Difference between revisions of "2025 AIME II Problems"

Latest revision as of 14:23, 23 February 2025

2025 AIME II (Answer Key) Printable version \| AoPS Contest Collections • PDF
Instructions 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. 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 way 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.

$[asy] size(4cm); defaultpen(linewidth(1.2)); draw((0, 0) -- (2, 0) -- (2, 1), red); draw((0, 1) -- (1, 1) -- (1, 2) -- (2,2), red); draw((0, 0) -- (0, 1),blue+dotted); draw((1, 0) -- (1, 1) -- (2, 1) -- (2, 2),blue+dotted); draw((0, 1) -- (0, 2) -- (1, 2),blue+dotted); [/asy]$ 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.

$[asy] import olympiad; size(6cm); defaultpen(fontsize(10pt)); pair B = (0, 0), A = (Cos(60), Sin(60)), C = (Cos(60)+Sin(60)/Tan(36), 0), D = midpoint(B--C), E = midpoint(A--C), F = midpoint(A--B); guide circ = circumcircle(D, E, F); pair G = intersectionpoint(B--D, circ), J = intersectionpoints(A--F, circ)[0], H = intersectionpoints(A--E, circ)[0]; draw(B--A--C--cycle); draw(D--E--F--cycle); draw(circ); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(J); label("$A$", A, (0, .8)); label("$B$", B, (-.8, -.8)); label("$C$", C, (.8, -.8)); label("$D$", D, (0, -.8)); label("$E$", E, (.8, .2)); label("$F$", F, (-.8, .2)); label("$G$", G, (0, .8)); label("$H$", H, (-.2, -1)); label("$J$", J, (.2, -.8)); [/asy]$

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 $m+n$ .

$[asy] size(5cm); defaultpen(fontsize(10pt)); pair A = (9, 0), B = (15, 0), C = (-15, 0), D = (9, 12), E = (9+12/sqrt(5), -6/sqrt(5)), F = (9+12/sqrt(5), 6/sqrt(5)), G = (9-12/sqrt(5), 6/sqrt(5)), H = (9-12/sqrt(5), -6/sqrt(5)); filldraw(G--H--C--cycle, lightgray); filldraw(D--G--F--cycle, lightgray); draw(B--C); draw(A--D); draw(E--F--G--H--cycle); draw(circle((0,0), 15)); draw(circle(A, 6)); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); label("$A$", A, (.8, -.8)); label("$B$", B, (.8, 0)); label("$C$", C, (-.8, 0)); label("$D$", D, (.4, .8)); label("$E$", E, (.8, -.8)); label("$F$", F, (.8, .8)); label("$G$", G, (-.8, .8)); label("$H$", H, (-.8, -.8)); label("$\omega_1$", (9, -5)); label("$\omega_2$", (-1, -13.5)); [/asy]$

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 $\textit{greedy algorithm}$ , 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 an $11$ -sided non-convex simple polygon with the following properties:

• For every integer $2 \leq i \leq 10$ , the area of $A_iA_1A_{i+1}$ is $1$ ,

• For every integer $2 \leq i \leq 10$ , $\cos(\angle A_iA_1A_{i+1})=\frac{12}{13}$ ,

• The perimeter of the $11$ -gon $A_1A_2\dots A_{11}$ is equal to $20$ .

If $A_1A_2+A_1A_{11}$ can be expressed as $\frac{m\sqrt{n}-p}{q}$ where $m,n,p,q$ are positive integers, $n$ is not divisible by any square, and no prime divides all of $m$ , $p$ , and $q$ . 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 $x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right)$ for all $k \leq 1$ . Then $x_{2025}$ 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

There are exactly three positive real numbers $k$ such that the function $f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}$ defined over the positive real numbers achieves its minimum value at exactly two positive real numbers $x$ . Find the sum of these three values of $k$ .

Solution

2025 AIME II (Problems • Answer Key • Resources)
Preceded by 2025 AIME I	Followed by 2026 AIME I
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Difference between revisions of "2025 AIME II Problems"

