Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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== Solution 3 == | == Solution 3 == | ||
− | <math>[ABC] = | + | By using <math>[ABC] = r\cdot s</math> we get: |
− | < | + | <cmath>3(a+b+c) = r \cdot \frac{a+b+c}{2}</cmath> |
+ | <cmath>r=6</cmath> | ||
+ | <cmath>r = \frac{a+b-c}{2}</cmath> | ||
+ | <cmath>a+b-c = 12</cmath> | ||
+ | <cmath>c = a + b - 12</cmath> | ||
+ | |||
+ | |||
+ | By the triangle's area we get: | ||
+ | |||
+ | <cmath>\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}</cmath> | ||
+ | <cmath>ab = 6(a+b+c)</cmath> | ||
+ | <cmath>\frac{ab}{6} = a+b+c</cmath> | ||
+ | <cmath>c = \frac{ab}{6} - a - b</cmath> | ||
+ | |||
+ | <math>a + b - 12 = \frac{ab}{6} - a - b</math>, <math>ab - 12a - 12b + 72 = 0</math>, <math>(a - 12)(b - 12) = 72</math> | ||
+ | |||
+ | As <math>72 = 2^3 \cdot 3^2</math>, there are <math>\frac{(3+1)(2+1)}{2} = 6</math> solutions, <math>\boxed{\textbf{(A) } 6}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:09, 5 December 2022
Contents
[hide]Problem
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters?
Solution 1
Let and be the two legs of the triangle.
We have .
Then .
We can complete the square under the root, and we get, .
Let and , we have .
After rearranging, squaring both sides, and simplifying, we have .
Putting back and , and after factoring using Simon's Favorite Factoring Trick, we've got .
Factoring 72, we get 6 pairs of and
And this gives us solutions .
Alternatively, note that . Then 72 has factors. However, half of these are repeats, so we have solutions.
Solution 2
We will proceed by using the fact that , where is the radius of the incircle and is the semiperimeter .
We are given .
The incircle of breaks the triangle's sides into segments such that , and . Since ABC is a right triangle, one of , and is equal to its radius, 6. Let's assume .
The side lengths then become , and . Plugging into Pythagorean's theorem:
We can factor to arrive with pairs of solutions: and .
Solution 3
By using we get:
By the triangle's area we get:
, ,
As , there are solutions,
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
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