Accepting Obsolescence
In which my kid is better at math than I am
Both SteelyKid and The Pip have always excelled at math in school, and are on the “accelerated” track in their respective grades. The Pip is good enough that he generally blows through the homework quickly enough that I never see it, but last night was an unusual occasion when he brought me a problem and asked how to do it.
They’re working on area problems involving circles, and the question (which had been designated as an optional bonus problem by his teacher) was to determine the area of the darker region in this figure:
That’s a quarter-circle with two overlapping half-circles in it, and my first reaction was “Yeah, Little Dude, that’s a tough one…” He went away to work on the rest of the required figures, but it kept bugging me, and I eventually figured out how to brute-force an answer by drawing a square:
Now you’ve got a big quarter-circle of radius 2r with two smaller quarter-circles of radius r removed from it (the light bits outside the square). Then you also need to remove the two light areas inside the square, but each of those has the same area, namely the area of the square with side length r minus a quarter-circle of radius r. So the calculation is:
I walked him through it, and he seemed a little dubious at the complexity of it, but got the basic idea. At bedtime, I showed it to SteelyKid, who enjoys these kinds of problems, and explained my reasoning. SteelyKid’s immediate reaction was “Ugh, I hate it. There has to be a more elegant way to do that.”
And, in less time than it took me to come up with the brute-force method, SteelyKid figured it out. The reasoning goes like this:
— The big quarter-circle has twice the radius of the small half-circles, which means that each half-circle has half the area of the big quarter-circle. If they didn’t overlap, they would completely cover the big quarter-circle.
— The darker region inside the square, where the two half-circles overlap, is one-half of the total dark region, because for every bit of area you “uncover” in the region outside the square, you “double-cover” inside the square.
— Then you find the area of the overlap region more or less as I did above, and double it:
And, I have to admit, that is more elegant than what I was doing.
This is pretty typical of the different relationships SteelyKid and I have to math. I’m very much a low-energy experimental physicist, comfortable with the idea of math as a tool to solve problems, but a little bit clumsy with it. I do a lot of brute-forcing of problems by breaking them into chunks that I can solve simply, and adding up a whole lot of dumb little solutions. (I have the same basic approach to data-intensive problems— when I have to crunch numbers for administrative purposes, I tend to bang everything into a big spreadsheet and do a lot of small tedious if/then operations generating loads of additional columns before assembling the final numbers.)
SteelyKid, on the other hand, has a much more intuitive approach to high-level math than I do— spotting immediately that the area of the overlap region in the figure above would be half of the total. It took me about three tries to follow the logic that leads to that conclusion. It’s much more like an actual mathematician, or a theoretical physicist.
So, in much the same way that I’ve had to adjust to the fact that The Pip is a vastly better baseball player than I ever was, I’ve had to accept that SteelyKid has a better feel for math than I do. Which I guess is an important parenting milestone: watching your kids find the areas where they can surpass you.
I can still whup both their asses in basketball, though. I’ve got that going for me, for at least a little while longer…
A bit of a parental #humblebrag, I guess, but that’s what you get. If you enjoyed this, here’s a button:
And if you know an even more elegant solution to this problem, or have other thoughts on math, the comments will be open:
I had to read both solutions and stare for a long time before I realized:
Take the lozenge shape and cut it in half along the long axis. Put each half in the chord of the smaller circles. Now the remaining shape is a right triangle with short sides of length 2r. πr² - 2r² = (π-2)r²
Fun post and neat problem. Great that your kids are doing so well in math. I did manage to see the overlap dark section is half the total dark area, but then got stuck trying to find its area in a geometric way. Really appreciated Chris's approach once I read it.