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

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(→‎Solution 8 (Heron's Formula))
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===Solution 8 (Heron's Formula) ===
 
===Solution 8 (Heron's Formula) ===
 
Since there is no explicit length, we can say that the lengths of <math>\triangle ABC</math> are 1, and that <math>AB'</math> and the other two are of length 3. Then, we can solve for the area of the obtuse triangles by using one of the sides of <math>\triangle ABC</math> plus it's extension as a base. For example, <math>AB'</math> will be the base, which is of length <math>3+1 = 4. A'A</math> is of length 3, so the altitude of the obtuse triangle is <math>\frac{3\sqrt{3}}{2}</math> because of <math>30:60:90</math> triangles. We do <math>\frac{3\sqrt{3}}{2}*\frac{4}{2}=3\sqrt{3}</math>. We then multiply by 3 to get the total area of the obtuse triangles, <math>9\sqrt{3}</math>. Now, using the Heron's Formula, we can find the area of <math>\triangle ABC</math>. <math>\frac{1+1+1}{2}=\frac{3}{2}=s</math>, <math>\sqrt{(\frac{3}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})}=\frac{\sqrt{3}}{4}</math>. Thus, the ratio is <math>9\sqrt{3}+\frac{\sqrt{3}}{4}:\frac{\sqrt{3}}{4}</math>, which can easily be simplified to <math>37:1</math>. Thus, the solution is <math>\boxed{\textbf{(E) } 37 : 1}</math>
 
Since there is no explicit length, we can say that the lengths of <math>\triangle ABC</math> are 1, and that <math>AB'</math> and the other two are of length 3. Then, we can solve for the area of the obtuse triangles by using one of the sides of <math>\triangle ABC</math> plus it's extension as a base. For example, <math>AB'</math> will be the base, which is of length <math>3+1 = 4. A'A</math> is of length 3, so the altitude of the obtuse triangle is <math>\frac{3\sqrt{3}}{2}</math> because of <math>30:60:90</math> triangles. We do <math>\frac{3\sqrt{3}}{2}*\frac{4}{2}=3\sqrt{3}</math>. We then multiply by 3 to get the total area of the obtuse triangles, <math>9\sqrt{3}</math>. Now, using the Heron's Formula, we can find the area of <math>\triangle ABC</math>. <math>\frac{1+1+1}{2}=\frac{3}{2}=s</math>, <math>\sqrt{(\frac{3}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})}=\frac{\sqrt{3}}{4}</math>. Thus, the ratio is <math>9\sqrt{3}+\frac{\sqrt{3}}{4}:\frac{\sqrt{3}}{4}</math>, which can easily be simplified to <math>37:1</math>. Thus, the solution is <math>\boxed{\textbf{(E) } 37 : 1}</math>
 +
 
~ Solution by lovemath1337
 
~ Solution by lovemath1337
  

Revision as of 16:20, 15 March 2020

Problem

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

[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 $3*4*3+1*1=37:1, \boxed{\text{E}}$. ~ Solution by mathchampion1

Solution 8 (Heron's Formula)

Since there is no explicit length, we can say that the lengths of $\triangle ABC$ are 1, and that $AB'$ and the other two are of length 3. Then, we can solve for the area of the obtuse triangles by using one of the sides of $\triangle ABC$ plus it's extension as a base. For example, $AB'$ will be the base, which is of length $3+1 = 4. A'A$ is of length 3, so the altitude of the obtuse triangle is $\frac{3\sqrt{3}}{2}$ because of $30:60:90$ triangles. We do $\frac{3\sqrt{3}}{2}*\frac{4}{2}=3\sqrt{3}$. We then multiply by 3 to get the total area of the obtuse triangles, $9\sqrt{3}$. Now, using the Heron's Formula, we can find the area of $\triangle ABC$. $\frac{1+1+1}{2}=\frac{3}{2}=s$, $\sqrt{(\frac{3}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})}=\frac{\sqrt{3}}{4}$. Thus, the ratio is $9\sqrt{3}+\frac{\sqrt{3}}{4}:\frac{\sqrt{3}}{4}$, which can easily be simplified to $37:1$. Thus, the solution is $\boxed{\textbf{(E) } 37 : 1}$

~ Solution by lovemath1337

See Also

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