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...) |
(solution) |
||
Line 3: | Line 3: | ||
==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 13:15, 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.