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Difference between revisions of "2025 AIME II Problems"

(→‎Problem 3)
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==Problem 4==  
 
==Problem 4==  
−
 
+
The product<cmath>\prod^{63}_{k=4} \frac{\log_k (5^{k^2 - 1})}{\log_{k + 1} (5^{k^2 - 4})} = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}</cmath>is equal to <math>\tfrac mn,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math>
  
 
[[2025 AIME II Problems/Problem 4|Solution]]
 
[[2025 AIME II Problems/Problem 4|Solution]]
  
 
==Problem 5==   
 
==Problem 5==   
−
 
+
Suppose <math>\triangle ABC</math> has angles <math>\angle BAC = 84^\circ, \angle ABC=60^\circ,</math> and <math>\angle ACB = 36^\circ.</math> Let <math>D, E,</math> and <math>F</math> be the midpoints of sides <math>\overline{BC}, \overline{AC},</math> and <math>\overline{AB},</math> respectively. The circumcircle of <math>\triangle DEF</math> intersects <math>\overline{BD}, \overline{AE},</math> and <math>\overline{AF}</math> at points <math>G, H,</math> and <math>J,</math> respectively. The points <math>G, D, E, H, J,</math> and <math>F</math> divide the circumcircle of <math>\triangle DEF</math> into six minor arcs, as shown. Find <math>\overarc{DE}+2\cdot \overarc{HJ} + 3\cdot \overarc{FG},</math> where the arcs are measured in degrees.
  
 
[[2025 AIME II Problems/Problem 5|Solution]]
 
[[2025 AIME II Problems/Problem 5|Solution]]
  
 
==Problem 6==   
 
==Problem 6==   
 +
Circle <math>\omega_1</math> with radius <math>6</math> centered at point <math>A</math> is internally tangent at point <math>B</math> to circle <math>\omega_2</math> with radius <math>15</math>. Points <math>C</math> and <math>D</math> lie on <math>\omega_2</math> such that <math>\overline{BC}</math> is a diameter of <math>\omega_2</math> and <math>\overline{BC} \perp \overline{AD}</math>. The rectangle <math>EFGH</math> is inscribed in <math>\omega_1</math> such that <math>\overline{EF} \perp \overline{BC}</math>, <math>C</math> is closer to <math>\overline{GH}</math> than to <math>\overline{EF}</math>, and <math>D</math> is closer to <math>\overline{FG}</math> than to <math>\overline{EH}</math>, as shown. Triangles <math>\triangle DGF</math> and <math>\triangle CHG</math> have equal areas. The area of rectangle <math>EFGH</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
  
Line 34: Line 35:
  
 
==Problem 7==   
 
==Problem 7==   
 +
Let <math>A</math> be the set of positive integer divisors of <math>2025</math>. Let <math>B</math> be a randomly selected subset of <math>A</math>. The probability that <math>B</math> is a nonempty set with the property that the least common multiple of its element is <math>2025</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
[[2025 AIME II Problems/Problem 7|Solution]]   
 
[[2025 AIME II Problems/Problem 7|Solution]]   
  
 
==Problem 8==   
 
==Problem 8==   
 +
From an unlimited supply of 1-cent coins, 10-cent coins, and 25-cent coins, Silas wants to find a collection of coins that has a total value of <math>N</math> cents, where <math>N</math> is a positive integer. He uses the so-called greedy algorithm, successively choosing the coin of greatest value that does not cause the value of his collection to exceed <math>N.</math> For example, to get 42 cents, Silas will choose a 25-cent coin, then a 10-cent coin, then 7 1-cent coins. However, this collection of 9 coins uses more coins than necessary to get a total of 42 cents; indeed, choosing 4 10-cent coins and 2 1-cent coins achieves the same total value with only 6 coins. In general, the greedy algorithm succeeds for a given <math>N</math> if no other collection of 1-cent, 10-cent, and 25-cent coins gives a total value of <math>N</math> cents using strictly fewer coins than the collection given by the greedy algorithm. Find the number of values of <math>N</math> between <math>1</math> and <math>1000</math> inclusive for which the greedy algorithm succeeds.
  
