Therefore, $z_1$ has a magnitude of 5. If $z_2$ has a magnitude of at most one, that means for each point on the segment given by $z_1$ , the bounds of the region $\\mathcal{R}$ could be at most 1 away. Alon... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | null | 170 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each numb... | [
"It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \\not\\equiv s-16 \\pmod{5},$ one term must... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_6 | null | 360 | Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits . Find the sum of the elements of $\mathcal{S}.$ | [
"Numbers of the form $0.\\overline{abc}$ can be written as $\\frac{abc}{999}$ . There are $10\\times9\\times8=720$ such numbers. Each digit will appear in each place value $\\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\\frac{45\\times72\\times111}{999}=... |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_8 | null | 25 | Let $\mathcal{T}$ be the set of ordered triples $(x,y,z)$ of nonnegative real numbers that lie in the plane $x+y+z=1.$ Let us say that $(x,y,z)$ supports $(a,b,c)$ when exactly two of the following are true: $x\ge a, y\ge b, z\ge c.$ Let $\mathcal{S}$ consist of those triples in $\mathcal{T}$ that support $\left(\frac ... | [
"This problem just requires a good diagram and strong 3D visualization.\n1999 AIME-8.png\nThe region in $(x,y,z)$ where $x \\ge \\frac{1}{2}, y \\ge \\frac{1}{3}$ is that of a little triangle on the bottom of the above diagram, of $y \\ge \\frac{1}{3}, z \\ge \\frac{1}{6}$ is the triangle at the right, and $x \\ge ... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | null | 276 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | [
"\\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \\frac{z^5-1}{z-1}\\\\ 0 &=& \\frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \\frac{(z^2-z+1)(z^5-1)}{z-1} \\end{eqnarray*}\nThus $z^5 = 1, z \\neq 1 \\Longrightarrow z = \\mathrm{cis}\\ 72, 144, 216, 288$\nor $z^2 - z + 1 = 0 \\Longrightarrow z = \\frac... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8 | null | 24 | Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | [
"For any $k\\in Z$ , we have, \\begin{align*} & \\left( \\omega^{3k} + \\omega^k + 1 \\right) \\left( \\omega^{3\\left( 7 - k \\right)} + \\omega^{\\left( 7 - k \\right)} + 1 \\right) \\\\ & = \\omega^{3 \\cdot 7} + \\omega^{2k + 7} + \\omega^{3k} + \\omega^{-2k + 3 \\cdot 7} + \\omega^7 + \\omega^k + \\omega^{3\\l... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_13 | null | 321 | Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when \[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\] is divided by 1000. | [
"\\[\\prod_{k=0}^{12} \\left(2- 2\\omega^k + \\omega^{2k}\\right) = \\prod_{k=0}^{12} \\left((1 - \\omega^k)^2 + 1\\right) = \\prod_{k=0}^{12} \\left((1 + i) - \\omega^k)((1 - i) - \\omega^k\\right)\\]\nNow, we consider the polynomial $x^{13} - 1$ whose roots are the 13th roots of unity. Taking our rewritten produc... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | null | 65 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersec... | [
" Let $O_1$ and $O_2$ be the centers of $\\omega_1$ and $\\omega_2$ , respectively. There is a homothety at $A$ sending $\\omega$ to $\\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$ , so $\\overline{OO_2}\\parallel\\overline{O_1P}$ . Similarly, $\\overline{OO_1}\\parallel\\overline{O_2P}$ , so $OO_1PO_2$ is a par... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | E | 100 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | [
"Let $O$ be the center of the circle, and $X$ be the midpoint of $\\overline{CD}$ . Let $CX=a$ and $EX=b$ . This implies that $DE = a - b$ . Since $CE = CX + EX = a + b$ , we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$ . Since $\\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\\s... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14 | null | 432 | Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $\overline{AB} = 18$ , and let $m$ denote the maximum possible length of seg... | [
"Let $x = OC$ . Since $OT, AP \\perp TC$ , it follows easily that $\\triangle APC \\sim \\triangle OTC$ . Thus $\\frac{AP}{OT} = \\frac{CA}{CO} \\Longrightarrow AP = \\frac{9(x-9)}{x}$ . By the Law of Cosines on $\\triangle BAP$ \\begin{align*}BP^2 = AB^2 + AP^2 - 2 \\cdot AB \\cdot AP \\cdot \\cos \\angle BAP \\en... |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_15 | null | 997 | Let $\overline{CH}$ be an altitude of $\triangle ABC$ . Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$ . If $AB = 1995\,$ $AC = 1994\,$ , and $BC = 1993\,$ , then $RS\,$ can be expressed as $m/n\,$ , where $m\,$ and $n\,$ are relativ... | [
