2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 5

Problem

Let $A$ and $B$ be two points in the plane. Describe the set $S$ of all points in the plane such that for any point $P$ in $S$ we have $|PA| = 3|PB| \text{.}$

Solution

WLOG, let $A = (0, 0)$ , and $B = (b, 0)$ . That means that we have that for any point $(x, y) \in S$ , $\sqrt{x^2 + y^2} = 3 \sqrt{(x - b)^2 + y^2} \implies x^2 + y^2 = 9((x - b)^2 + y^2) \implies x^2 + y^2 = 9x^2 - 18xb + 9b^2 + 9y^2 \implies$ $8x^2 - 18xb + 9b^2 + 8y^2 = 0$ . Conic sections written in the form $ax^2 + bx + cy^2 + dy + e = 0$ are circles if and only if $a = c$ , which is true in our equation. Therefore, S is a circle. ~Puck_0

2015 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
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2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 5

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