Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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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: | ||
− | (x + y)^2 = (x+6)^2 + (y + 6)^2 | + | <math>(x + y)^2 = (x+6)^2 + (y + 6)^2 |
x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36 | x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36 | ||
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(x - 6)(y - 6) - 36 = 36 | (x - 6)(y - 6) - 36 = 36 | ||
− | (x - 6)(y - 6) = 72 | + | (x - 6)(y - 6) = 72</math> |
<math>72 = 2^3 \cdot 3^2</math> so it has <math>(3 + 1)\cdot (2 + 1) = 4\cdot 3 = 12</math> factors, meaning <math>x - 6</math> and <math>y - 6</math> can take on 12 values. But for each pair of factors that multiply to 72, they produce one distinct triangle. Thus, the number of right triangles ABC that satisfy the given condition is <math>\frac{12}{2} = 6 \Rightarrow</math> A. | <math>72 = 2^3 \cdot 3^2</math> so it has <math>(3 + 1)\cdot (2 + 1) = 4\cdot 3 = 12</math> factors, meaning <math>x - 6</math> and <math>y - 6</math> can take on 12 values. But for each pair of factors that multiply to 72, they produce one distinct triangle. Thus, the number of right triangles ABC that satisfy the given condition is <math>\frac{12}{2} = 6 \Rightarrow</math> A. |
Revision as of 12:58, 29 November 2014
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 ABC breaks the triangle's sides into segments such that , and . Since ABC is a triangle, one of , and is equal to its radius, 6. Let's assume .
The side lengths then become , and . Plugging into Pythagorean's theorem:
$(x + y)^2 = (x+6)^2 + (y + 6)^2
x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36
2xy - 12x - 12y = 72
xy - 6x - 6y = 36
(x - 6)(y - 6) - 36 = 36
(x - 6)(y - 6) = 72$ (Error compiling LaTeX. Unknown error_msg)
so it has factors, meaning and can take on 12 values. But for each pair of factors that multiply to 72, they produce one distinct triangle. Thus, the number of right triangles ABC that satisfy the given condition is A.
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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