Springtime and Daffodils

Spring has come to East Anglia, and the Daffodils are popping up everywere.  Each time I see a patch I hope for a little breeze that will demonstrate their beautiful geometric construction.  I first learned about this a few  years ago right around Easter, with snow on the ground... Hoping that doesn't happen this year, but  here is that blog again anyway:
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I’ve been thinking about geometry a lot lately. Partly that is due to the fact that I’m going through Trig and Vectors in my Pre-calc class. Partly it is probably because it has popped up in science stories I have been reading lately. On the same day I wrote about the Daffodils in the Snow, I read a note from a researcher on why they respond differently to wind than other similar flowers. The short answer is geometry.

When you watch Tulips, for instance, they will lean away from the wind. Daffodils, on the other hand, remain almost completely erect, but turn their tilted heads away from the wind. William Wordsworth must have had this in mind when he wrote,

“ Ten thousand saw I at a glance
Tossing their heads in sprightly dance.”


The reason the daffodil twists like a weather vane and the tulip bends more in the wind is the geometry of the cross-section of the stem. A tulip stem is nearly round, and so it can bend, but not twist. The same effect causes your garden hose to crimp up and cut off the flow of water when it can’t twist. They are very good at bending, very poor at twisting. The engineering types call the twisting motion torsion, and it is related to the cumulative sum of the fourth power of the distances from the center to the edge of the stem. Circular things are far away in all directions. But a Daffodil has a cross section that is more elliptical. If you pick it up you can see the difference in the long and short axis easily with the naked eye. Since the sum of the fourth powers of the distances is lower, it is more able to turn away from the wind. Scientists studied one type of daffodil and reported that up to about 22 mph, the stems stayed essentially erect, and the trumpets all turned away from the wind. After that, the flowers both turned and bent some, but they cannot bend as low as a tulip in any wind… all geometry.


The geometry of sharks popped up too. It seems that shark geometry may be to thank for the engineering advances that will result in a few more swimming records falling at the Olympics in China this year, although I assume the swimmers will want some of the credit. It seems a shark has dimples on its scales that breaks up the flow of water, reducing the drag so it uses less energy. Speedo reckons it can make better swimsuits making its swimmers go faster using something similar. The new body suits are already in the pools and having an impact.

Another bit of research answers the question, with such a huge blue ocean to wander around in, how exactly do marine predators like sharks find their next meal? Yeah…. You guessed it, geometry. It has been known that many land animals search for food much like a shopper in the super market searches for a particular item. The math term is called a Levy walk (actually a Levy flight), a fractal type structure from geometry where the small parts are self-similar. Ok, the actual rules are a little technical but in essence it means that the animal undergoes lots of short-distance journeys interspersed with fewer longer-distance journeys. Just as you go to an area of the store where you think the item is located, then circle around in that area looking for it. If you don’t find it, you go off to another area and begin a close search there.


Geometry, making your life better… See, because if two sides and the included angle of one triangle….

Repeat of an Isoperimetric Discovery

It's that time of year again, so I want to repeat this to explain some ideas I laid out in Calculus today...

Recently we have been doing maxima/minima applications in calculus. By the time they get her most of them know that when they maximize the area of a rectangle with perimeter constraints, they get a square. In Pre-calc we do some basic problems with 2 adjacent pens, a single pen against a wall or river, and all the similar cases. I have written about the generalizations of some of these here, and here. The problem of two adjacent congruent pens is (or should be) easy for a calculus or precalc student. It turns out that the maximum occurs when the total width is one-fourth the perimeter and the height is one-sixth the perimeter, for a maximum area of one twenty-fourth of the perimeter value (in square units, of course). If you extend that to even more congruent rectangular pens placed end to end it turns out that the width stays the same?

