FW’S NOTEBOOK #2 – Fagnano Problem

Lurking in my recent research readings, this is a problem at the intersection of middle school geometry and quantum chaos … 

This week’s post was delayed such that certain subscribers taking final exams won’t be disturbed. Hope you all had fun.

Question: Inscribe a triangle of minimum perimeter inside an arbitrary acute triangle.

Demonstration I.

We take an arbitrary acute triangle, △ABC.

A random acute triangle.
Fig. 2.1(a): Triangle

The shape is colour-coded for convenience. We fold the triangle along its edge 5 times, in the fashion shown below.

This action generates a tiling of six identical triangles, which we shall study further.

Fig. 2.1(b): Unfolded Triangle

Observe that the last edge, A’B’ is just a translated version of edge AB. Why?

If you count the edges trapped in the middle of this shape, each represents a flip around a certain edge. We have 2 Green, 2 Red, and 2 Blue, corresponding to two flips along direction AB, AC, and BC, respectively.

Focusing on rotations for now, AB is transformed in the following fashion during each flip:

#1, ABStationary (Acting as Axis)
#2, AC2∠A, CW
#3, BC2∠B, CW
#4, ABStationary (Acting as Axis)
#5, AC2∠A, CCW
#6, BC2∠B, CCW

Pictorially, on the 2D plane, two flips usually correspond to a translation and a rotation… you can verify this easily by multiplying together two affine transformation operators. Anyway, in a triangle, the rotational effects all cancel each other out.

Why is this tile relevant?

Draw an arbitrary inscribed triangle, shall we?

Fig. 2.2(a): An inscribed triangle.

As the black triangle gets carried away in the flips, it traces a zig-zag from which we can compute its perimeter.

We can do so simply by following the yellow line. It contains two copies of every edge of the black triangle, and maps its starting point on edge AB to its exact clone on edge A’B’. Hence the length of the yellow zigzag is twice the perimeter of the black triangle. (Letters aren’t necessary yet, so we skip naming stuff).

Fig. 2.2(b): The inscribed triangle unfolded

This clearly shows that a lower bound on any perimeter exists. For any point on edge AB to be mapped to its clone on edge A’B’, the shortest path is the length of AA’. In other words, the lowest bound of any inscribed triangle’s perimeter is half of that of ||AA’||.

Fig. 2.2(c): All inscribed triangles have perimeter no smaller than ||AA’||/2.

Up to this stage, we have treated every event where the black edges cross the boundary of 
△ABC by reflecting the big triangle and continuing straight line segment(s). But there’s an equally valid way of seeing the same picture:

Fig. 2.3: Attempts to join the midpoints of AB and A’B’, as well as the same path mapped back.

These two correspond to the same mathematics, just represented differently… Annnd we have a problem. The thing on the right, although having a combined perimeter (not an apt word to use, as hinted by the white fillings) edge length of ||AA’||, it is NOT a triangle!

Can there ever be a triangle? In particular, can the two hit-points on all every edge converge with each other simultaneously?

Geometry says they can… I want to prove so formally, and must say good night for the time being.