Difference between revisions of "2024 AMC 12B Problems/Problem 21"
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<cmath>\tan{\theta}=\frac{33}{56}</cmath> | <cmath>\tan{\theta}=\frac{33}{56}</cmath> | ||
Hence the base side lengths of the third triangle are <math>33</math> and <math>56</math>. By the Pythagorean Theorem, the hypotenuse of the third triangle is <math>65</math>, so the perimeter is <math>33+56+65=\boxed{\textbf{(C) }154}</math>. | Hence the base side lengths of the third triangle are <math>33</math> and <math>56</math>. By the Pythagorean Theorem, the hypotenuse of the third triangle is <math>65</math>, so the perimeter is <math>33+56+65=\boxed{\textbf{(C) }154}</math>. | ||
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| + | ~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino] | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2024|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:03, 14 November 2024
Problem
The measures of the smallest angles of three different right triangles sum to 
. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are 
and 
. What is the perimeter of the third triangle?
Solution 1
Let 
and 
be the smallest angles of the and triangles respectively. We have
![\[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\]](http://latex.artofproblemsolving.com/0/a/c/0acb8bbc41955aa405118a7e132cc75b502bdd51.png)
Then
![\[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\]](http://latex.artofproblemsolving.com/b/4/1/b4176f0be70485b00eff7d4d09df62bd06a996a7.png)
Let 
be the smallest angle of the third triangle. Consider
![\[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\]](http://latex.artofproblemsolving.com/2/d/9/2d952abbeaa36e78b2a4c1cb715d270112249bcb.png)
In order for this to be undefined, we need
![\[1-\frac{56}{33}\cdot\tan{\theta}=0\]](http://latex.artofproblemsolving.com/5/3/5/535f290628d1b17d55134f453e77c91871a9a461.png)
so
![\[\tan{\theta}=\frac{33}{56}\]](http://latex.artofproblemsolving.com/d/7/a/d7afd82be617d6f0ad1cd9daf05682c0234db504.png)
Hence the base side lengths of the third triangle are 
and 
. By the Pythagorean Theorem, the hypotenuse of the third triangle is 
, so the perimeter is 
.
See also
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
