Proving That Disturbing Triangle Theorem That Isn’t Morley’s Somehow

I couldn’t leave people just hanging on that triangle theorem from the other day. Tthis was a compass-and-straightedge method to split a triangle into two shapes of equal area. The trick was you could split it along any point on one of the three legs of the triangle.

The theorem unsettled me, yes. But proving that it does work is not so bad and I thought to do that today.

The process: start with a triangle ABC. Pick a point P on one of the legs. We’ll say it’s on leg AB. Draw the line segment from the other vertex, C, to point P.

Now from the median point S on leg AB, draw the line parallel to PC and that intersects either leg AC or leg BC. Label that point R. The line segment RP cuts the triangle ABC into one triangle and another shape, normally a quadrilateral. Both shapes have the same area, half that of the original triangle.

To prove it nicely will involve one extra line, and the identification of one point. Construct the line SC. Lines SC and PC intersect at some point; call that Q. I’ve actually made a diagram of this, just below. I’ve put the intersection point R on the leg AC. All that would change if the point R were on BC instead would be some of the labels.

Here’s how the proof will go. I want to show triangle APR has half the area of triangle ABC. The area of triangle ARP has to be equal to the area of triangle ASC, plus the area of triangle SPQ, minus the area of triangle QCR. So the first step is proving that triangle ASC has half the area of triangle ABC. The second step is showing triangle SPQ has the same area as does triangle QCR. When that’s done, we know triangle APR has the same area as triangle ASC, which is half that of triangle ABC.

First. That ASC has half the area of triangle ABC. The area of a triangle is one-half times the length of a base times its height. The base is any of the three legs which connect two points. The height is the perpendicular distance from the third point to the line that first leg is on. Here, take the base of triangle ABC to be the line segment AC. Also take the base of triangle ASC to be the line segment AC. They have the same base. Point S is the median of the line segment AB. So point S is half as far from the base AC as the point B is. Triangle ASC has half the height of triangle ABC. Same base, half the height. So triangle ASC has half the area of triangle ABC.

Second. That triangle SPQ has the same area as triangle QCR. This is going to be most easily done by looking at two other triangles, SPC and PCR. They’re relevant to triangles SPQ and QCR. Triangle SPC has the same area as triangle PCR. Take as the base for both of them the leg PC. Point S and point R are both on the line SR. SR was created parallel to the line PC. So the perpendicular distance from point S to line PC has to be the same as the perpendicular distance from point R to the line PC. Triangle SPC has the same base and same height as does triangle PCR. So they have the same area.

Now. Triangle SPC is made up of two smaller triangles: triangle SPQ and triangle PCQ. Its area is split, somehow, between those two. Triangle PCR is also made of two smaller triangles: triangle PCQ and triangle QCR. Its area is split between those two.

The area of triangle SPQ plus the area of triangle PCQ is the same as the area of triangle SPC. This is equal to the area of triangle PCR. The area of triangle PCR is the area of triangle PCQ plus the area of triangle QCR.

And that all adds up only if the area of triangle SPQ is the same as the area of triangle QCR.

So. We had that area of triangle APR is equal to the area of triangle ASC plus the area of triangle SPQ minus the area of triangle QCR. That’s the area of triangle ASC plus zero. And that’s half the area of triangle ABC. Whatever shape is left has to have the remaining area, half the area of triangle ABC.

It’s still such a neat result.

Morley’s theorem, by the way, says this: take any triangle. Trisect each of its three interior angles. That is, for each vertex, draw the two lines that cut the interior angle into three equal spans. This creates six lines. Take the three points where these lines for adjacent angles intersect. (That is, draw the obvious intersection points.) This creates a new triangle. It’s equilateral. What business could an equilateral triangle possibly have in all this? Exactly.

In Which I Am Disturbed By A Triangle Theorem That Isn’t Morley’s Somehow

I’ve been reading Alfred S Posamentier and Ingmar Lehmann’s The Secrets of Triangles: A Mathematical Journey. It is exactly what you’d think: 365 pages, plus endnotes and an index, describing what we as a people have learned about triangles. It’s almost enough to make one wonder if we maybe know too many things about triangles. I admit letting myself skim over the demonstration of how, using straightedge and compass, to construct a triangle when you’re given one interior angle, the distance from another vertex to its corresponding median point, and the radius of the triangle’s circumscribed circle.

But there are a bunch of interesting theorems to find. I wanted to share one. When I saw it I felt creeped out. The process seemed like a bit of dark magic, a result starting enough that it seemed to come from nowhere. Here it is.

Start with any old triangle ABC. Without loss of generality, select a point along the leg AB (other than the vertices). Call that point P. (This same technique would work if you put your point on another leg, but I would have to change the names of the vertices and line segments from here on. But it doesn’t matter what the names of the vertices are. So I can suppose that I was lucky enough that whatever leg you put your point P on I happened to name AB.)

Now. Pick the midpoint of the leg AB. This median is a point we’ll label S.

Draw the line PC.

Draw the line parallel to the line PC and which passes through S. This will intersect either the line segment BC or the line segment AC. Whichever it is, label this point of intersection R.

Draw the line from R to P.

The line RP divides the triangle ABC into two shapes, a triangle and (unless your P was the median point S) a quadrilateral.

