Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 7: Easiest Solution using Scaling)
(→‎Solution 7: Easiest Solution using Scaling)
Line 50: Line 50:
  
 
== Solution 7: Easiest Solution using Scaling==
 
== Solution 7: Easiest Solution using Scaling==
−
Since equilateral triangles and <math>120^\circ-x^\circ-(60-x)^\circ</math> triangles have the same form for area formulas (and the answer to this problem is simply found by adding <math>\text{(former+3*latter)/former}</math> ), a lot can cancel out when considering the ratio. Suppose a net factor of <math>f</math> is common to the two formulas. Now scale the diagram so that <math>f=1</math>. The area of ABC is simply then of the form <math>k*k=k^2</math>, whereas the area of A'AB', etc... is of the form <math>3k*4k=12k^2</math>. We can now scale down by another factor of k^2 to get <math>1</math> and <math>12</math>, respectively. Therefore, the overall area of A'B'C' is 12*3+1=37, and the area of ABC is simply 1, which gives us <math>\boxed{37,E}</math>.
+
Since equilateral triangles and <math>120^\circ-x^\circ-(60-x)^\circ</math> triangles have the same form for area formulas (and the answer to this problem is simply found by adding <math>\text{(former+3*latter)/former}</math> ), a lot can cancel out when considering the ratio. Suppose a net factor of <math>f</math> is common to the two formulas. Now scale the diagram so that <math>f=1</math>. The area of <math>\Delta ABC</math> is simply then of the form <math>k*k=k^2</math>, whereas the area of <math>\Delta A'AB'</math>, etc... is of the form <math>3k*4k=12k^2</math>. We can now scale down by another factor of <math>k^2</math> to get <math>1</math> and <math>12</math>, respectively. Therefore, the overall area of A'B'C' is <math>12*3+1=37</math>, and the area of ABC is simply <math>1</math>, which gives us <math>\boxed{37,E}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:49, 25 July 2017

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

Solution 1

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 $\triangle ABC$

It's quite obvious that $A'=(4,0,-3), B'=(-3,4,0), C'=(0,-3,4)$

now, since the coordinates are homogenized ($-3+4=1$), we can directly apply the area formula:

$[A'B'C']=[ABC]\cdot\left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC]$

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, $[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC]$

this yields that $[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37 : 1}$


Solutions 5 and 6 by Blast_S1

Solution 7: Easiest Solution using Scaling

Since equilateral triangles and $120^\circ-x^\circ-(60-x)^\circ$ triangles have the same form for area formulas (and the answer to this problem is simply found by adding $\text{(former+3*latter)/former}$ ), a lot can cancel out when considering the ratio. Suppose a net factor of $f$ is common to the two formulas. Now scale the diagram so that $f=1$. The area of $\Delta ABC$ is simply then of the form $k*k=k^2$, whereas the area of $\Delta A'AB'$, etc... is of the form $3k*4k=12k^2$. We can now scale down by another factor of $k^2$ to get $1$ and $12$, respectively. Therefore, the overall area of A'B'C' is $12*3+1=37$, and the area of ABC is simply $1$, which gives us $\boxed{37,E}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_19&oldid=86592"