If you extend this, as many teachers do, to a shape like this one, it starts to get interesting to the more observant student. The height is now one-eighth of the total , and the width is one-sixth of the perimeter. Even slow students can figure out that the area of the total turns out to be one forty-eighth of the value of the perimeter. Some will even generalize the area of an array of mxn congruent pens to be the perimeter squared divided by [4(m+1)(n+1))].... but it seems to take forever for some one to finally realize that in all these problems, whether there was one rectangle, or one with one side adjacent to a river or barn, or rows of rectangles... there is one invariant...

Theorem one for rectangular arrays of congruent rectangular pens "The amount of fence going right to left is the same as the amount going up and down." If you want to maximize a rectangular array with five rows of three pens in each row, we will need 24 lines of fence, four going vertically, six going horizontally. Use half the fencing to make the four vertical fences, and the other half to make the other six going horizontally. (Prove..... dear reader) ..

I always thought this fifty-fifty use of fencing was a cute mathematical "teacher trick" but I have recently come to believe there is something deeper at work. I played around recently with this problem but loosened the constraints on the pens to be just equal area rectangles rather than congruent. This offered some nice areas to explore. Which would be better, three congruent pens in a row, or two pens with a third beneath... like so. The three adjacent pens turn out to have a max area when the width was p/4 (where p is the perimeter) and the height was p/8, giving a maximum area of p²/32... and as before, that mean that half the perimeter was used for the vertical sections, and half for the horizontal sections..and for the three non-congruent sections with equal area the width was p/6, and the length was 3P/16, giving a max area of....wait, that's also p²/32. (Ok, I hadn't expected that)... but when I checked the dimensions, it turns out that the horizontal sections use up exactly half the perimeter, and the vertical sections the other half in both cases.... hmmm...

Ok, but every kid knows that the area is maximized when the rectangle is closest to a square... so what if we just put three squares arranged as shown below and didn't require that the whole assembly forms a rectangle... I thought that would probably be even more area.. It turns out that it isn't, The equal distribution of left and right still works out, but the total area is only 3p²/100 or about 96% of the other shapes. (ok, I wasn't ready for that)...

So, I decided to move up to five pens... I passed over four absolutely sure that the division into four equal squares in a two by two array would be the most efficient, but five offered several variations, five in a row; two rows of two with a single in the third row, or three in one row and two in another... The five congruent squares in a row gave a max area of p²/48, but the set up with three pens in one row and two in another gave a slightly larger area of 5p²/216 or p²/43.2. The 2,2,1 set up (shown below) gave not quite the same area.. This set up gave a max area of p²/44.8.

From all of this I have a formed a totally unproven conjecture about the maximum area when n equal area rectangular pens are set out in a rectangular boundary with a given amount of fencing.
The Theorem One above for congruent rectangles seems to be a stronger property than I suspected, and is applicable to all equal area pens in a rectangular boundary, And the conjecture is that when the number of pens, n, is between x²-x and x²+x, the maximum area will be found when there are x-rows of pens and all of the pens have either x-1, x, or x+1 pens in the row.

The number of pens in any row will never differ by more than one from any other row. For values of n that are perfect squares or pronic (product of consecutive integers), the number of pens in each row will be the same. When n > x² there will n-x² rows that have x+1 pens, and the remaining rows will all have x pens. Similarly, if n is less than x² there will be x² - n rows that have x-1 pens and the remaining rows will have x pens each. As an example, if n= 23, then x=5 since 5²-5 is less than 23 which is less than 5²+5. And for 23 pens there will be five rows with x=five pens in three of them, and x-1=four pens in two of them.

I also have discovered that the theorem about equal lengths of fencing horizontally and vertically seems to be more important as a "rule of thumb" for maximizing than is equal sides. For example if five squares are set out as efficiently as possible (four in a square with one attached to a side), the area will be p² / 45 with 8/15 of the perimeter going one way(say horizontally) and 7/15 going the other. If we change the squares to rectangles with each having a height of p/16 and a width of p/14 the area is p²/44.8.. an improvement of about 1/2 %....
Even if you go to six pens (where there is no lost perimeter enclosing the odd pen), putting two rows of three squares gave me an area of P² / (48 1/3) but using the equal division of fencing gives the very slightly improved maximum of p²/48.

