Euler's Equipotent MinIntegral Triangle!

If the two non-diameter sides of the right-angled triangle have lengths $a,b$ , then $P = a+b + 2R$ and $A =\tfrac12ab$ are both integers. Thus we deduce that $s = a+b$ must be an integer. Then $4R^2 \; = \; a^2 + b^2 \; = \; (a + b)^2 - 2ab \; =\; s^2 - 4A$ so we deduce that $s = 2\sigma$ must be even, and that $R^2 \; = \; \sigma^2 - A$ Since the semiperimeter of the triangle is $\tfrac12(a+b+2R) = \sigma + R$ , we also have that $(\sigma + R)r \; = \; A$ We then observe that $r_{\mathrm{equi}} = \tfrac12R$ . The point of contact of the incircle splits the diameter/hypotenuse into segments of length $R + \tfrac12(a-b)$ , $R + \tfrac12(b-a)$ , and so we deduce that $\begin{aligned} r^2 + \tfrac14(a-b)^2 &= \; r_{\mathrm{equi}}^2 \; = \; \tfrac14R^2 \\ 4r^2 + a^2 + b^2 - 2ab & = \; R^2 \\ 4r^2 + (a + b)^2 - 4ab & = \; R^2 \\ 4r^2 + 4\sigma^2 - 8A & = \; R^2 \end{aligned}$ and hence we deduce that $R = 2\rho$ is even, and so we have the equations $4\rho^2 \; =\; \sigma^2 - A \hspace{1cm} (\sigma +2\rho)r \; = \; A \hspace{1cm} r^2 + \sigma^2 - 2A \; = \; \rho^2$ relating $\rho, \sigma, r, A$ , with $R = 2\rho$ and $P = 2\sigma + 4\rho$ . But then $A = \sigma^2 - 4\rho^2$ and hence $r = \sigma - 2\rho$ , and so we have $\begin{aligned} (\sigma - 2\rho)^2 + \sigma^2 -2(\sigma^2 - 4\rho^2) & = \; \rho^2 \\ 11\rho^2 - 4\rho\sigma & = \; 0 \\ 4\sigma & = \; 11\rho \end{aligned}$ and hence $\sigma = 11k$ , $\rho = 4k$ for some positive integer $k$ . But then $R = 8k$ and $P = 38k$ . Also $A = \sigma^2 - 4\rho^2 = 57k^2$ and hence $r = 3k$ . Since $R,P,A,r$ are coprime, it follows that $k=1$ , so that $R=8$ , $P=38$ , $A=\boxed{57}$ and $r=3$ . Note that $a,b = 11 \pm \sqrt{7}$ .

Since the inradius of an equilateral triangle with a circumradius of $R$ is $r_{\text{equi}} = \frac{1}{2}R$ , and since the distance between the circumcenter and the incenter of a triangle is $OI = \sqrt{R(R - r)}$ , and since the center of the incircle lies on the circle with a radius of $r_{\text{equi}}$ , we know that $OI = \sqrt{R(R - r)} = r_{\text{equi}} = \frac{1}{2}R$ , and $\sqrt{R(R - r)} = \frac{1}{2}R$ solves to $r = \frac{3}{8}R$ .

Since $r$ must be an integer and $r = \frac{3}{8}R$ , let $R = 8k$ for some integer $k$ . Then $r = \frac{3}{8}R = 3k$ and $r_{\text{equi}} = \frac{1}{2}R = 4k$ will also both be integers as required.

Let $a$ and $b$ be the legs of the right triangle and $c$ be the hypotenuse. By Thales's theorem the hypotenuse is also the diameter of its circumcircle, so $c = 2R = 16k$ , and its inradius is $r = \frac{1}{2}(a + b - c)$ , or $3k = \frac{1}{2}(a + b - 16k)$ , which rearranges to $b = 22k - a$ .

By the Pythagorean Theorem, $a^2 + b^2 = c^2$ , or $a^2 + (22k - a)^2 = (16k)^2$ , which solves to $a = (11 \pm \sqrt{7})k$ .

Then $b = 22k - a = 22k - (11 \pm \sqrt{7})k = (11 \mp \sqrt{7})k$ , so the perimeter is $P = a + b + c = (11 \pm \sqrt{7})k + (11 \mp \sqrt{7})k + 16k = 28k$ and the area is $A = \frac{1}{2}ab = \frac{1}{2} \cdot (11 \pm \sqrt{7})k \cdot (11 \mp \sqrt{7})k = 57k^2$ .

For the $\gcd(R, r, A, P) = 1$ , the $\gcd(8k, 3k, 57k^2, 28k) = 1$ , which means the common factor $k = 1$ , and if $k = 1$ , then $A = 57k^2 = \boxed{57}$ .