B3. Two right triangles $\triangle A X Y$ and $\triangle B X Y$ have a common hypotenuse $X Y$ and side lengths (in units) $A X=5, A Y=10$, and $B Y=2$. Sides $A Y$ and $B X$ intersect at $P$. Determine the area (in square units) of $\triangle P X Y$.
Solution 1:
From the right triangle $A X Y$ we find $X Y^{2}=X A^{2}+A Y^{2}=125$. Then from the right triangle $B X Y$, $B X=\sqrt{X Y^{2}-B Y^{2}}=11$.

Labelling by $x, y$ and $z$ the area of the triangles $P X Y, A X P$ and $B P Y$ respectively (see the diagram), we get
$$
\begin{array}{l}
x+y=\frac{1}{2}(5)(10)=25 \\
x+z=\frac{1}{2}(2)(11)=11
\end{array}
$$

As $\triangle P A X$ and $\triangle P B Y$ are right angled and $\angle A P X=\angle B P Y$, the triangles are similar. Then, since $\frac{B Y}{A X}=\frac{2}{5}$ we can deduce that $z=\left(\frac{2}{5}\right)^{2} y=\frac{4}{25} y$. Substituting $z=\frac{4}{25} y$ into the system of equations and solving it gives $x=25 / 3$ and $y=50 / 3$. Thus, the desired area is $\frac{25}{3}$ square units.

Answer: $25 / 3$.
Solution 2:
From the right triangle $A X Y$ we find $X Y^{2}=X A^{2}+A Y^{2}=125$. Then from the right triangle $B X Y$, $B X=\sqrt{X Y^{2}-B Y^{2}}=11$.

As $\triangle P A X$ and $\triangle P B Y$ are right angled and $\angle A P X=\angle B P Y$, the triangles are similar. We have $A P: P B=X P: P Y=5: 2$, so $A P=5 u, P B=2 u, X P=5 v, P Y=2 v$, for some $u$ and $v$. Thus,
$$
\begin{array}{c}
B X=2 u+5 v=11 \\
A Y=5 u+2 v=10
\end{array}
$$

Solving this system we obtain $u=\frac{4}{3}$ and $v=\frac{5}{3}$. Thus, $X P=\frac{25}{3}$.
Therefore, the desired area is $\frac{1}{2} X P \cdot B Y=\frac{25}{3}$ square units.
Answer: $25 / 3$.