14. In $\triangle A B C, D$ and $E$ are the midpoints of the sides $A B$ and $A C$ respectively, $C D$ and $B E$ intersect at $P$ with $\angle B P C=90^{\circ}$. Suppose $B D=1829$ and $C E=1298$. Find $B C$
14. Ans: 2006

Since $D$ and $E$ are the midpoints of the sides $A B$ and $A C$ respectively, we have $D E$ is parallel to $B C$ and $\triangle P D E$ is similar to $\triangle P C B$ with $P D: P C=P E: P B=D E$ : $C B=1: 2$.

Let $P E=x$ and $P D=y$. Then $P B=2 x$ and $P C=2 y$. By Pythagoras' theorem applied to $\triangle P B D$ and $\triangle P C E$, we get $(2 x)^{2}+y^{2}=1829^{2}$ and $(2 y)^{2}+x^{2}=1298^{2}$. Adding these two equations, we have $x^{2}+y^{2}=\left(1829^{2}+1298^{2}\right) / 5$. Thus $B C^{2}=$ $(2 x)^{2}+(2 y)^{2}=4\left(1829^{2}+1298^{2}\right) / 5=4024036$. Therefore $B C=\sqrt{4024036}=2006$.