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

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

− | ==Solution 2 | + | ==Solution 2: Inspection== |

Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>. | Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>. | ||

## Revision as of 21:56, 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 1: Law of Cosines

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 .

## Solution 2: Inspection

Note that the height and base of are respectively 4 times and 3 times that of . Therefore the area of is 12 times that of .

By symmetry, . Adding the areas of these three triangles and for the total area of gives a ratio of , or .