id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
5,501 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_1 | 2 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n... | Let $\angle{BAC}$ be $x$ in degrees. $\angle{ADE}=x$ .
By Exterior Angle Theorem on triangle $AED$ $\angle{BED}=2x$ .
By Exterior Angle Theorem on triangle $ADB$ $\angle{BDC}=3x$ .
This tells us $\angle{BCA}=\angle{ABC}=3x$ and $3x+3x+x=180$ .
Thus $x=\frac{180}{7}$ and we want $\angle{ABC}=3x=\frac{540}{7}$ to get an ... | 547 |
5,502 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_1 | 3 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n... | graph soon
We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better.
\begin{align*} \angle A &= y \\ \angle B &= x+z \\ \angle C &= \frac{180-x}{2} \\ \end{align*}
Then, using triangle sum of angles theorem, we find that
\begin{align*} ... | 547 |
5,503 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_3 | 1 | A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such... | From the given information, $121a+11b+c=512+64b+8c+a \implies 120a=512+53b+7c$ . Since $a$ $b$ , and $c$ have to be positive, $a \geq 5$ . Since we need to minimize the value of $n$ , we want to minimize $a$ , so we have $a = 5$ . Then we know $88=53b+7c$ , and we can see the only solution is $b=1$ $c=5$ . Finally, $51... | 621 |
5,504 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_5 | 5 | Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. | First count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is $1$ such permutation where all the cards appear in order: $123456.$ There are $5$ such permutations where two adjacent cards are interchanged, as in $124356.$ The other such per... | 52 |
5,505 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_5 | 6 | Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. | More generally, suppose there are $n \geq 4$ cards numbered $1, 2, 3, \dots, n$ arranged in ascending order. If any one of the $n$ cards is removed and placed in one of the $n$ positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ... | 52 |
5,506 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_6 | 1 | A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$... | [asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X... | 173 |
5,507 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_8 | 1 | A bug walks all day and sleeps all night. On the first day, it starts at point $O$ , faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the po... | [asy] size(8cm); pair O, A, B, C, D, F, G, H, I, P, X; O = (0, 0); A = (5, 0); X = (8, 0); P = (5, 5 / sqrt(3)); B = rotate(-120, A) * ((O + A) / 2); C = rotate(-120, B) * ((A + B) / 2); D = rotate(-120, C) * ((B + C) / 2); F = rotate(-120, D) * ((C + D) / 2); G = rotate(-120, F) * ((D + F) / 2); H = rotate(-120, G) * ... | 103 |
5,508 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_8 | 2 | A bug walks all day and sleeps all night. On the first day, it starts at point $O$ , faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the po... | We place the ant at the origin of the complex plane with its first move being in the positive real direction. Then the ant's journey can be represented as the infinite series \[5\left(1 + \frac{e^{\frac{i\pi}{3}}}{2} + \left(\frac{e^{\frac{i\pi}{3}}}{2}\right)^2 + \cdots\right)\] Using the formula for an infinite geom... | 103 |
5,509 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_8 | 3 | A bug walks all day and sleeps all night. On the first day, it starts at point $O$ , faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the po... | The ant goes in the opposite direction every $3$ moves, going $(1/2)^3=1/8$ the distance backwards. Using geometric series, he travels $1-1/8+1/64-1/512...=(7/8)(1+1/64+1/4096...)=(7/8)(64/63)=8/9$ the distance of the first three moves over infinity moves. Now, we use coordinates meaning $(5+5/4-5/8, 0+5\sqrt3/4+5\sqrt... | 103 |
5,510 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_8 | 4 | A bug walks all day and sleeps all night. On the first day, it starts at point $O$ , faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the po... | The bug goes forward and backward in three directions: straight east, northeast, and northwest. It travels $5-\frac{5}{8}+\frac{5}{64}-\cdots=\frac{40}{9}$ units east. Thus, it goes northeast $\frac{\frac{40}{9}}{2}=\frac{20}{9}$ units northeast and $\frac{10}{9}$ units northwest. Now, the bug travels a total of $\frac... | 103 |
5,511 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_8 | 5 | A bug walks all day and sleeps all night. On the first day, it starts at point $O$ , faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the po... | The bug's bearings on each traversal are $0^\circ, 60^\circ, 120^\circ,$ and so on; in general, the $n-$ th traversal has length $5\cdot (1/2)^{n-1}$ and bearing $60(n-1).$ This means that the $x$ and $y$ displacements for the $n-$ th traversal are
\[(\Delta x_n, \Delta y_n)=(5\cdot (1/2)^{n-1}\cos (60(n-1))^\circ,5\cd... | 103 |
5,512 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_10 | 2 | Let $m$ and $n$ be positive integers satisfying the conditions
$\quad\bullet\ \gcd(m+n,210)=1,$
$\quad\bullet\ m^m$ is a multiple of $n^n,$ and
$\quad\bullet\ m$ is not a multiple of $n.$