The punch line: both shapes have half the area of the original triangle.

I usually read while eating. This was one of those lines that made me put the fork down and stare, irrationally angry, until I could work through the proof. It didn’t help that you can use a technique like this to cut the triangle into any whole number you like of equal-area wedges.

I’m sure this is old news to a fair number of readers. I don’t care. I haven’t noticed this before. And yes, it’s not as scary weird magic as Morley’s Theorem. But I’ve seen that one before, long enough ago I kind of accept it.

The End 2016 Mathematics A To Z: Yang Hui’s Triangle

Today’s is another request from gaurish and another I’m glad to have as it let me learn things too. That’s a particularly fun kind of essay to have here.

Yang Hui’s Triangle.

It’s a triangle. Not because we’re interested in triangles, but because it’s a particularly good way to organize what we’re doing and show why we do that. We’re making an arrangement of numbers. First we need cells to put the numbers in.

Start with a single cell in what’ll be the top middle of the triangle. It spreads out in rows beneath that. The rows are staggered. The second row has two cells, each one-half width to the side of the starting one. The third row has three cells, each one-half width to the sides of the row above, so that its center cell is directly under the original one. The fourth row has four cells, two of which are exactly underneath the cells of the second row. The fifth row has five cells, three of them directly underneath the third row’s cells. And so on. You know the pattern. It’s the one that pins in a plinko board take. Just trimmed down to a triangle. Make as many rows as you find interesting. You can always add more later.

In the top cell goes the number ‘1’. There’s also a ‘1’ in the leftmost cell of each row, and a ‘1’ in the rightmost cell of each row.

What of interior cells? The number for those we work out by looking to the row above. Take the cells to the immediate left and right of it. Add the values of those together. So for example the center cell in the third row will be ‘1’ plus ‘1’, commonly regarded as ‘2’. In the third row the leftmost cell is ‘1’; it always is. The next cell over will be ‘1’ plus ‘2’, from the row above. That’s ‘3’. The cell next to that will be ‘2’ plus ‘1’, a subtly different ‘3’. And the last cell in the row is ‘1’ because it always is. In the fourth row we get, starting from the left, ‘1’, ‘4’, ‘6’, ‘4’, and ‘1’. And so on.

It’s a neat little arithmetic project. It has useful application beyond the joy of making something neat. Many neat little arithmetic projects don’t have that. But the numbers in each row give us binomial coefficients, which we often want to know. That is, if we wanted to work out (a + b) to, say, the third power, we would know what it looks like from looking at the fourth row of Yanghui’s Triangle. It will be $1\cdot a^4 + 4\cdot a^3 \cdot b^1 + 6\cdot a^2\cdot b^2 + 4\cdot a^1\cdot b^3 + 1\cdot b^4$. This turns up in polynomials all the time.

Look at diagonals. By diagonal here I mean a line parallel to the line of ‘1’s. Left side or right side; it doesn’t matter. Yang Hui’s triangle is bilaterally symmetric around its center. The first diagonal under the edges is a bit boring but familiar enough: 1-2-3-4-5-6-7-et cetera. The second diagonal is more curious: 1-3-6-10-15-21-28 and so on. You’ve seen those numbers before. They’re called the triangular numbers. They’re the number of dots you need to make a uniformly spaced, staggered-row triangle. Doodle a bit and you’ll see. Or play with coins or pool balls.

The third diagonal looks more arbitrary yet: 1-4-10-20-35-56-84 and on. But these are something too. They’re the tetrahedronal numbers. They’re the number of things you need to make a tetrahedron. Try it out with a couple of balls. Oranges if you’re bored at the grocer’s. Four, ten, twenty, these make a nice stack. The fourth diagonal is a bunch of numbers I never paid attention to before. 1-5-15-35-70-126-210 and so on. This is — well. We just did tetrahedrons, the triangular arrangement of three-dimensional balls. Before that we did triangles, the triangular arrangement of two-dimensional discs. Do you want to put in a guess what these “pentatope numbers” are about? Sure, but you hardly need to. If we’ve got a bunch of four-dimensional hyperspheres and want to stack them in a neat triangular pile we need one, or five, or fifteen, or so on to make the pile come out neat. You can guess what might be in the fifth diagonal. I don’t want to think too hard about making triangular heaps of five-dimensional hyperspheres.

There’s more stuff lurking in here, waiting to be decoded. Add the numbers of, say, row four up and you get two raised to the third power. Add the numbers of row ten up and you get two raised to the ninth power. You see the pattern. Add everything in, say, the top five rows together and you get the fifth Mersenne number, two raised to the fifth power (32) minus one (31, when we’re done). Add everything in the top ten rows together and you get the tenth Mersenne number, two raised to the tenth power (1024) minus one (1023).

Or add together things on “shallow diagonals”. Start from a ‘1’ on the outer edge. I’m going to suppose you started on the left edge, but remember symmetry; it’ll be fine if you go from the right instead. Add to that ‘1’ the number you get by moving one cell to the right and going up-and-right. And then again, go one cell to the right and then one cell up-and-right. And again and again, until you run out of cells. You get the Fibonacci sequence, 1-1-2-3-5-8-13-21-and so on.

