Mutual I’m Sure

Let the radii of the three outer circles be $a<b<c$ , so that the sides of the right-angled triangle are $a+b$ , $a+c$ and $b+c$ . Let $P=329$ be the perimeter of this triangle; let $T$ be its area; and let $r$ be the radius of the interior circle. We're told that $r$ and $T$ are integers; and we want to find $T$ .

The idea will be to find a relationship between $r$ , $P$ and $T$ by eliminating $a,b,c$ from various equations. (This is useful as we don't know if $a,b,c$ are integers or not.)

Adding up the sides, ${\color{#3D99F6} 2(a+b+c)=P}$

By Pythagoras, $(a+b)^2+(a+c)^2=(b+c)^2 \Rightarrow {\color{#333333} a(a+b+c)=bc}$

Using the expression for the perimeter, this is ${\color{#3D99F6} aP=2bc}$

The area of the triangle is ${\color{#3D99F6} T=\frac{(a+b)(a+c)}{2}}$

Now, the blue equations are symmetric in $b$ and $c$ (we can swap them without changing the equations). So, let $u=bc$ and $v=b+c$ ; the equations become $2(a+v)=P \;\;\;\;\; aP=2u \;\;\;\;\; 2T=a^2+av+u$

We get $v=\frac{P}{2}-a$ and $u=\frac{aP}{2}$ ; substituting, $2T=a^2+a \left(\frac{P}{2}-a \right)+\frac{aP}{2}=aP$

so $a=\frac{2T}{P}$ .

Plugging back in, we get $u=T$ and $v=\frac{P}{2}-\frac{2T}{P}$ .

By Descartes' circle theorem, $\frac{1}{r}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\sqrt{\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}}$

Substituting from above, things cancel out nicely to give $r=\frac{PT}{2P^2-2T}$

Rearranging, $2rP^2-2rT=PT$

Note that, for a given perimeter, the largest right-angled triangle is an isosceles one, with area $T_{\max} = \frac14 \left(3 - 2\sqrt2 \right) P^2$ . In this case, that's around $4642.8$ .

The only integer solution to the final equation satisfying this constraint is $r=7$ and $T=\boxed{4418}$ .

Let the radii of the three exterior circles be $r_1$ , $r_2$ , and $r_3$ , so that the legs of the right triangle are $a = r_1 + r_2$ and $b = r_1 + r_3$ and the hypotenuse is $c = r_2 + r_3$ .

The semiperimeter is then $s = \frac{1}{2}(a + b + c) = \frac{1}{2}((r_1 + r_2) + (r_1 + r_3) + (r_2 + r_3)) = r_1 + r_2 + r_3$ , so $s - a = (r_1 + r_2 + r_3) - (r_1 + r_2) = r_3$ .

Similarly, $s - b = r_2$ and $s - c = r_1$ .

The semiperimeter is also half the perimeter, so $s = \frac{1}{2}P = \frac{329}{2}$ .

By Heron's formula the area of the triangle is $T = \sqrt{s(s - a)(s - b)(s - c)}$ . Substituting $s$ , $s - a$ , $s- b$ , and $s - c$ from above gives $T = \sqrt{\cfrac{329r_1r_2r_3}{2}}$ , and rearranging gives $r_1r_2r_3 = \cfrac{2T^2}{329}$ .

By the Pythagorean Theorem, $(r_1 + r_2)^2 + (r_1 + r_3)^2 = (r_2 + r_3)^2$ , which rearranges to $r_1r_2 + r_1r_3 + r_2r_3 = 2r_2r_3 - r_1^2$ .

By the area of the triangle equation, $T = \frac{1}{2}(r_1 + r_2)(r_1 + r_3)$ , which rearranges to $r_1r_2 + r_1r_3 + r_2r_3 = 2T - r_1^2$ .

That means $2T = r_1r_2 + r_1r_3 + r_2r_3 + r_1^2 = 2r_2r_3$ , so $T = r_2r_3$ .

Substituting $T = r_2r_3$ into $r_1r_2r_3 = \cfrac{2T^2}{329}$ gives $r_1 = \cfrac{2T}{329}$ .

Substituting $r_1 = \cfrac{2T}{329}$ into $r_1r_2 + r_1r_3 + r_2r_3 = 2T - r_1^2$ gives $r_1r_2 + r_1r_3 + r_2r_3 = 2T - \cfrac{4T^2}{329^2}$

By Descartes' Theorem , the reciprocal of the radius $d$ of the interior circle is $\cfrac{1}{d} = \cfrac{1}{r_1} + \cfrac{1}{r_2} + \cfrac{1}{r_3} + 2\sqrt{\cfrac{1}{r_1r_2} + \cfrac{1}{r_1r_3} + \cfrac{1}{r_2r_3}}$ .

