The Square Problem

Published: January 01, 2025

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

Fig. 1. Find the value of \(PD\) in the construction above, given that the side of the square \(AB\) and the segment \(PR\) have the same length. Use only tools a high-school student would know.

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

PD^2 + DR^2 = PR^2, \tag{1}

and the similarity between the triangles (PDR) and (ABR) yields

PD \cdot AR = DR \cdot AB. \tag{2}

These conditions are enough to specify a single equation for (PD). However, continuing down this road leads to a fourth-order polynomial, which a high-school student would not know how to solve, at least I could not, and still cannot off the top of my head.

By stubborn pride I refused to hear the solution when I had the chance, but the simplicity of the problem taunts me, and I keep coming back to it occasionally. While I have not found a satisfactory solution, I thought it would be nice to recap the things I have learned along the way.

Cheating

In the introduction of the book Abel’s Theorem in Problems and Solutions we are shown a derivation of the well-known formulas for the roots of polynomials up to fourth degree. This book is based on lectures of the mathematician Vladimir Arnold to high-school children in Moscow. We may then argue that, at least in 1960s Moscow, a high-schooler would know how to find the roots of the polynomial

y^4 - 2y^3 + y^2 + 2y - 1 = 0, \tag{3}

resulting from Eq. (1) and Eq. (2), where (y) is the ratio (PD/AB).

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.

Fig. 2. The circle centered at \(D\) with radius \((\sqrt{2}-1)AB\) 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} PR \cdot (\sqrt{2}-1)AB = \frac{1}{2} PD \cdot DR. \tag{6}

Now scale again so that (AB = PR = 1). Then Eq. (6) says

PD \cdot DR = \sqrt{2}-1. \tag{7}

Together with the Pythagorean theorem, this yields the biquadratic equation

PD^2\left(1-PD^2\right) = (\sqrt{2}-1)^2. \tag{8}

Therefore

\frac{PD}{AB} = \sqrt{\frac{1-\sqrt{8\sqrt{2}-11}}{2}} \approx 0.46899. \tag{9}

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 circle radius at that point. To construct this right angle we use Thales’s theorem and an appropriately constructed circle.

Fig. 3. The tangent construction. The point \(H\) is chosen with Thales's theorem, so \(DH\) is perpendicular to \(BG\).

Following the construction in algebraic language, for clarity, 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{10}

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{11}

The two circles intersect at the positive point (H=(x_1,y_1)), 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{12}

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{13}

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{14}

and the (x)-axis at

r = - \frac{2\sqrt{8\sqrt{2}-11}} {-2\sqrt{2}+1+\sqrt{8\sqrt{2}-11}}. \tag{15}

One can check that

p^2+r^2=1. \tag{16}

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.

% Eventually talk about constructible numbers in origami and Euclidean geometry

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 random-looking circle, and then do a lot of work to show that the result coincides with the problem statement. For this reason, I do not consider this solution to be in the spirit of the problem.

A Purely Geometric Solution

I wish I had it.

References

This page is in the style of Oliver Byrne’s The First Six Books of the Elements of Euclid, which is now available througth (Sergey Slyusarev’s birne LaTeX package)[https://github.com/jemmybutton/byrne-latex].