BdMO National Secondary 2007#11

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
samiul_samin
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Joined:Sat Dec 09, 2017 1:32 pm
BdMO National Secondary 2007#11

Unread post by samiul_samin » Sat Feb 23, 2019 9:03 am

Find the area of the largest square inscribed in a triangle of sides $5,6$ and $7$.

Ahmed Ashhab Mahir
Posts:4
Joined:Mon Aug 16, 2021 9:10 pm

Re: BdMO National Secondary 2007#11

Unread post by Ahmed Ashhab Mahir » Thu Oct 20, 2022 9:14 pm

There are 3 possibles squares for 3 different bases. Let the base BC be $m$ and side of the square be $x$ .

Now by heron's formula, $[ABC]=6 \sqrt{6}$

Let the vertex of the square B'C'LF in $AB$ be $B'$ and in $AC$ be $C'$ (and F,L in base BC) Now $\angle ABC=90-\angle BB'F =\angle AB'C'$ . Simillarly $\angle ACB= \angle AC'B'$. So $\triangle AB'C' \sim \triangle ABC$

So $\frac{[AB'C']}{[ABC]}=\frac{B'C'^2}{BC^2}=\frac{x^2}{m^2}$. Thus $[ABC]=\frac{6x^2 \sqrt{6}}{m^2}$.
Now, drop a altitude from A to $BC$ and let the feet be $X$. Let the intersection of $B'C'$ and $AX$ be $A'$. As $ B'C' \parallel BC$ and AX is the height of $\triangle ABC$ , AA' is the height of $\triangle AB'C'$ .

Now $\frac{AA'.x}{2}=\frac{6x^2 \sqrt{6}}{m^
2}$

Or,$ AA'= \frac{12x \sqrt{6}}{m^2}$

Simillarly $AX=\frac{12 \sqrt{6}}{m}$.

Now,$AA'+A'X=AX$

Or, $ \frac{12x\sqrt{6}}{m^2}+x=\frac{12 \sqrt{6}}{m}$

Or, $ x= \frac{12m \sqrt{6}}{12\sqrt{6}+m^2}$

Checking m=5,6 and 7 we get m=5 for x being highest and for m=5 , $x^2$ is 7.3 (approx) which is the answer. :)

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