2017 AMC 10B Problems/Problem 19

Problem

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

$\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1$

Solution

Note that by symmetry, $\triangle A'B'C'$ is also equilateral. Therefore, we only need to find one of the sides of $A'B'C'$ to determine the area ratio. WLOG, let $AB = BC = CA = 1$ . Therefore, $BB' = 3$ and $BC' = 4$ . Also, $\angle B'BC' = 120^{\circ}$ , so by the Law of Cosines, $B'C' = \sqrt{37}$ . Therefore, the answer is $(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37 : 1}$

Solution 2 (no trig)

As mentioned in the first solution, $\triangle A'B'C'$ is equilateral. WLOG, let $AB=2$ . Let $D$ be on the line passing through $AB$ such that $A'D$ is perpendicular to $AB$ . Note that $\triangle A'DA$ is a 30-60-90 with right angle at $D$ . Since $AA'=6$ , $AD=3$ and $A'D=3\sqrt{3}$ . So we know that $DB'=11$ . Note that $\triangle A'DB'$ is a right triangle with right angle at $D$ . So by the Pythagorean theorem, we find $A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.$ Therefore, the answer is $(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2017 AMC 10B Problems/Problem 19

Contents

Problem

Solution

Solution 2 (no trig)

See Also