Find the least possible value of $m+n.$ | Assume for the sake of contradiction that $n$ is a multiple of a single digit prime number, then $m$ must also be a multiple of that single digit prime number to accommodate for $n^n | m^m$ . However that means that $m+n$ is divisible by that single digit prime number, which violates $\gcd(m+n,210) = 1$ , so contradict... | 407 |
5,513 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_11 | 2 | For integers $a,b,c$ and $d,$ let $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d.$ Find the number of ordered triples $(a,b,c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2))=g(f(4))=0.$ | Define $h(x)=x^2+cx$ . Since $g(f(2))=g(f(4))=0$ , we know $h(f(2))=h(f(4))=-d$ . Plugging in $f(x)$ into $h(x)$ , we get $h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)$ . Setting $h(f(2))=h(f(4))$ \[16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc\] . Simplifying and cancelling terms, \[240+... | 510 |
5,514 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_12 | 1 | Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$ | As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$ Lifting the Exponent shows that \[3=v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\] so thus, $3^2$ divides $n$ . It also shows that \[7=v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\] so thus, $7^5$ divides $n$
Now, setting $n = 4c$ (necessitated... | 270 |
5,515 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_12 | 2 | Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$ | Note that for all $n$ $149^n - 2^n$ is divisible by $149-2 = 147$ by difference of $n$ th powers. That is $3\cdot7^2$ , so now we can clearly see that the smallest $n$ to make the expression divisible by $3^3$ is just $3^2$ . Similarly, we can reason that the smallest $n$ to make the expression divisible by $7^7$ is ju... | 270 |
5,516 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_12 | 3 | Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$ | As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$ . For divisibility by $3^3$ , notice that $v_3(149^3 - 2^3) = 2$ as $149^3 - 2^3 =$ $(147)(149^2 + 2\cdot149 + 2^2)$ , and upon checking mods, $149^2 + 2\cdot149 + 2^2$ is divisible by $3$ but not $9$ . In addition, $149^9 - 2^9$ is divisible by ... | 270 |
5,517 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_13 | 8 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | Trig values we use here:
$\cos A = \frac{9}{16}$
$\cos \frac{A}{2} = \frac{5}{4\sqrt2}$
$\sin \frac{A}{2} = \frac{\sqrt7}{4\sqrt2}$
$\cos \frac{B}{2} = \frac{3}{4}$
$\cos \frac{C}{2} = \frac{\sqrt7}{2\sqrt2}$
First let the incenter be $I$ . Let $M$ be the midpoint of minor arc $BC$ on $(ABC)$ and let $K$ be the foot of... | 36 |
5,518 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_13 | 9 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | Firstly, it is easy to find $BD=2,CD=3$ with angle bisector theorem.
Using LOC and some trig formulas we get all those values: $cos\angle{B}=\frac{1}{8},sin\angle{B}=\frac{3\sqrt{7}}{8},cos\angle{\frac{B}{2}}=\frac{3}{4},cos\angle{\frac{{ACB}}{2}}=\frac{\sqrt{14}}{4}$ Now we find the coordinates of points $A,E,F$ and w... | 36 |
5,519 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_15 | 3 | 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... | Note that $\overline{BC}$ bisects $\overline{OO'}$ . Using the same method from Solution 3 we find $R = \frac{21}{2\sqrt{5}}$ . Let the midpoint of $\overline{BC}$ be $N$ , then by the Pythagorean Theorem we have $BN^2 = OB^2 - ON^2 = R^2 - \frac{3}{2}^2$ , so $BN = \frac{3\sqrt{11}}{\sqrt{5}}$ . Since $h_a = 5$ we hav... | 58 |
5,520 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_1 | 1 | Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$ | In this problem, we want to find the number of ordered pairs $(m, n)$ such that $m^2n = 20^{20}$ . Let $x = m^2$ . Therefore, we want two numbers, $x$ and $n$ , such that their product is $20^{20}$ and $x$ is a perfect square. Note that there is exactly one valid $n$ for a unique $x$ , which is $\tfrac{20^{20}}{x}$ . T... | 231 |
5,521 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_4 | 2 | Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ $B(0,12)$ $C(16,0)$ $A'(24,18)$ $B'(36,18)$ $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ | [asy] /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point st... | 108 |
5,522 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_4 | 3 | Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ $B(0,12)$ $C(16,0)$ $A'(24,18)$ $B'(36,18)$ $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ | For the above reasons, the transformation is simply a $90^\circ$ rotation. Proceed with complex numbers on the points $C$ and $C'$ . Let $(x, y)$ be the origin. Thus, $C \rightarrow (16-x)+(-y)i$ and $C' \rightarrow (24-x)+(2-y)i$ . The transformation from $C'$ to $C$ is a multiplication of $i$ , which yields $(16-x)+(... | 108 |
5,523 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_4 | 4 | Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ $B(0,12)$ $C(16,0)$ $A'(24,18)$ $B'(36,18)$ $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ | We know that the rotation point $P$ has to be equidistant from both $A$ and $A'$ so it has to lie on the line that is on the midpoint of the segment $AA'$ and also the line has to be perpendicular to $AA'$ . Solving, we get the line is $y=\frac{-4}{3}x+25$ . Doing the same for $B$ and $B'$ , we get that $y=-6x+123$ . S... | 108 |