We can even make an astounding picture from this. Take the cells of Yang Hui’s triangle. Color them in. One shade if the cell has an odd number, another if the cell has an even number. It will create a pattern we know as the Sierpiński Triangle. (Wacław Sierpiński is proving to be the surprise special guest star in many of this A To Z sequence’s essays.) That’s the fractal of a triangle subdivided into four triangles with the center one knocked out, and the remaining triangles them subdivided into four triangles with the center knocked out, and on and on.

By now I imagine even my most skeptical readers agree this is an interesting, useful mathematical construct. Also that they’re wondering why I haven’t said the name “Blaise Pascal”. The Western mathematical tradition knows of this from Pascal’s work, particularly his 1653 Traité du triangle arithmétique. But mathematicians like to say their work is universal, and independent of the mere human beings who find it. Constructions like this triangle give support to this. Yang lived in China, in the 12th century. I imagine it possible Pascal had hard of his work or been influenced by it, by some chain, but I know of no evidence that he did.

And even if he had, there are other apparently independent inventions. The Avanti Indian astronomer-mathematician-astrologer Varāhamihira described the addition rule which makes the triangle work in commentaries written around the year 500. Omar Khayyám, who keeps appearing in the history of science and mathematics, wrote about the triangle in his 1070 Treatise on Demonstration of Problems of Algebra. Again so far as I am aware there’s not a direct link between any of these discoveries. They are things different people in different traditions found because the tools — arithmetic and aesthetically-pleasing orders of things — were ready for them.

Yang Hui wrote about his triangle in the 1261 book Xiangjie Jiuzhang Suanfa. In it he credits the use of the triangle (for finding roots) as invented around 1100 by mathematician Jia Xian. This reminds us that it is not merely mathematical discoveries that are found by many peoples at many times and places. So is Boyer’s Law, discovered by Hubert Kennedy.

The End 2016 Mathematics A To Z: Jordan Curve

I realize I used this thing in one of my Theorem Thursday posts but never quite said what it was. Let me fix that.

Jordan Curve

Get a rubber band. Well, maybe you can’t just now, even if you wanted to after I gave orders like that. Imagine a rubber band. I apologize to anyone so offended by my imperious tone that they’re refusing. It’s the convention for pop mathematics or science.

Anyway, take your rubber band. Drop it on a table. Fiddle with it so it hasn’t got any loops in it and it doesn’t twist over any. I want the whole of one edge of the band touching the table. You can imagine the table too. That is a Jordan Curve, at least as long as the rubber band hasn’t broken.

This may not look much like a circle. It might be close, but I bet it’s got some wriggles in its curves. Maybe it even curves so much the thing looks more like a kidney bean than a circle. Maybe it pinches so much that it looks like a figure eight, a couple of loops connected by a tiny bridge on the interior. Doesn’t matter. You can bring out the circle. Put your finger inside the rubber band’s loops and spiral your finger around. Do this gently and the rubber band won’t jump off the table. It’ll round out to as perfect a circle as the limitations of matter allow.

And for that matter, if we wanted, we could take a rubber band laid down as a perfect circle. Then nudge it here and push it there and wrinkle it up into as complicated a figure as you like. Either way is as possible.

A Jordan Curve is a closed curve, a curve that loops around back to itself. And it’s simple. That is, it doesn’t cross over itself at any point. However weird and loopy this figure is, as long as it doesn’t cross over itself, it’s got in a sense the same shape as a circle. We can imagine a function that matches every point on a true circle to a point on the Jordan Curve. A set of points in order on the original circle will match to points in the same order on the Jordan Curve. There’s nothing missing and there’s no jumps or ambiguous points. And no point on the Jordan Curve matches to two or more on the original circle. (This is why we don’t let the curve to cross over itself.)

When I wrote about the Jordan Curve Theorem it was about how to tell how a curve divides a plane into two pieces, an inside and an outside. You can have some pretty complicated-looking figures. I have an example on the Jordan Curve Theorem essay, but you can make your own by doodling. And we can look at it as a circle, as a rubber band, twisted all around.

This all dips into topology, the study of how shapes connect when we don’t care about distance. But there are simple wondrous things to find about them. For example. Draw a Jordan Curve, please. Any that you like. Now draw a triangle. Again, any that you like.

There is some trio of points in your Jordan Curve which connect to a triangle the same shape as the one you drew. It may be bigger than your triangle, or smaller. But it’ll look similar. The angles inside will all be the same as the ones you started with. This should help make doodling during a dull meeting even more exciting.

There may be four points on your Jordan Curve that make a square. I don’t know. Nobody knows for sure. There certainly are if your curve is convex, that is, if no line between any two points on the curve goes outside the curve. And it’s true even for curves that aren’t complex if they are smooth enough. But generally? For an arbitrary curve? We don’t know. It might be true. It might be impossible to find a square in some Jordan Curve. It might be the Jordan Curve you drew. Good luck looking.

Reading the Comics, November 12, 2016: Frazz and Monkeys Edition

Two things made repeat appearances in the mathematically-themed comics this week. They’re the comic strip Frazz and the idea of having infinitely many monkeys typing. Well, silly answers to word problems also turned up, but that’s hard to say many different things about. Here’s what I make the week in comics out to be.

