Ratio boxes

I worked up a short new chapter for my Geometry book. It's about a device I'm calling ratio boxes, for want of a better word. When we have similar triangles, we have equal ratios of sides.

An example:

Above we have three similar right triangles, so we write down the sides in order from smallest to largest, and then repeat, going through each triangle in order.

The trick is that any four entries making a rectangle are a valid ratio from this data.

In particular, I'm hoping you may be able to see a quick proof of Pythagoras's Theorem.

There are several more examples. The most complicated is one from Inversive Transformation in a circle. The rule for the transformation is OA times OA' = r^2, where r is the radius of the circle with the solid line.

As we work through the example, you should be able to see how the ratio boxes dramatically simplify the bookkeeping involved in the proof. The chapter is on my Dropbox as a pdf.

The theorem is one of my very favorites.

Napoleon's Theorem

Napoleon's Theorem is a theorem some attribute (naturally enough) to Napoleon.

It says that if you take any triangle and draw equilateral triangles on each side, then the incenters of those triangles form a fourth equilateral triangle.

There is a variant in which the new triangles are drawn as reflections of the other ones, that is, inside the original triangle.

There is a terrific vector proof that I diagram here. (I think I got the idea for the proof from Alexander Bogomolny, but I can't find it at the moment. Wonderful site).

Define vectors for paths to and from the incenters based on the following.

Then apply a simple test for the adjacent sides of an equilateral triangle:

The details depend on the definition of the direction of rotation, and the path taken around the putative equilateral triangle. Details in the links below. Here is a variant of the problem:

My write-up is here. Probably the neatest thing is we get the variant basically for free, once the setup is done. I also (finally) got a proof on ProofWiki here as well as the variant (here)

Archimedes and the Broken chord

The theorem of the "broken chord" is ascribed to Archimedes, although his original work has been lost. It was analyzed by the Arabic mathematician Al-Biruni in his Book on the Derivation of Chords in a Circle.

[UPDATE: I have made a translation and commentary of a German translation of this book. It is here. ]

Here is the general setup:

Let A and C be any two points on a circle. Let M be equidistant from both so that arc AM is equal to arc MC. Let B be another point on the circle, lying between A and M, so that AB < BC.

Drop the perpendicular from M to F on BC.

We claim that AB + BF = FC.

I will not spoil the fun by giving the proofs here. But these are eight constructions I know about.

Draw E such that AB = EC. (As an alternative approach, draw E such that BF = FE).

Draw the rectangle such that H is on the circle.

Extend BC such that DF = FC.

Draw E such that BF = FE and D such that BM = MD.

Draw E such that BF = FE and extend ME to G.

Extend BC and MF and draw DAG colinear.

At this point, I discovered a German translation of al Biruni's book (by Suter, link below). Therefore, I switched notation to match his figures. I can select the text in Preview, then Google Translate does a good job with it.

Extend BG as shown.

Extend the perpendicular DE as shown. Draw AG. Draw the diameter DK. (Hint: DK is perpendicular to AG).

Sources: Drakaki, al Biruni, Suter. There is a chapter in my geometry book on this. The chapter as a pdf is here, and the github repo for the book is here.

Killing geometry

I think it's fair to say that math is not my granddaughter's favorite subject. The whole debate about whether some people are inherently good at math and some are not, is for another day. It is probably relevant that she is using online materials to learn.

I've been excited because she's starting geometry, and I really like the subject. So I am presented with this (I'm reconstructing) as the first problem.

There are so many things wrong with this that it's hard to know where to start. The biggest one is that she has not previously seen a problem like this being solved. The idea seems to be that students learn best when they figure everything out for themselves. Naturally, she's lost.

The second major issue is that whoever designed this curriculum thinks that in studying geometry, the student should spend most of the time practicing the skills from previous years. Hence the injection of algebra and arithmetic into this problem, where it really does not belong.

Beyond that, there is a misplaced emphasis on exact calculation, as if the measure of angles is the heart of the subject.

And there is a pedantic distinction between the name of an angle and its measure. Granted, this is a distinction worth being made, but then, move on. There is no harm and great simplification in using the name to refer to both things.

