It’s Not 3-4-5?

Let $A$ , $P$ be the area and the perimeter of the right triangle $\triangle ABC$ and $a$ , $b$ , $c$ its hypotenuse and legs.
Then $A=\dfrac{bc}{2}\Rightarrow c=\dfrac{2A}{b}$ , $A\ge 1$ .

$P=b+c+a=b+\dfrac{2A}{b}+\sqrt{{{b}^{2}}+\dfrac{4{{A}^{2}}}{{{b}^{2}}}}$

The sum we want to minimise is $S=A+P=A+b+\dfrac{2A}{b}+\sqrt{{{b}^{2}}+\dfrac{4{{A}^{2}}}{{{b}^{2}}}}$ .
For any fixed value of the area $A$ , the sum becomes a function of $b$ and ${S}'\left( b \right)=\dfrac{\left( {{b}^{2}}-2A \right)\left[ \sqrt{{{b}^{4}}+4{{A}^{2}}}+{{b}^{2}}+2A \right]}{{{b}^{2}}\sqrt{{{b}^{4}}+4{{A}^{2}}}}$ .
Thus, ${S}'\left( b \right)=0\Leftrightarrow {{b}^{2}}-2A=0\overset{b>0}{\mathop{\Leftrightarrow }}\,b=\sqrt{2A}$ .

This means that, for any constant value of $A$ , the sum is minimised when ${S}'\left( b \right)=0\Leftrightarrow {{b}^{2}}-2A=0\overset{b>0}{\mathop{\Leftrightarrow }}\,b=\sqrt{2A}$ .
The minimum value is $S\left( \sqrt{2A} \right)=\ldots =A+2\left( \sqrt{2}+1 \right)\sqrt{A}$ .

In terms of $A$ , the function $y=A+2\left( \sqrt{2}+1 \right)\sqrt{A}$ is increasing, thus, for $A\ge 1$ , it gives its minimum when $A=1$ .
Then $S$ becomes ${{S}_{\min }}=S\left( 1 \right)=2\sqrt{2}+3\approx 5.828$ .

We conclude that the minimum possible integer value of $S$ can be 6.
In this case, we have: $\begin{aligned} S=6& \overset{A=1}{\mathop{\Leftrightarrow }}\,1+b+\dfrac{2}{b}+\sqrt{{{b}^{2}}+\dfrac{4}{{{b}^{2}}}}=6 \\ & \Leftrightarrow \sqrt{{{b}^{4}}+4}=-{{b}^{2}}+5b+20 \\ & \Leftrightarrow \ldots \\ & \Leftrightarrow b=\dfrac{29\pm \sqrt{41}}{20} \\ \end{aligned}$

With this value, we get:
$c=\dfrac{2}{b}=\dfrac{29\mp \sqrt{41}}{20}$ and hypotenuse $\sqrt{{{b}^{2}}+\dfrac{4}{{{b}^{2}}}}=\dfrac{21}{10}.$

Summing up the elements of our triangle:

$b=\dfrac{29\pm \sqrt{41}}{20}$ , $c=\dfrac{29\mp \sqrt{41}}{20}$ , $a=\dfrac{21}{10}$ , $A=1$ and $P=5$ .

For the answer, $p=21$ , $q=10$ , so $p+q=\boxed{31}$ .

Let the area of the triangle be $A$ and its perimeter $P$ .

Over all right-angled triangles, the ratio $\frac{P^2}{A}$ is minimised by the right-angled isosceles triangle, in which it is $2(2+\sqrt2)^2 \approx 23.31$ . (This is a fairly intuitive result but also a nice thing to prove as a side-problem.)

We're interested in minimising $A+P$ for positive integers $A,P$ subject to $\frac{P^2}{A}\ge 2(2+\sqrt2)^2$ ; the pair we're after is $A=1,P=5$ .

To find the triangle with these properties, say the triangle has legs $x,y$ and hypotenuse $z$ , so $x^2+y^2=z^2$ . Also $A=\frac{xy}{2}$ and $P=x+y+z$ .

We have $z=P-x-y$ and $y=\frac{2A}{x}$ . Putting these together, $\begin{aligned} x^2+\frac{4A^2}{x^2} &= \left(P-x-\frac{2A}{x} \right)^2 \\ x^2+\frac{4A^2}{x^2} &= P^2+x^2+\frac{4A^2}{x^2}-2Px-\frac{4AP}{x}+4A \\ 0 &= P^2-2Px-\frac{4AP}{x}+4A \\ 2Px^2-(4A+P^2)x+4AP &=0 \end{aligned}$

We can substitute in and solve here, but it's neater to note that the two roots of this quadratic are just the two legs of the triangle; so - by Vieta - $x+y=\frac{4A+P^2}{2P}$ and $z=P-x-y$ so that $z=\frac{P^2-4A}{2P}$

Plugging in $A=1,P=5$ we get $z=\frac{21}{10}$ so the answer is $\boxed{31}$ .

Note that in this case the legs $x,y$ are not rational; in fact $x,y=\frac{29\pm \sqrt{41}}{20}$ .