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2022 AIME I Problems/Problem 8

Revision as of 19:54, 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); dot(0,0); label("$O$",(0,0),N); 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 triangle 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 $12$.

Let $O$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$. By rotating this point $120^{\circ}$ around $O$, we get the center of $\omega_B$. This means that the magnitude of vector $\overrightarrow{O\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\sqrt{3},3)$ and by symmetry the coordinates of the center of $\omega_C$ is $(-3\sqrt{3},3)$.

The upper left and right circles intersect at two points, the lower of which is $X$. The equations of these two circles are: \[(x+3\sqrt3)^2 + (y-3)^2 = 12^2\] \[(x-3\sqrt3)^2 + (y-3)^2 = 12^2\]

We solve this system by subtracting to get $x = 0$. Plugging back in to the first equation, we have $(3\sqrt{3})^2 + (y-3)^2 = 144 \implies (y-3)^2 = 117 \implies y-3 = \pm sqrt{117} \implies y = 3 \pm sqrt{117}$. Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\sqrt{117})$.

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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