Actually, the "first period" (zero period) class is Ocean Science. Many students are with the regular teacher on a Beach Cleanup field trip. Those remaining are watching a video documentary about a tragedy involving a young surfer nearly a century ago -- along with subsequent efforts to make the beach safer for all. The other classes are all Honors Biology. Since these are honors classes, there are very few problems with behavior.
I've been discussing some science classes in more detail -- especially middle school in order to compare to my "science class" at the old charter school. It's trickier with high school, since there's no guarantee that anything from high school Bio belongs in middle school science. At any rate, here's what the students are working on -- a cell lab. The students have previously looked at cheek cells and red onion cells under a microscope and described how they change in a saline solution. I'm not sure whether I should have given such a lab in seventh grade (life) science, but notice that there were indeed microscopes in that classroom.
As I usually do for life science or Bio classes, I sing "Meet Me in Pomona." During tutorial, I also add my new version of Vi Hart's "Twelve Days of Christmath" -- a simpler version that omits some of Hart's advanced topics. (I posted this song on the blog last year but never sang it in class until today.)
Today on her Mathematics Calendar 2019, Theoni Pappas writes:
Line L is perpendicular to line M. Line L's equation is 28.6y + 1 = 3 - 2.2x. Line M passes through point (-10, 2). What is line M's slope?
This really is more of an Algebra I problem than Geometry, but since it does appear in Lesson 3-5 of the U of Chicago Geometry text, I might as well post it on the blog:
28.6y + 1 = 3 - 2.2x
28.6y = -2.2x + 2
y = (-2.2/28.6)x + 2/28.6
So the slope of line L is -2.2/28.6 = -22/286 = -1/13. We know that the slopes of perpendicular lines are opposite reciprocals, and thus the slope of line M is 13. Notice that the problem doesn't ask for the equation of M, so the point (-10, 2) is irrelevant. The desired slope is 13, and of course, today's date is the thirteenth -- Friday the thirteenth, that is. I apologize to those suffering with triskaidekaphobia -- I know we just have the unlucky day in September, and three months later we have it again.
Since I didn't write much last year about today's finals review worksheet, let's instead wrap up Putnam week with a discussion of a few more problems from the B session. We begin with B1:
https://artofproblemsolving.com/community/c7h1966348_2019_putnam_b1
Denote by Z^2 the set of all points (x, y) in the plane with integer coordinates. For each integer n > 0, let P_n be the subset of Z^2 consisting of the point (0, 0) together with all points (x, y) such that x^2 + y^2 = 2^k for some integer k > n. Determine, as a function of n, the number of four-point subsets of P_n whose elements are the vertices of a square.
There are several interesting things happening in this problem. First of all, the U of Chicago gives a special name to the set Z^2 of points with integer coordinates -- lattice points. (This term is defined in the Exploration Questions in Lesson 1-3.) And of course, we have x^2 + y^2 = 2^k, which is the equation of a circle centered at the origin with radius sqrt(2^k) (that is, 2^(k/2)). Thus first and foremost, we must locate and count the lattice points that lie on each such circle.
Interestingly enough, in Lesson 11-3 on circle equations, there is a question about finding lattice points on a circle. In that example, the equation was x^2 + y^2 = 25. This circle contains 12 lattice points because 25 is the sum of non-trivial squares: 9 + 16 = 25. But for this problem we need to find squares that sum to a power of two.
For the smallest powers of two, 0 + 1 = 1 and 1 + 1 = 2 are the only ways to partition them into two perfect squares. Thus the zeroth circle (that is, k = 0) contains four points (1, 0), (0, 1), (-1, 0), (0, -1), while the first circle also contains four points (1, 1), (-1, 1), (-1, -1), (1, -1). We notice that this generalizes into two patterns based on whether k is even or odd -- if k is even, then k is itself a perfect square and so the circle has four lattice points on the x- and y-axes. But if k is odd, then the four lattice points are on the diagonals y = x and y = -x. For example, 128 = 64 + 64 and 256 = 0 + 256 -- but as far as we know, there might be other squares summing to 128, 256, and higher powers.
Earlier this week, I mentioned the principle of induction and how it's often used to solve Putnam problems involving positive integers (natural numbers). It turns out that a form of induction can be used to show that for each k, the circle has only four lattice points. The method we will use is called infinite descent. (In the thread at the link above, "trumpeter" just calls it "descent." Technically, our proof of the irrationality of sqrt(2) from Dolciani is an infinite descent proof -- and indeed, this proof will be similar.)
Here's how we do it -- suppose we have x^2 + y^2 = 2^k for integers x, y, k. Except for the initial case 2^0, 2^k must be even, and so x^2 and y^2 must be either both even or both odd.