  
 
[[2025 AIME II Problems/Problem 8|Solution]]
 
[[2025 AIME II Problems/Problem 8|Solution]]
  
−
==Problem 9==
+
==Problem 9==  
−
 
+
There are <math>n</math> values of <math>x</math> in the interval <math>0<x<2\pi</math> where <math>f(x)=\sin(7\pi\cdot\sin(5x))=0</math>. For <math>t</math> of these <math>n</math> values of <math>x</math>, the graph of <math>y=f(x)</math> is tangent to the <math>x</math>-axis. Find <math>n+t</math>.
−
 
 
  
 
[[2025 AIME II Problems/Problem 9|Solution]]
 
[[2025 AIME II Problems/Problem 9|Solution]]
  
 
==Problem 10==   
 
==Problem 10==   
 +
Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let <math>N</math> be the number of subsets of <math>16</math> chairs that could be selected. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  
 
[[2025 AIME II Problems/Problem 10|Solution]]   
 
[[2025 AIME II Problems/Problem 10|Solution]]   
  
 
==Problem 11==
 
==Problem 11==
−
 
+
Let <math>S</math> be the set of vertices of a regular <math>24</math>-gon. Find the number of ways to draw <math>12</math> segments of equal lengths so that each vertex in <math>S</math> is an endpoint of exactly one of the <math>12</math> segments.
  
 
[[2025 AIME II Problems/Problem 11|Solution]]   
 
[[2025 AIME II Problems/Problem 11|Solution]]   
  
 
==Problem 12==   
 
==Problem 12==   
 +
Let <math>A_1A_2\dots A_{11}</math> be a non-convex <math>11</math>-gon such that
 +
\begin{itemize}
 +
\item The area of <math>A_iA_1A_{i+1}</math> is <math>1</math> for each <math>2 \le i \le 10</math>,
 +
\item <math>\cos(\angle A_iA_1A_{i+1})=\frac{12}{13}</math> for each <math>2 \le i \le 10</math>,
 +
\item The perimeter of <math>A_1A_2\dots A_{11}</math> is <math>20</math>.
 +
\end{itemize}
 +
If <math>A_1A_2+A_1A_{11}</math> can be expressed as <math>\frac{m\sqrt{n}-p}{q}</math> for positive integers <math>m,n,p,q</math> with <math>n</math> squarefree and <math>\gcd(m,p,q)=1</math>, find <math>m+n+p+q</math>.
  
 
[[2025 AIME II Problems/Problem 12|Solution]]   
 
[[2025 AIME II Problems/Problem 12|Solution]]   
  
 
==Problem 13==   
 
==Problem 13==   
 +
Let the sequence of rationals <math>x_1,x_2,\dots</math> be defined such that <math>x_1=\frac{25}{11}</math> and
 +
<cmath>x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right).</cmath><math>x_{2025}</math> can be expressed as <math>\frac{m}{n}</math> for relatively prime positive integers <math>m</math> and <math>n</math>. Find the remainder when <math>m+n</math> is divided by <math>1000</math>.
  
 
[[2025 AIME II Problems/Problem 13|Solution]]   
 
[[2025 AIME II Problems/Problem 13|Solution]]   
  
 
==Problem 14==   
 
==Problem 14==   
 +
Let <math>{\triangle ABC}</math> be a right triangle with <math>\angle A = 90^\circ</math> and <math>BC = 38.</math> There exist points <math>K</math> and <math>L</math> inside the triangle such<cmath>AK = AL = BK = CL = KL = 14.</cmath>The area of the quadrilateral <math>BKLC</math> can be expressed as <math>n\sqrt3</math> for some positive integer <math>n.</math> Find <math>n.</math>
  
 
[[2025 AIME II Problems/Problem 14|Solution]]   
 
[[2025 AIME II Problems/Problem 14|Solution]]   
  
 
==Problem 15==   
 
==Problem 15==   
−
 
+
Let
 +
<cmath>f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}.</cmath>There exist exactly three positive real values of <math>k</math> such that <math>f</math> has a minimum at exactly two real values of <math>x</math>. Find the sum of these three values of <math>k</math>.
  
 
[[2025 AIME II Problems/Problem 15|Solution]]
 
[[2025 AIME II Problems/Problem 15|Solution]]

Revision as of 21:01, 13 February 2025

2025 AIME II (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, rulers and compasses are permitted. In particular, graph paper, protractors, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

Six points $A, B, C, D, E,$ and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,$ and $DG=30.$ Find the area of $\triangle BGE.$

Solution

Problem 2

Find the sum of all positive integers $n$ such that $n + 2$ divides the product $3(n + 3)(n^2 + 9)$.

Solution

Problem 3

Four unit squares form a $2 \times 2$ grid. Each of the $12$ unit line segments forming the sides of the squares is colored either red or blue in such a say that each unit square has $2$ red sides and $2$ blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.

