# Properties of triangles that have 3 equal 120 degree angles around an internal point

Where is the point in a triangle, that when connected to the three vertices forms equal angles of 120 degrees? Does this point exist in all triangles?

For a triangle with side lengths of 2,4,5, this point can be found, but with a triangle with side lengths of 3,6,8, there is no such point that exists. How would I prove this for the general case and find the necessary conditions for this point to exist?

Many Thanks

The point you described is called the "first isogonal center" of the triangle. Also called the "Fermat point." To construct it: construct an outward-pointing equilateral triangle on two of the sides of the given triangle, then connect each outward-pointing point with the opposite vertex of the given triangle. Where the two lines cross is the center. As was noted by one of the other correspondents above the construction is not feasible if the given triangle has an angle greater than or equal to 120 degrees.

If the equilateral triangles are constructed pointing inward, the result is the "second isogonal center."

• thanks very much for your reply. I understand what you are saying about constructing equilateral triangles to find the point, but I am not sure why this works? I have found this point, by constructing a circle on each side of the given triangle so that any point on the circumference and inside the triangle will form a 120 degrees angle, and so the intersection of these three circles will be the "first isogonal center", and as you say it only works for triangle with all angles less than 120 degrees. So could you please explain why constructing equilateral triangles will fin this point. – Nik.Ram Aug 14 '17 at 10:56

There is an elegant mechanical solution. Draw the triangle on a flat table and drill a small hole at each vertex. Pass a string through each hole. Above the table, bring the strings together and knot them. Below the table, attach an object of the same weight to each string. Let everything settle into equilibrium. Because the tension in each string is the same, the angles between the strings must the same, so they meet at $120^o$.

The minimal property follows because the system of weights and strings will settle into the configuration that has least potential energy. Therefore the sum of the lengths of the strings below the table must be maximal. So the sum of the lengths of the strings above the table must be minimal.

• thank you for replying. I hadn't thought to use physics to visualize the problem... this is a very interesting way to think about it. how did you come up with this particular method? have you worked on this problem before? – Nik.Ram Aug 14 '17 at 10:45
• It is something that I recall from long ago, and I think it might have been a Martin Gardner article. I wish I could take some credit, but I cant. – Philip Roe Aug 14 '17 at 20:57

You are looking at Fermat-Toricelli Point whose construction is made by 3 equilateral triangles built on each side. The obtuse angle triangle case is also shown as Case 1 (one or two triangles looks folded inwards) given here and also in many such references.

The geometrical property is "Sum of three distances is minimum". Using minimal surface area principle of soap films it can be realized in a physics experiment in the Motorway problem last para as well. It shows how/why $120^0$ junctions form. A hollow triangular prism dipped in soap solution and taken out actually locates the Fermat-Toricelli point as a consequence of minimal length sum which is also constitutes minimal area in this case.. Such beautiful examples are given in Cyril Isenberg's book reference.

• thanks for your reply -- it was very helpful. I understand what you mean about the Fermat point, but can you please clarify what you are saying in the last sentence, about surface area. – Nik.Ram Aug 14 '17 at 10:38
• Hope it is clearer in link.It is the physics law of equilibrium of forces that is minimization ... by variational principle. – Narasimham Aug 14 '17 at 11:40