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

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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 $A'B'=B'C'=C'A'=y$ , $y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) =$ $(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24x(cos120)=25x^2+12x^2=37x^2.$ 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 $\boxed{\textbf{(E) } 37 : 1}$ .

Solution 3: Coordinates

First we note that $A'B'C'\sim ABC$ due to symmetry. WLOG, let $B = (0, 0)$ and $AB = 1$ Therefore, $C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)$ . Using the condition that $CC' = 3$ , we get $C' = (4, 0)$ and $B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)$ . It is easy to check that $B'C' = \sqrt{37}$ . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is $\boxed{\textbf{(E) } 37 : 1}$

Solution by mathwiz0803

Solution 4

Note that angle $C'BB'$ is $120$ °, as it is supplementary to the equilateral triangle. Then, using area $= \frac{1}{2}ABsin\theta$ and calling side $AB = 1$ for ease, we get: $4*3*\frac{sin120}{2} = 3\sqrt{3}$ as the area of $C'BB'$ . Then, the area of $ABC$ is $\frac{\sqrt{3}}{4}$ , so the ratio is $\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}$

Solution by Aadileo

2017 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

2017 AMC 10B (Problems • Answer Key • Resources)
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@@ Line 31: / Line 31: @@
 ==Solution 4==
-Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= AB\frac{1}{2}sin\theta</math> and calling side <math>AB = 1</math> for ease, we get: <math>4*3*\frac{sin120}{2} = 3\sqrt{3}</math> as the area of <math>C'BB'</math>. Then, the area of <math>ABC</math> is <math>\frac{\sqrt{3}}{4}</math>, so the ratio is <math>\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}</math>
+Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= \frac{1}{2}ABsin\theta</math> and calling side <math>AB = 1</math> for ease, we get: <math>4*3*\frac{sin120}{2} = 3\sqrt{3}</math> as the area of <math>C'BB'</math>. Then, the area of <math>ABC</math> is <math>\frac{\sqrt{3}}{4}</math>, so the ratio is <math>\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}</math>
 Solution by Aadileo

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