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>, <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>. | 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>. | Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>. |
Revision as of 17:15, 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 . 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 .