"\nFrom the Pythagorean Theorem $AH^2+CH^2=1994^2$ , and $(1995-AH)^2+CH^2=1993^2$\nSubtracting those two equations yields $AH^2-(1995-AH)^2=3987$\nAfter simplification, we see that $2*1995AH-1995^2=3987$ , or $AH=\\frac{1995}{2}+\\frac{3987}{2*1995}$\nNote that $AH+BH=1995$\nTherefore we have that $BH=\\frac{1995}... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_15 | null | 14 | Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoi... | [
"Let $V = \\overline{NM} \\cap \\overline{AC}$ and $W = \\overline{NM} \\cap \\overline{BC}$ . Further more let $\\angle NMC = \\alpha$ and $\\angle MNC = 90^\\circ - \\alpha$ . Angle chasing reveals $\\angle NBC = \\angle NAC = \\alpha$ and $\\angle MBC = \\angle MAC = 90^\\circ - \\alpha$ . Additionally $NB = \\f... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_9 | null | 540 | Let $\tau(n)$ denote the number of positive integer divisors of $n$ . Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$ | [
"In order to obtain a sum of $7$ , we must have:\nSince both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square-like $3^2$ with $3$ divisors, or a fourth power like $2^4$ with $5$ divisors. We then find the smallest su... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_12 | null | 18 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are re... | [
"We have $\\angle BCE = \\angle ECD = \\angle ECA = \\tfrac 13 \\cdot 90^\\circ = 30^\\circ$ . Drop the altitude from $D$ to $CB$ and call the foot $F$\nLet $CD = 8a$ . Using angle bisector theorem on $\\triangle CDB$ , we get $CB = 15a$ . Now $CDF$ is a $30$ $60$ $90$ triangle, so $CF = 4a$ $FD = 4a\\sqrt{3}$ , an... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13 | C | 10 | Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$
$\textbf{(A) } 8 \qquad \tex... | [
"Suppose that $\\overline{BD}$ intersects $\\overline{AP}$ and $\\overline{AC}$ at $X$ and $Y,$ respectively. By Angle-Side-Angle, we conclude that $\\triangle ABX\\cong\\triangle AYX.$\nLet $AB=AY=2x.$ By the Angle Bisector Theorem, we have $AC=3x,$ or $YC=x.$\nBy alternate interior angles, we get $\\angle YAD=\\a... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | null | 717 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ | [
"Let $O$ be the circumcenter of $\\triangle ABC$ ; say $OT$ intersects $BC$ at $M$ ; draw segments $XM$ , and $YM$ . We have $MT=3\\sqrt{15}$\nSince $\\angle A=\\angle CBT=\\angle BCT$ , we have $\\cos A=\\tfrac{11}{16}$ . Notice that $AXTY$ is cyclic, so $\\angle XTY=180^{\\circ}-A$ , so $\\cos XTY=-\\cos A$ , an... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_15 | null | 58 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written i... | [
"The following is a power of a point solution to this menace of a problem:\n Let points be what they appear as in the diagram below. Note that $3HX = HY$ is not insignificant; from here, we set $XH = HE = \\frac{1}{2} EY = HL = 2$ by PoP and trivial construction. Now, $D$ is the reflection of $A$ over $H$ . Note $A... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 | null | 75 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\a... | [
"By Miquel's theorem, $P=(AEF)\\cap(BFD)\\cap(CDE)$ (intersection of circles). The law of cosines can be used to compute $DE=42$ $EF=35$ , and $FD=13$ . Toss the points on the coordinate plane; let $B=(-7, 0)$ $D=(0, 0)$ , and $C=(48, 0)$ , where we will find $\\tan^{2}\\left(\\measuredangle CDP\\right)$ with $P=(B... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | D | 110 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagon... | [
"\nDrawing it out, we see $\\angle BDC$ and $\\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\\angle ABC=70^{\\circ}$ . We can find $\\angle ECB=20^{\\circ}$ and $\\angle DBC=50^{\\circ}$ by the triangle angle sum on $\\triangle ECB$ an... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | null | 110 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagon... | [
"$\\triangle{ABC}$ is isosceles so $\\angle{CAB}=70^{\\circ}$ . Since $CB$ is a diameter, $\\angle{CDB}=\\angle{CEB}=90^{\\circ}$ . Quadrilateral $ADFE$ is cyclic since $\\angle{ADF}+\\angle{AEF}=180^{\\circ}$ . Therefore $\\angle{BFC}=\\angle{DFE}=180^{\\circ}-\\angle{CAB}=\\boxed{110}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | null | 250 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | [
"This solution refers to the Diagram section.\nLet $\\angle PAB = \\angle PBC = \\angle PCA = \\theta,$ from which $\\angle PAC = 90^\\circ-\\theta,$ and $\\angle APC = 90^\\circ.$\nMoreover, we have $\\angle PBA = \\angle PCB = 45^\\circ-\\theta,$ as shown below: Note that $\\triangle PAB \\sim \\triangle PBC$ by... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_10 | null | 468 | Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$ , circumradius $13$ , and inradius $6$ . Find $AB\cdot AC$ | [
