Black and White

healey chess problem

By Francis Healey. White to mate in two moves.

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1. Qh1! and the queen will mate. From A Collection of Two Hundred Chess Problems, 1866.

February 22, 2019February 21, 2019 | Puzzles

A Pretty Puzzle

I don’t know who came up with this; I found it on r/mathpuzzles. What’s the area of the red region?

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Draw a line from each vertex to the interior point where the four regions meet. This divides each region into two triangles. On each side of the large figure (north, south, east, and west), two of these triangles are neighbors and have the same area (because they have the same base and height) (here’s a figure). The four triangles that make up the northeast and southwest regions have the same area as the four triangles that make up the northwest and southeast regions. This means that each pair of diagonally opposite regions makes up half the total area of the figure. So the missing area is 28 cm².

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February 21, 2019 | Puzzles · Science & Math

Loose Change

penny puzzle

You’re holding a penny, and you’re standing on an infinite plane. The plane bears a grid of squares, each of which is twice the width of the penny. If you roll the penny out onto the grid, what is the probability that it will come to rest entirely within a square? (Assume the lines are of negligible thickness.)

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If the penny is to lie entirely within a square, then its center must fall within a square region one radius from the square’s edge. That region is the width of a penny, or one-fourth the square’s total area. So the probability is 1/4. (From Ann Hirst, “Roll-a-Penny Probabilities,” Mathematical Gazette 73:464 [June 1989], 101-106.)

February 17, 2019 | Puzzles

A Memorial Puzzle

When violinist Hugh Gordon Langton was killed in World War I, the Commonwealth War Graves Commission inscribed a phrase of music (below, click to enlarge) on his headstone in Belgium’s Poelcapelle British Cemetery.

What the notes represent has been a mystery for more than a century. “They are not part of a tune that anyone has been able to discern and several have tried,” writes Sarah Wearne in Epitaphs of the Great War: Passchendaele. “In particular they are not from the song ‘After the Ball’ as some have suggested.”

Langton is the only casualty who was commemorated with a piece of music — the CWGC maintains more than a million graves and memorials worldwide, and only this one bears musical notes as an epitaph. If you can identify the tune, please contact the Commission.

February 9, 2019 | History · Puzzles

A Problem From 1725

archimedes problem

Suppose that when Marcellus besieged Syracuse, Archimedes was standing at a corner of the city wall. A ditch runs parallel to the wall, separated from it by a distance a. To Archimedes’ left at distance b along the wall stands a catapult, which is distance c from a line perpendicular to the ditch. If Archimedes’ line of sight to the camp runs perpendicular to the wall and the ditch, show that he stood a distance ab/c from the camp.

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In the diagram, (y + a)/b = a/c, so multiplying both sides by b we get y + a = ab/c. (From The Ladies’ Diary: or, Woman’s Almanack, 1725, via J.L. Heilbron’s Geometry Civilized, 1998.)

February 4, 2019February 4, 2019 | Puzzles

Rubik’s Clock

Hungarian sculptor and architect Ernő Rubik presented this puzzle in 1988; it was originally created by Christopher C. Wiggs and Christopher J. Taylor. The puzzle has two sides, with nine clocks on each side, and the goal is to set all the clocks to 12 o’clock simultaneously.

There are two ways to adjust the clocks. Turning a wheel at any of the four corners will adjust the clock at that corner on both sides of the puzzle. And turning a wheel will also adjust the three clocks adjacent to that corner on one side of the puzzle or the other; which side is determined by the four buttons surrounding the central clock.

So, for example, pressing the northwest button “in” and then turning the northwest wheel will adjust the northwestern quartet of clocks and the corresponding corner clock on the other side of the puzzle. Pulling the northwest button “out” and turning the same wheel will adjust the northwestern clock on the front of the puzzle, its counterpart on the back, and the three clocks adjacent to it on that side.

This is more intuitive than it sounds. Here’s a simulator.

Since there are 14 independent clocks, with 12 settings each, there are a total of 12¹⁴ = 1,283,918,464,548,864 possible configurations. It turns out that no configuration requires more than 12 moves to solve; for comparison, in the “worst case” solving a Rubik’s cube can take 20 moves. The trouble, of course, is knowing how to go about it.

February 2, 2019 | Puzzles · Science & Math

Black and White

muthuswami chess problem

By N.S. Muthuswami. White to mate in two moves.

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1. Qe8! and the queen will mate.

February 1, 2019 | Puzzles

Outside the Box

An old puzzle asks: Without lifting your pencil from the paper, can you draw a series of four straight lines that passes through all nine points in this grid?

outside the box 1

The trick is to realize that the lines can extend beyond the grid’s area:

outside the box 2

In 1970 Solomon Golomb and John Selfridge found a way to draw a closed path of eight segments that passes through all 25 points in this grid:

outside the box 3

Can you?

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“This remarkable construction contains a line of slope 2, and has turning points for the path which not only are not lattice points, but do not lie on the grid lines (horizontal or vertical) through the lattice points.” (Solomon W. Golomb and John L. Selfridge, “Unicursal Polygonal Paths and Other Graphs on Point Lattices,” Pi Mu Epsilon Journal 5:3 [Fall 1970], 107-117.)

January 24, 2019January 23, 2019 | Puzzles

Threes and Fours

https://commons.wikimedia.org/wiki/File:Euclid_Tetrahedron_4.svg — Image: Wikimedia Commons

A problem from the Tenth International Mathematical Olympiad, 1968:

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.

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Well, three segments can make a triangle if and only if the longest of them is shorter than the other two combined. So consider a tetrahedron with vertices A, B, C, D, and let AB be the longest side. Now suppose there isn’t a vertex such that the three edges meeting there can form the sides of a triangle. The three edges that meet at vertex A are AB, AC, and AD. If those three edges won’t form a triangle, that means that AB ≥ AC + AD. Similarly, considering vertex B, we can conclude that BA ≥ BC + BD. Adding two those inequalities, we get 2AB ≥ AC + BC + AD + BD. But we also know that two of the tetrahedron’s faces are ABC and ABD, and each of these is a triangle, so AB < AC + BC and AB < AD + BD. Adding those two inequalities gives 2AB < AC + BC + AD + BD, which contradicts the inequality we just reached. So it can’t be the case that no vertex has edges that will form a triangle.

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