A Fika Project in Geometry

The Square Problem

In which coloured diagrams and symbols are used for the greater ease of learners

Published:

In the first year of my Ph.D., I was given the following geometry problem:

A square construction with coloured segments and no point labels.
The square problem Given that and both have unit length, obtain the length of using only tools a high-school student would know.

The problem seems trivial at first glance. The Pythagorean theorem implies

and the similarity between the triangles and yields

These conditions are enough to specify a single equation for . However, continuing down this road leads to the fourth-order polynomial (here we use = = (1))

which a high-school student would not know how to solve, allegedly.

I refused to hear the solution when I had the chance, but the simplicity of the problem taunted me, and I kept coming back to it occasionally. If you want to try it for yourself, stop reading here; below are three increasingly credible solutions, with the final one being the simplest.

Avoiding the problem

In the introduction of the book Abel’s Theorem in Problems and Solutions, the reader is shown the derivations of the formulas for the roots of polynomials of degree up to four. This book is based on lectures by the mathematician Vladimir Arnold for high-school students in Moscow. We may then argue that—at least in 1960s Moscow—a high-schooler would know how to solve Eq. 3 and call it a day.

Constructible Numbers

To solve the problem, we may try to approach the question constructively. Given a square, how can the problem setup be constructed? After a lot of exploration using GeoGebra, I found the following observation.

The tangent observation: a circle centered at D appears tangent to BR. A B C D P R E
Construction I The circle centered at \(D\), with radius \((\sqrt{2}-1)\times AB\), appears tangent to the line \(BR\).

A circle of radius equal to the square diagonal minus its side, centered at the bottom-right corner of the square, seems tangent to the line \(BR\). If this is the case, we are done. Using the formula for the area of triangle \(PDR\), once with base \(PR\) and once with the two perpendicular sides, we have

\[ \frac{1}{2} \times PR \times (\sqrt{2}-1) \times AB = \frac{1}{2} \times PD \times DR. \tag{4} \]

Now scale again so that \(AB = PR = 1\). Then Eq. (4) says

\[ PD \times DR = \sqrt{2}-1. \tag{5} \]

Together with the Pythagorean theorem, this yields the biquadratic equation

\[ PD^2 \times \left(1-PD^2\right) = (\sqrt{2}-1)^2. \tag{6} \]

Therefore

\[ \frac{PD}{AB} = \sqrt{\frac{1-\sqrt{8\sqrt{2}-11}}{2}} \approx 0.46899. \tag{7} \]

How do we show the tangency property we have claimed? The easiest way is to construct the tangent line and then show that it gives \(PR = AB\), as specified in the original problem statement. The crucial property of a tangent is that it touches the circle at a single point, where it makes a right angle with the radius to that point. To construct this right angle we use Thales’s theorem and an appropriately constructed circle.

Construction of the tangent line using a Thales circle. A B C D F G H O
Construction II The point \(H\) is chosen with Thales's theorem, so \(DH\) is perpendicular to \(BG\).

For clarity, translating the construction into algebra, set \(AB=1\) and place the origin at \(D\). Then \(B=(-1,1)\), and the circle centered at \(D\) has equation

\[ x^2+y^2 = (\sqrt{2}-1)^2. \tag{8} \]

The Thales circle has center \(O=(-1/2,1/2)\), radius \(\sqrt{2}/2\), and equation

\[ \left(x+\frac{1}{2}\right)^2 + \left(y-\frac{1}{2}\right)^2 = \left(\frac{\sqrt{2}}{2}\right)^2. \tag{9} \]

The two circles intersect at the point \(H=(x_1,y_1)\) in the first quadrant, where

\[ x_1 = \frac{2\sqrt{2}-3+\sqrt{8\sqrt{2}-11}}{2}, \qquad y_1 = \frac{-2\sqrt{2}+3+\sqrt{8\sqrt{2}-11}}{2}. \tag{10} \]

The tangent line \(BG\) passes through \(H=(x_1,y_1)\) and \(B=(-1,1)\), so its equation is

\[ y-1 = \frac{y_1-1}{x_1+1}(x+1). \tag{11} \]

This line intersects the \(y\)-axis at

\[ p = \frac{2\sqrt{8\sqrt{2}-11}} {2\sqrt{2}-1+\sqrt{8\sqrt{2}-11}}, \tag{12} \]

and the \(x\)-axis at

\[ r = - \frac{2\sqrt{8\sqrt{2}-11}} {-2\sqrt{2}+1+\sqrt{8\sqrt{2}-11}}. \tag{13} \]

One can check that

\[ p^2+r^2=1. \tag{14} \]

In other words, the segment cut out by the tangent line between the two axes has length \(1\), which is exactly the condition \(PR=AB\). This completes the verification.

Although I have been calling this a proof by construction, I really should call it a proof by verification. One needs to start with the unexpected idea of a tangent to a seemingly random circle, and then do a lot of work to show that the result aligns with the problem statement. For this reason, I do not consider this solution to be in the spirit of the problem. On the other hand, the process is quite mechanical, and it leads one to think about the kind of numbers that can be derived using only a straightedge and compass. Such numbers are called constructible, and they depend on the kind of geometry one is considering. An interesting note is that the geometry resulting from origami is stronger than Euclidean geometry, allowing traditionally unsolvable Euclidean problems-like the trisection of an angle-to be solved using paper folding.

A Geometric Solution

One of the first things one may look for is an intermediate variable that absorbs some of the complexity of the algebra. Making this approach work is what took me the most time, but I finally found the idea below (in hindsight, pretty obvious).

The square problem with a dashed perpendicular drawn from the lower-right corner to the sloping blue segment.
The height construction and both have unit length. The dashed black segment is perpendicular to the blue segment.

The area of a triangle is half the product of its base and height. Calculating it first with and , and then with and , yields

The Pythagorean theorem gives the second relation:

The similarity between the triangles and yields the final relation we need:

Expanding the square of the sum and the difference, then using Eq. (15) and Eq. (16), gives

Combining Eq. (15), Eq. (17), and the minus case of Eq. (18) yields

Squaring both sides results in a quadratic, rather than a quartic, equation for the dashed height:

With this, we directly obtain the final result:

References

This page takes its visual language from Oliver Byrne’s The First Six Books of the Elements of Euclid. The diagrams and colour system were informed by Sergey Slyusarev’s Byrne LaTeX project.