5,524 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_6 | 2 | Define a sequence recursively by $t_1 = 20$ $t_2 = 21$ , and \[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\] for all $n \ge 3$ . Then $t_{2020}$ can be expressed as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$ , so now we are able to determine the numerical value of $t_3$ using this information: \[t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac... | 626 |
5,525 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_7 | 1 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$ , ... | Using the diagram above, we notice that the desired length is simply the distance between the point $C$ and $\overline{AB}$ . We can mark $C$ as $(3,3)$ since it is $3$ units away from each of the bases. Point $B$ is $(8,3)$ . Thus, line $\overline{AB}$ is $y = \frac{3}{8}x \Rightarrow 3x + 8y = 0$ . We can use the dis... | 298 |
5,526 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_8 | 1 | Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ | First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \dots, {a + n(n-1)}$ , where $a = n - \frac{n(n-1)}2.$
This is certainly true for $n=1$ . Suppose that it is true for $n = m-1 \ge 1$ , and note that the zeros of $f_m$ are the solutions of $|x - m| = k$ , where $k$ is a nonn... | 101 |
5,527 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_8 | 2 | Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ | Starting from $f_1(x)=|x-1|$ , we can track the solutions, the number of solutions, and their sum.
\[\begin{array}{c|c|c|c} n&Solutions&number&sum\\ 1&1&1&1\\ 2&1,3&2&4\\ 3&0,2,4,6&4&12\\ 4&-2,0,2...10&7&28\\ 5&-5,-3,-1...15&11&55\\ \end{array}\]
It is clear that the solutions form arithmetic sequences with a dif... | 101 |
5,528 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_10 | 1 | Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ | The formula for the sum of cubes, also known as Nicomachus's Theorem, is as follows: \[1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2=\left(\frac{k(k+1)}{2}\right)^2\] for any positive integer $k$
So let's apply this to this problem.
Let $m=n+5$ . Then we have \begin{align*} 1^3+2^3+3^3+\dots+(m-5)^3&\equiv 17 \mod m \\ \left... | 239 |
5,529 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_10 | 2 | Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ | Using the formula for $\sum_{k=1}^n k^3$ \[1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}\] Since $1^3 + 2^3 + 3^3 + ... + n^3$ divided by $n + 5$ has a remainder of $17$ \[\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}\] Using the rules of modular arithmetic, \[n^2(n+1)^2 \equiv 68\pmod {n + 5}\] \[n^2(n+1)^2 - 68\e... | 239 |
5,530 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_10 | 3 | Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ | As before, we note that $(5+a)^3 + (n-a)^3 \equiv (5+a)^3 - (n+5 - (n-a))^3 \equiv 0 \pmod {n+5}.$ Thus, we can pair up the terms from $5^3$ to $n^3$ and cancel them. We have to deal with two cases:
If $n$ is even, then $5^3+6^3 + \cdots + n^3 \equiv 0 \pmod {n+5},$ as there are an even number of terms and they pair an... | 239 |
5,531 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_12 | 1 | Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$ , those in the second row are numbered left to right with the integers $n + 1$ through $2n$ , and so on. Square $200$ is in ... | Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$ $5$ $7$ , or $9$ . Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$ . Therefore, $m < 1800 \mod n < 1800-m$
If $m=3$ $n$ ca... | 248 |
5,532 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | 1 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | Assume the incircle touches $AB$ $BC$ $CD$ $DE$ $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$ $ET=y=ES=CR=CQ$ $AP=AT=z$ . So we have $x+y=6$ $x+z=5$ and $y+z$ =7, solve it we have $x=2$ $z=3$ $y=4$ . Let the center of the incircle be $I$ , by SAS we can proof triangle $BIQ$ is congruent to triangle $DIS$ ,... | 60 |
5,533 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | 2 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | Suppose that the circle intersects $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ at $P$ $Q$ $R$ $S$ , and $T$ respectively. Then $AT = AP = a$ $BP = BQ = b$ $CQ = CR = c$ $DR = DS = d$ , and $ES = ET = e$ . So $a + b = 5$ $b + c = 6$ $c + d = 6$ $d + e = 6$ , and $e + a = 7$ . Th... | 60 |
5,534 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | 3 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | This pentagon is very close to a regular pentagon with side lengths $6$ . The area of a regular pentagon with side lengths $s$ is $\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}$ $5-2\sqrt{5}$ is slightly greater than $\frac{1}{2}$ given that $2\sqrt{5}$ is slightly less than $\frac{9}{2}$ $4\sqrt{5-2\sqrt{5}}$ is then slightly grea... | 60 |
