# 1969 AHSME Problems/Problem 30

## Problem

Let $P$ be a point of hypotenuse $AB$ (or its extension) of isosceles right triangle $ABC$. Let $s=AP^2+PB^2$. Then:

$\text{(A) } s<2CP^2 \text{ for a finite number of positions of P}\quad\\ \text{(B) } s<2CP^2 \text{ for an infinite number of positions of P}\quad\\ \text{(C) } s=2CP^2 \text{ only if P is the midpoint or an endpoint of AB}\quad\\ \text{(D) } s=2CP^2 \text{ always}\quad\\ \text{(E) } s>2CP^2 \text{ if P is a trisection point of AB}$

## Solution

$[asy] pair A=(0,50),B=(50,0),C=(0,0); draw(A--B--C--A); dot(A); label("A",A,NE); dot(B); label("B",B,NE); dot(C); label("C",C,SW); pair P=(14,36); dot(P); label("P",P,NE); draw((0,36)--P,dotted); draw((14,0)--P,dotted); label("a",(7,36),S); label("a",(0,43),W); label("x-a",(14,18),E); label("x-a",(32,0),S); [/asy]$

Consider the case where $P$ is on the hypotenuse of $AB$. Draw perpendicular lines from $P$ towards the sides. Using the Pythagorean Theorem, $$AP^2 = a^2 + a^2$$ $$BP^2 = (x-a)^2 + (x-a)^2$$ $$CP^2 = a^2 + (x-a)^2$$ This means $$s = 4a^2 - 4ax + 2x^2$$ $$2 \cdot CP^2 = 4a^2 - 4ax + 2x^2$$ Thus, $s = 2 \cdot CP^2$ when $P$ is on the hypotenuse of $AB$.

$[asy] pair A=(0,50),B=(50,0),C=(0,0); draw(A--B--C--A); dot(A); label("A",A,NE); dot(B); label("B",B,NE); dot(C); label("C",C,SW); pair P=(-10,60); dot(P); label("P",P,NE); draw(P--A); draw(P--(-10,50)--A,dotted); draw(P--(-10,0)--C,dotted); pair D=(-10,50); dot(D); label("D",D,SW); label("a",(-10,55),W); label("a",(-5,50),S); label("x",(0,25),E); label("x",(25,0),S); [/asy]$

Consider the case where $P$ is on the extension of $AB$. WLOG, let point $A$ be between point $P$ and point $B$. Extend $BC$ and draw perpendicular line from $P$. Also, draw point $D$, where $PD \parallel AC$ and $DA \parallel CB$.

Using the Pythagorean Theorem again, $$AP^2 = a^2 + a^2$$ $$BP^2 = (a+x)^2 + (a+x)^2$$ $$CP^2 = (a+x)^2 + a^2$$ That means $$s = 4a^2 + 4ax + 2x^2$$ $$2 \cdot CP^2 = 4a^2 + 4ax + 2x^2$$ Thus, $s = 2 \cdot CP^2$ when $P$ is outside the hypotenuse.

In summary, $AP^2 + BP^2 = 2 \cdot CP^2$, so the answer is $\boxed{\textbf{(D)}}$.