"By Euler's formula $OI^{2}=R(R-2r)$ , we have $OI^{2}=13(13-12)=13$ . Thus, by the Pythagorean theorem, $AI^{2}=13^{2}-13=156$ . Let $AI\\cap(ABC)=M$ ; notice $\\triangle AOM$ is isosceles and $\\overline{OI}\\perp\\overline{AM}$ which is enough to imply that $I$ is the midpoint of $\\overline{AM}$ , and $M$ itsel... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13 | null | 126 | Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ | [
"First note that \\[\\angle I_1AI_2 = \\angle I_1AX + \\angle XAI_2 = \\frac{\\angle BAX}2 + \\frac{\\angle CAX}2 = \\frac{\\angle A}2\\] is a constant not depending on $X$ , so by $[AI_1I_2] = \\tfrac12(AI_1)(AI_2)\\sin\\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$ . Let $a = BC$ $b = AC$ $c = AB$ , and ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_25 | E | 15 | 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_... | [
"For all nonnegative integers $n$ , let $\\angle C_nA_nB_n=x_n$ $\\angle A_nB_nC_n=y_n$ , and $\\angle B_nC_nA_n=z_n$\nNote 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 ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_22 | B | 26 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | [
"2002 12B AMC-20.png\nLet $OM = x$ $ON = y$ . By the Pythagorean Theorem on $\\triangle XON, MOY$ respectively, \\begin{align*} (2x)^2 + y^2 &= 19^2\\\\ x^2 + (2y)^2 &= 22^2\\end{align*}\nSumming these gives $5x^2 + 5y^2 = 845 \\Longrightarrow x^2 + y^2 = 169$\nBy the Pythagorean Theorem again, we have\n\\[(2x)^2 +... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_22 | null | 26 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | [
"Let $XO=x$ and $YO=y.$ Then, $XY=\\sqrt{x^2+y^2}.$\nSince $XN=19$ and $YM=22,$ \\[XN^2=19^2=x^2+(\\dfrac{y}{2})^2)=\\dfrac{x^2}{4}+y^2\\] \\[YM^2=22^2=(\\dfrac{x}{2})^2+y^2=\\dfrac{x^2}{4}+y^2.\\]\nAdding these up: \\[19^2+22^2=\\dfrac{4x^2+y^2}{4}+\\dfrac{x^2+4y^2}{4}\\] \\[845=\\dfrac{5x^2+5y^2}{4}\\] \\[3380=5x... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_20 | B | 26 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | [
"2002 12B AMC-20.png\nLet $OM = x$ $ON = y$ . By the Pythagorean Theorem on $\\triangle XON, MOY$ respectively, \\begin{align*} (2x)^2 + y^2 &= 19^2\\\\ x^2 + (2y)^2 &= 22^2\\end{align*}\nSumming these gives $5x^2 + 5y^2 = 845 \\Longrightarrow x^2 + y^2 = 169$\nBy the Pythagorean Theorem again, we have\n\\[(2x)^2 +... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_20 | null | 26 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | [
"Let $XO=x$ and $YO=y.$ Then, $XY=\\sqrt{x^2+y^2}.$\nSince $XN=19$ and $YM=22,$ \\[XN^2=19^2=x^2+(\\dfrac{y}{2})^2)=\\dfrac{x^2}{4}+y^2\\] \\[YM^2=22^2=(\\dfrac{x}{2})^2+y^2=\\dfrac{x^2}{4}+y^2.\\]\nAdding these up: \\[19^2+22^2=\\dfrac{4x^2+y^2}{4}+\\dfrac{x^2+4y^2}{4}\\] \\[845=\\dfrac{5x^2+5y^2}{4}\\] \\[3380=5x... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7 | null | 725 | Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ ... | [
"Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\\triangle PQR$ , where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$ ), $s = \\frac{PQ + QR + RP}{2} = 180$ is the semiperimeter , and $A = \\frac 12 bh = 5400$ is the area, we find $r_{1} = \\frac As = 30$ . Or,... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_23 | D | 78 | Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$ | [
"When $m=1$ $a_{n+1}=1+a_n+n$ . Hence, \\[a_{2}=1+a_1+1\\] \\[a_{3}=1+a_2+2\\] \\[a_{4}=1+a_3+3\\] \\[\\dots\\] \\[a_{12}=1+a_{11}+11\\] Adding these equations up, we have that $a_{12}=12+(1+2+3+...+11)=\\boxed{78}$",
"Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$ $a_{m+1}=a_m+a_{1}+m$\nSince $a_1 = 1$ $a_{m+1}=a_... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_24 | C | 1,341 | Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$ , and in general,
\[a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{... | [
"First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$ (decreasing exponential function), and $g(x) = x^k$ for $k > 0$ (increasing power function for positive $x$ ). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_17 | null | 1 | Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ . What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac... | [
"Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\\frac a{1-r_1}$ and $\\frac a{1-r_2}$\nHence we have $\\frac a{1-r_1} = r_1$ and $\\frac a{1-r_2} = r_2$ .\nThis can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$\nAs we are given that $r_1$ and $r_2$ are dist... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_8 | null | 251 | Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer. | [
"By the product-to-sum identities , we have that $2\\cos a \\sin b = \\sin (a+b) - \\sin (a-b)$ . Therefore, this reduces to a telescope series: \\begin{align*} \\sum_{k=1}^{n} 2\\cos(k^2a)\\sin(ka) &= \\sum_{k=1}^{n} [\\sin(k(k+1)a) - \\sin((k-1)ka)]\\\\ &= -\\sin(0) + \\sin(2a)- \\sin(2a) + \\sin(6a) - \\cdots - ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_25 | B | 315 | Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$ | [
