Question

Suppose that $\triangle ABC$ is an equilateral triangle of side length $s,$ with the property that there is a unique point $P$ inside the triangle such that $AP = 1,$ $BP = \sqrt{3},$ and $CP = 2.$ What is $s?$

$1 + \sqrt{2}$

$\sqrt{7}$

$\dfrac{8}{3}$

$\sqrt{5 + \sqrt{5}}$

$2\sqrt{2}$

Solution:

A point at distances $p, q, r$ from the vertices of an equilateral triangle of side $s$ satisfies $3(p^4 + q^4 + r^4 + s^4) = (p^2 + q^2 + r^2 + s^2)^2.$

With $p^2 = 1,$ $q^2 = 3,$ $r^2 = 4,$ letting $S = s^2$ gives $3(26 + S^2) = (8 + S)^2,$ so $S^2 - 8S + 7 = 0$ and $S = 1$ or $S = 7.$

A triangle of side $1$ cannot contain a point at distance $2$ from a vertex, so $S = 7$ and $s = \sqrt{7}.$

Thus, B is the correct answer.

2020 AMC 12A Problem 24

Solution:

Problem 24 in Other Years