Bounds for the Sum of Distances to the Vertices


Consider complex numbers $a,$ $b,$ $c$ that satisfy


Prove that

$3\le |z-a|+|z-b|+|z-c|\le 4,$

for any $z,$ with $|z|\le 1.$


Let's agree that - for simplicity - for any expression $f(x),$ $\sum f(a)$ denotes $f(a)+f(b)+f(c).$ Since $|a|=1,$ so $|\overline{a}|=1$ and $\overline{a}a=1,$ etc., we have,

$\begin{align} \sum |z-a| &= \sum \overline{a}|z-a| = \sum |\overline{a}z - 1|\\ &\ge |(\overline{a}z - 1)+(\overline{b}z - 1)+(\overline{b}z - 1)|\\ &= |z\sum\overline{a}-3|=3, \end{align}$

because $\sum\overline{a}=\overline{\sum a}=0.$

Now observe that the two conditions $a+b+c=0$ and $|a|=|b|=|c|=1$ imply that points $a,b,c$ on the unit circle form an equilateral triangle. Indeed, place the origin at the center of the circle and let $\gamma$ be the angle between $a$ and $b$ (as vectors from the origin) then, by the Law of Cosines,

$\displaystyle\begin{align} |a+b|^{2} &= |a|^{2}+|b|^{2}-2|a||b|\cos(180^{\circ}-\gamma )\\ &= 2+2\cos\gamma\\ &= 4\cos^{2}\frac{\gamma}{2}. \end{align}$

To insure $a+b+c=0,$ it is necessary to have $|a+b|=1$ which is achieved only when $\displaystyle\cos\frac{\gamma}{2}=\pm\frac{1}{2},$ i.e., when $\displaystyle\frac{\gamma}{2}=\pm 60^{\circ},$ or, $\gamma =\pm 120^{\circ}.$ Similarly, we verify that $a$ and $b$ form angles of $120^{\circ}$ with $c.$

For a given point (complex number) $z,$ let $pq$ be the diameter passing through $z.$ There is a real number $\alpha ,$ $0\le\alpha\le 1$ such that $z=\alpha p+(1-\alpha)q.$ This helps prove the right inequality $|z-a|+|z-b|+|z-c|\le 4:$

$\begin{align} \sum |z-a| &= \sum |\alpha p+(1-\alpha)q - a|\\ &= \sum |\alpha (p-a)+(1-\alpha)(q - a)|\\ &\le \sum (\alpha |(p-a)|+(1-\alpha)|q - a|)\\ &= \alpha\sum |p-a|+(1-\alpha)\sum |q-a|\\ &\le \max\{\sum |p-a|, \sum |q-a|\}. \end{align}$

All five points $a,b,c,p,q$ lie on the unit circle - the circumcircle of $\Delta ABC.$ It then follows that $\max\{|p-a|,|p-b|,|p-c|\}=2.$ The same of course holds for $q.$ By van Schooten's theorem, one of the three terms $|p-a|,|p-b|,|p-c|$ is the sum of the other two, implying that the sum of the three does not exceed $4!$. So, finally:

$\begin{align} \sum |z-a| &\le \max\{\sum |p-a|, \sum |q-a|\}\\ &\le 4. \end{align}$

Now, the left inequality - $3\le\sum |z-a|$ - only holds when $z$ is the center of the circle, i.e., $z=0.$ The right inequality - $\sum |z-a|\le 4$ - only holds when $z$ is one of the three points opposite the vertices $a,b,c.$


The left inequality - $3\le\sum |z-a|$ - expresses a property of the Fermat point of a triangle with all angles smaller than $120^{\circ}.$ Namely, the Fermat point $F$ minimizes the sum $AF+BF+CF$ for any three points $A,B,C.$ In an equilateral triangle the Fermat point - as all other centers of the triangle, the circumcenter in particular - coincides with the centroid.


The problem has been offered at the 2012 Romanian Mathematical Olympiad for grade 10 (Problem 2). I am grateful to Leo Giugiuc for pointing me to the source.

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

Copyright © 1996-2018 Alexander Bogomolny