Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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Then <math>ab=6\cdot (a+b+\sqrt {a^2 + b^2})</math>. | Then <math>ab=6\cdot (a+b+\sqrt {a^2 + b^2})</math>. | ||
− | We can complete the square under the root, and we get, <math>ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})</math>. | + | We can complete the square under the root, and we get, <math>ab=6\cdot \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)</math>. |
− | Let <math>ab=p</math> and <math>a+b=s</math>, we have <math>p=6\cdot (s+ \sqrt {s^2 - 2p})</math>. | + | Let <math>ab=p</math> and <math>a+b=s</math>, we have <math>p=6\cdot \left(s+ \sqrt {s^2 - 2p}\right)</math>. |
After rearranging, squaring both sides, and simplifying, we have <math>p=12s-72</math>. | After rearranging, squaring both sides, and simplifying, we have <math>p=12s-72</math>. |
Revision as of 13:34, 1 January 2016
Contents
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:
and
.
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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