Distances to Three Points on a Circle

Problem

Distances to Three Points on a Circle, problem

Solution

The two conditions $|a|=|b|=|c|=1\,$ and $a+b+c=0\,$ insure that the three points, $a,b,c,\,$ form an equilateral triangle inscribed into the unit circle. The problem is about the sum of distances from points inside and on the unit circle to the vertices of the triangle. We have to prove that the minimum of the sum is $3\,$ and the maximum is $4.$

Thus we shall consider an equilateral $\Delta ABC\,$ with the circumradius $R=1\,$ and point $D\,$ in or inside its circumcircle. The task is to prove that $3\le AD+BD+CD\le 4.$

  • $\mathbf{4\,is\,the\,maximum}$

    First observe that the reflections of the segments cut off the circumcircle in the sides of the triangle cover completely (even with overlaps) the whole triangle. If point $D\,$ is in one of such reflections, its reflection, say, $D_1,\,$ in the middle of the corresponding side/chord will fall into the originating circular segment:

    Serbanescu, maximum, 1

    With the reference to the above configuration, $AD+BD=AD_1+BD_1,\,$ but $CD_1\ge CD.\,$ So it follows that $AD+BD+CD\le AD_1+BD_1+CD_1,\,$ implying that the maximum of the sum of distances to the vertices could not be attained at the point within the triangle.

    Serbanescu, maximum, 2

    Thus let point $D\,$ be located in the segment of the circumcircle with $AB\,$ as a chord. If $D\,$ is not on the circle, then define $D_2\,$ as the second intersection of $CD\,$ with the circle. Obviously, $CD_2\gt CD.\,$ Also $AD_2+BD_2\gt AD+BD,\,$ because $D\,$ is in the interior of $\Delta ABD_2.$

    Thus the only place where the maximum is attained is the circumcircle of the triangle. In the above configuration, for $D\in\overset{\frown}{AB},\,$ by Pompeiu's theorem, $AD+BD=CD\,$ so that $AD+BD+CD=2\cdot CD.\,$ The maximum of the sum is attained together with the maximum of the chord $CD,\,$ i.e., when $CD$ is a diameter and, hence, has the length of $2.\,$

  • $\mathbf{3\,is\,the\,minimum}$

    As we saw earlier, the sum of distances can only achieve the minimum for $D\,$ in the interior of the triangle. So, let $D\,$ be such a point. Consider an equilateral $\Delta DD'D''\,$ with the same center as that of $\Delta ABC.\,$

    Serbanescu, minimum

    Then $AD+BD+CD=AD+AD'+AD''.\,$ Now,

    $\begin{align} AD+AD'+AD''&\ge |\overset{\rightarrow}{AD}+\overset{\rightarrow}{AD'}+\overset{\rightarrow}{AD''}|\\ &=3|\overset{\rightarrow}{AE}|\\ &=3\cdot AE=3. \end{align}$

Acknowledgment

I am grateful to Leo Giugiuc for posting the above problem at the CutTheKnotMath facebook page. The problem is due to Dinu Serbanescu (Romania).

 

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