Triangle ABC has side lengths $AB = 12$, $BC = 24$, and $AC = 18$. The line through the incenter of triangle ABC parallel to $\overline{BC}$ intersects $\overline{AB}$ at M and $\overline{AC}$ at N. What is the perimeter of triangle AMN?

While there is a very elegant solution posted here, I went a more convoluted route:

I found the inradius of the triangle as $\frac{K}{s}$, which after using Heron's formula to find the area $K$ gave the inradius as $\sqrt{15}$.

I then attempted to set up a ratio between the heights of the two similar triangles.

First I found the height of $\triangle ABC$ as its area divided by the length of $BC$, which gave $\frac{27 \sqrt{15}}{24}$. The height of $\triangle AMN$ is then $\frac{27 \sqrt{15}}{24} - \sqrt{15} = \frac{\sqrt{15}}{8}$.

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Thus the ratio between the two heights is $9$, which should be the ratio of the perimeters also — however, this is incorrect according to the posted solution, which states that the similarity constant is instead $\frac{9}{5}$.

I can't see where I messed up though.

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    $\begingroup$ You mislabeled your side lengths, so it looks like you solved for the wrong triangle. $\endgroup$ – onetoinfinity Jan 29 '16 at 1:09
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    $\begingroup$ @onetoinfinity Oops swap $B$ and $C$ in the diagram. $\endgroup$ – 1110101001 Jan 29 '16 at 1:10
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    $\begingroup$ @Arthur That just involves flipping $B$ and $C$. the line is still parallel to the bottom. $\endgroup$ – 1110101001 Jan 29 '16 at 1:11
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    $\begingroup$ Recheck your calculations. The height from $A$ should be more than twice the inradius, which yours isn't. $\endgroup$ – dxiv Jan 29 '16 at 1:13

Your error comes from the fact that the area of the triangle is $$|\triangle ABC| = rs = \frac{1}{2}bh,$$ so there is a missing factor of 2. We know $|\triangle ABC| = 27 \sqrt{15}$, $r = \sqrt{15}$, $s = 27$, $b = 24$, so $h = \frac{9}{4}\sqrt{15}$. Consequently, the ratio of similitude of the larger to the smaller triangle is $$\frac{h}{h-r} = \frac{9/4}{9/4-1} = \frac{9}{5},$$ as claimed.

  • $\begingroup$ Oh duh. Not sure how I missed that.. $\endgroup$ – 1110101001 Jan 29 '16 at 1:16

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