HMMT-Nov 2010 gen, Problem 3

Triangle $A B C$ has $A B=5, B C=7$, and $C A=8$. New lines not containing but parallel to $A B$, $B C$, and $C A$ are drawn tangent to the incircle of $A B C$. What is the area of the hexagon formed by the sides of the original triangle and the newly drawn lines?

Solution: From the law of cosines we compute $\measuredangle A=\cos ^{-1}\left(\frac{5^{2}+8^{2}-7^{2}}{2(5)(8)}\right)=60^{\circ}$. Using brackets to denote the area of a region, we find that \[ [A B C]=\frac{1}{2} A B \cdot A C \cdot \sin 60^{\circ}=10 \sqrt{3} \] The radius of the incircle can be computed by the formula \[ r=\frac{2[A B C]}{A B+B C+C A}=\frac{20 \sqrt{3}}{20}=\sqrt{3} \] Now the height from $A$ to $B C$ is $\frac{2[A B C]}{B C}=\frac{20 \sqrt{3}}{7}$. Then the height from $A$ to $D E$ is $\frac{20 \sqrt{3}}{7}-2 r=\frac{6 \sqrt{3}}{7}$. Then $[A D E]=\left(\frac{6 \sqrt{3} / 7}{20 \sqrt{3} / 7}\right)^{2}[A B C]=\frac{9}{100}[A B C]$. Here, we use the fact that $\triangle A B C$ and $\triangle A D E$ are similar. Similarly, we compute that the height from $B$ to $C A$ is $\frac{2[A B C]}{C A}=\frac{20 \sqrt{3}}{8}=\frac{5 \sqrt{3}}{2}$. Then the height from $B$ to $H J$ is $\frac{5 \sqrt{3}}{2}-2 r=\frac{\sqrt{3}}{2}$. Then $[B H J]=\left(\frac{\sqrt{3} / 2}{5 \sqrt{3} / 2}\right)^{2}[A B C]=\frac{1}{25}[A B C]$. Finally, we compute that the height from $C$ to $A B$ is $\frac{2[A B C]}{5}=\frac{20 \sqrt{3}}{5}=4 \sqrt{3}$. Then the height from $C$ to $F G$ is $4 \sqrt{3}-2 r=2 \sqrt{3}$. Then $[C F G]=\left(\frac{2 \sqrt{3}}{4 \sqrt{3}}\right)^{2}[A B C]=\frac{1}{4}[A B C]$. Finally we can compute the area of hexagon $D E F G H J$. We have \[ \begin{gathered} {[D E F G H J]=[A B C]-[A D E]-[B H J]-[C F G]=[A B C]\left(1-\frac{9}{100}-\frac{1}{25}-\frac{1}{4}\right)=[A B C]\left(\frac{31}{50}\right)=} \\ 10 \sqrt{3}\left(\frac{31}{50}\right)=\frac{31}{5} \sqrt{3} \end{gathered} \] $\fbox{\frac{31}{5} \sqrt{3}}$.

Answer: \frac{31}{5} \sqrt{3}