Let there be an equilateral triangle $ABC$ and a point $P$ in its plane such that $AP<BP<CP.$ Suppose that the lengths of segments $AP,BP$ and $CP$ uniquely determine the side of $ABC$. Prove that $P$ lies on the circumcircle of triangle $ABC.$
Problem
Source: Original RMM 2019 P4
Tags: geometry, Equilateral Triangle, circumcircle, contest problem
22.06.2020 00:33
22.06.2020 00:48
Call a quadruple of positive real numbers $(a, b, c, d)$ riveting if there exists an equilateral triangle with side $d$ and a point in the plane with distances $a$, $b$, $c$ to the vertices. The following result is folklore. Claim. Any permutation of a riveting quadruple is also riveting. Now let $(a, b, c) = (PA, PB, PC)$ and assume that $P$ does not lie on the circumcircle of $\triangle ABC$ (so no two of $a$, $b$, $c$ sum to the third). Let $d$ be the side length of $\triangle ABC$, so $(a, b, c, d)$ is riveting. Then $(b, c, d, a)$ is riveting: let $XYZ$ be an equilateral triangle with side $a$ and let $Q$ be a point with $QX = b$, $QY = c$, $QZ = d$. Since no two of $a$, $b$, $c$ sum to the third, $Q$ does not lie on line $XY$. Let $Q'$ be the reflection of $Q$ over $\overline{XY}$, so $Q'X = QX = b$, $Q'Y = QY = c$, and let $Q'Z = d'$, say. Clearly $d' \ne d$. Then $(b, c, d', a)$ is riveting, so $(a, b, c, d')$ is riveting but $d' \ne d$, as wanted.
22.06.2020 02:26
We prove the contrapositive: if $P$ does not lie on the circumcircle then there exist at least two possible side lengths for $ABC$.
22.06.2020 15:32
Suppose, for the sake of contradiction, that $P$ does not lie on $(ABC)$. Let $\Gamma=(ABC)$, $O$ its center and $R$ its radius. Clearly $P \neq O$. Let $P'$ be the inverse of $P$ with respect to $\Gamma$. By inversion distance formula, $\frac{AP'}{AP}=\frac{R}{OP}=\frac{BP'}{BP}=\frac{CP'}{CP}$. Consider the homothety with center $O$ and factor $\frac{OP}{R} \neq 1$. It maps $\triangle ABC$ to a certain triangle $\triangle A'B'C'$ and point $P'$ to a point $P''$ such that $A'P''=AP$, $B'P''=BP$ and $C'P''=CP$, yielding the desired contradiction.