"We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$\nThus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polyn... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_21 | B | 315 | Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$ | [
"We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$\nThus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polyn... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_14 | null | 896 | Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$ | [
"The first condition implies\n\\[a^{128} = \\log_a\\log_a 2 + \\log_a 24 - 128\\]\n\\[128+a^{128} = \\log_a\\log_a 2^{24}\\]\n\\[a^{a^{128}a^{a^{128}}} = 2^{24}\\]\n\\[\\left(a^{a^{128}}\\right)^{\\left(a^{a^{128}}\\right)} = 2^{24} = 8^8\\]\nSo $a^{a^{128}} = 8$\nPutting each side to the power of $128$\n\\[\\left(... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14 | B | 671 | Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$ | [
"We want to try make $a+b$ $b+c$ $c+d$ , and $d+e$ as close as possible so that $M$ , the maximum of these, is smallest.\nNotice that $2010=670+670+670$ . In order to express $2010$ as a sum of $5$ numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers a... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14 | null | 671 | Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$ | [
"Since $a + b \\le M$ $d + e \\le M$ , and $c < b + c \\le M$ , we have that $2010 = a + b + c + d + e < 3M$ . Hence, $M > 670$ , or $M \\ge 671$\nFor the values $(a,b,c,d,e) = (669,1,670,1,669)$ $M = 671$ , so the smallest possible value of $M$ is $\\boxed{671}$ . The answer is (B)."
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_16 | D | 18 | Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$ | [
"From $|a-b|=2$ we get that $a=b\\pm 2$\nSimilarly, $b=c\\pm3$ and $c=d\\pm4$\nSubstitution gives $a=d\\pm 4\\pm 3\\pm 2$ . This gives $|a-d|=|\\pm 4\\pm 3\\pm 2|$ . There are $2^3=8$ possibilities for the value of $\\pm 4\\pm 3\\pm2$\n$4+3+2=9$\n$4+3-2=5$\n$4-3+2=3$\n$-4+3+2=1$\n$4-3-2=-1$\n$-4+3-2=-3$\n$-4-3+2=-5... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | null | 994 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | [
"We draw the altitude $h$ to $c$ , to get two right triangles\nThen $\\cot{\\alpha}+\\cot{\\beta}=\\frac{c}{h}$ , from the definition of the cotangent\nLet $K$ be the area of $\\triangle ABC.$ Then $h=\\frac{2K}{c}$ , so $\\cot{\\alpha}+\\cot{\\beta}=\\frac{c^2}{2K}$\nBy identical logic, we can find similar express... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_18 | C | 16 | Let $a$ $b$ , and $c$ be digits with $a\ne 0$ . The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qq... | [
"Let $k$ be the lesser of the two integers. Then the squares of the integers are $k^2$ and $k^2+2k+1$ , and the distance between them is $2k+1$ . Let this be equivalent to $3d$ , so that the one-third of the distance between the squares is equivalent to $d$ . The numbers $abc$ and $acb$ are one-third and two-thirds... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_24 | E | 253 | Let $a$ $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$
What is $a$
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$ | [
"Add the two equations.\n$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$\nNow, this can be rearranged and factored.\n$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$\n$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$\n$a$ $b$ , and $c$ are all integers, so the three terms on the left side of the equation must al... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_21 | E | 253 | Let $a$ $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$
What is $a$
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$ | [
"Add the two equations.\n$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$\nNow, this can be rearranged and factored.\n$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$\n$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$\n$a$ $b$ , and $c$ are all integers, so the three terms on the left side of the equation must al... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_20 | B | 1 | Let $a$ $b$ , and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$ | [
"Rearranging, we get $a+8c=7b+4$ and $8a-c=7-4b$\nSquaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.\nAdding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$ , so $a^2-b^2+c^2=\\boxed{1}$",
"The easiest way is to assume a value for $a$ and then solve the syste... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_20 | null | 1 | Let $a$ $b$ , and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$ | [