Sandra Bell-Lundy’s Between Friends for the 6th introduces the infinite monkeys problem. I wonder sometimes why the monkeys-on-typewriters thing has so caught the public imagination. And then I remember it encourages us to stare directly into infinity and its intuition-destroying nature from the comfortable furniture of the mundane — typewriters, or keyboards, for goodness’ sake — with that childish comic dose of monkeys. Given that it’s a wonder we ever talk about anything else, really.

Monkeys writing Shakespeare has for over a century stood as a marker for what’s possible but incredibly improbable. I haven’t seen it compared to finding a four-digit PIN. It has got me wondering about the chance that four randomly picked letters will be a legitimate English word. I’m sure the chance is more than the one-in-a-thousand chance someone would guess a randomly drawn PIN correctly on one try. More than one in a hundred? I’m less sure. The easy-to-imagine thing to do is set a computer to try out all 456,976 possible sets of four letters and check them against a dictionary. The number of hits divided by the number of possibilities would be the chance of drawing a legitimate word. If I had a less capable computer, or were checking even longer words, I might instead draw some set number of words, never minding that I didn’t get every possibility. The fraction of successful words in my sample would be something close to the chance of drawing any legitimate word.

If I thought a little deeper about the problem, though, I’d just count how many four-letter words are already in my dictionary and divide that into 456,976. It’s always a mistake to start programming before you’ve thought the problem out. The trouble is not being able to tell when that thinking-out is done.

Richard Thompson’s Poor Richard’s Almanac for the 7th is the other comic strip to mention infinite monkeys. Well, chimpanzees in this case. But for the mathematical problem they’re not different. I’ve featured this particular strip before. But I’m a Thompson fan. And goodness but look at the face on the T S Eliot fan in the lower left corner there.

Jeff Mallet’s Frazz for the 6th gives Caulfield one of those flashes of insight that seems like it should be something but doesn’t mean much. He’s had several of these lately, as mentioned here last week. As before this is a fun discovery about Roman Numerals, but it doesn’t seem like it leads to much. Perhaps a discussion of how the subtractive principle — that you can write “four” as “IV” instead of “IIII” — evolved over time. But then there isn’t much point to learning Roman Numerals at all. It’s got some value in showing how much mathematics depends on culture. Not just that stuff can be expressed in different ways, but that those different expressions make different things easier or harder to do. But I suspect that isn’t the objective of lessons about Roman Numerals.

Frazz got my attention again the 12th. This time it just uses arithmetic, and a real bear of an arithmetic problem, as signifier for “a big pile of hard work”. This particular problem would be — well, I have to call it tedious, rather than hard. doing it is just a long string of adding together two numbers. But to do that over and over, by my count, at least 47 times for this one problem? Hardly any point to doing that much for one result.

Patrick Roberts’s Todd the Dinosaur for the 7th calls out fractions, and arithmetic generally, as the stuff that ruins a child’s dreams. (Well, a dinosaur child’s dreams.) Still, it’s nice to see someone reminding mathematicians that a lot of their field is mostly used by accountants. Actuaries we know about; mathematics departments like to point out that majors can get jobs as actuaries. I don’t know of anyone I went to school with who chose to become one or expressed a desire to be an actuary. But I admit not asking either.

Mike Thompson’s Grand Avenue started off a week of students-resisting-the-test-question jokes on the 7th. Most of them are hoary old word problem jokes. But, hey, I signed up to talk about it when a comic strip touches a mathematics topic and word problems do count.

Zach Weinersmith’s Saturday Morning Breakfast Cereal reprinted the 7th is a higher level of mathematical joke. It’s from the genre of nonsense calculation. This one starts off with what’s almost a cliche, at least for mathematics and physics majors. The equation it starts with, $e^{i Pi} = -1$, is true. And famous. It should be. It links exponentiation, imaginary numbers, π, and negative numbers. Nobody would have seen it coming. And from there is the sort of typical gibberish reasoning, like writing “Pi” instead of π so that it can be thought of as “P times i”, to draw to the silly conclusion that P = 0. That much work is legitimate.

From there it sidelines into “P = NP”, which is another equation famous to mathematicians and computer scientists. It’s a shorthand expression of a problem about how long it takes to find solutions. That is, how many steps it takes. How much time it would take a computer to solve a problem. You can see why it’s important to have some study of how long it takes to do a problem. It would be poor form to tie up your computer on a problem that won’t be finished before the computer dies of old age. Or just take too long to be practical.

Most problems have some sense of size. You can look for a solution in a small problem or in a big one. You expect searching for the solution in a big problem to take longer. The question is how much longer? Some methods of solving problems take a length of time that grows only slowly as the size of the problem grows. Some take a length of time that grows crazy fast as the size of the problem grows. And there are different kinds of time growth. One kind is called Polynomial, because everything is polynomials. But there’s a polynomial in the problem’s size that describes how long it takes to solve. We call this kind of problem P. Another is called Non-Deterministic Polynomial, for problems that … can’t. We assume. We don’t know. But we know some problems that look like they should be NP (“NP Complete”, to be exact).

