“The Pythagorean Curiosity”

waterhouse pythagorean curiosity

Here’s the item I mentioned in Episode 99 of the podcast — New York City engineer John Waterhouse published it in July 1899. It’s not a proof of the Pythagorean theorem, as I’d thought, but rather a related curiosity. It made a splash at the time — the Proceedings of the American Society of Civil Engineers said it “interested instructors of geometry all over the country, bringing many letters of commendation to him from prominent teachers.” Listener Colin Beveridge has been immensely helpful in devising the diagram above and making sense of Waterhouse’s proof as it appears on page 252 of Elisha Scott Loomis’ 1940 book The Pythagorean Proposition. Click the diagram to enlarge it a bit further.

  1. Red squares BN = AI + CE — Pythagoras’s theorem
  2. Blue triangles AEH, CDN, BMI are all equal in area to ABC, reasoning via X and Y and base sides.
  3. Green angles GHI and IBM are equal and green triangle GHI is congruent to IBM (side angle side), so IG = IK = IM. IH′K is congruent to IHK as angle HIK = angle HIG and the adjacent sides correspond. This means G and K are the same distance from the line HH′, so GK is parallel to HI. Similarly, DE is parallel to PF and MN is parallel to LO.
  4. GK = 4HI, because TU=HI, TG = AH (HTG congruent to EAH) and UK = UG (symmetry). Similarly, PF = 4DE. Dark blue triangles IVK and LWM are equal, so WM = VK. Similarly, OX = QD (dark green triangles PQD and NXO are congruent). Also, WX=MJ and XN=NJ, so M and N are the midpoints of WJ and XJ. That makes WX=2MN, so LO = 4MN.
  5. Each of the trapezia we just looked at (HIKG, OLMN and PFED) have five times the area of ABC.
  6. The areas of orange squares MK and NP are together five times EG. This is because:
    • the square on MI is (the square on MY) + (the square on IY) = (AC2) + (2AB)2 = 4AB2 + AC2.
    • the square on ND is (the square on NZ) + (the square on DZ) = (AB2) + (2AC)2 = 4AC2 + AB2
    • the sum of these is 5(AB2 + AC2) = 5BC2, and BC = HE.
  7. A′S = A′T, so A′SAT is a square and the bisector of angle B′A′C′ passes through A. However, the bisectors of angle A′B′C′ and A′C′B′ do not pass through B and C (resp.) [Colin says Waterhouse’s reasoning for this is not immediately clear.]
  8. Square LO = square GK + square FP, as LO = 4AC, GK = 4AB and FP = 4BC.
  9. [We’re not quite sure what Waterhouse means by “etc. etc.” — perhaps that one could continue to build squares and triangles outward forever.]

Corner Reflectors

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

An arrangement of three mutually perpendicular planes, like those in the corner of a cube, have a pleasing property: They’ll reflect a ray of light back in the direction that it came from. This happy fact is exploited in a variety of technologies, from laser resonators to radar reflectors; the taillights on cars and bicycles contain arrays of tiny corner reflectors.

“A more dramatic application is to reflect laser rays from the Moon, where many such devices have been in place since the 1969 Apollo mission, which sent men to the Moon for the first time,” note mathematicians Juan A. Acebrón and Renato Spigler. “Among other things, the Earth-Moon distance can be measured by firing a laser beam from the Earth to the Moon, and measuring the travel time it takes for the beam to reflect back. This has allowed an estimate of the distance to within an accuracy of 3 cm.”

(Juan A. Acebrón and Renato Spigler, “The Magic Mirror Property of the Cube Corner,” Mathematics Magazine 78:4 [October 2005], 308-311.)

Moving Constants

circle theorems

If you mark two points on a circle, A and B, and a third point T, then angle ATB remains constant as T moves along the segment between A and B. (If you mark a point S in the circle’s other segment then you get another constant angle, ASB, and ASB = 180 – ATB.)

If two circles intersect at A and B and we move T as before along the segment opposite the second circle, and we extend TA and TB to P and Q on the second circle, then the length of chord PQ remains constant as T moves.

(From David Wells, The Penguin Dictionary of Curious and Interesting Geometry, 1992.)

The Falling Chain

Here are two identical rope ladders with slanting rungs. One falls to the floor, the other onto a table. The ladders are released at the same time and fall freely, but the one on the left falls faster, as if the table is “sucking” it downward. Why does this happen?