This creeps into the discussion in other ways. In the next problem, the phrase linear pair is insisted upon, as if distinguishing the case where two angles add up to two right angles (I mean, 180°) really matters. It's that misplaced emphasis on calculation again.

They insist on using the classical notation invented by the Greeks. As everyone knows, it's confusing to constantly refer back to a diagram and then say, now was that angle ABC or CBD? It is so much better to use θ and φ, or even, gasp, s and t. Having the right notation frees the mind to think about what's important.

The geometry content of this question can be reduced to restating the definition: to bisect an angle means to cut it in half. The two resulting parts have equal measure. Even better, show how the construction can be done, and then, have a discussion about why it works.

Now, that's worth talking about.

Acheson's Geometry

One of my favorite books is David Acheson's The Wonder Book of Geometry (Amazon here).

I especially enjoyed the proof that similar right triangles have equal ratios of sides. Here is how I might expand it in a slightly.

Draw a rectangle and then add one of the diagonals.

This forms two right triangles which are congruent (by SSS or SAS).

Any rectangle is divided by its diagonal into equal areas above and below the line.

Next, introduce a point on the diagonal and draw two lines, one vertical and one horizontal. This forms more rectangles. Divide each of them along their diagonals (which lie along the original one).

All six triangles in the figure are similar, having all three angles equal. (Prove this using some combination of vertical and complementary angles and the alternate interior angles theorem).

As before, the two triangles shaded blue have equal area, as do the two shaded red.

The key step is to realize that since the whole area above the diagonal in the original rectangle is equal to that below, light blue is equal to dark blue, and light red is equal to dark red, the remaining areas are also equal.

tall skinny rect + light blue + light red = short fat rect + red + blue.

Coloring one of those sub-rectangles for clarity, we have shown (in other words) that white is equal to gray in the figure below.

The area in white is Ab and that in gray is aB. Equate them to obtain Ab = aB, and then divide, giving A/a = B/b.

Also, since A/a + 1 = (A + a)/a = B/b + 1 = (B + b)/b, either of the smaller triangles has the same ratios as the big ones that span the entire original rectangle.

Here is his figure:

From there, it is not too difficult to derive the Pythagorean theorem.

Except that first we need the converse theorem, which seems a bit tricky.

However, playing with the ratio above, we see that if Ab = aB, then not only A/a = B/b but also A/B = a/b.

Let A/a = k = B/b. Then we have that A = ka, and B = kb.

Place two right triangles with equal ratios of sides next to each other, and grow the small one by a factor of k by extending the base, preserving the acute angle at the base. Say we set the new length of the base to be ka.

Then, by the forward theorem we still have equal ratios, meaning that the height is equal to kb, which as we saw is equal to B. Therefore the top vertices superimpose at the same point.

Therefore, the two triangles are congruent, and equality of angles follows. Since we maintain equality of two of three angles in growing the triangle, we preserve all three.

Here is a proof without words for the Pythagorean theorem, growing triangles in the same fashion:

By a simple extension, the general result can be proved, all angles equal means equal ratios of sides, for any triangle.

Update: I realize now that the last two examples depend on extending the equal ratios result to the hypotenuse of the similar right triangles. Of course, one can use the Pythagorean theorem and do some algebra with k^2. (Relying on Euclid's famous I.47 which uses SAS).

Alternatively, here's a nice simple proof. In any right triangle, drop the altitude to the hypotenuse, h. This forms two more similar triangles.

Form the ratio of the longer side (not the hypotenuse) to the shorter side in each of the three similar triangles: h/x = y/h = b/a. But b/a is also the ratio of the hypotenuse when comparing the medium and small triangles, and the equality says that this ratio is the same as h/x the ratio of the short sides comparing the same triangles. This completes the proof.

Pizza theorem

In Acheson's The Wonder Book of Geometry, I came across the "pizza theorem."

The basic form of the problem says to take any point within a disk, and construct a grid of chords which all cross at that point. There are two pairs of perpendicular chords, evenly spaced at 45 degrees.

