2017 AMC 12B Problems/Problem 15

Revision as of 21:56, 16 February 2017 by Hydroquantum (talk | contribs) (Solution 2: Inspection)

Problem 15

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3AB$. Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3BC$, and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3CA$. What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$?

$\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1$


Solution 1: Law of Cosines

Solution by HydroQuantum

Let $AB=BC=CA=x$.


Recall The Law of Cosines. Letting $A'B'=B'C'=C'A'=y$, \[y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) =\] \[(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.\] Since both $\triangle ABC$ and $\triangle A'B'C'$ are both equilateral triangles, they must be similar due to $AA$ similarity. This means that $\frac{A'B'}{AB}$ $=$ $\frac{B'C'}{BC}$ $=$ $\frac{C'A'}{CA}$ $=$ $\frac{[\triangle A'B'C']}{[\triangle ABC]}$ $=$ $\frac{37}{1}$.


Therefore, our answer is $\boxed{\textbf{(E) }37:1}$.

Solution 2: Inspection

Note that the height and base of $\triangle A'CC'$ are respectively 4 times and 3 times that of $\triangle ABC$. Therefore the area of $\triangle A'CC'$ is 12 times that of $\triangle ABC$.

By symmetry, $\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'$. Adding the areas of these three triangles and $\triangle ABC$ for the total area of $\triangle A'B'C'$ gives a ratio of $(12 + 12 + 12 + 1) : 1$, or $\boxed{\textbf{(E) } 37 : 1}$.


2017 AMC 10B Problem 19 Solution

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