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A triangle has inradius $4$ cm and a circumradius of $\frac{65}{8}$ cm. Find its area.

I let the sides of the triangle be $a$, $b$, and $c$. After using their corresponding formulas and some manipulation, I got $65 = \frac{abc}{a+b+c}$. I'm kind of stuck here and I don't know how to proceed.

suklay
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  • $13,14,15$ cm work, but how to derive it?.. – Oleg567 Oct 01 '19 at 13:37
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    You can let one of your sides be anything from $2\sqrt{49-\sqrt{65}}\approx 12.8$ to $2\sqrt{49+\sqrt{65}}\approx 15.1$ and find other 2 sides. So the problem is underdetermined. – Vasily Mitch Oct 01 '19 at 15:38
  • If you fix $A$ and incircle, you can draw infinitely many segments $BC$ with different $R/r$. I don't see how this helps – Vasily Mitch Oct 01 '19 at 16:11
  • Based on @VasilyMitch comment, the area of triangle can be any real number from the segment $$\left[\sqrt{\dfrac{14113-\sqrt{65}}{2}},\sqrt{\dfrac{14113+\sqrt{65}}{2}} \right]$$ or (approximately) $$[;83.9789787454...;;, ;; 84.026966676622...;].$$ – Oleg567 Oct 02 '19 at 13:06

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You are correct in the formula you found and in the particular example you found for the sides. But @VasilyMitch is correct in his comment that the area is not uniquely determined.

The diagram below may help to show what is going on.

Let's use the usual notation of $R$ for circumradius and $r$ for inradius, and let's call the circumcentre $O$ and the incentre $I$. Euler's formula gives the distance $d=OI$ as satisfying: $$d^2=R(R-2r)$$

In this case $d$ is much smaller than $r$ or $R$, so we get two circles which are nearly concentric. Hence the triangle doesn't vary much from equilateral. In fact, $\angle A$ is smallest, about $52^o$ when $A$ lies on the ray $IO$ and largest, about $68^o$ when it lies diametrically opposite on the ray $OI$.

enter image description here

The only case when all the sides are integral is the one you found with sides $13,14,15$. You might enjoy figuring out $\angle AIO$ in that case.

The area can take any value in a narrow range around 84. It cannot vary much, but to be unique we would need $d=0$, in which case the circles would be concentric and the triangle would be equilateral.

almagest
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  • The smallest area of triangle: $$\sqrt{\dfrac{14113-\sqrt{65}}{2}} \approx 83.9789787454...$$ can be reached on isosceles triangle with sides $$ 2\sqrt{49+\sqrt{65}}, \quad\frac{1}{8}\sqrt{130(97-\sqrt{65})}, \quad\frac{1}{8}\sqrt{130(97-\sqrt{65})}. $$ The largest area of triangle: $$\sqrt{\dfrac{14113+\sqrt{65}}{2}} \approx 84.026966676622...$$ can be reached on isosceles triangle with sides $$ 2\sqrt{49-\sqrt{65}}, \quad\frac{1}{8}\sqrt{130(97+\sqrt{65})}, \quad\frac{1}{8}\sqrt{130(97+\sqrt{65})}. $$ – Oleg567 Oct 02 '19 at 13:01