# 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?

• The name I've seen for this is the Fermat point.
– kate
Commented Dec 3, 2014 at 5:24
• Specifically, first Fermat point. It also has the property of minimizing sum of (L1) distances to sides.
– smci
Commented Dec 3, 2014 at 7:43
• Think of $A$, $B$ and $C$ as three cities (on a flat Earth with no topographic obstacles). How can we connect the three cities with roads such that the the total road length is minimal? Answer: Case 1: If one angle of $\triangle ABC$ is $2\pi /3$ or greater, the city at that angle is the "middle" city. The solution to the shortest-road problem is then simply to connect the middle city to each of the other two cities. The roads look like a very open letter V. Case 2: If all angles in $\triangle ABC$ are less than $2\pi /3$, connect the Fermat point to each of the cities, in a Y-shape. Commented Dec 3, 2014 at 9:58

## 4 Answers

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:

• If all the angles of the original triangle are less than $120^\circ$ then the Fermat point is inside the triangle, with the other isogonic centre outside the triangle. Otherwise both isogonic centres are outside the triangle. If an isogonic centre is outside the triangle, then the angles there are $60^\circ$, $60^\circ$ and $120^\circ$. Your circles make up a vesica piscis, and Wikipedia has a nice picture of the intersections Commented Dec 3, 2014 at 8:34
• @Henry. Thanks. I updated my answer. Commented Dec 3, 2014 at 9:00
• Another way to construct the point: Construct equilateral $\triangle$s $BAC', ACB', CBA'$ outwards on the sides of $\triangle ABC$. Then the lines $AA', BB', CC'$ concur at the Fermat point. Jack D'Aurizio pointed this out. Those triangles' circumcircles are your circles. Commented Jun 29, 2018 at 8:11

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 $C$.

• Can we say there always exists such a point? Commented Dec 3, 2014 at 5:01
• I do follow this intuition. One can also imagine what happens (goes wrong) if $A$ is 120° or greater, and if $C$ is 120° or greater. In that case the Y-shaped stick configuration will not fit (if a triangle angle is exactly 120°, the Y will fit if we allow the triangle vertex to be in the center of the Y, not on the arm). Commented Dec 3, 2014 at 9:49

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?

• what do $x$ and $y$ represent? Commented Dec 3, 2014 at 5:03
• $x$ and $y$ represent a simultaneous choice of two of $A$, $B$, $C$. I didn't want to write out three very similar equations, so I wrote one. It has to hold true in the following three cases: when $x = A$ and $y = B$, when $x = B$ and $y = C$, and when $x = A$ and $y = C$.
– Unit
Commented Dec 3, 2014 at 5:37

It is called the first Fermat point.

From MathWorld, it seems it only exists when all angles < 120º