maths
posted by ramesh reddy on .
How many non congruent right triangles with positive integer leg lengths have areas that are numerically equal to 3 times their perimeters?

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 $p=12s72$.
Putting back $a$ and $b$, and after factoring using $SFFT$, we've got $(a12)(b12)=72$.
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 there 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 $(14, 15) \Rightarrow \mathrm{(A)}$.