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

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<math>Insert formula here</math>==Problem==
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==Problem==
 
Two sides of a triangle have lengths <math>10</math> and <math>15</math>.  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?
 
Two sides of a triangle have lengths <math>10</math> and <math>15</math>.  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?
  
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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>
  
==Solution 1 (cheap way)==
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==Solution 1 (Process of Elimination)==
*Credit to the Infuzion17 for this solution
 
  
 
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>.
 
'''
 
'''
==Solution 2 (actually solving)==
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[[Category: Introductory Geometry Problems]]
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==Solution 2==
 
'''
 
'''
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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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>.
  
 
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
 
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>.
  
<math>h_{3} = 1.25h_{1}</math>
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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:
  
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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<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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==Video Solution==
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https://youtu.be/27TpizTnMeM
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 01:12, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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