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2022 AIME I Problems/Problem 8

Revision as of 15:45, 18 February 2022 by Kingravi (talk | contribs) (→‎Solution 1)

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 $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$.

[asy] unitsize(0.3cm); draw(circle((0,0),18)); pair C = (9 * sqrt(3), -9); pair B = (-9 * sqrt(3), -9); pair B2 = (-12 * sqrt(3), -18); pair C2 = (12 * sqrt(3), -18); pair A = (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); label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$B'$",B2,W); label("$C'$",C2,E); label("$X$",X,S); label("$Y$",Y,E); label("$Z$",Z,W); draw(X--Y--Z--cycle, dashed); [/asy]


Since the diameter of the circle is the height of this triangle, the height of this triangle is 36. We can use inradius or equilateral triangle properties to get the inradius of this trianglee is 12 (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is 12.

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

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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