Question

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 $m + n.$

Answer

Solution:

Center $\omega_2$ at the origin with $B = (15, 0).$ Internal tangency at $B$ puts $A = (9, 0),$ and $C = (-15, 0).$ Since $\overline{AD} \perp \overline{BC}$ and $D$ is on $\omega_2,$ we get $D = (9, 12)$ (taking $D$ above the line). Because $\overline{EF} \perp \overline{BC},$ the rectangle has vertical sides, so its vertices are $(9 \pm a, \pm b)$ with $a^2 + b^2 = 36.$ The conditions on $C$ and $D$ make $\overline{GH}$ the left side and $\overline{FG}$ the top side: $F = (9 + a, b),$ $G = (9 - a, b),$ $H = (9 - a, -b),$ $E = (9 + a, -b).$

Triangle $DGF$ has base $GF = 2a$ and height $12 - b,$ so its area is $a(12 - b).$ Triangle $CHG$ has base $GH = 2b$ and height $(9 - a) - (-15) = 24 - a,$ so its area is $b(24 - a).$ Setting these equal, $12a - ab = 24b - ab,$ so $a = 2b,$ and then $a^2 + b^2 = 5b^2 = 36.$

The area of the rectangle is $2a \cdot 2b = 8b^2 = \frac{288}{5},$ so $m + n = 288 + 5 = 293.$

2025 AIME II Problem 6

Solution:

Problem 6 in Other Years