plane-geometrytriangles
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
Let $a$ be the hypotenuse and $b,c$ the others sides, then by Pythagorean Theorem $$a^2=b^2+c^2.$$ Then $$a^2>b^2,\quad a^2>c^2.$$ Therefore $$a>b\quad a>c.$$
Experimentally it appears that seems that something close to Hagen von Eitzen's comment of the construction using a regular pentagon
works for all obtuse triangles. Here is an example, with two isosceles triangles cutting off the sharp angles, and then choosing a suitable point inside the remaining pentagon to give $7$ acute angled triangles in all. Not all pairs of isosceles triangles allow there to be a suitable interior point, but I could not find an obtuse angled triangle without some solution.
I would be surprised if there was a solution with fewer acute angled triangles.
Best Answer
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."