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I tried to calculate the length of the side $a'$ of the contact triangle of $\Delta ABC$ (the side corresponding the the side $BC$), and I found $(-a+b+c)\sin \frac {A}2$ (if we denote the incenter by $I$, the ends of $a'$ by $E$ and $F$, and $M$ the midpoint of $EF$, we have $a'=EF=2FM=2\cdot AF\cdot \frac{FM}{AF}=2\cdot (s-a)\cdot \sin \angle FAM=(-a+b=c)\sin\frac{A}2$).

But Wolfram say this is $(-a+b+c)\cos \frac{A}2$. Am I wrong, or Wolfram is?

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    $\begingroup$ Please share your working on this problem. If you have made a mistake, this will allow us to point out where you error arises $\endgroup$ Jun 10, 2022 at 15:32

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The formula given by Wolfram is indeed wrong. Here is a counterexample: if the triangle is equilateral with sidelengths $a=b=c=1$ (with a contact triangle with sidelengths $a'=b'=c'=1/2$), using their formula gives

$$a'=(1+1-1)\cos(\pi/6)=\sqrt{3}/2$$

instead of $a'=1/2$.

Your proof is correct ; another one is given here.

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