Alternate radial sections are shaded. The result is that the pizza is always evenly divided between light and dark pieces, even though the pieces themselves are oddly shaped.

Here is Acheson's figure.

If the grid coincides with the center of the circle, the theorem is obviously correct. And if the grid moves by sliding along a diagonal of the circle, for example vertically, the result is still correct. That's because the figure has mirror-image symmetry.

After a vertical move to move to the desired final radial position, the grid can be either rotated or moved "horizontally", parallel to a chord but not on a diagonal of the circle. I worked out a geometric proof that this movement does not change the relative areas.

The basic idea is that the areas gained and lost by the movement go like the difference t-s for the two parts of each chord. (Except that the area covered by the vertical chord is fatter by sqrt(2)).

But that difference is closely related to the distance from the center of the chord to the grid center. The sum of those is easily shown to be invariant, because the center of each chord lies on its vertical bisector, which passes through the center of the circle.

I also found on the web a nice geometric proof for invariance under rotation (first answer at the link). I expanded it to make it (hopefully) even clearer (pdf).

I also added a chapter to my geometry book (38 MB). There are shorter versions elsewhere that I linked previously, but they are now outdated.

Sum of angles: Ptolemy

 The "sum of angles" theorems are incredibly useful in calculus.  These are formulas for the sine (or cosine) of the sum (or difference) of two angles.  There are a number of derivations, of which my favorite is a proof-without-words (here).

Recently, I got interested in Ptolemy's theorem, which is illustrated in this graphic:

There are proofs based on similar triangles (a bit complicated), and one based on areas with a wonderful trick in the middle.  Another terrific proof-without-words for that is here.

It turns out that there are easy proofs for the trigonometric theorems starting from Ptolemy's theorem, and I want to look at them here.  The first one is for sin s+t:

The proof depends on the following details.  First, the black line in the middle is a diameter of the circle, scaled to have length 1.  As a result, the two triangles are right triangles, justifying the labels on the sides.  Ptolemy says to multiply two pairs of opposite sides and add them.

In addition, the dotted line is the chord of the sum of the two angles, and its length is:  2R sin s+t = L.  But 2R is scaled to be 1, so finally we have the formula for the sine of s+t.  It basically writes itself.

The second one is for the difference.  This is extra, since the easiest way to get from the sum to the difference is to use the fact that cosine is an even function and sine is odd:  sin(x) = -sin(-x).

If you're trying to set this up yourself, a big clue is that since the formula has a minus sign in it (sin s cos t minus sin t cos s), the sin s-t term is going to be one of the sides, and naturally, it is the side opposite the diameter.

As for the cosine formulas, there is yet another trick.  We will still have that the dotted line is going to be the sine of some angle θ, but it is also the cosine of the angle that is complementary to θ!  And again, cosine s+t is one of the sides.

The last one isn't quite as pretty (I don't know if there are others, I came up with this one myself).  The hard part is writing an expression for the angle marked with the black dot, but its complement is just s-t.

Pentagons, again

 It's been some years, but I took another look at pentagons.  It is fun to see the connection between φ,  the "golden ratio", and the internal triangles in a pentagon.

First of all. the figure has five-fold rotational symmetry.

The marked angles in the panel on the left are equal by the isosceles triangle theorem, so the angles on either side of the central angle at each vertex are equal.  

Then we do some addition:  3 black + 2 magenta = 180 = 4 magenta + 1 black.  We conclude that black = mageneta.  So the vertex angle is divided into 3 equal parts.

The small triangle at the top is isosceles, because the triangles on either side are congruent to each other (by rotational symmetry, or ASA), so both sides are equal. Now, we see that the horizontal chord is parallel to the bottom side, by alternate interior angles.  We have a bunch of rhombi!

There are two types of similar triangles:  one tall and skinny, the other short and squat.

In the right-hand panel, a triangle of the second type is shown in light red.  It has equal sides of length x and a base of length x + 1.  The red one has sides of length 1 and a base of length x.  We form the equal ratios:  (x+1) /x = x.

One can also use the tall skinny triangles.

