Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.3 AMC 10/12 Find the answer

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:

Pick one

Solution

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

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.