## Archive for May, 2008

### More Math Magic

May 16, 2008

Create a chart on a piece of paper like the one below. It could be any NxN square; there’s nothing special about the fact that N=7 in this example.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

Give the chart to a Member of the Audience (or a student in your class, or your Great-Aunt Mabel, or…). Without you watching, this person should circle a number, cross off all the other numbers in that row and column, and then pass the paper to another person. The next person circles one of the remaining numbers, crosses off everything else in that number’s row and column, and passes it along. This continues until everything on the chart has either been circled or crossed off. At this point, the last person to circle a number adds up all the circled numbers and think very hard about the sum.

You, meanwhile, know that people are circling numbers but don’t know which numbers are being circled. Once the final person has added up the numbers and telepathically thinks of the sum, you will use your Amazing Mindreading Powers to announce the sum!

Which will be 175.

Or, in general for an NxN square, will be (N3+N)/2. Pretty cool, isn’t it? Of course, if you do this trick more than once for the same crowd, you might want to vary the size of the square to mask the fact that the same size square always leads to the same answer.

Here’s a brief justification for the total. Because numbers in the same row and column as a circled number are crossed out, it turns out exactly one number in each row is circled and, at the same time, exactly one number in each column is circled.

• The first number will be between 1 and 7 (inclusive). Call it A, where 1≤A≤7.
• The second number will be between 8 and 14 (inclusive). Call it 7+B, where 1≤B≤7.
• The third number will be between 15 and 21 (inclusive). Call it 2·7+C, where 1≤C≤7.
• Continue all the way to the seventh number, which will be between 43 and 49. Call it 6·7+G, where 1≤G≤7.

When you add these all up you get (A)+(7+B)+(2·7+C)+…+(6·7+G). Reordering gives (7+2·7+…+6·7)+(A+B+…+G), which can be simplified as (1+2+…+6)·7+(A+B+…+G).

Because you don’t know which numbers were circled, you don’t know what A, B, etc. are. But you do know that one number was picked in each column! Each number in the first column is 1 more than a multiple of 7, each number in the second column is 2 more than a multiple of 7, etc.

Since one number is picked out of each column, the numbers A, B, …, G are 1, 2, …, 7 in some order. This means that A+B+…+G is just 1+2+…+7, and the total amount is (1+2+…+6)·7+(1+2+…+7). In general, for an NxN square, the total will be (1+2+…+[N-1])·N+(1+2+…+N). The first summation is (N-1)N/2 and the second summation is N(N+1)/2, so the total becomes {(N-1)N/2}·N+N(N+1)/2. Factoring out N/2 gives (N/2)(N2-N+(N+1)), which becomes N(N2+1)/2 or (N3+N)/2.

But of course, you don’t have to give away your secret.

I found this trick on curiousmath.

### And or Or

May 15, 2008

I was looking over car seat laws recently, and my main discovery was that it was confusing. All states have age restrictions for carseats/booster seats (younger than 3-8, depending on the state), some have height restrictions (40″-57″), and others have weight restrictions (20-80lbs). The best overview I could find came from the Insurance Institute for Highway Safety, but at times the distinction between AND or OR was subtle, indicated only by a semicolon.

For example, let’s look at Massachusetts. Up through July 9 this year, the law is

No child under age five and no child weighing forty pounds or less shall ride as a passenger in a motor vehicle on any way unless such child is properly fastened and secured, according to the manufacturer’s instructions, by a child passenger restraint as defined in section one.

So children under age five OR under 40 lbs need a carseat/booster seat, while children over age five AND over 40 lbs can go without. But starting July 10:

A passenger in a motor vehicle on any way who is under the age of 8 shall be fastened and secured by a child passenger restraint, unless such passenger measures more than 57 inches in height.

At that point, children over age 8 OR over 57″ can ride without a special seat.

I found as I surfed the web that the shorter the state-by-state summary, the more confusing it was to know if AND or OR should be used, but if I could track down something Official from the State Government itself (as above), it was usually pretty clear. This and/or problem did, however, remind me of back when the boys were young and the carseat restrictions were actually impossible to cover.

Babies are supposed to ride rear-facing until they are 1 year of age AND 20 lbs. That meant that our boys, who reached 20 pounds way before a year, still were supposed to ride rear-facing until they turned one. No problem so far.

