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int64 1
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|---|---|---|---|---|---|
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 answer of $\boxed{547}$
| 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*} \angle A + \angle B + \angle C = x+y+z+\frac{180-x}{2}=180 \\ \end{align*}
Now we just need to find the variables.
\begin{align*} (180-2y)+z = 180& \\ (180-2z)+y+\frac{180-x}{2} = 180& \\ \end{align*}
Notice how all the equations equal 180. We can use this to write
\begin{align*} (180-2y)+z = (180-2z)+y+\frac{180-x}{2}=x+y+z+\frac{180-x}{2} \\ \end{align*}
Simplifying, we get
\begin{align*} (180-2y)+z=(180-2z)+y+\frac{180-x}{2} \\ 360-4y+2z=360-4z+2y+180-x \\ \end{align*}
\begin{align*} 6z=6y+180-x \\ x=6y-6z+180 \\ \end{align*}
\begin{align*} (180-2y)+z=6y-6z+180+y+z+\frac{180-(6y-6+180)}{2} \\ 360-4y+2z=12y-12z+360+2y+2z+180-6y+6z-180 \\ \end{align*}
\begin{align*} 6z=12y& \\ z=2y& \\ \end{align*}
Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation
\begin{align*} \frac{180-x}{2}=x+z \\ \end{align*}
Substituting $z$ with $2y$ , we get
\begin{align*} \frac{180-x}{2}=x+2y \\ 180-x=2x+4y \\ \end{align*} \begin{align*} 180-(6y-6z+180)=2(6y-6z+180)+4y& \\ 180-6y+12y-180=12y-24y+360+4y& \\ \end{align*} \begin{align*} 6y=-8y+360& \\ \end{align*}
With this, we get
\begin{align*} y=\frac{180}{7} \\ x=\frac{180}{7} \\ z=\frac{360}{7} \\ \end{align*}
And a final answer of $\frac{180}{7}+\frac{360}{7} = \frac{540}{7} = \boxed{547}$
| 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 $N$ expressed in base ten.
|
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, $515_{11} = 621_{10}$ , so our answer is $\boxed{621}$
| 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 permutations arise from removing one card from $123456$ and placing it in a position at least two away from its starting location. There are $4$ such positions to place each of the cards numbered $1$ and $6,$ and $3$ such positions for each of the cards numbered $2, 3, 4,$ and $5.$ This accounts for $2\cdot4 + 4\cdot3 =20$ permutations. Thus there are $1 + 5 + 20 = 26$ permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total $26 + 26 = \boxed{52}$
| 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 ascending order. This accounts for $n\cdot n = n^2$ permutations. However, the original ascending order has been counted $n$ times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is $n^2-(n-1)-(n-1)=(n-1)^2+1.$ When $n = 6$ , this is $(6-1)^2+1 = 26$ , and the final answer is $2\cdot26 = \boxed{52}$
| 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 $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
[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, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label("$r$", O -- Y, N); label("$r$", Y -- P, N); label("$r$", O -- A, NW); label("$r$", P -- D, NE); label("$1$", A -- H, N); label("$2$", L -- D, N); label("$7$", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); [/asy] Set the common radius to $r$ . First, take the cross section of the sphere sitting in the hole of radius $1$ . If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$ . Therefore, the height of this circle outside of the hole is $\sqrt{r^2-1}$
The other circle follows similarly for a height (outside the hole) of $\sqrt{r^2-4}$ . Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$ , as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\sqrt{r^2-1} - \sqrt{r^2-4}$ . Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \[\left(\sqrt{r^2-1} - \sqrt{r^2-4}\right)^2 + 7^2 = (2r)^2.\] Simplifying a few times, \begin{align*} r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 &= 4r^2 \\ 2r^2-44 &= -2\left(\sqrt{(r^2-1)(r^2-4)}\right) \\ 22-r^2 &= \left(\sqrt{r^4 - 5r^2 + 4}\right) \\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\ 39r^2&=480 \\ r^2&=\frac{480}{39} = \frac{160}{13}. \end{align*} Therefore, our answer is $\boxed{173}$
| 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 point $P$ . Then $OP^2=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
[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) * ((F + G) / 2); I = rotate(-120, H) * ((G + H) / 2); draw(O -- A -- B -- C -- D -- F -- G -- H -- I); draw(A -- X, dashed); markscalefactor = 0.05; path angle = anglemark(X, A, B); draw(angle); dot(P); dot(O); dot(A); dot(B); dot(C); dot(D); label("$O$", O, W); label("$P$", P, E); label("$A$", A, S); label("$B$", B, E); label("$C$", C, E); label("$D$", D, W); label("$60^\circ$", angle, ENE*3); [/asy]
We notice that the moves cycle every 6 moves, so we plot out the first 6 moves on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately. Then, we will combine them and account for the cycling using the formula for an infinite geometric series.
First move: The ant moves right $5$ .
Second move: We use properties of a $30-60-90$ triangle to get $\frac{5}{4}$ right, $\frac{5\sqrt{3}}{4}$ up.
Third move: $\frac{5}{8}$ left, $\frac{5\sqrt{3}}{8}$ up.
Fourth move: $\frac{5}{8}$ left.
Fifth move: $\frac{5}{32}$ left, $\frac{5\sqrt{3}}{32}$ down.
Sixth move: $\frac{5}{64}$ right, $\frac{5\sqrt{3}}{64}$ down.
Total of x-coordinate: $5 + \frac{5}{4} - \frac{5}{8} - \frac{5}{8} - \frac{5}{32} + \frac{5}{64} = \frac{315}{64}$ .
Total of y-coordinate: $0 + \frac{5\sqrt{3}}{4} + \frac{5\sqrt{3}}{8} + 0 - \frac{5\sqrt{3}}{32} - \frac{5\sqrt{3}}{64} = \frac{105\sqrt{3}}{64}$
After this cycle of six moves, all moves repeat with a factor of $(\frac{1}{2})^6 = \frac{1}{64}$ . Using the formula for a geometric series, multiplying each sequence by $\frac{1}{1-\frac{1}{64}} = \frac{64}{63}$ will give us the point $P$
Now, knowing the initial $x$ and $y,$ we plug this into the geometric series formula ( $\frac{a}{1-r}$ ), and we get $\frac{315}{64} \cdot \frac{64}{63} = 5$ $\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}$ .
Therefore, the coordinates of point $P$ are $(5,\frac{5\sqrt{3}}{3})$ , so using the Pythagorean Theorem, $OP^2 = \frac{100}{3}$ , for an answer of $\boxed{103}$
| 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 point $P$ . Then $OP^2=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
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 geometric series, this is equal to \[\frac{5}{1 - \frac12e^{\frac{i\pi}{3}}} = \frac{5}{1 - \frac{1 + i\sqrt{3}}{4}} = \frac{20}{3 - i\sqrt{3}} = 5 + \frac{5i\sqrt{3}}{3}\] We are looking for the square of the modulus of this value: \[\left|\frac{5 + 5i\sqrt{3}}{3}\right|^2 = 25 + \frac{25}{3} = \frac{100}{3}\] so the answer is $100 + 3 = \boxed{103}$
| 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 point $P$ . Then $OP^2=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
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 $(5+5/4-5/8, 0+5\sqrt3/4+5\sqrt3/8)$ or $(45/8, 15\sqrt3/8)$ . Multiplying these by $8/9$ , we get $(5, 5\sqrt3/3)$ $\implies$ $\boxed{103}$
| 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 point $P$ . Then $OP^2=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
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 $\frac{40}{9}+\frac{10}{9}-\frac{5}{9}=\frac{45}{9}=5$ units east and a total of $\frac{10\sqrt{3}}{9}+\frac{5\sqrt{3}}{9}=\frac{15\sqrt{3}}{9}=\frac{5\sqrt{3}}{3}$ units north because of the 30-60-90 right triangles formed. Now, $OP^2=5^2+\frac{5^2}{3}=\frac{100}{3}$ by the Pythagorean Theorem, and the answer is $\boxed{103}$
| 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 point $P$ . Then $OP^2=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
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
\[(\Delta x_n, \Delta y_n)=(5\cdot (1/2)^{n-1}\cos (60(n-1))^\circ,5\cdot (1/2)^{n-1}\sin (60(n-1))^\circ).\]
Summing this over all the displacements, we get
\[x_P=\sum_{n=1}^{\infty} 5\cdot (1/2)^{n-1}\cos (60(n-1))^\circ, y_P=\sum_{n=1}^{\infty} 5\cdot (1/2)^{n-1}\sin (60(n-1))^\circ.\]
We then have
\begin{align} OP^2 &= x_P^2+y_P^2 \\ &=\sum_{n=1}^{\infty} (5\cdot (1/2)^{n-1}\cos ^2(60(n-1))^\circ) + (5\cdot (1/2)^{n-1}\sin ^2(60(n-1))^\circ) \\ &= \sum_{n=1}^{\infty} (5\cdot (1/2)^{n-1})^2(\cos ^2 (60(n-1))^\circ+\sin ^2 (60(n-1))^\circ) \\ &= \sum_{n=1}^{\infty} (25\cdot (1/4)^{n-1}) \\ &= \dfrac{25}{1-1/4} \\ &= 100/3. \end{align}
Thus, the answer is $100+3=\boxed{103}.$ --MenuThreeOne
| 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 contradiction.
$n$ is also not 1 because then $m$ would be a multiple of it.
Thus, $n$ is a multiple of 11 and/or 13 and/or 17 and/or...
Assume for the sake of contradiction that $n$ has at most 1 power of 11, at most 1 power of 13...and so on...
Then, for $n^n | m^m$ to be satisfied, $m$ must contain at least the same prime factors that $n$ has. This tells us that for the primes where $n$ has one power of, $m$ also has at least one power, and since this holds true for all the primes of $n$ $n|m$ . Contradiction.
Thus $n$ needs more than one power of some prime.
The obvious smallest possible value of $n$ now is $11^2 =121$ .
