Dear Professor Greitzer

Author(s): 
Joe Richards and Don Crossfield

Authors' Note: Professor Sam Greitzer was a giant among the U.S. math community, as an author, as a teacher, and as a coach for the US Math Olympiad team for much of the second half of the 20th century. We knew him through the Arbelos, a mathematics journal “produced for precollege philomaths” which he published and for which he wrote most of the articles. Although the journal was filled with an amazingly wide spectrum of mathematics jewels, his favorite field of play was Geometry, and every issue had something interesting in that valley. His death was a great loss for the philomath community. This article is in tribute to those contributions, the best way we know how to show him our respect. (For more about Professor Greitzer, click here.)

Dear Professor Greitzer,

In your articles you lament the state of Geometry education nowadays; and perhaps you’re right, Professor. Our classes may not know the difference between Ceva’s Theorem and Ptolemy’s Theorem. But we’re explorers, and that’s central to mathematics. We are writing to tell you a tale that will warm your heart, and a tale that will perhaps give us all some insight into the thought processes of some creative minds of the mathematical past. We do so at your request, quoting you on page 15 of the January 1984 edition of the Arbelos in an article demonstrating methods of capturing the value of \(\pi\):

Other mathematicians have used other formulas [to approximate \(\pi\)]. For example, Dase based his evaluation on the formula

$${{\pi}\over{4}}=\arctan{\frac{1}{2}}+\arctan{\frac{1}{5}} +\arctan{\frac{1}{8}}\quad {\rm !}$$

... and Machin used the formula

$${\frac{\pi}{4}}=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{239}} \quad {\rm !!!}$$

One wonders how these mathematicians derived their initial formulas (hence the exclamation points).

One can find biographies of Zacharias Dase and of John Machin in the MacTutor History of Mathematics Archive.

Dase was not a mathematician, but rather a savant, hired to utilize formulae like that above to generate useful tables of values. Machin was a renowned mathematician of the early 1700s, and very likely discovered the series formula ascribed to him above, although proofs of its correctness are attributed instead to Abraham De Moivre, Johann Bernoulli, and Jakob Hermann.

We’d like to show you a path our Geometry classes walked that effectively discovers these formulae, and uses only elementary geometric tools, which we know will appeal to you, since you frequently eschewed the Calculus when simpler tools would get the job done.

Dear Professor Greitzer - Introduction

Author(s): 
Joe Richards and Don Crossfield

Authors' Note: Professor Sam Greitzer was a giant among the U.S. math community, as an author, as a teacher, and as a coach for the US Math Olympiad team for much of the second half of the 20th century. We knew him through the Arbelos, a mathematics journal “produced for precollege philomaths” which he published and for which he wrote most of the articles. Although the journal was filled with an amazingly wide spectrum of mathematics jewels, his favorite field of play was Geometry, and every issue had something interesting in that valley. His death was a great loss for the philomath community. This article is in tribute to those contributions, the best way we know how to show him our respect. (For more about Professor Greitzer, click here.)

Dear Professor Greitzer,

In your articles you lament the state of Geometry education nowadays; and perhaps you’re right, Professor. Our classes may not know the difference between Ceva’s Theorem and Ptolemy’s Theorem. But we’re explorers, and that’s central to mathematics. We are writing to tell you a tale that will warm your heart, and a tale that will perhaps give us all some insight into the thought processes of some creative minds of the mathematical past. We do so at your request, quoting you on page 15 of the January 1984 edition of the Arbelos in an article demonstrating methods of capturing the value of \(\pi\):

Other mathematicians have used other formulas [to approximate \(\pi\)]. For example, Dase based his evaluation on the formula

$${{\pi}\over{4}}=\arctan{\frac{1}{2}}+\arctan{\frac{1}{5}} +\arctan{\frac{1}{8}}\quad {\rm !}$$

... and Machin used the formula

$${\frac{\pi}{4}}=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{239}} \quad {\rm !!!}$$

One wonders how these mathematicians derived their initial formulas (hence the exclamation points).