5,535 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | 4 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | Let $\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$ , so the problem is reduced to finding $r$ . Let $a$ be the length of the tangent segment from $A$ to $\omega$ , and analogously define $b$ $c$ $d$ , and $e$ . Then $a+b=5$ ... | 60 |
5,536 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_14 | 1 | For a real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$ , and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$ . For example, $\{3\} = 0$ and $\{4.56\} = 0.56$ . Define $f(x)=x\{x\}$ , and let $N$ be the number of real-valued solutions to the equation $f(f(f... | To solve $f(f(f(x)))=17$ , we need to solve $f(x) = y$ where $f(f(y))=17$ , and to solve that we need to solve $f(y) = z$ where $f(z) = 17$
It is clear to see for some integer $a \geq 17$ there is exactly one value of $z$ in the interval $[a, a+1)$ where $f(z) = 17$ . To understand this, imagine the graph of $f(z)$ on ... | 10 |
5,537 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | 1 | 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 $MT=3\sqrt{15}$
Since $\angle A=\angle CBT=\angle BCT$ , we have $\cos A=\tfrac{11}{16}$ . Notice that $AXTY$ is cyclic, so $\angle XTY=180^{\circ}-A$ , so $\cos XTY=-\cos A$ , and the cosine law ... | 717 |
5,538 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | 2 | 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 $M$ denote the midpoint of $\overline{BC}$ . The critical claim is that $M$ is the orthocenter of $\triangle AXY$ , which has the circle with diameter $\overline{AT}$ as its circumcircle. To see this, note that because $\angle BXT = \angle BMT = 90^\circ$ , the quadrilateral $MBXT$ is cyclic, it follows that \[\ang... | 717 |
5,539 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | 3 | 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 $H$ be the orthocenter of $\triangle AXY$
Lemma 1: $H$ is the midpoint of $BC$
Proof: Let $H'$ be the midpoint of $BC$ , and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \cap BA=E$ and $H'X \cap AC=F$ , then note that: \[\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.\] That implies that $\a... | 717 |
5,540 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | 4 | 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 $XY^2$ | Using the Claim (below) we get $\triangle ABC \sim \triangle XTM \sim \triangle YMT.$
Corresponding sides of similar $\triangle XTM \sim \triangle YMT$ is $MT,$ so
$\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT$ – parallelogram.
\[4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.\] The formula fo... | 717 |
5,541 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_1 | 1 | Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\] Find the sum of the digits of $N$ | Let's express the number in terms of $10^n$ . We can obtain $(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)$ . By the commutative and associative property, we can group it into $(10+10^2+10^3+\cdots+10^{321})-321$ . We know the former will yield $1111....10$ , so we only have to figure out what the last few digits are. T... | 342 |
5,542 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_1 | 2 | Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\] Find the sum of the digits of $N$ | Observe how adding results in the last term but with a $1$ concatenated in front and also a $1$ subtracted ( $09$ $108$ $1107$ $11106$ ). Then for any index of terms, $n$ , the sum is $11...10-n$ , where the first term is of length $n+1$ . Here, that is $\boxed{342}$ | 342 |
5,543 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_3 | 1 | In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ | We know the area of the hexagon $ABCDEF$ to be $\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED$ . Since $PR^2+RQ^2=PQ^2$ , we know that $\triangle PRQ$ is a right triangle. Thus the area of $\triangle PQR$ is $150$ . Another way to compute the area is \[\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cd... | 120 |
5,544 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_3 | 4 | In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ | Knowing that $\triangle{PQR}$ has area $150$ and is a $3$ $4$ $5$ triangle, we can find the area of the smaller triangles $\triangle{DRE}$ $\triangle{APF}$ , and $\triangle{CQB}$ and subtract them from $\triangle{PQR}$ to obtain our answer. First off, we know $\triangle{DRE}$ has area $12.5$ since it is a right triangl... | 120 |
5,545 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_4 | 1 | A soccer team has $22$ available players. A fixed set of $11$ players starts the game, while the other $11$ are available as substitutes. During the game, the coach may make as many as $3$ substitutions, where any one of the $11$ players in the game is replaced by one of the substitutes. No player removed from the game... | There are $0-3$ substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for $0$ subs is $1$ , and the ways to reorganize after $n$ subs is the product of the number of new subs ( $12-n$ ) and the players that can be ejected ( $11$ ). ... | 122 |
5,546 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_5 | 1 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The p... | One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\] for $x,y \geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero.