Since $121^{121}=11^{242}$ , we need $m$ to be a multiple of 11 at least $242$ that is not divisible by $121$ and most importantly, $\gcd(m+n,210) = 1$ $242$ is divisible by $121$ , out. $253+121$ is divisible by 2, out. $264+121$ is divisible by 5, out. $275+121$ is divisible by 2, out. $286+121=37\cdot 11$ and satisfies all the conditions in the given problem, and the next case $n=169$ will give us at least $169\cdot 3$ , so we get $\boxed{407}$
| 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, \[240+112a+24b+12a^2+4ab+12c+2ac=0\] \[120+56a+12b+6a^2+2ab+6c+ac=0\] \[6a^2+2ab+ac+56a+12b+6c+120=0\] \[6a^2+2ab+ac+20a+36a+12b+6c+120=0\] \[a(6a+2b+c+20)+6(6a+2b+c+20)=0\] \[(a+6)(6a+2b+c+20)=0\]
Therefore, either $a+6=0$ or $6a+2b+c=-20$ . The first case is easy: $a=-6$ and there are $441$ tuples in that case. In the second case, we simply perform casework on even values of $c$ , to get $77$ tuples, subtracting the $8$ tuples in both cases we get $441+77-8=\boxed{510}$
| 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 by $149^n \equiv 2^n \pmod 5$ in order to set up LTE), we see \[v_5(149^{4c}-2^{4c}) = v_5(149^{4c}-16^{c})\] and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4c}-2^{4c})=v_5(149^4-16)+v_5(c)=1+v_5(c)$ meaning that we have that by LTE, $5^4 | c$ and $4 \cdot 5^4$ divides $n$
Since $3^2$ $7^5$ and $4\cdot 5^4$ all divide $n$ , the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$ . Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$
| 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 just $7^5$
Finally, for $5^5$ , take $\pmod {5}$ and $\pmod {25}$ of each quantity (They happen to both be $-1$ and $2$ respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum $n$ for divisibility by $5$ is $4$ , and other values are factors of $4$ . Testing all of them(just $1$ $2$ $4$ using mods-not too bad), $4$ is indeed the smallest value to make the expression divisible by $5$ , and this clearly is NOT divisible by $25$ .
Therefore, the smallest $n$ to make this expression divisible by $5^5$ is $2^2 \cdot 5^4$
Calculating the LCM of all these, one gets $2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$ . Using the factor counting formula,
the answer is $3\cdot3\cdot5\cdot6$ $\boxed{270}$
| 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 $3^3$ because $149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\cdot2^3 + 2^6)$ , and the rightmost factor equates to $1 + 1 + 1 \pmod{3} \equiv 0 \pmod{3}$ . In fact, $n = 9 = 3^2$ is the least possible choice to ensure divisibility by $3^3$ because if $n = a \cdot 3^b$ , with $3 \nmid a$ and $b < 2$ , we write \[149^{a \cdot 3^b} - 2^{a \cdot 3^b} = (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\cdot2^{3^b}+\cdots2^{3^b(a - 1)}).\] Then, the rightmost factor is equivalent to $\pm a \pmod{3} \not\equiv 0 \pmod{3}$ , and $v_3(149^{3^b} - 2^{3^b}) = b + 1 < 3$
For divisibility by $7^7$ , we'll induct, claiming that $v_7(149^{7^k} - 2^{7^k}) = k + 2$ for whole numbers $k$ . The base case is clear. Then, \[v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k}).\] By the induction hypothesis, $v_7(149^{7^k} - 2^{7^k}) = k + 2$ . Then, notice that \[S(k) = 149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{7} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{49}.\] This tells us that $S(k)$ is divisible by $7$ , but not $49$ so that $v_7\left(S(k)\right) = 1$ , completing our induction. We can verify that $7^5$ is the least choice of $n$ to ensure divisibility by $7^7$ by arguing similarly to the $3^3$ case.
Finally, for $5^5$ , we take the powers of $149$ and $2$ in mod $5$ and mod $25$ . Writing out these mods, we have that $149^n \equiv 2^n \pmod{5}$ if and only if $4 | n$ , in which $149^n \equiv 2^n \equiv 1 \pmod{5}$ . So here we claim that $v_5(149^{4\cdot5^k} - 2^{4\cdot5^k}) = k + 1$ and perform yet another induction. The base case is true: $5 | 149^4 - 2^4$ , but $149^4 - 2^4 \equiv 1 - 16 \pmod{25}$ . Now then, assuming the induction statement to hold for some $k$ \[v_5(149^{4\cdot5^{k+1}} - 2^{4\cdot5^{k+1}}) = (k+1) + v_5(149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}).\] Note that $S'(k) = 149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}$ equates to $S''(k) = 1 + 2^{4\cdot5^k} + \cdots + 2^{16\cdot5^k}$ in both mod $5$ and mod $25$ . We notice that $S''(k) \equiv 0 \pmod{5}$ . Writing out the powers of $2$ mod $25$ , we have $S''(0) \equiv 5 \pmod{25}$ . Also $2^n \equiv 1 \pmod{25}$ when $n$ is a multiple of $20$ . Hence for $k > 0$ $S''(k) \equiv 5 \mod{25}$ . Thus, $v_5\left(S'(k)\right) = 1$ , completing our induction. Applying the same argument from the previous two cases, $4\cdot5^4$ is the least choice to ensure divisibility by $5^5$
Our answer is the number of divisors of $\text{lcm}(3^2, 7^5, 2^2\cdot5^4) = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$ . It is $(2 + 1)(2 + 1)(4 + 1)(5 + 1) = \boxed{270}$
| 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 written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$
|
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 $M$ to $BC$
We can find $AD$ using Stewart's Theorem: from Angle Bisector Theorem $BD = 2$ and $CD = 3$ . Then it is easy to find that $AD = 3\sqrt3$
Now we trig bash for $DI = MI - MD$ . Notice that $MI = MB$ from the Incenter Excenter Lemma. We obtain that $MB = \frac{BK}{\cos \frac{A}{2}} = \frac{\frac{5}{2}}{\frac{5}{4\sqrt2}}=2\sqrt2$ . To get $MD$ we angle chase to get $\angle KDM = \frac{A}{2}+C$ . Then \[\cos(\frac{A}{2}+C) = \cos\frac{A}{2}\cos C - \sin\frac{A}{2}\sin C = \frac{1}{2\sqrt2} = \frac{\frac{1}{2}}{MD}\] gives $MD = \sqrt{2}$ . This means $DI = \sqrt2$
Now let $AI \cap EF = G$ . It is easy to angle chase $\angle GIE = 90- \frac{B}{2}$ and $\angle GIF = 90- \frac{C}{2}$ . Since $GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}$ , we compute that \[EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}\] which implies \[[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}\] which gives an answer of $\boxed{36}$ . ~Leonard_my_dude~
| 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 written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$
|
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 we apply shoelace theorem later. Point A's coordinates is $(4*\frac{1}{8},4*\frac{3\sqrt{7}}{8})=(\frac{1}{2},\frac{3\sqrt{7}}{2})$ , Let $AJ$ is perpendicular to $BC$ $tan\angle{JAD}=\frac{2-\frac{1}{2}}{\frac{3\sqrt{7}}{2}}=\frac{\sqrt{7}}{7}$ , which means the slope of $FE$ is $\frac{\sqrt{7}}{7}$ . Find the coordinate of $M$ , it is easy, $(\frac{5}{4},\frac{3\sqrt{7}}{4})$ , the function $EF$ is $y=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$ . Now find the intersection of $EF,EB$ $\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}=\frac{\sqrt{7}x}{3}$ , getting that $x=3,E(3,\sqrt{7})$ Now we look at line segment $CF$ , since $cos\angle{\frac{ACB}{2}}=\frac{\sqrt{14}}{4},tan\angle{\frac{ACB}{2}}=\frac{\sqrt{7}}{7}$ . Since the line passes $(5,0)$ , we can set the equation to get the $CF:y=-\frac{\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$ , find the intersection of $CF,EF$ $-\frac{\sqrt{7}x}{7}+\frac{5\sqrt{7}}{7}=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$ , getting that $F:(\frac{1}{2},\frac{9\sqrt{7}}{14})$ and in the end we use shoelace theorem with coordinates of $A,F,E$ getting the area $\frac{15\sqrt{7}}{14}$ leads to the final answer $\boxed{36}$
| 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 in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$
|
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 have that the area of ABC is $3\sqrt{55}$ so the answer is $3 + 55 = \boxed{58}$
| 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 $\tfrac{20^{20}}{x}$ . This reduces the problem to finding the number of unique perfect square factors of $20^{20}$
$20^{20} = 2^{40} \cdot 5^{20} = \left(2^2\right)^{20}\cdot\left(5^2\right)^{10}.$ Therefore, the answer is $21 \cdot 11 = \boxed{231}.$
| 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 style */ real xmin = -8.0451801958033, xmax = 47.246151591238494, ymin = -10.271454747548662, ymax = 21.426040258770957; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqffff = rgb(0,1,1); draw((16,0)--(0,0)--(0,12)--cycle, linewidth(2) + zzttqq); draw((24,2)--(24,18)--(36,18)--cycle, linewidth(2) + blue); draw((16,0)--(21,-3)--(24,2)--cycle, linewidth(2) + qqwuqq); draw((21.39134584768662,-2.3477569205223032)--(20.73910276820892,-1.9564110728356852)--(20.347756920522304,-2.608654152313382)--(21,-3)--cycle, linewidth(2) + qqffff); /* draw figures */ draw((16,0)--(0,0), linewidth(2) + zzttqq); draw((0,0)--(0,12), linewidth(2) + zzttqq); draw((0,12)--(16,0), linewidth(2) + zzttqq); draw((24,2)--(24,18), linewidth(2) + blue); draw((24,18)--(36,18), linewidth(2) + blue); draw((36,18)--(24,2), linewidth(2) + blue); draw((16,0)--(24,2), linewidth(2)); draw((16,0)--(21,-3), linewidth(2) + qqwuqq); draw((21,-3)--(24,2), linewidth(2) + qqwuqq); draw((24,2)--(16,0), linewidth(2) + qqwuqq); draw((21,-3)--(20,1), linewidth(2.8) + qqffff); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$A$", (-0.6228029714727868,0.12704474547474198), NE * labelscalefactor); dot((0,12),dotstyle); label("$B$", (0.1301918194013232,12.354245873478124), NE * labelscalefactor); dot((16,0),dotstyle); label("$C$", (16.15822379657881,0.34218611429591583), NE * labelscalefactor); dot((24,18),dotstyle); label("$A'$", (24.154311337765787,18.342347305667463), NE * labelscalefactor); dot((24,2),dotstyle); label("$C'$", (23.186175178070503,1.95574638045472), NE * labelscalefactor); dot((36,18),dotstyle); label("$B'$", (36.13051420214449,18.342347305667463), NE * labelscalefactor); dot((21,-3),dotstyle); label("$P$", (21.35747354309052,-3.458644734878156), NE * labelscalefactor); dot((20,1),linewidth(4pt) + dotstyle); label("$D$", (20.13833911977053,1.2744653791876692), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
We first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$ . Note that $PC=PC'$ because $P$ lies on the perpendicular bisector of $CC'$ (it must be equidistant from $C$ and $C'$ by properties of a rotation).
