Difference between revisions of "1992 USAMO Problems/Problem 4"

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== Resources ==
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{{USAMO box|year=1992|num-b=3|num-a=5}}
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks]
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[[Category:Olympiad Algebra Problems]]

Revision as of 10:52, 22 April 2010

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.

$\mathbb{QED.}$

Resources

1992 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions
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