Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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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\ | + | By using <math>[ABC] = \frac{r}{2} (a+b+c)</math>, where <math>r</math> is the in-radius of <math>\triangle ABC</math>, we get: |
− | <cmath>3(a+b+c) = | + | <cmath>3(a+b+c) = \frac{r}{2} (a+b+c)</cmath> |
<cmath>r=6</cmath> | <cmath>r=6</cmath> | ||
Revision as of 11:29, 5 December 2022
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 , where
is the in-radius of
, 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 |
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