Since $AB$ is vertical while $A'B'$ is horizontal, we have that the angle of rotation must be $90^{\circ}$ , and therefore $\angle P = 90^{\circ}$ . Therefore, $CPC'$ is a 45-45-90 right triangle, and $CD=DP$
We calculate $D$ to be $(20,1)$ . Since we translate $4$ right and $1$ up to get from point $C$ to point $D$ , we must translate $1$ right and $4$ down to get to point $P$ . This gives us $P(21,-3)$ . Our answer is then $90+21-3=\boxed{108}$ . ~Lopkiloinm & samrocksnature
| 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 $(16-x)+(-y)i=(y-2)+(24-x)i$ . Equating the real and complex terms results in the equations $16-x=y-2$ and $-y=24-x$ . Solving, $(x, y) : (21, -3) \rightarrow 90+21-3=\boxed{108}$
| 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 $y=-6x+123$ . Since the point $P$ of rotation must lie on both of these lines, we set them equal, solve and get: $x=21$ $y=-3$ . We can also easily see that the degree of rotation is $90$ since $AB$ is initially vertical, and now it is horizontal. Also, we can just sketch this on a coordinate plane and easily realize the same. Hence, the answer is $21-3+90 = \boxed{108}$
| 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: \[t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}\] \[t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}\] \[t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}\] \[t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20\]
Alas, we have figured this sequence is period 5! But since $2020 \equiv 5 \pmod 5$ , we can state that $t_{2020} = t_5 = \frac{101}{525}$ . According to the original problem statement, our answer is $\boxed{626}$ . ~ nikenissan
| 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 $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
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 distance from point to line formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ , where $x_0$ and $y_0$ are the coordinates of the point, and A, B, and C are the coefficients of the line in form $Ax + By + C = 0$ . Plugging everything in, we get \[\frac{|3(3) - 8(3)|}{\sqrt{8^2 + 3^2}} = \frac{15}{\sqrt{73}} \Rightarrow \frac{225}{73} \Rightarrow \boxed{298}\]
| 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 nonnegative zero of $f_{m-1}$ . Because the zeros of $f_{m-1}$ form an arithmetic sequence with common difference $2,$ so do the zeros of $f_m$ . The greatest zero of $f_{m-1}$ is \[m-1+\frac{(m-1)(m-2)}2 =\frac{m(m-1)}2,\] so the greatest zero of $f_m$ is $m+\frac{m(m-1)}2$ and the least is $m-\frac{m(m-1)}2$
It follows that the number of zeros of $f_n$ is $\frac{n(n-1)}2+1=\frac{n^2-n+2}2$ , and their average value is $n$ . The sum of the zeros of $f_n$ is \[\frac{n^3-n^2+2n}2.\] Let $S(n)=n^3-n^2+2n$ , so the sum of the zeros exceeds $500000$ if and only if $S(n) > 1000000 = 100^3\!.$ Because $S(n)$ is increasing for $n > 2$ , the values $S(100) = 1000000 - 10000 + 200 = 990200$ and $S(101)=1030301 - 10201 + 202 = 1020302$ show that the requested value of $n$ is $\boxed{101}$
| 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 difference of 2, and the negative solutions cancel out all but $n$ of the $1+\frac{n(n-1)}{2}$ solutions. Thus, the sum of the solutions is $n \cdot [1+\frac{n(n-1)}{2}]$ , which is a cubic function.
$n \cdot [1+\frac{n(n-1)}{2}]>500,000$
Multiplying both sides by $2$
$n \cdot [2+n(n-1)]>1,000,000$
1 million is $10^6=100^3$ , so the solution should be close to $100$
100 is slightly too small, so $\boxed{101}$ works.
| 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(\frac{(m-5)(m-4)}{2}\right)^2&\equiv 17 \mod m \\ \left(\dfrac{m(m-9)+20}2\right)^2&\equiv 17\mod m \\ \left(\dfrac{20}2\right)^2&\equiv 17\mod m \\ \frac{400}{4}&\equiv 17 \mod m \\ 332 &\equiv 0 \mod m \\ \end{align*} So, $m\in\{83,166,332\}$ . Testing the cases, only $332$ fails. This leaves $78+161=\boxed{239}$
| 239
|
5,529
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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, \[n^2(n+1)^2 \equiv 68\pmod {n + 5}\] \[n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}\] Expanding the left hand side, \[n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}\] This means that $n^4 + 2 n^3 + n^2 - 68$ is divisible by ${n + 5}$
\[(n + 5) | (n^4 + 2 n^3 + n^2 - 68)\] Dividing polynomials, \[\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}\] \[= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}\] $(n + 5)$ $|$ $(n^4 + 2 n^3 + n^2 - 68)$ $\iff$ $\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$ $\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$ $\iff$ $(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332$ Note that $n$ $\in$ $\mathbb{N}$ and $n + 5 > 17$ (because the remainder when dividing by $n + 5$ is $17$ , so $n + 5$ must be greater than $17$ ), so all options $\leq 17$ can be eliminated. \[(n + 5) = 83, 166, 332\] \[n = 78, 161, 327\] Checking all 3 cases, $n = 78$ and $n = 161$ work; $n = 327$ fails. Therefore, the answer is $78 + 161 = \boxed{239} ~
| 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 and cancel. We thus get $1^2+2^3+3^3+4^3 = 100 \equiv 17 \pmod {n+5},$ or $(n+5) | 83,$ which yields $n=78.$
If $n$ is odd, then $1^3+2^3+\cdots + n^3 \equiv 1^3+2^3+3^3+4^3+\left( \frac{n+5}{2} \right)^3 \equiv 17 \pmod {n+5}.$ Letting $k = \frac{n+5}{2}$ yields $k^2 + 83 \equiv 0 \pmod {2k}.$ However, this means that $83$ is divisible by $k,$ so $k=1,83.$ Plugging this back into $n$ yields $n=2(83)-5 = 161$ in the latter case.
Thus, the sum of all possible $n$ is just $78+161 = \boxed{239}.$
| 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 the top row, and square $2000$ is in the bottom row. Find the number of ordered pairs $(m,n)$ of odd integers greater than $1$ with the property that, in the $m\times n$ rectangle, the line through the centers of squares $200$ and $2000$ intersects the interior of square $1099$
|
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 $m=3$ $n$ can range from $667$ to $999$ . However, $900$ divides $1800$ , so looking at mods, we can easily eliminate $899$ and $901$ . Now, counting these odd integers, we get $167 - 2 = 165$
Similarly, let $m=5$ . Then $n$ can range from $401$ to $499$ . However, $450|1800$ , so one can remove $449$ and $451$ . Counting odd integers, we get $50 - 2 = 48$
Take $m=7$ . Then, $n$ can range from $287$ to $333$ . However, $300|1800$ , so one can verify and eliminate $299$ and $301$ . Counting odd integers, we get $24 - 2 = 22$
Let $m = 9$ . Then $n$ can vary from $223$ to $249$ . However, $225|1800$ . Checking that value and the values around it, we can eliminate $225$ . Counting odd integers, we get $14 - 1 = 13$
Add all of our cases to get \[165+48+22+13 = \boxed{248}\]
| 248
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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 $DIS$ , and triangle $CIR$ is congruent to triangle $SIE$ . Then we have $\angle AED=\angle BCD$ $\angle ABC=\angle CDE$ . Extend $CD$ , cross ray $AB$ at $M$ , ray $AE$ at $N$ , then by AAS we have triangle $END$ is congruent to triangle $BMC$ . Thus $\angle M=\angle N$ . Let $EN=MC=a$ , then $BM=DN=a+2$ . So by law of cosine in triangle $END$ and triangle $ANM$ we can obtain \[\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}\] , solved it gives us $a=8$ , which yield triangle $ANM$ to be a triangle with side length 15, 15, 24, draw a height from $A$ to $NM$ divides it into two triangles with side lengths 9, 12, 15, so the area of triangle $ANM$ is 108. Triangle $END$ is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is $108-48=\boxed{60}$
| 60
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5,533
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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 $e + a = 7$ . Then $2a + 2b + 2c + 2d + 2e = 30$ , so $a + b + c + d + e= 15$ . Then we can solve for each individually. $a = 3$ $b = 2$ $c = 4$ $d = 2$ , and $e = 4$ . To find the radius, we notice that $4 \arctan(\frac{2}{r}) + 4 \arctan(\frac{4}{r}) + 2 \arctan (\frac{3}{r}) = 360 ^ \circ$ , or $2 \arctan(\frac{2}{r}) + 2 \arctan(\frac{4}{r}) + \arctan (\frac{3}{r}) = 180 ^ \circ$ . Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \cdot (r + 4i)^2 \cdot (r + 3i)$ is real. Expanding, we get:
\[(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)\]
\[(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)\]
On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0.
\[15ir^4 - 252ir^2 + 192i = 0\]
\[5r^4 - 84r^2 + 64 = 0\]
\[(5r^2 - 4)(r^2 - 16) = 0\]
$r$ must equal 4, as r cannot be negative or be approximately equal to 1.
