Difference between revisions of "2022 AIME I Problems/Problem 8"

Revision as of 15:30, 18 February 2022

Problem

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 $a+b$ .

Diagram

$[asy] unitsize(0.3cm); draw(circle((0,0),18)); pair A = (9 * sqrt(3), -9); pair B = (-9 * sqrt(3), -9); pair C = (0,18); draw(A--B--C--cycle); draw(circle((0,-6),12), gray); draw(circle((3*sqrt(3),3),12), gray); draw(circle((-3*sqrt(3),3),12), gray); pair X = (0, 3-sqrt(117)); pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 ); pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 ); dot(X); dot(Y); dot(Z); draw(X--Y--Z--cycle, dashed); [/asy]$

Solution 1

We can extend $AB$ and $AC$ to $B'$ and $C'$ such that circle $\omega_A$ is the incircle of $\triangle AB'C'$ .

$[asy] unitsize(0.3cm); draw(circle((0,0),18)); pair A = (9 * sqrt(3), -9); pair B = (-9 * sqrt(3), -9); pair B2 = (-12 * sqrt(3), -18); pair C2 = (12 * sqrt(3), -18); pair C = (0,18); draw(A--B--C--cycle); draw(circle((0,-6),12), gray); draw(circle((3*sqrt(3),3),12), gray); draw(circle((-3*sqrt(3),3),12), gray); draw(B--B2,dashed); draw(C--C2,dashed); draw(B2--C2,dashed); dot(B2); dot(C2); pair X = (0, 3-sqrt(117)); pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 ); pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 ); dot(X); dot(Y); dot(Z); draw(X--Y--Z--cycle, dashed); [/asy]$

Solution 2

Let bottom left point as the origin, the radius of each circle is $36/3=12$ , note that three centers for circles are $(9\sqrt{3},3),(12\sqrt{3},12),(6\sqrt{3},12)$

It is not hard to find that one intersection point lies on $\frac{\sqrt{3}x}{3}$ since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation $(x-9\sqrt{3})^2+(\frac{\sqrt{3}x}{3}-3)^2=12^2$ , getting that $x=\frac{15\sqrt{3}+3\sqrt{39}}{2}$ , the length is $2*(\frac{15\sqrt{3}+3\sqrt{39}-18\sqrt{3}}{2}=3\sqrt{39}-3\sqrt{3}$ , leads to the answer $378$

~bluesoul

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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