Identical Incircles

Let's start by finding a family of solutions, Suppose that $BC = 2pq$ , $AC = p^2- q^2$ and $AB = p^2 + q^2$ where $p,q$ are coprime positive integers with $p > q$ . Then the semiperimeter of $ABC$ is $p(p+q)$ , and using the result to be found here , we have $BD^2 \; = \; s(s - AC) \; = \; p(p+q) \times q(p+q) \; = \; pq(p+q)^2$ so that $BD = \sqrt{pq}(p+q)$ . Since $p,q$ are coprime positive integers, we deduce that $p=u^2$ , $q=v^2$ where $u,v$ are coprime positive integers with $u > v$ . But then we have $AB \; = \; u^4 + v^4 \hspace{1cm} BC \; = \; 2u^2v^2 \hspace{1cm} AC \; = \; u^4 - v^4 \hspace{1cm} BD \; = \; uv(u^2 + v^2)$ But then we see that $CD \;=\; uv(u^2-v^2) \hspace{2cm} AD \;=\; (u^2-v^2)(u^2+uv+v^2)$ are also integers. Putting $u=2$ , $v=1$ we obtain a solution $AB = 17$ , $BC = 8$ , $CD = 6$ , $DA = 9$ and $BD = 10$ , and the perimeter of $\Delta ABC$ is then $40$ .

Since it is a finite problem, we can set a computer to check that there is no solution to the equations $\begin{aligned} a^2 + b^2 & = \; c^2 \\ a^2 + (b+d)^2 & = \; e^2 \\ 4c^2 & = \; (a+e)^2 - (b+d)^2 \end{aligned}$ in positive integers $a,b,c,d,e$ with $a+b+d+e < 40$ . This means that the solution given above is optimal, and so the answer is $\boxed{40}$ .

Let $a = BC$ , $b = AC$ , $c = AB$ , and $x = CD$ . Then $AD = b - x$ and by Pythagorean's Theorem $BD = \sqrt{a^2 + x^2}$ .

Since the radius of an incircle is the ratio of the area of its triangle and its semiperimeter, the radius of the incircle of $\triangle BCD$ is $r_1 = \frac{ax}{a + x + \sqrt{a^2 + x^2}}$ and the radius of the incircle of $\triangle BDA$ is $r_2 = \frac{a(b - x)}{b - x + c + \sqrt{a^2 + x^2}}$ .

Equating $r_1$ and $r_2$ and simplifying (using $a^2 + b^2 = c^2$ ) leads to $(x - b)(4x^2 - (2ac - 2a^2)) = 0$ , and solving for $0 < x < b$ gives $x = \frac{1}{2}\sqrt{2a(c - a)}$ .

Since $a$ and $c$ are sides of a right triangle, if $c = m^2 + n^2$ , then $a = 2mn$ or $a = m^2 - n^2$ for integers and $m$ and $n$ such that $m > n$ . If $a = 2mn$ , then $x = (m - n)\sqrt{mn}$ , so for $x$ to be an integer, $mn$ is a square, and the smallest integers that satisfy the conditions are $m = 3$ and $n = 1$ , which leads to $a = BC = 8$ , $c = AB = 17$ , $x = CD = 6$ , $b = AC = 15$ , $DA = 9$ , and $BD = 10$ , all integers lengths for a perimeter of $P = 40$ . If $a = m^2 - n^2$ , then $x = n\sqrt{m^2 - n^2}$ , so for $x$ to be an integer, $m^2 - n^2$ is a square, for another Pythagorean triple, and the smallest Pythagorean triple to satisfy the conditions is $(3, 4, 5)$ , so $m = 5$ and $n = 3$ , which leads to $a = BC = 16$ , $c = AB = 34$ , $x = CD = 12$ , $b = AC = 30$ , $DA = 18$ , and $BD = 20$ , all integers lengths for a perimeter of $P = 80$ . Out of the two possibilities, the smallest perimeter is $P = \boxed{40}$ .