The Birthday Problem and DNA

Keith Devlin does a variation of the birthday problem involving DNA and finds out how many people must be in a database before there are two whose DNA would match on a 9 locus match (the common judicial standard) as well as pointing out that the Body Mass Index is ....ummm...crap, stupid, not a very good predictor of obesity. 

Good lines.....

"Now, I have no problem with people disagreeing with me. Heavens, I was a mathematics department chair for four years and a dean for eight, and before that a father of two daughters growing up through adolescence to adulthood, so I have had my fair share of disagreements. What bothers me is that some folks seem (1) all too willing to accept something simply because its looks scientific and some august body advocates it, and (2) unable to adjust their intuitions when faced with evidence that they mislead us, as they do on occasion.
It bothers me because I am in the education business, and the primary goal of education is to help people develop the ability to think for themselves and to reach conclusions based on evidence and rational thought.
It bothers me particularly because, as the ancient Greeks recognized, when taught well, mathematics is one of the best mental disciplines to develop analytic thinking skills."


"The fact is, in the era of DNA identification, judges and juries simply cannot avoid getting to grips with the relevant math. Identification hinges on those calculations. There may be no way of avoiding bringing mathematicians into court to explain how the calculations are done. But for that to be effective, those judges and juries need first to learn (and accept) that human intuitions about probabilities are hopelessly unreliable. That can prepare the way, not for mathematical laypersons to learn how to do the calculations themselves - the experts can do that part - rather how to follow the calculations and evaluate the answers. For it is on those answers that justice will ultimately depend."

Midpoint Madness

I have written a few times about the medians of triangles here, and here. Here is a nice problem about midpoints suitable for really bright HS students that I adapted from an old Math Central problem at the Univ of Regina.
Given the three midpoints of a triangle, find the vertices of the triangle.

Beware, the solution will follow:
S
P
O
I
L
E
R
S
P
A
C
E

I think the easiest way to solve this, in fact, the only way I can think of at the moment, is to use vectors...which are almost totally ignored in US Geometry classes...

If we label the midpoints of sides a, b and c respectively as Ma, Mb, and Mc then we know that the vector from the origin to Ma (I'll label vectors and points with the same notation, for convenience) and the vectors B and C will obey B+C=2Ma. In the same way A+B=2Mc and A+C=2Mb.(Students should draw a segment AB and its midpoint and see why this vector addition must always be true by creating a parallelogram from two copies of the vectors OA and OB)

From these three equations we may show that since
A+B=2Mc THEN 2Mc-B= A

And since B+C=2Ma we know that B=2Ma-C... substituting that into 2Mc-B= A
gives us 2Mc-(2Ma-C)= A = 2Mc-2Ma+C

Now we use c=2Mb-A to get A= 2Mc-2Ma+(2Mb-A) or A= 2Mc-2Ma+2Mb - A...

Sooooo

2A = 2Mc-2Ma+2Mb or A = Mc-Ma+Mb...
Now we can just construct the three vectors to find the vertex A.

In a similar way we could get B=Ma-Mb+Mc and C=Mb-Mc+Ma.. but once you have one of the points, you can generate the others by construcing the segments in question.

All of this can be done by conventional compass and straightedge construction. The image shows the three vectors used to construct point A (in red) and adding the two blue vectors to the vector Mc will give Point B. I have ommitted vectors for point C to avoid clutter.

I think one big idea for students is that you can place the midpoints on a blank sheet of paper and put the origin anywhere you want and it will construct exactly the same triangle. This is a nice idea for them to grasp. With interactive geometry software you can actually create the whole thing on a page without coordinates, then grab the point you called the origin and move it around and only the vectors move with it.. the midpoints, vertices and sides of the triangle stay fixed.

It is interesting that you can do a similar approach for ANY odd sided polygon, but not for even sided ones. For the even sided polygons there may not be any solution, and if there is, then there are an infinite number of them. (if you try this for four sided figures, remember that the four midpoints of a quadrilateral must form a parallelogram.... you probably proved that in HS geometry)

Using the same vector approach as above, the substitution leads to a string of vectors that equal zero..... Ma-Mb+Mc-Md+.....=0 . Thus you can pick any point on the plane to be point A, and generate the n-gons from there.

If the number of sides is odd and greater than three, it is not even necessary that the midpoints all lie on a plane. The same method will work to build a two or three space solution for any odd number of vertices.

