The Art Thief’s Window and a Toy Boat’s Return Journey

Sawyer introduces a geometry challenge in Real Numbers: an art museum has a broken window lock, and the curator wants to know which paintings an art thief could slide through the open window.

The window is a square with one-metre sides, topped by a semicircle whose diameter is the square’s upper edge. The framed rectangular paintings are three metres wide and have different heights. Treat the frames as two-dimensional and assume the paintings cannot be removed, bent or folded. What is the tallest frame that can pass through the window? Sawyer leaves this week’s problem for viewers to solve.

Before returning to that question, Sawyer solves the previous episode’s boat problem. Two parallel riverbanks are 100 metres apart. A friend points the toy boat Wavey directly across from their dock, but the current carries it to a point 60 metres downstream of your dock.

On the return trip the motor and current have the same respective speeds. The first trip combines 100 metres of motion across the river with 60 metres of downstream drift, so the motor speed is 5/3 times the current speed. The goal is to aim upstream so the current carries the boat exactly to the friend’s dock.

Call your location A and the aiming point B, which is x metres upstream of the friend. The motor would carry the boat along AB while the current carries it downstream by x. By the Pythagorean theorem, the speed ratio gives 5/3 = AB/x = √(100² + (x+60)²)/x.

Clearing denominators and squaring gives 25x² = 9(10,000 + x² + 120x + 3,600), or 16x² − 1,080x − 122,400 = 0. The quadratic formula yields x = 255/2 or x = −60. The negative root is extraneous: the relevant distance is positive, and squaring can introduce a solution that fails the earlier equation.

The valid answer is to aim 255/2, or 127.5 metres, upstream of the friend’s dock. Sawyer then looks for a geometric route that reveals more about the construction, without using the Pythagorean theorem.

Label your dock C, the friend’s dock D, and the intersection of the motor trajectories E. Because the riverbanks are parallel, triangles ACE and BDE are similar. Thus CE/AC = DE/BD and AE/AC = BE/BD.

The motor-to-current speed ratio also gives CD/AC = AB/BD. Writing CD as CE+DE and AB as AE+BE, substituting the similar-triangle ratios and factoring shows DE = AE. Triangle ADE is therefore isosceles, so angles EAD and EDA are equal; call each θ.

Draw AF perpendicular to the bank, with F directly across from A. Parallel lines show angle DAF is also θ, so angle BAF is 2θ. Since tan θ = AC/CD = 60/100 = 3/5, the double-angle formula gives tan(2θ) = 15/8.

In right triangle ABF, tan(2θ) = (x+60)/100. Equating this to 15/8 again gives x = 255/2 = 127.5 metres. Sawyer closes by returning to the arched-window puzzle and inviting viewers to submit their solutions.