"Notice that the coefficients of $a$ and $c$ are pretty similar (15s for reading and noticing), so let $b=0$ gives $a+8c=4$ , and $8a-c=7$ (10s writing). Since the desired quantity simplifies to $a^2+c^2$ , the $ac$ term of the quadratics after squaring gets canceled by adding up the squares of the two equations be... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_14 | D | 16.5 | Let $a$ $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$
$\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17$ | [
"Expanding the equation and combining like terms results in $2x^2-(a+2b+c)x+(ab+bc)=0$ . By Vieta's formula the sum of the roots is $\\dfrac{-[-(a+2b+c)]}{2}=\\dfrac{a+2b+c}{2}$ . To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$ . So let ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_4 | E | 0.8 | Let $a$ and $b$ be distinct real numbers for which \[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]
Find $\frac{a}{b}$
$\text{(A) }0.4 \qquad \text{(B) }0.5 \qquad \text{(C) }0.6 \qquad \text{(D) }0.7 \qquad \text{(E) }0.8$ | [
"For sake of speed, WLOG, let $b=1$ . This means that the ratio $\\frac{a}{b}$ will simply be $a$ because $\\frac{a}{b}=\\frac{a}{1}=a.$ Solving for $a$ with some very simple algebra gives us a quadratic which is $5a^2 -9a +4=0.$ Factoring the quadratic gives us $(5a-4)(a-1)=0$ . Therefore, $a=1$ or $a=\\frac{4}{5}... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_8 | null | 36 | Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$ . The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | [
"Let us call the quantity $\\frac{a^3b^3+1}{a^3+b^3}$ as $N$ for convenience. Knowing that $a$ and $b$ are positive integers, we can legitimately rearrange the given inequality so that $a$ is by itself, which makes it easier to determine the pairs of $(a, b)$ that work. Doing so, we have \\[\\frac{ab+1}{a+b} < \\fr... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_14 | null | 7 | Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac... | [
"Notice that the given equation implies\nWe have $2by \\ge y^2$ , so $2ax \\le a^2 \\implies x \\le \\frac {a}{2}$\nThen, notice $b^2 + x^2 = a^2 + y^2 \\ge a^2$ , so $b^2 \\ge \\frac {3}{4}a^2 \\implies \\rho^2 \\le \\frac {4}{3}$\nThe solution $(a, b, x, y) = \\left(1, \\frac {\\sqrt {3}}{2}, \\frac {1}{2}, 0\\ri... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_21 | C | 2 | Let $a$ and $b$ be real numbers greater than $1$ for which there exists a positive real number $c,$ different from $1$ , such that
\[2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.\]
Find the largest possible value of $\log_a b.$
$\text{(A) }\sqrt{2} \qquad \text{(B) }\sqrt{3} \qquad \text{(C) }2 \qquad \text{(D) }\sqrt{6} \... | [
"We may rewrite the given equation as \\[2(\\frac {\\log c}{\\log a} + \\frac {\\log c}{\\log b}) = \\frac {9\\log c}{\\log a + \\log b}\\] Since $c \\neq 1$ , we have $\\log c \\neq 0$ , so we may divide by $\\log c$ on both sides. After making the substitutions $x = \\log a$ and $y = \\log b$ , our equation becom... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | C | 3 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | [
"Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\\frac{73}{3}$ , the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$ . Looking at the answer choices, the only multiple of $3$ is $\\boxed{3}$",
... |
https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_28 | D | 1 | Let $a, b, c, d$ be real numbers. Suppose that all the roots of $z^4+az^3+bz^2+cz+d=0$ are complex numbers lying on a circle
in the complex plane centered at $0+0i$ and having radius $1$ . The sum of the reciprocals of the roots is necessarily
$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textb... | [
"Let's denote the roots of the polynomial as $z_1, z_2, z_3, z_4$ . We know that the magnitudes of these 4 roots are 1 as given in the problem statement. Therefore, we have $z_1 \\overline{z_1}, z_2 \\overline{z_2}, z_3 \\overline{z_3}, z_4 \\overline{z_4} = 1$ . We want to find $\\frac{1}{z_1} + \\frac{1}{z_2} + \... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22 | D | 13 | Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{... | [
"The GCD information tells us that $24$ divides $a$ , both $24$ and $36$ divide $b$ , both $36$ and $54$ divide $c$ , and $54$ divides $d$ . Note that we have the prime factorizations: \\begin{align*} 24 &= 2^3\\cdot 3,\\\\ 36 &= 2^2\\cdot 3^2,\\\\ 54 &= 2\\cdot 3^3. \\end{align*}\nHence we have \\begin{align*} a &... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | null | 145 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | [
"From the fourth equation we get $d=\\frac{30}{abc}.$ Substitute this into the third equation and you get $abc + \\frac{30(ab + bc + ca)}{abc} = abc - \\frac{120}{abc} = 14$ . Hence $(abc)^2 - 14(abc)-120 = 0$ . Solving, we get $abc = -6$ or $abc = 20$ . From the first and second equation, we get $ab + bc + ca = ab... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_12 | null | 23 | Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that \[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\] Find the least possible value of $a+b.$ | [
