2022 AIME I Problems/Problem 8

Problem

Equilateral triangle is inscribed in circle with radius Circle is tangent to sides and and is internally tangent to . Circles and are defined analogously. Circles , , and meet in six pointstwo points for each pair of circles. The three intersection points closest to the vertices of are the vertices of a large equilateral triangle in the interior of , and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of . The side length of the smaller equilateral triangle can be written as , where and are positive integers. Find .

Diagram

Solution 1

We can extend and to and such that circle is the incircle of .

unitsize(0.3cm);
draw(circle((0,0),18));
pair A = (9 * sqrt(3), -9);
pair B = (-9 * sqrt(3), -9);
pair B' = (-12 * sqrt(3), -18);
pair C' = (12 * sqrt(3), -18);
draw(B--B',dashed);
draw(C--C',dashed);
draw(B'--C',dashed);
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);
 (Error making remote request. Unknown error_msg)

Solution 2

Let bottom left point as the origin, the radius of each circle is , note that three centers for circles are

It is not hard to find that one intersection point lies on since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation , getting that , the length is , leads to the answer

~bluesoul

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.