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

SUMMARY

In 30 seconds

Sawyer presents a new challenge about sliding a rectangular painting through an arched window, leaving it open for viewers to solve. He then solves the previous toy-boat problem with a quadratic equation and a geometric argument using similar triangles and the tangent double-angle formula.

1An arched-window challengeThe window is a one-metre square topped by a semicircle; the framed paintings are three metres wide.
2Boat and currentThe original crossing gives a motor-speed-to-current-speed ratio of 100:60, or 5:3.
3A quadratic solutionSolving the distance ratio gives 127.5 metres upstream; the negative root is extraneous.
4A geometric alternativeSimilar triangles and the speed ratio imply an isosceles triangle, leading to the same result via a double-angle formula.

Takeaways

The new challenge asks for the tallest three-metre-wide framed painting that can pass through a specified arched window.
The boat’s first crossing reveals a 5:3 ratio between its motor speed and the river speed.
A quadratic calculation gives an upstream aiming point 127.5 metres from the friend’s dock.
Squaring an equation can introduce an extraneous solution that must be checked.
Similar triangles reveal an isosceles triangle and a second route using a double-angle identity.

TopicsMathematics Education

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A new museum puzzle

Sawyer presents a window made from a one-metre square and a semicircle. Viewers are asked to find the maximum height of a three-metre-wide painting that can pass through it.

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.

SLIDE 1 / 6The painting and the arched windowThe diagram shows the three-metre-wide paintings and the one-metre square topped by a semicircle.

The boat problem by algebra

Sawyer revisits the previous river-crossing problem and sets up a speed ratio. A quadratic yields the valid upstream aiming distance.

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.

SLIDE 2 / 6Setting up the boat equationThe diagram and equations relate the aiming distance to the motor and current speed ratio.

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.

SLIDE 3 / 6Solving the quadraticThe displayed calculation shows the valid 255/2 result and the extraneous negative root.

A geometric solution

Similar triangles lead to an isosceles triangle and a double-angle calculation. Sawyer returns to the unsolved window challenge.

SLIDE 4 / 6The geometric constructionPoints A through E identify the riverbanks, docks and crossing of the trajectories.

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.

SLIDE 5 / 6Similar-triangle relationshipsThe proportions lead to equality of AE and DE.

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.

SLIDE 6 / 6The double-angle solutionThe geometric diagram and tangent identity recover the same upstream distance.

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.