Thus, the area of $ABCDE$ is $4 \cdot (a + b + c + d + e) = 4 \cdot 15 = \boxed{60}$
| 60
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5,534
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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 greater than $2\sqrt{2}$ . We will approximate that to be $2.9$ . The area is now roughly $\frac{180}{2.9}$ , but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$ , we can safely say that the answer is most likely $\boxed{60}$
| 60
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5,535
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https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13
| 4
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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 $a+b=5$ $b+c= c+d=d+e=6$ , and $e+a=7$ , with a total of $a+b+c+d+e=15$ . Hence $a=3$ $b=d=2$ , and $c=e=4$ . It follows that $\angle B= \angle D$ and $\angle C= \angle E$ . Let $Q$ be the point where $\omega$ is tangent to $\overline{CD}$ . Then $\angle IAE = \angle IAB =\frac{1}{2}\angle A$ . Now we claim that points $A, I, Q$ are collinear, which can be proved if $\angle{AIQ}=\angle{QIA}=180^{\circ}$ . The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA$ , which implies that $\angle AIQ$ must be $180^\circ$ . Therefore points $A$ $I$ , and $Q$ are collinear. [asy] defaultpen(fontsize(8pt)); unitsize(0.025cm); pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; string[] labels = {"$A$", "$B$", "$C$", "$D$", "$E$"}; pair[] dirs = {SW, SE,E, N, NW}; string[] smallLabels = {"$a$", "$b$", "$c$", "$d$", "$e$"}; pair I = (3,4); real rad = 4; pair Q = foot(I, vertices[2], vertices[3]); pair[] interpoints = {}; for(int i =0; i<vertices.length; ++i){ interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length])); } for(int i = 0; i< vertices.length; ++i){ draw(vertices[i]--vertices[(i+1)%vertices.length]); dot(labels[i],vertices[i],dirs[i]); draw(I--vertices[i]); } draw(Circle(I, rad)); dot("$I$", I, dir(200)); draw(I--Q); dot("$Q$", Q, NE); for(int i = 0; i < vertices.length; ++i){ label(smallLabels[i], vertices[i] --interpoints[i]); //dot(interpoints[i], blue); label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]); } [/asy] Because $\overline{AQ} \perp \overline{CD}$ , it follows that \[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\] Another expression for $AC^2-AD^2$ can be found as follows. Note that $\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}$ and $\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}$ , so \[\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}\] and \[\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.\] Applying the Law of Cosines to $\triangle ABC$ and $\triangle AED$ gives \[AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}\] and \[AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.\] Hence
\[12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},\] yielding \[2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;\] equivalently \[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\] Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$ , with solutions $\frac{42 - 38}{5}=\frac45$ , and $\frac{42 + 38}{5}= 16$ . The solution $r^2=\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$ , then $\tan \left(\frac{\angle A}{2}\right)$ $\tan \left(\frac{\angle C}{2}\right)$ , and $\tan \left(\frac{\angle E}{2}\right)$ are less than $1,$ implying that $\angle A$ $\angle C$ , and $\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$ . The requested area is $15\cdot4 = \boxed{60}$
| 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 $f(f(f(x)))=17$ for $0\leq x\leq 2020$ . Find the remainder when $N$ is divided by $1000$
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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 the interval $[a, a+1)$ The graph starts at $0$ , is continuous and increasing, and approaches $a+1$ . So as long as $a+1 > 17$ , there will be a solution for $z$ in the interval.
Using this logic, we can find the number of solutions to $f(x) = y$ . For every interval $[a, a+1)$ where $a \geq \left \lfloor{y}\right \rfloor$ there will be one solution for $x$ in that interval. However, the question states $0 \leq x \leq 2020$ , but because $x=2020$ doesn't work we can change it to $0 \leq x < 2020$ . Therefore, $\left \lfloor{y}\right \rfloor \leq a \leq 2019$ , and there are $2020 - \left \lfloor{y}\right \rfloor$ solutions to $f(x) = y$
We can solve $f(y) = z$ similarly. $0 \leq y < 2020$ to satisfy the bounds of $x$ , so there are $2020 - \left \lfloor{z}\right \rfloor$ solutions to $f(y) = z$ , and $0 \leq z < 2020$ to satisfy the bounds of $y$
Going back to $f(z) = 17$ , there is a single solution for z in the interval $[a, a+1)$ , where $17 \leq a \leq 2019$ . (We now have an upper bound for $a$ because we know $z < 2020$ .) There are $2003$ solutions for $z$ , and the floors of these solutions create the sequence $17, 18, 19, ..., 2018, 2019$
Lets first look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$ . Then $f(y) = z$ would have $2003$ solutions, and the floors of these solutions would also create the sequence $17, 18, 19, ..., 2018, 2019$
If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 17$ , there would be $2003$ solutions for $f(x) = y$ . If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 18$ , there would be $2002$ solutions for $x$ , and so on. So for the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$ , there will be $2003 + 2002 + 2001 + ... + 2 + 1 = \binom{2004}{2}$ solutions for $x$
If we now look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 18$ , there would be $\binom{2003}{2}$ solutions for $x$ . If we looked at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 19$ , there would be $\binom{2002}{2}$ solutions for $x$ , and so on.
The total number of solutions to $x$ is $\binom{2004}{2} + \binom{2003}{2} + \binom{2002}{2} + ... + \binom{3}{2} + \binom{2}{2}$ . Using the hockey stick theorem, we see this equals $\binom{2005}{3}$ , and when we take the remainder of that number when divided by $1000$ , we get the answer, $\boxed{10}$
| 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 in $\triangle TXY$ gives \[1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.\]
Since $\triangle BMT \cong \triangle CMT$ , we have $TM\perp BC$ , and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$ ) be the midpoint of $BT$ (resp. $CT$ ). So $P$ (resp. $Q$ ) is the center of $(BXTM)$ (resp. $CYTM$ ). Then $\theta=\angle ABC=\angle MTX$ and $\phi=\angle ACB=\angle YTM$ . So $\angle XPM=2\theta$ , so \[\frac{\frac{XM}{2}}{XP}=\sin \theta,\] which yields $XM=2XP\sin \theta=BT(=CT)\sin \theta=TY$ . Similarly we have $YM=XT$
Ptolemy's theorem in $BXTM$ gives \[16TY=11TX+3\sqrt{15}BX,\] while Pythagoras' theorem gives $BX^2+XT^2=16^2$ . Similarly, Ptolemy's theorem in $YTMC$ gives \[16TX=11TY+3\sqrt{15}CY\] while Pythagoras' theorem in $\triangle CYT$ gives $CY^2+YT^2=16^2$ . Solve this for $XT$ and $TY$ and substitute into the equation about $\cos XTY$ to obtain the result $XY^2=\boxed{717}$
| 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 \[\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,\] implying that $\overline{MX} \perp \overline{AC}$ . Similarly, $\overline{MY} \perp \overline{AB}$ . In particular, $MXTY$ is a parallelogram. [asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label("$\omega$", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot("$X$", X, W); dot("$Y$", Y, E); dot("$O$", O, W); dot("$T$", T, S); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, E); dot("$M$", M, N); [/asy] Hence, by the Parallelogram Law, \[TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).\] But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$ . Therefore \[XY^2 = \frac13(2 \cdot 1143-135) = \boxed{717}.\]
| 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 $\angle H'XB=\angle H'YC=90^\circ-\angle A$ $\angle CH'Y=\angle EH'B=90^\circ-\angle B$ , and $\angle BH'Y=\angle FH'C=90^\circ-\angle C$ . Thus $YH'\perp AX$ and $XH' \perp AY$ $H'$ is indeed the same as $H$ , and we have proved lemma 1.
Since $AXTY$ is cyclical, $\angle XTY=\angle XHY$ and this implies that $XHYT$ is a paralelogram.
By the Law of Cosines: \[XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)\] \[XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)\] \[HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)\] \[HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).\] We add all these equations to get: \[HT^2+XY^2=2(XT^2+TY^2) \qquad (1).\] We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \perp BC$ , so by the Pythagorean Theorem, it follows that $HT^2=135$ . We were also given that $XT^2+TY^2=1143-XY^2$ , which we multiply by $2$ to use equation $(1)$ \[2(XT^2+TY^2)=2286-2 \cdot XY^2\] Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$ , we have \[135+XY^2=2286-2 \cdot XY^2\] \[3 \cdot XY^2=2151.\] Therefore, $XY^2=\boxed{717}$ . ~ MathLuis
| 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 for median $DT$ of triangle $XYT$ is \[2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},\] \[3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,\] \[3 \cdot XY^2 = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}.\]
| 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. There are currently $321$ 1's. We know the last four digits are $1110$ , and that the others will not be affected if we subtract $321$ . If we do so, we get that $1110-321=789$ . This method will remove three $1$ 's, and add a $7$ $8$ and $9$ . Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$
| 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 \[\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cdot 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.\] Then the area of $\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}$ . Preceding in a similar fashion for $\triangle PAF$ , the area of $\triangle PAF$ is $10$ . Since $\angle ERD = 90^{\circ}$ , the area of $\triangle RED=\frac{25}{2}$ . Thus our desired answer is $150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}$
| 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 triangle. To the find the areas of $\triangle{APF}$ and $\triangle{CQB}$ , we can use Law of Cosines ( $c^2 = a^2 + b^2 - 2ab\cos C$ ) to find the lengths of $AF$ and $CB$ , respectively. Computing gives $AF = \sqrt{20}$ and $CB = \sqrt{10}$ . Now, using Heron's Formula, we find $\triangle{APF} = 10$ and $\triangle{CQB} = 7.5$ . Adding these and subtracting from $\triangle{PQR}$ , we get $150 - (10 + 7.5 + 12.5) = \boxed{120}$ -Starsher
| 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 may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let $n$ be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when $n$ is divided by $1000$
|
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 ( $11$ ). The formula for $n$ subs is then $a_n=11(12-n)a_{n-1}$ with $a_0=1$
Summing from $0$ to $3$ gives $1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9$ . Notice that $10+9\cdot11\cdot10=10+990=1000$ . Then, rearrange it into $1+11^2+11^3\cdot (10+11\cdot10\cdot9)= 1+11^2+11^3\cdot (1000)$ . When taking modulo $1000$ , the last term goes away. What is left is $1+11^2=\boxed{122}$
| 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 probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$
|
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 recursively find $P(4,4) = \frac{245}{2187}$ so the answer is $245 + 7 = \boxed{252}$
| 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 probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$
|
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 gives a probability of $1 \cdot \frac{1}{27} + \frac{4!}{2!} \cdot \frac{1}{81} + \frac{5!}{2! \cdot 2!} \cdot \frac{1}{243} +\frac{6!}{3! \cdot 3!} \cdot \frac{1}{729}=\frac{245}{729}$ to get to $(1,1)$ . The probability of reaching $(0,0)$ is $\frac{245}{3^7}$ . This gives $245+7=\boxed{252}$
| 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 probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$
|
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
(3) 1 move diagonally, 2 moves left and 2 moves down.
(4) 3 moves left, 3 moves down.
The probability of (1) is $\frac{1}{3^3}$ . The probability of (2) is $\frac{\frac{4!}{2!}}{3^4} = \frac{12}{3^4}$ . The probability of (3) is $\frac{\frac{5!}{2!2!}}{3^5} = \frac{30}{3^5}$ . The probability of (4) is $\frac{\frac{6!}{3!3!}}{3^6} = \frac{20}{3^6}$ . Adding all of these together, we obtain a total probability of $\frac{245}{3^6}$ that our particle will hit $(1,1)$ . Trivially, there is a $\frac{1}{3}$ chance our particle will hit $(0,0)$ from $(1,1)$ . So our final probability will be $\frac{245}{3^6} \cdot \frac{1}{3} = \frac{245}{3^7} \implies m = 245, n = 7 \implies \boxed{252}$
| 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, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy] Note that since $\angle KLN = \angle KMN$ , quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\] so $\triangle KLO \sim \triangle KML$ , giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$
| 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, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]
By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$ . Thus, $LN^2 = KM^2 + 65^2 - 28^2$
By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$ , and $PN^2 + LP^2 = LN^2$
\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]
It follows that \[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]
\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2\]
Since $KP^2 + LP^2 = 28^2$ \[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2\]
\[-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2\]
\[KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}\]
$\angle OKP = \angle NKM$ (it's the same angle) and $\angle OPK = \angle KMN = 90^{\circ}$ . Thus, $\triangle KOP \sim \triangle KNM$
Thus,
\[\frac{KO}{KN} = \frac{KP}{KM}\]
\[\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}\]
Multiplying both sides by $\sqrt{KM^2 + 65^2}$
\[8 = \frac{28^2}{KM}\]
\[KM = 98\]
Therefore, $OM = 98-8 = \boxed{90}$
| 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 $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$
|
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 previously mentioned theorem, we can get the equation $3\log(xy)=630$ .
