# 1969 AHSME Problems/Problem 28

## Problem

Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1$, such that the sum of squares of the distances from $P$ to the endpoints of a given diameter is $3$. Then $n$ is: $\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 4\quad \text{(E) } \infty$

## Solution $[asy] draw(circle((0,0),50)); draw((-50,0)--(-10,20)--(50,0)--(-50,0)); draw((-10,20)--(30,40)--(50,0),dotted); dot((-50,0)); label("A",(-50,0),W); dot((50,0)); label("B",(50,0),E); dot((-10,20)); label("P",(-10,20),S); dot((30,40)); label("C",(30,40),NE); [/asy]$

Let $A$ and $B$ be points on diameter. Extend $AP$, and mark intersection with circle as point $C$.

Because $AB$ is a diameter, $\angle ACB = 90^\circ$. Also, by Exterior Angle Theorem, $\angle ACB + \angle CBP = \angle APB$, so $\angle APB > \angle ACB$, making $\angle APB$ an obtuse angle.

By the Law of Cosines, $AP^2 + BP^2 - 2 \cdot AP \cdot BP \cdot \cos{\angle APB} = 4$. Since $AP^2 + BP^2 = 3$, substitute and simplify to get $\cos{\angle APB} = -\frac{1}{2 \cdot AP \cdot BP}$. This equation has infinite solutions because for every $AP$ and $BP$, where $AP + BP \ge 2$ and $AP$ and $BP$ are both less than $2$, there can be an obtuse angle that satisfies the equation, so the answer is $\boxed{\textbf{(E)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 