The problem was that our carseats had height restrictions. We used the Britax Roundaboud, which could be rear-facing for little ones and then turn around and be forward facing for a couple more years. Here’s what the manual said (this is the updated one, but our version had the same kind of info):

Use rear-facing only with children:

• who weigh between 5 and 35 lbs. (2.3 – 15.9 kg.) and
• the top of the child’s head is 1” (2.5 cm.) or more below the top of the child seat (Fig. A) and
• the harness straps are in the nearest slots at or below the child’s shoulders.

If the child cannot be secured within these requirements because the child exceeds height or weight requirements, review the forward-facing guidelines on page 7.

But here’s how those forward-facing guildelines start:

Use forward-facing only with children who are at least one year of age….

So our boys couldn’t be rear-facing because they were too tall, but couldn’t be forward facing because they were too young. And in talking with other parents, there was no way around this contradiction, even with other seats that we looked at. I picture a Venn diagram of age, weight, and height, and not all of the combinations were covered.

At least the contradiction was short-lived: eventually the boys did turn one and the problem (for us) went away.

Picture taken by Tracybenn, and used under the GNU Free Documentation License.

### Can anyone explain this?

May 14, 2008

We went to a ballgame today (go Red Wings!) and saw this sign at the parking garage:

Can anyone tell me how much it costs to park for, say, 4 hours?

Update 6/5: The next time we went to a game, we noticed that they had a new sign up at the ticket booth giving exactly the minutes and prices that Joel gave in the comments below.  Good call, Joel!

### Lorenzo Mascheroni and the Euler-Mascheroni Constant

May 13, 2008

May 13th is Lorenzo Mascheroni’s birthday. According to the MacTutor History of Mathematics website, he was born in 1750 in Bergamo, Lombardo-Veneto. After training for the priesthood (and ordination at age 17), he taught rhetoric, physics and mathematics.

Mascheroni is primarily remembered for proving that every ruler and compass construction can in fact be done using compasses alone (a result that was found independently by Georg Mohr in 1672). Mascheroni’s name is also associated with one of the fundamental constants of mathematics….

The story of the Euler-Mascheroni Constant

In 1734, Leonhard Euler presented a paper to the Academy of Sciences in St. Petersburg on properties of harmonic progressions. In that paper, he notes that the harmonic series diverges, and indeed that $1 + \frac12 + \frac13 + \cdots + \frac1n$ approaches the function $C + \ln(1+n)$ (for a particular constant $C$) as $n$ goes to infinity. That constant $C$ is now known as the Euler-Mascheroni Constant, also known as gamma.

The relevant portion of Euler’s paper goes as follows:

Step 1. We’ll define a function $s(n) = 1 + \frac12 + \frac13 + \cdots + \frac1{n}$, and note that if $n$ increases by 1, then $s(n)$ will increase by $\frac1{n+1}$. From this, Euler concludes that as $n$ goes to infinity the function $s(n)$ satisfies the differential equation $\frac{ds}{dn} = \frac1{n+1}$, and thus $1+\frac12+\frac13+\cdots+\frac1n = C + \ln(1+n)$ for some constant $C$.

Step 2. Since (via Taylor Series) we know that $\ln(1+\frac1k) = \frac1k - \frac12 (\frac1k)^2 + \frac13 (\frac1k)^3 - \cdots$, if we solve for $\frac1k$ we find that $\frac1k = \ln( \frac{k+1}{k} ) + \frac{1}{2k^2} - \frac{1}{3k^3} + \frac{1}{4k^4} - \cdots$

Step 3. We will substitute the identity found in Step 2 for $\frac1k$ for each term of the harmonic series in the identity from Step 1. This gives

$1 = \ln 2 + \frac12 - \frac13 + \frac14 - \frac15 + \cdots$

$\frac12 = \ln\frac32 + \frac1{2\cdot4} - \frac1{3\cdot8} + \frac1{4\cdot16} - \frac1{5\cdot32} + \cdots$

$\frac13 = \ln\frac43 + \frac1{2\cdot9} - \frac1{3\cdot27} + \frac1{4\cdot81} - \frac1{5\cdot243} + \cdots$

$\frac14 = \ln\frac54 + \frac1{2\cdot16} - \frac1{3\cdot64} + \frac1{4\cdot256} - \frac1{5\cdot1024} + \cdots$

etcetera, concluding with

$\frac1n = \ln\frac{n+1}n + \frac1{2\cdot n^2} - \frac1{3\cdot n^3} + \frac1{4\cdot n^4} - \frac1{5\cdot n^5} + \cdots$