One can find biographies of Zacharias Dase and of John Machin in the MacTutor History of Mathematics Archive.

Dase was not a mathematician, but rather a savant, hired to utilize formulae like that above to generate useful tables of values. Machin was a renowned mathematician of the early 1700s, and very likely discovered the series formula ascribed to him above, although proofs of its correctness are attributed instead to Abraham De Moivre, Johann Bernoulli, and Jakob Hermann.

We’d like to show you a path our Geometry classes walked that effectively discovers these formulae, and uses only elementary geometric tools, which we know will appeal to you, since you frequently eschewed the Calculus when simpler tools would get the job done.

Dear Professor Greitzer - Slopes of Lines and Angles of Inclination

Author(s): 
Joe Richards and Don Crossfield

We began with an exploration of the relationship between the slopes of lines and the angles of inclination that they make with a horizontal axis. After some rather non-precise measurements with a protractor, we used Geometer’s Sketchpad to explore further (see Figure 1), and built a table of values precise to the nearest tenth of a degree. Relevant parts of the table are found in Chart 1.

Figure 1

Anytime we collect data, we stare at it, looking for patterns. For some in our classes, the observation that the 45o entries are collinear is a revelation. But, every now and again, we get deeper results, and one year a student remarked that the angles associated with slopes of 1/2 and of 1/3 (26.6o and 18.4o, respectively) appeared to give a sum of 45 degrees. We changed the preferences on Sketchpad to explore more significant digits and this conjecture seemed to hold. That gave us the open door from the inductive living room to the deductive kitchen, and in our best Euclid voice, we asked, “Why?”

Figure 2

Figure 2 illustrates why. The triangle is clearly a right triangle (legs have negative reciprocal slopes) and is isosceles as well (leg lengths are each the square root of 10). The 45-degree angle at the origin has been split into two angles by the x-axis. The segment below the x-axis, connecting A(0,0) and B(3,-1) has a slope of negative 1/3. Similarly, the segment above, connecting A(0,0) and C(4,2) has a slope of 2/4 = 1/2. Done.

Dear Professor Greitzer - The First Identity

Author(s): 
Joe Richards and Don Crossfield

Calculus teachers, and their students, know that the relationship

$${{\pi}\over{4}}=\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}$$

is historically famous, as well as a quickly converging way to calculate the digits of \(\pi\), by using the Taylor series for the arctangent function:

$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots ,$$

but our Geometry classes know only that the two angles that go with slopes of 1/2 and 1/3 have a sum of 45 degrees. That’s OK ... we’re just glad they’re looking for patterns. Here is another one they noticed.

Hey, it looks like the angles that go with slopes of 1/5 and 1/8 (11.3o and 7.1o) have a sum of 18.4 degrees, the angle that goes with the slope of 1/3.

Slightly varying Figure 2 into Figure 3 gives us assurance that this relationship is also true.

This triangle is also clearly a right triangle, since the slopes of the legs are -1/5 and 5. The short leg is 1/3 the length of the long leg, since we drew three 5 x 1 segments, and then turned 90 degrees and drew only one 5 x 1 segment. The angle at the origin must, therefore, be the one associated with a slope of 1/3, and it has also been split into two angles by the x-axis. The segment below, connecting A(0,0) and B(15,-5), has a slope of -1/5. The segment above, connecting A(0,0) and C(16,2), has a slope of 1/8. Again, done.

Professor Greitzer, our Geometry kids may not understand anything about the radius of convergence of the Taylor series for the arctan function, but based on their exploring skills, we’re thinking that any mathematician of Dase’s era could have measured with a protractor, built a table, conjectured our conjectures, and built the coordinate plane arguments to verify that

$${{\pi}\over{4}}=\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}$$

and that

$$\arctan{\frac{1}{3}}=\arctan{\frac{1}{5}} +\arctan{\frac{1}{8}},$$

hence,

$${{\pi}\over{4}}=\arctan{\frac{1}{2}}+\arctan{\frac{1}{5}} +\arctan{\frac{1}{8}}\quad {\rm !}$$

so we conclude that the single exclamation mark identity is decidedly within the range of a mathematician (or a Geometry class), using only elementary mathematical tools. Cool, huh?

