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.

Standard Deviations of Sums of Distributions

A week or so ago I was at a textbook selection conference with a couple of really good teachers, and one of them (thanks, Dru) pulled out a copy of Robert Hayden's, "Advice to Mathematics Teachers on Evaluating Statistics Textbooks." I mention it now because it has two good pieces of advice. (Ok, it has way more pieces of good advice than that, but I'm mentioning these two in particular)

The first, I hope I follow, "...make sure the textbook mentions assumptions and teaches students to check them rather than make them." One of the ways I try to get students to check assumptions is to make them understand, as much as possible in the limited time of a AP course, the WHY. In order to do that, I frequently violate one of Professor Hayden's other pieces of wisdom; "Be wary of an author who is not familiar with enough real data sets to illustrate a textbook."

Ok, I'm gonna claim some "weasel" room here. First, I'm thinking more along the lines of an exercise to help the students understand why checking independence is so important, and not writing a textbook. Second, even Professor Bob himself says "While there may be places (such as Anscombe’s regression examples , in which a skillfully fabricated batch of numbers illustrates a pedagogical point,.." Ok, so the "skillfully" may not apply to what follows, but I hope the fabricated data at least help drive home a "pedagogical point".

I begin with two simple data populations, X= {1,1,1,2,2,2,3,3,3} and Y= {1,1,1,3,3,3,5,5,5}. Students who have learned the "Standard Deviation as Distance" approach can quickly check and find the standard deviation of the X population (or using a calculator) is sqrt(2/3)or appx .8165. For Y the std. dev. is 1.633. Perhaps for what we will be doing, we remind them that the variance of each is the square of the standard deviation, so Var(X)=2/3 and Var(Y)= 8/3.

So what happens if we add or subtract the populations? It all depends! If the populations are independent, then any X and any Y may (must?) be associated with equal probability. I illustrate this by pairing one of each X value with one of each Y.. (is it possible to have two distributions be independent without this type of each x with each y association?)

X___1___1___1___2___2___2___3___3___3

Y___1___3___5___1___3___5___1___3___5.

and the sum and differences are then

X+Y =2___4___6___3___5___7___4___6___8 and

X-Y =0__-2__-4___1__-1__-3___2___0__-2

I think it is worth drawing the two resulting distributions because many students will NOT see that these are distributions are reflections of each other. So they should have exactly the same standard deviations (this takes a moments reflection for some students).




Wow, that's good news. If the populations items are independent of each other in the way they are combined, it doesn't matter if you add them or subtract them, the spread is the same since the two distributions are symmetric, which means the standard deviations should (and are) the same, about 1.8257. Even better, we can point out that the variance, 10/3, is simply the sum of the original variances, 2/3 + 8/3. For me it is worth pointing out this "Pythagorean" relationship, [StDev(X+Y)]²=[StDev(X)]²+[StDev(Y)]², IFF X and Y are independently associated.....(oops, I have been called out on this mistake... The statement is true IF x and y are independent, but also in any situation in which the correlation coefficient is zero... which does not necessarily require independence...see comment from "gasstationwithoutpumps" below... "mia culpa" and thanks to "gas..."

BUT... what if the original populations were NOT independent. (quick, think of two data sets that you would really combine in real life that are totally independent...better yet, send your ideas in the comments)

Well they might have a positive or a negative correlation, so we slightly rearrange our data sets and group lower numbers somewhat together (no Ones with the fives) like this..

X___1___1___1___2___2___2___3___3___3

Y___1___3___1___1___3___5___5___3___5.

Now our sums and differences are

X+Y =2___3___2___3___4___5___6___5___6 and

X-Y =0__-1___0___1___0__-1___0___1___0

We recognize quickly that the sets no longer have the same shapes. The distribution of sums is almost uniform with the peaks at the ends, while the difference distribution has two peaks closer to the center .

So what are the spread measures now. The standard deviation of the summation distribution is 2.26 or the square root of the variance of 46/9. The differences have a standard deviation of 1.247, the square root of a variance of 14/9, a really big difference. In fact, we help the students notice that the variances are the same distance from the equal variance of 10/3 = 30/9 when the populations were combined independently. The distribution of sums variance is 16/9 higher, the differences are 16/9 lower. Is this just a curious coincidence...(by now my students know that almost NOTHING I bring up is a "curious coincidence" ).