This rearranges to $d = \cfrac{r_1r_2r_3}{r_1r_2 + r_1r_3 + r_2r_3 + 2\sqrt{(r_1 + r_2 + r_3)r_1r_2r_3}}$ .

Substituting $r_1 + r_2 + r_3 = \cfrac{329}{2}$ , $r_1r_2r_3 = \cfrac{2T^2}{329}$ , and $r_1r_2 + r_1r_3 + r_2r_3 = 2T - \cfrac{4T^2}{329^2}$ into $d$ gives $d = \cfrac{\frac{2T^2}{329}}{2T - \frac{4T^2}{329^2} + 2\sqrt{\frac{329}{2} \cdot \frac{2T^2}{329}}} = \cfrac{329T}{2(329^2 - T)}$ .

The equation $d = \cfrac{329T}{2(329^2 - T)}$ can be rearranged to $(329^2 - T)(2d + 329) = 329^3$ .

Therefore, both $329^2 - T$ and $2d + 329$ are positive factors of $329^3$ , so $329^2 - T > 0$ , or $T < 329^2$ .

Let $f$ be the factor of $329^3$ such that $f = 329^2 - T > 0$ . Then $T = 329^2 - f$ , and since $T$ is positive, $f < 329^2$ , or $f < 108241$ .

Now for a right triangle, $P = a + b + \sqrt{a^2 + b^2}$ and $T = \frac{1}{2}ab$ . Solving for $a$ in terms of $P$ and $T$ gives $a = \frac{1}{4P}(P^2 + 4T \pm \sqrt{P^4 - 24P^2T + 16T^2})$ , so from the discriminant $P^4 - 24P^2T + 16T^2 > 0$ we know that either $T < \frac{1}{4}(3 - 2\sqrt{2})P^2$ or $T > \frac{1}{4}(3 + 2\sqrt{2})P^2$ .

So for $P = 329$ , either $T < \frac{1}{4}(3 - 2\sqrt{2})329^2$ or $T > \frac{1}{4}(3 + 2\sqrt{2})329^2$ . However, since $T < 329^2$ , $T \ngtr \frac{1}{4}(3 + 2\sqrt{2})329^2$ , so $T < \frac{1}{4}(3 - 2\sqrt{2})329^2$ . In other words, $T < 4643$ .

Since $T = 329^2 - f$ , that means $329^2 - f < 4643$ , and this solves to $103598 < f$ .

Therefore, $103598 < f < 108241$ . Testing the $(3 + 1)(3 + 1) = 16$ possible factors of $329^3 = 7^3 47^3$ ( $1$ , $7$ , $47$ , $49$ , $329$ , $343$ , $2209$ , $2303$ , $15463$ , $16121$ , $103823$ , $108241$ , $726761$ , $757687$ , $5087327$ , and $35611289$ ), only $103823$ is within this range, so $f = 103823$ .

Therefore, $T = 329^2 - f = 329^2 - 103823 = \boxed{4418}$ .

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Also:

$d = \cfrac{329T}{2(329^2 - T)} = \cfrac{329 \cdot 4418}{2(329^2 - 4418)} = 7$

$a = \cfrac{1}{4P}(P^2 + 4T - \sqrt{P^4 - 24P^2T + 16T^2}) = \cfrac{1}{4 \cdot 329}(329^2 + 4T - \sqrt{329^4 - 24 \cdot 329^2 \cdot 4418 + 16 \cdot 4418^2}) = \cfrac{47}{28}(57 - \sqrt{113})$

$b = \cfrac{1}{4P}(P^2 + 4T + \sqrt{P^4 - 24P^2T + 16T^2}) = \cfrac{1}{4 \cdot 329}(329^2 + 4T + \sqrt{329^4 - 24 \cdot 329^2 \cdot 4418 + 16 \cdot 4418^2}) = \cfrac{47}{28}(57 + \sqrt{113})$

$c = \sqrt{a^2 + b^2} = \sqrt{(\frac{47}{28}(57 - \sqrt{113}))^2 + (\frac{47}{28}(57 + \sqrt{113}))^2} = \cfrac{1927}{14}$

$r_1 = s - c = \cfrac{329}{2} - \cfrac{1927}{14} = \cfrac{188}{7}$

$r_2 = s - b = \cfrac{329}{2} - \cfrac{47}{28}(57 + \sqrt{113}) = \cfrac{47}{28}(41 - \sqrt{113})$

$r_3 = s - a = \cfrac{329}{2} - \cfrac{47}{28}(57 - \sqrt{113}) = \cfrac{47}{28}(41 + \sqrt{113})$