Difference between revisions of "1981 AHSME Problems/Problem 11"
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Revision as of 13:43, 28 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
Solution
Let the three sides be 
and 
Because of the Pythagorean Theorem,
![\[(a-d)^2 + a^2 = (a+d)^2\]](http://latex.artofproblemsolving.com/b/c/5/bc5880d763c71647a88ec7d028568ce18e987765.png)
This can be simplified to
![\[a(a-4d) = 0.\]](http://latex.artofproblemsolving.com/7/1/9/719be2ceff702b9b7f58e08746213d84854d3696.png)
So, 
is a multiple of 
and the triangle has sides 
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.
The answer is 
-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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