But as it turns out, x^2 and y^2 can't both be odd. This is because the square of an odd number isn't merely odd -- it must be one more than a multiple of four. The sum of two such numbers must be two more than a multiple of four. But except for the initial cases 2^0 and 2^1, 2k must be a multiple of four, not two more than such a multiple. This is a contradiction -- thus x^2, y^2 are both even.
(Using Dozenal Forum terminology, here's what we'd say: the square of an odd number isn't merely odd -- it must be overquadrate. The sum of two overquadrate numbers must be counterquadrate. But except for the initial cases 2^0 and 2^1, 2^k must be quadrate, not counterquadrate. This is a contradiction -- thus x^2 + y^2 are both even.)
OK, so now we have x^2 + y^2 = 2^k with x^2, y^2, and 2^k all even. Indeed, they must all be multiples of four (quadrate), since the square of a even number is a multiple of four. In that case, divide both sides by four. Now we have (x/2)^2 + (y/2)^2 = 2(k - 2) -- that is, we just found a lattice point for a smaller circle (k - 2 instead of k).
Then the same logic applies to this circle -- x/2 and y/2 can't be odd, so they're even. Then we would be able to divide by four again. This keeps going -- hence the name infinite descent. Actually, the descent ends when we reach one of the initial cases -- k = 0 (for k even) and k = 1 (for k odd). This tells us that for all k, the number of lattice points must the same as one of the initial cases -- and we already found exactly four points on the zeroth and first circles.
OK, so now we know how many points there are for each set P_n -- the origin, plus four points for each k from 0 up to n. But we're asked to find the number of squares, not points. Let's take it up from here with "awesomemath," who uses induction again. The conclusion is that P_0 has one square (the four lattice points on that circle form a square), and that each time n goes up by one, we add five more squares:
Note that we now add 5 squares. The one formed by the set we just added and the 4 if we cut that one in quarters. Any other new square has to include one of the new points but if an edge emanating from a new point ended with a vertex not from the aforementioned case then the other edge emanating from that vertex would not lie inside the convex hull of the points in our set and hence we obtain a contradiction.
And so the final answer is 5n - 1. The concepts behind this problem are simple, but making the proof fully rigorous is quite difficult. (How do we know there aren't any extra points or squares beyond the ones we know of.)
Let's peek at Question B3 now:
https://artofproblemsolving.com/community/c7h1966307_putnam_2019_b3
This question is on matrices and Linear Algebra (which we mentioned briefly back when we were reading Ian Stewart's book). But as it turns out, there is a connection to Geometry.
The question is about n * n matrices. It turns out that these correspond to transformations if the space is n-dimensional. (In the U of Chicago Algebra II text, Chapter 4 describes how matrices correspond to 2D transformations.)
The question mentions the word "orthogonal." It turns out that orthogonal matrices don't merely correspond to transformations -- they represent isometries. In particular, reflections (in a mirror passing through the origin) and rotations (centered at the origin, or with an axis passing through the origin in higher dimensions) have orthogonal matrices.
Read the rest of the thread linked above. The solvers discover that the P mentioned in the problem represents a certain reflection. The phrase "1 is an eigenvalue" has a geometric interpretation as well, namely that the isometry has a fixed point. Thus the statement to be proved becomes:
If Q (the original isometry) doesn't have a fixed point, then PQ (which turns out to be the composite of the reflection P following Q) must have a fixed point.
The U of Chicago text tells us that the only isometries that have matrices must fix the origin, and so by "has a fixed point" we mean "has a fixed point other than the origin." In the 2D case, Q ends up being a rotation (since the only isometry with the origin as its only fixed point is a rotation), and so PQ is the composite of a reflection and a rotation. Since the mirror of the reflection passes through the origin, the composite must also be a reflection -- and this reflection obviously has many fixed points besides the origin. (In particular, every point on the mirror is a fixed point.)
Ordinarily, problems near the end of the B session (like B5 and B6) involve Calculus or higher math that's too difficult to explain on the blog. But some participants were surprised to find that B5 and B6 are easier this year than in the past (while A1 was harder than usual). Indeed, this year's B5 is all about Fibonacci numbers and polynomials. The year 2019 makes a second appearance in B5.
Thus concludes Putnam Week on the blog and my discussion of the competition.
Let's continue our review for the final exam. This is what I wrote last year about the second half of the review worksheet:
Most of the questions on this half, which cover Chapters 5 through 7 of the U of Chicago text, are mostly self-explanatory. Notice that in Question 47, we are given that ABCD is a trapezoid with one pair of opposite angles congruent and we are to prove that it is a parallelogram. In other words, we are using the inclusive definition where a parallelogram is a trapezoid. If teachers prefer the exclusive definition, they can change the Given section to:
Here is the Review for Final, Part II.
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