A request from Nathan Kraft tied together a bunch math ideas from my day today

I had a nice time teaching the kids this morning.  We were covering some exciting material – stuff that is  fun all by itself while also providing great examples of mathematical thinking.  Unfortunately I goofed up one of the examples with my older son, but who knows, maybe the lesson was made more memorable because of the goof up 🙂

Anyway, these fun ideas were kicking around in my mind all day and  then a fun question that Nathan Kraft put on twitter tonight just happened to tie them all together.  Love it when that happens. Here is Nathan’s question:

The interesting thing about this question is that you don’t have to actually find the value of the shaded area to find the answer.  In fact, trying to find that area makes the problem much harder.  (I’m not going to give away the answer, but a hint would be that this sure looks an awful lot like a Venn diagram problem, doesn’t it?)    It is pretty amazing that you can find the answer to this problem without knowing the area of the shaded regions – but that shows how powerful some ideas in math can be.    By coincidence, I touched on this point in three different ways with my kids this morning.

My younger son and I started a new chapter in our Number Theory book this morning.  The first topic in the chapter is counting divisors.   A typical problem would be how many different divisors does the number 100 have?  Prior to this chapter he would have solved that problem by making a list and counting each element.  This new chapter shows a different approach – one that allows us to say how many divisors there are without actually knowing what all of them are.  It is pretty amazing that you can count the divisors without knowing what they are!!  Quite a beautiful illustration of the power of mathematics:

Next, my older son had a little difficulty with this problem from an old AMC 8:

http://www.artofproblemsolving.com/Wiki/index.php/1992_AJHSME_Problems/Problem_21%20

The difficulty he was having was that he didn’t see how to solve this problem without checking the percent increase on all 5 pairs of bar charts.   But, a little bit of mathematical thinking helps you see that you don’t need to do every computation.   There are 3 examples were the difference between the two sales charts is 2 boxes and there are 2 examples where the difference between the two sales charts is 3 boxes.    Since you are looking for the largest percent increase, you only need to compare the largest percent increase from each of these two sets.    The amazing thing here is that you can know which chart has the largest percent increase in each of those two sets without actually doing any computation since the percent increase will be largest when the starting point is the smallest.  So, just eyeballing the charts now we can see that February’s percent increase (an increase of about 66%) is the one that we need to compare to March’s (an increase of 50%) and we see that February had the largest percent increase.  Again, we can see that February has the largest percent increase without calculating what the percent increase was in January, April, or May.    An amazing idea when you are seeing it for the first time – or at least that’s the impression I got this morning!

Finally an old V.I. Arnold problem that I saw on Tanya Kovanova’s blog a while ago.  Unfortunately her blog has been damaged by a spam attack so the problem isn’t up right now.  Hopefully she’s able to save the blog.

Now, this problem shows basically the exact opposite of what I’ve been talking about up to now.  In the prior three examples we found (or asserted) solutions without doing lots of calculation.  Here we’ll see the danger of computing without thinking.  The problem itself is easy to state:  Suppose you have a right triangle with a hypotenuse of 10 and that the altitude to that hypotenuse has a length of 6.  What is the area of this triangle?

It is funny sometimes how little seemingly unrelated ideas can suddenly tie a bunch of other ideas together.  Thanks to Nathan Kraft for providing that little spark tonight.

4 thoughts on “A request from Nathan Kraft tied together a bunch math ideas from my day today”

1. O.. says:

Mike, I really like the first two problems as they match my world-view that if you’re doing complicated math then it’s a good chance you don’t understand the problem well enough to simplify the solution or to ask a simpler problem. Nathan’s problem is especially fun because of the Gordian knot type simple one line solution. But I’m philosophically confused by your response to the last problem. Why isn’t the answer whatever crazy imaginary number pops out of the quadratic formula? To me, the rules of math problems are: Somebody asks the question. Somebody else answers the question. Does it have to make sense? No, that’s what makes it fun. If the question is “Suppose you have two ropes each of which takes 30 minutes to burn from end-to-end. Now time 45 minutes.” And an acceptable answer is “Ah screw you question asker, I’m just going to set a 45 minute reminder on my iPhone.” Doesn’t that destroy the whole soul of the math problem? Is this because you haven’t covered imaginary numbers with your kids yet?

1. I have covered imaginary numbers a little bit with both boys, and even the quadratic formula with my older son.

I guess it comes down to whether or not you think the length of the side of a triangle can be an imaginary number. That idea would certainly be fun -> please draw a right triangle with sides 5, 4i, and hypotenuse 3, for example.

I think triangles in plane geometry would normally be thought to have lengths that were positive real numbers, though, and the main point of this example is that if you apply the formulas without considering the geometry, you’ll run into a problem.

1. yes, a length is defined to be a non-negative real number. As a more general concept, it is a metric on a topological space, a function from pairs of points in the space where the function has to satisfy certain criteria. As with all things in math, you get to interesting places if you eliminate or relax some of the criteria and study what comes out.

Another angle for this problem is to consider other ways to see the collection of right triangles with hypotenuse of length 10. In particular, draw a circle with radius 5 centered on the midpoint of the hypotenuse. You can choose any third point on the circle as the other vertex of a right triangle. It is pretty intuitive that the largest height you can get to the hypotenuse is the isosceles right triangle. As an added bonus, because the height is a radius of the circle, you see right away it is length 5, so a longer length is impossible.

BTW, really enjoyed the moment when you son said “maybe it doesn’t exist.”

2. Another thought: the approach outlined could be considered a proof by contradiction where some assumption has been shown to be false. We have a bunch of choices for which assumption, not just the assumption that this triangle exists. O suggested this might show that the assumption that lengths must be real numbers is false and this might be worth pursuing to see what other implications it might have and whether there is an interesting model of such a geometry.

Another place to look for a “false” assumption is within the axioms of geometry. One standard postulate to pick is the parallel postulate (given a line and a point not on that line, there is a unique line through that point parallel to the first line). If we assume that no such parallel line exists, then a right triangle with hypotenuse 10 and height 6 is possible.