Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it United States

Problem:

Let $A$ and $B$ be two points on the plane with $AB = 7$ . What is the set of points $P$ such that $PA^2 = PB^2 - 7$ ?

Solution

Solution:

If we let $K$ be the point on $AB$ with $AK = 4$ , $BK = 3$ , then the answer is the line through $K$ perpendicular to $AB$ . To see this, set $A = (0, 0)$ and $B = (7, 0)$ . Then the points $P = (x, y)$ are exactly those satisfying
$(x - 0)^2 + (y - 0)^2 = (x - 7)^2 + (y - 0)^2 - 7$
which rearranges to $9 = -14x + 42$ , i.e., $x = 3$ .

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