Another Pretty Geometry Problem

Here is another nice problem from Math Central at the Univ of Regina. This one is related to the Log Spiral I recently wrote about.  It combines a little simple 30/60/90 right triangle geoemtry with a geometric series...A nice problem for letting the students create a geometric series (geometrically) rather than throwing out a formula. And at the end, a really cute solution by geometric transformations.

The unit circle is divided into twelve equal parts, and the twelve dividing points are joined to the circle's centre, producing twelve rays. Starting from one of the dividing points a segment is drawn perpendicular to the next ray in the clockwise sense; from the foot of this perpendicular another perpendicular segment is drawn to the next ray, and so on to the center of the circle (note the diagram only shows part of the total curve). What is the limit of the sum of the lengths of these segments?

The  classic solution approach is to see that each central angle is 30^o and so the hypotenuse of each succeeding triangle is the cosine of 30^o  times the previous.  So the largest of the spiraling segments is opposite a 30^o with a hypotenuse of one, and must have a lenth of 1/2.  Each succeeding length is the previous length times .    These form a geometric series with the n^th  segment being .

The sum of an infinite sequence with a common ratio less than one is given by and in this case that becomes .  

A really pretty solution was given that can be illustrated with some simple rotations..

If we look at just the first two triangles to make it simple...

And then rotate the second (30 degrees in this cae) about the connection with the previous segment we can align them in a vertical column...

This can be continued with each following segment and aligning the segments to produce a triangle like the one below.  [Students see this better, it seems, if you describe it as happening from the center out.  Note that we get a right triangle with the right angle at top of the first segment created by drawing a one unit segment parallel side from the origin.  The sum of the segments is now shown to be one leg of a triangle whose other leg is one.  So the lengths must sum to the tan(75^o).  I call that pretty cool geometry. 

Thomas Harriot and the Roanoak Colony

It was on August 17, 1585 that the Colony of Roanoke Island was established by the landing of Sir Walter Raleigh's agents led by Ralph Lane(Raleigh actually never visited North America). This Colony would become known as the "Lost Colony". But before it got lost, it was part of the little known story of one of the better English Mathematicians of the period. Thomas Harriot's name was once synonymous with a common method of solving quadratics taught in nearly every high school. Once commonly called Harriot's Method, today it is simply referred to as factoring.

From my article on "Twenty Ways to Solve a Quadratic."

"For most students the first method of solving quadratic equations that they learn is by factoring. I have written (too often say some) that I think this is a pedagogical mistake, and that probably a graphic solution should be first. Vera Sanford points out in her Short History of Mathematics, 1930 that “In view of the present emphasis given to the solution of quadratic equations by factoring, it is interesting to note that this method was not used until Harriot’s work of 1631. Even in this case, however, the author ignores the factors that give rise to negative roots.” Harriot died in 1621, and like all his books, this one, Artis Analyticae Praxis ad Aequationes Algebraicas Resolvendas , was published after his death. An article on Harriot at the Univ of Saint Andrews math history web site says that in his personal writing on solving equations Harriot did use both positive and negative solutions, but his editor, Walter Warner, did not present this in his book."

And how did he come to be in the exploration of Virginia?? Here is the story from Encyclopedia Virginia, 2010:

Thomas Hariot (often spelled Harriot) was an English mathematician, astronomer, linguist, and experimental scientist. During the 1580s, he served as Sir Walter Raleigh's primary assistant in planning and attempting to establish the English colonies on Roanoke Island off the coast of present-day North Carolina. He taught Raleigh's sea captains to sail the Atlantic Ocean using sophisticated navigational methods not well understood in England at the time. He also learned the Algonquian language from two Virginia Indians, Wanchese and Manteo. In 1585, Hariot joined the expedition to Roanoke, which failed and returned to England the next year. During his stay in America, Hariot helped to explore the present-day Outer Banks region and, farther north, the Chesapeake Bay. He also collaborated with the artist John White in producing several maps notable at the time for their accuracy. Although Hariot left extensive papers, the only work published during his lifetime was "A Briefe and True Report of the New Found Land of Virginia", which evaluated the economic potential of Virginia. The report appeared most impressively in Theodor de Bry's 1590 edition that included etchings based on the White-Hariot maps and White's watercolors of Indian life. After a brief imprisonment in connection to the Gunpowder Plot (1605), Hariot calculated the orbit of Halley's Comet, sketched and mapped the moon, and observed sunspots. He died in 1621.

Harriott was not actually involved with the gunpowder plot, and was only held and questioned briefly because one of his financial sponsors, Henry Percy, the Ninth Earl of Northumberland, was a second cousin to Thomas Percy, one of the conspirators.