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

(Solution 11 (Quick Guess))
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<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>
 
<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 ==
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== Diagram ==
 
<asy>
 
<asy>
 
size(12cm);
 
size(12cm);
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</asy>
 
</asy>
===Solution 1 (Uses Trig) ===
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==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) } 37:1}</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 ===
+
==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) } 37:1}</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 ===
+
==Solution 3 ==
 
Let <math>AB=BC=CA=x</math>. We start by noting that we can just write <math>AB'</math> as just <math>AB+BB'=4AB</math>.
 
Let <math>AB=BC=CA=x</math>. We start by noting that we can just write <math>AB'</math> as just <math>AB+BB'=4AB</math>.
 
Similarly <math>BC'=4BC</math>, and <math>CA'=4CA</math>. We can evaluate the area of triangle <math>ABC</math> by simply using Heron's formula,  
 
Similarly <math>BC'=4BC</math>, and <math>CA'=4CA</math>. We can evaluate the area of triangle <math>ABC</math> by simply using Heron's formula,  
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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>
 
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) ===
+
==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) } 37:1}</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
 
Solution by sp1729
  
===Solution 5 (Barycentric Coordinates) ===
+
==Solution 5 (Barycentric Coordinates) ==
 
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
 
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| 403340034 \right| = (64-27)[ABC]=37[ABC],</cmath>
 
<cmath>[A'B'C']=[ABC]\cdot\left| 403340034 \right| = (64-27)[ABC]=37[ABC],</cmath>
 
so the answer is <math>\boxed{\textbf{(E) } 37:1}</math>
 
so the answer is <math>\boxed{\textbf{(E) } 37:1}</math>
  
===Solution 6 (Area Comparison) ===
+
==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,  
 
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>
 
<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) } 37:1}</math>
 
so <math>[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37:1}</math>
  
=== Solution 7 (Quick Proportionality) ===
+
== Solution 7 (Quick Proportionality) ==
 
Scale down the figure so that the area formulas for the <math>120^\circ</math> 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 <math>3\cdot4\cdot3+1\cdot1=37, \boxed{\text{E}}</math>.
 
Scale down the figure so that the area formulas for the <math>120^\circ</math> 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 <math>3\cdot4\cdot3+1\cdot1=37, \boxed{\text{E}}</math>.
  
 
~ solution by mathchampion1
 
~ solution by mathchampion1
  
===Solution 8 (Sin area formula) ===
+
==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) }37:1}</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>
  
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===Solution 9 (Same as Solution 8 but faster)===
+
==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 <math>\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}</math>. Now simply multiply by three and add <math>\frac{\sqrt{3}}{4}</math> (the area of the small triangle) we get <math>\frac{37\sqrt{3}}{4}</math> and so the ratio is <math>37:1</math>. <math>\boxed{\textbf{(E) }37:1}</math>
 
WLOG, let the side length of the smaller triangle be 1. The area of the big portion (A'B'A) is then <math>\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}</math>. Now simply multiply by three and add <math>\frac{\sqrt{3}}{4}</math> (the area of the small triangle) we get <math>\frac{37\sqrt{3}}{4}</math> and so the ratio is <math>37:1</math>. <math>\boxed{\textbf{(E) }37:1}</math>
  
===Solution 10 (Area Ratios)===
+
==Solution 10 (Area Ratios)==
 
Connect <math>BA', CB',</math> and <math>AC'</math>. Let <math>[\triangle ABC]=k</math>. <math>\frac{[\triangle AC'C]}{[\triangle ABC]}=\frac{CC'}{BC}=3</math>. So the area of <math>\triangle ACC'</math> is equal to <math>3k</math>. We can do the same thing for <math>\triangle AA'C'</math>. <math>\frac{[\triangle A'AC']}{[\triangle ACC']}=\frac{A'A}{AC}=3</math>. Thus, the area of <math>\triangle AA'C'</math> is equal to <math>9k</math>. We will now find the area of <math>\triangle A'B'C'</math> in terms of <math>k</math>. <math>[\triangle A'B'C']=[\triangle ABC]+[\triangle A'CC']+[\triangle B'AA']+[\triangle C'BB']=[\triangle ABC]+3[\triangle A'CC']</math>. The area of <math>\triangle A'CC'</math> is equal to the sum of the areas of <math>\triangle ACC'</math> and <math>\triangle AA'C'</math>, which is <math>12k</math>. So the area of <math>\triangle A'B'C'</math> is equal to <math>37k</math> and the area of <math>\triangle ABC</math> is equal to <math>k</math> so the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math> is equal to <math>\boxed{\textbf{(E) } 37:1}</math>
 
Connect <math>BA', CB',</math> and <math>AC'</math>. Let <math>[\triangle ABC]=k</math>. <math>\frac{[\triangle AC'C]}{[\triangle ABC]}=\frac{CC'}{BC}=3</math>. So the area of <math>\triangle ACC'</math> is equal to <math>3k</math>. We can do the same thing for <math>\triangle AA'C'</math>. <math>\frac{[\triangle A'AC']}{[\triangle ACC']}=\frac{A'A}{AC}=3</math>. Thus, the area of <math>\triangle AA'C'</math> is equal to <math>9k</math>. We will now find the area of <math>\triangle A'B'C'</math> in terms of <math>k</math>. <math>[\triangle A'B'C']=[\triangle ABC]+[\triangle A'CC']+[\triangle B'AA']+[\triangle C'BB']=[\triangle ABC]+3[\triangle A'CC']</math>. The area of <math>\triangle A'CC'</math> is equal to the sum of the areas of <math>\triangle ACC'</math> and <math>\triangle AA'C'</math>, which is <math>12k</math>. So the area of <math>\triangle A'B'C'</math> is equal to <math>37k</math> and the area of <math>\triangle ABC</math> is equal to <math>k</math> so the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math> is equal to <math>\boxed{\textbf{(E) } 37:1}</math>
 
-Heavytoothpaste
 
-Heavytoothpaste
  
  
===Solution 11 (Quick Guess If You Have No Time)===
+
==Solution 11 (Quick Guess If You Have No Time)==
 
 
 
 
 
 
 
<math>(A)</math>, <math>(B)</math>, <math>(C)</math>, and <math>(D)</math> are all perfect squares, which makes them seem unlikely, so we can guess that the answer probably is <math>\boxed{(E)}</math>.
 
<math>(A)</math>, <math>(B)</math>, <math>(C)</math>, and <math>(D)</math> are all perfect squares, which makes them seem unlikely, so we can guess that the answer probably is <math>\boxed{(E)}</math>.
  

Revision as of 08:18, 21 August 2022

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 $\triangle ABC$?

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

Diagram

[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


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 $\boxed{(E)}$.

Video Solution (Law of Cosines)

https://youtu.be/4_x1sgcQCp4?t=5373

~ pi_is_3.14

Video Solution (Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=710

~ 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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