10. In triangle $A B C, A B=13, B C=14, C A=15$. Squares $A B B_{1} A_{2}, B C C_{1} B_{2}, C A A_{1} C_{2}$ are constructed outside the triangle. Squares $A_{1} A_{2} A_{3} A_{4}, B_{1} B_{2} B_{3} B_{4}, C_{1} C_{2} C_{3} C_{4}$ are constructed outside the hexagon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2}$. Squares $A_{3} B_{4} B_{5} A_{6}, B_{3} C_{4} C_{5} B_{6}, C_{3} A_{4} A_{5} C_{6}$ are constructed outside the hexagon $A_{4} A_{3} B_{4} B_{3} C_{4} C_{3}$. Find the area of the hexagon $A_{5} A_{6} B_{5} B_{6} C_{5} C_{6}$.
Answer:
19444

Solution 1.
We can use complex numbers to find synthetic observations. Let $A=a, B=b, C=c$. Notice that $B_{2}$ is a rotation by $-90^{\circ}$ (counter-clockwise) of $C$ about $B$, and similarly $C_{1}$ is a rotation by $90^{\circ}$ of $B$ about $C$. Since rotation by $90^{\circ}$ corresponds to multiplication by $i$, we have $B_{2}=(c-b) \cdot(-i)+b=b(1+i)-c i$ and $C_{1}=(b-c) \cdot i+c=b i+c(1-i)$. Similarly, we get $C_{2}=c(1+i)-a i, A_{1}=c i+a(1-i)$, $A_{2}=a(1+i)-b i, B_{1}=a i+b(1-i)$. Repeating the same trick on $B_{1} B_{2} B_{3} B_{4}$ et. al, we get $C_{4}=-a+b(-1+i)+c(3-i), C_{3}=a(-1-i)-b+c(3+i), A_{4}=-b+c(-1+i)+a(3-i)$, $A_{3}=b(-1-i)-c+a(3+i), B_{4}=-c+a(-1+i)+b(3-i), B_{3}=c(-1-i)-a+b(3+i)$. Finally, repeating the same trick on the outermost squares, we get $B_{6}=-a+b(3+5 i)+c(-3-3 i), C_{5}=$ $-a+b(-3+3 i)+c(3-5 i), C_{6}=-b+c(3+5 i)+a(-3-3 i), A_{5}=-b+c(-3+3 i)+a(3-5 i), A_{6}=$ $-c+a(3+5 i)+b(-3-3 i), B_{5}=-c+a(-3+3 i)+b(3-5 i)$.
From here, we observe the following synthetic observations.
S1. $B_{2} C_{1} C_{4} B_{3}, C_{2} A_{1} A_{4} C_{3}, A_{2} B_{1} B_{4} A_{3}$ are trapezoids with bases of lengths $B C, 4 B C ; A C, 4 A C ; A B, 4 A B$ and heights $h_{a}, h_{b}, h_{c}$ respectively (where $h_{a}$ is the length of the altitude from $A$ to $B C$, and likewise for $h_{b}, h_{c}$ )

S2. If we extend $B_{5} B_{4}$ and $B_{6} B_{3}$ to intersect at $B_{7}$, then $B_{7} B_{4} B_{3} \cong B B_{1} B_{2} \sim B_{7} B_{5} B_{6}$ with scale factor 1:5. Likewise when we replace all $B$ 's with $A$ 's or $C$ 's.

