# A Combination Locus

The problem below is due to Leo Giugiuc and Dan Marinescu. Leo has posted the problem at the CutTheKnotMath facebook page and communicated its solution via private mail.

Let $B,$ $C$ be two fixed distinct points in the plane $\alpha$ and let $M$ be the midpoint $BC.$ Find the locus of points $A\in\alpha,$ $A\notin BC,$ for which the following relation holds:

$4R\cdot AM = AB^{2} + AC^{2}.$

where $R$ is the circumradius of $\Delta ABC.$

### Solution

Let $A$ be a point in the locus. We choose a Cartesian system so that $A=(0, 1),$ $B=(-b, 0),$ $C=(c, 0).$ Obviously, $\displaystyle M=\bigg(\frac{c-b}{2},0\bigg).$

By the Pythagorean Theorem, $AM^{2}=\frac{1}{4}(4+(c-b)^{2}),$ $AB^{2}=1+b^{2},$ $AC^{2}=1+c^{2},$ and from $4RS=abc,$ $R^{2}=\frac{1}{4}(1+b^{2})(1+c^{2}).$ Thus, the imposed identity (after squaring) is expressed as

$(4+(c-b)^{2})(1+b^{2})(1+c^{2})=(2+b^{2}+c^{2})^{2}.$

This is manipulated into $(c-b)^{2}(1-bc)^{2}=0.$ It follows that the initial requirement splits into two cases:

$b=c$ or $bc=1.$

The first case corresponds to points $A$ equidistant from $B$ and $C.$ The second case is equivalent to $AB\perp AC,$ because the slopes of those lines are equal $-b$ and $c,$ respectively. Let $d$ be the perpendicular bisector of $BC$ and $\omega$ the circle with $BC$ as a diameter. Then the locus in question is $(d\cup\omega)\setminus\{A,M,B\}.$

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny

63414612 |