2017 AMC 10B Problem 19

P16-20~5 minPrint

Triangles: Area & Pythagorean Angles & Polygons Place the figure on coordinates and compute Cut the figure into known shapes (triangles, rectangles, sectors)

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ ?

Show solution without answering

Key insight

Cut the big triangle into ABC plus three outside triangles. Each has sides 4s and 3s with a 120-degree angle, so area 12 times ABC.

Solution

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.

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 $3s$ .
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 $180^\circ - 60^\circ = 120^\circ$ .
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 $\frac{\sqrt3}{2} \cdot 3s = \frac{3\sqrt3}{2}s$ .
Area of $AA'B'$ : $\frac12 \cdot 4s \cdot \frac{3\sqrt3}{2}s = 3\sqrt3\,s^2$ .
Area of $ABC$ : $\frac{\sqrt3}{4}s^2$ . So $AA'B'$ is $3\sqrt3 \div \frac{\sqrt3}{4} = 12$ times as big as $ABC$ .
Total: $1 + 3 \cdot 12 = 37$ times the area of $ABC$ .

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

Why this works

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.

Alternative approach

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 $37 : 1$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 17: Analytic Geometry (§17.4, §17.6)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)