From PUMaC:
In right triangle $\triangle ABC$, a square $W XY Z$ is inscribed such that vertices $W$ and $X$ lie on hypotenuse $AB$, vertex $Y$ lies on leg $BC$, and vertex $Z$ lies on leg $CA$. Let $AY$ and $BZ$ intersect at some point $P$. If the length of each side of square $W XY Z$ is $4$, the length of the hypotenuse $AB$ is $60$, and the distance between point $P$ and point $G$, where $G$ denotes the centroid of the triangle, is $\frac{a}{b}$, compute the value of $a + b$.
Initially I thought thought this would be entirely analytical, both because of the given centroid and that $P$ and $G$ don't really "fit", but I thought for a while and got nowhere. I looked at the solution, and it seems that one key insight is that by Ceva's and letting $M$ be the midpoint of $AB$, the segment $CP$ is part of the segment $CM$.
Then the solution says this following, where $s=WX=4$ and $c=AB$.
Hence, by mass points, we know that the ratio of CP to CM, where M is the midpoint of hypotenuse AB must be $$\frac{2(\frac{s}{c})}{\frac{s}{c}+1}.$$
I'm not very familiar with mass points, but with the result given I still haven't been able to prove it, or how to start. Any hints or full explanations would be appreciated.
The full solution is linked, #12: https://static1.squarespace.com/static/570450471d07c094a39efaed/t/5c32c3811ae6cfe5e9e7b630/1546830721291/Team+Round+Solutions.pdf.
In barycentric coordinates (or mass points), we have $C=(0,0,1)$, $Y=(0,s/c,(c-s)/c)$, $Z=(s/c,0,(c-s)/c)$. So:
So $\alpha=\beta=c/(c+s)$. Since $w_A(P)=w_B(P)$, we have $P$ lies on the median $CM$, where $M$ is the mid-point of $AB$. The ratio of distances $$\frac{MP}{MC}=\frac{\operatorname{area}[\triangle PAB]}{\operatorname{area}[\triangle CAB]}=w_C(P)=\frac{c-s}{c+s}$$ and so $\frac{CP}{CM}=1-\frac{MP}{MC}=\frac{2s}{c+s}$.