Difference between revisions of "1996 IMO Problems/Problem 5"
(→Solution) |
(→Solution) |
||
Line 12: | Line 12: | ||
Let <math>\alpha_{1}=\angle FAB,\;\alpha_{2}=\angle ABC,\;\alpha_{3}=\angle BCD,\;\alpha_{4}=\angle CDE,\;\alpha_{5}=\angle DEF,\;\alpha_{6}=\angle EFA</math> | Let <math>\alpha_{1}=\angle FAB,\;\alpha_{2}=\angle ABC,\;\alpha_{3}=\angle BCD,\;\alpha_{4}=\angle CDE,\;\alpha_{5}=\angle DEF,\;\alpha_{6}=\angle EFA</math> | ||
+ | |||
+ | From the parallel lines on the hexagon we get: | ||
+ | |||
+ | $\alpha_{1}=\alpha_{4},\;\alpha_{2}=\alpha_{5},\;\alpha_{3}=\alpha_{6},$ | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 12:34, 13 November 2023
Problem
Let be a convex hexagon such that is parallel to , is parallel to , and is parallel to . Let , , denote the circumradii of triangles , , , respectively, and let denote the perimeter of the hexagon. Prove that
Solution
Let
Let
Let
From the parallel lines on the hexagon we get:
$\alpha_{1}=\alpha_{4},\;\alpha_{2}=\alpha_{5},\;\alpha_{3}=\alpha_{6},$
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.