Difference between revisions of "1988 AHSME Problems/Problem 22"
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==Solution== | ==Solution== | ||
| − | + | We first notice that the sides <math>10</math> and <math>24</math>, can be part of <math>2</math> different right triangles, one with sides <math>10,24,26</math>, and the other with a leg | |
| − | + | somewhere between <math>21</math> and <math>22</math>. We now notice that if <math>x</math> is less than <math>21</math>, one of the angles is obtuse, and that the same is the same for | |
| + | any value of <math>x</math> above <math>26</math>. Thus the only integer values of <math>x</math> that fit the conditions, are <math>x=22, 23, 24, \text{and }25.</math> | ||
| + | So, the answer is <math>\boxed{\text{A}}</math> | ||
== See also == | == See also == | ||
Revision as of 06:03, 31 August 2015
Problem
For how many integers 
does a triangle with side lengths 
and have all its angles acute?
Solution
We first notice that the sides 
and 
, can be part of 
different right triangles, one with sides 
, and the other with a leg
somewhere between 
and 
. We now notice that if is less than , one of the angles is obtuse, and that the same is the same for
any value of above 
. Thus the only integer values of that fit the conditions, are 
So, the answer is 
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