Click for Answer

The Hollowood Function

California high school student Derek Hollowood created this function after considering recurrence relations:

h(x)=\frac{10^{x+1}-9x-10}{81}
h(-10) = 0.987654321
h(-9)  =  0.87654321
h(-8)  =   0.7654321
h(-7)  =    0.654321
h(-6)  =     0.54321
h(-5)  =      0.4321
h(-4)  =       0.321
h(-3)  =        0.21
h(-2)  =         0.1
h(-1)  =           0
h(0)   =           0
h(1)   =           1
h(2)   =           12
h(3)   =           123
h(4)   =           1234
h(5)   =           12345
h(6)   =           123456
h(7)   =           1234567
h(8)   =           12345678
h(9)   =           123456789

(Thanks to Chris Smith for the tip.)

Magic Space

knecht most-perfect square

Craig Knecht, whose “terraformed” magic squares we explored in 2013, has begun to experiment with applying “magic” properties to David Hilbert’s space-filling curve.

The Hilbert curve finds its way to every cell in the square above by following the pattern shown at the lower left. Knecht divided that path into eight-cell segments, as shown in (a), and then sought solutions in which each colored eight-cell panel produced the magic sum of 260 while each of the eight ordinal positions across the eight panels did so as well. For example, in (b), a large red digit 1 marks the “first” position in each panel; the hope was to find values for these eight cells that would sum to 260, and likewise for all the “second” cells, the “third” ones, and so on.

The result, shown in (c), is a “most-perfect” magic square: Each colored panel sums to 260, and every set of cells that are 8 spaces apart on the Hilbert curve also sum to 260.

The next step was to apply this idea in three dimensions, and recently Knecht made the breakthrough shown below — a 4×4×4 “most perfect” number cube. The 64 numbers in the cube can be broken into 36 2×2 subsquares in each dimension, as shown. In all 108 of these subsquares, the four constituent numbers total 130. And as with the two-dimensional square above, a Hilbert curve can be drawn through the cube that visits each cell once, and cells that are eight cells apart on this curve sum to 260.

One of Knecht’s correspondents pointed out that the cube is even magicker than he had supposed: The “wraparound” subsquares (for example, 5, 28, 44, and 53 on the top of the cube) also sum to 130, as does each set of four corners, making a total of 192 2×2 subsquares that sum to 130.

“So in summary … making the Hilbert space-filling curve path have this magic property of values 8 spaces summing to the magic constant + this 2×2 planar criteria produces a very interesting cube!”

knecht most-perfect cube

A Reflexive Rainbow

sallows color table

From Lee Sallows: The international color code is used to mark the values of electronic components such as resistors. It assigns a distinct color to each of the 10 decimal digits, as seen in the center column of the table at right: 0 = BLACK, 1 = BROWN, …, 9 = WHITE.

Lee’s table has an ingenious reflexive property. The letters in the left-hand column are associated with the values -1 to -9, and those in the right-hand column with the values 1 to 9.

Now spelling the name of each color produces a sum that matches the number represented by that color:

sallows color sums

“This is more remarkable that it may seem,” Lee writes, “because the numbers assigned to the letters are now restricted to single-digit values only.”

Lanchester’s Laws

https://commons.wikimedia.org/wiki/File:Batalla_de_Arica.jpg

In 1916 English engineer Frederick Lanchester set out to find a mathematical model to describe conflicts between two armies. In ancient times, he reasoned, each soldier engaged with one enemy at a time, so the number of soldiers who survived a battle was simply the difference in size between the two armies. But the advent of modern combat, including long-range weapons such as firearms, changes things. Suppose two armies, A and B, are fighting. A and B represent the number of soldiers in each army, and a and b represent the number of enemy fighters that each soldier can kill per unit time. Now the equations

dA/dt = -bB
dB/dt = -aA
,

show us the rate at which the size of each army is changing at a given instant. And these give us

bB2aA2 = C,

where C is a constant.

This is immediately revealing. It shows that the strength of an army depends more on its bare size than on the sophistication of its weapons. In order to meet an army twice your size you’d need weapons (or fighting skills) that are four times as effective.

Simple as they are, these ideas shed light on the historic choices of leaders such as Nelson, who sought to divide his enemies into small groups, and Lanchester himself illustrated his point by referring to the British and German navies then at war. Today his ideas (and their descendants) inform the rules behind tabletop and computer wargames.