"Denote $P = \\left( x , y \\right)$\nBecause $\\frac{x^2}{a^2}+\\frac{y^2}{a^2-16} = 1$ $P$ is on an ellipse whose center is $\\left( 0 , 0 \\right)$ and foci are $\\left( - 4 , 0 \\right)$ and $\\left( 4 , 0 \\right)$\nHence, the sum of distance from $P$ to $\\left( - 4 , 0 \\right)$ and $\\left( 4 , 0 \\right)$ ... |
https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_1 | null | 16 | Let $a,b,c,d$ be real numbers such that $b-d \ge 5$ and all zeros $x_1, x_2, x_3,$ and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take. | [
"Using the hint we turn the equation into $\\prod_{k=1} ^4 (x_k-i)(x_k+i) \\implies P(i)P(-i) \\implies (b-d-1)^2 + (a-c)^2 \\implies \\boxed{16}$ . This minimum is achieved when all the $x_i$ are equal to $1$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_20 | C | 34 | Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30 \qquad \mathrm{(B)}\ 32 \qquad \mathrm{(C)}\ 34 \qquad \mathrm{(D)}\ 40 \qquad \mathrm{(E)}\ 50$ | [
"The sum of the set is $-7-5-3-2+2+4+6+13=8$ , so if we could have the sum in each set of parenthesis be $4$ then the minimum value would be $2(4^2)=32$ . Considering the set of four terms containing $13$ , this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be $1... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 | null | 289 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | [
"To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} \\geq \\frac{1}{7\\cdot8\\cdot9}.$\nIf we minimize the numerator, then $a \\cdot b \\cdot c - d \\cdot e \\cdot f = 1.$ Note that $a \\cd... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_18 | E | 3 | Let $a_1 , a_2 , ...$ be a sequence for which $a_1=2$ $a_2=3$ , and $a_n=\frac{a_{n-1}}{a_{n-2}}$ for each positive integer $n \ge 3$ . What is $a_{2006}$
$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3$ | [
"Looking at the first few terms of the sequence:\n$a_1=2 , a_2=3 , a_3=\\frac{3}{2}, a_4=\\frac{1}{2} , a_5=\\frac{1}{3} , a_6=\\frac{2}{3} , a_7=2 , a_8=3 , ....$\nClearly, the sequence repeats every 6 terms.\nSince $2006 \\equiv 2\\bmod{6}$\n$a_{2006} = a_2 = \\boxed{3}$",
"$a_n = \\frac{a_{n-1}}{a_{n-2}} = \\f... |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_9 | C | 10,000 | Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be arithmetic progressions such that $a_1 = 25, b_1 = 75$ , and $a_{100} + b_{100} = 100$ .
Find the sum of the first hundred terms of the progression $a_1 + b_1, a_2 + b_2, \ldots$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 10,000 \qquad \textbf{(D)... | [
"Notice that $a_{100}$ and $b_{100}$ are $25+99k_1$ and $75+99k_2$ , respectively. Therefore $k_2 = -k_1$ . Now notice that $a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100$ . The sum of the first $100$ terms is $100\\cdot100 = \\boxed{10,000}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_16 | E | 4 | Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | [
"Verify that $a^3 \\equiv a \\pmod{6}$ manually for all $a\\in \\mathbb{Z}/6\\mathbb{Z}$ . We check: $0^3 \\equiv 0 \\pmod{6}$ $1^3 \\equiv 1 \\pmod{6}$ $2^3 \\equiv 8 \\equiv 2 \\pmod{6}$ $3^3 \\equiv 27 \\equiv 3 \\pmod{6}$ $4^3 \\equiv 64 \\equiv 4 \\pmod{6}$ , and $5^3 \\equiv 125 \\equiv 5 \\pmod{6}$ . We conc... |
https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_5 | null | 1,700 | Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$ , define $b_m=\min\{n: a_n \ge m\}$ , that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$ . If $a_{19}=85$ , determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$ | [
"We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the $i$ th column we fill the first $a_i$ cells with one dot each. Then the $19$ th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row $j$... |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_6 | null | 35 | Let $a_n=6^{n}+8^{n}$ . Determine the remainder upon dividing $a_ {83}$ by $49$ | [
"Firstly, we try to find a relationship between the numbers we're provided with and $49$ . We notice that $49=7^2$ , and both $6$ and $8$ are greater or less than $7$ by $1$\nThus, expressing the numbers in terms of $7$ , we get $a_{83} = (7-1)^{83}+(7+1)^{83}$\nApplying the Binomial Theorem , half of our terms can... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_2 | null | 112 | Let $a_{0} = 2$ $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$ $a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$ ) is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$ | [
"When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.\nAfter computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$\n$a_{0} = 2$ $a_{1} = 5$ $a_{2} = 8$ $a_{3} = 5$ $a_{4} = 6$ ... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | null | 45 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | [
