Sanity Check

From Raymond Smullyan: Every inhabitant of this island either a knight or a knave. Knights always tell the truth, and knaves always lie. Further, every inhabitant is either mad or sane. Sane inhabitants always answer questions correctly, and mad ones always answer incorrectly. You meet an inhabitant of the island and want to know whether he’s sane or mad. How can you determine this with a single yes-or-no question?

	Select Click for Answer
Ask, “Are you a knight?” A sane knight will say yes, a mad knight will say no, a sane knave will say yes, and a mad knave will say no. So a sane inhabitant will answer yes and a mad inhabitant will answer no. From Logical Labyrinths, 2009. Obligatory John Finnemore sketch:

March 5, 2019 | Puzzles

For Pi Day

From the prolifically interesting Catriona Shearer: The red line is perpendicular to the bases of the three semicircles. What’s the total area shaded in yellow?

	Select Click for Answer
At first it appears there’s not enough information, but there is! No matter the relative sizes of the two smaller semicircles, the area of the shaded region is π. There are any number of ways of proving this; here’s a pretty “proof without words” that relies on the fact that the Pythagorean theorem works with similar figures of any shape (here, semicircles). (Thanks to Bill Bowser for the illustration.)

Click for Answer

March 2, 2019March 3, 2019 | Puzzles · Science & Math

E Pluribus Unum

Replace each * with a different digit 1-9 to make this equation true:

$\displaystyle \frac{*}{**} + \frac{*}{**} + \frac{*}{**} = 1$

	Select Click for Answer
$\displaystyle \frac{9}{12} + \frac{5}{34} + \frac{7}{68} = 1$ It’s easy to solve this by writing a program to try all possibilities, but reasoning it out is tricky. Donald Aucamp gives one solution in MIT Technology Review, March 2002. 03/18/2019 UPDATE: I should have explained that the double asterisks in the denominators are two-digit numbers. Interestingly, there are two other interpretations that both yield answers! Reader Nick Berry sent this: $\displaystyle \frac{1}{6 \times 3} + \frac{7}{4 \times 2} + \frac{5}{9 \times 8} = 1$ And Paul Wagner sent: $\displaystyle \frac{6}{3 + 9} + \frac{5}{7 + 8} + \frac{1}{2 + 4} = 1$ Thanks to both of them.

Click for Answer

February 27, 2019March 18, 2019 | Puzzles

The THOG Problem

https://commons.wikimedia.org/wiki/File:THOG.png — Image: Wikimedia Commons

I have picked one color (black or white) and one shape (square or circle). A symbol that possesses exactly one of the properties I have picked is called a THOG. The black circle is a THOG. For each of the other symbols, is it (a) definitely a THOG, (b) undecidable, or (c) definitely not a THOG?

Cognitive psychologist Peter Wason invented this puzzle in 1979 to demonstrate some weaknesses in human thinking. In pilot studies, 0 of 10 student barristers were able to solve it correctly, with one arguing for more than an hour against the correctness of Wason’s solution. Seven of 14 medical students solved it, taking an average of 6.3 minutes. (“This is quite an impressive result.”) One young doctor solved it in his head in about a minute and said, “I would not let any doctor near me who couldn’t solve that problem.” What’s the answer?

	Select Click for Answer
In saying “I have picked one color and one shape,” in effect I am saying, “I have picked one of the symbols.” I can’t have chosen the black circle, because we know that symbol has only one of the required attributes. The symbol I’ve chosen must be either the black square or the white circle. Either way, the white square would match my choice in exactly one particular, so it’s definitely a THOG. If I’ve chosen the black square then the black square has both chosen attributes and the white circle has neither; equally, if I’ve chosen the white circle then the white circle has both chosen attributes and the black square has neither. Either way, neither of these symbols is a THOG. So the white square is definitely a THOG and the other two symbols definitely aren’t. “In one sense the THOG problem is utterly trivial,” writes Wason. “And yet it is non-trivial in the sense that it is astonishingly easy for some individuals and astonishingly difficult for others. At the very least it does seem a cogent test of formal operational thought in an abstract sphere.” See the link below for more details. (Peter C. Wason and Philip G. Brooks, “THOG: The Anatomy of a Problem,” Psychological Research 41:1 [1979], 79-90.)

Click for Answer

February 24, 2019 | Puzzles

Black and White

healey chess problem

By Francis Healey. White to mate in two moves.

	Select Click for Answer
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?

	Select Click for Answer
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. 02/24/2019 Reader Bill Bowser sent this diagram (thanks, Bill):

Click for Answer

February 21, 2019February 24, 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.)

	Select Click for Answer
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.

	Select Click for Answer
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