Hard geometry problem: given a triangle with sides 16, 30, 34, find the area of the triangle introduced by the inscribed circle You are given a triangle $\triangle PQR$ with sides $16, 30, 34$. Let the incircle touch the sides of $\triangle PQR$ at $X,Y,$ and $Z$. Given that the ratio $[XYZ]/[PQR]$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, find $m+n$.
What I did first was to draw the diagram and I noticed that $PX=PZ, QX=QY, RY=RZ$ due to two intercepting lines both being tangent to a circle. Since triangles $PXZ, QXY, RYZ$ are isosceles I bisected them and got another right triangle after I tried to do similar triangles but got a little stuck.
My other friend used "barycentric coordinates", which I thought was completely unnecessary.
Any help would be appreciated.
 A: Let $PZ=PX=p$, $QX=QY=q$, $RY=RZ=r$. Then
$$\begin{aligned}
&p+q=PX+QX=PQ=16\\
&q+r=QY+RY=QR=30\\
&r+p=RZ+PZ=RP=34\\
\end{aligned}$$
So we have
$$\begin{aligned}
&p=\frac{(p+q)+(r+p)-(q+r)}2=10\\
&q=\frac{(q+r)+(p+q)-(r+p)}2=6\\
&r=\frac{(r+p)+(q+r)-(p+q)}2=24\\
\end{aligned}$$
As you claimed, in order to deal with the ratio between the areas of triangles, we do not have to use "barycentric coordinates".
$$\begin{aligned}
&\frac{[PXZ]}{[PQR]}=\frac{PX\cdot PZ}{PQ\cdot PR}=\frac{10^2}{16\cdot34}=\frac{25}{8\cdot17}\\
&\frac{[QYX]}{[QRP]}=\frac{QY\cdot QX}{QR\cdot QP}=\frac{6^2}{30\cdot16}=\frac{3}{5\cdot8}\\
&\frac{[RZY]}{[RPQ]}=\frac{RZ\cdot RY}{RP\cdot RQ}=\frac{24^2}{34\cdot30}=\frac{48}{5\cdot17}\\
\end{aligned}$$
Note that I write $[PQR]$ also as $[QRP]$ or $[RPQ]$ so that all three equalities are symmetric.
$$\frac{[XYZ]}{[PQR]}=1-\frac{[PZX]}{[PQR]}-\frac{[QXY]}{[PQR]}-\frac{[RYZ]}{[PQR]}=\frac{120}{5\cdot8\cdot17}=\frac3{17}$$
$m=3$, $n=17$, $m+n=20$.

You might wonder why $\frac{[PXZ]}{[PQR]}=\frac{PX\cdot PZ}{PQ\cdot PR}$.
Let us consider $\triangle PXZ$ and $\triangle PQZ$.
Let $ZF$ be an altitude of $\triangle PXZ$. So $[PXZ]=\frac12PX\cdot ZF$.
Since $ZF$ is also an altitude of $\triangle PQZ$,
$[PQZ]=\frac12PQ\cdot ZF$.
$$\frac{[PXZ]}{[PQZ]}=\frac{\frac12PX\cdot ZF}{\frac12PQ\cdot ZF}=\frac{PX}{PQ}$$
Similarly, we have
$$\frac{[PZQ]}{[PRQ]}=\frac{PZ}{PR}$$
Multiplying the two equalities above, we get $\frac{[PXZ]}{[PRQ]}=\frac{PX}{PQ}\frac{PZ}{PR}$, which is what we wanted.
Similarly, we have $\frac{[QYX]}{[QRP]}=\frac{QY\cdot QX}{QR\cdot QP}$ and $\frac{[RZY]}{[RPQ]}=\frac{RZ\cdot RY}{RP\cdot RQ}$.
A: John Omeilan has provided an excellent solution. However the answer can be simpler if we observe that $\Delta PQR$ is a right angled triangle.

Since $16^2+30^2=34^2$, $\Delta PQR$ is a right angled triangle and hence  $ [PQR] = \frac{16 \times 30}{2}=240$
$s=\frac{p+q+r}{2}=\frac{30+16+34}{2}=40$
Let the radius of the inscribed circle be $x$
From $xs=[PQR] =240$, we have
$$x=\frac{240}{40}=6$$
Notice that $\sin R=\frac{16}{34}=\frac{8}{17}, 
\sin P=\frac{15}{17}$
and $\sin Q=\frac{34}{34}=1$
Also note that $\angle YOZ=180^o-R$.
Hence $ [YOZ]=\frac{1}{2}x^2\sin(180^o-R)=\frac{1}{2}\times 6^2 \times \sin R=18 \times \frac{8}{17}$
Similarly $[ZOX]=\frac{1}{2}x^2 \times \frac{15}{17}=18 \times \frac{15}{17}$
and $[XOY]=\frac{1}{2}x^2=\frac{1}{2}\times 6^2=18$
Hence $[XYZ]=18 \times \left(\frac{8}{17}+\frac{15}{17}+1 \right)=\frac{720}{17}$
Thus $\frac{[XYZ]}{[PQR]}=\frac{\frac{720}{17}}{240}=\frac{3}{17}$