Dear Professor Greitzer - The Second Identity

Author(s): 
Joe Richards and Don Crossfield

On to the second identity, which was that

$${\frac{\pi}{4}}=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{239}} \quad {\rm !!!}$$

Looking back at Chart 1, we observed that the angle associated with a slope of 4/3 appeared to be twice that of the angle that goes with a slope of 1/2. Exploring that possibility led to the picture in Figure 4.

Figure 4

Triangle OAB is a 3-4-5 triangle, with segment OB having a slope of 4/3. Segment OD has slope 1/2 and our conjecture was that it bisected angle AOB. We extended OD until its endpoint, called E, was horizontally to the right of B, and observed happily that the length of segment BE was also 5. Therefore triangle BEO was isosceles. The students also noted that angles BEO and EOA were alternate interior angles of parallel line segments. Since angle BEO was equal to each of angles BOE and AOE, the latter two angles were equal and we were done.

Figure 5

Figure 5 is the generalized version of Figure 4. Notice that angle EOF is half of angle BOA, and that the slope of angle EOF is the fraction

EF

OF
= rise

run + hypotenuse
.

This let us build Chart 2.

While the students happily constructed entries, a few clever ones noticed from the top row that 2arctan(1/3) = arctan(3/4) and from the bottom row that 2arctan(3/4) = arctan(24/7), from which one may conclude that 4arctan(1/3) = arctan(24/7).

This struck us (teachers) as looking a bit like the identity (!!!). If only we wanted the value of arctan(24/7). What we really wanted, to make some headway in (!!!), was the value of arctan 1 ... and we didn’t actually want it; we knew it was 45o. What we wanted was an expression equivalent to it, in terms of slopes of smaller angles. It occurred to us to bisect, twice, the 45-degree angle. It then occurred to us that we tried that in Chart 2 and since 1-1-\(\sqrt 2\) isn’t a Pythagorean Triple, we got messy numbers very quickly. A happy compromise would be to find a practically isosceles right triangle, whose sides form a Pythagorean triple. Twice bisecting this practically 45-degree angle will give us practically what we want, and perhaps we can adjust later (do you hear the sound of arctan(1/239)?).

Practically isosceles right triangles, hmmm... the only ones with legs that are consecutive integers that are commonly known in our math circles are 3-4-5, 20-21-29, and 119-120-169. One can use natural number values of m and n in the expressions m2 - n2, 2mn, and m2 + n2 to generate Pythagorean Triples, and the pairs (m=2, n=1), (m=5, n=2), and (m=12, n=5) generate those triples above.

Dear Professor Greitzer - Practically Isosceles Triangles

Author(s): 
Joe Richards and Don Crossfield

In Figure 6 we show how to rotate a practically isosceles right triangle and create a larger REALLY isosceles right triangle.

Figure 6

Practically isosceles right triangle OAB, whose legs are a and a+1, rotated 90o around O, creates OA’B’. Triangle OBB’ is a right isosceles triangle, because we rotated 90 degrees and OB=OB’. Since angle OB’A’ was just slightly over 45o (note that side OA’ is the longer leg), and angle OB’B is exactly 45o, angle A’B’B must be the little adjuster, the amount we subtract from OB’A’ to make the slightly over 45o into the exactly 45o OB’B. What’s very neat is that the slope of segment B’B is easy to read. To go from B’ to B, you drop 1 (because the legs were consecutive integers), and you run an amount equal to the sum of the legs.

We now know the adjuster angles for each of the practically isosceles triangles listed on page 4.

