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Difference between revisions of "2020 AIME I Problems"

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==Problem 1==
 
==Problem 1==
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Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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In <math>\triangle ABC</math> with <math>AB=AC,</math> point <math>D</math> lies strictly between <math>A</math> and <math>C</math> on side <math>\overline{AC},</math> and point <math>E</math> lies strictly between <math>A</math> and <math>B</math> on side <math>\overline{AB}</math> such that <math>AE=ED=DB=BC.</math> The degree measure of <math>\angle ABC</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
 
[[2020 AIME I Problems/Problem 1 | Solution]]
 
[[2020 AIME I Problems/Problem 1 | Solution]]
  
 
==Problem 2==
 
==Problem 2==
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Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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There is a unique positive real number <math>x</math> such that the three numbers <math>\log_8(2x),\log_4x,</math> and <math>\log_2x,</math> in that order, form a geometric progression with positive common ratio. The number <math>x</math> can be written as <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
 
[[2020 AIME I Problems/Problem 2 | Solution]]
 
[[2020 AIME I Problems/Problem 2 | Solution]]
  
 
==Problem 3==
 
==Problem 3==
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Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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A positive integer <math>N</math> has base-eleven representation <math>\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}</math> and base-eight representation <math>\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},</math> where <math>a,b,</math> and <math>c</math> represent (not necessarily distinct) digits. Find the least such <math>N</math> expressed in base ten.
  
 
[[2020 AIME I Problems/Problem 3 | Solution]]
 
[[2020 AIME I Problems/Problem 3 | Solution]]
  
 
==Problem 4==
 
==Problem 4==
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Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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Let <math>S</math> be the set of positive integers <math>N</math> with the property that the last four digits of <math>N</math> are <math>2020,</math> and when the last four digits are removed, the result is a divisor of <math>N.</math> For example, <math>42{,}020</math> is in <math>S</math> because <math>4</math> is a divisor of <math>42{,}020.</math> Find the sum of all the digits of all the numbers in <math>S.</math> For example, the number <math>42{,}020</math> contributes <math>4+2+0+2+0=8</math> to this total.
  
 
[[2020 AIME I Problems/Problem 4 | Solution]]
 
[[2020 AIME I Problems/Problem 4 | Solution]]
  
 
==Problem 5==
 
==Problem 5==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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Six cards numbered <math>1</math> through <math>6</math> 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.
  
 
[[2020 AIME I Problems/Problem 5 | Solution]]
 
[[2020 AIME I Problems/Problem 5 | Solution]]
  
 
==Problem 6==
 
==Problem 6==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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A flat board has a circular hole with radius <math>1</math> and a circular hole with radius <math>2</math> such that the distance between the centers of the two holes is <math>7</math>. 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 <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
[[2020 AIME I Problems/Problem 6 | Solution]]
 
[[2020 AIME I Problems/Problem 6 | Solution]]
  
 
==Problem 7==
 
==Problem 7==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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A club consisting of <math>11</math> men and <math>12</math> women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as <math>1</math> member or as many as <math>23</math> members. Let <math>N</math> be the number of such committees that can be formed. Find the sum of the prime numbers that divide <math>N.</math>
  
 
[[2020 AIME I Problems/Problem 7 | Solution]]
 
[[2020 AIME I Problems/Problem 7 | Solution]]
  
 
==Problem 8==
 
==Problem 8==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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A bug walks all day and sleeps all night. On the first day, it starts at point <math>O,</math> faces east, and walks a distance of <math>5</math> units due east. Each night the bug rotates <math>60^\circ</math> 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 <math>P.</math> Then <math>OP^2=\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
 
[[2020 AIME I Problems/Problem 8 | Solution]]
 
[[2020 AIME I Problems/Problem 8 | Solution]]
  
 
==Problem 9==
 
==Problem 9==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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Let <math>S</math> be the set of positive integer divisors of <math>20^9.</math> Three numbers are chosen independently and at random with replacement from the set <math>S</math> and labeled <math>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> divides <math>a_3</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m.</math>
  
 
[[2020 AIME I Problems/Problem 9 | Solution]]
 
[[2020 AIME I Problems/Problem 9 | Solution]]
  
 
==Problem 10==
 
==Problem 10==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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Let <math>m</math> and <math>n</math> be positive integers satisfying the conditions
 +
 
 +
<math>\quad\bullet\ \gcd(m+n,210)=1,</math>
 +
 
 +
<math>\quad\bullet\ m^m</math> is a multiple of <math>n^n,</math> and
 +
 
 +
<math>\quad\bullet\ m</math> is not a multiple of <math>n.</math>
 +
 
 +
Find the least possible value of <math>m+n.</math>
  
 
[[2020 AIME I Problems/Problem 10 | Solution]]
 