I can't explain why, but the equal division of fencing theorem is stronger than I first suspected, and if someone out there can shed more light on it than my feeble exploration, please do.

I haven't had time to expand the formula for the max area of all n when a rectangle is divided into congruent area rectangles, (not too difficult, I think).. but for square or pronic numbers of pens, it seems the max area will always be p²/(4 (x+1)(y+1)) where xy=n. As a limit as n approaches infinity, it seems the max area for rectangles divided into n equal area rectangles approaches

Folding a cube???

Found this at Math Recreations by Dan MacKinnon. Fold a cube into..well,some interesting stuff.
The origninal work comes from Eric Demaine

click here, then wait, it takes a minute to animate...but well worth it.

Why Didn't I know This Already?

I was playing with a problem that Dave Renfro sent to the AP Calculus EDG and got curious about something else, and stumbled upon a simple relationship about tangents to circles that I never knew (or don't remember??? )

The problem Dave sent was about finding the common tangent to a circle centered on the origin and a parabola with a vertex at the origin. I wandered over to playing around with a general solution, and along the way I remembered a cool idea that was known way back to Apollonius, I think. If you have a parabola centered at the origin, like y=ax², and you draw a tangent to some point, call it (p,q), then the y-intercept of that tangent line will be at (0,-q)... no matter what the value of a is. And it seems clear that a straight line that goes from (p,q) to (0,-q) must have an x-intercept at (p/2,0)

For example, if you used y=x² at the point (2,4) the tangent would have an equation of y-4 = 4(x-2) and when x= 0, the y-intercept is at (0,-4). If instead you pick y= 3x², when x=2, y= 12. The equation of the tangent will by y-12 = 12(x-2) and the y-intercept will be at (0,-12), the x-intercept at (1,0)in both cases.

It occurred to me that probably all the conics might have simple relationships that would predict intercepts of the tangent. So I started with a circle.... say x²+y²= r². What happens if we pick a point (p,q)? As I wrote in a recent blog, an easy way to find the tangent to a conic is by the use of polars. If we want the tangent to a circle at a point (p,q) we need only to replace one x and one y in the equation with the values p and q. In the case of x²+y²= r², the tangent through the point (p,q) would be px+qy=r². Pretty easy (so why had I played with this all these years and never noticed that the x-intercept would be r²/p, and the y-intercept would be r²/q. So why had I never noticed this relationship??? (don't answer, the truth hurts)

So a simple hyperbola or ellipse could be done the same way... tooooo easy...

The Geometry of the Babylonian Square Root Method

Physical evidence exists that the Babylonians had a method of calculating the square root of some numbers as early as 2000 years before the birth of Christ. In the Yale collection there exist an artifact that shows the calculation of the square root of two to five decimal places accurately. One image of the item, called YBC7289, can be seen above. What may surprise many students is that the ancient Babylonian method seems to be the same as the method frequently taught in school text books. The method is frequently attributed to Heron (or Hero) because it appears in his Metrica, and is also called Newton’s method, but most students who know the method, know it as the "divide-and-average" method.

I was recently reminded of this while reading a nice post about it from Brent over at The Math Less Traveled. He did a nice job of showing why it worked...except.. It was all algebraic.. and I don't think the Babylonians had a very developed synthetic algebra at all. So I wondered how they might have convinced themselves of its authenticity by geometric methods.
This is a flight of whimsey, or maybe two, because I have no evidence they thought these ideas anymore than they thought algebraic ones, but it was a good exercise in geometry, kept me out of trouble for awhile, and that's not all bad....

