Pi from High School Maths
theartofmachinery.com
theartofmachinery.com
The best I could come up with is related to a student using the inscribed n-gon method for computing a lower bound.
The two obvious choices for an inscribed n-gon are a 6-gon and a 12-gon.
Computing their perimeters involves computing cos(pi/3) or cos(pi/6) respectivly.
Both of these are generally found the same way. By bisecting an equalateral triangle, you get a right angle triangle with angles pi/3 and pi/6, from which sin/cos/tan can easily be read.
Once you have the relevent cos, it is straightforward trig and algebra to find
With a 6-gon: pi>3
With a 12-gon: pi>6sqrt(2-sqrt(3))
Using a calculator, the 12-gon lower bound is pi>3.1058...
Computing by hand (which I did), you end up having enough wiggle room to get away with some further approximations; but still need to go through some somewhat tedious computation.
Maybe they were concerned that some students solution for pi>3 that I didn't, but as far as I have been able to see, the extra .05 just makes the problem more messy.
I.e - doing 100 problems with Pi = 3 probably doesn't even require a calculator, whereas doing the same problems with a decimal approximation will require students to pull out their calculators, and therefore students can do more problems in a shorter time, using pi = 3. This would certainly make sense in countries where rote learning is the king, and students are pretty much churning through problems.
Or, it could be that the vast majority of people will never need the use of an accurate pi value, and it's more beneficial to make them remember some easy value like 3, rather than a confusing decimal or fraction. The people that are going to use pi in any serious manner, will obviously know its value and derivation.
That's robbing them of getting an (intuitive) understanding of a whole class of numbers. And limits too!
This minor change in the official curriculum came at a time when the Japanese government was trying to de-emphasize rote memorization and promote critical thinking in school education. Because that policy also reduced class hours and the amount of material covered, it was widely criticized for supposedly dumbing down school education. The notion that children were being taught “pi = 3” was an exaggeration, but it resonated with other objections to educational policy and apparently led the university to pose that problem on its entrance exam in protest.
I still think it's an interesting problem. I wonder how far I could have gotten with it if, at the age of 18 or 19, I had encountered it on an exam and had just twenty or thirty minutes to try to come up with a proof.
> I wonder how far I could have gotten with it if, at the age of 18 or 19, I had encountered it on an exam and had just twenty or thirty minutes to try to come up with a proof.
It's a nice one. Especially because since pi shows up in so many places there are multiple ways of doing it.
Would you care to compile and share them?
Basically, a lot of things can be thought of as something vaguely geometric, and then enumerated discretely using some method of enumeration (e.g. a grid) to form an approximation to the value, which is a known bound. It feels like the general idea can be used for things other than approximating real values (something to do with property membership and graphs?) but I can't give any concrete examples of that.
One of the ideas I got from this, which I think was what spurred me to write that comment, was doing it in reverse. You can think of certain discrete systems as being, in some nebulous way stronger than a bound, inextricably linked to a continuous function. Kind of like Pisot–Vijayaraghavan numbers,[0] but more so. There was something about the resource requirements of an algorithm and applying this technique to compiler optimisation, but that's an incomplete thought.
Lots of things.
[0]: https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_n...
You can basically make a few experiments to illustrate that the perimeter of a circle is proportional to its radius and then ask them to calculate the perimeter another circle knowing only its radius.
pi is approximately: (2 x the stick length * the number dropped) divided by (the distance between lines * sticks crossing a line)
import random
n=1000000
print(4.*sum(random.random()**2 + random.random()**2 < 1 for i in range(n))/n)
Both this version and the Buffon's Needle version are a lot less efficient than the version in this article, because the article's version is making educated guesses about where the boundary is, while the randomized versions are making uneducated guesses. :-)https://lee-phillips.org/pidaydarts/
I think it was my first Svelte application.
We did the paper airplane version in class, which was fun.