So how can we explain this difference. Slowly you lead their thinking...."If the distributions are NOT independent, they must be dependent,.... and there must be some relationship,..... some measure of how UN-independent they are." Eventually they will think of the correlation coefficient, r. In this association between X and Y they have a positive correlation of 2/3 ... can that help. If the relationship when the association was independent is "Pythagorean", maybe we can look for some extension of the Pythagorean theorem to help... Can we find something like the Law of Cosines that would tie the package together? After all, we need something that will add 16/9 to the sum distribution, and subtract the same amount for the differences... I can't imagine that I would have kids who would see this, and will probably lead them to observe that StDev(X+Y)=[StDev(X)]²+[StDev(Y)]²+2 r [StDev(X)][StDev(Y)]. They can quickly test that the change of sign leads to
StDev(X-Y)=[StDev(X)]²+[StDev(Y)]² - 2 r [StDev(X)][StDev(Y)].

I hope before I get to this point I have laid a foundation for this by giving a short presentation based on a blog from John D Cook at "The Endeavor" that shows this geometrical relation between the correlation coefficient and the cosine of an angle. I hope to write a blog about this relationship in a more vector sense later.

All of this follows in the wake of a warning about non-real data from Professor Hayden, so it is important to follow up with real data that should bare this out. I'm thinking something simple like their own age in months and height. If it is true for all data sets, it should be true with the measures we have about them; but I am very willing to consider suggestions about a more appropriate data base.

Conics with Vectors

Almost every Alg II class in America ( and most geometry classes) contains problems such as "Write the equation of a circle with a radius of 5 and a center at (3,1)." The assumption is that the circle lies in the x-y plane.

I stood in an Alg II classroom recently as students worked through a worksheet of such problems and wondered how hard it would be to extend this idea to write the equation of a circle with a given radius and center in ANY plane. The response is that it should be amazingly easy if we are familiar with the simpler ideas of using vectors.

The first step would be to develop the vector equivalent of the standard circle
equation, something like (x,y) = (h, k) + r cos(t) (1,0) + r sin(t) (0,1) where t has a domain from 0 to 2 Pi (or the equivalent in degrees). Then in any other plane, all that would be needed would be to replace the two unit vectors in the x and y direction with two perpendicular unit vectors in the desired plane.

Since each of the desired vectors would be perpendicular to each other and the normal to the plane we could pick one arbitrarily, then find the second using the cross product of the first with the normal to the plane.

Here is an example: " Find the circle with radius 5 and center at (3,1,2) in the plane x + y + 2z = 8. " Any vector perpendicular to (1,1,2) will work for the first direction vector, so (2,0,-1) is an easy choice. The cross product of (1,1,2) and (2,0,-1) will then be perpendicular to (2,0,-1) and lie in the desired plane. The cross product gives (1,-5, 2) for the perpendicular direction vector. Now we just write the equation using unit vectors in the directions of (2,0,-1) and (1,-5, 2). The equation for the circle requested is (x,y,z) = (3,1,2) + 5 cos(t) (2,0,-1)/sqrt(5) + 5 sin(t) (1,-5,2)/sqrt(30).

Checking that a given value of t meets all the requested, I pick pi/6 as an easy
choice. The point will be (x,y,z) = (3,1,2) + 5 sqrt(3)/2(2,0,-1)/sqrt(5) + (5/2)(1,-5,2)/sqrt(30). So x=7.3294 (appx) and y=-1.2822 and z= 0.9764.

If we are close this point should lie in the correct plane, and also be five units from the center, (3,1,2).

One more thing that starts to happen when you use vectors and write an equation in the x-y plane, such as a circle, is that you begin to imagine pulling it out in the z-direction. If you do this as a function of t, then you create a helix that coils around the z-axis as it climbs. A unit circle climbing at a rate of t/pi will climb one unit up the z-axis for each half turn of the circle. Using the equations x,y,z= cos(t)(1,0,0) + sin(t) (0,1,0) + t/pi (0,0,1) is an example.


Other conics: It is simple to extend the circle to an ellipse by just replacing the radius in the appropriate places with the semi-major/minor axis lengths, a and b. The Ellipse centered at (3,1) with a=5, b=4 would be (x,y) = (3, 1) + 5 cos(t) (1,0) + 4 sin(t) (0,1).

As with quadratics, it is easy to write the ellipse rotated to lie along any line through the center. For example, if we wanted to rotate the ellipse above so that it made a 45 degree angle, we simply replace the two unit vectors in the x and y direction with unit vectors [sqr(2)/2, sqr(2)/2] and [-sqr(2)/2, sqr(2)/2].
Using decimal approximations for ease of writing, the equation becomes (x,y) = (3, 1) + 5 cos(t) (.707, .707) + 4 sin(t) (-.707,.707). The Winplot graph looks like this:


For fans of Geogebra, you can download a ggb sketch that will allow you to rotate a conic through any radian value.

For the hyperbola, simply replace the cosine and sine with tangent and secant (or secant and tangent to have the two branches open the other way).