In Brief

In 1962, botanist Reid Moran published a note in the journal Madroño recounting his collection of a bush rue on a mountaintop in Baja California.

The note’s title was “Cneoridium dumosum (Nuttall) Hooker f. Collected March 26, 1960, at an Elevation of About 1450 Meters on Cerro Quemazón, 15 Miles South of Bahía de Los Angeles, Baja California, México, Apparently for a Southeastward Range Extension of Some 140 Miles.”

The text read, “I got it there then.”

This was followed by a 28-line acknowledgment section in which Moran thanked the person who had reviewed the text, his college professors, and the person who had mailed the manuscript.

Misc

http://commons.wikimedia.org/wiki/File:Parrish_Hall.jpg

  • SWARTHMORE is an anagram of EARTHWORMS.
  • The sum of the reciprocals of the divisors of any perfect number is 2.
  • We recite at a play and play at a recital.
  • Is sawhorse the past tense of seahorse?
  • “Things ’twas hard to bear ’tis pleasant to recall.” — Seneca

In Book II, Chapter 9, of H.G. Wells’ novel The War of the Worlds, a sentence begins “For a time I stood regarding …” These words contain 3, 1, 4, 1, 5, and 9 letters.

(Thanks, Dheeraj.)

Memorable Experiments

hanging boy experiment

In 1730 Stephen Gray found that an orphan suspended by insulating silk cords could hold an electrostatic charge and attract small objects.

In 1845, C.H.D. Buys Ballot tested the Doppler effect by arranging for an orchestra of trumpeters to play a single sustained note on an open railroad car passing through Utrecht.

In 1746 Jean-Antoine Nollet arranged 200 Carthusian monks in a circle, each linked to his neighbor with an iron wire. Then he connected the circuit to a rudimentary electric battery.

“It is singular,” he noted, “to see the multitude of different gestures, and to hear the instantaneous exclamation of those surprised by the shock.”

Project Management

How can six people be organized into four committees so that each committee has three members, each person belongs to two committees, and no two committees have more than one person in common?

It’s possible to work this out laboriously, but it yields immediately to a geometric insight:

project management diagram

If each line represents a committee and each intersection is a person, then the problem is solved.

Bon Appétit

http://commons.wikimedia.org/wiki/File:Omelet_With_Fixings.jpg
Image: Wikimedia Commons
  1. It is now true that Clarence will have a cheese omelette for breakfast tomorrow. [Premise]
  2. It is impossible that God should at any time believe what is false, or fail to believe anything that is true. [Premise: divine omniscience]
  3. Therefore, God has always believed that Clarence will have a cheese omelette for breakfast tomorrow. [From 1, 2]
  4. If God has always believed a certain thing, it is not in anyone’s power to bring it about that God has not always believed that thing. [Premise: the unalterability of the past]
  5. Therefore, it is not in Clarence’s power to bring it about that God has not always believed that he would have a cheese omelette for breakfast. [From 3, 4]
  6. It is not possible for it to be true both that God has always believed that Clarence would have a cheese omelette for breakfast, and that he does not in fact have one. [From 2]
  7. Therefore, it is not in Clarence’s power to refrain from having a cheese omelette for breakfast tomorrow. [From 5, 6]

So Clarence’s eating the omelette tomorrow is not an act of free choice.

From William Hasker, God, Time, and Knowledge, quoted in W. Jay Wood, God, 2011.

The Ballot Box Problem

ballot box lattice diagram

In 1878 W. A. Whitworth imagined an election between two candidates. A receives m votes, B receives n votes, and A wins (m>n). If the ballots are cast one at a time, what is the probability that A will lead throughout the voting?

The answer, it turns out, is given by the pleasingly simple formula

ballot box formula

Howard Grossman offered the proof above in 1946. We start at O, where no votes have been cast. Each vote for A moves us one point east and each vote for B moves us one point north until we arrive at E, the final count, (m, n). If A is to lead throughout the contest, then our path must steer consistently east of the diagonal line OD, which represents a tie score. Any path that starts by going north, through (0,1), must cut OD on its way to E.

If any path does touch OD, let it be at C. The group of such paths can be paired off as p and q, reflections of each other in the line OD that meet at C and continue on a common track to E.

This means that the total number of paths that touch OD is twice the number of paths p that start their journey to E by going north. Now, the first segment of any path might be up to m units east or up to n units north, so the proportion of paths that start by going north is n/(m + n), and twice this number is 2n/(m + n). The complementary probability — the probability of a path not touching OD — is (mn)/(m + n).

(It’s interesting to consider what this means. If m = 2n then p = 1/3 — even if A receives twice as many votes as B, it’s still twice as likely that B ties him at some point as that A leads throughout.)

Theory and Practice

http://commons.wikimedia.org/wiki/File:Richard_Feynman_-_Fermilab.jpg

In 1982 Richard Feynman and his friend Tom Van Sant met in Geneva and decided to visit the physics lab at CERN. “There was a giant machine that was going to be rolled into the line of the particle accelerator,” Van Sant remembered later. “The machine was maybe the size of a two-story building, on tracks, with lights and bulbs and dials and scaffolds all around, with men climbing all over it.

“Feynman said, ‘What experiment is this?’

“The director said, ‘Why, this is an experiment to test the charge-change something-or-other under such-and-such circumstances.’ But he stopped suddenly, and he said, ‘I forgot! This is your theory of charge-change, Dr. Feynman! This is an experiment to demonstrate, if we can, your theory of 15 years ago, called so-and-so.’ He was a little embarrassed at having forgotten it.

“Feynman looked at this big machine, and he said, ‘How much does this cost?’ The man said, ‘Thirty-seven million dollars,’ or whatever it was.

“And Feynman said, ‘You don’t trust me?'”

(Quoted in Christopher Sykes, No Ordinary Genius, 1994.)

Pick’s Theorem

http://commons.wikimedia.org/wiki/File:Gitterpolygon.svg

Georg Alexander Pick found a useful way to determine the area of a simple polygon with integer coordinates. If i is the number of lattice points in the interior and b is the number of lattice points on the boundary, then the area is given by

pick's theorem

There are 40 lattice points in the interior of the figure above and 12 on the boundary, so its area is 40 + 12/2 – 1 = 45.

(Thanks, Pål.)

Misc

  • Only humans are allergic to poison ivy.
  • GUNPOWDERY BLACKSMITH uses 20 different letters.
  • New York City has no Wal-Marts.
  • (5/8)2 + 3/8 = (3/8)2 + 5/8
  • “Ignorance of one’s misfortunes is clear gain.” — Euripides

For any four consecutive Fibonacci numbers a, b, c, and d, ad and 2bc form the legs of a Pythagorean triangle and cdab is the hypotenuse.

(Thanks, Katie.)