A Problem in Three Squares:
What Is This About?
A Mathematical Droodle

13 October 2015, Created with GeoGebra

13 October 2015, Created with GeoGebra

An interesting fact holds in a configuration of three equal squares placed side by side. If the lines drawn as shown then ∠BDC + ∠BFC = ∠BEC. There are many proofs of this result. These are so simple, the problem has been included in many a puzzle book.

First note that ∠BEC = 45°. The task is then to show that ∠BDC + ∠BFC = 45°. Let's introduce α = ∠BDC and β = ∠BFC. We are to prove that α + β = 45°.

Proof 1

Add another row of squares below the given three. ΔBKD is isosceles (BK = DK) and, in addition, ∠K = 90°. Therefore, ∠BDK = 45°. However, since triangles DEK and FCB are equal, we also have ∠KDE = β.

Proof 2

Now with some trigonometry (arctan = tan^-1),

α + β = arctan(1/3) + arctan(1/2)

= arctan((1/3 + 1/2)/(1 - 1/3·1/2))

= arctan(1),

where I used the standard formula for the tangent of a sum

tan(α + β) = (tanα + tanβ)/(1 - tanα·tanβ).

Since arctan(1) = 45°, α + β = 45°.

Proof 3

Triangles BDE and FEB are similar. Indeed, assuming the side of the square equals 1, then

DE = 2, BE = √2, DE/BE = √2,

BE = √2, EF = 1, BE/EF = √2,

BD = √10, BF = √5, BD/BF = √2,

which implies that ∠BDC = ∠FBE = a. Finally, the exterior angle BEC of ΔFEB, equals the sum of the opposite interior angles, i.e., a + b.

In short, we proved that

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

Observe that this implies another identity:

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

Proof 4

Following a proof without words by Hasan Unal (Math Magazine, v 82, n 1, February 2009 , p. 56) we shall prove a more general statement. Let x and y be positive numbers satisfying x² + y² = 1. Then

arctan
1 - x

y
+ arctan
1 - y

x
=
π

4

The identity arctan(1/2) + arctan(1/3) = arctan(1) is obtained from the above by letting x = 3/5 and y = 4/5, making it yet another appearance of the Egyptian 3-4-5 triangle. The proof is based on the following diagram.

a trigonometric identity

Set α = (1 - y) / x, and β = (1 - x) / y. We see at the origin that 2α + 2β = π/2, implying α + β = π/4 = arctan(1).

Elsewhere we prove another curiosity: arctan(1) + arctan(2) + arctan(3) = π.

References

B. Richardson, Three Squares Theorem, in The Changing Shape of Geometry, edited by C. Pritchard, Cambridge University Press, 2003
A Decade of the Berkeley Mathematical Circle, The American Experience, Volume I, Z. Stankova, Tom Rike (eds), AMS/MSRI, 2008
A Decade of the Berkeley Mathematical Circle, The American Experience, Volume II, Z. Stankova, Tom Rike (eds), AMS/MSRI, 2015, pp. 173-174

Trigonometry

49551962

α + β	= arctan(1/3) + arctan(1/2)
	= arctan((1/3 + 1/2)/(1 - 1/3·1/2))
	= arctan(1),

DE = 2,	BE = √2,	DE/BE = √2,
BE = √2,	EF = 1,	BE/EF = √2,
BD = √10,	BF = √5,	BD/BF = √2,

A Problem in Three Squares:What Is This About? A Mathematical Droodle