Does a triangle always have a point where each side subtends equal 120° angles?

Is there a point $O$ inside a triangle $\triangle ABC$ (any triangle) such that the angle $\angle{AOB} = \angle{BOC} = \angle{AOC}$? What do we call this point?

Solution 1:

This is not the case for every triangle, $1^\circ-1^\circ-178^\circ$ triangle, for example, is one of the counterexamples to this claim. However, if all angles are less then $120^\circ$, then the claim is true.

To construct such a point; Take any side $[AB]$, find two intersections of perpendicular bisector and circle with radius $\dfrac{|AB|}{2\sqrt 3}$ centered at middle point of $[AB]$. Call this points $A'$ and $B'$. Draw two circles contains points $A'AB$ and $B'AB$. All the $120^\circ$ angles that see $[AB]$ are on these circles. If you apply these procedure to other sides and take intersection points of these circles, you can see combinations of intersection points such that three circles intersect, gave you two points. One of these points are always outside of the triangle and you can see other point could be outside or inside of the triangle.

Update: Apparently; these two intersection points are named Fermat points; point on always outside is called second Fermat point, and the other is called first Fermat point. Also, above circles which have these points on are called Vesica piscis.

Here is a picture of these Fermat points and circles:

Solution 2:

All the angles are $120^\circ$ as they add to a full circle. Yes there is such a point as long as the triangle angles are less than $120^\circ$. Imagine having a Y shaped set of sticks at $120^\circ$ angles. If you put one arm through $A$ with the vertex very close to $A$ and a second arm through $B$, the third arm will almost extend $AB$. Put the third arm on the side of $AB$ that $ C$ is on. If you then slide the vertex away from $A$ and nearer $B$, when it gets very close to $B$ the third arm will extend $AB$ the other direction. Somewhere in between it will go through $C$. I don't know what the point is called. It also gives you the minimum length network that connects $A$, $B$, and $C$.

Solution 3:

I've never considered this before and I don't know what such a point is called. But if such a point $O$ exists inside $\triangle ABC$, then $$\angle AOB = \angle BOC = \angle COA = 120^\circ.$$ By the law of cosines, $|Ox|^2 + |Ox||Oy| + |Oy|^2 = |xy|^2$ for every pair $\{x, y\} \subseteq \{A, B, C\}$. (I'm using the bars to mean length.)

I'm not sure about the converse: if there is a point $O$ for which these three algebraic conditions are met, are all three angles $\angle AOB$, $\angle BOC$, $\angle COA$ equal?