This allows the base to have length 1.  Then, the sides of the tall skinny triangle in the middle are x.  The small, tall skinny triangle at the top has sides x - 1 and base x - 2(x - 1) so the ratios of similar triangles are x = (x - 1)/(-x + 2), which simplifies to the same expression as before.

The apex angle of a tall skinny triangle is equal to 1/3 of the interior angle of the pentagon, which is 108/3 = 36.

We will find an expression for the sine of 36°, and this will give us another way to calculate the side length for the pentagon in terms of the chords.

We start by working with 18°, because 90/18 = 5 exactly.  The trick is that (using θ for 18°):  5θ = 90° so 2θ = 90 - 3θ.  Taking the sine of both sides and then converting to the complementary angle on the right, we have sin 2θ = cos 3θ.

One can also use de Moivre's formula and take the real part.

[Since φ - 1 = 1/φ, the above result can be written more simply, but it makes the calculation below more complicated].

There is something else.  If one draws the central angles of the pentagon, then the triangle containing a central angle of 72° and a base of one side length has a half angle of 36°.

So one-half the side length is the sine of 36°, which we found above.  The circle containing the five vertices for a pentagon of side length 1, has a radius of r = 1/2 sin36° = 1/sqrt(3 - φ).

The pdf of the write-up is here (Latex one level down).

Pythagorean triples

A Pythagorean triple is a set of integers a, b, and c such that

a^2 + b^2 = c^2

We want primitive triples, those where a, b and c do not have a common factor.

Euclid presents a formula for that:  take any two integers n > m > 0, then

b = 2mn
a = n^2 - m^2
c = n^2 + m^2

One more restriction:  m and n must not have any common factor, they should be co-prime, otherwise the triple will not be primitive.

It is easy to show that when a, b, and c are found in this way, they satisfy the Pythagorean equation.

The trick is to see where this comes from, and more importantly, to show that it describes all such triples.

Derivation

I found a very nice derivation in Maor (The Pythagorean theorem).  We start by investigating the properties of a, b and c.  

For example a and b cannot both be even, for in that case, c will also be even, and then there is a common factor.

Recall that any even number can be written as 2k, and any odd number as 2k + 1 for positive integer k (or k = 0 for n = 1), so it is easy to see that if n is even, so is n^2.  And if n is odd then so is n^2 as well.  Therefore an even perfect square implies an even integer square root, and an odd square implies an odd root.

In the case where a and b are both odd, c must be even, since odd plus odd equals even.  To see a problem with this re-write the equation 

(2i + 1)^2 + (2j + 1)^2 = (2k)^2

The left-hand side is not a multiple of 4, but the right-hand side is.  This is impossible.  Hence one of a or b must be even and the other odd.  Let a be odd.  Then b can be written as b = 2t for some integer t.  We have

(2t)^2 = c^2 - a^2 = (c + a)(c - a)

Every a, b and c must satisfy this equation.  Furthermore, we see that there is a factor of 4 on the left-hand side, hence we can get a divisor of 2 for each factor on the right:

t^2 = (c + a)/2 . (c - a)/2

a and c are both odd:

a = 2i + 1
c = 2k + 1

so their sum and difference are even, since

2k + 1 + 2i + 1 = 2(k + i) + 2
2k + 1 - 2i - 1 = 2(k - i)

After canceling the factor of 2 we get

(c + a)/2 = k + i + 1
(c - a)/2 = k - i

Thus both of these factors on the right-hand side are integers, while the left-hand side (t^2) is a perfect square.

Key step

Here's the last step.  We claim that these two terms do not have a common factor.  

Whenever x and y have a common factor, so do their sum and difference since x + y = fp + fq = f(p + q).  and so on.  The converse is also true.  

But the sum and difference for these terms are:

(c + a)/2 + (c - a)/2 = c
(c + a)/2 - (c - a)/2 = a

On the supposition that (c+a)/2 and (c-a)/2 did have a common factor, then they would share that common factor with both c and a.   Since we know that a and c (at least the particular ones we're interested in) do not have a common factor, neither do these two terms.

