Difference between revisions of "2017 AMC 10B Problems/Problem 19"

Revision as of 21:55, 27 November 2020

Problem

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 $\triangle ABC$ ?

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

Solution

$[asy] size(12cm); dot((0,0)); dot((2,0)); dot((1,1.732)); dot((-6,0)); dot((5,-5.196)); dot((4,6.928)); draw((0,0)--(2,0)--(1,1.732)--cycle); draw((-6,0)--(5,-5.196)--(4,6.928)--cycle); draw((-6,0)--(0,0)); draw((5,-5.196)--(2,0)); draw((4,6.928)--(1,1.732)); [/asy]$

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, $\angle B'BC' = 120^{\circ}$ , so by the Law of Cosines, $B'C' = \sqrt{37}$ . 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

@@ Line 2: / Line 2: @@
 Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3 \cdot AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3 \cdot BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3 \cdot CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?
-<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37</math>
+<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math>
 == Solution ==
 <asy>
@@ Line 20: / Line 21: @@
 </asy>
 ===Solution 1 (Uses Trig) ===
-Note that by symmetry, <math>\triangle A'B'C'</math> is also equilateral. Therefore, we only need to find one of the sides of <math>A'B'C'</math> to determine the area ratio. WLOG, let <math>AB = BC = CA = 1</math>. Therefore, <math>BB' = 3</math> and <math>BC' = 4</math>. Also, <math>\angle B'BC' = 120^{\circ}</math>, so by the Law of Cosines, <math>B'C' = \sqrt{37}</math>. Therefore, the answer is <math>(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37}</math>
+Note that by symmetry, <math>\triangle A'B'C'</math> is also equilateral. Therefore, we only need to find one of the sides of <math>A'B'C'</math> to determine the area ratio. WLOG, let <math>AB = BC = CA = 1</math>. Therefore, <math>BB' = 3</math> and <math>BC' = 4</math>. Also, <math>\angle B'BC' = 120^{\circ}</math>, so by the Law of Cosines, <math>B'C' = \sqrt{37}</math>. Therefore, the answer is <math>(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37:1}</math>
 ===Solution 2 ===
-As mentioned in the first solution, <math>\triangle A'B'C'</math> is equilateral. WLOG, let <math>AB=2</math>. Let <math>D</math> be on the line passing through <math>AB</math> such that <math>A'D</math> is perpendicular to <math>AB</math>. Note that <math>\triangle A'DA</math> is a <math>30-60-90</math> with right angle at <math>D</math>. Since <math>AA'=6</math>, <math>AD=3</math> and <math>A'D=3\sqrt{3}</math>. So we know that <math>DB'=11</math>. Note that <math>\triangle A'DB'</math> is a right triangle with right angle at <math>D</math>. So by the Pythagorean theorem, we find <math>A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.</math> Therefore, the answer is <math>(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37}</math>.
+As mentioned in the first solution, <math>\triangle A'B'C'</math> is equilateral. WLOG, let <math>AB=2</math>. Let <math>D</math> be on the line passing through <math>AB</math> such that <math>A'D</math> is perpendicular to <math>AB</math>. Note that <math>\triangle A'DA</math> is a <math>30-60-90</math> with right angle at <math>D</math>. Since <math>AA'=6</math>, <math>AD=3</math> and <math>A'D=3\sqrt{3}</math>. So we know that <math>DB'=11</math>. Note that <math>\triangle A'DB'</math> is a right triangle with right angle at <math>D</math>. So by the Pythagorean theorem, we find <math>A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.</math> Therefore, the answer is <math>(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37:1}</math>.
 ===Solution 3 ===
@@ Line 35: / Line 36: @@
 Similarly from above we know <math>AB'</math> and <math>A'A</math> to be <math>4x</math>, and <math>3x</math> respectively. Thus the area of <math>\Delta{AA'B'}</math> is <math>\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}</math>. Likewise we can find <math>BB'C', \text{ and } CC'A'</math> to also be <math>3x^2\cdot\sqrt{3}</math>.
