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 "1967 AHSME Problems/Problem 15"
Line 11: Line 11:
 
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>
 
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>
  
−
Plugging in the second equation into the first leads to <math>(k^2 - 1)A_2 = 18</math>.  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>.  Since <math>A_1 = A_2 + 18</math>, this would lead to <math>A_1 = 19, 20, 21, 24, 27, 36</math>.  This would lead to <math>\frac{A_1}{A_2} = 19, 10, 7, 4, \frac{8}{3}, \frac{3}{2}</math>, and the only square is <math>4</math>.   
+
Plugging in the second equation into the first leads to <math>(k^2 - 1)A_2 = 18</math>.   
 +
 
 +
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>.   
 +
 
 +
Since <math>A_1 = A_2 + 18</math>, this would lead to <math>A_1 = 19, 20, 21, 24, 27, 36</math>.   
 +
 
 +
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>.   
  
 
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>.
 
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>.

Revision as of 01:49, 13 July 2019

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 AHSME (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
All AHSME 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=1967_AHSME_Problems/Problem_15&oldid=107593"