Difference between revisions of "1996 IMO Problems/Problem 5"
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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> | ||
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+ | From the parallel lines on the hexagon we get: | ||
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+ | $\alpha_{1}=\alpha_{4},\;\alpha_{2}=\alpha_{5},\;\alpha_{3}=\alpha_{6},\$ | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 13:34, 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.