Problem 1723

Solution:

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 $B_1B_2B_3B_4$ 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 $B_6 = -a + b(3+5i) + c(-3-3i)$, $C_5 = -a + b(-3+3i) + c(3-5i)$, $C_6 = -b + c(3+5i) + a(-3-3i)$, $A_5 = -b + c(-3+3i) + a(3-5i)$, $A_6 = -c + a(3+5i) + b(-3-3i)$, $B_5 = -c + a(-3+3i) + b(3-5i)$.

From here, we observe the following synthetic observations.

S1. $B_2C_1C_4B_3$, $C_2A_1A_4C_3$, $A_2B_1B_4A_3$ are trapezoids with bases of lengths $BC, 4BC$; $AC, 4AC$; $AB, 4AB$ and heights $h_a, h_b, h_c$ respectively (where $h_a$ is the length of the altitude from $A$ to $BC$, and likewise for $h_b, h_c$).

S2. If we extend $B_5B_4$ and $B_6B_3$ to intersect at $B_7$, then $B_7B_4B_3 \cong BB_1B_2 \sim B_7B_5B_6$ 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 $B_2C_1 \parallel B_3C_4$ and $B_3C_4 = 4 B_2C_1$. Furthermore, since translation preserves properties of trapezoids, we can translate $B_2C_1C_4B_3$ such that $B_2$ coincides with $A$. Being a translation of $a - B_2$, we see that $B_3$ maps to $B_3' = 2b - c$ and $C_4$ maps to $C_4' = -2b + 3c$. Both $2b - c$ and $-2b + 3c$ lie on the line determined by $b$ and $c$ (since $-2 + 3 = 2 - 1 = 1$), so the altitude from $A$ to $BC$ is also the altitude from $A$ to $B_3'C_4'$. Thus $h_a$ equals the length of the altitude from $B_2$ to $B_3C_4$, which is the height of the trapezoid $B_2C_1C_4B_3$. This proves S1 for $B_2C_1C_4B_3$; the other trapezoids follow similarly.

Proof of S2. Notice a translation of $-a + 2b - c$ maps $B_1$ to $B_4$, $B_2$ to $B_3$, and $B$ to a point $B_8 = -a + 3b - c$. This means $B_8B_3B_4 \cong BB_1B_2$. We can also verify that $4B_8 + B_6 = 5B_3$ and $4B_8 + B_5 = 5B_4$, showing that $B_8B_5B_6$ is a dilation of $B_8B_4B_3$ with scale factor $5$. We also get $B_8$ lies on $B_3B_6$ and $B_5B_4$, so $B_8 = B_7$. This proves S2 for $B_3B_4B_5B_6$, and similar arguments prove the likewise part.

Now we are ready to attack the final computation. By S2, $[B_3B_4B_5B_6] + [BB_1B_2] = [B_7B_5B_6] = [BB_1B_2]$. But by the $\frac{1}{2}ac\sin B$ formula, $[BB_1B_2] = [ABC]$ (since $\angle B_1BB_2 = 180^{\circ} - \angle ABC$). Hence,

$[B_3B_4B_5B_6] + [BB_1B_2] = 25[ABC]$. Similarly, $[C_3C_4C_5C_6] + [CC_1C_2] = 25[ABC]$ and $[A_3A_4A_5A_6] + [AA_1A_2] = 25[ABC]$. Finally, the formula for area of a trapezoid shows $[B_2C_1C_4B_3] = \frac{5BC}{2} \cdot h_a = 5[ABC]$, and similarly the other small trapezoids have area $5[ABC]$. The trapezoids thus contribute area $(75 + 3 \cdot 5) = 90[ABC]$. Finally, $ABC$ contributes area $[ABC] = 84$.

By S1, the outside squares have side lengths $4BC, 4CA, 4AB$, so the sum of areas of the outside squares is $16(AB^2 + AC^2 + BC^2)$. Furthermore, a Law of Cosines computation shows $A_1A_2^2 = AB^2 + AC^2 + 2 \cdot AB \cdot AC \cdot \cos \angle BAC = 2AB^2 + 2AC^2 - BC^2$, and similarly $B_1B_2^2 = 2AB^2 + 2BC^2 - AC^2$ and $C_1C_2^2 = 2BC^2 + 2AC^2 - AB^2$. Thus the sum of the areas of $A_1A_2A_3A_4$ et al. is $3(AB^2 + AC^2 + BC^2)$. Finally, the small squares have area add up to $AB^2 + AC^2 + BC^2$. Aggregating all contributions from trapezoids, squares, and triangle, we get

$$[A_5A_6B_5B_6C_5C_6] = 91[ABC] + 20(AB^2 + AC^2 + BC^2) = 7644 + 11800 = 19444$$

Solution:

Let $a = BC$, $b = CA$, $c = AB$. We can prove S1 and S2 using some trigonometry instead.

Proof of S1. The altitude from $B_3$ to $B_2C_1$ has length $B_2B_3 \sin \angle BB_2B_1 = B_1B_2 \sin \angle BB_2B_1 = BB_1 \sin \angle B_1BB_2 = AB \sin \angle ABC = h_a$ using Law of Sines. Similarly, we find the altitude from $C_4$ to $B_2C_1$ equals $h_a$, thus proving $B_2C_1C_4B_3$ is a trapezoid. Using $B_1B_2 = \sqrt{2a^2 + 2c^2 - b^2}$ from end of Solution 1, we get the length of the projection of $B_2B_3$ onto $B_3C_4$ is $B_2B_3 \cos BB_2B_1 = \frac{(2a^2 + 2c^2 - b^2) + a^2 - c^2}{2a} = \frac{3a^2 + c^2 - b^2}{2a}$, and similarly the projection of $C_1C_4$ onto $B_3C_4$ has length $\frac{3a^2 + b^2 - c^2}{2a}$. It follows that $B_3C_4 = \frac{3a^2 + c^2 - b^2}{2a} + a + \frac{3a^2 + b^2 - c^2}{2a} = 4a$, proving S1 for $B_2C_1C_4B_3$; the other cases follow similarly.

Proof of S2. Define $B_8$ to be the image of $B$ under the translation taking $B_1B_2$ to $B_4B_3$. We claim $B_8$ lies on $B_3B_6$. Indeed, $B_8B_4B_3 \cong BB_1B_2$, so $\angle B_8B_3B_4 = \angle BB_2B_1 = 180^{\circ} - \angle B_3B_2C_1 = \angle B_2B_3C_4$. Thus $\angle B_8B_3C_4 = \angle B_4B_3B_2 = 90^{\circ}$. But $\angle B_6B_3C_4 = 90^{\circ}$, hence $B_8, B_3, B_6$ are collinear. Similarly we can prove $B_5B_4$ passes through $B_8$, so $B_8 = B_7$. Finally, $\frac{B_7B_3}{B_7B_6} = \frac{B_7B_4}{B_7B_5} = \frac{1}{5}$ (using $B_3B_6 = 4a$, $B_4B_5 = 4c$, $B_7B_3 = a$, $B_7B_4 = c$) shows $B_7B_4B_3 \sim B_7B_5B_6$ with scale factor $1:5$, as desired. The likewise part follows similarly.

$$[A_5A_6B_5B_6C_5C_6] = 91[ABC] + 20(AB^2 + AC^2 + BC^2) = 7644 + 11800 = 19444$$