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 $\boxed{\textbf{(E)}}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
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1969 AHSME Problems/Problem 28

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