We then recu... | 252 |
5,547 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_5 | 2 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The p... | Obviously, the only way to reach (0,0) is to get to (1,1) and then have a $\frac{1}{3}$ chance to get to (0,0). Let x denote a move left 1 unit, y denote a move down 1 unit, and z denote a move left and down one unit each. The possible cases for these moves are $(x,y,z)=(0,0,3),(1,1,2),(2,2,1)$ and $(3,3,0)$ . This giv... | 252 |
5,548 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_5 | 3 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The p... | Since the particle stops at one of the axes, we know that the particle most pass through $(1,1)$ . Thus, it suffices to consider the probability our particle will reach $(1,1)$ . Then the only ways to get to $(1,1)$ from $(4,4)$ are the following:
(1) 3 moves diagonally
(2) 2 moves diagonally, 1 move left, 1 move down
... | 252 |
5,549 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | 8 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, ... | 90 |
5,550 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | 9 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, ... | 90 |
5,551 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | 1 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $... | Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630$ .
Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$ .
Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$ , along with the previousl... | 880 |
5,552 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | 2 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $... | First simplifying the first and second equations, we get that
\[\log_{10}(x\cdot\text{gcd}(x,y)^2)=60\] \[\log_{10}(y\cdot\text{lcm}(x,y)^2)=570\]
Thus, when the two equations are added, we have that \[\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630\] When simplified, this equals \[\log_{10}(x^3y^3)=630\] so ... | 880 |
5,553 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | 3 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $... | Let $x=10^a$ and $y=10^b$ and $a<b$ . Then the given equations become $3a=60$ and $3b=570$ . Therefore, $x=10^{20}=2^{20}\cdot5^{20}$ and $y=10^{190}=2^{190}\cdot5^{190}$ . Our answer is $3(20+20)+2(190+190)=\boxed{880}$ | 880 |
5,554 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | 5 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $... | Let $x=d\alpha, y=d\beta, (\alpha, \beta)=1$ . Simplifying, $d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}$ . Notice that since $\alpha, \beta$ are coprime, and $\alpha < 5^{90}$ (Prove it yourself !) , $\alpha=1, \beta = 10^{170}$ . Hence, $x=10^{20}, y=10^{190}... | 880 |
5,555 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | 6 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $... | Add the two equations and use the fact that $\gcd\left(x,y\right)\cdot\mathrm{lcm}\left(x,y\right)=xy$ to find that $xy=10^{210}$ . So let $x=2^a5^b$ and $y=2^{210-a}5^{210-b}$ for $0\leq a,b\leq210$ . If $a\geq105$ then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2=10^{60}$ is $a+2\left(210-a\right)=420-a$ , so... | 880 |
5,556 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_9 | 1 | 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:
Since 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 such val... | 540 |
5,557 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | 1 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | Notice that we must have \[\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .\] However, $f(t)-t=t(t-20)$ , so \begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\ &=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\ &=(z-19)(z+1)\\ &=(z-9)^2-100. \end{align*} Then, the real part ... | 230 |
5,558 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | 2 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | We will use the fact that segments $AB$ and $BC$ are perpendicular in the complex plane if and only if $\frac{a-b}{b-c}\in i\mathbb{R}$ . To prove this, note that when dividing two complex numbers you subtract the angle of one from the other. Therefore, if the two complex numbers are perpendicular, the difference betwe... | 230 |
5,559 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | 4 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | The arguments of the two complex numbers differ by $90^\circ$ if the ratio of the numbers is a pure imaginary number. Thus three distinct complex numbers $A,\,B,$ and $C$ form a right triangle at $B$ if and only if $\tfrac{C-B}{B-A}$ has real part equal to $0.$ Hence \begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z... | 230 |
5,560 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | 5 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | Firstly, the angle between the three complex numbers is equivalent to the angle between $f(z)-z,0,$ and $f(f(z))-f(z)$
Using $f(z)=z(z-19)$ to help expand,
$z-f(z)=20z-z^2$ and $f(f(z))-f(z)=(z^2-19z)(z^2-19z-19)-(z^2-19z)$
The second equation can be rewritten as $(z^2-19z)(z^2-19z-20)=z(z-19)(z-20)(z+1)$
Note that the... | 230 |
5,561 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | 9 | 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... | Like Solution 7, let $Q'$ be the altitude from $O$ to $XY$ . And, let $M$ be the intersection of $O_1O_2$ and $PQ$ . Construct $P'$ on line $AO$ such that $PP' \parallel O_2O_1$ . First, because of isosceles triangles $OAB$ $O_1AP$ , and $O_2BP$ , we have $\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}$ , wh... | 65 |
