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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 12:34, 13 November 2023

Problem

Let $ABCDEF$ be a convex hexagon such that $AB$ is parallel to $DE$, $BD$ is parallel to $EF$, and $CD$ is parallel to $FA$. Let $R_{A}$, $R_{C}$, $R_{E}$ denote the circumradii of triangles $FAB$, $BCD$, $DEF$, respectively, and let $P$ denote the perimeter of the hexagon. Prove that

$R_{A}+R_{C}+R_{E} \ge \frac{P}{2}$

Solution

Let $s_{1}=\left| AB \right|,\;s_{2}=\left| BC \right|,\;s_{3}=\left| CD \right|,\;s_{4}=\left| DE \right|,\;s_{5}=\left| EF \right|,\;s_{6}=\left| FA \right|$

Let $d_{1}=\left| FB \right|,\;d_{2}=\left| BD \right|,\;d_{1}=\left| DF \right|$

Let $\alpha_{1}=\angle FAB,\;\alpha_{2}=\angle ABC,\;\alpha_{3}=\angle BCD,\;\alpha_{4}=\angle CDE,\;\alpha_{5}=\angle DEF,\;\alpha_{6}=\angle EFA$

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.

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