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 18:13, 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 .