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I was soving geometry problem. And found this question , i tried to solve it .But i am unable to get the answer.

The hypotenuse of a right triangle has length of 5 cm.

Determine its maximum possible area of its incircle.

I try to solve it by assuming that for maximum area of incircle is when the other two sides of triangle are equal.

But this i am unable to get the answer. Where am i wrong?

Thanks in advance.

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2 Answers

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If $C$ is the right angle, $c=5$ CM

We know from the Law of Sines, $\frac c {\sin C}=2R\implies R=\frac 5 2$

$A+B=\pi-C=\pi-\frac \pi 2=\frac \pi 2 $

Using this, $r=\frac{a+b-c}2=\frac{2R(\sin A+\sin B-\sin C)}2=\frac 52(\sin A+\cos A-1)=\frac 52(\sqrt2 \cos(A-\frac{\pi}4)-1)$

$r$ will be maximum if $\cos(A-\frac{\pi}4)=1,A=2m\pi+\frac{\pi}4$ where $m$ is any integer.

But, $0<A,B<\frac \pi 2$ So, $A=\frac{\pi}4$

So, $B=\frac \pi 2-A=\frac{\pi}4=A$


Form this, the in-radius $r=4R\sin \frac A 2 \sin \frac B 2 \sin \frac C 2$

$r=4\frac 5 2\sin \frac A 2 \sin \frac B 2 \frac 1{\sqrt 2}$ $=\frac5{\sqrt2}2\sin \frac A 2 \sin \frac B 2$ $=\frac5{\sqrt2}\left(\cos(\frac{A-B}2)-\cos(\frac{A+B}2)\right)$

$=\frac5{\sqrt2}\left(\cos(\frac{A-B}2)-\frac 1{\sqrt2}\right)$

This will be maximum if $\cos(\frac{A-B}2)=1$ or

if $\frac{A-B}2=2n\pi$ where $n$ is any integer.

$\implies A=B+4n\pi,$

But $0<A,B<\frac \pi 2$ So, the difference of $A,B$ can not be $>\frac \pi 2$

So, $n=0,A=B\implies a=2R\sin A=2R\sin B=b$

$r_{max}=\frac{5(\sqrt 2-1)}2$

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So the other two sides will be equal? –  vikiiii Nov 4 '12 at 11:53
    
Ya, they must be equal for the maximum value of in-radius. –  lab bhattacharjee Nov 4 '12 at 11:54
    
@vikiiii, I've added another simpler method. –  lab bhattacharjee Nov 4 '12 at 12:12
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The inradius of a right triangle is equal to $(b+c-a)/2$ where $a$ is the hypothenuse. Therefore the maximal radius is when $b+c$ is maximal.

On the other hand you have the inequality $$ b+c \leq \sqrt{2(b^2+c^2)}=a\sqrt{2} $$ with equality if and only if $c=b$.

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IF a = b , then hypotenuse will be equal to other side –  vikiiii Nov 4 '12 at 11:40
    
sorry. I've mixed up the notations, but the ideas were clear. –  Beni Bogosel Nov 4 '12 at 12:22
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