This can easily be simplified to $\log(xy)=210$ , or $xy = 10^{210}$
$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$ , and $m+n$ equals to the sum of the exponents of $2$ and $5$ , which is $210+210 = 420$ .
Multiply by two to get $2m +2n$ , which is $840$ .
Then, use the first equation ( $\log x + 2\log(\gcd(x,y)) = 60$ ) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$ , which is a contradiction to the restrains we set before). Therefore, $\gcd(x,y)=x$ . Then, turn the equation into $3\log x = 60$ , which yields $\log x = 20$ , or $x = 10^{20}$ .
Factor this into $2^{20} \cdot 5^{20}$ , and add the two 20's, resulting in $m$ , which is $40$ .
Add $m$ to $2m + 2n$ (which is $840$ ) to get $40+840 = \boxed{880}$
| 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 $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$
|
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 this means that \[x^3y^3=10^{630}\] so \[xy=10^{210}.\]
Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x<y$ and $x$ and $y$ are both powers of $10$ . This means the first equation would simplify to \[x^3=10^{60}\] and \[y^3=10^{570}.\] Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so \[3\cdot 40 + 2\cdot 380 = \boxed{880}.\]
| 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 $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$
|
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 $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$
|
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, $x=10^{20}, y=10^{190}$ giving the answer $\boxed{880}$
| 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 $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$
|
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 $a=360$ , contradiction. So $a<105$ . Then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2$ is $a+2a=3a$ , so $a=20$ . Similarly, $b=20$ . Then $3m+2n=3\left(a+b\right)+2\left(420-a-b\right)=\boxed{880}$ as desired.
| 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 values by hand.
Having computed the working possibilities, we take the sum of the corresponding values of $n$ $8+9+16+25+121+361 = \boxed{540}$ . ~Kepy.
| 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 of $(z-9)^2$ is $100$ . Since $\text{Im}(z-9)=\text{Im}(z)=11$ , let $z-9=a+11i$ . Then, \[100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.\] It follows that $z=9+\sqrt{221}+11i$ , and the requested sum is $9+221=\boxed{230}$
| 230
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5,558
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12
| 2
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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 $m+n$
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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 between their arguments will be 90 degrees, so subtracting the angles will yield an imaginary number with no real part(an argument of 90 degrees puts a complex number on the imaginary axis).
Now to apply this: \[\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}\] \[\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}\] \[\frac{z^2-20z}{z^4-38z^3+341z^2+380z}\] \[\frac{z(z-20)}{z(z+1)(z-19)(z-20)}\] \[\frac{1}{(z+1)(z-19)}\in i\mathbb{R}\]
The factorization of the nasty denominator above is made easier with the intuition that $(z-20)$ must be a divisor for the problem to lead anywhere. Now we know $(z+1)(z-19)\in i\mathbb{R}$ so using the fact that the imaginary part of $z$ is $11i$ and calling the real part r,
\[(r+1+11i)(r-19+11i)\in i\mathbb{R}\] \[r^2-18r-140=0\]
solving the above quadratic yields $r=9+\sqrt{221}$ so our answer is $9+221=\boxed{230}$
| 230
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5,559
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12
| 4
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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 $m+n$
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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^2-19z)^2-19(z^2-19z)-(z^2-19z)}{(z^2-19z)-z}\\ &=\frac{(z^2-19z)(z^2-19z-19-1)}{z^2-20z}\\ &=\frac{z(z-19)(z+1)(z-20)}{z(z-20)} \\ &=z^2-18z-19 \end{align*} must have real part equal to $0.$ If $z=x+11i,$ the real part of $z^2-18z-19$ is $x^2-11^2-18x-19,$ which is $0$ when $x=9\pm\sqrt{221}.$ The requested sum is $9+221=\boxed{230}.$
| 230
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5,560
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12
| 5
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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 $m+n$
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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 angle between $(a+bi), 0,$ and $(a+bi)(c+di)$ is the same as the angle between $1, 0,$ and $c+di$ .
The proof of this is as follows:
Treating the complex numbers like vectors (e.g. $a+bi$ turns into $\left[\begin{array}{c} a \\ b \end{array}\right]$ ) we have $cos \theta = \frac{a \cdot (ac-bd) + b \cdot (ad+bc)}{2\sqrt{a^2+b^2}\sqrt{a^2+b^2}\sqrt{c^2+d^2}}$
$cos \theta = \frac{a^2 c - abd + abd + b^2 c}{2(a^2+b^2)\sqrt{c^2+d^2}}$
$cos \theta = \frac{(a^2+b^2)c}{2(a^2+b^2)\sqrt{c^2+d^2}}=\frac{c}{2\sqrt{c^2+d^2}}$
Using the dot product formula for the cosine of the angle between the other two vectors (1, c + di) we get the same result. Thus, it is proven.
Now, we see that the three resulting complex numbers have the same angle as $-1,0,$ and $(z-19)(z+1)$ as a factor of $z(z-20)$ can be taken out of both expressions. Note that the value of $z$ clearly is not 0, nor is it 20 as it has a value of $11i$
Expanding $z^2-18z-19$ with $z=a+11i$ yields
$a^2-121+22ai-18a-198i-19=a^2-18a-140+22ai-198i$ . Turning these into vectors again, the resulting relation is
$cos(90^\circ)=0=\frac{-(a^2-18a-140)}{2\sqrt{(a^2-18a-140)^2 + (22a-198)^2} \cdot 1}$
In order for this thing to be 0, the numerator must be 0, so setting it equal to 0 yields
$a^2-18a-140=0 \Rightarrow a=9\pm \sqrt{81+140}$ . Neither of these make the denominator in the $cos(90^\circ)$ expression 0, so they are valid. The requested result is the positive root, so the answer is $\boxed{230}$
| 230
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5,561
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15
| 9
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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 $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
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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 $\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}$ , which means $OO_1PO_2$ is a parallelogram. So, $O_2P = OO_1$ . It is also clear that $PP'O_1O_2$ is a parallelogram by virtue of our definition. Thus, $O_2P = O_1P' = OO_1$ . Since $OQ' \parallel O_1O_2 \parallel P'P$ (because of the right angles), $\frac{Q'M}{MP} = \frac{OO_1}{O_1P'} = 1 \implies Q'M = MP$ . And, because $QM = MP$ $Q = Q'$ . From Power of a Point on $P$ , we have $XP(11-XP) = 15$ , giving us $XP = \frac{11 - \sqrt{61}}{2}$ . Since $OQ$ is perpendicular to $XY$ $Q$ is the midpoint of $XY$ , so $XQ = \frac{11}{2}$ . Thus, $PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies {PQ}^2 = \frac{61}{4}$ . Therefore, our answer is $\boxed{65}$
| 65
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5,562
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1
| 6
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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 $m+n$
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[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)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); label("$O$",O,dir(90)); label("$K$",K,dir(-120)); draw(A--K,dotted); draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); label("$E$",E,dir(-135)); [/asy]
Since $\triangle ABC \cong \triangle BAD$ $\angle ADB = \angle BCA$ . Thus, $A$ $B$ $C$ $D$ are concyclic.
By Ptolemy's Theorem on $ABCD$ \[(AD)(BC) + (AB)(DC) = (BD)(AC)\] \[10^2 + 9(DC) = 17^2\] \[DC = 21\]
The altitudes dropped from $C$ and $D$ onto the extension of AB are equal, meaning that $DC \parallel AB$ . Therefore, $\triangle DCO \sim \triangle ABO$ . It follows that \[\frac{OB}{17 - OB} = \frac{AB}{DC} = \frac{9}{21} = \frac{3}{7}\] Solving yields $OB = \frac{51}{10}$
In $\triangle ABD$ , drop an altitude from $A$ to $BD$ . Call the intersection of this altitude and $BD$ $K$
The area of $\triangle ABD$ is $\frac{1}{2}(AB)(DE) = 36$ . Thus, $\frac{1}{2}(AK)(BD) = 36$ , and $AK = \frac{72}{17}$
Therefore, the area of $\triangle AOB$ is $\frac{1}{2}(OB)(AK) = \frac{1}{2}(\frac{51}{10})(\frac{72}{17}) = \frac{54}{5}$
The requested answer is $54 + 5 = \boxed{59}$
| 59
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5,563
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2
| 1
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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 $p$ and $q$ are relatively prime positive integers. Find $p+q$
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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 \[P_5 = \frac{3}{4}\] \[P_4 = \frac{5}{8}\] \[P_3 = \frac{11}{16}\] \[P_2 = \frac{21}{32}\] \[P_1 = \frac{43}{64}\] $43 + 64 = \boxed{107}$
| 107
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5,564
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2
| 2
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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 $p$ and $q$ are relatively prime positive integers. Find $p+q$
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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) $\binom{4}{2} \cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$
Case 4: (3 two jumps) $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$
Summing the probabilities gives us $\frac{43}{64}$ so the answer is $\boxed{107}$
| 107
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5,565
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2
| 3
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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 $p$ and $q$ are relatively prime positive integers. Find $p+q$
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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} \cdot 1=\dfrac{1}{2}\\ P_3&=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}\\ P_4&=\dfrac{5}{8}\\ P_5&=\dfrac{11}{16}\\ P_6&=\dfrac{21}{32}\\ P_7&=\dfrac{43}{64}\\ \end{align*} We calculate $P_7$ to be $\frac{43}{64}$ , meaning that our answer is $\boxed{107}$
| 107
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5,566
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2
| 4
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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 $p$ and $q$ are relatively prime positive integers. Find $p+q$
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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 $n-1$ , the recursion is just $P_n = \frac{P_{n-2}+P_{n-1}}{2}$ . Using the fact that $P_1 = 1$ , and $P_2 = \frac{1}{2}$ , we find that $P_7 = \frac{43}{64}$ $43 + 64 = \boxed{107}$
| 107
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5,567
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4
| 1
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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 $m+n$
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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.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$ , let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.