5,562 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | 6 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair O = (4.5,2.4); pair E = (-6,0); pair K = (1.993,3.737); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); labe... | 59 |
5,563 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | 1 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where... | Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$ . Then $P_7 = 1$ $P_6 = \frac12$ , and $P_n = \frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \leq n \leq 5$ . Working our way down, we find \[P_5 = \frac{3}{4}\] \[P_4 = \frac{5}{8}\] \[P_3 = \frac{11}{16}\] \[P_2 = \frac{21}{32}\] \[P_1 = ... | 107 |
5,564 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | 2 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where... | Define a one jump to be a jump from $k$ to $k + 1$ and a two jump to be a jump from $k$ to $k + 2$
Case 1: (6 one jumps) $\left (\frac{1}{2} \right)^6 = \frac{1}{64}$
Case 2: (4 one jumps and 1 two jumps) $\binom{5}{1} \cdot \left(\frac{1}{2}\right)^5 = \frac{5}{32}$
Case 3: (2 one jumps and 2 two jumps) $\binom{4}{2} ... | 107 |
5,565 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | 3 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where... | Let $P_n$ be the probability that the frog lands on lily pad $n$ . The probability that the frog never lands on pad $n$ is $\frac{1}{2}P_{n-1}$ , so $1-P_n=\frac{1}{2}P_{n-1}$ . This rearranges to $P_n=1-\frac{1}{2}P_{n-1}$ , and we know that $P_1=1$ , so we can compute $P_7$ \begin{align*} P_1&=1\\ P_2&=1-\dfrac{1}{2}... | 107 |
5,566 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | 4 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where... | For any point $n$ , let the probability that the frog lands on lily pad $n$ be $P_n$ . The frog can land at lily pad $n$ with either a double jump from lily pad $n-2$ or a single jump from lily pad $n-1$ . Since the probability when the frog is at $n-2$ to make a double jump is $\frac{1}{2}$ and same for when it's at $... | 107 |
5,567 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | 1 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).
Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.
Case 2: Two 5's are rolled.
Cas... | 187 |
5,568 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | 2 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square).
Probability ... | 187 |
5,569 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | 3 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:
If there are four 1/4's, then there are $2^4=1... | 187 |
5,570 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | 5 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can do recursion on the number of rolls to find the number of ways we can get $4$ rolls to multiply to a square.
After $n$ rolls, let us say that the product is $p = 2^a3^b5^c$
Define the following:
$A_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is odd, and $b$ $c$ are even
$B_{n} =$ the nu... | 187 |
5,571 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | 6 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Consider all the distinct "fundamental" groups of integers from $1$ to $6$ whose product is a perfect square. A "fundamental" group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, $\{2,2\}$ is a fundamental group, while $\{3,3,4\}$ is not, because it can be bro... | 187 |
5,572 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | 7 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | There are a total of $2^4=1296$ possible die rolls.
We use casework:
Case 1 : All 4 numbers are the same.
There are obviously $6$ ways.
Case 2 : Sets of 2 different numbers.
A set of two different numbers is basically $(x,x,y,y)$ . There are a total of $\frac{4!}{2!\cdot 2!}=6$ ways to arrange the numbers.
By listing t... | 187 |
5,573 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_5 | 1 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table und... | Let us first consider the $4$ ambassadors and the $6$ even-numbered chairs. If we consider only their relative positions, they can sit in one of $3$ distinct ways: Such that the $2$ empty even-numbered chairs are adjacent, such that they are separated by one occupied even-numbered chair, and such that they are opposite... | 520 |
5,574 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_5 | 2 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table und... | In the diagram, the seats are numbered 1...12. Rather than picking seats for each person, however, each ambassador/assistant team picks a gap between the seats (A...L) and the ambassador sits in the even seat while the assistant sits in the odd seat. For example, if team 1 picks gap C then Ambassador 1 will sit in se... | 520 |
5,575 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_5 | 4 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table und... | We see that for every 2 adjacent spots on the table, there is exactly one way for an ambassador and his or her partner to sit. There are 2 cases:
For the first case, there are 3 pairs and 4 empty spaces, which results in $\frac{7!}{3!4!} = 35$ ways to arrange the pairs.
For the second case, there are 4 pairs and 4 empt... | 520 |
5,576 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | 1 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | Using change of base on the second equation to base b, \[\frac{\log x}{\log \log_{b}{x}}=54\] \[\log x = 54 \cdot \log\log_{b}{x}\] \[x = (\log_{b}{x})^{54}\] Note by dolphin7 - you could also just rewrite the second equation in exponent form.
