ID stringlengths 8 10 | Problem stringlengths 82 1.6k | Solution stringlengths 11 3.02k | Answer int64 16 907 |
|---|---|---|---|
2025-I-1 | Find the sum of all integer bases $b > 9$ for which $17_b$ is a divisor of $97_b$. | We have that $a(b+7)=9b+7$ where $a$ is a natural number. Rearranging we get $(a-9)(b+7)=-56$. Since $b>9$, $b=49,21$. Thus the answer is $49+21=\boxed{70}$ | 70 |
2025-I-2 | In $\triangle ABC$ points $D$ and $E$ lie on $\overline{AB}$ so that $AD < AE < AB$, while points $F$ and $G$ lie on $\overline{AC}$ so that $AF < AG < AC$. Suppose $AD = 4,$ $DE = 16,$ $EB = 8,$ $AF = 13,$ $FG = 52,$ $GC = 26.$ Let $M$ be the reflection of $D$ through $F$, and let $N$ be the reflection of $G$ through ... | Note that the triangles outside $\triangle ABC$ have the same height as the unshaded triangles in $\triangle ABC$. Since they have the same bases, the area of the heptagon is the same as the area of triangle $ABC$. Therefore, we need to calculate the area of $\triangle ABC$. Denote the length of $DF$ as $x$ and the alt... | 588 |
2025-I-3 | The 9 members of a baseball team went to an ice-cream parlor after their game. Each player had a single scoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was grea... | Let $c$ be the number of players who choose chocolate, $v$ be the number of players who choose vanilla, and $s$ be the number of players who choose strawberry ice cream. We have $c,v,s\ge 1$ and $c + v + s = 9.$\newline
\newline
By inspection, the only solutions for $(c,v,s)$ are $(2,3,4),(1,2,6),(1,3,5).$\newline
\new... | 16 |
2025-I-4 | Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ are integers between $-100$ and $100$ inclusive, such that $12x^2 - xy - 6y^2 = 0$. | We begin by factoring, $12x^2 - xy - 6y^2 = (3x+2y)(4x-3y)=0.$ Since the RHS is $0$ we have two options,\newline
\newline
\underline{Case 1:}\, $3x+2y=0$.\newline
In this case we have $y = \frac{-3x}{2}.$ Using the bounding on $y$ we get,\[-100\le \frac{-3x}{2}\le 100\]\[\frac{200}{3} \ge x \ge \frac{-200}{3}\]\newline... | 117 |
2025-I-5 | There are $8!=40320$ eight-digit positive integers that use each of the digits $1, 2, 3, 4, 5, 6, 7, 8$ exactly once. Let $N$ be the number of these integers that are divisible by $22$. Find the difference between $N$ and $2025$. | Notice that if the 8-digit number is divisible by $22$, it must have an even units digit. Therefore, we can break it up into cases and let the last digit be either $2, 4, 6,$ or $8$. Due to symmetry, upon finding the total count for the last digit $2$, we multiply the result by $4$ for the other possible last digits.
... | 279 |
2025-I-6 | An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is 3, and the area of the trapezoid is 72. Let the parallel sides of the trapezoid have lengths $r$ and $s$, with $r \neq s$. Find $r^2 + s^2$. | To begin, since a circle is tangent to the trapezoid’s four sides, the height of the trapezoid is twice the radius, i.e. $6$. The area formula $\tfrac{h(b_1 + b_2)}{2}=72$ gives:
\[
\frac{6(r+s)}{2} = 72\quad\Longrightarrow\quad r + s = 24.
\]
Next, by Pitot’s Theorem for a tangential quadrilateral, $AB + CD = BC + AD$... | 504 |
2025-I-7 | The twelve letters $A,B,C,D,E,F,G,H,I,J,K,L$ are randomly grouped into six pairs of letters. The two letters in each pair are placed next to each other in alphabetical order to form six two-letter words, and then those six words are listed alphabetically. For example, a possible result is $AB, CJ, DG, EK, FL, HI$. The ... | Note that when pairing the 12 letters into 6 two-letter words, each word is written in alphabetical order, and then the resulting 6 words are themselves listed in alphabetical order. Because of this, we treat different permutations that lead to the same final listing as identical.
