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Difference between revisions of "1996 IMO Problems/Problem 5"

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==Solution==
 
==Solution==
  
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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|$
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Let <math>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|</math>
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 13:29, 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 $s_{1}=\left| AB \right|,\;s_{2}=\left| BC \right|,\;s_{3}=\left| CD \right|,\;s_{4}=\left| DE \right|,\;s_{5}=\left| EF \right|,\;s_{6}=\left| FA \right|$

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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