2017 AMC 10B Problem 19

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

The three extensions rotate the picture by $120^\circ$ into itself, so a single outer triangle determines everything. Each outer triangle shares a vertex of $ABC$ and has a $120^\circ$ angle there, the supplement of $60^\circ$; its area follows from half base times height with a $30$-$60$-$90$ altitude. Recognizing that $A'B'C'$ is not similar to $ABC$ in a simple scaled way is the main hurdle.

Coordinates with $s = 1$: $A = (0,0)$, $B = (1,0)$, $C = (\frac12, \frac{\sqrt3}{2})$. Then $B' = (4, 0)$, $C' = B + 4(C - B) = (-1, 2\sqrt3)$, $A' = C + 4(A - C) = (-\frac32, -\frac{3\sqrt3}{2})$. The shoelace formula gives $[A'B'C'] = \frac{37\sqrt3}{4}$, and $[ABC] = \frac{\sqrt3}{4}$.

• Assuming the big triangle is a scaled copy with side ratio 4 (or 6) and answering 16:1 or 36:1.
• Treating the $3s$ and $4s$ sides of an outer triangle as perpendicular (area $6s^2$), or forgetting to add the central triangle and answering $36:1$.