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solve:enter image description here

I tried dropping the projections from I to the three sides, from here it's clear that the area of the triangle is $rs$ where $s$ is the semiperimeter and $r$ the inradius.

But by Heron's formula the area is just $80 \sqrt{12 383}$. We can then find: $$r=20 \sqrt{\dfrac{203}{61}}.$$

Then I got stuck, I was trying to get the ratio of the similarity between MAN and BAC but I have been unable to do so.

Thanks for the help

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Given that $P_A$ is the intersection between $AI$ and $BC$, then $\triangle AMN$ and $\triangle ABC$ are similar, and the ratio between the two perimeters is just $\frac{AI}{AP}$. Using Van Obel's theorem and the bisector theorem we have: $$\frac{AI}{IP}=\frac{AP_C}{P_C B}+\frac{AP_B}{P_B C}=\frac{AB+AC}{BC},\tag{1}$$ where $P_B$ and $P_C$ are the feet of the angle bisector on $AC$ and $AB$. Now $(1)$ gives: $$\frac{AI}{AP}=\frac{AB+AC}{AB+AC+BC}\tag{2}$$ hence the perimeter of $\triangle AMN$ is just $AB+AC = 272$.

We can achieve the same by proving that $\triangle MIB$ and $\triangle NIC$ are isosceles triangles.

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  • $\begingroup$ An upvote would not be offensive ;) $\endgroup$ – Jack D'Aurizio Sep 2 '14 at 21:53

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