Adding in columns, and using the fact that $\ln(2) + \ln(\frac32) + \ln(\frac43) + \cdots + \ln(\frac{n+1}{n}) = \ln(2 \cdot \frac32 \cdot \frac43 \cdots \frac{n+1}{n}) = \ln(n+1)$ we find that

$1 + \frac12 + \frac13 + \cdots +\frac1n = \ln(n+1)$

$\qquad \mbox{.} \qquad\qquad\qquad\qquad\qquad + \frac12 (1 + \frac14 + \frac19 + \frac1{16} + \cdots )$

$\qquad \mbox{.} \qquad\qquad\qquad\qquad\qquad - \frac13 (1+\frac18 + \frac1{27} + \frac1{64} + \cdots )$

$\qquad\mbox{.} \qquad\qquad\qquad\qquad\qquad + \frac14 (1+\frac1{16} + \frac1{81} + \frac1{256} + \cdots)$

$\qquad\mbox{.} \qquad\qquad\qquad\qquad\qquad - \cdots$

Thus Euler has found a way to approximate $C$, using the alternating sum on the right hand side of this identity, expressed in terms of series whose values Euler knew approximations for (values of the zeta function). Euler does this calculation, and publishes the approximation $C \approx 0.577218$.

Euler returned to this constant throughout his career, and using an approach similar to that outlined above was able to compute its first sixteen decimal places.

In 1790, Lorenzo Mascheroni published a commentary on Euler’s integral calculus texts, in which he gave a 32 digit approximation to the constant. Mascheroni’s record was eclipsed in 1809 when Johann von Soldner redid the calculation to 40 decimal places, and got a different value in the 19th place. Much consternation ensued, until 1812 when Gauss and F. Nicolai evaluated $C$ to 40 decimal places, confirming von Soldner’s result. Throughout the subsequent decades, both estimates to $C$ remained in circulation, and a number of mathematicians undertook the task of recomputing its value.

Mascheroni, through his digit error, may very well have generated extra buzz for this constant. It is also Mascheroni who coined the name “gamma” for it, and henceforth gamma has been known as the Euler-Mascheroni constant.

In honor of Mascheroni’s 258th birthday, a little musical offering….

### “Green guilt”, a study in statistical significance

May 12, 2008

An article dated May 7 2008 in the on-line edition of USA Today describes the results of the second annual “green guilt” survey, conducted at the behest of the Rechargeable Battery Recycling Commission. (“Green guilt” is a reaction to the belief that one ought to live more ecologically than one does or attempts to do.) As the USA Today article puts it, people’s guilt levels are rising! And we’re recycling more! Unless you’re male, that is.

### Mathematician of the Week: John Charles Fields

May 11, 2008

Welcome to a new (weekly) series: the Mathematician of the Week. We’ll use this space to highlight upcoming anniversaries associated with prominent mathematicians of the present and past, and feature biographical notes on one distinguished individual.

John Charles Fields was born on May 14 1863 in Hamilton, Ontario, Canada. While his research area was the theory of algebraic functions, Fields is best remembered as the namesake of the Fields Medal, given quadrannually quadrennially at the International Congress of Mathematicians, and often viewed as the mathematical equivalent of the Nobel Prize.

Fields played a central role in ensuring the success of the 1924 ICM (held in Toronto), and in 1931 proposed the creation of an international medal to recognize significant achievement in mathematics, using funds left over from the 1924 ICM to endow the award.

### How much is enough?

May 10, 2008

If you notice a pattern, how many times do you have to check that it works before being certain. Six? Twenty? Two thousand?

Two favorite examples of mine that demonstrate that patterns can continue for a long time before going awry:

The first example is the polynomial f(n)=n2+41n+41. If you plug in any whole number for n from 1 to 40, then f(n) is a prime number: f(1)=83, f(2)=127, all the way to f(40)=3281. But f(41)=3403=41·83 is not prime. So something that works 40 times in a row might fail.

