### Freese found the {10/3} dissection

In Ernest Freese's diagrams (from the 1950's), which I got access to late in 2002, there is a diagram of the same 10-piece dissection of a {10/3} to two {5/2}s as in Figure 16.2, which I had (implicitly) attributed to Harry Lindgren. Freese also had an 8-piece dissection of an {8/2} to two {4/1.5}s, but did not make a claim about there being an infinite class of such instances.

### Typos, typos, typos

There is a typographical error in the first paragraph on page 173. In the sentence that begins, "Lindgren split ...", it should read "decagon" rather than "pentagon".
There are several typographical errors in the paragraph on page 173 that begins: "An equally masterful dissection ..." The third sentence should read:
"The ratio s{5/2}/r{10/3} = sqrt(2) between the side of the pentagram and the short radius of a {10/3} of equal area makes this dissection possible."
The last sentence of that paragraph should read:
"... such as two {3}s to one {6}, and two {7/3}s to one {14/5}."
(The last sentence is corrected in the paperback edition.)
There is a related typographical error in the last paragraph on page 179 that begins: "Galaxy (16.7) has auspicious dissections of two ..." The next expression should read "{(4n+2)/(2n-1)}s" rather than "{(4n+2)/(n+1)}s". This makes it consistent with line (16.7) of the first table.
(This was supposed to be corrected in the paperback edition, but the change was unfortunately made to the sentence that begins "Galaxy (16.5) has auspicious dissections of ...", thus rendering both sentences incorrect!)
In the second paragraph on page 177, replace "simple two-to-one dissections" by "simple one-to-one dissections".

### Who reads tables carefully enough to find typos?

Koji Miyazaki, Hirohisa Hioki, and Naoki Odaka examined my tables on pages 174-175 and 181 and found a number of mistakes:
Equation (16.7) on page 175 should have the numerator and denominator interchanged.
Equation (16.8) on page 175 should have R{(8n-2)/(2n-1)} rather than R{(8n-2)/(2n)}.
Equation (16.11) on page 175 should have an extra factor of 2 on the righthand side.
Equation (16.13) on page 181 should have the 2 replaced by sqrt(2) on the righthand side.
Equation (16.18) on page 181 should not have that factor of 2 on the righthand side.
Equation (16.19) on page 181 should have an additional factor of 2 on the righthand side.

### An earlier attribution

On page 179, I note that Harry Lindgren (1964) gave a 4-piece dissection of a hexagon to two triangles. This appears to have been known much earlier, because it seems to have been described briefly by Leonardo da Vinci (1452-1519) in one of his notebooks. On Foglio 122 (volume 2) of the Codex Atlanticus, Leonardo sketched a hexagon, inscribed a triangle inside of it, and cut the triangle into three pieces identical to the portions of the hexagon outside of the triangles. He labeled this drawing with the following (in Italian):
El triangolo che con li sua angoli è contingente alli 3 angoli dello esagone, vale la metà d'esso esagone.
which translates as:
The triangle that with its angles is contingent on the 3 angles of the hexagon is the same as half of this hexagon.
See the updates to my preface for the full reference to the transcription and facsimile reproduction of Codex Atlanticus.

### A newly-discovered galaxy!

On page 181 I listed those trigonometric relationships between stars (and polygons) that seemed to be of no general pattern. One of those was the relationship (16.13), corrected in an earlier entry on this webpage as:
r{10/4} / r{5/2} = sqrt(2) cos(π /5)
On the next page I mentioned that Harry Lindgren gave an 18-piece dissection of two {5/2}s to one {10/4}, which I reproduced as level 2 of the prism dissection in Figure 22.8.

It turns out that the relationship between one {10/4} and two {5/2}s is not an isolated curiosity but the first in an infinite sequence. For every positive n, we have that
s{ (8n+2) / (2n+2) } / s{ (4n+1) / (n+1) } = sqrt(2) cos( (π / 2)(2n+1) / (4n+1) )
Note that I express the relationship in terms of the ratio of the side lengths, rather than the ratio of the r parameters, the shorter of the two radii for a star. This seems more natural, allowing us to use the law of sines and the formula for the sine of a double angle to simplify the ratio.

We can then plug in the first several values of n to identify the first few cases. For n = 1, we get Harry's two {5/2}s to one {10/4}. For n = 2, we have two {9/3}s to one {18/6}. For n = 3, it's two {13/4}s to one {26/8}. And so on.

There are natural dissections for each case, with the nth dissection using 16n+2 pieces. For the case n = 2 we see below the 34-piece dissection:

### More on Harry Lindgren

Project Runeberg maintains a database on Nordic authors. The biographical information for Harry Lindgren focuses on his work related to English spelling reform.

In 2017 Amanda Laugesen published an article, "Lindgren, Harry (1912-1992)", for the Australian Dictionary of Biography. Follow the link to Harry's biography.

Copyright 1997-2014, Greg N. Frederickson.
Permission is granted to any purchaser of Dissections: Plane & Fancy to print a copy of this page for his or her own personal use.

Last updated March 4, 2017.