 <math>[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)</math>.
-Therefore the ratio of <math>[A'B'C']</math> to <math>[ABC]</math> is <math>\frac{\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37}</math>
+Therefore the ratio of <math>[A'B'C']</math> to <math>[ABC]</math> is <math>\frac{\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37:1}</math>
 ===Solution 4 (Elimination) ===
-Looking at the answer choices, we see that all but <math>{\textbf{(E)}}</math> 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 <math>\boxed{\textbf{(E) } 37}</math>.
+Looking at the answer choices, we see that all but <math>{\textbf{(E)}}</math> 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 <math>\boxed{\textbf{(E) } 37:1}</math>.
 Solution by sp1729
@@ Line 45: / Line 46: @@
 We use barycentric coordinates wrt <math>\triangle ABC</math>, to which we can easily obtain that <math>A'=(4,0,-3)</math>, <math>B'=(-3,4,0)</math>, and <math>C'=(0,-3,4)</math>. Now, since the coordinates are homogenized (<math>-3+4=1</math>), we can directly apply the area formula to obtain that
 <cmath>[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],</cmath>
-so the answer is <math>\boxed{\textbf{(E) } 37}</math>
+so the answer is <math>\boxed{\textbf{(E) } 37:1}</math>
 ===Solution 6 (Area Comparison) ===
 First, comparing bases yields that <math>[BA'B']=3[AA'B]=9[ABC]\implies [AA'B']=12</math>. By congruent triangles,
 <cmath>[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC],</cmath>
-so <math>[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37}</math>
+so <math>[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37:1}</math>
 === Solution 7 (Quick Proportionality) ===
@@ Line 58: / Line 59: @@
 ===Solution 8 (Sin area formula) ===
-Drawing the diagram, we see that the large triangle, <math>A'B'C'</math>, is composed of three congruent triangles with the triangle <math>ABC</math> at the center. Let each of the sides of triangle <math>ABC</math> be <math>x</math>. Therefore, using the equilateral triangle area formula, the <math>[ABC] = \frac{x^2\sqrt{3}}{4}</math>. We also know now that the sides of the triangles are <math>3x</math> and <math>3x + x</math>, or <math>4x</math>. We also know that since <math>BB'</math> are collinear, as are the others, angle <math>C'BB'</math> is <math>180 - 60</math>, which is <math>120</math> degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are <math>\frac{12x^2\sin120}{2} \cdot 3</math>. Simplifying that yields <math>\frac{36x^2\sqrt{3}}{4}</math>. Adding that to the <math>[ABC]</math> yields <math>\frac{37x^2\sqrt{3}}{4}</math>. From this, we can compare the ratios by canceling everything out except for the <math>37</math>, so the answer is <math>\boxed{\textbf{(E) }37}</math>
+Drawing the diagram, we see that the large triangle, <math>A'B'C'</math>, is composed of three congruent triangles with the triangle <math>ABC</math> at the center. Let each of the sides of triangle <math>ABC</math> be <math>x</math>. Therefore, using the equilateral triangle area formula, the <math>[ABC] = \frac{x^2\sqrt{3}}{4}</math>. We also know now that the sides of the triangles are <math>3x</math> and <math>3x + x</math>, or <math>4x</math>. We also know that since <math>BB'</math> are collinear, as are the others, angle <math>C'BB'</math> is <math>180 - 60</math>, which is <math>120</math> degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are <math>\frac{12x^2\sin120}{2} \cdot 3</math>. Simplifying that yields <math>\frac{36x^2\sqrt{3}}{4}</math>. Adding that to the <math>[ABC]</math> yields <math>\frac{37x^2\sqrt{3}}{4}</math>. From this, we can compare the ratios by canceling everything out except for the <math>37</math>, so the answer is <math>\boxed{\textbf{(E) }37:1}</math>
 ~Solution by EricShi1685

2017 AMC 10B (Problems • Answer Key • Resources)
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