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Difference between revisions of "1981 AHSME Problems/Problem 11"

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==Solution==
 
==Solution==
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Let the three sides be <math>a-d,</math> <math>a,</math> and <math>a+d.</math>  Because of the Pythagorean Theorem,
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Let the three sides be <math>a-d</math>, <math>a</math>, and <math>a+d</math>. By the Pythagorean Theorem,
 
<cmath>(a-d)^2 + a^2 = (a+d)^2</cmath>
 
<cmath>(a-d)^2 + a^2 = (a+d)^2</cmath>
 
This can be simplified to
 
This can be simplified to
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<cmath>a(a-4d) = 0.</cmath>
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<cmath>a(a-4d) = 0</cmath>
 +
which has solutions
 +
<cmath>a = 0, a = 4d</cmath>
  
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So, <math>a</math> is a multiple of <math>4d</math> and the triangle has sides <math>3d, 4d, 5d.</math> We check the answer choices for anything divisible by 3, 4, or 5. The only one that works is 81, which is divisible by 3.
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Since <math>a=0</math> is invalid, <math>a=4d</math> and the triangle has sides <math>3d</math>, <math>4d</math>, <math>5d</math>.  Therefore, the correct answer must be divisible by <math>3</math>, <math>4</math>, or <math>5</math>. The only valid answer choice is <math>\boxed{\textbf{(C)}\ 81}</math>, since it is divisible by <math>3</math>.
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The answer is <math>\textbf{(C)}\ 81.</math>
 
  
 
-edited by coolmath34
 
-edited by coolmath34

Latest revision as of 14:11, 29 June 2025

Problem 11

The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length

$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 81\qquad\textbf{(D)}\ 91\qquad\textbf{(E)}\ 361$

Solution

Let the three sides be $a-d$, $a$, and $a+d$. By the Pythagorean Theorem, \[(a-d)^2 + a^2 = (a+d)^2\] This can be simplified to \[a(a-4d) = 0\] which has solutions \[a = 0, a = 4d\]

Since $a=0$ is invalid, $a=4d$ and the triangle has sides $3d$, $4d$, $5d$. Therefore, the correct answer must be divisible by $3$, $4$, or $5$. The only valid answer choice is $\boxed{\textbf{(C)}\ 81}$, since it is divisible by $3$.

-edited by coolmath34

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

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