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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.

3 Answers 3

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Given the three sides of a triangle will result in only one incircle. Radius of incircle is the largest circle that can be inscribed in any triangle. Maximum radius of incircle can only occur if only two sides are given, because it will result into an infinitely number of third sides.

Suppose the question is this, "What is the maximum radius of incircle of a triangle with sides $3$ cm and $4$ cm?" You cannot assume that it is a right triangle and its maximum inradius is $1$ cm because there are so many oblique triangles with sides $3$ cm and $4$ cm with different radii of incircles.

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The inradius of a right triangle is equal to $\frac{b+c-a}2$ 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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    sorry. I've mixed up the $n$otations, but the ideas were clear.2012-11-04
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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 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 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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    @vikiiii, I've added another simpler method.2012-11-04