[[2020 AIME I Problems/Problem 10 | Solution]]
  
 
==Problem 11==
 
==Problem 11==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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For integers <math>a,b,c</math> and <math>d,</math> let <math>f(x)=x^2+ax+b</math> and <math>g(x)=x^2+cx+d.</math> Find the number of ordered triples <math>(a,b,c)</math> of integers with absolute values not exceeding <math>10</math> for which there is an integer <math>d</math> such that <math>g(f(2))=g(f(4))=0.</math>
 +
 
 
[[2020 AIME I Problems/Problem 11 | Solution]]
 
[[2020 AIME I Problems/Problem 11 | Solution]]
  
 
==Problem 12==
 
==Problem 12==
−
Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
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Let <math>n</math> be the least positive integer for which <math>149^n-2^n</math> is divisible by <math>3^3\cdot5^5\cdot7^7.</math> Find the number of positive integer divisors of <math>n.</math>
  
 
[[2020 AIME I Problems/Problem 12 | Solution]]
 
[[2020 AIME I Problems/Problem 12 | Solution]]
  
 
==Problem 13==
 
==Problem 13==
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Point <math>D</math> lies on side <math>\overline{BC}</math> of <math>\triangle ABC</math> so that <math>\overline{AD}</math> bisects <math>\angle BAC.</math> The perpendicular bisector of <math>\overline{AD}</math> intersects the bisectors of <math>\angle ABC</math> and <math>\angle ACB</math> in points <math>E</math> and <math>F,</math> respectively. Given that <math>AB=4,BC=5,</math> and <math>CA=6,</math> the area of <math>\triangle AEF</math> can be written as <math>\tfrac{m\sqrt{n}}p,</math> where <math>m</math> and <math>p</math> are relatively prime positive integers, and <math>n</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math>
+
Point <math>D</math> lies on side <math>\overline{BC}</math> of <math>\triangle ABC</math> so that <math>\overline{AD}</math> bisects <math>\angle BAC.</math> The perpendicular bisector of <math>\overline{AD}</math> intersects the bisectors of <math>\angle ABC</math> and <math>\angle ACB</math> in points <math>E</math> and <math>F,</math> respectively. Given that <math>AB=4,BC=5,</math> and <math>CA=6,</math> the area of <math>\triangle AEF</math> can be written as <math>\tfrac{m\sqrt{n}}p,</math> where <math>m</math> and <math>p</math> are relatively prime positive integers, and <math>n</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p</math>.
  
 
[[2020 AIME I Problems/Problem 13 | Solution]]
 
[[2020 AIME I Problems/Problem 13 | Solution]]

Latest revision as of 16:51, 5 February 2022

2020 AIME I (Answer Key)
Printable version | AoPS Contest Collections • PDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

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 $m+n.$

Solution

Problem 2

There is a unique positive real number $x$ such that the three numbers $\log_8(2x),\log_4x,$ and $\log_2x,$ in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

Problem 3

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.

Solution

Problem 4

Let $S$ be the set of positive integers $N$ with the property that the last four digits of $N$ are $2020,$ and when the last four digits are removed, the result is a divisor of $N.$ For example, $42{,}020$ is in $S$ because $4$ is a divisor of $42{,}020.$ Find the sum of all the digits of all the numbers in $S.$ For example, the number $42{,}020$ contributes $4+2+0+2+0=8$ to this total.

Solution

Problem 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.

Solution

Problem 6

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 $m+n$.

Solution

Problem 7

A club consisting of $11$ men and $12$ women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as $1$ member or as many as $23$ members. Let $N$ be the number of such committees that can be formed. Find the sum of the prime numbers that divide $N.$

Solution

Problem 8

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 $m+n.$

Solution

Problem 9

Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$

Solution

Problem 10

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 $m+n.$

Solution

Problem 11

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 $g(f(2))=g(f(4))=0.$

Solution

Problem 12

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 $n.$

Solution

Problem 13

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 $m+n+p$.

Solution

Problem 14

Let $P(x)$ be a quadratic polynomial with complex coefficients whose $x^2$ coefficient is $1.$ Suppose the equation $P(P(x))=0$ has four distinct solutions, $x=3,4,a,b.$ Find the sum of all possible values of $(a+b)^2.$

Solution

Problem 15

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 $m+n.$


Solution

2020 AIME I (Problems • Answer Key • Resources)
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