The Asymmetric Propeller

asymmetric propeller theorem

Arrange three congruent equilateral triangles so that their corners meet at a point, like the red triangles above. The arrangement doesn’t have to be symmetric; the triangles can even overlap. Now draw lines BC, DE, and FA to complete a hexagon inscribed in a circle. The midpoints of these three lines will form the vertices of an equilateral triangle.

That’s called the asymmetric propeller theorem, and it’s been known since the 1930s. But in 1979 Beverly Hills dentist and geometry enthusiast Leon Bankoff told Martin Gardner of some further discoveries. Bankoff never found time to write them up, so after the dentist’s death in 1997 Gardner published them in the College Mathematics Journal:

  • The three equilateral triangles need not be congruent. Each can be of any size and the theorem still holds.
  • The triangles need not meet at a point. They can meet at the corners of any equilateral triangle.
  • They need not even be equilateral! If three similar triangles of any sizes meet at a point, the midpoints of the three added lines will form a triangle similar to each of the “propellers.”
  • The similar triangles need not meet at a point! If they meet at the corners of a fourth triangle (of any size) that’s similar to each propeller, then the midpoints of the added lines will form a triangle similar to each propeller, provided that the vertices of the central triangle touch the corresponding corners of the propellers.

Given all this flexibility, Gardner asked, do the propellers even have to be triangles? It turns out that the answer is yes. Still, the discoveries above form a fitting tribute to Bankoff, whom Gardner called “one of the most remarkable mathematicians I have been privileged to know.”

(Martin Gardner, “The Asymmetric Propeller,” College Mathematics Journal 30:1 [January 1999], 18-22.)

The Uncounted,_Rainbow.JPG

It was a good answer that was made by one who when they showed him hanging in a temple a picture of those who had paid their vows as having escaped shipwreck, and would have him say whether he did not now acknowledge the power of the gods, — ‘Aye,’ asked he again, ‘but where are they painted that were drowned after their vows?’ And such is the way of all superstition, whether in astrology, dreams, omens, divine judgments, or the like; wherein men, having a delight in such vanities, mark the events where they are fulfilled, but where they fail, though this happens much oftener, neglect and pass them by.

— Francis Bacon, Novum Organum, 1620

Parrondo’s Paradox

Imagine a staircase with 1001 stairs, numbered -500 to 500. You’re standing in the middle, on stair 0, and you want to reach the top. On each step you can play either of two coin-flipping games — if the result is heads then you move up a step; if it’s tails then you move down a step:

  • In game 1 you flip coin A, which is slightly biased: It comes up heads 49.5 percent of the time and tails 50.5 percent.
  • In game 2 you use two coins, B and C. Coin B produces heads 9.5 percent of the time and tails 90.5 percent. Coin C produces heads 74.5 percent of the time and tails 25.5 percent. In game 2 if the number of the stair you’re on is a multiple of 3 then you flip coin B; otherwise you flip coin C.

Both of these are losing games — if you played either game 1 or game 2 exclusively, you’d eventually find yourself at the bottom of the staircase. But in 1996 Spanish physicist Juan Parrondo found that if you play the two games in succession in random order, keeping your place on the staircase as you switch between them, you’ll rise to the top of the staircase. It’s not, properly speaking, a paradox, but it’s certainly counterintuitive.

This example is from David Darling’s Universal Book of Mathematics. (Thanks, Nick.)

Making Pi

We’ve mentioned before that you can estimate π by dropping needles on the floor. (Reader Steven Karp also directed me to this remarkable solution, from Daniel A. Klain and Gian-Carlo Rota’s Introduction to Geometric Probability [1997].)

Here’s a related curiosity. If a circle of diameter L is placed at random on a pattern of circles of unit diameter, which are arranged hexagonally with centers C apart, then the probability that the placed circle will fall entirely inside one of the fixed circles is

circle and scissel pi estimate 1

If we put k = C/(1 – L), we get

circle and scissel pi estimate 1

And a frequency estimate of P will give us an estimate of π.

Remarkably, in 1933 A.L. Clarke actually tried this. In Scripta Mathematica, N.T. Gridgeman writes:

His circle was a ball-bearing, and his scissel a steel plate. Contacts between the falling ball and the plate were electrically transformed into earphone clicks, which virtually eliminated doubtful hits. With student help, a thousand man-hours went into the accumulation of N = 250,000. The k was about 8/5, and the final ‘estimate’ of π was 3.143, to which was appended a physical error of ±0.005.

“This is more or less the zenith of accuracy and precision,” Gridgeman writes. “It could not be bettered by any reasonable increase in N — even if the physical error could be reduced, hundreds of millions of falls would be needed to establish a third decimal place with confidence.”

(N.T. Gridgeman, “Geometric Probability and the Number π,” Scripta Mathematica 25:3 [November 1960], 183-195.)

Three of a Kind

This trick seems to have been invented independently by Martin Gardner and Karl Fulves. A blindfolded magician asks a spectator to lay three pennies on a table, in any arrangement of heads and tails. The magician’s goal is to put all three coins into the same state, all heads or all tails.

If the three coins already match, then the trick is done. If not, then the magician gives three instructions: Flip the left coin, flip the middle coin, flip the left coin. After each step he asks whether the three coins now match. By the third flip, they will.

“It’s no surprise that the magician can eventually equalize all the coins,” writes MIT computer scientist Erik Demaine, “but it’s impressive that it always takes at most three moves.” The technique exploits a principle used in Gray codes, which are used to reduce errors when using analog signals to represent digital data. Demaine relates a similar trick involving four coins in the November-December 2010 issue of American Scientist.

See Lincoln Seeks Equality.

All Roads

A self-working card trick by New York magician Henry Christ:

Shuffle a deck thoroughly and deal out nine cards in a row, face down. Choose a card, look at it, and assemble the nine cards into a stack face down, with the chosen card at the top. Add this stack to the bottom of the deck.

Now deal cards one at a time from the top of the deck into a pile, face up, counting backward from 10 as you do so. If at some point the card’s rank matches the number said, then begin dealing into a new pile at that point, counting again backward from 10. If you reach 1 without a match occurring, then “close” that pile by dealing a face-down card onto it, and start a new pile.

Keep this up until you’ve created four piles. Now add the values of any face-up cards on top of the piles, count down through the remaining cards until you’ve reached this position, and you’ll find your chosen card.

This works because it always leads to the 44th card in the deck, but it takes some thinking to see this. You can put a sealed deck into a stranger’s hands and direct him to perform the trick himself, with mystifying results.

Two in One

Look at this image closely and you’ll see the features of Albert Einstein.

But look at it from across a room and you’ll see Marilyn Monroe.

It’s a “hybrid image,” created using a technique developed by Aude Oliva of MIT and Philippe Schyns of the University of Glasgow. The image combines the low spatial frequencies of one picture with the high spatial frequencies of another, so that it’s processed differently at different viewing distances.

See their paper for the details, and this gallery for more examples.

Star Power

A puzzle by A. Korshkov, from the Russian science magazine Kvant:

It’s easy to show that the five acute angles in the points of a regular star, like the one at left, total 180°.

Can you show that the sum of these angles in an irregular star, like the one at right, is also 180°?

Click for Answer