It’s an open question whether P and NP are the same thing. It’s possible that everything we think might be NP actually can be solved by a P-class algorithm we just haven’t thought of yet. It would be a revolution in our understanding of how to find solutions if it were. Most people who study algorithms think P is not NP. But that’s mostly (as I understand it) because it seems like if P were NP then we’d have some leads on proving that by now. You see how this falls short of being rigorous. But it is part of expertise to get a feel for what seems to make sense in light of everything else we know. We may be surprised. But it would be inhuman not to have any expectations of a problem like this.

Mark Anderson’s Andertoons for the 8th gives us the Andertoons content for the week. It’s a fair question why a right triangle might have three sides, three angles, three vertices, and just the one hypotenuse. The word’s origin, from Greek, meaning “stretching under” or “stretching between”. It’s unobjectionable that we might say this is the stretch from one leg of the right triangle to another. But that leaves unanswered why there’s just the one hypothenuse, since the other two legs also stretch from the end of one leg to another. Dr Sarah on The Math Forum suggests we need to think of circles. Draw a circle and a diameter line on it. Now pick any point on the circle other than where the diameter cuts it. Draw a line from one end of the diameter to your point. And from your point to the other end of the diameter. You have a right triangle! And the hypothenuse is the leg stretching under the other two. Yes, I’m assuming you picked a point above the diameter. You did, though, didn’t you? Humans do that sort of thing.

I don’t know if Dr Sarah’s explanation is right. It sounds plausible and sensible. But those are weak pins to hang an etymology on. But I have no reason to think she’s mistaken. And the explanation might help people accept there is the one hypothenuse and there’s something interesting about it.

The first (and as I write this only) commenter, Kristiaan, has a good if cheap joke there.

Reading the Comics, May 28, 2016: Visual Interest Will Never Reappear Edition

OK, that’s three weeks in a row in which all my mathematically-themed comic strips are from Gocomics. Maybe I should commission some generic Reading The Comics art from the cartoonists and artists I know. It could make things more exciting on a visually dull week like this.

Mark Anderson’s Andertoons got its entry on the 25th. We draw the name “exponents” from the example of Michael Stifel, a 16th-century German theologian/mathematician. He’d described them as exponents in his influential 1544 book Arithmetica Integra. But I don’t know why he picked the name “exponent” rather than some other word.

Nate Fakes’s Break of Day for the 25th is the anthropomorphic numerals gag for this week.

Dave Blazek’s Loose Parts for the 25th is not quite the anthropomorphic shapes joke for this week. The word “isosceles” does trace back to Greek, of course. The first part comes from “isos”, meaning equal; you see the same root in terms like “isobar” and “isometric view”. The “sceles” part comes from “skelos”, meaning leg. Say what you will about an isosceles triangle, and you may as they’ve got poor hearing, but they do have two legs with the same length. If you want to say an equilateral triangle, which has three legs the same length, is an isosceles triangle you can do that. You’ll be right. But you will look like you’re trying a little too hard to make a point, the way you do if you point to a square and start off by calling it a rhombus.

Donna A Lewis’s Reply All Lite for the 25th tries doing a joke about doing mathematics by hand being a sign of old age. If we’re talking about arithmetic … I could go along with that, grudgingly. Calculator applications are so reliable and so quick that it’s hard to justify doing arithmetic by hand unless it’s a very simple problem. If you have fun doing that, good.

But if we’re doing real mathematics, the working out of a model and the implications of that, or working out calculus or group theory or graph theory or the like? There are surely some people who can do all this work in their heads and I am impressed by that. But much of real mathematics is working out implications of ideas, and that’s done so very well by hand. I haven’t found a way of typing in strings of expressions which makes it easier for me to think about the mathematics rather than the formatting. And I would believe in a note-taking program that was as sensitive and precise as pen on paper. I haven’t seen one yet, though. (I have small handwriting, and the applications I’ve tried turn all my writing into tiny, disconnected dots and scribbles.)

Ralph Hagen’s The Barn for the 27th is superficially about Olbers’s Paradox. If there’s an infinitely large, infinitely old universe, then how can the night sky by dark? The light of all those stars should come together to make night even more brilliantly blazing than the daytime sun. This is a legitimate calculus problem. The reasoning is sound. The light of a star trillions upon trillions of light-years away may be impossibly faint. But there are so many stars that would be that far away that they would be, on average, about as bright as the sun is. Integral calculus tells us what happens when we have infinitely large numbers of impossibly tiny things added together. In the case of stars, infinitely many impossibly faint stars would come together to an infinitely bright night sky. That night is dark tells us: the universe can’t be infinitely large and infinitely old. There must be limits to how far away anything can be.

The Barn reappears in my attention on the 28th, with a subverted word problem joke.

Spaghetti Mathematics

Let’s ease into Monday. Win Smith with the Well Tempered Spreadsheet blog encountered one of those idle little puzzles that captures the imagination and doesn’t let go. It starts as many will with spaghetti.

I’m sure you’re intrigued too. It’s not the case that any old splitting of a strand of spaghetti will give you three pieces you can make into a triangle. You need the lengths of the three pieces to satisfy what’s imaginatively called the Triangle Inequality. That inequality demands the lengths of any two sides have to be greater than the length of the third side. So, suppose we start with spaghetti that’s 12 inches long, and we have it cut into pieces 5, 4, and 3 inches long: that’s fine. If we have it cut into pieces 9, 2, and 1 inches long, we’re stuck.

