Difference between revisions of "1992 USAMO Problems/Problem 3"

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-==Problem==
-Chords <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> of a sphere meet at an interior point <math>P</math> but are not contained in the same plane.  The sphere through <math>A</math>, <math>B</math>, <math>C</math>, and <math>P</math> is tangent to the sphere through <math>A'</math>, <math>B'</math>, <math>C'</math>, and <math>P</math>.  Prove that <math>AA'=BB'=CC'</math>.
-==Solution==
-Consider the plane through <math>A,A',B,B'</math>.  This plane, of course, also contains <math>P</math>. We can easily find the <math>\triangle APB</math> is isosceles because the base angles are equal. Thus, <math>AP=BP</math>. Similarly, <math>A'P=B'P</math>. Thus, <math>AA'=BB'</math>. By symmetry, <math>BB'=CC'</math> and <math>CC'=AA'</math>, and hence <math>AA'=BB'=CC'</math> as desired.
-<math>\mathbb{QED.}</math>

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