To find $a_{n+1}$ given $a_n$ (where $n \ge 1$ ), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ( $5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$
Computing $a_2$ $a_3$ $a_4$ gives $a_2 = 7$ $a_3 = 32$ , and $a_4 = 157$ . Thus for Case 3, there are 157 outcomes. For case 2, we multiply $a_2$ by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is
\[\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187}\]
| 187
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5,568
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4
| 2
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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 $m+n$
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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 of rolling 4 duds: $\left(\frac{1}{3}\right)^4$
Probability of rolling 3 duds: $4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3}$
Probability of rolling 2 duds: $6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2$
Probability of rolling 1 dud: $4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3$
Probability of rolling 0 duds: $\left(\frac{2}{3}\right)^4$
Now we will find the probability of a square product given we have rolled each amount of duds
Probability of getting a square product given 4 duds: 1
Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)
Probability of getting a square product given 2 duds: $\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 dud: $\frac{3!}{4^3}$ $\frac{3}{32}$ (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: $\frac{40}{4^4}$ $\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).
We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values. \[\left(\frac{1}{3}\right)^4 * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}.\]
$25+162$ $\boxed{187}$
| 187
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5,569
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4
| 3
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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 $2^4=16$ combinations.
If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square.
If there are two 1/4's, there are $2^2=4$ ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and $\frac{4!}{2!2!}=6$ ways to arrange, so there are $4\cdot 4\cdot 6=96$ combinations for this case.
If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\cdot 4!=48$ combinations for this case.
Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\binom{4}{2}$ to choose the numbers and $\frac{4!}{2!2!}=6$ ways to arrange them, so $6\cdot 6=36$ . If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case.
Hence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\frac{200}{1296}=\frac{25}{162}$ , yielding an answer of $25+162=\boxed{187}$
| 187
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5,570
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4
| 5
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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 $m+n$
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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 number of ways to have a product after $n$ rolls where $b$ is odd, and $a$ $c$ are even
$C_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is odd, and $a$ $b$ are even
$D_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is even, and $a$ $b$ are odd
$E_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is even, and $a$ $c$ are odd
$F_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is even, and $b$ $c$ are odd
$G_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all odd
$S_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all even (square!)
We have the following equations after considering the possible values of the nth roll:
\[A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}\]
\[B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}\]
\[C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}\]
\[D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}\]
\[E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}\]
\[F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}\]
\[G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}\]
\[S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}\]
We have the following values after considering the possible values of the 1st roll:
\[A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2\]
After applying recursion twice, we get:
\[A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8\]
\[A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38\]
Finally, we have $S_4 = 200$ $\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}$ meaning our answer is $\boxed{187}$
| 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 broken up into $\{3,3\}$ and $\{4\}$
$1$ and $4$ are already perfect squares, so they each form a "fundamental" group and cannot belong in any other group. Pairs of the other $4$ numbers ( $\{2,2\}$ $\{3,3\}$ , etc. ) form "fundamental" groups as well. The last "fundamental" group is $\{2,3,6\}$ . It can be easily seen that no more groups exist.
Thus, we have the "fundamental" groups $\{1\}$ $\{4\}$ $\{2,2\}$ $\{3,3\}$ $\{5,5\}$ $\{6,6\}$ , and $\{2,3,6\}$
We now consider the ways to use these groups to form a sequence of $4$ numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size $2$ or $3$ and fill in any remaining spots with $1$ s and $4$ s. We can do this in one of $5$ ways: Using only $1$ s and $4$ s, using one group of size $2$ , using one group of size $3$ , using two different groups of size $2$ , and using the same group of size $2$ twice.
If we only use $1$ s and $4$ s, each of the $4$ slots can be filled with one of the $2$ numbers, so there are $2^4=16$ possibilities.
If we use one group of size $2$ , there are $4$ options for the group to use, $\binom{4}{2}$ ways to place the two numbers (since they are identical), and $2^2$ ways to fill in the remaining slots with $1$ s and $4$ s, so there are $4\cdot\binom{4}{2}\cdot2^2=96$ possibilities.
If we use one group of size $3$ , there is only $1$ option for the group to use, $4\cdot3\cdot2$ ways to place the three numbers (since they are distinct), and $2$ ways to fill in the remaining slot, so there are $4\cdot3\cdot2\cdot2=48$ possibilities.
If we use two different groups of size $2$ , there are $\binom{4}{2}$ options for the groups to use and $\binom{4}{2}$ ways to place the four numbers (since there are $2$ groups of identical numbers, and one group's placement uniquely determines the other's), so there are $\binom{4}{2}\cdot\binom{4}{2}=36$ possibilities.
If we use the same group of size $2$ twice, there are $4$ options for the group to use and $1$ way to place the four numbers (since they are all identical), so there are $4=4$ possibilities.
This gives us a total of $16+96+48+36+4=200$ possibilities, and since there are $6^4=1296$ total sequences that can be rolled, the probability is equal to $\frac{200}{1296}=\frac{25}{162}$ , so the answer is $25+162=\boxed{187}$ . ~ emerald_block
| 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 these cases, we quickly see a pattern:
$(1,1,2,2)$
$(1,1,3,3)$
$...$
$(1,1,6,6)$
$(2,2,3,3)$
$...$
$(2,2,6,6)$
$...$
$(5,5,6,6)$
There are a total of $5+4+3+2+1=15$ cases. Multiplying this by $6$ yields $15\cdot 6=90$ ways.
Case 3 : Sets of numbers in the form of $(x,x,1,4)$
A special case must be made for the number $4$ because $4$ itself is a perfect square.
$(1,1,1,4)$ $4$
$(2,2,1,4)$ $12$
$(3,3,1,4)$ $12$
$(4,4,1,4)$ $4$
$(5,5,1,4)$ $12$
$(6,6,1,4)$ $12$
Summing these up yields a total of $4+12+12+4+12+12=56$ ways.
Case 4 : Sets with all 4 numbers different
Note that the sets
$(1,2,3,6)$
$(2,3,4,6)$
Multiply to perfect square. The total of these cases are $24+24=48$
Adding all these cases together yields $6+90+56+48=200$ ways that the product of the values of the die can be a perfect square.
Therefore the probability is
$\frac{200}{1296}=\frac{25}{162}$
$m+n = 25+162 = \boxed{187}$
| 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 under these conditions. Find the remainder when $N$ is divided by $1000$
|
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 each other. For each way, there are $4!=24$ ways to assign the $4$ ambassadors to the $4$ selected seats.
In the first way, there are $6$ distinct orientations. The $4$ advisors can be placed in any of the $5$ odd-numbered chairs adjacent to the ambassadors, and for each placement, there is exactly one way to assign them to the ambassadors. This means that there are $24\cdot6\cdot\binom{5}{4}=720$ total seating arrangements for this way.
In the second way, there are $6$ distinct orientations. $3$ advisors can be placed in any of the $4$ chairs adjacent to the "chunk" of $3$ ambassadors, and $1$ advisor can be placed in either of the $2$ chairs adjacent to the "lonely" ambassador. Once again, for each placement, there is exactly one way to assign the advisors to the ambassadors. This means that there are $24\cdot6\cdot\binom{4}{3}\cdot\binom{2}{1}=1152$ total seating arrangements for this way.
In the third way, there are $3$ distinct orientations. For both "chunks" of $2$ ambassadors, $2$ advisors can be placed in any of the $3$ chairs adjacent to them, and once again, there is exactly one way to assign them for each placement. This means that there are $24\cdot3\cdot\binom{3}{2}\cdot\binom{3}{2}=648$ total seating arrangements for this way.
Totalling up the arrangements, there are $720+1152+648=2520$ total ways to seat the people, and the remainder when $2520$ is divided by $1000$ is $\boxed{520}$ . ~ emerald_block
| 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 under these conditions. Find the remainder when $N$ is divided by $1000$
|
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 seat 2 while assistant 1 will sit in seat 3. No two teams can pick adjacent gaps. For example, if team 1 chooses gap C then team 2 cannot pick gaps B or D. In the diagram, the teams have picked gaps C, F, H and J. Note that the gap-gaps - distances between the chosen gaps - (in the diagram, 2, 1, 1, 4) must sum to 8. So, to get the number of seatings, we:
So, the total is $12\cdot35\cdot6=2520$
Therefore, the remainder is $\boxed{520}$
| 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 under these conditions. Find the remainder when $N$ is divided by $1000$
|
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 empty spaces, which results in $\frac{8!}{4!4!} = 70$ ways to arrange the pairs.
Finally, we have to assign ambassadors and their advisors to the pairs, which has $4!=24$ ways, so $N = (35+70)\cdot 24 = 2520$ , or $\boxed{520}$ ~PenguinMoosey
| 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, \[3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56\] \[3\log_{b}{(\log_{b}{x})^{28}} = 56\] \[\log_{b}{(\log_{b}{x})^{84}} = 56\]
We can manipulate this equation to be able to substitute $x = (\log_{b}{x})^{54}$ a couple more times: \[\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}\] \[\log_{b}{x} = 36\] \[(\log_{b}{x})^{54} = 36^{54}\] \[x = 6^{108}\]
However, since we found that $\log_{b}{x} = 36$ $x$ is also equal to $b^{36}$ . Equating these, \[b^{36} = 6^{108}\] \[b = 6^3 = \boxed{216}\]
| 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 \[\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] \[\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))\] \[x=\log_b^{54}x\] Substituting this into the first equation gives \[\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}\] \[x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}\] Plugging this into $x=\log_b^{54}x$ gives \[b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}\] \[b^{\frac{2}{3}}=36\] \[b=36^{\frac{3}{2}}=6^3=\boxed{216}\] -ktong
| 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: \[3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}\] Substituting \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] into the equation $\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$ yields: \[\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}\] So \[\log_b(x)=36.\] Substituting this back in to \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] yields \[\frac{36}{54}=\log_b(36).\] So, \[b^{\frac{2}{3}}=36\Rightarrow \boxed{216}\]
| 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 \[\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}\] \[\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}\] This means that $\frac{14}{27}a=\frac{56}{3}$ , so \[a=36\] \[b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}\]
| 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 the first equation to see what we get: $\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3$ , and simplifying, $\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}$ , so $b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}$
Combining this new finding with what we had above $a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}$
Now that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get $\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies$ $\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36$
Finally, that gives us that $\frac{b^{\frac{56}3}}{b^{18}}=36\implies b^{\frac{56}{3}-18}=b^{\frac{56}{3}-\frac{54}{3}}=b^{\frac 23}=36\implies b=36^{\frac 32}=6^3$ . Thus, $b=\boxed{216}$
| 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 $n = \log_{b} \log_{b} x$ and we have
\[\frac{3}{2} m + 3n = 56\] \[\frac{m}{n} = 54\]
Solving this yields $m = 36$ $n = \frac{2}{3}$
Now, $n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36$ , so we have $\log_{b} 36 = \frac{2}{3}$ . Solving this yields $b = \boxed{216}$
| 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 \[b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.\] Because $b^{36}=x,$ $\log_b x=36$ by the definition of a logarithm.