The second example are the cyclotomic polynomials. Look at polynomials of the form xn-1 that have been factored:

x-1=(x-1)
x2-1=(x-1)(x+1)
x3-1=(x-1)(x2+x+1)
x4-1=(x-1)(x+1)(x2+1)
x5-1=(x-1)(x4+x3+x2+x+1)
x6-1=(x-1)(x+1)(x2+x+1)(x2-x+1)

The last polynomial in each case is the cyclotomic polynomial of order n. [It has a much more technical definition using imaginary numbers and the product of primitive roots of unity]. And at first glance it looks like the coefficients are 0, 1, or -1. Even at second glance or sixth glance, since it’s true for the first 104 cyclotomic polynomials. But not the 105th.

x105-1=(x-1)•(x2+x+1)•(x4+x3+x2+x+1)•
(x6+x5+x4+x3+x2+x+1)•(x8-x7+x5-x4+x3-x+1)•
(x12-x11+x9-x8+x6-x4+x3-x+1)•
(x24-x23+x19-x18+x17-x16+x14-x13+x12-x11
+x10-x8+x7-x6+x5-x+1)•
(x48+x47+x46-x43-x422x41-x40-x39+x36+x35
+x34+x33+x32+x31-x28-x26-x24-x22-x20+x17+x16
+x15+x14+x13+x12-x9-x82x7-x6-x5+x2+x+1)

See those two coefficients of 2 in that last polynomial? So the pattern of coefficients being only 0, 1, or -1 fails. Interestingly, the reason for this initial failure occurring so late in the game is that 105 is the smallest number that has three distinct odd prime factors (105=3·5·7). The integer 385 is also a product of three distinct odd primes (385=3·5·11) and the 385th cyclotomic polynomial is the first one to have a 3 as a coefficient (see Wolfram Mathworld).

Patterns. You just can’t trust them.

### Decagon Mug

May 9, 2008

TwoPi and I went out for lunch to the Cheesecake Factory the other day, and noticed partway through that the mugs were regular decagons! So he snuck a photo while pretending to check text messages.

Is it hard to see? Here’s the outline on top:

Cool, isn’t it? I’m not sure what the most common shape would be, especially since our own glasses are all circles (though I noticed our department chair has octagon glasses). If anyone else has any neat polygonal glasses, send me a photo* and I’ll post them here. I love finding real life polygons!

* hlewis5 following by the @ sign and then after that @ sign put naz.edu

### Toilet Paper Math

May 8, 2008

Really, is there a more appropriate follow-up to yesterday’s featured theorem?

Last night young Quentin, age 4½, went to get some toilet paper to clean toothpaste out of the sink after brushing his teeth (because — get this — he likes to clean up after himself. I can hardly believe it.). As he pulled off a strip of TP, he suddenly held it against himself and got all excited: “This is as big as my belly!” I pointed out that his belly was three squares big, and asked how long his arm was. He measured, and exclaimed, “My arm is three squares long!” When he tried to measure his leg, it fell short so I suggested he might need one more square. He immediately went to the roll, counted off a strip four squares long, and held it against his leg. Yup, four squares worked.

The sink stayed dirty for a while after that while he went around measuring his hand (one square), our arms, etc. The nice thing about toilet paper is that he could take strips of various sizes and just pick the one that seemed best. His measurements weren’t exact (I’m not going to hire him to build a bookcase, for example) but he did seem to have the basic idea of measurement and that’s a topic that several K-6 teachers I’ve talked to say is the one that students need the most help with after number sense. (Speaking of which, Denise on Let’s Play Math had a great post Tuesday about helping kids learn number sense.) And I think non-standard measurement is one of the NYS math standards. [Quick check — yup, it’s 1.M.2, 1.M.11, 2.M.1, 2.M.10, and 3.M.10. I spent a while last year putting all the NYS math standards into Excel worksheets for easy searching and posted them here if anyone would find that useful.]

Thinking about blogging this, I googled “Toilet Paper Math” and found some other interesting ways to use toilet paper to do math. You can determine the least expensive choice of TP at the grocery store. You can fold it in half twelve times. You can find the thickness of a sheet of TP (although it seems like density — aka fluffiness — might make that inexact). You can calculate how much text you can print on a roll of toilet paper.

And finally, you can read about how Sir Roger Penrose sued the Kimberly Clark Corporation back in 1997 because one of the designs that was printed on Kleenex quilted toilet paper looked like Penrose tiles (see Wolfram’s Mathworld or this more detailed summary from Professor Richard H. Stern’s Computer Law Class at George Washington University.)