Latest revision as of 14:23, 23 February 2025

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also

@@ Line 3: / Line 3: @@
 == Problem 1 ==
-Six points <math>A, B, C, D, E,</math> and <math>F</math> lie in a straight line in that order. Suppose that <math>G</math> is a point not on the line and that <math>AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,</math> and <math>DG=30.</math> Find the area of <math>\triangle BGE.</math>
+Six points <math>A, B, C, D, E,</math> and <math>F</math> lie in a straight line in that order. Suppose that <math>G</math> is a point not on the line and that <math>AC=26</math>, <math>BD=22</math>, <math>CE=31</math>, <math>DF=33</math>, <math>AF=73</math>, <math>CG=40</math>, and <math>DG=30.</math> Find the area of <math>\triangle BGE.</math>
 [[2025 AIME II Problems/Problem 1|Solution]]
 == Problem 2 ==
-Find the sum of all positive integers <math>n</math> such that <math>n + 2</math> divides the product <math>3(n + 3)(n^2 + 9)</math>.
+Find the sum of all positive integers <math>n</math> such that <math>n+2</math> divides the product <math>3(n+3)(n^2+9).</math>
 [[2025 AIME II Problems/Problem 2|Solution]]
 ==Problem 3==
-Four unit squares form a <math>2 \times 2</math> grid. Each of the <math>12</math> unit line segments forming the sides of the squares is colored either red or blue in such a say that each unit square has <math>2</math> red sides and <math>2</math> blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.
+Four unit squares form a <math>2 \times 2</math> grid. Each of the <math>12</math> unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has <math>2</math> red sides and <math>2</math> blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.
+<asy>
+size(4cm);
+defaultpen(linewidth(1.2));
+draw((0, 0) -- (2, 0) -- (2, 1), red);
+draw((0, 1) -- (1, 1) -- (1, 2) -- (2,2), red);
+draw((0, 0) -- (0, 1),blue+dotted);
+draw((1, 0) -- (1, 1) -- (2, 1) -- (2, 2),blue+dotted);
+draw((0, 1) -- (0, 2) -- (1, 2),blue+dotted);
+</asy>
 [[2025 AIME II Problems/Problem 3|Solution]]
@@ Line 24: / Line 34: @@
 ==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.
+<asy>
+        import olympiad;
+        size(6cm);
+        defaultpen(fontsize(10pt));
+        pair B = (0, 0), A = (Cos(60), Sin(60)), C = (Cos(60)+Sin(60)/Tan(36), 0), D = midpoint(B--C), E = midpoint(A--C), F = midpoint(A--B);
+        guide circ = circumcircle(D, E, F);
+        pair G = intersectionpoint(B--D, circ), J = intersectionpoints(A--F, circ)[0], H = intersectionpoints(A--E, circ)[0];
+        draw(B--A--C--cycle);
+        draw(D--E--F--cycle);
+        draw(circ);  dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(J);
+        label("$A$", A, (0, .8));
+        label("$B$", B, (-.8, -.8));
+        label("$C$", C, (.8, -.8));
+        label("$D$", D, (0, -.8));
+        label("$E$", E, (.8, .2));
+        label("$F$", F, (-.8, .2));
+        label("$G$", G, (0, .8));
+        label("$H$", H, (-.2, -1));
+        label("$J$", J, (.2, -.8));
+</asy>
 [[2025 AIME II Problems/Problem 5|Solution]]
 ==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>.
+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>.
+<asy>
+        size(5cm);
+        defaultpen(fontsize(10pt));
+        pair A = (9, 0), B = (15, 0), C = (-15, 0), D = (9, 12), E = (9+12/sqrt(5), -6/sqrt(5)), F = (9+12/sqrt(5), 6/sqrt(5)), G = (9-12/sqrt(5), 6/sqrt(5)), H = (9-12/sqrt(5), -6/sqrt(5));
+        filldraw(G--H--C--cycle, lightgray);
+        filldraw(D--G--F--cycle, lightgray);
+        draw(B--C);
+        draw(A--D);
+        draw(E--F--G--H--cycle);
+        draw(circle((0,0), 15));
+        draw(circle(A, 6));
+        dot(A);
+        dot(B);
+        dot(C);
+        dot(D);
+        dot(E);
+        dot(F);
+        dot(G);
+        dot(H);
+        label("$A$", A, (.8, -.8));
+        label("$B$", B, (.8, 0));
+        label("$C$", C, (-.8, 0));
+        label("$D$", D, (.4, .8));
+        label("$E$", E, (.8, -.8));
+        label("$F$", F, (.8, .8));
+        label("$G$", G, (-.8, .8));
+        label("$H$", H, (-.8, -.8));
+        label("$\omega_1$", (9, -5));
+        label("$\omega_2$", (-1, -13.5));
+</asy>
 [[2025 AIME II Problems/Problem 6|Solution]]
@@ Line 40: / Line 100: @@
 ==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.
+From an unlimited supply of <math>1</math>-cent coins, <math>10</math>-cent coins, and <math>25</math>-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 <math>\textit{greedy algorithm}</math>, 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 <math>42</math> cents, Silas will choose a <math>25</math>-cent coin, then a <math>10</math>-cent coin, then <math>7</math> <math>1</math>-cent coins. However, this collection of <math>9</math> coins uses more coins than necessary to get a total of <math>42</math> cents; indeed, choosing <math>4</math> <math>10</math>-cent coins and <math>2</math> <math>1</math>-cent coins achieves the same total value with only <math>6</math> coins.
+In general, the greedy algorithm succeeds for a given <math>N</math> if no other collection of <math>1</math>-cent, <math>10</math>-cent, and <math>25</math>-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.
@@ Line 61: / Line 123: @@
 ==Problem 12==
-Let <math>A_1A_2\dots A_{11}</math> be a non-convex <math>11</math>-gon such that
+Let <math>A_1A_2\dots A_{11}</math> be an <math>11</math>-sided non-convex simple polygon with the following properties:
-\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>,
+• For every integer <math>2 \leq i \leq 10</math>, the area of <math>A_iA_1A_{i+1}</math> is <math>1</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>.
+• For every integer <math>2 \leq i \leq 10</math>, <math>\cos(\angle A_iA_1A_{i+1})=\frac{12}{13}</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>.
+• The perimeter of the <math>11</math>-gon <math>A_1A_2\dots A_{11}</math> is equal to <math>20</math>.
+If <math>A_1A_2+A_1A_{11}</math> can be expressed as <math>\frac{m\sqrt{n}-p}{q}</math> where <math>m,n,p,q</math> are positive integers, <math>n</math> is not divisible by any square, and no prime divides all of <math>m</math>,<math>p</math>, and <math>q</math>. Find <math>m+n+p+q</math>.
 [[2025 AIME II Problems/Problem 12|Solution]]
 ==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>.
+<cmath>x_{k+1}=\frac{1}{3}\left(x_k+\frac{1}{x_k}-1\right)</cmath> for all <math>k \leq 1</math>. Then <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]]
 ==Problem 14==
@@ Line 83: / Line 148: @@
 ==Problem 15==
-Let
+There are exactly three positive real numbers <math>k</math> such that the function
-<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>.
+<cmath>f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}</cmath> defined over the positive real numbers achieves its minimum value at exactly two positive real numbers <math>x</math>. Find the sum of these three values of <math>k</math>.
 [[2025 AIME II Problems/Problem 15|Solution]]