2022 AIME I Problems/Problem 10

Problem

Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ .

solution 1

Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately. $a-11,b-13,c-19$ . According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2;(11+13)^2-(b-a)^2=560$ . After solving we have $b-a=4$ , plug this back to $11^2-a^2=13^2-b^2; a=4,b=8,c=16$

The desired value is $(11+19)^2-(16-4)^2=\boxed{756}$

~bluesoul

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2022 AIME I Problems/Problem 10

Problem

solution 1

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