Substituting this into the second equation, \[\log_{36} x=54 \implies x=36^{54}.\] Finally, \[b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.\]
| 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 $55,45$ , and $15$ , respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$ , and $\ell_C$
|
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 perimeter is $XE+EF+FY+YG+GH+HZ+ZI+ID+DX=(DX+XE)+(FY+YG)+(HZ+ZI)+115$ since $HG=55$ $EF=15$ , and $ID=45$
Note that $\triangle AHG\sim \triangle BID\sim \triangle EFC\sim \triangle ABC$ , so using similar triangle ratios, we find that $BI=HA=30$ $BD=HG=55$ $FC=\frac{45}{2}$ , and $EC=\frac{55}{2}$
We also notice that $\triangle EFC\sim \triangle YFG\sim \triangle EXD$ and $\triangle BID\sim \triangle HIZ$ . Using similar triangles, we get that \[FY+YG=\frac{GF}{FC}\cdot \left(EF+EC\right)=\frac{225}{45}\cdot \left(15+\frac{55}{2}\right)=\frac{425}{2}\] \[DX+XE=\frac{DE}{EC}\cdot \left(EF+FC\right)=\frac{275}{55}\cdot \left(15+\frac{45}{2}\right)=\frac{375}{2}\] \[HZ+ZI=\frac{IH}{BI}\cdot \left(ID+BD\right)=2\cdot \left(45+55\right)=200\] Hence, the desired perimeter is $200+\frac{425+375}{2}+115=600+115=\boxed{715}$ -ktong
| 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 $55,45$ , and $15$ , respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$ , and $\ell_C$
|
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 $\bigtriangleup IHZ$ from $Z$ is half the altitude of $\bigtriangleup ABC$ from $C$ , say $\frac{h}{2}$ . Also since $\frac{EF}{AB}=\frac{1}{8}$ , the distance from $\ell_C$ to $AB$ is $\frac{7}{8}h$ . Therefore the altitude of $\bigtriangleup XYZ$ from $Z$ is \[\frac{1}{2}h+\frac{7}{8}h=\frac{11}{8}h\]
By triangle scaling, the perimeter of $\bigtriangleup XYZ$ is $\frac{11}{8}$ of that of $\bigtriangleup ABC$ , or \[\frac{11}{8}(220+180+120)=\boxed{715}\]
| 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 $55,45$ , and $15$ , respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$ , and $\ell_C$
|
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, \[\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 - k_1 - k_2,\] \[\frac{FY}{BC} = \frac{FE'}{AC} = \frac{AC - AF - CE'}{AC} = 1 - k_1 - k_3.\] \[k = \frac{ZY}{BC} = \frac{ZE + EF + FY}{BC} = (1 - k_1 - k_2) + k_1 + (1 - k_1 - k_3)\] \[k = 2 - k_1 - k_2 - k_3 = 2 - \frac{1}{4} - \frac{1}{4} - \frac{1}{8} = \frac{11}{8}.\] \[\frac{ZY+YX +XZ}{BC +AB + AC} = k \implies ZY + YX + XZ =\frac{11}{8} (220 + 120 + 180) = \boxed{715}.\] vladimir.shelomovskii@gmail.com, vvsss
| 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 are not many cases to check.
If $a+1 = 4$ , then $b+1 = 5$ . But this leads to no solutions, as $(a,b) = (3,4)$ gives $2^3 5^4 > 2019$
If $a+1 = 5$ , then $b+1 = 2$ or $4$ . The first case gives $n = 2^4 \cdot 5^1 \cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23$ . The sum of all such $n$ is $80(3+7+11+13+17+19+23) = 7440$ . In the second case $b+1 = 4$ and $d(k) = 1$ , and there is one solution $n = 2^4 \cdot 5^3 = 2000$
If $a+1 = 10$ , then $b+1 = 2$ , but this gives $2^9 \cdot 5^1 > 2019$ . No other values for $a+1$ work.
Then we have $\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}$
| 472
|
5,587
|
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_9
| 2
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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 only one , and the only integer divisible by $20$ that has the second form is $2^{9}5$ , which is 2560, greater than $2019$
For the third form, the only $20$ -pretty numbers are $2^45^3=2000$ and $2^35^4=5000$ , and only $2000$ is small enough.
For the fourth form, any number of the form $2^45r$ where $r$ is a prime other than $2$ or $5$ will satisfy the $20$ -pretty requirement. Since $n=80r<2019$ $r\le 25$ . Therefore, $r$ can take on $3, 7, 11, 13, 17, 19,$ or $23$
Thus, $\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}$
| 472
|
5,588
|
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10
| 1
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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 integers. Find $p+q$
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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: \[2^0\theta \equiv 2^3\theta \equiv 2^6\theta \equiv ... \pmod{180^{\circ}}.\] (This is because if it isn't the same value, the terminal angle will gradually shift from the first quadrant into different quadrants, making the condition for positive tan untrue. This must also be true in order for $\theta$ to be unique.)
This is the case if $2^3\theta \equiv 2^0\theta \pmod{180^{\circ}}$ , so $7\theta \equiv 0^{\circ} \pmod{180^{\circ}}$ . Therefore, recalling that $0^{\circ}<\theta<90^{\circ},$ the possible $\theta$ are: \[\frac{180}{7}^{\circ}, \frac{360}{7}^{\circ}, \frac{540}{7}^{\circ}.\]
$\frac{180}{7}^{\circ}$ does not work since $\tan\left(2 \cdot \frac{180}{7}^{\circ}\right)$ is positive.
$\frac{360}{7}^{\circ}$ does not work because $\tan\left(4 \cdot \frac{360}{7}^{\circ}\right)$ is positive.
Thus, $\theta = \frac{540}{7}^{\circ}$ , and a quick check verifies that it does work. Our desired answer is $540 + 7 = \boxed{547}$
| 547
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5,589
|
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10
| 2
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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 integers. Find $p+q$
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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 $\theta\in(0,90)$ . We can keep track of the range of $2^n\theta$ for each $n$ by considering the portion in the desired quadrants, which gives \[n=1 \implies (90,180)\] \[n=2\implies (270,360)\] \[n=3 \implies (180,270)\] \[n=4 \implies (90,180)\] \[n=5\implies(270,360)\] \[n=6 \implies (180,270)\] \[\cdots\] at which point we realize a pattern emerging. Specifically, the intervals repeat every $3$ after $n=1$ . We can use these repeating intervals to determine the desired value of $\theta$ since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.
Initially, the lower bound is $0$ (at $n=0$ ), then increases to $\frac{90}{2}=45$ at $n=1$ . This then becomes $45+\frac{45}{2}$ at $n=2$ $45+\frac{45}{2}$ at $n=3$ $45+\frac{45}{2}+\frac{45}{2^3}$ at $n=4$ $45+\frac{45}{2}+\frac{45}{2^3}+\frac{45}{2^4}$ at $n=5$ , etc. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as $n$ approaches infinity, the lower bound converges to \[\sum_{k=0}^{\infty}\left(45+\frac{45}{2}\right)\cdot \left(\frac{1}{8}\right)^k=\frac{45+\frac{45}{2}}{1-\frac{1}{8}}=\frac{\frac{135}{2}}{\frac{7}{8}}=\frac{540}{7}\implies p+q=540+7=\boxed{547}\] -ktong
| 547
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5,590
|
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10
| 3
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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 integers. Find $p+q$
|
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 $\tan\left(8\theta\right) > 0$ $\theta$ has to be in the first half of (67.5, 90). Inductively, the pattern repeats: $\theta$ is in the second half of the second half of the first half of the second half of the second half... of the interval (0, 90). Consider the binary representation of numbers in the interval (0, 1). Numbers in the first half of the interval start with 0.0... and numbers in the second half start with 0.1... . Similarly, numbers in the second half of the second half start with 0.11... etc. So if we want a number in the first half of the second half of the second half... of the interval, we want its binary representation to be $0.11011011011..._2 = \frac{6}{7}_{10}$ . So we want the number which is 6/7 of the way through the interval (0, 90) so $\theta = \frac{6}{7}\cdot 90 = \frac{540}{7}$ and $p+q = 540 + 7 = \boxed{547}$
| 547
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5,591
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10
| 5
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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 integers. Find $p+q$
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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 ${\frac{720}{7}}^\circ$ ${\frac{900}{7}}^\circ$ , and ${\frac{1080}{7}}^\circ$ are negative. With a bit of trial and error, we find $\theta$ is ${\frac{540}{7}}^\circ$ and $540+7$ is $\boxed{547}$
| 547
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5,592
|
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11
| 3
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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 $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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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 \[\angle AO_{1}B = \angle AO_{2}C = 2x\] Now we first find $\cos x.$ We use law of cosines on $\bigtriangleup ABC$ to obtain \[64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}\] \[\implies \cos{x} =\frac{11}{21}\] \[\implies \sin{x} =\frac{8\sqrt{5}}{21}\] Then applying law of sines on $\bigtriangleup AO_{1}B$ we obtain \[\frac{7}{\sin{2x}} =\frac{O_{1}B}{\sin{90^{\circ}-x}}\] \[\implies\frac{7}{2\sin{x}\cos{x}} =\frac{O_{1}B}{\cos{x}}\] \[\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}\] Using similar logic we obtain $O_{2}A =\frac{189}{16\sqrt{5}}.$
Now we know that $\angle O_{1}AO_{2}=180^{\circ}-x.$ Thus using law of cosines on $\bigtriangleup O_{1}AO_{2}$ yields \[O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}\] While this does look daunting we can write the above expression as \[\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}\] Then factoring yields \[\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}\] The area \[[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] Now $AK$ is twice the length of the altitude of $\bigtriangleup O_{1}AO_{2}$ so we let the altitude be $h$ and we have \[\frac{1}{2} \cdot h \cdot\frac{147}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] \[\implies h =\frac{9}{4}\] Thus our desired length is $\frac{9}{2} \implies m+n = \boxed{11}.$
| 11
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5,593
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https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11
| 4
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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 $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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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 \[AK = \frac{AM'}{2} = \frac{AC\cdot AB}{2AM} = \frac{7\cdot 9}{2\sqrt{\frac{9^2\cdot 4 + 7^2\cdot 4 - 4^2\cdot 8}{8}}} = \frac{7\cdot 9}{2\sqrt{49}} = \frac{9}{2}\implies m + n = \boxed{11}.\]
| 11
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5,594
|
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_12
| 1
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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 $360$
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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 $N$ as follows:
\begin{align*}N &= a(359) + a(179) + a(119) + a(89) + a(71) + a(59) + a(44) + a(39) \\ &+ a(35) + a(29) + a(23) + a(19) + a(17) + a(14) + a(11) + a(9) \\ &+ a(8) + a(7) + a(5) + a(4) + a(3) + a(2) + a(1) + 1 \end{align*}
The $+1$ at the end accounts for the sequence whose only term is 360. Fortunately, many of these numbers are prime; we have $a(p) = 1$ for primes $p$ as the only such sequence is " $p$ " itself. Also, $a(1) = 0$ . So we have
\[N = 15 + a(119) + a(44) + a(39) + a(35) + a(14) + a(9) + a(8) + a(4)\]
For small $k$ $a(k)$ is easy to compute: $a(4) = 1$ $a(8) = 2$ $a(9) = 2$ . For intermediate $k$ (e.g. $k=21$ below), $a(k)$ can be computed recursively using previously-computed values of $a(\cdot)$ , similar to dynamic programming. Then we have \begin{align*} a(14) &= 1+a(6) = 1+2 = 3\\ a(35) &= 1+a(6)+a(4) = 1 + 2 + 1 = 4 \\ a(39) &= 2 + a(12) = 2+4 = 6 \\ a(44) &= 2 + a(21) + a(10) = 2+4+2=8 \\ a(119) &= 1 + a(16) + a(6) = 1 + 3 + 2 = 6 \end{align*} Thus the answer is $N = 15+6+8+6+4+3+2+2+1 = \boxed{47}$
| 47
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5,595
|
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_13
| 1
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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 $\overline{PA_3},\overline{PA_4},$ and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac{1}{9}.$ There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6},\overline{PA_7},$ and the minor arc $\widehat{A_6A_7}$ of the circle is equal to $\tfrac{1}{8}-\tfrac{\sqrt2}{n}.$ Find $n.$
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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, we need to find the "D" shape, the small area enclosed by one side of the octagon and \(\frac{1}{8}\) of the circumference of the circle: \[D= \frac{1}{8} \pi r^2 - [A_1 A_2 O]=\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\]
Let \(PU\) be the height of \(\triangle A_1 A_2 P\), \(PV\) be the height of \(\triangle A_3 A_4 P\), \(PW\) be the height of \(\triangle A_6 A_7 P\). From the \(\frac{1}{7}\) and \(\frac{1}{9}\) condition we have \[\triangle P A_1 A_2= \frac{\pi r^2}{7} - D= \frac{1}{7} \pi \left(4+2\sqrt{2}\right)-\left(\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\right)\] \[\triangle P A_3 A_4= \frac{\pi r^2}{9} - D= \frac{1}{9} \pi \left(4+2\sqrt{2}\right)-\left(\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\right)\] which gives \(PU= \left(\frac{1}{7}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\) and \(PV= \left(\frac{1}{9}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\).
