So i've tried this question for a while now, but can't seem to get an answer. I tried to equate z but don't know how to proceed. Can someone help?
In△XYZ, the measure of∠XZY is 90. Also, YZ=x cm, XZ=y cm, and hypotenuse XY has length z cm. Further, the perimeter of △XYZ is P cm and the area of △XYZ is A cmsq .
Determine all possible integer values of x, y and z for which A = 3P .
On
We begin with Euclid's formula as $\qquad A=m^2-k^2\quad B=2mk \quad C=m^2+k^2\quad$
$$R=\frac{area}{perimeter}=\frac{AB}{2P}=\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}$$
We can solve the ratio formula for $k$ and test a finite range of $m$-values to see which, if any yield integers.
\begin{equation} R=\frac{mk-k^2}{2}\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\\\text{for}\quad\big\lceil\sqrt{8R}\big\rceil\le m \le (2R+1) \end{equation} The lower limit insures that $k\in \mathbb{N}$ and the upper limit ensures that $m> k$.
$$R=3\implies \lceil\sqrt{8\cdot3}\space\rceil=5\le m \le (6+1)=7 \\ \text{and we find}\quad m\in\{ 5\}\implies k\in\{ 3,2\} \quad\land \quad m\in\{ 7\}\implies k\in\{ 6,1\}$$
$$ f(5,3)=(16,30,34)\qquad \frac{16\cdot30}{2(16+30+34)}=\frac{480}{2(80)}=3\\ f(5,2)=(21,20,29)\qquad \frac{21\cdot20}{2(21+20+29)}=\frac{420}{2(70)}=3\\ f(7,6)=(13,84,85)\qquad\frac{13\cdot84}{2(13+84+85)}=\frac{1092}{2(182)}=3\\ f(7,1)=(48,14,50)\qquad\frac{48\cdot14}{2(48+14+50)}=\frac{672}{2(112)}=3$$
Using $A=rs$, $P=2s$, where $s=$ semiperimeter and $r=$ inradius, given condition reduces to $$rs=3\cdot 2s\Rightarrow r=6$$
From the formula, $x+y-z=2r$, we get $$x+y=z+12$$
which leads to $$(x-12)(y-12)=72$$
for which there are six integral solutions.