We construct two main cases, focusing... | 821 |
2025-I-8 | Let $k$ be a real number such that the system
has exactly one complex solution $z$. The sum of all possible values of $k$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Here $i = \sqrt{-1}$. | We want exactly one complex solution $z$ satisfying the circle $|z-(25+20i)|=5$ and $|z-(4+k)|=|z-(3i+k)|$. Geometrically, $z$ must lie on the perpendicular bisector of the segment from $(4+k,0)$ to $(k,3)$, and that line must be tangent to the circle centered at $(25,20)$ with radius $5$. That yields two possible $k$ ... | 77 |
2025-I-9 | The parabola with equation $y = x^2 - 4$ is rotated $60^{\circ}$ counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has $y$-coordinate $\frac{a - \sqrt{b}}{c}$, where $a, b,$ and $c$ are positive integers, and $a$ and $c$ are relatively prime.... | To begin with, notice that a $60^{\circ}$ counterclockwise rotation about the origin on the $y$-axis can be viewed as a reflection over the line $y = -\sqrt{3}\,x$ (per the argument given). Since the parabola $y = x^2 - 4$ is symmetric about the $y$-axis, reflecting it over that line suffices to find intersections that... | 62 |
2025-I-10 | The $27$ cells of a $3 \times 9$ grid are filled in using the numbers $1$ through $9$ so that each row contains $9$ different numbers, and each of the three $3\times 3$ blocks heavily outlined in the example below contains $9$ different numbers, as in the first three rows of a Sudoku puzzle.
[asy]
unitsize(20);
add(gr... | We fill out the $3\times 9$ grid row by row. There are $9!$ ways to place distinct digits $1\dots 9$ in the first row. (Observe $9! = 2^7\cdot 3^4\cdot 5\cdot 7.$)
Then the second row is placed under certain restrictions so each $3\times 3$ block contains $1\dots 9.$ A condensed case analysis (with partial symmetry ar... | 81 |
2025-I-11 | A piecewise linear function is defined by
\[
f(x) = \begin{cases}
x & \text{if } -1 \leq x < 1,\\
2 - x & \text{if } 1 \le x < 3,
\end{cases}
\]
and $f(x + 4) = f(x)$ for all real numbers $x$. The graph of $f(x)$ has the sawtooth pattern depicted below.
The parabola $x^2 = 34y$ intersects the graph of $f(x)$ a... | Note that $f(x)$ is made of line segments, e.g. $y=x -4k$ or $y=(4k+2)-x,$ each repeating every $4$ units. Then $x=34y^2$ intersects each line in at most two points, whose $y$-coordinates sum to $\pm\tfrac{1}{34}$ by Vieta. Typically those pairs cancel out in the sum of $y$-coordinates, except near the boundary where t... | 259 |
2025-I-12 | The set of points in $3$-dimensional coordinate space that lie in the plane $x+y+z=75$ whose coordinates satisfy the inequalities
\[
x - yz < y - zx < z - xy
\]
forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form $a\sqrt{b},$ w... | Consider the inequalities $x - yz < y - zx < z - xy$ subject to $x+y+z=75$. Through pairwise comparisons (like $z(y - x) > x - y$), we deduce specific sign constraints for $x,y,z$. Most resulting cases allow infinite adjustments (making them unbounded), except one arrangement yields $z>y>x>-1.$
Thus the feasible regio... | 510 |
2025-I-13 | Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws $25$ more lines segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting these two points. Find the expect... | We partition the disk into 4 quadrants with perpendicular diameters. We then randomly select two distinct quadrants for each line segment’s endpoints.