The Triangle Inequality is often known as the Cauchy Inequality, or the Cauchy-Schwarz Inequality, or the Cauchy-Bunyakovsky-Schwarz Inequality, or if that’s getting too long the CBS Inequality. And some pranksters reorder it to the Cauchy-Schwarz-Bunyakovsky Inequality. The Cauchy (etc) Inequality isn’t quite the same thing as the Triangle Inequality. But it’s an important and useful idea, and the Cauchy (etc) Inequality has the Triangle Inequality as one of its consequences. The name of it so overflows with names because mathematics history is complicated. Augustin-Louis Cauchy published the first proof of it, but for the special case of sums of series. Viktor Bunyakovsky proved a similar version of it for integrals, and has a name that’s so very much fun to say. Hermann Amandus Schwarz first put the proof into its modern form. So who deserves credit for it? Good question. If it influences your decision, know that Cauchy was incredibly prolific and has plenty of things named for him already. He’s got, without exaggeration, about eight hundred papers to his credit. Collecting all his work into a definitive volume took from 1882 to 1974.

Back to the spaghetti. The problem’s a fun one and if you follow the Twitter link above you’ll see the gritty work of mathematicians reasoning out the problem. As with every probability problem ever, the challenge is defining exactly what you’re looking for the probability of. This we call finding the “sample space”, the set of all the possible outcomes and how likely each of them are. Subtle changes in how you think of the problem will change whether you are right.

Smith cleans things up a bit, but preserves the essence of how the answer worked out. The answer that looks most likely correct was developed partly by reasoning and partly by numerical simulation. Numerical simulation is a great blessing for probability problems. Often the easiest way to figure out how likely something is will be trying it. But this does require working out the sample space, and what parts of the sample space represent what you’re interested in. With the information the numerical simulation provided, Smith was able to go back and find an analytic, purely reason-based, answer that looks plausible.

The Short, Unhappy Life Of A Doomed Conjecture

So last month amongst the talk about the radius of a circle inscribed in a Pythagorean right triangle I mentioned that I had, briefly, floated a conjecture that might have spun off it. It didn’t, though I promised to describe the chain of thought I had while exploring it, on the grounds that the process of coming up with mathematical ideas doesn’t get described much, and certainly doesn’t get described for the sorts of fiddling little things that make up a trifle like this.

The point from which I started was a question about the radius of a circle inscribed in the right triangle with legs of length 5, 12, and 13. This turns out to have a radius of 2, which is interesting because it’s a whole number. It turns out to be simple to show that for a Pythagorean right triangle, that is, a right triangle whose legs are a Pythagorean triple — like (3, 4, 5), or (5, 12, 13), any where the square of the biggest number is the same as what you get adding together the squares of the two smaller numbers — the inscribed circle has a radius that’s a whole number. For example, the circle you could inscribe in a triangle of sides 3, 4, and 5 would have radius 1. The circle inscribed in a triangle of sides 8, 15, and 17 would have radius 3; so does the circle inscribed in a triangle of sides 7, 24, and 25.

Since I now knew that (and in multiple ways: HowardAt58 had his own geometric solution, and you can also do this algebraically) I started to wonder about the converse. If a Pythagorean right triangle’s inscribed circle has a whole number for a radius, can does knowing a circle has a whole number for a radius tell us anything about the triangle it’s inscribed in? This is an easy way to build new conjectures: given that “if A is true, then B must be true”, can it also be that “if B is true, then A must be true”? Only rarely will that be so — it’s neat when it is — but we might be able to patch something up, like, “if B, C, and D are all simultaneously true, then A must be true”, or perhaps, “if B is true, then at least E must be true”, where E resembles A but maybe doesn’t make such a strong claim. Thus are tiny little advances in mathematics created.

My Math Blog Statistics, November 2014

October 2014 was my fourth-best month in the mathematics blog here, if by “best” we mean “has a number of page views”, and second-best if by “best” we mean “has a number of unique visitors”. And now November 2014 has taken October’s place on both counts, by having bigger numbers for both page views and visitors, as WordPress reveals such things to me. Don’t tell October; that’d just hurt its feelings. Plus, I got to the 19,000th page view, and as of right now I’m sitting at 19,181; it’s conceivable I might reach my 20,000th viewer this month, though that would be a slight stretch.

But the total number of page views grew from 625 up to 674, and the total number of visitors from 323 to 366. The number of page views is the highest since May 2014 (751), although this is the greatest number of visitors since January 2014 (473), the second month when WordPress started revealing those numbers to us mere bloggers. I like the trends, though; since June the number of visitors has been growing at a pretty steady rate, although steadily enough I can’t say whether it’s an arithmetic or geometric progression. (In an arithmetic progression, the difference between two successive numbers is about constant, for example: 10, 15, 20, 25, 30, 35, 40. In a geometric progression, the ratio between two successive numbers is about constant, for example: 10, 15, 23, 35, 53, 80, 120.) Views per visitor dropped from 1.93 to 1.84, although I’m not sure even that is a really significant difference.