From the 3-4-5 triangle we get 45o = arctan(4/3) - arctan(1/(4+3)) or

$${\frac{\pi}{4}}=\arctan{\frac{4}{3}}-\arctan{\frac{1}{7}},$$

and from the observations that produced Chart 2, where we bisected angles, we know that arctan(4/3) = 2arctan(1/2), so we may write

$${\frac{\pi}{4}}=2\arctan{\frac{1}{2}}-\arctan{\frac{1}{7}}.$$

This starts to look like (!!!). From the 20-21-29 triangle, we get 45o = arctan(21/20) - arctan(1/(21+20)) or

$${\frac{\pi}{4}}=\arctan{\frac{21}{20}}-\arctan{\frac{1}{41}},$$

and, from Chart 2, we know that arctan(21/20) = 2 arctan(3/7), so we may write

$${\frac{\pi}{4}}=2\arctan{\frac{3}{7}}-\arctan{\frac{1}{41}}.$$

From the 119-120-169 triangle, we get 45o = arctan(120/119) - arctan(1/(120+119)) or

$${\frac{\pi}{4}}=\arctan{\frac{120}{119}}-\arctan{\frac{1}{239}},$$

and we look to Chart 2 to find the slope that goes with an angle half as large as that whose associated slope is 120/119.

It’s not there; we didn’t build the chart that large. But we know how to find it, and we are encouraged that the result will be rational, because 119-120-169 is a Pythagorean triple. So we add an additional row to the chart, which you can read in Chart 3.

Aha! We already bisected the angle associated with a 5/12 slope in that same chart, and its terminal side has slope 1/5. So arctan(120/119) = 2arctan(5/12), and arctan(5/12) = 2arctan(1/5). Therefore we have arctan(120/119) = 4arctan(1/5) and, since 45o = arctan(120/119) - arctan(1/239), we have our result:

$${\frac{\pi}{4}}=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{239}} \quad {\rm !!!}$$

Did our classes walk the entire road? Truthfully, no ... but they walked a good deal of it and there’s nothing they couldn’t have walked; it’s just elementary geometry that gradually became couched in higher mathematics vocabulary. How did Machin find his (!!!)? We don’t know, but in the spirit of exploration, this road gets there pretty quickly, with fruitful side roads, and needs no “higher mathematics”. Professor Greitzer, we enjoyed reading your articles for many years. We hope you get the opportunity to enjoy reading this one.

P.S. The Arbelos article also contained this formula:

$${\frac{\pi}{4}}=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{70}}+\arctan{\frac{1}{99}}\quad {\rm !!}$$

The interested reader will find all of the necessary tools to “discover” this formula in the present article.

Dear Professor Greitzer - Sam Greitzer

Author(s): 
Joe Richards and Don Crossfield

Dr. Sam Greitzer was a professor of mathematics at Rutgers University in the mid 1900s. He started the USAMO, a sequel to the American High School Mathematics Exam, in 1971, and was the guiding force behind the organization of the Mathematical Olympiad Program in 1974. Rutgers currently offers the Samuel Greitzer Award to an outstanding graduating mathematics major at their Newark campus.

Greitzer wrote numerous works, perhaps the most famous being one co-authored with H.M.S. Coxeter on Geometry. A Google search will link his name with a large, but finite, number of publications. One of two that deserve a special mention here is "Credit where credit is due," Mathematics Teacher, 60 (1967), pp. 155-156. It is a quick summary of many historical events that a teacher could talk about when introducing a new topic to a class. Besides the 32 historical tidbits, the article also gives the reader sources where more information can be found concerning each event. A peer review called it “A good article for future high school teachers.”

The second of the two is a magazine, called Arbelos, which was published during the 1980s, for “Precollege Philomaths”. Contact MAA American Mathematics Competitions (MAA AMC: http://amc.maa.org/) to inquire about purchasing copies.

The six volumes in this series are of special interest to students preparing for free-response examinations in mathematics. A peer review of these issues says that “They especially help to make USA students competitive in geometry within the international community.” We have found numerous intriguing mathematical tidbits within its pages, one of which prompted our article.