So the two terms have no common factor and yet multiply together to give a perfect square.

t cannot be prime itself (because there would not be two different factors).

So clearly there are two factors m^2 and n^2 (m and n not necessarily prime), and both terms are perfect squares!  Write:

n^2 = (c + a)/2
m^2 = (c - a)/2

n^2 + m^2 = c
n^2 - m^2 = a

And

m^2n^2 = t^2
mn = t

Since b^2 = 4t^2, b = 2t = 2mn.

These properties of m and n follow from the standard rules about factors, applied to a, b and c.  Therefore, any triple a, b, and c can be written in terms of an integer m and n by this method.

I checked this by using a Python script to search the entire space of squares below an arbitrary limit, for those which sum to give another square, keeping only those triples of squares with no common factors.

I tested each triple by taking the even member, dividing by 2, and then finding its factors.  This gives candidate pairs m and n with 2mn = b.  Then, just check for whether n^2 - m^2 = a.

The gist is here.

It is interesting to see which patterns among the triples come from which choices of m and n.

Pi again

 I was working on a post about Archimedes calculation for pi, and I saw a reference that said Ptolemy used 377/120 as a rational approximation to the value.  

And then I also read that 355/113 is an even better approximation, and it is credited to the Chinese mathematician Zu Chongzhi (5th century CE).

There is a whole article in wikipedia about this subject.  Recall that Archimedes established upper and lower bounds on pi of 22/7 (3.14285714) and 3-10/71 (3.14084507)

In decimal form, Ptolemy's value is 

377/120 = 3.14166...

This is correct to the third digit after the decimal point.  The error is 7401 parts in ten million.  And really, it is accurate enough for any practical work.

I found it curious that the other value has a smaller denominator, but was not found by Ptolemy, and yet it is better, much better.

355/113 = 3.14159292

This is correct to the sixth digit after the decimal point.  The error is 27 parts in ten million.  That's about 275 times better than Ptolemy's value.

There is some text from Pi: A Source Book, by Berggren et al online.  In a chapter by Tam Lay-Young and And Tian-Se, I found a calculation ascribed to Liu Hui (3rd century CE).

The calculation does not seem to be any kind of a technical breakthrough, it is just a basic usage of the Pythagorean theorem.  (However, there are some values given that I don't see where they came from, so maybe I'm missing something).

Start with a circle of radius 10, and inscribe a hexagon.  One of the six equilateral triangles is shown.

DC = 5.17638.

Repeat until you're satisfied.  Keep track of n, the total number of sides.

At the end, compute the area of the polygon as 1/2 times the radius times n times the chord.  Since r = 10, the area of the circle is pi times 100.

The estimate based on the dodecagon is 6 x 10 x 5.17638 divided by 100 = 3.10582854.

In the end, they obtain 314-64/625, 314-169/625 as the bounds on pi (it's not clear to me how an upper bound was obtained).

In decimal, this means that 3.141024 < pi < 3.142704, which is slightly better than Archimedes.  

Apparently, Zu Chongzhi (5th century CE) went farther, far enough to recognize that 355/113 matches an improved estimate very well.  

The wikipedia article on Zu Chongzhi claims that he is "most notable for calculating pi as between 3.1415926 and 3.1415927".  It claims that his works show he calculated pi to 6 digits using a "12,288-gon".  

That's obtained by doubling a hexagon 9 times.  Using Python the other day, we obtained the sixth digit (after the decimal point) on round 9 of hexagon doubling as well. 

The answer as to how Ptolemy missed it seems to be that he didn't know the true value of pi well enough to believe that 355/113 was any better than 377/120.  The other possibility is that he was led to his value by some logic, and didn't check nearby values systematically.

I wrote a Python program to find rational approximations to (irrational) numbers.  The gist is here.

355/113 is a spectacular success.  

No other value comes close for a long time.  Aside from 355/113, every other fraction that becomes a better approximation is a very gradual improvement on what came before.  The first fraction to be a better approximation for pi than 355/113 is 52163/16604.

A Dedekind cut has to land somewhere.  I suppose 355/113 just lands particularly close to pi by accident.