Update 5/10: I think these are the kinds of pictures that mbork was referring to below. In the first two examples the triangle has a right angle, but in the third the angle a bit larger.

### The Geometry Van

May 7, 2008

I taught Geometry this spring, and we spent about half the semester working through Euclid (Book I and a smattering of some others). We proved SAS (Side Angle Side), ASA, AAS, but not ASS (Angle Side Side). Because there is no ASS in Geometry.

Here <A=<A’, AB=A’B’, and BC=B’C’ but the two triangles are not congruent. All we get are bad jokes.

Except that’s not quite true! If <A=<A’, AB=A’B’, BC=B’C’, and BC≥AB, then the two triangles are congruent! In class, I referred to this theorem as ASS.

(The Hypotenuse-Leg Theorem is a special case of this, since the hypotenuses, as the sides across from the corresponding right angles, are certainly longer than the corresponding legs of the triangle).

We didn’t refer to this theorem very often, but it is, well, memorable. And so Adele, one of our majors, was thrilled when she noticed ASS written on a van this past weekend! She snuck a photo of it:

According to Adele and everyone who was with her, this is exactly how I would write the shorthand. We have no idea why it’s on this van. I like to think that this is a Secret Geometry Van, coming out to help students everywhere by providing them with extra theorems.

### Pi Day Sudoku Solution

May 6, 2008

Remember Pi Day Sudoku from Brainfreeze Puzzles?

If you were wanting the answer, it’s been posted!  You can find the solution here.

### The Spring Newsletter is here!

May 6, 2008

We’ve just posted the Spring 2008 issue of Our Newsletter (the seasonal newsletter of the Nazareth College Math Department).   This issue is named The Rudin in honor of Mary Ellen Rudin, according to our tradition of naming each issue after a mathematician (which would have been a great idea to come up with from the start, but in fact only evolved three issues in when we STILL didn’t have any ideas for a Newsletter name).  The lead articles on Extracting Square Roots and Cube Roots come straight from TwoPi’s posts here; other articles are about conferences and competitions our students were in, Alumni News, Toilet Seat Gymnastic (which was in fact our second post ever here, although the article is a little longer than the blog entry), Math Club news, and Problems.  And if any of you submit solutions to problems, we promise to post your name for laud and admiration in the next issue!

### Cinco de Mayo Math

May 5, 2008

Today is the day many people (mostly in the US) celebrate Mexican Independence Day (nope that’s in September) the day in 1862 that the Mexican Army beat the French at Puebla, about 70-80 miles east of Mexico City. As part of this celebration, you might do some Cinco de Mayo Math by Paula J. Maida, which includes ideas like finding the proportion of colors on a Mexican Flag or writing the eleven letters CINCODEMAYO on wooden cacti and letting kids choose a letter randomly as a way to explore both probability and fractions.

Or, expanding beyond Mexico (as the celebration has in many places), you might read “Spanish Colonial Mathematics: A Window on the Past” by Ed Sandifer, which was published in the College Math Journal in September 2002 and is available as a .pdf file here from Ed’s homepage. As Ed explains in this interview:

[S]ome Latino students feel disconnected to math. But we can help make a connection by teaching about Spanish-colonial math and pointing to facts such as there were 11 math books published in Spanish in the New World before there were any in English, with the first being published in 1556, only 100 years after the printing press was invented.

The article above examines 7 of those math books. There’s the 1556 Sumario Compendioso de las quentas de plata y oro que in los reynos del Piru son necessarias a los mercadores: y todo genero de tratantes. Los algunas reglas tocantes al Arithmetica. Fecho por Juan Diez freyle. (Isn’t that a GREAT title!?! It translates as Compendious summary of the counting of silver and gold that are necessary in the kingdoms of Peru to merchants and all kinds of traders. The other rules touching on Arithmetic. Made by Juan Diez, friar.) which was full of all sorts of tables to help you out of you were a tax collector (Hey! More Tax Math!) and includes among other things the following example of how to multiply 875 by 978:

Isn’t this wild? Some of it is explained in the article: the initial 8×9 of 800×900 becomes the 72 of the upper left. Then 8×7 of 800×70 is 56, but this is written with the 5 below the 2 in 72 and the 6 next to the 72, making the 72 look like 726. Then 8×8 of 800×8 is 64, and the 6 of 64 is put next to the 5 while the 4 is next to the 726.