The countries sending me the most readers were just about the same set as last month: the United States at 458; Canada recovering from a weak October with 27 viewers; Argentina at 20; Austria and the United Kingdom tied at 19; Australia at 17; Germany at 16 and Puerto Rico at 14.

Sending only one reader this month were: Belgium, Bermuda, Croatia, Estonia, Guatemala, Hong Kong, Italy, Lebanon, Malaysia, the Netherlands, Norway, Oman, the Philippines, Romania, Singapore, South Korea, and Sweden. (Really, Singapore? I’m a little hurt. I used to live there.) The countries repeating that from October were Estonia, the Netherlands, Norway, and Sweden; Sweden’s going on three months with just a single reader each. I don’t know what’s got me only slightly read in Scandinavia and the Balkans.

My most-read articles for November were pretty heavily biased towards the comics, with a side interest in that Pythagorean triangle problem with an inscribed circle. Elke Stangl had wondered about the longevity of my most popular posts, and I was curious too, so I’m including in brackets a note about the number of days between the first and the last view which WordPress has on record. This isn’t a perfect measure of longevity, especially for the most recent posts, but it’s a start.

As ever there’s no good search term poetry, but among the things that brought people here were:

• trapezoid
• how many grooves are on one side of an lp record?
• origin is the gateway to your entire gaming universe.
• cauchy funny things done
• trapezoid funny
• yet another day with no plans to use algebra

Won’t lie; that last one feels a little personal. But the “origin is the gateway” thing keeps turning up and I don’t know why. I’d try to search for it but that’d just bring me back here, leaving me no more knowledgeable, wouldn’t it?

Another Reason Why It’s Got To Be 2

To circle back around that inscribed circle problem, about what the radius of the circle that just fits inside a right triangle with sides of length 5, 12, and 13: I’d had an approach for solving it different from HowardAt58’s geometric answer. This isn’t to imply that his answer’s wrong, I should point out: problems can often be solved by several different yet equally valid approaches. (It might almost be the definition of cutting-edge research if it’s a problem there’s only one approach for.)

So here’s another geometry-based approach to finding what the radius of the circle that just fits inside the triangle has to be. We started off with the right triangle, and sides a and b and c; and there’s a circle inscribed in it. This is the biggest circle that’ll fit within the triangle. The circle has some radius, and we’ll just be a little daring and original and use the symbol “r” to stand for that radius. We can draw a line from the center of the circle to the point where the circle touches each of the legs, and that line is going to be of length r, because that’s the way circles work. My drawing, Figure 1, looks a little bit off because I was sketching this out on my iPad and being more exact about all this was just so, so much work.

The next step is to add three more lines to the figure, and this is going to make it easier to see what we want. What we’re adding are liens that go from the center of the circle to each of the corners of the original triangle. This divides the original triangle into three smaller ones, which I’ve lightly colored in as amber (on the upper left), green (on the upper right), and blue (on the bottom). The coloring is just to highlight the new triangles. I know the figure is looking even sketchier; take it up with how there’s no good mathematics-diagram sketching programs for a first-generation iPad, okay?

If we can accept my drawings for what they are already, then, there’s the question of why I did all this subdividing, anyway? The good answer is: looking at this Figure 2, do you see what the areas of the amber, green, and blue triangles have to be? Well, the area of a triangle generally is half its base times its height. A base is the line connecting two of the vertices, and the height is the perpendicular distane between the third vertex and that base. So, for the amber triangle, “a” is obviously a base, and … say, now, isn’t “r” the height?

It is: the radius line is perpendicular to the triangle leg. That’s how inscribed circles work. You can prove this, although you might convince yourself of it more quickly by taking the lid of, say, a mayonnaise jar and a couple of straws. Try laying down the straws so they just touch the jar at one point, and so they cross one another (forming a triangle), and try to form a triangle where the straw isn’t perpendicular to the lid’s radius. That’s not proof, but, it’ll probably leave you confident it could be proven.

So coming back to this: the area of the amber triangle has to be one-half times a times r. And the area of the green triangle has to be one-half times b times r. The area of the blue triangle, yeah, one-half times c times r. This is great except that we have no idea what r is.

But we do know this: the amber triangle, green triangle, and blue triangle together make up the original triangle we started with. So the areas of the amber, green, and blue triangles added together have to equal the area of the original triangle, and we know that. Well, we can calculate that anyway. Call that area “A”. So we have this equation:

$\frac12 ar + \frac12 br + \frac12 cr = A$

Where a, b, and c we know because those are the legs of the triangle, and A we may not have offhand but we can calculate it right away. The radius has to be twice the area of the original triangle divided by the sum of a, b, and c. If it strikes you that this is twice the area of the circle divided by its perimeter, yeah, that it is.

Incidentally, we haven’t actually used the fact that this is a right triangle. All the reasoning done would work if the original triangle were anything — equilateral, isosceles, scalene, whatever you like. If the triangle is a right angle, the area is easy to work out — it’s one-half times a times b — but Heron’s Formula tells us the area of a triangle knowing nothing but the lengths of its three legs. So we have this:

(Right triangle)

$r = \frac{1}{a + b + c} \cdot \left(a\cdot b\right)$

(Arbitrary triangle)

$r = \frac{1}{a + b + c} \cdot 2 \sqrt{p\cdot(p - a)\cdot(p - b)\cdot(p - c)} \mbox{ where } p = \frac12\left(a + b + c\right)$.

