## Task 24/85

Let there be a rectangle $A B C D$ with an internal point $P$. Determine $P A$ in terms of $P B, P C$, and $P D$.

What value is obtained for $P B=33 \mathrm{LE}, P C=28 \mathrm{LE}, P D=41 \mathrm{LE}$?
We denote the perpendiculars from $P$ to the sides $A B, B C, C D$, and $D A$ by $m$, $n$, $o$, and $p$ respectively. Then, according to the Pythagorean theorem,

$$
A P^{2}=m^{2}+p^{2} ; \quad P B^{2}=m^{2}+n^{2} ; \quad P C^{2}=n^{2}+o^{2} ; \quad P D^{2}=o^{2}+p^{2}
$$

and therefore

$$
A P^{2}+P C^{2}=\left(m^{2}+p^{2}\right)+\left(n^{2}+o^{2}\right)=\left(m^{2}+n^{2}\right)+\left(p^{2}+o^{2}\right)=P B^{2}+P D^{2}
$$

so $A P^{2}=P B^{2}+P D^{2}-P C^{2}$ and

$$
A P=\sqrt{P B^{2}+P D^{2}-P C^{2}}=\sqrt{33^{2}+41^{2}-28^{2}}=\sqrt{1996} \text{ units}
$$

### 2.26 Problems and Solutions 1986