Difference between revisions of "2007 AMC 12B Problems/Problem 23"
Isabelchen (talk | contribs) (→Solution 3) |
Isabelchen (talk | contribs) m (→Solution 3) |
||
Line 58: | Line 58: | ||
== Solution 3 == | == Solution 3 == | ||
− | Let the right triable be <math>\triangle ABC</math>, the two legs be <math>a</math> and <math>b</math>, the hypotenuse be <math>c</math> | + | Let the right triable be <math>\triangle ABC</math>, the two legs be <math>a</math> and <math>b</math>, the hypotenuse be <math>c</math>. |
By using <math>[ABC] = r\cdot s</math> we get: | By using <math>[ABC] = r\cdot s</math> we get: | ||
Line 64: | Line 64: | ||
<cmath>3(a+b+c) = r \cdot \frac{a+b+c}{2}</cmath> | <cmath>3(a+b+c) = r \cdot \frac{a+b+c}{2}</cmath> | ||
<cmath>r=6</cmath> | <cmath>r=6</cmath> | ||
− | < | + | |
+ | In a right triangle, <math>r = \frac{a+b-c}{2}</math> | ||
<cmath>a+b-c = 12</cmath> | <cmath>a+b-c = 12</cmath> | ||
<cmath>c = a + b - 12</cmath> | <cmath>c = a + b - 12</cmath> | ||
Line 80: | Line 81: | ||
<cmath>(a - 12)(b - 12) = 72</cmath> | <cmath>(a - 12)(b - 12) = 72</cmath> | ||
− | 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> | + | 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] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 10:24, 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
Let the right triable be , the two legs be and , the hypotenuse be .
By using we get:
In a right triangle,
By the triangle's area we get:
By substituting in:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.