Finding the no. of possible right angled triangle.

Question

How many right-angled triangles with integer sides have an incircle with radius $2013$?

Answer 1

I get there are 27 solutions with $a \le b$.

Ultimate result:

If the inradius is $r$ and the legs $a$ and $b$, all solutions with $a \le b$ are: for each divisor $d$ of $2r^2$ such that $1 \le d \lt r\sqrt{2} $, $a = 2r+d, b = 2r+\frac{2r^2}{d} $.

Therefore, the number is $\frac12\tau(2r^2)$, where $\tau(n)$ is the number of divisors of $n$.

If $n = \prod_{p_i|n} p_i^{a_i} $, then $\tau(n) =\prod (a_i+1) $.

If $r$ is odd, then $\tau(2r^2) =2\prod_{p_i|r} (2a_i+1) $ so the number is $\prod_{p_i|r} (2a_i+1) $.

Since $2013 = 3\cdot 11\cdot 17$, the number is $3^3 = 27$.

If $r$ is even, $r = 2^em$, where $m$ is odd, $\tau(2r^2) =\tau(2^{2e+1}m^2) =(2e+2)\tau(r^2) $ so the number of solutions is $(e+1)\tau(r^2) $.

The derivation:

If the inradius and sides of a right triangle are $r,a,b,c$, then $r(a+b+c) =ab$.

Assume $a \le b$.

We have $r(a+b+\sqrt{a^2+b^2}) =ab$.

Since $a \le b$, $ab \ge r(2a+a\sqrt{2}) = ra(2+\sqrt{2}) $ or $b \ge r(2+\sqrt{2}) $.

Similarly, $ab \le r(2b+b\sqrt{2}) = rb(2+\sqrt{2}) $ or $a \le r(2+\sqrt{2}) $.

Therefore $\dfrac{a}{r} \le 2+\sqrt{2} \le \dfrac{b}{r} $.

Rearrange to get $r\sqrt{a^2+b^2} =ab-r(a+b) $.

Squaring, $r^2(a^2+b^2) =a^2b^2-2rab(a+b) +r^2(a^2+2ab+b^2) $.

Simplifying, $0 =a^2b^2-2rab(a+b) +2abr^2 $.

Canceling $ab$, $0 =ab-2r(a+b) +2r^2 =ab-2ra-2rb +2r^2 =b(a-2r)-2r(a-r) $, or $b =\dfrac{2r(a-r)}{a-2r} =r\dfrac{2(a/r-1)}{a/r-2} $.

Since this is symmetric in $a$ and $b$, $a =\dfrac{2r(b-r)}{b-2r} =r\dfrac{2(b/r-1)}{b/r-2} $.

Also $b =\dfrac{2r(a-r)}{a-2r} =a\dfrac{2(r/a)(1-r/a)}{1-2r/a} $.

Since $b \ge a$, if $x = r/a$, we must have $\dfrac{2x(1-x)}{1-2x} \ge 1 $. These are equal when $2x(1-x) = 1-2x$ or $2x^2-4x+1=0 $ or $x =\dfrac{4\pm\sqrt{16-8}}{4} =\dfrac{2\pm\sqrt{2}}{2} = 1\pm\sqrt{1/2} $.

Let $x_1 = 1-\sqrt{1/2}$ and $x_2 = 1+\sqrt{1/2}$. Thenfore we must have $x > x_2$ or $x_1 \le x \lt \frac12$. Note that $x_1x_2 = \frac12$ so that $\frac1{x_1} = 2x_2$ and $\frac1{x_2} = 2x_1$.

Since $x_1 < \frac{r}{a} < \frac12$ or $ \frac{r}{a} > x_2$, we have $2 <\frac{a}{r} \lt \frac1{x_1} =2x_2 =2+\sqrt{2} $.

Therefore $b =\dfrac{2r(a-r)}{a-2r} $.

We have

$\begin{array}\\ b-2r &=\dfrac{2r(a-r)}{a-2r}-2r\\ &=\dfrac{2r(a-r)-2r(a-2r)}{a-2r}\\ &=\dfrac{2r(a-r-a+2r)}{a-2r}\\ &=\dfrac{2r(r)}{a-2r}\\ &=\dfrac{2r^2}{a-2r}\\ \text{so that}\\ b &=2r+\dfrac{2r^2}{a-2r}\\ \end{array} $

Therefore, we must have $(a-2r) | 2r^2$ and $2r \lt a < r(2+\sqrt{2})$ so that $0 \lt a-2r \lt r\sqrt{2} $.

Therefore, for each divisor $d$ of $2r^2$ such that $1 \le d \lt r\sqrt{2} $, a solution is $a = 2r+d, b = 2r+\frac{2r^2}{d} $.

Since $a \le b$, $d \le \frac{2r^2}{d}$ so $d \le r\sqrt{2}$.

As a check,

$\begin{array}\\ a^2+b^2 &=(2r+d)^2+(2r+\dfrac{2r^2}{d})^2\\ &=\dfrac{(d^2+2dr+2r^2)^2}{d^2} \qquad\text{(according to Wolfy)}\\ &=(d+2r+\dfrac{2r^2}{d})^2\\ &=(a+b-2r)^2\\ \end{array} $

so

$\begin{array}\\ r(a+b+c) &=r(a+b+(a+b-2r))\\ &=2r(a+b-r)\\ &=2r(3r+d+\frac{2r^2}{d})\\ &=2r(\frac{d(3r+d)+2r^2}{d})\\ &=2r(\frac{2r^2+3rd+d^2}{d})\\ &=2r(\frac{(2r+d)(r+d)}{d})\\ \text{and}\\ ab &=(2r+d)( 2r+\frac{2r^2}{d})\\ &=2r(2r+d )\frac{d+r}{d}\\ &=2r \frac{(2r+d)(d+r)}{d}\\ \end{array} $

We know that $d=1, 2, r$ are always divisors. The values of $a$ and $b$ are $(a, b)= (2r+1, 2r+2r^2), (2r+2, 2r+r^2), (3r, 4r) $.

If $r = 2013$, $r\sqrt{2} = 2846...$ and the divisors of $2r^2$ up to this (according to Wolfy) are 1 | 2 | 3 | 6 | 9 | 11 | 18 | 22 | 33 | 61 | 66 | 99 | 121 | 122 | 183 | 198 | 242 | 363 | 366 | 549 | 671 | 726 | 1089 | 1098 | 1342 | 2013 | 2178 .

There are 27 of these.