2007 AMC 12B Problems/Problem 23

Problem

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution 1

Let $a$ and $b$ be the two legs of the triangle.

We have $\frac{1}{2}ab = 3(a+b+c)$ .

Then $ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)$ .

We can complete the square under the root, and we get, $ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)$ .

Let $ab=p$ and $a+b=s$ , we have $p=6 \left(s+ \sqrt {s^2 - 2p}\right)$ .

After rearranging, squaring both sides, and simplifying, we have $p=12s-72$ .

Putting back $a$ and $b$ , and after factoring using Simon's Favorite Factoring Trick, we've got $(a-12)(b-12)=72$ .

Factoring 72, we get 6 pairs of $a$ and $b$

$(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$

And this gives us $6$ solutions $\Rightarrow \mathrm{(A)}$ .

Alternatively, note that $72 = 2^3 \cdot 3^2$ . Then 72 has $(3+1)(2+1) = (4)(3) = 12$ factors. However, half of these are repeats, so we have $\frac{12}{2} = 6$ solutions.

Solution 2

We will proceed by using the fact that $[ABC] = r\cdot s$ , where $r$ is the radius of the incircle and $s$ is the semiperimeter $\left(s = \frac{p}{2}\right)$ .

We are given $[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6$ .

The incircle of $ABC$ breaks the triangle's sides into segments such that $AB = x + y$ , $BC = x + z$ and $AC = y + z$ . Since ABC is a right triangle, one of $x$ , $y$ and $z$ is equal to its radius, 6. Let's assume $z = 6$ .

The side lengths then become $AB = x + y$ , $BC = x + 6$ and $AC = y + 6$ . Plugging into Pythagorean's theorem:

$(x + y)^2 = (x+6)^2 + (y + 6)^2$

$x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36$

$2xy - 12x - 12y = 72$

$xy - 6x - 6y = 36$

$(x - 6)(y - 6) - 36 = 36$

$(x - 6)(y - 6) = 72$

We can factor $72$ to arrive with $6$ pairs of solutions: $(7, 78), (8,42), (9, 30), (10, 24), (12, 18),$ and $(14, 15) \Rightarrow \mathrm{(A)}$ .

Solution 3

Let $a$ and $b$ be the two legs of the triangle, and $c$ be the hypotenuse.

By using $Area = \frac{r}{2} (a+b+c)$ , where $r$ is the in-radius, we get:

$3(a+b+c) = \frac{r}{2} (a+b+c)$ $r=6$

In right triangle, $r = \frac{a+b-c}{2}$ $a+b-c = 12$ $c = a + b - 12$

By the triangle's area we get:

$\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}$ $ab = 6(a+b+c)$

By substituting $c$ in:

$ab = 6(a+b+a + b - 12)$ $ab - 12a - 12b + 72 = 0$ $(a - 12)(b - 12) = 72$

As $72 = 2^3 \cdot 3^2$ , there are $\frac{(3+1)(2+1)}{2} = 6$ solutions, $\boxed{\textbf{(A) } 6}$ .

~isabelchen

Solution 4

All pythagorean triples can be parametrized in the form $(a, b, c) = k(r^2 - s^2), k(2rs), k(r^2 + s^2)$ for positive integers $k, r, s$ . The area being triple the perimeter implies that $k^2(r^2 - s^2)rs = 3(k(r^2 - s^2) + k(2rs) + k(r^2 + s^2)).$ This can be simplified to get $ks(r - s) = 6.$ Now, we get the triples $(k, r, s) = (1, 7, 1), (1, 5, 2), (1, 5, 3), (1, 7, 6), (2, 4, 1), (2, 4, 3), (3, 3, 1), (3, 3, 2), (6, 2, 1).$ However, the ones where $r$ and $s$ are not different signs and relatively prime are redundant, so we get $6$ triples total.

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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