Question

Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points — two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a + b.$

Answer

Solution:

Let $O$ be the center of $\omega.$ The center of $\omega_A$ lies on line $AO$ (the bisector of $\angle A$ ) at some distance $d$ from $A;$ since $\overline{AB}$ makes a $30^\circ$ angle with $AO,$ the radius is $r = d \sin 30^\circ = \frac{d}{2}.$ Internal tangency to $\omega$ requires the center to be $18 - r$ from $O,$ which forces the center past $O:$ $d - 18 = 18 - \frac{d}{2},$ so $d = 24,$ $r = 12,$ and the center is $6$ beyond $O.$

Place $O$ at the origin with $A = (0, 18).$ Then the three centers are $O_A = (0, -6)$ and $O_B, O_C = (\pm 3\sqrt{3}, 3),$ all with radius $12.$ The intersections of $\omega_B$ and $\omega_C$ lie on the $y$ -axis: $27 + (y - 3)^2 = 144$ gives $y = 3 \pm \sqrt{117}.$ The point $(0, 3 + \sqrt{117})$ is closer to $A$ and belongs to the larger triangle, so the smaller triangle has vertex $(0, 3 - \sqrt{117}),$ at distance $\sqrt{117} - 3$ from $O.$

By symmetry the smaller triangle is equilateral with circumradius $\sqrt{117} - 3,$ so its side is $\sqrt{3}\left(\sqrt{117} - 3\right) = \sqrt{351} - \sqrt{27}.$ Thus $a + b = 351 + 27 = 378.$

2022 AIME I Problem 8

Solution:

Problem 8 in Other Years