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

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<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math> | <math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math> | ||

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+ | ==Solution== | ||

+ | Let <math>AB=BC=CA=x</math>. Then, the area of the small (inside) equilateral triangle is <math>\frac{x^2\sqrt{3}}{4}</math>. Therefore the denominator of the ratio must be <math>\frac{x^2\sqrt{3}}{4}</math>. | ||

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

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+ | Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>. | ||

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[[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]] | [[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]] |

## Revision as of 17:13, 16 February 2017

## Problem 15

Let be an equilateral triangle. Extend side beyond to a point so that . Similarly, extend side beyond to a point so that , and extend side beyond to a point so that . What is the ratio of the area of to the area of ?

## Solution

Let . Then, the area of the small (inside) equilateral triangle is . Therefore the denominator of the ratio must be .

Recall The Law of Cosines. Letting , . This simplifies to . Since both and are both equilateral triangles, they must be similar due to similarity. This means that .

Therefore, our answer is .