2017 AMC 10B Problem 19

Let each side of $ABC$ be $s$. Then $BB' = CC' = AA' = 3s$, and $AB' = BC' = CA' = s + 3s = 4s$.

The big triangle $A'B'C'$ is made of $ABC$ plus three outside triangles: $AA'B'$, $BB'C'$ and $CC'A'$. The picture looks the same after a one-third turn, so the three outside triangles have equal areas. Find one of them.

1. Look at triangle $AA'B'$. Its side $AB'$ lies along line $AB$ and has length $4s$. Its side $AA'$ lies along line $CA$ (past $A$) and has length $3s$.
2. The angle between them at $A$: the angle of $ABC$ at $A$ is $60^\circ$, and $AA'$ goes the opposite way from $AC$, so the angle is $180^\circ - 60^\circ = 120^\circ$.
3. Height from $A'$ to line $AB'$: the $120^\circ$ angle leaves a $60^\circ$ angle on the other side of $A$, so dropping the perpendicular from $A'$ makes a $30$-$60$-$90$ triangle with long side $3s$. Its height is $\frac{\sqrt3}{2} \cdot 3s = \frac{3\sqrt3}{2}s$.
4. Area of $AA'B'$: $\frac12 \cdot 4s \cdot \frac{3\sqrt3}{2}s = 3\sqrt3\,s^2$.
5. Area of $ABC$: $\frac{\sqrt3}{4}s^2$. So $AA'B'$ is $3\sqrt3 \div \frac{\sqrt3}{4} = 12$ times as big as $ABC$.
6. Total: $1 + 3 \cdot 12 = 37$ times the area of $ABC$.

The answer is $\boxed{\textbf{(E)}\ 37:1}$.

Since the construction is the same at all three corners, one outside triangle tells you everything. Each outside triangle has a $120^\circ$ angle (the angle on a straight line next to a $60^\circ$ angle), and half base times height with a $30$-$60$-$90$ triangle gives its area. The big triangle is not a scaled copy of the small one, so splitting it into pieces is the safe way.

Use coordinates with $s = 1$: $A = (0, 0)$, $B = (1, 0)$, $C = \left(\frac12, \frac{\sqrt3}{2}\right)$. Going $4$ side-lengths from each vertex along its side:

• $B' = (4, 0)$,
• $C' = B + 4(C - B) = \left(-1, 2\sqrt3\right)$,
• $A' = C + 4(A - C) = \left(-\frac32, -\frac{3\sqrt3}{2}\right)$.

The shoelace formula gives the area of $A'B'C'$ as $\frac{37\sqrt3}{4}$, and $ABC$ has area $\frac{\sqrt3}{4}$, a ratio of $37 : 1$.

Let $AB = BC = CA = s$, so $[ABC] = \frac{\sqrt3}{4}s^2$. Then $AB' = 4s$, $BC' = 4s$, $CA' = 4s$, while $BB' = CC' = AA' = 3s$.

Triangle $A'B'C'$ decomposes into $\triangle ABC$ and three outer triangles: $\triangle AA'B'$, $\triangle BB'C'$, $\triangle CC'A'$. Consider $\triangle AA'B'$: it has $AB' = 4s$ and $AA' = 3s$, and the angle at $A$ between ray $AB'$ (along $AB$) and ray $AA'$ (along $CA$ extended past $A$) is $180^\circ - 60^\circ = 120^\circ$.

Drop the perpendicular from $A'$ to line $AB'$. It makes a $30$-$60$-$90$ triangle with hypotenuse $3s$, so the height is $\frac{\sqrt3}{2} \cdot 3s$. Hence
$$[AA'B'] = \frac12 \cdot 4s \cdot \frac{3\sqrt3}{2}s = 3\sqrt3\, s^2 = 12 \cdot \frac{\sqrt3}{4}s^2 = 12\,[ABC].$$
By the rotational symmetry of the configuration, the other two outer triangles have the same area. Therefore
$$[A'B'C'] = [ABC] + 3 \cdot 12\,[ABC] = 37\,[ABC].$$

The answer is $\boxed{\textbf{(E)}\ 37:1}$.