Here is a plot of the errors.  The x-axis is the denominator.  The log of the absolute value of the difference from pi is plotted for all fractions that are closest to pi for a given denominator and also in lowest terms.  (gist).  You can see how special 355/113 is, and it stays that way (not shown).

I've written up the math for the area and perimeter methods for estimating pi and compared it to Archimedes (here).

Archimedes and pi

In this post we take a look at Archimedes' approximation of the value of pi.  I'll try to make things as simple as I possibly can, but no simpler.

Here is the fundamental idea.  We will approximate the distance around a circle, called its circumference, by the distance around a regular polygon that just fits inside the circle, called its perimeter.  The points or vertices of the polygon will lie on the circle.

                                                 

It doesn't actually matter what kind of a polygon it is (i.e. how many sides there are), but it has to be regular, with sides all of the same length.  

For an inscribed polygon, the perimeter will be smaller than the circumference of the circle.

But the trick is, there is a way to compute the new perimeter for a polygon with twice the number of sides, 2n, given the original perimeter for n sides.  

Let's look at that what that means physically.  Actually, I'm going to show below just a small arc of the circle.  One side of our initial polygon will be colored red, and the chord it makes is the red line in the figure below.  The white area shows that it doesn't track very closely with the circle, and its length will be smaller than the length of this arc of the circle.

But then, when we double the number of sides, we get a polygon of 2n sides, and its perimeter follows the two blue lines.  This is clearly more like the circumference, there is less white space.  One side has become 2.

Double the number of sides again, to 4n.  The result is the magenta perimeter.  I hope you can see that as the number of sides gets larger, the perimeter of the polygon becomes a better and better approximation to the circumference of the circle.

This method is great, if you can find a way to express the new perimeter in terms of the old one each time. This is called the method of exhaustion and it was probably invented by a Greek mathematician named Eudoxus who lived before Archimedes.

To make things easier, we will only compute the lower limit.  That is, we will use inscribed polygons and compute their perimeters as we double the number of sides again and again.  

Python will do the actual computation.  Archimedes has this estimate in the end as 3-10/71 or about 3.140845.   Let's see how we compare.

Remember that pi is the ratio of the circumference of a circle, C, to its diameter.  If we use a circle whose diameter is equal to 1, then the circumference C will be equal to pi.  

The procedure we follow here is to inscribe a hexagon into a circle of diameter 1, and then calculate the length of the perimeter of the hexagon.  We do this because a hexagon's perimeter is particularly easy to find.

We will then derive a simple method to calculate the length of the perimeter after the number of sides is doubled (to give a 12-sided polygon).  This method will be applied repeatedly.  Archimedes reached 96 sides.  That is 4 doublings (6 - 12 - 24 - 48 - 96).

What he achieved is not just an estimate for the value of pi, it is an algorithm to compute the value of pi to any desired accuracy.

Begin with the hexagon.  A hexagon can be viewed as a group of six equilateral triangles.  Since the diameter of the circle will be 1, we choose the length of each triangle's side as 1/2.  This matches the diameter, made from two sides, and it results in a total perimeter of 6 times 1/2 or just 3.  

3 is our first lower bound for pi.

I introduced the side length of 1/2 above in order to see, even before we really get started, what the perimeter of the hexagon is and give a first estimate for the value of pi.  I must ask you to set that value aside for a minute.

We will need two ratios of sides in a right triangle, and it will make the math slightly easier if we scale the next equilateral triangle up to have a side length of 2 (we'll scale it back later, all we need are the ratios).  

Obtain a right triangle by cutting an equilateral triangle of side length 2 in half, so the base of the halved triangle is 1.  Pythagoras gives us the other side as the square root of 3, since 1^2 + 3 = 2^2.

Since the original triangle was equilateral, with angles of 60 degrees each, the halved triangle has one angle of 30 degrees.  Archimedes would call that measure one-third of a right angle, and the essential ratios for him would be 2 to 1 (hypotenuse to short side), and the square root of 3 to 1.

We have the angle that we're going to start with, 30 degrees, and we have two essential ratios.

