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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | Note the 3 smallest angles as A, B and C | ||
| + | sinA = 3/5 cosA = 4/5 | ||
| + | sinB = 5/13 cosB = 12/13 | ||
| + | The angle C would be the third triangle's smallest angle in the satisfying condition. | ||
| + | cos(A+B) = cosAcosB-sinAsinB | ||
| + | = (4/5)(12/13)-(3/5)(5/13) | ||
| + | = 48/65 - 15/65 | ||
| + | = 33/65 | ||
| + | We can also conclude that sin(A+B) = 56/65 (By sin^2(x)+cos^2(x) = 1) | ||
| + | Method 1: Notice that C = 90-(A+B) | ||
| + | We could quickly conclude the triangle is 33-56-65 | ||
| + | Method 2: | ||
| + | Because A + B + C = 90 degrees, therefore cos((A+B)+C) = 0 | ||
| + | This leads to cos(A+B)cos(C)-sin(A+B)sin(C) = 0 | ||
| + | Which (33/65)cos(C)=(56/65)sin(C) | ||
| + | Therefore 33cos(C) = 56sin(C) | ||
| + | tan(C) = 33/56 | ||
| + | Therefore the third triangle is 33-56-65 | ||
| + | |||
| + | Overall, Answer = 33+56+65 = (C)154 | ||
~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino] | ~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino] | ||
Revision as of 01:05, 14 November 2024
Contents
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 
.
Solution 2
Note the 3 smallest angles as A, B and C sinA = 3/5 cosA = 4/5 sinB = 5/13 cosB = 12/13 The angle C would be the third triangle's smallest angle in the satisfying condition. cos(A+B) = cosAcosB-sinAsinB
= (4/5)(12/13)-(3/5)(5/13)
= 48/65 - 15/65
= 33/65
We can also conclude that sin(A+B) = 56/65 (By sin^2(x)+cos^2(x) = 1) Method 1: Notice that C = 90-(A+B) We could quickly conclude the triangle is 33-56-65 Method 2: Because A + B + C = 90 degrees, therefore cos((A+B)+C) = 0 This leads to cos(A+B)cos(C)-sin(A+B)sin(C) = 0 Which (33/65)cos(C)=(56/65)sin(C) Therefore 33cos(C) = 56sin(C) tan(C) = 33/56 Therefore the third triangle is 33-56-65
Overall, Answer = 33+56+65 = (C)154
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