Substituting this into the $\sqrt x$ of the first equation, \[3\log_{b}{((\l... | 216 |
5,577 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | 2 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | We start by simplifying the first equation to \[3\log_{b}{(\sqrt{x}\log x)}=\log_{b}{(x^{\frac{3}{2}}\log^3x)}=56\] \[x^\frac{3}{2}\cdot \log_b^3x=b^{56}\] Next, we simplify the second equation to \[\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] \[\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))\] \[x=\lo... | 216 |
5,578 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | 3 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | Apply change of base to \[\log_{\log x}(x)=54\] to yield: \[\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] which can be rearranged as: \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] Apply log properties to \[3\log(\sqrt{x}\log x)=56\] to yield: \[3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(... | 216 |
5,579 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | 4 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | 1st equation: \[\log (\sqrt{x}\log x)=\frac{56}{3}\] \[\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}\] 2nd equation: \[x=(\log x)^{54}\] So now substitute $\log x=a$ and $x=b^a$ \[b^a=a^{54}\] \[b=a^{\frac{54}{a}}\] We also have that \[\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}\] \[\frac... | 216 |
5,580 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | 5 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | This system of equations looks complicated to work with, so we let $a=\log_bx$ to make it easier for us to read.
Now, the first equation becomes $3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3$
The second equation, $\log_{\log(x)}(x)=54$ gives us $\underline{a^{54} = x}$
Let's plug this back into... | 216 |
5,581 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | 6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | Using change of base on the second equation, we have
\[\frac{\log_{b} x}{\log_{b} \log_{b} x} = 54\]
Using log rules on the first equation, we have
\[\frac{3}{2} \log_{b} x + 3 \log_{b} \log_{b} x = 56\]
We notice that $\log_{b} x$ and $\log_{b} \log_{b} x$ are in both equations. Thus, we set $m = \log_{b} x$ and $n = ... | 216 |
5,582 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | 7 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | The second equation implies that \[\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}\] The first equation implies that \[3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x\] Substituting the first result into the second gives us \[b^{\frac{56}{3}}=x^{\frac{1}{2}}\c... | 216 |
5,583 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_7 | 1 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,4... | Let the points of intersection of $\ell_A, \ell_B,\ell_C$ with $\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$ . Furthermore, let the desired triangle be $\triangle XYZ$ , with $X$ closest to side $BC$ $Y$ closest to side $AC$ , and $Z$ closest to side $AB$ . Hence, the desired p... | 715 |
5,584 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_7 | 2 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,4... | Let the diagram be set up like that in Solution 1.
By similar triangles we have \[\frac{AH}{AB}=\frac{GH}{BC}\Longrightarrow AH=30\] \[\frac{IB}{AB}=\frac{DI}{AC}\Longrightarrow IB=30\] Thus \[HI=AB-AH-IB=60\]
Since $\bigtriangleup IHZ\sim\bigtriangleup ABC$ and $\frac{HI}{AB}=\frac{1}{2}$ , the altitude of $\bigtriang... | 715 |
5,585 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_7 | 3 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,4... | Notation shown on diagram. By similar triangles we have \[k_1 = \frac{EF}{BC} = \frac{AE}{AB} = \frac {AF}{AC} = \frac {1}{4},\] \[k_2 = \frac{F''E''}{AC} = \frac {BF''}{AB} = \frac{1}{4},\] \[k_3 = \frac{E'F'}{AB} = \frac{E'C }{AC} = \frac{1}{8}.\] So, \[\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 ... | 715 |
5,586 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_9 | 1 | Call a positive integer $n$ $k$ pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ | Every 20-pretty integer can be written in form $n = 2^a 5^b k$ , where $a \ge 2$ $b \ge 1$ $\gcd(k,10) = 1$ , and $d(n) = 20$ , where $d(n)$ is the number of divisors of $n$ . Thus, we have $20 = (a+1)(b+1)d(k)$ , using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there... | 472 |
5,587 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_9 | 2 | Call a positive integer $n$ $k$ pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ | For $n$ to have exactly $20$ positive divisors, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$ where $p,q,r$ are primes. No number that is a multiple of $20$ can be expressed in the first form because 20 has two primes in its prime factorization, while the first form has o... | 472 |
5,588 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | 1 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integ... | Note that if $\tan \theta$ is positive, then $\theta$ is in the first or third quadrant, so $0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}$
