Home > Calculus of Variations, Geometry > Torricelli point and angles of 120 degrees

Torricelli point and angles of 120 degrees

Denote ${ABC}$ a triangle with angles smaller than ${120^\circ}$. The point ${T}$ which minimizes the sum ${TA+TB+TC}$ is called the Torricelli point of the triangle ${ABC}$. One interesting property of the Toricelli point, besides the fact that it minimizes the above sum is that all the angles formed around ${T}$ are equal and have ${120^\circ}$.

I will prove here that the fact that the angles around ${T}$ are of ${120^\circ}$ can be derived without any geometric considerations, just from the fact that ${T}$ is the solution of the problem

$\displaystyle \min_{T \in \Delta ABC } TA+TB+TC.$

The usual method of construction for the Torricelli point is to build outside the triangle ${ABC}$ three equilateral triangles ${BCX,CAY,ABZ}$. After that, draw the circumcircles of these equliateral triangles and they will meet at a point ${T}$. Using Ptolemy’s theorem we can find that ${TB+TC=TX}$ (and the other two relations), and furthermore, using the properties of the angles of cyclic quadrilaterals, ${T}$ is the intersection of ${AX,BY,CZ}$. The fact that ${T}$ minimizes ${TA+TB+TC}$ is just a consequence of the triangle inequality now, and the angles around ${T}$ are of ${120^\circ}$ by construction.

The trouble with this proof is that it is constructive, i.e. it builds directly the object it tries to prove it exists. This does not usually happen when we encounter more general problems where the explicit solution is not obvious, and so the method presented below can turn out to be useful in more general problems.

First, it is obvious that the problem

$\displaystyle \min_{T \in \Delta ABC} TA+TB+TC$

has a solution ${T}$. The conditions made on the angles of the triangle ${ABC}$ are to avoid the degenerated case where ${T}$ is one of the vertices of the triangle.

Denote ${\alpha=\angle BTC,\ \beta = \angle CTA,\ \gamma = \angle ATB}$. Consider ${S \in (BT)}$ such that ${ST=t}$ (for ${t}$ small enough). Then we can calculate ${AS,BS,CS}$ using only ${t}$ and ${\alpha,\beta,\gamma}$ (and the cosine theorem) in the following way:

$\displaystyle AS+BS+CS=BT-t+\sqrt{TA^2+t^2-2tTA\cos \gamma}+\sqrt{TC^2+t^2-2tTC\cos \alpha}.$

The point ${S}$ can be considered as a function of ${t}$, and so can be considered the sum ${AS+BS+CS}$. The fact that for ${t=0}$ the above sum is minimum implies that the derivative of

$\displaystyle g: t \mapsto AS(t)+BS(t)+CS(t)$

in ${t=0}$ is equal to ${0}$. But we have

$\displaystyle g'(t)=-1+\frac{2t-2TA\cos \gamma}{2\sqrt{TA^2+t^2-2tTA\cos \gamma}}+\frac{2t-2TC\cos \alpha}{2\sqrt{TC^2+t^2-2tTC\cos \alpha}}$

and ${g'(t)=0}$ is equivalent to ${\cos\alpha+\cos \gamma=-1}$.

Doing the same thing on the segments ${(TA),(TC)}$ we obtain the other two similar relations which combined give ${\cos\alpha=\cos \beta =\cos\gamma=-1/2}$, and so all the angles around ${T}$ have ${120^\circ}$.

This proof can be used in more general cases: if we have a partition of some set in a finite number of polygons such that at every vertex at most three segments meet, the optimal partition (if it exists) which minimizes the sum of (relative) perimeters of the given polygons then at each joint point the angles will be of ${120^\circ}$.