Proof of S1. Observe $C_{1}-B_{2}=c-b$ and $C_{4}-B_{3}=4(c-b)$, hence $B_{2} C_{1} \| B_{3} C_{4}$ and $B_{3} C_{4}=4 B_{2} C_{1}$. Furthermore, since translation preserves properties of trapezoids, we can translate $B_{2} C_{1} C_{4} B_{3}$ such that $B_{2}$ coincides with $A$. Being a translation of $a-B_{2}$, we see that $B_{3}$ maps to $B_{3}^{\prime}=2 b-c$ and $C_{4}$ maps to $C_{4}^{\prime}=-2 b+3 c$. Both $2 b-c$ and $-2 b+3 c$ lie on the line determined by $b$ and $c$ (since $-2+3=2-1=1$ ), so the altitude from $A$ to $B C$ is also the altitude from $A$ to $B_{3}^{\prime} C_{4}^{\prime}$. Thus $h_{a}$ equals the length of the altitude from $B_{2}$ to $B_{3} C_{4}$, which is the height of the trapezoid $B_{2} C_{1} C_{4} B_{3}$. This proves $\mathrm{S} 1$ for $B_{2} C_{1} C_{4} B_{3}$; the other trapezoids follow similarly.
Proof of S2. Notice a translation of $-a+2 b-c$ maps $B_{1}$ to $B_{4}, B_{2}$ to $B_{3}$, and $B$ to a point $B_{8}=-a+3 b-c$. This means $B_{8} B_{3} B_{4} \cong B B_{1} B_{2}$. We can also verify that $4 B_{8}+B_{6}=5 B_{3}$ and $4 B_{8}+B_{5}=5 B_{4}$, showing that $B_{8} B_{5} B_{6}$ is a dilation of $B_{8} B_{4} B_{3}$ with scale factor 5 . We also get $B_{8}$ lies on $B_{3} B_{6}$ and $B_{5} B_{4}$, so $B_{8}=B_{7}$. This proves $\mathrm{S} 2$ for $B_{3} B_{4} B_{5} B_{6}$, and similar arguments prove the likewise part.
Now we are ready to attack the final computation. By S2, $\left[B_{3} B_{4} B_{5} B_{6}\right]+\left[B B_{1} B_{2}\right]=\left[B_{7} B_{5} B_{6}\right]=$ $\left[B B_{1} B_{2}\right]$. But by the $\frac{1}{2} a c \sin B$ formula, $\left[B B_{1} B_{2}\right]=[A B C]$ (since $\left.\angle B_{1} B B_{2}=180^{\circ}-\angle A B C\right)$. Hence,
$\left[B_{3} B_{4} B_{5} B_{6}\right]+\left[B B_{1} B_{2}\right]=25[A B C]$. Similarly, $\left[C_{3} C_{4} C_{5} C_{6}\right]+\left[C C_{1} C_{2}\right]=25[A B C]$ and $\left[A_{3} A_{4} A_{5} A_{6}\right]+$ $\left[A A_{1} A_{2}\right]=25[A B C]$. Finally, the formula for area of a trapezoid shows $\left[B_{2} C_{1} C_{4} B_{3}\right]=\frac{5 B C}{2} \cdot h_{a}=$ $5[A B C]$, and similarly the other small trapezoids have area $5[A B C]$. The trapezoids thus contribute area $(75+3 \cdot 5)=90[A B C]$. Finally, $A B C$ contributes area $[A B C]=84$.
By S1, the outside squares have side lengths $4 B C, 4 C A, 4 A B$, so the sum of areas of the outside squares is $16\left(A B^{2}+A C^{2}+B C^{2}\right)$. Furthermore, a Law of Cosines computation shows $A_{1} A_{2}^{2}=A B^{2}+A C^{2}+$ $2 \cdot A B \cdot A C \cdot \cos \angle B A C=2 A B^{2}+2 A C^{2}-B C^{2}$, and similarly $B_{1} B_{2}^{2}=2 A B^{2}+2 B C^{2}-A C^{2}$ and $C_{1} C_{2}^{2}=2 B C^{2}+2 A C^{2}-A B^{2}$. Thus the sum of the areas of $A_{1} A_{2} A_{3} A_{4}$ et. al is $3\left(A B^{2}+A C^{2}+B C^{2}\right)$. Finally, the small squares have area add up to $A B^{2}+A C^{2}+B C^{2}$. Aggregating all contributions from trapezoids, squares, and triangle, we get
$$
\left[A_{5} A_{6} B_{5} B_{6} C_{5} C_{6}\right]=91[A B C]+20\left(A B^{2}+A C^{2}+B C^{2}\right)=7644+11800=19444 .
$$

Solution 2. Let $a=B C, b=C A, c=A B$. We can prove $\mathrm{S} 1$ and $\mathrm{S} 2$ using some trigonometry instead. Proof of $S 1$. The altitude from $B_{3}$ to $B_{2} C_{1}$ has length $B_{2} B_{3} \sin \angle B B_{2} B_{1}=B_{1} B_{2} \sin \angle B B_{2} B_{1}=$ $B B_{1} \sin \angle B_{1} B B_{2}=A B \sin \angle A B C=h_{a}$ using Law of Sines. Similarly, we find the altitude from $C_{4}$ to $B_{2} C_{1}$ equals $h_{a}$, thus proving $B_{2} C_{1} C_{4} B_{3}$ is a trapezoid. Using $B_{1} B_{2}=\sqrt{2 a^{2}+2 c^{2}-b^{2}}$ from end of Solution 1, we get the length of the projection of $B_{2} B_{3}$ onto $B_{3} C_{4}$ is $B_{2} B_{3} \cos B B_{2} B_{1}=$ $\frac{\left(2 a^{2}+2 c^{2}-b^{2}\right)+a^{2}-c^{2}}{2 a}=\frac{3 a^{2}+c^{2}-b^{2}}{2 a}$, and similarly the projection of $C_{1} C_{4}$ onto $B_{3} C_{4}$ has length $\frac{3 a^{2}+b^{2}-c^{2}}{2 a}$. It follows that $B_{3} C_{4}=\frac{2 a a^{2}+c^{2}-b^{2}}{2 a}+a+\frac{3 a^{2}+b^{2}-c^{2}}{2 a}=4 a$, proving S1 for $B_{2} C_{1} C_{4} B_{3}$; the other cases follow similarly.
Proof of S2. Define $B_{8}$ to be the image of $B$ under the translation taking $B_{1} B_{2}$ to $B_{4} B_{3}$. We claim $B_{8}$ lies on $B_{3} B_{6}$. Indeed, $B_{8} B_{4} B_{3} \cong B B_{1} B_{2}$, so $\angle B_{8} B_{3} B_{4}=\angle B B_{2} B_{1}=180^{\circ}-\angle B_{3} B_{2} C_{1}=\angle B_{2} B_{3} C_{4}$. Thus $\angle B_{8} B_{3} C_{4}=\angle B_{4} B_{3} B_{2}=90^{\circ}$. But $\angle B_{6} B_{3} C_{4}=90^{\circ}$, hence $B_{8}, B_{3}, B_{6}$ are collinear. Similarly we can prove $B_{5} B_{4}$ passes through $B_{8}$, so $B_{8}=B_{7}$. Finally, $\frac{B_{7} B_{3}}{B_{7} B_{6}}=\frac{B_{7} B_{4}}{B_{7} B_{5}}=\frac{1}{5}$ (using $B_{3} B_{6}=4 a, B_{4} B_{5}=$ $4 c, B_{7} B_{3}=a, B_{7} B_{4}=c$ ) shows $B_{7} B_{4} B_{3} \sim B_{7} B_{5} B_{6}$ with scale factor $1: 5$, as desired. The likewise part follows similarly.