In square $ABCD$, points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$, respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$, with $BR=12$ and $PR=14$. What is the area of the square?
Here $BP=BR+PR=26$. Let $s$ be the side of the square and $\theta=\angle ABP$. In right triangle $ABP$, $\cos\theta=\frac s{26}$. Since $CQ\perp BP$, $\angle RBC=90^\circ-\theta$, and in right triangle $BRC$, $\cos(90^\circ-\theta)=\sin\theta=\frac{12}s$. From $\cos^2\theta+\sin^2\theta=1$: $\frac{s^2}{676}+\frac{144}{s^2}=1$, i.e. $s^4-676s^2+97344=0$, so $s^2=468$ or $s^2=208$. Since $P$ lies on side $AD$, $AP=\sqrt{676-s^2}\le s$, which rules out $s^2=208$. The area is $s^2=468$.