Since we started out with a Pythagorean right triangle, with sides 5, 12, and 13, then: a = 5, b = 12, c = 13; a times b is 60; a plus b plus c is 30; and therefore the radius of the inscribed circle is 60 divided by 30, or, 2.

I’m not precisely sure how to embed this, so, let me just point over to the Five Triangles blog (on blogspot) where there’s a neat little puzzle. It starts with Pythagorean triplets, the sets of numbers (a, b, c) so that a2 plus b2 equals c2. Pretty much anyone who knows the term “Pythagorean triplet” knows the set (3, 4, 5), and knows the set (5, 12, 13), and after that knows that there’s more if you really have to dig them up but who can be bothered?

Anyway, the problem at Five Triangle’s “Inscribed Circle” here draws that second Pythagorean triplet triangle, the one with sides of length 5, 12, and 13, and inscribes a circle within it. The problem: find the radius of the circle?

I’m embarrassed to say how much time I took to work it out, but that’s because I was looking for purely geometric approaches, when casting it over to algebra turns this into a pretty quick problem. I do feel like there should be an obvious geometric solution, though, and I’m sure I’ll wake in the middle of the night feeling like an idiot for not having that before I talked about this.

January 2014’s Statistics

So how does the first month of 2014 compare to the last month of 2013, in terms of popularity? The raw numbers are looking up: I went from 176 unique visitors looking at 352 pages in December up to 283 unique visitors looking at 498 pages. If WordPress’s statistics are to be believed that’s my greatest number of page views since June of 2013, and the greatest number of visitors since February. This hurt the ratio of views per visitor a little, which dropped from 2.00 to 1.76, but we can’t have everything unless I write stuff that lots of people want to read and they figure they want to read a lot more based on that, which is just crazy talk. The most popular articles, though, were:

1. Something Neat About Triangles, this delightful thing about forming an equilateral triangle starting from any old triangle.
2. How Many Trapezoids I Can Draw, with my best guess for how many different kinds of trapezoids there are (and despite its popularity I haven’t seen a kind not listed here, which surprises me).
3. Factor Finding, linking over to IvaSallay’s quite interesting blog with a great recreational mathematics puzzle (or educational puzzle, depending on how you came into it) that drove me and a friend crazy with this week’s puzzles.
4. What’s The Worst Way To Pack? in which I go looking for the least-efficient packing of spheres and show off these neat Mystery Science Theater 3000 foam balls I got.
5. Reading The Comics, December 29, 2013, the old year’s last bunch of mathematics-themed comic strips.

The countries sending me readers the most often were the United States (281), Canada (52), the United Kingdom (25), and Austria (23). Sending me just a single reader each this past month were a pretty good list:
Bulgaria, France, Greece, Israel, Morocco, the Netherlands, Norway, Portugal, Romania, Russia, Serbia, Singapore, South Korea, Spain, and Viet Nam. Returning on that list from last month are Norway, Romania, Spain, and Viet Nam, and none of those were single-country viewers back in November 2013.

I was reading a biography of Donald Coxeter, one of the most important geometers of the 20th century, and it mentioned in passing something Coxeter referred to as Morley’s Miracle Theorem. The theorem was proved in 1899 by Frank Morley, who taught at Haverford College (if that sounds vaguely familiar that’s because you remember it’s where Dave Barry went) and then Johns Hopkins (which may be familiar on the strength of its lacrosse team), and published this in the first issue of the Transactions of the American Mathematical Society. And, yes, perhaps it isn’t actually important, but the result is so unexpected and surprising that I wanted to share it with you. The biography also includes a proof Coxeter wrote for the theorem, one that’s admirably straightforward, but let me show the result without the proof so you can wonder about it.

First, start by drawing a triangle. It doesn’t have to have any particular interesting properties other than existing. I’ve drawn an example one.

The next step is to cut into three equal pieces each of the interior angles of the triangle, and draw those lines. I’m doing that in separate diagrams for each of the triangle’s three original angles because I want to better suggest the process.

I should point out, this trisection of the angles can be done however you like, which is probably going to be by measuring the angles with a protractor and dividing the angle by three. I made these diagrams just by sketching them out, so they aren’t perfect in their measure, but if you were doing the diagram yourself on a sheet of scratch paper you wouldn’t bother getting the protractor out either. (And, famously, you can’t trisect an angle if you’re using just compass and straightedge to draw things, but you don’t have to restrict yourself to compass and straightedge for this.)

Now the next bit is to take the points where adjacent angle trisectors intersect — that is, for example, where the lower red line crosses the lower green line; where the upper red line crosses the left blue line; and where the right blue line crosses the upper green line. Draw lines connecting these points together and …

This new triangle, drawn in purple on my sketch, is an equilateral triangle!

(It may look a little off, but that’s because I didn’t measure the trisectors when I drew them in and just eyeballed it. If I had measured the angles and drawn the new ones in carefully, it would have been perfect.)

I’ve been thinking back on this and grinning ever since reading it. I certainly didn’t see that punch line coming.