Solution

Problem 4

The product\[\prod^{63}_{k=4} \frac{\log_k (5^{k^2 - 1})}{\log_{k + 1} (5^{k^2 - 4})} = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}\]is equal to $\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

Problem 5

Suppose $\triangle ABC$ has angles $\angle BAC = 84^\circ, \angle ABC=60^\circ,$ and $\angle ACB = 36^\circ.$ Let $D, E,$ and $F$ be the midpoints of sides $\overline{BC}, \overline{AC},$ and $\overline{AB},$ respectively. The circumcircle of $\triangle DEF$ intersects $\overline{BD}, \overline{AE},$ and $\overline{AF}$ at points $G, H,$ and $J,$ respectively. The points $G, D, E, H, J,$ and $F$ divide the circumcircle of $\triangle DEF$ into six minor arcs, as shown. Find $\overarc{DE}+2\cdot \overarc{HJ} + 3\cdot \overarc{FG},$ where the arcs are measured in degrees.

Solution

Problem 6

Circle $\omega_1$ with radius $6$ centered at point $A$ is internally tangent at point $B$ to circle $\omega_2$ with radius $15$. Points $C$ and $D$ lie on $\omega_2$ such that $\overline{BC}$ is a diameter of $\omega_2$ and $\overline{BC} \perp \overline{AD}$. The rectangle $EFGH$ is inscribed in $\omega_1$ such that $\overline{EF} \perp \overline{BC}$, $C$ is closer to $\overline{GH}$ than to $\overline{EF}$, and $D$ is closer to $\overline{FG}$ than to $\overline{EH}$, as shown. Triangles $\triangle DGF$ and $\triangle CHG$ have equal areas. The area of rectangle $EFGH$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


Solution

Problem 7

Let $A$ be the set of positive integer divisors of $2025$. Let $B$ be a randomly selected subset of $A$. The probability that $B$ is a nonempty set with the property that the least common multiple of its element is $2025$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 8

From an unlimited supply of 1-cent coins, 10-cent coins, and 25-cent coins, Silas wants to find a collection of coins that has a total value of $N$ cents, where $N$ is a positive integer. He uses the so-called greedy algorithm, successively choosing the coin of greatest value that does not cause the value of his collection to exceed $N.$ For example, to get 42 cents, Silas will choose a 25-cent coin, then a 10-cent coin, then 7 1-cent coins. However, this collection of 9 coins uses more coins than necessary to get a total of 42 cents; indeed, choosing 4 10-cent coins and 2 1-cent coins achieves the same total value with only 6 coins. In general, the greedy algorithm succeeds for a given $N$ if no other collection of 1-cent, 10-cent, and 25-cent coins gives a total value of $N$ cents using strictly fewer coins than the collection given by the greedy algorithm. Find the number of values of $N$ between $1$ and $1000$ inclusive for which the greedy algorithm succeeds.


Solution

Problem 9

There are $n$ values of $x$ in the interval $0<x<2\pi$ where $f(x)=\sin(7\pi\cdot\sin(5x))=0$. For $t$ of these $n$ values of $x$, the graph of $y=f(x)$ is tangent to the $x$-axis. Find $n+t$.

Solution

Problem 10

Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let $N$ be the number of subsets of $16$ chairs that could be selected. Find the remainder when $N$ is divided by $1000$.

Solution

Problem 11

Let $S$ be the set of vertices of a regular $24$-gon. Find the number of ways to draw $12$ segments of equal lengths so that each vertex in $S$ is an endpoint of exactly one of the $12$ segments.

Solution

Problem 12

Let $A_1A_2\dots A_{11}$ be a non-convex $11$-gon such that \begin{itemize} \item The area of $A_iA_1A_{i+1}$ is $1$ for each $2 \le i \le 10$, \item $\cos(\angle A_iA_1A_{i+1})=\frac{12}{13}$ for each $2 \le i \le 10$, \item The perimeter of $A_1A_2\dots A_{11}$ is $20$. \end{itemize} If $A_1A_2+A_1A_{11}$ can be expressed as $\frac{m\sqrt{n}-p}{q}$ for positive integers $m,n,p,q$ with $n$ squarefree and $\gcd(m,p,q)=1$, find $m+n+p+q$.

Solution

Problem 13

Let the sequence of rationals $x_1,x_2,\dots$ be defined such that $x_1=\frac{25}{11}$ and \[x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right).\]$x_{2025}$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Find the remainder when $m+n$ is divided by $1000$.

Solution

Problem 14

Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$

Solution

Problem 15

Let \[f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}.\]There exist exactly three positive real values of $k$ such that $f$ has a minimum at exactly two real values of $x$. Find the sum of these three values of $k$.

Solution

See also

2025 AIME II (Problems • Answer Key • Resources)
Preceded by
2025 AIME I 		Followed by
2026 AIME I
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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