I believe there is a mathematic approach and a geometric approach. I would be interested in any feedback.

cmdavidson
mdavidson@canberra.com

I'm not sure if this is the easiest way to solve the puzzle, but it looks like it works. The sum of the three line segments extending from the vertices of an equilateral triangle to an interior point is 12 (= 3 units + 4 units + 5 units). If the point of intersection is shifted from inside the triangle to the mid-point of one of the sides, it can be said that the same three line segments (a, b, c) extending from the vertices now have lengths where a = h (height of the triangle), b = s/2 (half the side length), and c = s/2. Since the sum of the line segments is fixed at 12, h + s = 12.

Given an equilateral triangle, it is possible to express the height in terms of the side length in the following way. The height of the triangle corresponds to the line segment perpendicular to a side of the triangle which bisects that side (a line from a vertex to the mid-point of the opposite side). The height line bisects the equilateral triangle into two congruent right triangles with legs x = s/2, and y = h, and hypotenuse z = s. Since x2 + y2 = z2 it is therefore possible to express the height of the triangle in terms of its side length as h = sqrt(s2 - s2/4).

The sum of the line segments can now be written as s + sqrt(s2 - s2/4) = 12. Solve for s, and you get s = 6.4308. I'm not sure how to solve for s algebraically, or if it is even possible, so I just had my computer approximate s.

On that page, there's another discussion on that same problem. A link to mathsoft.com is broken, though, as they have recently changed the layout of their site. Try this:

https://www.mathcad.com/library/LibraryContent/puzzles/soln15/soln15.shtml