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

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− | Recall The Law of Cosines. Letting <math>A'B'=B'C'=C'A'=y</math>, < | + | Recall The Law of Cosines. Letting <math>A'B'=B'C'=C'A'=y</math>, <cmath>y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = </cmath> <cmath>(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.</cmath> 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>. |

## Revision as of 19:07, 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

Solution by HydroQuantum

Let .

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

Therefore, our answer is .