\textbf{Probability that two segments intersect.}\newline
- \(\textbf{Case 1:}\) A segment $A$ has endpoints in opposite quadrants (probability $\tfrac13$). Suppose tho... | 204 |
2025-I-14 | Let $ABCDE$ be a convex pentagon with $AB=14,$ $BC=7,$ $CD=24,$ $DE=13,$ $EA=26,$ and $\angle B=\angle E=60^{\circ}.$ For each point $X$ in the plane, define $f(X)=AX+BX+CX+DX+EX.$ The least possible value of $f(X)$ can be expressed as $m+n\sqrt{p},$ where $m$ and $n$ are positive integers and $p$ is not divisible by t... | Assume $AX=a,\, BX=b,\, CX=c$. By Ptolemy's inequality we have $a + 2b \geq \sqrt{3}\,XE$ and $a + 2c \geq \sqrt{3}\,BX,$ while the equality is reached when both $CXAB$ and $AXDE$ are concyclic. Since $\angle BCA=\angle BXA=\angle EXA=\angle ADE=90^{\circ}$, it follows that $B,X,E$ lie on the same line. Thus, the desir... | 60 |
2025-I-15 | Let $N$ denote the number of ordered triples of positive integers $(a, b, c)$ such that $a, b, c \leq 3^6$ and $a^3 + b^3 + c^3$ is a multiple of $3^7$. Find the remainder when $N$ is divided by $1000$. | \boxed{735} | 735 |
2025-II-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 $\triangle BGE.$ | Let $AB=a$, $BC=b$, $CD=c$, $DE=d$ and $EF=e$. Then we know that $a+b+c+d+e=73$, $a+b=26$, $b+c=22$, $c+d=31$ and $d+e=33$. From this we can easily deduce $c=14$ and $a+e=34$ thus $b+c+d=39$. Using Heron's formula we can calculate the area of $\triangle{CGD}$ to be $\sqrt{(42)(28)(12)(2)}=168$, and since the base of $\... | 468 |
2025-II-2 | Find the sum of all positive integers $n$ such that $n + 2$ divides the product $3(n + 3)(n^2 + 9).$ | \[
\textbf{Rewritten Solution:}
\]
\[
\text{Consider the expression }\; \frac{3(n+3)(n^2+9)}{n+2}.\;\text{We seek all positive integers }n\text{ for which this is an integer.}
\]
\[
\text{Observe that }n+3 = (n+2) + 1.\;\text{Hence we can rewrite:}
\]
\[
\frac{3(n+3)(n^2+9)}{n+2} \;=\; \frac{3\bigl((n+2) + 1\bigr)(n^2... | 49 |
2025-II-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.
\[
\begi... | Since all four edges of the center are common to both squares, we consider five distinct cases:
Case 1: All center edges are used. There is only one way to do this.
Case 2: Three center edges are used, meaning two squares are missing an edge. For each square, there are 2 ways to choose an edge, resulting in $2 \times... | 82 |
2025-II-4 | The product
\[
\prod_{k=4}^{63} \frac{\log_k\bigl(5^{k^2 - 1}\bigr)}{\log_{k+1}\bigl(5^{k^2 - 4}\bigr)}
\;=\; \frac{\log_4\bigl(5^{15}\bigr)}{\log_5\bigl(5^{12}\bigr)} \;\cdot\; \frac{\log_5\bigl(5^{24}\bigr)}{\log_6\bigl(5^{21}\bigr)} \;\cdot\; \frac{\log_6\bigl(5^{35}\bigr)}{\log_7\bigl(5^{32}\bigr)} \;\cdots\; \frac... | The product is equal to $\displaystyle\prod_{k=4}^{63} \frac{(k-1)(k+1)\log_k 5}{(k-2)(k+2)\log_{k+1} 5}$ by applying the difference of squares and properties of logarithms.\n\n\[
\prod_{k=4}^{63}\frac{(k-1)(k+1)\log_k 5}{(k-2)(k+2)\log_{k+1} 5}\;=\;\prod_{k=4}^{63}\frac{\log_k 5}{\log_{k+1} 5} \cdot \frac{3\cdot5\cdot... | 106 |
2025-II-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{A... | Notice that due to midpoints, $\triangle DEF\sim\triangle FBD\sim\triangle AFE\sim\triangle EDC\sim\triangle ABC$. As a result, the angles and arcs are directly related.
Due to inscribed angles,
\[
\widehat{DE} = 2\angle DFE = 2\angle ACB = 2\cdot36 = 72^\circ.
\]
Similarly,
\[
\widehat{FG} = 2\angle FDB = 2\angle ACB... | 336 |
2025-II-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 ... | Let $GH = 2x$ and $GF = 2y$. First, observe that since $\overline{BC}$ is perpendicular to $\overline{GH}$ (using angle chasing) and $\overline{BC}$ is a diameter of $\omega_2$, it follows that $\overline{CB}$ is the perpendicular bisector of $\overline{GH}$. Similarly, because $\overline{AD}$ is perpendicular to $\ove... | 293 |
2025-II-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 elements is $2025$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | We split into different conditions:
Note that for the set to have all elements' $\mathrm{lcm} = 2025$, it must include at least one number that is a multiple of $3^4$ and at least one number that is a multiple of $5^2$.
\[
\text{Multiples of }3^4:\quad 81,\;405,\;2025\quad\quad
\text{Multiples of }5^2:\quad 25,\;75,\... | 237 |
2025-II-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 ... | First, note that every value of $N$ up to 25 behaves correctly under the greedy algorithm (all solutions match the greedy solution). We then analyze larger values of $N$ by looking at patterns where the greedy algorithm fails.