The next graphic is a proof without words of Thales' famous circle theorem:  any angle inscribed in a semicircle is a right angle.  Since PQ is a diameter, the angle PRQ is a right angle.  

Proof.  the two smaller triangles formed by the radius OR are both isosceles, so by Euclid I.5 they have equal base angles.  Since the total of all the angles in a triangle is equal to two right angles and angle PRQ is one-half that, the result follows.

We're going to choose the angle P to have the particular value of 30 degrees.  Then the complementary angle Q is 60 degrees, and RQ can be viewed as one side of an inscribed hexagon.  (We won't prove it, but the peripheral angle P is one-half the measure of the central angle ROQ which cuts off the same arc of the circle).  That makes ROQ equilateral.

The ratios that we had above will apply, because P is a 30 degree angle in a right triangle.  So the ratio of PQ to RQ is 2, and the ratio of PR to RQ is the square root of 3.

We said that we will inscribe a hexagon into circle.  I don't actually have a picture of that, but I do have a picture of an octagon, so I'll show that, and ask you to use your imagination.

Let us just focus on one of the six sides.  Suppose ROQ is an equilateral triangle, as we had before, so that RQ is one of the six sides of an inscribed hexagon.  We approximated the value of pi by finding the length of RQ (in diameter units) and then multiplying by n = 6.

Our procedure is going to be as follows:  at each step we will halve the angle at P, then compute the new length SQ.  After one cycle, SQ is one side of a 12-sided polygon, so we multiply by 12 to get an approximation for pi.

To carry out this program, we need one more preliminary result.  It is called the angle bisector theorem.  This theorem provides a way to start with the special ratios for any angle, and find the same ratios for one-half that angle.

The line segment b bisects the angle at the left, so the two half-angles labeled theta are equal.  In the upper triangle, we draw the altitude to side c and label it d.  We can do this because the larger of the two triangles formed by the altitude is congruent to the triangle in red with sides a,b,d.  

Next, the small triangle with sides e and d at the top right is similar to the whole starting triangle with sides a,f,c (both are right triangles with one smaller angle gamma).  Similar triangles gives us this ratio:

e/d = c/a

Now add 1 to both sides

e/d + d/d = c/a + a/a

f/d = c/a + a/a

a/d = c/f + a/f

On the right-hand side we have what are called in trigonometry the cosecant and cotangent of the angle 2 theta.  For the case of 2 theta equal to 30 degrees, these are the ratios from above:  c/f = 2 and a/f = square root of 3.  The left-hand side is the cotangent of the half-angle, theta, which is obtained by simple addition.  So the cotangent of 15 degrees is 2 + square root of 3.

Let's just check:

To get the other ratio for the half-angle (b/d), recall that our new triangle has sides a,b,d, and we know the ratio a/d.  But we also know from the Pythagorean theorem that

a^2 + d^2 = b^2 

so

b/d = sqrt(1 + a^2/d^2)

And that's it!  

Going back to this figure

We have that PR/RQ = square root of 3 and PQ/RQ = 2.  We compute the ratios for the half-angle, PS/SQ and PQ/SQ using the method laid out above. 

To get an estimate for pi, invert the last ratio to get SQ/PQ, recall that PQ is equal to 1, and multiply by the number of sides.  We've doubled the number of sides so n = 12. 

The essential code  is:

I set it up as a generator, which is initialized with the values of sine and cosine for 30 degrees.  A gist for the whole program is here.

Here's a screenshot of the results:

The fifth entry gives the result from four doublings as 3.141032.  Archimedes has this estimate in the end as 3-10/71 or about 3.140845.  His lower bound is smaller than ours, because we've used a powerful computer to do very accurate calculations.

One interesting aspect of the whole story is to think about how he may have obtained a rational approximation to the square root of three (all of his calculations were with rational numbers).  That's for another day.

Many thanks to the author of this page and to Neil Carothers.  (His pages have since been disappeared by his university, which hosted them, you may be able to find them by using the wayback machine).  The  post you're reading does things the way Archimedes did them, but there is a simpler formula using the values for the perimeter itself (see my pdf).