# Lines Crossing Circles at Vertices of Equilateral Triangle

Below we give an analytic solution to the following problem (already discussed elsewhere):

Circles $(A),$ $(B),$ $(C)$ centered at the vertices of an equilateral triangle $ABC$ with side length $4$ all have radius $\sqrt{3}.$

Describe all straight lines that have common points with all three circles.

The problem and the solution are due to Leo Giugiuc.

The solution is based on the following

### Lemma

Let $d$ be a line in the plane of $\Delta ABC;$ $A',$ $B',$ $C'$ the projections of the vertices onto $d.$ Then

$[ABC]\le[A'B'C]+[AB'C']+[A'BC'],$

where $[X]$ denotes the area of figure $X.$

Either two of the projections coincide or one of the three lies between the other two. In any case, one of the segments $A'B',$ $B'C',$ $A'C'$ is the sum of the other two. Without loss of generality assume that $B'C' = A'B'+A'C'.$ More importantly, $B'C'\ge A'B'$ and $B'C'\ge A'C'.$ So, assume line $d$ has common points with all three circles. We then have:

$\begin{align}\displaystyle 2\sqrt{3} &= \frac{2[ABC]}{4}\\ &\le\frac{2([A'B'C]+[AB'C']+[A'BC'])}{4}\\ &=\frac{CC'\cdot A'B'+AA'\cdot B'C'+BB'\cdot A'C'}{4}\\ &=\frac{\sqrt{3}}{4}2B'C'\\ &\le\frac{\sqrt{3}}{4}2\sqrt{3}=2\sqrt{3}. \end{align}$

It therefore follows that for a line $d$ that meets all three circles and has $A'$ between $B'$ and $C'$, $B'C'=BC$ and $B'C'\parallel BC.$ There is just one such line. There is also one line parallel to each of the remaining two sides.

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

Copyright © 1996-2018 Alexander Bogomolny71674056