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Difference between revisions of "1967 AHSME Problems/Problem 15"

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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math>
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If the areas of the two triangles are <math>A_1</math> and <math>A_2</math> with <math>A_1 > A_2</math>, we are given that <math>A_1 - A_2 = 18</math> and <math>\frac{A_1}{A_2} = k^2</math>
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Plugging in the second equation into the first leads to <math>(k^2 - 1)A_2 = 18</math>. 
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If <math>A_2</math> is an integer, it can only be a factor of <math>18</math> - namely <math>1, 2, 3, 6, 9, 18</math>. 
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Since <math>A_1 = A_2 + 18</math>, this would lead to <math>A_1 = 19, 20, 21, 24, 27, 36</math>. 
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This would lead to <math>\frac{A_1}{A_2} = 19, 10, 7, 4, 3, 2</math>, and the only square is <math>4</math>. 
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Thus, the ratio of the areas is <math>4</math>, and hence the ratio of the sides is <math>\sqrt{4} = 2</math>.  The corresponding side has a length of <math>3 \cdot 2 = 6</math>, which is option <math>\fbox{D}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=14|num-a=16}}   
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{{AHSME 40p box|year=1967|num-b=14|num-a=16}}   
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:39, 16 August 2023

Problem

The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is:

$\textbf{(A)}\ 12\quad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 6\sqrt{2}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 3\sqrt{2}$

Solution

If the areas of the two triangles are $A_1$ and $A_2$ with $A_1 > A_2$, we are given that $A_1 - A_2 = 18$ and $\frac{A_1}{A_2} = k^2$

Plugging in the second equation into the first leads to $(k^2 - 1)A_2 = 18$.

If $A_2$ is an integer, it can only be a factor of $18$ - namely $1, 2, 3, 6, 9, 18$.

Since $A_1 = A_2 + 18$, this would lead to $A_1 = 19, 20, 21, 24, 27, 36$.

This would lead to $\frac{A_1}{A_2} = 19, 10, 7, 4, 3, 2$, and the only square is $4$.

Thus, the ratio of the areas is $4$, and hence the ratio of the sides is $\sqrt{4} = 2$. The corresponding side has a length of $3 \cdot 2 = 6$, which is option $\fbox{D}$.

See also

1967 AHSC (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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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