Difference between revisions of "2017 AMC 12B Problems/Problem 15"

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Problem 15

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 1: Law of Cosines

Solution by HydroQuantum

Let $AB=BC=CA=x$ .

Recall The Law of Cosines. Letting $A'B'=B'C'=C'A'=y$ , $y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) =$ $(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.$ Since both $\triangle ABC$ and $\triangle A'B'C'$ are both equilateral triangles, they must be similar due to $AA$ similarity. This means that $\frac{A'B'}{AB}$ $=$ $\frac{B'C'}{BC}$ $=$ $\frac{C'A'}{CA}$ $=$ $\frac{[\triangle A'B'C']}{[\triangle ABC]}$ $=$ $\frac{37}{1}$ .

Therefore, our answer is $\boxed{\textbf{(E) }37:1}$ .

Solution 2: Inspection

Note that the height and base of $\triangle A'CC'$ are respectively 4 times and 3 times that of $\triangle ABC$ . Therefore the area of $\triangle A'CC'$ is 12 times that of $\triangle ABC$ .

By symmetry, $\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'$ . Adding the areas of these three triangles and $\triangle ABC$ for the total area of $\triangle A'B'C'$ gives a ratio of $(12 + 12 + 12 + 1) : 1$ , or $\boxed{\textbf{(E) } 37 : 1}$ .

Solution 3: Coordinates

First we note that $A'B'C'\sim ABC$ due to symmetry. WLOG, let $B = (0, 0)$ and $AB = 1$ Therefore, $C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)$ . Using the condition that $CC' = 3$ , we get $C' = (4, 0)$ and $B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)$ . It is easy to check that $B'C' = \sqrt{37}$ . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is $\boxed{\textbf{(E) } 37 : 1}$

Solution by [i]mathwiz0803[/i]

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 12 Problems and Solutions

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.

@@ Line 25: / Line 25: @@
 First we note that <math>A'B'C'\sim ABC</math> due to symmetry. WLOG, let <math>B = (0, 0)</math> and <math>AB = 1</math> Therefore, <math>C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)</math>. Using the condition that <math>CC' = 3</math>, we get <math>C' = (4, 0)</math> and <math>B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)</math>. It is easy to check that <math>B'C' = \sqrt{37}</math>. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is <math>\boxed{\textbf{(E) } 37 : 1}</math>
+Solution by [i]mathwiz0803[/i]
 {{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}}
 {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}}
 {{MAA Notice}}

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