So let's pretend our ancient Babylonian scholar is trying to reason out the square root of eight. He knows that means he is looking for the sides of a square with an area of eight. So.... he picks an unlikely strating point, and guesses four. Then he creates a rectangle with one side of four that has an area of eight. (I have kept these images small for space, but if you click on them, I think they can be expanded to accomodate somewhat better viewing)

Knowing full well that if he bisected the angle in a corner of a square, he would also bisect the opposite corner, he constructs the angle bisector (y=x in my image) and observes that it does NOT hit the opposite corner... a 2x4 rectangle is NOT a square.. Now he drops a vertical line to the base of the rectangle (x-axis to we moderns) and can see the difference between his two rectangle sides... SO, let's just average the two sides and try again.  We make our base equal to the average of 2 and 4 (I got 3) he sets his base equal to 3. Dividing eight by this new length, he creates another rectangle with an area of eight that has sides of 3 and 2 ²/₃, closer to equal than the last. Noticing that it is Still not a square, he constructs another base averaging 3 and 2 ²/₃ and repeats until the accuracy of the construction meets his needs.








A second excursion of fantasy adopts a more modern idea that is too little known to students and teachers today. If you pick two points (a, a²) and (b, b²) on the curve y=x²), the line segment connecting them will cross the y-axis at the point (0, -ab)... This gives an easy way to find the square root of eight.  I constructed a point at (0,2) and used a horizontal line (easy on graph paper) to find the point (1.414,2), then the point (2,4) falls out on grid paper quite easily.  The line will now intersect the y-axis at the point x=0 and y= -sqrt(8). Of course a clever student might just draw a horizontal line at (0,8) and that would intersect at (sqrt(8), 8) also.

I wanted to use the property to illustrate what happens when we use the divide and average method. Working backwards, we can start with a point at (4, 16) and drew a line through (0,8) to the opposite side of the parabola it would intersect at (-2,4) [because -(-2)x4=8].


If our line through (0,8) was level, it would mean we had equal sides, and hence sides of a square of eight. when we average them (2 and 4 again) and plot a point on the new x value (x=3 in this case), we are at (3,9) (red)
Then another ray through (0,8) to the opposite side and we hit (8/3, 64/9)to give the other side of our second rectangle. Repeat as needed....

So two geometric approaches to the divide and average method of finding square roots. Hope they may be useful to someone.

Geometry and Understanding

During a conversation with a fellow teacher, he made the statement that he really didn't like geometry, and went on to suggest it might not be a useful course for students. I was somewhat taken aback, and came out with what is probably more an article of faith than a tested fact... "Mathematically, if you really understand it, you can demonstrate it with geometry". Don't be misled; I love the power that comes from algebraic abstraction, but I find that algebraic "proofs" just lack a little something for convincing most students. Something visual seems to have more impact.

The teacher responded by asking (challenging?) for an example, and my first thought was the Pythagorean theorem since it is so ubiquitous in all areas of mathematics. You remember, a² + b² = c²... that one. Now there may be more existing proofs of this theorem than any other in existence. So I drew a right triangle, made three copies and formed them into a square as shown in the figure. Then, with the power of modern educational technology, I moved two of the triangles to get the second figure. Q.E.D. as we say; thus it is shown. Any student who knows the geometric idea of the area of a square can see that the white area in both pictures is the same since the same amount has been removed from it (four congruent triangles). In the first the white area is a square whose sides are formed by the hypotenuses (hypotenii?), c, of the four triangles. In the second it is divided into two smaller squares, one with sides of length a, one with sides of length b. The area then can be expressed as a^2 + b^2 or as c^2, and it is the same area. Somehow I envision a good sixth grader understanding this (and the next sixth grader I see is in for a sit down/talk to so I can find out).