"Writing out the recursive statement for $a_n, a_{n-1}, \\dots, a_{10}$ and summing them gives \\[a_n+\\dots+a_{10}=100(a_{n-1}+\\dots+a_{10})+n+\\dots+10\\] Which simplifies to \\[a_n=99(a_{n-1}+\\dots+a_{10})+\\frac{1}{2}(n+10)(n-9)\\] Therefore, $a_n$ is divisible by 99 if and only if $\\frac{1}{2}(n+10)(n-9)$ i... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_13 | E | 1 | Let $c = \frac{2\pi}{11}.$ What is the value of \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{1... | [
"Plugging in $c$ , we get \\[\\frac{\\sin 3c \\cdot \\sin 6c \\cdot \\sin 9c \\cdot \\sin 12c \\cdot \\sin 15c}{\\sin c \\cdot \\sin 2c \\cdot \\sin 3c \\cdot \\sin 4c \\cdot \\sin 5c}=\\frac{\\sin \\frac{6\\pi}{11} \\cdot \\sin \\frac{12\\pi}{11} \\cdot \\sin \\frac{18\\pi}{11} \\cdot \\sin \\frac{24\\pi}{11} \\cd... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_22 | A | 4.5 | Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ $z_2$ $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible v... | [
"Because $c$ is real, $z_2 = \\bar z_1$ .\nWe have \\begin{align*} 10 & = z_1 z_2 \\\\ & = z_1 \\bar z_1 \\\\ & = |z_1|^2 , \\end{align*} where the first equality follows from Vieta's formula.\nThus, $|z_1| = \\sqrt{10}$\nWe have \\begin{align*} c & = z_1 + z_2 \\\\ & = z_1 + \\bar z_1 \\\\ & = 2 {\\rm Re}(z_1), \\... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_22 | null | 4.5 | Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ $z_2$ $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible v... | [
"Since $c$ , which is the sum of roots $z_1$ and $z_2$ , is real, $z_1=\\overline{z_2}$\nLet $z_1=a+bi$ . Then $z_2=a-bi$ . Note that the product of the roots is $10$ by Vieta's, so $z_1z_2=(a+bi)(a-bi)=a^2+b^2=10$\nThus, $\\frac{1}{z_1}=\\frac{1}{a-bi}\\cdot\\frac{a+bi}{a+bi}=\\frac{a+bi}{a^2+b^2}=\\frac{a+bi}{10}... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_5 | B | 0 | Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$ | [
"Using factoring:\n$2x^{2}+3x-5=0$\n$(2x+5)(x-1)=0$\n$x = -\\frac{5}{2}$ or $x=1$\nSo $d$ and $e$ are $-\\frac{5}{2}$ and $1$\nTherefore the answer is $\\left(-\\frac{5}{2}-1\\right)(1-1)=\\left(-\\frac{7}{2}\\right)(0)=\\boxed{0}$",
"We can use the sum and product of a quadratic (a.k.a Vieta):\n$(d-1)(e-1)=de-(d... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_25 | E | 9 | Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N... | [
"We consider the prime factorization of $n:$ \\[n=\\prod_{i=1}^{k}p_i^{e_i}.\\] By the Multiplication Principle, we have \\[d(n)=\\prod_{i=1}^{k}(e_i+1).\\] Now, we rewrite $f(n)$ as \\[f(n)=\\frac{d(n)}{\\sqrt [3]n}=\\frac{\\prod_{i=1}^{k}(e_i+1)}{\\prod_{i=1}^{k}p_i^{e_i/3}}=\\prod_{i=1}^{k}\\frac{e_i+1}{p_i^{{e_... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_24 | B | 19 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | [
"Let $y = \\frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -\\frac{1}{3}, \\frac{2}{9}$ . These sum up to $\\boxed{19}$",
"This is quite trivially solved, as $3x = \\dfrac{9x}{3}$ , so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$ $81x^2+9x-6 = 0$ has sol... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_15 | B | 19 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | [
"Let $y = \\frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -\\frac{1}{3}, \\frac{2}{9}$ . These sum up to $\\boxed{19}$",
"This is quite trivially solved, as $3x = \\dfrac{9x}{3}$ , so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$ $81x^2+9x-6 = 0$ has sol... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_9 | C | 52 | Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$ | [
"Letting $x = 500$ and $y = \\dfrac65$ in the given equation, we get $f(500\\cdot\\frac65) = \\frac3{\\frac65} = \\frac52$ , or $f(600) = \\boxed{52}$",
"The only function that satisfies the given condition is $y = \\frac{k}{x}$ , for some constant $k$ . Thus, the answer is $\\frac{500 \\cdot 3}{600} = \\boxed{52... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_20 | B | 2,000 | Let $f$ be a real-valued function such that
\[f(x) + 2f(\frac{2002}{x}) = 3x\]
for all $x>0.$ Find $f(2).$
$\text{(A) }1000 \qquad \text{(B) }2000 \qquad \text{(C) }3000 \qquad \text{(D) }4000 \qquad \text{(E) }6000$ | [
"Setting $x = 2$ gives $f(2) + 2f(1001) = 6$ .\nSetting $x = 1001$ gives $2f(2) + f(1001) = 3003$\nAdding these 2 equations and dividing by 3 gives $f(2) + f(1001) = \\frac{6+3003}{3} = 1003$\nSubtracting these 2 equations gives $f(2) - f(1001) = 2997$\nTherefore, $f(2) = \\frac{1003+2997}{2} = \\boxed{2000}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22 | B | 96 | 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.\nThen, 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.\nNow consider $f(p^k)$ ,... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11 | E | 2 | Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$ | [
