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In the triangle ABC, $MN \parallel BC$, $BO$ and $CO$ are angle bisectors of $\angle MBC$ and $\angle NCB$ respectively. If $AB=12$, $BC=24$ and $AC=18$, then what is the perimeter of $\triangle AMN$ ?

I made an aproximated drawing

enter image description here

I tried with similarity but got nothing there. Maybe i'm not seeing the similar triangles in the correct way. Any hints in this kind of problem?

Thanks.

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    $\begingroup$ Hint: triangles $\triangle AMN$ and $\triangle ABC$ are similar with ratio = (height - inradius) / height from $\,A\,$. $\endgroup$ – dxiv Jun 23 '18 at 3:09
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    $\begingroup$ Hint: O is the incenter, and triangle AMN is similar to triangle ABC $\endgroup$ – user547075 Jun 23 '18 at 3:13
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It's better the following way: $$P_{\Delta AMN}=AM+MO+AN+NO=AM+MB+AN+NC=12+18=30.$$

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  • $\begingroup$ Thanks,i found another way to do it and got 30 too!. But thanks anyway. $\endgroup$ – Rodrigo Pizarro Jun 23 '18 at 3:30
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jun 23 '18 at 3:31
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    $\begingroup$ @Rodrigo Pizarro I added something better. $\endgroup$ – Michael Rozenberg Jun 23 '18 at 3:37
  • $\begingroup$ Exactly the way i did it. Nice. $\endgroup$ – Rodrigo Pizarro Jun 23 '18 at 3:43
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The semi-perimeter is $s = {12+18+24 \over 2} = 27$

The area of $\triangle$ ABC is, by Heron, $\sqrt{27 . 15 . 9 . 3 } = 27 \sqrt{15} $

The inradius is $ r = {Area \over s} = \sqrt{15} $ = the elevation of O.

The elevation of A is $ {2Area\over 24 } = {54\over 24 }\sqrt{15} $

The similarity ratio comes out from the ratio of the two elevations $= {54\over (54-24) }$

The semi-perimeter of $\triangle$ AMN is $= {30 \over 54} 27 = 15 $

The required perimeter is then $30$.

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