2017 AMC 10B Problems/Problem 19


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

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


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Solution 1 (Uses Trig)

Note that by symmetry, $\triangle A'B'C'$ is also equilateral. Therefore, we only need to find one of the sides of $A'B'C'$ to determine the area ratio. WLOG, let $AB = BC = CA = 1$. Therefore, $BB' = 3$ and $BC' = 4$. Also, $\angle B'AC' = 120^{\circ}$, so by the Law of Cosines, $B'C' = \sqrt{4^2 + 3^2 - 2\circ(4)(3)\cos(120^\circ)} = \sqrt{16+9 - 24(-1/2)} = \sqrt{37}$. Therefore, the answer is $(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37:1}$

Solution 2

As mentioned in the first solution, $\triangle A'B'C'$ is equilateral. WLOG, let $AB=2$. Let $D$ be on the line passing through $AB$ such that $A'D$ is perpendicular to $AB$. Note that $\triangle A'DA$ is a $30-60-90$ with right angle at $D$. Since $AA'=6$, $AD=3$ and $A'D=3\sqrt{3}$. So we know that $DB'=11$. Note that $\triangle A'DB'$ is a right triangle with right angle at $D$. So by the Pythagorean theorem, we find $A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.$ Therefore, the answer is $(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37:1}$.

Solution 3

Let $AB=BC=CA=x$. We start by noting that we can just write $AB'$ as just $AB+BB'=4AB$. Similarly $BC'=4BC$, and $CA'=4CA$. We can evaluate the area of triangle $ABC$ by simply using Heron's formula, $[ABC]=\sqrt{\frac{3x}{2}\cdot {\Bigg(\frac{3x}{2}-x\Bigg)}^3}=\frac{x^2\sqrt{3}}{4}$. Next in order to evaluate $A'B'C'$ we need to evaluate the area of the larger triangles $AA'B',BB'C', \text{ and } CC'A'$. In this solution we shall just compute $1$ of these as the others are trivially equivalent. In order to compute the area of $\Delta{AA'B'}$ we can use the formula $[XYZ]=\frac{1}{2}xy\cdot\sin{z}$. Since $ABC$ is equilateral and $A$, $B$, $B'$ are collinear, we already know $\angle{A'AB'}=180-60=120$ Similarly from above we know $AB'$ and $A'A$ to be $4x$, and $3x$ respectively. Thus the area of $\Delta{AA'B'}$ is $\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}$. Likewise we can find $BB'C', \text{ and } CC'A'$ to also be $3x^2\cdot\sqrt{3}$. $[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x^2\sqrt{3}}{4}=\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)$. Therefore the ratio of $[A'B'C']$ to $[ABC]$ is $\frac{\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37:1}$

Solution 4 (Elimination)

Looking at the answer choices, we see that all but ${\textbf{(E)}}$ has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick $\boxed{\textbf{(E) } 37:1}$.

Solution by sp1729

Solution 5 (Barycentric Coordinates)

We use barycentric coordinates wrt $\triangle ABC$, to which we can easily obtain that $A'=(4,0,-3)$, $B'=(-3,4,0)$, and $C'=(0,-3,4)$. Now, since the coordinates are homogenized ($-3+4=1$), we can directly apply the area formula to obtain that \[[A'B'C']=[ABC]\cdot\left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC],\] so the answer is $\boxed{\textbf{(E) } 37:1}$

Solution 6 (Area Comparison)

First, comparing bases yields that $[BA'B']=3[AA'B]=9[ABC]\implies [AA'B']=12$. By congruent triangles, \[[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC],\] so $[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37:1}$

Solution 7 (Quick Proportionality)

Scale down the figure so that the area formulas for the $120^\circ$ and equilateral triangles become proportional with proportionality constant equivalent to the product of the corresponding sides. By the proportionality, it becomes clear that the answer is $3\cdot4\cdot3+1\cdot1=37, \boxed{\text{E}}$.

~ solution by mathchampion1

Solution 8 (Sin area formula)

Drawing the diagram, we see that the large triangle, $A'B'C'$, is composed of three congruent triangles with the triangle $ABC$ at the center. Let each of the sides of triangle $ABC$ be $x$. Therefore, using the equilateral triangle area formula, the $[ABC] = \frac{x^2\sqrt{3}}{4}$. We also know now that the sides of the triangles are $3x$ and $3x + x$, or $4x$. We also know that since $BB'$ are collinear, as are the others, angle $C'BB'$ is $180 - 60$, which is $120$ degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are $\frac{12x^2\sin120}{2} \cdot 3$. Simplifying that yields $\frac{36x^2\sqrt{3}}{4}$. Adding that to the $[ABC]$ yields $\frac{37x^2\sqrt{3}}{4}$. From this, we can compare the ratios by canceling everything out except for the $37$, so the answer is $\boxed{\textbf{(E) }37:1}$

~Solution by EricShi1685

Solution 9 (Same as Solution 8 but faster)

WLOG, let the side length of the smaller triangle be 1. The area of the big portion (A'B'A) is then $\frac{1}{2}\cdot3\cdot4\cdot\sin\left(120\right)=\frac{1}{2}\cdot3\cdot4\cdot\sin\left(60\right)=\frac{1}{2}\cdot3\cdot4\cdot\frac{\sqrt{3}}{2}=3\sqrt{3}$. Now simply multiply by three and add $\frac{\sqrt{3}}{4}$ (the area of the small triangle) we get $\frac{37\sqrt{3}}{4}$ and so the ratio is $37:1$. $\boxed{\textbf{(E) }37:1}$

Solution 10 (Area Ratios)

Connect $BA', CB',$ and $AC'$. Let $[\triangle ABC]=k$. $\frac{[\triangle AC'C]}{[\triangle ABC]}=\frac{CC'}{BC}=3$. So the area of $\triangle ACC'$ is equal to $3k$. We can do the same thing for $\triangle AA'C'$. $\frac{[\triangle A'AC']}{[\triangle ACC']}=\frac{A'A}{AC}=3$. Thus, the area of $\triangle AA'C'$ is equal to $9k$. We will now find the area of $\triangle A'B'C'$ in terms of $k$. $[\triangle A'B'C']=[\triangle ABC]+[\triangle A'CC']+[\triangle B'AA']+[\triangle C'BB']=[\triangle ABC]+3[\triangle A'CC']$. The area of $\triangle A'CC'$ is equal to the sum of the areas of $\triangle ACC'$ and $\triangle AA'C'$, which is $12k$. So the area of $\triangle A'B'C'$ is equal to $37k$ and the area of $\triangle ABC$ is equal to $k$ so the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ is equal to $\boxed{\textbf{(E) } 37:1}$ -Heavytoothpaste

Solution 11 (Quick Guess If You Have No Time)

$(A)$, $(B)$, $(C)$, and $(D)$ are all perfect squares, which makes them seem unlikely, so we can guess that the answer probably is $\boxed{(E)}$ as it is the only one with a side length that is implied to not be an integer.

Video Solution by OmegaLearn (Law of Cosines)


~ pi_is_3.14

Video Solution by OmegaLearn (Meta-Solving Technique)


~ pi_is_3.14

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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