Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>. | We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>. | ||
− | The incircle of <math>ABC</math> breaks the triangle's sides into segments such that <math>AB = x + y</math>, <math>BC = x + z</math> and <math>AC = y + z</math>. Since ABC is a triangle, one of <math>x</math>, <math>y</math> and <math>z</math> is equal to its radius, 6. Let's assume <math>z = 6</math>. | + | The incircle of <math>ABC</math> breaks the triangle's sides into segments such that <math>AB = x + y</math>, <math>BC = x + z</math> and <math>AC = y + z</math>. Since ABC is a right triangle, one of <math>x</math>, <math>y</math> and <math>z</math> is equal to its radius, 6. Let's assume <math>z = 6</math>. |
The side lengths then become <math>AB = x + y</math>, <math>BC = x + 6</math> and <math>AC = y + 6</math>. Plugging into Pythagorean's theorem: | The side lengths then become <math>AB = x + y</math>, <math>BC = x + 6</math> and <math>AC = y + 6</math>. Plugging into Pythagorean's theorem: |
Revision as of 19:41, 25 January 2015
Contents
[hide]Problem 23
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters?
Solution
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 , we've got .
Factoring 72, we get 6 pairs of and
And this gives us 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: $(7, 78), (8,42), (9, 30), (10, 24), (12, 18) & (14, 15) \Rightarrow \mathrm{(A)}$ (Error compiling LaTeX. Unknown error_msg).
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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