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If the inradius in a right angled triangle with integer sides is $r$. Prove that

Q1) If $r= 4$, the greatest perimeter (in units) is $90$

Q2) If $r=5$, the greatest area (in sq. units ) is $330$

I tried to solve the question by

$=(s-a)\tan(A/2)$

where, $s$=semi perimeter, $a$ and $A$ are side length and angle corresponding to vertex $A$.

So I assumed vertex $A$ to be right angled and get $$r=(s-a)$$ But to maximise $s$ we need one more equation in $r$ and $a$.

Novice
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Advait Danade
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2 Answers2

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Let the legs of the right-angled triangle be $a,b$, and the hypotenuse be $c$. For question $1$, we known that $r = \frac{1}{2}(a+b-c)\Rightarrow 8 = a+b-c$.

Now use the fact that any Pythagorean triple can be uniquely written in the form:

$$a = k(m^2 -n^2), \ b = k(2mn), \ c = k(m^2+n^2)$$

where $m,n,k$ are positive integers, $m > n$, $m$ and $n$ are coprime and are not both odd (Wikipedia).

Then $a+b-c = 8$ gives:

$$k(m^2 - n^2 + 2mn - m^2 - n^2) = 8$$ $$k(-2n^2+2mn) = 8$$ $$kn(-n+m) = 4$$

There are only a few cases to check, where each case corresponds to a factor pair of $4$. Can you continue?

Toby Mak
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We begin with a variation of Euclid's formula where $(m=2n-1+k).\quad$ Expanded: \begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+2k^2 \end{align*}

Observation:$\quad \text{inradius}=r=(2n -1)k$

Proof: We let inradius(r)=area(a)/semiperimeter(s) and we let $j=(2n-1)$

$a=\dfrac{AB}{2} =\dfrac{(j^2+2jk)(2jk+2k^2)}{2}\\ \quad=jk(j^2+3jk+2k^2)$

$s=\dfrac{A+B+C}{2} =\dfrac{(j^2+2jk)+(2jk+2k^2)+(j^2+2jk+2k^2)}{2}\\ \quad =j^2+3jk+2k^2$

$r=\dfrac{a}{s} =\dfrac{jk(j^2+3jk+2k^2)}{(j^2+3jk+2k^2)}=jk\\ \quad =(2n-1)k$

poetasis
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