It continues in this fashion, with 7×9=63 of 70×900 put underneath the 56 and with all the numbers crammed together like Galley Division of the same time period. David Smith write in his book History of Mathematics (p. 119) that this method is essentially The Method of the Cup (per copa) because it looks like a goblet.

Several books followed the Sumario Compendioso…, including King Philip of Spain’s Pragmática sobre los diez días del año in 1584, which wasn’t so much about math as about how to deal with the fact that switching from the Julian to the Gregorian calendar involved skipping ten days, and several books that look at military mathematics and formations. In the latter category is the Breve aritmética por el mas sucinto modo, que hasta oy se ha visto. Trata en las quentas que se pueden ofrecer para formar campos y esquadrones by Benito Fernandez de Belo— another fabulous title which means Brief arithmetic for the most succinct method which has been seen up to today. Treating calculations that one can do for the formation of camps and squadrons — which shows how to align 278 men in a squadron into a regular pentagon and contains a woodcut doodle in the back. The final book in the article is the 1696 Cubus, et sphaera geometrice duplicata by Juan Ramón Coninkius about straightedge and compass constructons. Granted, the constructions (doubling the volume of a cube or sphere) were impossible, but that was still unknown at the time.

Happy Cinco de Mayo!

### Music to Study By

May 4, 2008

Or to work by.  Or to procrastinate by.  Here’s a video by starkravenmadd of Doctor Steel’s Fibonacci Sequence:

### Kepler’s First Attempt

May 3, 2008

When it comes to orbits, Johannes Kepler knew his stuff. He’s the one who in 1602 realized that planets orbit in ellipses rather than circles, which became the first of his Three Planetary Laws. But no one is perfect, and these were not his first attempts at describing the motions of the Heavens. In 1596 he published Mysterium Cosmographicum (The Mysteries of the Cosmos), in which he proposed the following model for the solar system:

In this model, the six known planets were envisioned as traveling in circles, along the equators of six giant spheres. The six giant spheres were separated by the five platonic solids. Saturn and Jupiter were separated by a giant cube, and Jupiter and Mars by a giant tetrahedron. It’s harder to see the interior planets in the drawing above, so here’s a close up:

Mars and Earth were separated by a giant dodecahedron, Earth and Venus by a giant icosahedron, and, finally, Venus and Mercury by a giant octahedron. And then, in center of all the orbits, was the Sun.

Let’s see how accurate this model is. If you start with a giant platonic solid, like a cube, you can circumscribe a sphere on the outside and inscribe a sphere on the inside, and then compare the ratio of the radii of the two spheres. It turns out to be √3≈1.73. And lo, if you look at the average radius of Saturn’s orbit (9.021 Astronomical Units) and divide it by the average radius of Jupiter’s orbit (5.20336 AU), it rounds to 1.73. Let’s see how the other ratios match up:

Giant Polyhedron Ratio of spheres in model Ratio of Actual Planet Orbits
Saturn to Jupiter cube 1.73 1.73
Jupiter to Mars tetrahedron 3.00 3.42
Mars to Earth dodecahedron 1.26 1.52
Earth to Venus icosahedron 1.26 1.38
Venus to Mercury octahedron 1.73 1.87

Not too shabby! Plus, as a bonus, you can see that the cube and the octahedron, which are dual polyhedra, have the same ratios of the radii of the circumscribed and inscribed spheres (√3≈1.73); likewise, the dodecahedron and the icosahedron (which are also duals of each other) have the same ratio of the radii of circumscribed and inscribed sphere ($\frac{3\sqrt{10+2\sqrt{5}}}{3\sqrt{3}+\sqrt{15}}$≈1.26). And unlike the Titius-Bode law, the big gap between Jupiter and Mars didn’t really cause any problems since the tetrahedron fit nicely in there. But a few years later Kepler realized it was wrong, and Uranus’s discovery later would have sealed the deal in any case. Poor Kepler. But it’s still an impressive idea, and was deemed important enough even recently to put on a 2002 commemorative 10-Euro coin in Austria (designed by Thomas Pesendorfer).

Yeah Kepler!

The planet data came from NASA; the data on the radii of circumscribed and inscribed spheres came from Wolfram MathWorld. It’s not clear if the coin is copyrighted or even copyrightable or not; it seems to fall under fair-use guidelines, however. You can find the coin at the Austrian Mint.