Furthermore, the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ that are multiples of $3$ is when: \[2^0\theta \equiv 2^3\theta \equiv 2^6\theta \equiv ... | 547 |
5,589 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | 2 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integ... | As in the previous solution, we note that $\tan \theta$ is positive when $\theta$ is in the first or third quadrant. In order for $\tan\left(2^n\theta\right)$ to be positive for all $n$ divisible by $3$ , we must have $\theta$ $2^3\theta$ $2^6\theta$ , etc to lie in the first or third quadrants. We already know that $\... | 547 |
5,590 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | 3 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integ... | Since $\tan\left(\theta\right) > 0$ $0 < \theta < 90$ . Since $\tan\left(2\theta\right) < 0$ $\theta$ has to be in the second half of the interval (0, 90) ie (45, 90). Since $\tan\left(4\theta\right) < 0$ $\theta$ has to be in the second half of that interval ie (67.5, 90). And since $\tan\left(8\theta\right) > 0$ $... | 547 |
5,591 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | 5 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integ... | Since $7$ is the only number n such that f(x) = $2^{\lfloor x\rfloor}$ $\text{(mod 7)}$ has a period of 3, we find that $\theta$ is a multiple of ${\frac{180}{7}}^\circ$ . Note that the tangents of ${\frac{180}{7}}^\circ$ ${\frac{360}{7}}^\circ$ ${\frac{540}{7}}^\circ$ are positive while those of ${\frac{720}{7}}^\circ... | 547 |
5,592 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | 3 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ an... | Let the centers of the circles be $O_{1}$ and $O_{2}$ where the $O_{1}$ has the side length $7$ contained in the circle. Now let $\angle BAC =x.$ This implies \[\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x\] by the angle by by tangent. Then we also know that \[\angle AO_{1}B = \angle... | 11 |
5,593 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | 4 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ an... | By the definition of $K$ , it is the spiral center mapping $BA\to AC$ , which means that it is the midpoint of the $A$ -symmedian chord. In particular, if $M$ is the midpoint of $BC$ and $M'$ is the reflection of $A$ across $K$ , we have $\triangle ABM'\sim\triangle AMC$ . By Stewart's Theorem, it then follows that \[A... | 11 |
5,594 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_12 | 1 | For $n \ge 1$ call a finite sequence $(a_1, a_2 \ldots a_n)$ of positive integers progressive if $a_i < a_{i+1}$ and $a_i$ divides $a_{i+1}$ for all $1 \le i \le n-1$ . Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360$ | If the first term is $x$ , then dividing through by $x$ , we see that we can find the number of progressive sequences whose sum is $\frac{360}{x} - 1$ , and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the answer $N$... | 47 |
5,595 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_13 | 1 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, \(1\), and assume the side length of the octagon is \(2\).
Let \(r\) denote the radius of the circle, \(O\) be the center of the circle. Then: \[r^2= 1^2 + \left(\sqrt{2}+1\right)^2= 4+2\sqrt{2}.\]
Now... | 504 |
5,596 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_13 | 2 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point $P$ from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram, where the octagon is oriented so as $A_1A_2$ is ho... | 504 |
5,597 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | 2 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can b... | Let $BC=a$ $AC=b$ , and $AB=c$ . Let $\cos\angle A=k$ . Then $AP=bk$ and $AQ=ck$
By Power of a Point theorem, \begin{align} AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\ AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 \end{align} Thus $bck = (bk)^2+400=(ck)^2+525 = u$... | 574 |
5,598 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | 3 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can b... | Let $AP=p$ $PB=q$ $AQ=r$ , and $QC=s$ . By Power of a Point, \begin{align} AP\cdot PB=XP\cdot YP \quad &\Longrightarrow \quad pq=400\\ AQ\cdot QC=YQ\cdot XQ \quad &\Longrightarrow \quad rs=525 \end{align} Points $P$ and $Q$ lie on the circle, $\omega$ , with diameter $BC$ , and pow $(A,\omega) = AP\cdot AB = AQ\cdo... | 574 |
5,599 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | 4 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can b... | This solution is directly based of @CantonMathGuy's solution.
We start off with a key claim.
Claim. $XB \parallel AC$ and $YC \parallel AB$
Proof.
Let $E$ and $F$ denote the reflections of the orthocenter over points $P$ and $Q$ , respectively. Since $EF \parallel XY$ and \[EF = 2 PQ = XP + PQ + QY = XY,\] we have that... | 574 |
5,600 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | 1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Let the linear factors be $(x+c)(x+d)$
Then, $a=c+d$ and $b=cd$
We know that $1\le a\le 100$ and $b\ge 0$ , so $c$ and $d$ both have to be non-negative
However, $a$ cannot be $0$ , so at least one of $c$ and $d$ must be greater than $0$ , ie positive.
Also, $a$ cannot be greater than $100$ , so $c+d$ must be less than ... | 600 |
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