\textbf{1. Initial Cases (\(N\le 25\)):}\newline
All integers $1\le N\le 25$ automatically w... | 610 |
2025-II-9 | There are $n$ values of $x$ in the interval $0<x<2\pi$ for which
\[
f(x) = \sin\bigl(7\pi \cdot \sin(5x)\bigr) = 0.
\]
For $t$ of these $n$ values of $x$, the graph of $y = f(x)$ is tangent to the $x$-axis. Find $n + t$. | We want \(\sin\bigl(7\pi\sin(5x)\bigr)=0\) over the interval \(0<x<2\pi.\)
\[
\sin\bigl(7\pi\sin(5x)\bigr)=0\quad\Longrightarrow\quad 7\pi\sin(5x)=k\pi\quad\Longrightarrow\quad \sin(5x)=\frac{k}{7},\;\text{where}\; k\in\mathbb{Z},\;-7\le k\le7.
\]
Because \(-1\le\sin(5x)\le1\), only integer \(k\) in \([-7,7]\) are rel... | 149 |
2025-II-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 \emph{could} be selected. Find the remainder when $N$ is divided by $1000$. | We split the seat selections into cases based on how the 8 people might form single seats (no neighbor on either side) or pairs of adjacent seats, ensuring no person is adjacent to two others:
\noindent\textbf{Case 1:}\; All 8 people sit singly (no pairs).\newline
The count is given as \(\binom{8}{0}\times\binom{9}{1}... | 907 |
2025-II-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. | We must pair the 24 vertices two by two with segments of the same length, effectively choosing a fixed step size \(k\) around the circle. Each edge is formed by connecting a vertex to the vertex \(k\) steps away.
\textbf{Key Observation:}\newline
The 24-gon can be viewed as a cycle of length 24. When we choose a step ... | 113 |
2025-II-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 ex... | Since all the areas $[A_1A_iA_{i+1}]$ (for $2\le i\le 10$) are the same, let $A_1A_{11}=A_1A_9=\dots=A_1A_3=x$ and $A_1A_2=A_1A_4=\dots=A_1A_{10}=y.$ Because $\angle (A_iA_1A_{i+1})$ is constant, each $A_iA_{i+1}$ (for $2\le i\le 10$) is congruent; call that common side $d.$
Hence we obtain the relationships:
\[
\begi... | 19 |
2025-II-13 | Let the sequence of rationals $x_1, x_2, \dots$ be defined such that $x_1=\tfrac{25}{11}$ and
\[
x_{k+1} = \frac{1}{3}\bigl(x_k + \tfrac{1}{x_k} - 1\bigr).
\]
$x_{2025}$ can be expressed as $\tfrac{m}{n}$ for relatively prime positive integers $m$ and $n$. Find the remainder when $m+n$ is divided by $1000$. | Note that \(x_{k+1} = \tfrac{1}{3}\bigl(\tfrac{x_k^2 - x_k + 1}{x_k}\bigr).\) Rewriting and factoring cleverly reveals:
\[
x_{k+1} = \frac{x_k^3 + 1}{3\,x_k\,(x_k + 1)}.
\]
Let \(x_k = \tfrac{m_k}{n_k}\). Substituting gives
\[
\frac{m_{k+1}}{n_{k+1}} \;=\; \frac{m_k^3 + n_k^3}{3\,m_k\,n_k\,(m_k + n_k)}.
\]
Because \... | 248 |
2025-II-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 that
\[
AK = AL = BK = CL = KL = 14.
\]
The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt{3}$ for some positive integer $n.$ Find $n.$ | We have a right triangle \(\triangle ABC\) with \(\angle A = 90^\circ\) and \(BC = 38.\) Points \(K\) and \(L\) are located such that \(AK = AL = BK = CL = KL = 14.\)
\textbf{Angle Observations:}\newline
From the conditions, one can deduce \(\angle LAK = 60^\circ\) and that \(\triangle LCA\) and \(\triangle BAK\) are ... | 104 |
2025-II-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 $k$. | Since $f(x)$ attains its minimum at two points, for some constant $n$ we can write \[ f(x)-n=\frac{(x-\alpha)^2(x-\beta)^2}{x}. \] Comparing coefficients gives \[ \alpha+\beta=94+\frac{k}{2}\quad\text{and}\quad (\alpha\beta)^2=18\cdot72\cdot98\,k, \] so that \[ \alpha\beta=252\sqrt{2k}. \] Matching the $x^2$ terms lead... | 240 |
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