Now, let \(A_1 A_2\) intersects \(A_3 A_4\) at \(X\), \(A_1 A_2\) intersects \(A_6 A_7\) at \(Y\),\(A_6 A_7\) intersects \(A_3 A_4\) at \(Z\).
Clearly, \(\triangle XYZ\) is an isosceles right triangle, with right angle at \(X\) and the height with regard to which shall be \(3+2\sqrt2\). Now \(\frac{PU}{\sqrt{2}} + \frac{PV}{\sqrt{2}} + PW = 3+2\sqrt2\) which gives:
\begin{align*}
PW&= 3+2\sqrt2-\frac{PU}{\sqrt{2}} - \frac{PV}{\sqrt{2}} \\
&=3+2\sqrt{2}-\frac{1}{\sqrt{2}}\left(\left(\frac{1}{7}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1+\left(\frac{1}{9}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\right)\\
&=1+\sqrt{2}- \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\pi\left(4+2\sqrt{2}\right)
\end{align*}
Now, we have the area for \(D\) and the area for \(\triangle P A_6 A_7\), so we add them together:
\begin{align*}
\text{Target Area} &= \frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right) + \left(1+\sqrt{2}\right)- \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\pi\left(4+2\sqrt{2}\right)\\
&=\left(\frac{1}{8} - \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\right)\text{Total Area}
\end{align*}
The answer should therefore be \(\frac{1}{8}- \frac{\sqrt{2}}{2}\left(\frac{16}{63}-\frac{16}{64}\right)=\frac{1}{8}- \frac{\sqrt{2}}{504}\). The answer is \(\boxed{504}\).
| 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 $\overline{PA_3},\overline{PA_4},$ and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac{1}{9}.$ There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6},\overline{PA_7},$ and the minor arc $\widehat{A_6A_7}$ of the circle is equal to $\tfrac{1}{8}-\tfrac{\sqrt2}{n}.$ Find $n.$
|
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 horizontal (and therefore $A_3A_4$ is vertical). Note that the area bounded by $\overline{A_iA_j}$ and the arc $\widehat{A_iA_j}$ is fixed, so we only need to consider the relevant triangles.
[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label("$P$", P, dir(-75)); label("$A_1$", dir(112.5), dir(112.5)); label("$A_2$", dir(112.5-45), dir(112.5-45)); label("$A_3$", dir(112.5-90), dir(112.5-90)); label("$A_4$", dir(112.5-135), dir(112.5-135)); label("$A_5$", dir(112.5-180), dir(112.5-180)); label("$A_6$", dir(112.5-225), dir(112.5-225)); label("$A_7$", dir(112.5-270), dir(112.5-270)); label("$A_8$", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]
Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\triangle PA_1A_2$ by $1$ . We can see that $P$ was moved down by $\tfrac{1}{7}-\tfrac{1}{8}=\tfrac{1}{56}$ units to make the area defined by $P$ $A_1$ , and $A_2$ $\tfrac{1}{7}$ . Similarly, $P$ was moved right by $\tfrac{1}{8}-\tfrac{1}{9}=\tfrac{1}{72}$ to make the area defined by $P$ $A_3$ , and $A_4$ $\tfrac{1}{9}$ . This means that $P$ has coordinates $(\tfrac{1}{72},-\tfrac{1}{56})$
Now, we need to consider how this displacement in $P$ affected the area defined by $P$ $A_6$ , and $A_7$ . This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\tfrac{\sqrt{2}}{2}(\tfrac{1}{56}-\tfrac{1}{72})=\tfrac{\sqrt{2}}{504}$
Remembering the definition of our unit, this yields a final area of \[\frac{1}{8}-\frac{\sqrt{2}}{\boxed{504}.\]
| 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 be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$
|
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 $bck = (bk)^2+400=(ck)^2+525 = u$ . Then $bk=\sqrt{u-400}$ $ck=\sqrt{u-525}$ , and \[k=\sqrt{\frac{(u-400)(u-525)}{u^2}}\] Use the Law of Cosines in $\triangle APQ$ to get $25^2=b^2k^2+c^2k^2-2bck^3 = 2bck-925-2bck^3$ , which rearranges to \[775=bck - k^2\cdot bck = u-\frac{(u-400)(u-525)}{u}\] Upon simplification, this reduces to a linear equation in $u$ , with solution $u=1400$ . Then \[AB\cdot AC = bc = \frac 1{k}\cdot bck = \frac{u^2}{\sqrt{(u-400)(u-525)}}=560 \sqrt{14}\] So the final answer is $560 + 14 = \boxed{574}$
| 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 be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$
|
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 $(A,\omega) = AP\cdot AB = AQ\cdot AC$ , so \[p(p+q)=r(r+s)\quad \Longrightarrow \quad p^2-r^2=125\] Use Law of Cosines in $\triangle APQ$ to get $25^2=p^2+r^2-2pr\cos A$ ; since $\cos A = \frac r{p+q}$ , this simplifies as \[500 \ =\ 2r^2-\frac{2pr^2}{p+q} \ =\ 2r^2-\frac{2p^2r^2}{p^2+400} \ =\ \frac{800r^2}{r^2+525}\] We get $r=5\sqrt{35}$ and thus \[r=5\sqrt{35}, \quad p = \sqrt{r^2+125} = 10\sqrt{10}, \quad q = \frac{400}{p} =4\sqrt{10}, \quad s= \frac{525}{r} = 3\sqrt{35}.\] Therefore $AB\cdot AC = (p+q)\cdot(r+s) = 560\sqrt{14}$ . So the answer is $560 + 14 = \boxed{574}$
| 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 be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$
|
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 $E X Y F$ is a rectangle. Then, since $\angle XYF = 90^\circ$ we obtain $\angle XBF = 90^\circ$ (which directly follows from $XBYF$ being cyclic); hence $\angle XBQ = \angle AQB$ , or $XB \parallel AQ \Rightarrow XB \parallel AC$
Similarly, we can obtain $YC \parallel AB$ $\ \blacksquare$
A direct result of this claim is that $\triangle BPX \sim \triangle APQ \sim \triangle CYQ$
Thus, we can set $AP = 5k$ and $BP = 2k$ , then applying Power of a Point on $P$ we get $10 \cdot 40 = 10k^2 \implies k = 2\sqrt{10} \implies AB = 14 \sqrt{10}$ . Also, we can set $AQ = 5l$ and $CQ = 3l$ and once again applying Power of a Point (but this time to $Q$ ) we get
$\phantom{...................}15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}$
Hence,
$\phantom{...................}AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}$
and the answer is $560 + 14 = \boxed{574}$ . ~rocketsri
| 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 or equal to $100$
Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices $(0,0), (0, 100),$ and $(100,0)$ . Remember that $(0,0)$ does not work, so there is a square with the top right corner $(1,1)$
Note that $c$ and $d$ are interchangeable since they end up as $a$ and $b$ in the end anyways. Thus, we simply draw a line from $(1,1)$ to $(50,50)$ , designating one of the halves as our solution (since the other side is simply the coordinates flipped).
We note that the pattern from $(1,1)$ to $(50,50)$ is $2+3+4+\dots+51$ solutions and from $(51, 49)$ to $(100,0)$ is $50+49+48+\dots+1$ solutions, since we can decrease the $y$ -value by $1$ until $0$ for each coordinate.
Adding up gives \[\dfrac{2+51}{2}\cdot 50+\dfrac{50+1}{2}\cdot 50.\] This gives us $2600$ , and $2600\equiv 600 \bmod{1000}.$
Thus, the answer is: \[\boxed{600}.\]
| 600
|
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