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Uniquely determined side

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Source: Original RMM 2019 P4

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#1 h Jun 21, 2020, 2:18 PM

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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 $ABC.$

This post has been edited 1 time. Last edited by huricane, Jun 21, 2020, 2:22 PM
Reason: fix

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#2 h Jun 21, 2020, 9:33 PM • 3 Y

Y by Inconsistent, IvoBucata, Mango247

Bad Writeup with Motivation

Good Writeup

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#3 h Jun 21, 2020, 9:48 PM • 2 Y

Y by mijail, aopsuser305

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.

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#4 h Jun 21, 2020, 11:26 PM

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We prove the contrapositive: if $P$ does not lie on the circumcircle then there exist at least two possible side lengths for $ABC$ .

sketch

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#5 h Jun 22, 2020, 12:32 PM • 1 Y

Y by khina

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.

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