Difference between revisions of "1992 USAMO Problems/Problem 3"
ZzZzZzZzZzZz (talk | contribs) (New page: ==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 throug...) |
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==Solution== | ==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> | ||
Revision as of 14:15, 26 December 2008
Problem
Chords 
, 
, and 
of a sphere meet at an interior point 
but are not contained in the same plane. The sphere through 
, 
, 
, and is tangent to the sphere through 
, 
, 
, and . Prove that 
.
Solution
Consider the plane through 
. This plane, of course, also contains . We can easily find the 
is isosceles because the base angles are equal. Thus, 
. Similarly, 
. Thus, 
. By symmetry, 
and 
, and hence as desired.
