Good Math Problems

Algebraic Manipulation Is Overrated

An intuition question.

Look at the function below. It may surprise you that it is a constant! For any value of x, the function g will have the same value. I’m wondering, now that you know this, if you can get a sense of why it would be a constant, without (a) using your graphing calculator, or (b) taking the derivative to show that it is 0 [that is what I did, and as a side note, I have to use this on a test or homework next year].

g(x)=\frac{\sin(x)+\sin(x+a)}{\cos(x)-\cos(x+a)}

Can you find some geometric way to see that?

It took me somewhere between a half hour and an hour of playing around to get it. I can post my solution in a couple days, but right now I don’t have the energy to find a program to draw my solution [1]. But let me just tell you: it’s beautiful. You’ll be stunned when you first do it. Yeah, the calculus way tells you it is a constant, but seeing the “why” is still a mystery. The geometric way takes a bit, but whoa nellie, you won’t regret spending the time!

[1] Or maybe I should claim there is no room in the margin! (JK)

Update: I did finally write up my solution. I quickly did something I never have done before: do my work in powerpoint. It worked fine.

Update: Mr. K solved the problem in 3 minutes and found a way to show the geometric solution. Head over to his very excellent blog to see it in all it’s glory.

Update: Besides mine and Mr. Ks, a third and perhaps more elegant solution is up at 11011110.

Of the three, I think I like Mr. K’s visualization best, even though it might not be a proof in the formal sense.

6MMM^3

The 6th Monday Math Madness is online now. This week is actually not so hard, even though there are two different problems… I was able to get both of the answers in about 30 minutes (assuming I didn’t make any huge errors). I especially like the first question, because it can be easily transposed into a slightly more difficult and fun problem…

The first question reads:

  • Start with 500 gallons of mayonaise.
    1) Mix in 10 gallons of ketchup. Stir until completely mixed.
    2) Remove 10 gallons of the mixture.
    3) Repeat steps 1 and 2 until the mixture is approximately 50% mayonaise and 50% ketchup.How many iterations will it take to do this?

Here’s the slightly modified problem:

  • Start with 500 gallons of mayonaise.
    1) Mix in 3 gallons of mayonaise and 7 gallons of ketchup. Stir until completely mixed.
    2) Remove 10 gallons of the mixture.
    3) Repeat steps 1 and 2 until the mixture is approximately 40% mayonaise and 60% ketchup.How many iterations will it take to do this? 

That slight change makes it more difficult! But fun fun fun!

Two cows are in a field…

In math club this past week, we didn’t have anything to work on explicitly. So we just made up a problem, based on a problem we encountered in the previous week.

Without further ado, here it is. You have a circular field, enclosed by a fence. Two cows Antonio and Barry graze in the field. They are each tethered to some place on the circle, tied with ropes of lengths r_A and r_B respectively.

The problem is: come up with a formula for the area of the region that both cows can graze together.

I love that we came up with the problem, and that we’re exploring it ourselves. It’s great that it’s so simply stated, and that it has a pretty tough solution. I love that it’s a generalization of something we did earlier. And I love that even this problem can be generalized further (e.g. we have n cows).

What we did in 15 minutes:

We know we’re going to have a piecewise function of three variables. To start the problem, we make the circle a unit circle, we place Antonio at the point (1,0) and we place Barry at (\cos \theta, \sin \theta).

By the end of our math club meeting, we had one part of the piecewise function f(r_A, r_B, \theta). We found where there would be no overlapping grazing area, where the function would be zero.

I have some sketches of the problem and the bit of solution we got together. I’ll put them below in a bit.

Cavalieri’s Principle

Even though I don’t mince words when I proclaim my dislike of geometry, there are times when its simplicity and elegance strike me. Today the geometry teacher at my school was talking about Cavalieri’s Principle. Coincidentally, in my 7th grade math class, we are learning about volumes of prisms, cones, pyramids, and maybe spheres.

It got me thinking: without calculus, where do we get the formula for spheres? I mean, with calculus, it’s cake, but let’s assume we don’t have calculus.

These two things in mind led me to this great blog post on Cavalieri. It’s pretty much a proof without words, so I’m going to crib the picture from the post,

but the post has words explaining the picture. So click on the link and check it out if you are still scratching your head.

Generating Fibonacci: Part II

We left off in our quest for an explicit formula for the nth Fibonacci number having created this amazing generating function: g(x)=\frac{1}{1-x-x^2}.

To do what we’re about to do, I need to remind you of two precalculus things:

First, that \frac{1}{1-pq}=1+(pq)+(pq)^2+(pq)^3.... If you don’t know why, I suggest doing long division!

Second, partial fractions.

I’m going to go through this explanation assuming you know these two things.

So let’s look at the denominator of our g(x) and factor it. Okay, okay, you got me. There isn’t a nice factoring with integers. But it can still be factored, of course.

g(x)=\frac{1}{(1-x\phi)(1-x\overline{\phi})}, where \phi=\frac{1+\sqrt{5}}{2} and \overline{\phi}=\frac{1-\sqrt{5}}{2}. Using partial fractions, we get:

g(x)=\frac{a}{1-x\phi}+\frac{b}{1-x\overline{\phi}}=a(\frac{1}{1-x\phi})+b(\frac{1}{1-x\overline{\phi}})

We’ve made good headway, but what we don’t know are a and b! However, noticing that we can use the first precalculus topic above, that

g(x)=a(1+x\phi+x^2\phi^2+...)+b(1+x\overline{\phi}+x^2\overline{\phi}^2+...).

Rewriting this as a simple polynomial, we get:

g(x)=(a+b)+(a\phi+b\overline{\phi})x+(a\phi^2+b\overline{\phi}^2)x^2+...

Now we’re almost done! We use the initial conditions to find out a and b.

Since we know F_0=1 and F_1=1, we can say a+b=1 and a\phi+b\overline{\phi}=1. Solving these two equations simultaneously yields a=\frac{1}{\sqrt{5}} and b=-\frac{1}{\sqrt{5}}.

So the nth Fibonacci number is: \frac{1}{\sqrt{5}}(\phi^n)-\frac{1}{\sqrt{5}}(\overline{\phi}^n)

which simplifies to: \frac{1}{\sqrt{5}}(\phi^n-\overline{\phi}^n).

Which is what we set out to show! Huzzah! What’s also nice about this (besides the fact that it’s an integer, which is surprising) is that it shows that the Fibonacci sequence grows exponentially!