A triangle and a point outside of it. Draw a line through the point
that will divide triangle into 2 equal areas.

Given an arbitrary triangle (defined by vertices at points A, B, C) and a point P outside of the triangle, first draw two lines from P to the two vertices (A, B) such that neither line intersects the triangle, and the sum of their Euclidean distances is the maximum possible, given the relative locations of ABC and P in the plane. Next draw a line through P such that it bisects he angle APB. This line also divides the area of triangle ABC in half.

I didn’t actually prove this method, but it appears to be empirically and intuitively correct. I believe it could be proved (or disproved) using a system of linear equations to define the lines forming the triangle and the bisecting line, and their corresponding integrals to measure the relevant areas. Please let me know if this is correct, thanks!

NJZ

Hi NJZ,

Thanks for the reply.

In your solution the position of the third vertex C is irrelevant.
But the result will change significantly with change in C relative
to points A and B.

Golland.

You are absolutely right. Obviously, my solution only works in special cases. I should have given this one a bit more thought before posting my answer. I'll have work on this one some more, and see if I can come up with a correct general solution.

NJZ