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

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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>. | 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>. | ||

## 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 by HydroQuantum

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 .