Vietnam tst 2004

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SANZEED
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Joined: Wed Dec 28, 2011 6:45 pm
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Vietnam tst 2004

Unread post by SANZEED » Wed May 23, 2012 1:25 am

It was number 5.
Let $A_{1},B_{1},C_{1},D_{1},E_{1}, F_{1}$ be the midpoints of the sides $AB, BC, CD, DE, EF, FA$ respectively of a hexagon $ABCDEF$. Let $p$ be the perimeter of hexagon $ABCDEF$ and $p_{1}$ be that of $A_{1}B_{1}C_{1}D_{1}E_{1} F_{1}$ .Suppose that the hexagon $A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}$ has equal angles. Prove that $p ≥\frac {2}{\sqrt 3}p_{1}$. When does equality hold?.

My $100th$ post!! :) :) :)
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Ragib Farhat Hasan
Posts: 62
Joined: Sun Mar 30, 2014 10:40 pm

Re: Vietnam tst 2004

Unread post by Ragib Farhat Hasan » Mon Nov 11, 2019 1:51 pm

The equality holds only when $ABCDEF$ is also equiangular.

Taking the $\triangle A_1BB_1$, it can be observed that when the perpendicular is dropped from $B$ on to $A_{1}B_{1}$, two right-angled triangles are formed, where the other two angles in each triangle are $30^\circ$ and $60^\circ$.

Hence, we obtain the ratio $=>$

$p:p_1=2:\sqrt3$
$p=\frac {2}{\sqrt3} p_1$.

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