jmc

Recall that if two triangles have their bases along the same straight line and they share a common vertex that is not on this line, then the ratio of their areas is equal to the ratio of the lengths of their bases. We will use this fact extensively throughout the proof.

Let the area of $△PYV$, $△PZU$, $△UXP$, and $△XVP$, be $a$, $b$, $c$, and $d$, respectively. Since $$\frac{|△PYV|}{|△PYW|}=\frac{VY}{YW}=\frac{4}{3},$$then $$a=|△PYV|=\frac{4}{3}\times |△PYW|=\frac{4}{3}(30)=40.$$Also, $\frac{|△VZW|}{|△VZU|}=\frac{ZW}{ZU}=\frac{|△PZW|}{|△PZU|}$ or $|△VZW|\times |△PZU|=|△PZW|\times |△VZU|$. Thus, $$\frac{|△VZU|}{|△PZU|}=\frac{|△VZW|}{|△PZW|}=\frac{35+30+40}{35}=\frac{105}{35}=\frac{3}{1}.$$Therefore, $\frac{|△VZU|}{|△PZU|}=\frac{3}{1}$, or $\frac{b+c+d}{b}=\frac{3}{1}$ or $b+c+d=3b$ and $c+d=2b$.

Next, $$\frac{|△UVY|}{|△UYW|}=\frac{VY}{YW}=\frac{4}{3},$$so $$\frac{40+c+d}{30+35+b}=\frac{4}{3}.$$Since $c+d=2b$, we have $3(40+2b)=4(65+b)$, so $120+6b=260+4b$, then $2b=140$ and $b=70$.

Next, $$\frac{|△UXW|}{|△XVW|}=\frac{UX}{XV}=\frac{|△UXP|}{|△XVP|},$$or $$\frac{35+b+c}{30+a+d}=\frac{c}{d}.$$Since $b=70$ and $a=40$, $\frac{105+c}{70+d}=\frac{c}{d}$, or $d(105+c)=c(70+d)$. Thus, $105d+cd=70c+cd$ or $105d=70c$, and $\frac{d}{c}=\frac{70}{105}=\frac{2}{3}$ or $d=\frac{2}{3}c$.

Since $c+d=2b=2(70)=140$, we have $$c+d=c+\frac{2}{3}c=\frac{5}{3}c=140,$$or $c=\frac{3}{5}(140)=84$. Therefore, the area of $△UXP$ is $\boxed{84}$.