Afterwards it struck me that this was not the best of examples. The Pythagorean theorem is essentially a geometric idea, its about triangles after all. What about an idea from math that has nothing obvious to connect it to geometry . Partitions seemed to fit the bill, and I thought of a simple example that I was sure elementary kids could play with and confirm, and I hope be convinced by a simple geometric argument.
Ok, so for the uninitiated, a partition of a number is just writing it out as a sum of smaller numbers. In order to count them we usually write them from largest addend to smallest. For example 6 = 3 + 2 + 1, so that is one partition of six. There are obviously others, 2+2+2, or 3 + 1 + 1 + 1 would be two more. A common, and not trivial, question is what is the total number of partitions of six? Most elementary students can take an organized attack and come up with the answer. If we include 6 itself, there are eleven of them. 6; 5+1; 4+2; 4+1+1; 3+3; 3+2+1; 3+1+1+1; 2+2+2; 2+ 2+ 1+ 1; 2+1+1+1+1; and of course 1+1+1+1+1+1.
But an interesting thing pops out if we look at some special limitations. There are exactly 3 ways to partition six into three terms; 4+1+1; 3+2+1; 2+2+2. There are also exactly three partitions that use three as the largest number; 3+3; 3+2+1; 3+1+1+1 . Coincidence? Not a chance. The same thing would happen if you partitioned 25 into four terms (or five or six or pick your favorite number). There are exactly the same number of partitions of a number into n differen terms as there are partitions with n as the biggest value. If you have never seen this proof, try to make it clear to yourself why it is true before you read on.
OK, it seems true, but how do we relate that to geometry, and a visual proof? To the rescue comes a 19th Century British Mathematician, Norman M Ferrers, who was a fellow at Gonville and Caius (pronounced Keys) College just down the road here at Cambridge. He seems to have been the first to make a simple discovery about partitions using dots. Here is the Ferrer diagram for two of the unique partitions of six into three groups.







Reading the number of dots in each row we see that 6 = 4 + 1+1; but if we read down the columns we see 3+1+1+1. In the second diagram we see rows of 2+2+2; but columns of 3+3. Can you see that this would apply to any diagram? If there are three rows that add up to some number, then the columns will give us the same total with a three as the largest number. The same thing would be as clear with any other number of rows.

See, Geometry makes it easy to understand, and to explain. Geometry is good. Now go do your homework!

OOPS, The New York State Geometry Experiment,

Well, if you live in New York State, it may well happen that your kid will get a geometry teacher this year, who has NEVER taken a geometry class in his life… Honest…


A couple of decades ago, the schools in New York went to an “integrated” approach that covered a little algebra, a little geometry, a little something else each year, and broke the traditional cycle of Algebra I, Geometry, Algebra II, Pre-Calculus so common in much of the world. But in 2003, according to the NY Times, “State education officials created a committee to begin rethinking the math standards .., after two-thirds of the students who took the Math A exam failed, prompting a flood of complaints and criticism from parents and teachers.”

To go back to the old way would almost certainly cause someone to lose face after telling the parents, teachers, and world in general that the “integrated” approach would be the salvation of math education. So the New York system has come up with an even better way, It begins with “Integrated Algebra I”, proceeds to “Integrated Geometry”, and then on to ….are you ready for this??? “Integrated Algebra II.” Now THAT is a creative solution. “Never look back”, Satchel Paige often said, “somebody may be gaining on you.”…


There is a hitch. Many of the math teachers, like math teachers in most places, have been teaching for less than four years , at least according to Alfred S. Posamentier, dean of the School of Education at the City College of New York. So they mostly can be assumed to have graduated from high school within the last ten years. Fact one, they never took geometry as a proof based course in high school. Fact two, most teacher training institutions do NOT offer traditional Euclidean geometry in their colleges. Most likely conclusion, there will be some kind of patch up training sessions combined with lots of geometry teachers trying to stay one day ahead of the kids in the teachers manual. Not exactly a recipe for success, so you gotta believe they thought things had REALLY gone downhill.

So if your kid comes home next year and needs help in Geometry, perhaps you shoud not have them follow the tradtional advice of "Ask the teacher."