I discovered this “magic trick” while painfully solving a related geometry puzzle.

First, some background. Given an arbitrary triangle ABC, draw squares on each side. Label the centers of the squares D, E, F. This forms another triangle DEF. Figure 1 indicates the basic construction (draw a perpendicular bisector half the length of the side ). ABC does not need to be a right angle triangle.

**Figure 1: Constructing square centers**

Now, it is fascinating to find that if squares are drawn on the second triangle DEF, but the *squares are drawn inward* rather than outward, the centers land on the mid points of the original triangle ABC. This is shown in Figure 2: JG is the inward perpendicular bisector of side EF, with JG = JE = JF. G is found to bisect side BC.

**Figure 2: Outer squares -> Inner Squares -> midpoints**

Now, let’s ask what would happen if we started with triangle ABC, but constructed the inner squares instead of outer squares, on it. Thus we have triangle DEF in figure 3 below, constructed on centers of inner squares.

Figure 3: Centers of Inner Squares on Triangle ABC

Now, it is natural to expect that the centers of outer squares drawn on DEF will fall on the midpoints of the original triangle ABC. That is, inner squares -> outer squares -> midpoints. Thus we have figure 4 below, and see that the square centers indeed fall on the midpoints of triangle ABC.

**Figure 4: inner squares -> outer squares -> midpoints**

So where is the magic trick, the sleight of hand? Well, examine Figure 4 again. Actually, we found the midpoints thus: inner squares -> inner squares -> midpoints.

And that was really amazing to discover, that the constructions break a natural symmetry:

outer squares -> inner squares -> midpoints

inner squares -> inner squares -> midpoints.

*Note: the figures are drawn using a beautiful geometry construction software called C.A.R. (Compass and Ruler)*.

## The puzzle

For the reader who wondered what the original puzzle was, it is a set of related puzzles:

(a) Construct outward squares on a triangle ABC of any shape. Let the centers of the squares be DEF. Now, given only DEF, reconstruct the original triangle ABC.

(b) Instead of squares, draw equilateral triangles on sides of ABC. Given the centers of the equilateral triangle, reconstruct the original triangle ABC.

(c) Generalizing further: On side BC with midpoint G, draw a perpendicular GD such that GD = m * GC, where m is a real number. This generates point D. Similarly draw points E and F using sides AC and AB respectively, using the same value of m.Given DEF, reconstruct the original triangle ABC. (setting m = 1 gives puzzle (a), and so on.)

An alternative approach to draw points D E F is as follows: Draw two lines from the ends of each side at a constant angle to the side. The intersection of these two lines gives the points D E F. Then 45 degree angles gives case (a); 60 degree angles give case (b).