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Difference between revisions of "2013 AMC 10A Problems/Problem 15"

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<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 </math>
 
<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 </math>
  
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==Solution 1 (cheap way)==
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==Solution 1 (Process of Elimination)==
  
 
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
 
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
 
therefore the length of the side perpendicular to that altitude will be between <math>10</math> and <math>15</math>. The only answer choice that meets this requirement is <math>\boxed{\textbf{(D) }12}</math>.
 
therefore the length of the side perpendicular to that altitude will be between <math>10</math> and <math>15</math>. The only answer choice that meets this requirement is <math>\boxed{\textbf{(D) }12}</math>.
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'''
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[[Category: Introductory Geometry Problems]]
  
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==Solution 2==
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'''
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Let the height to the side of length <math>15</math> be <math>h_{1}</math>, the height to the side of length 10 be <math>h_{2}</math>, the area be <math>A</math>, and the height to the unknown side be <math>h_{3}</math>.
  
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-credit goes to infuzion17
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Because the area of a triangle is <math>\frac{bh}{2}</math>, we get that <math>15(h_{1}) = 2A</math> and <math>10(h_{2}) = 2A</math>, so, setting them equal, <math>h_{2} = \frac{3h_{1}}{2}</math>.  From the problem, we know that <math>2h_{3} = h_{1} + h_{2}</math>.  Substituting, we get that
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<cmath>h_{3} = 1.25h_{1}.</cmath> 
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Thus, the side length is going to be <math>\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}</math>.
  
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==Solution 2 (actually solving)==
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==Solution 3 (Answer Choices)==
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Let <math>x</math> be the height of triangle when the base is <math>10</math> and <math>y</math> is the height of the triangles when the base is <math>15</math>. This means the height for when the triangles has the <math>3</math>rd side length, the height would be <math>\dfrac{(x+y)}{2}</math> giving us the following equation:
  
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Let the height to the side of length 15 be <math>h_{1}</math>, the height to the side of length 10 be <math>h_{2}</math>, the area be <math>A</math>, and the height to the unknown side be <math>h_{3}</math>.
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<math>\dfrac{10x}{2}=\dfrac{15y}{2}=\dfrac{n(x+y)}{2}</math>
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For <math>n</math>, we can plug in the answer choices and check when the ratio of <math>\dfrac{x}{y}</math> for <math>\dfrac{10x}{2}=\dfrac{n(x+y)}{2}</math> and <math>\dfrac{15y}{2}=\dfrac{n(x+y)}{2}</math> is the same because we can observe that only for one value, the ratio remains constant. From brute force we get that <math>\boxed{n=12}</math> meaning that <math>12</math> is the <math>3</math>rd base.
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~ Batmanstark
  
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Because the area of a triangle is <math>\frac{bh}{2}</math>, we get that <math>15(h_{1}) = 2A</math> and <math>10(h_{2}) = 2A</math>, so, setting them equal, <math>h_{2} = \frac{3h_{1}}{2}</math>. Now, we know that <math>2h_{3} = h_{1} + h_{2}</math>.  Substituting, we get that
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==Video Solution==
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https://youtu.be/27TpizTnMeM
  
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<math>h_{3} = 1.25h_{1}</math>. 
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~savannahsolver
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Thus, the side length is going to be <math>\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}</math>.
 
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2013|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2013|ab=A|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 00:12, 22 September 2022

Problem

Two sides of a triangle have lengths $10$ and $15$. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?


$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Solution 1 (Process of Elimination)

The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, therefore the length of the side perpendicular to that altitude will be between $10$ and $15$. The only answer choice that meets this requirement is $\boxed{\textbf{(D) }12}$.

Solution 2

Let the height to the side of length $15$ be $h_{1}$, the height to the side of length 10 be $h_{2}$, the area be $A$, and the height to the unknown side be $h_{3}$.

Because the area of a triangle is $\frac{bh}{2}$, we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$, so, setting them equal, $h_{2} = \frac{3h_{1}}{2}$. From the problem, we know that $2h_{3} = h_{1} + h_{2}$. Substituting, we get that \[h_{3} = 1.25h_{1}.\] Thus, the side length is going to be $\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}$.

Solution 3 (Answer Choices)

Let $x$ be the height of triangle when the base is $10$ and $y$ is the height of the triangles when the base is $15$. This means the height for when the triangles has the $3$rd side length, the height would be $\dfrac{(x+y)}{2}$ giving us the following equation:

$\dfrac{10x}{2}=\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ For $n$, we can plug in the answer choices and check when the ratio of $\dfrac{x}{y}$ for $\dfrac{10x}{2}=\dfrac{n(x+y)}{2}$ and $\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ is the same because we can observe that only for one value, the ratio remains constant. From brute force we get that $\boxed{n=12}$ meaning that $12$ is the $3$rd base. ~ Batmanstark

Video Solution

https://youtu.be/27TpizTnMeM

~savannahsolver

See Also

2013 AMC 10A (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 10 Problems and Solutions

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