HMMT-Feb 2017 geo, Problem 9

Let $A B C$ be a triangle, and let $B C D E, C A F G, A B H I$ be squares that do not overlap the triangle with centers $X, Y, Z$ respectively. Given that $A X=6, B Y=7$, and $C Z=8$, find the area of triangle $X Y Z$.

Solution: By the degenerate case of Von Aubel's Theorem we have that $Y Z=A X=6$ and $Z X=B Y=7$ and $X Y=C Z=8$ so it suffices to find the area of a $6-7-8$ triangle which is given by $\frac{21 \sqrt{15}}{4}$. To prove that $A X=Y Z$, note that by $\mathrm{LoC}$ we get \[ Y X^{2}=\frac{b^{2}}{2}+\frac{c^{2}}{2}+b c \sin \angle A \] and \[ \begin{aligned} A X^{2} & =b^{2}+\frac{a^{2}}{2}-a b(\cos \angle C-\sin \angle C) \\ & =c^{2}+\frac{a^{2}}{2}-a c(\cos \angle B-\sin \angle B) \\ & =\frac{b^{2}+c^{2}+a(b \sin \angle C+c \sin \angle B)}{2} \\ & =\frac{b^{2}}{2}+\frac{c^{2}}{2}+a h \end{aligned} \] where $h$ is the length of the $A$-altitude of triangle $A B C$. In these calculations we used the well-known fact that $b \cos \angle C+c \cos \angle B=a$ which can be easily seen by drawing in the $A$-altitude. Then since $b c \sin \angle A$ and $a h$ both equal twice the area of triangle $A B C$, we are done. $\fbox{\frac{21 \sqrt{15}}{4}}$.

Answer: \frac{21 \sqrt{15}}{4}