INTcircles

Let the center of the incircle be $O$ , its radius be $R$ , and the center of the radius $r$ circle be $A$ . Draw a line parallel to side $a$ through $O$ and a line parallel to side $b$ through $A$ , and let their point of intersection be $B$ . Then $\Delta OAB$ is a right triangle with hypotenuse length $R + r$ , vertical leg length $R - r$ and horizontal leg length $2\sqrt{Rr}$ .

Now $\angle OAB$ is one-half the lower right interior angle $\theta$ of the triangle, and thus

$\tan(\angle OAB) = \tan(\frac{1}{2}\theta) = \dfrac{1 - \cos(\theta)}{\sin(\theta)} = \dfrac{1 - \frac{b}{c}}{\frac{a}{c}} = \dfrac{c - b}{a}$ .

But $\tan(\angle OAB) = \dfrac{R - r}{2\sqrt{Rr}}$ , so

$\dfrac{(c - b)^{2}}{a^{2}} = \dfrac{(R - r)^{2}}{4Rr} \Longrightarrow 4Rr \times \dfrac{(c - b)^{2}}{a^{2}} = R^{2} + r^{2} - 2Rr \Longrightarrow$

$r^{2} - 2R\left(1 + \dfrac{2(c - b)^{2}}{a^{2}}\right)r + R^{2} = 0$ .

By the quadratic equation we then have that, (after simplifying),

$r = R\left(1 + 2\dfrac{(c - b)^{2}}{a^{2}} - 2\dfrac{(c - b)}{a}\sqrt{1 + \dfrac{(c - b)^{2}}{a^{2}}}\right)$ ,

where we took the negative root as we should have $r \lt R$ . Noting that $a^{2} + b^{2} = c^{2}$ we then have that

$r = \dfrac{R}{a^{2}}(3c^{2} + b^{2} - 4bc - 2(c - b)\sqrt{2c(c - b)})$ .

Now for a right triangle $R = \dfrac{ab}{a + b + c}$ , so we finally have that

$r = \dfrac{b}{a(a + b + c)}(3c^{2} + b^{2} - 4bc - 2(c - b)\sqrt{2c(c - b)})$ .

Now for $r$ to be an integer we will require that $2c(c - b)$ be a perfect square. At this point we will note that the side lengths of a Pythagorean triangle can be represented as $k(m^{2} - n^{2}), 2kmn$ and $k(m^{2} + n^{2})$ for positive integers $k,m,n$ . Now if

• $b = k(m^{2} - n^{2}), c = k(m^{2} + n^{2}) \Longrightarrow 2c(c - b) = 4k^{2}n^{2}(m^{2} + n^{2})$ , and if

• $b = 2kmn, c = k(m^{2} + n^{2}) \Longrightarrow 2c(c - b) = 2k^{2}(m - n)^{2}(m^{2} + n^{2})$ .

In the first case, $2c(c - b)$ will be a perfect square if and only if $m^{2} + n^{2}$ is itself a perfect square. The smallest option is $m = 4, n = 3, k = 1$ , which corresponds to a $(a,b,c) = (24,7,25)$ right triangle, which from our formula yields $r = \dfrac{3}{4}$ . The next smallest option is $m = 8, n = 6, k = 1$ , (the same as $m = 4, n = 3, k = 4$ ), which corresponds to a $(a,b,c) = (96,28,100)$ right triangle and a value of $r = 3$ . Thus the smallest possible perimeter for which $r$ is integral is $96 + 28 + 100 = \boxed{224}$ .

Notes: For the $b = 2kmn$ case we would require that $2(m^{2} + n^{2})$ be a perfect square, the first occurrence for which is $m = 7, n = 1, k = 1$ , which corresponds to a $(48, 14, 50)$ right triangle, yielding $r = \frac{3}{2}$ . The next smallest option would be with $m = 7, n = 1, k = 2$ which corresponds to the $(96,28,100)$ solution triangle found above. The next option we could have looked at is $m = 12, n = 5, k = 1$ , which corresponds to a $(120,119,169)$ right triangle, which has a greater perimeter.

I did this by trial and error but, of course, I'm aware that this isn't the solution expected. Beginning with the primal (3,4,5) type of right triangle say (36, 48,60) which has an in-circle of radius=12, I realized that the answer for the smaller radius always contains a radical. So it was important for the hypotenuse to be a square itself. This was my inference from the trials & I do not know whether it is right or wrong. Nonetheless, I discarded the (3.4.5) or (5,12,13) type and turned to (7, 24, 25) and set up the following equation for the radius 'r' of the smaller circle: (7n-12 -2(12r)^(1/2))/(7n-12)=r/12, considering the sides to be (7n, 24n, 25n) where 'n' is an integer and the smaller circle is between the side that is 7n and the hypotenuse. At n=4, we get, r=3 and thus we have the triangle (28,96,100) with a perimeter of 224.