1992 USAMO Problems/Problem 4

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Problem

Chords $AA'$ , $BB'$ , and $CC'$ of a sphere meet at an interior point $P$ but are not contained in the same plane. The sphere through $A$ , $B$ , $C$ , and $P$ is tangent to the sphere through $A'$ , $B'$ , $C'$ , and $P$ . Prove that $AA'=BB'=CC'$ .

Solution

Consider the plane through $A,A',B,B'$ . This plane, of course, also contains $P$ . We can easily find the $\triangle APB$ is isosceles because the base angles are equal. Thus, $AP=BP$ . Similarly, $A'P=B'P$ . Thus, $AA'=BB'$ . By symmetry, $BB'=CC'$ and $CC'=AA'$ , and hence $AA'=BB'=CC'$ as desired.

$\mathbb{QED.}$

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1992 USAMO Problems/Problem 4

From AoPSWiki

Problem

Solution