Trigonometry by Paper Folding

We need to prove the following:

arctan(1) + arctan(2) + arctan(3) = π.

With the notation as in the diagram below, make three folds: BD, EF (E and F are the midpoints of AB and CD, respectively), and EC.

From ΔBCE, ∠BEG = ∠ECF = arctan(2). From ΔABD, ∠EBG = ∠ABD = arctan(1). To establish the assertion, we only need to show that ∠BGE = arctan(3), for then in ΔBEG the angles are exactly arctan(1), arctan(2, arctan(3).

Imagine adding a diagonal AC and mark the center of the square H. H lies on AC, BD, and EF.

In ΔABC, BH and CE are two medians and G is the centroid. G divides the medians in the ratio 2:1. It follows that GH = BH/3. In ΔCGH, CH = BH = 3×GH, implying ∠BGE = ∠CGH = arctan(3).