Question 1205030
<br>
Don't use "\" -- it might have a special meaning in some area(s) of mathematics.<br>
I assume the side lengths are p/sqrt(6), q/sqrt(3), and p/sqrt(2).<br>
The squares of the side lengths are then p^2/6, q^2/3, and p^2/2.<br>
On first glance, with the denominators 6, 3, and 2, I immediately see that, if p and q are both equal to the same number x, then I have x^2/6+x^2/3=x^2/2, which is true for all values of x.<br>
So the problem has an infinite number of solutions in which p=q.<br>
But there might be other solutions hiding somewhere, so lets' look at the problem more closely.<br>
We know that p/sqrt(2) is greater than p/sqrt(6); but q/sqrt(3) could be less than or greater than p/sqrt(2).  So there are two cases to consider: the longest side (hypotenuse) can be either p/sqrt(2) or q/sqrt(3).<br>
Case 1: the hypotenuse is p/sqrt(2)<br>
(Note this is the case discussed informally above.)<br>
{{{(p/sqrt(6))^2+(q/sqrt(3))^2=(p/sqrt(2))^2}}}
{{{p^2/6+q^2/3=p^2/2}}}
{{{p^2+2q^2=3p^2}}}
{{{2q^2=2p^2}}}
{{{p^2=q^2}}}
{{{p=q}}}<br>
Case 2: The hypotenuse is q/sqrt(3)<br>
{{{(p/sqrt(2))^2+(p/sqrt(6))^2=(q/sqrt(3))^2}}}
{{{p^2/2+p^2/6=q^2/3}}}
{{{3p^2+p^2=2q^2}}}
{{{4p^2=2q^2}}}
{{{2p^2=q^2}}}
{{{p*sqrt(2)=q}}}<br>
This case also has an infinite number of solutions, where p is any number x and q is x*sqrt(2).<br>
ANSWER:
p = any number;
q = p OR q=p*sqrt(2)<br>