"Converting both summands to exponential form, \\begin{align*} -1 + i\\sqrt{3} &= 2e^{\\frac{2\\pi i}{3}}, \\\\ -1 - i\\sqrt{3} &= 2e^{-\\frac{2\\pi i}{3}} = 2e^{\\frac{4\\pi i}{3}}. \\end{align*} Notice that both are scaled copies of the third roots of unity.\nWhen we replace the summands with their exponential fo... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15 | null | 258 | Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n... | [
"Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$ . Note that this formula also returns the correct value when $n=(k+1)^2$ , but not when $n=k^... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_13 | null | 400 | Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$ | [
"This is a pretty easy problem just to bash. Since the max number we can get is $7$ , we just need to test $n$ values for $1.5,2.5,3.5,4.5,5.5$ and $6.5$ . Then just do how many numbers there are times $\\frac{1}{\\lfloor n \\rfloor}$ , which should be $5+17+37+65+101+145+30 = \\boxed{400}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4 | null | 2,047 | Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd. | [
"First of all, note that $f(n)$ $\\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \\le n$ . We let $f(0) = 1$ for convenience.\nFrom here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \\in \\mathbb... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_17 | B | 16,089 | 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 $h_{2011}(1)$
$\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099$ | [
"$g(x)=\\log_{10}\\left(\\frac{x}{10}\\right)=\\log_{10}\\left({x}\\right) - 1$\n$h_{1}(x)=g(f(x))\\text{ = }g(10^{10x}=\\log_{10}\\left({10^{10x}}\\right){ - 1 = 10x - 1}$\nProof by induction that $h_{n}(x)\\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$\nFor $n=1$ $h_{1}(x)=10x - 1$\nAssume $h_{n}(x)=10^n x ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_24 | B | 12 | Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ ... | [
"The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.\nWe note that since all of the $\\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_17 | E | 2 | Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is
$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$
Solution | [
"By the Pythagorean identity we can rewrite the given expression as follows. \\[\\sqrt{\\sin^4{x} + 4 \\cos^2{x}} - \\sqrt{\\cos^4{x} + 4 \\sin^2{x}} = \\sqrt{\\sin^4{x} + 4(1 - \\sin^2{x})} - \\sqrt{\\cos^4{x} + 4(1 - \\cos^2{x})}\\]\nExpanding each bracket gives \\[\\sqrt{\\sin^4{x} - 4\\sin^2{x} + 4} - \\sqrt{\\... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_25 | E | 25 | Let $f(x) = x^2 + 6x + 1$ , and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that \[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\] The area of $R$ is closest to
$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$ | [
"The first condition gives us that \\[x^2 + 6x + 1 + y^2 + 6y + 1 \\le 0 \\Longrightarrow (x+3)^2 + (y+3)^2 \\le 16\\]\nwhich is a circle centered at $(-3,-3)$ with radius $4$ . The second condition gives us that\n\\[x^2 + 6x + 1 - y^2 - 6y - 1 \\le 0 \\Longrightarrow (x^2 - y^2) + 6(x-y) \\le 0 \\Longrightarrow (x... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_8 | A | 0 | Let $f(x) = x^{2}(1-x)^{2}$ . What is the value of the sum \[f \left(\frac{1}{2019} \right)-f \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]
$\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}... | [
"First, note that $f(x) = f(1-x)$ . We can see this since \\[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\\left(1-\\left(1-x\\right)\\right)^{2} = f(1-x)\\] Using this result, we regroup the terms accordingly: \\[\\left( f \\left(\\frac{1}{2019} \\right) - f \\left(\\frac{2018}{2019} \\right) \\right) + \\left( f \\... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14 | null | 676 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | [
"If the leading term of $f(x)$ is $ax^m$ , then the leading term of $f(x)f(2x^2) = ax^m \\cdot a(2x^2)^m = 2^ma^2x^{3m}$ , and the leading term of $f(2x^3 + x) = 2^max^{3m}$ . Hence $2^ma^2 = 2^ma$ , and $a = 1$ . Because $f(0) = 1$ , the product of all the roots\nof $f(x)$ is $\\pm 1$ . If $f(\\lambda) = 0$ , then... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | null | 72 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | [
"Let $f(x)$ $ax^3+bx^2+cx+d$ .\nSince $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up.\nBy drawing a coordinate axis, and two lines representing $12$ and $-12$ , it is easy to see that $f(1)=f(5)=f(6)$ , and $f(2)=f(3)=f(7)$ ; otherwise more... |
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