A great morning talking math with the boys

Some days working with the boys in the morning goes really well and I’m left super charged up for the rest of the day. Love days like this.

Today I started a new chapter in Art of Problem Solving’s
Introduction to Number theory book with my younger son. This chapter is about modular arithmetic, which is a topic that we’ve touched on a little here and there but never in any formal way.

The book introduces modular arithmetic in a pretty natural way – with clock arithmetic and also by looking at rows of integers and pointing out which numbers are in the same columns. The only real speed bump in this introductory section came when we looked at negative numbers. It wasn’t immediately obvious to my son that, say, -1 was the same as 7 modulo 8, but the analogies to clocks helped get us over this hump. I really enjoy watching kids digest new math topics.

While I was working through this chapter with my younger son, my older son was working on some challenge problems from an old AMC10. Problem #14 from the 2000 AMC10 gave him some trouble:

2000 AMC10 Problem #14

Here’s the problem:

“Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?”

So, by funny coincidence, my older son had a modular arithmetic problem, too. In talking through this problem with him we noted that:

(1) There’s nothing to learn from the first number,
(2) The second number has to have the same parity as the first number, and
(3) Then things get hard šŸ™‚

I showed him that looking at the five numbers modulo three helped with the third step. The numbers become 2, 1, 2, 1, and 1. There’s only one way to have the numbers add up to be divisible by 3 – you have to take the three numbers that are congruent to 1 mod 3. The interesting thing is that you don’t know the order, but you do know the first three numbers.

The same approach works for the 4th number since 76, 82, and 91 add up to be 1 mod 4, we need a number that is 3 mod 4 to be the 4th number. The two choices are 80 and 71, so 71 is the only choice. That means the last number must be 80.

So, wonderful start to the day talking introductory modular arithmetic with my younger son, and then some slightly more advanced ideas in the same topic with my older son.

With that problem behind us we moved on to our Introduction to Geometry book. We are finishing up the review section in the chapter about triangles and stumbled on an absolutely wonderful problem:


Problem 7.46
: Medians AX and BY of triangle ABC are perpendicular at point O. AX = 12 and BC = 10.

(a) Find AO and BY.
(b) Find the length of median CZ.

Part (a) led to a good discussion about properties of medians, but did not require a deep dive into the math. Several of the other review problems have been about medians, so basic properties of medians are pretty fresh in his mind.

Part (b), though . . . great question! His approach was really interesting to me, but it was pretty difficult algebraically. He understands that the three medians chop up the original triangle into 6 smaller triangles with equal area. He used this fact, the fact that he knew two legs of one of the triangles that had CX as a third leg, and Heron’s formula to try to solve for CX.

This is obviously some messy algebra, though I think it is important to be able to learn to work through algebra like this. We went through it slowly and did eventually arrive at the answer.

An interesting geometric solution involves not looking at CX, but at XZ and the side AB of the original triangle. In this approach you can use the idea that the median to the hypotenuse of a right triangle has length equal to half the length of the hypotenuse to solve the problem. This fact is a slightly advanced geometry fact, but one that was fun to talk through.

The combination of these two solutions made for a great 30 minutes of talking through this problem.

This whole morning left me really happy and serves as another fun example of why I love teaching my kids math.

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