## Saturday, September 03, 2011

### enjoy

I came up with the first version of this problem while waiting for my students to complete some exercises. One of my students saw the original version and came up with this slightly more complicated version, which I like better. _

Surprises Aplenty said...

I haven't worked on the answer yet, but the question seems Archemedian. To determine Pi, he put various polygons inside and outside of a circle. Here, the small square is the lower limit for the circumference and the large square is the upper limit. Do this again for a pentagon, hexagon, octagon... and the upper and lower limits converge, giving you some info on Pi.

Kevin Kim said...

I vaguely remember hearing/reading about that, but I couldn't remember exactly how the polygon thing went, and I'd completely forgotten that this was attributed to Archimedes. Thanks for reminding me; I'll use that on my students.

Ever read the novel Contact? Sagan posits (fictionally) a very interesting property of pi at the end of his story.

Horace Jeffery Hodges said...

This problem is so simple-minded that I refuse to do it in case I get it wrong . . .

Jeffery Hodges

* * *

Kevin Kim said...

A brave choice.

Horace Jeffery Hodges said...

Okay, I'll be even braver.

We're asked to combine 1, 5, and 6. Mere addition, quite simple:

1 + 5 + 6 = 12

QED (i.e., Latin for "Quite Easily Done")

Basically, every math problem is about arithmetic and can be solved by addition . . .

Jeffery Hodges

* * *

Kevin Kim said...

This is uncannily like dealing with one of my high schoolers.

Horace Jeffery Hodges said...

I actually do know the answer and how to get it.

(I looked it up on the internet, so I know I'm right.)

Jeffery Hodges

* * *

Kevin Kim said...

I suspect you do know the answer. Perhaps you're being gallant and allowing others to answer first.

Horace Jeffery Hodges said...

I don't know what "gallant" means, but if you're suggesting that I'm being arrogant and letting others make fools of themselves first . . .

Actually, that's a good idea.

Jeffery Hodges

* * *

Kevin Kim said...

It's Labor Day weekend, so I'm going to assume, based on the silence, that the usual crowd isn't around to try this problem out-- in which case I may as well explain the answer here.

It's given that ABCD and EFGH are squares, and that the area of Circle Q is 16π. The simplest and most obvious conclusion we can draw is that the radius of the circle is 4. From this, we can conclude that the larger square has sides measuring 8 units, since the distance from the midpoint of Segment AD to the midpoint of Segment BC must be 2r, i.e., [2 * 4], which is 8. The area of Square ABCD is 64.

The inscribed square, Square EFGH, has diagonals that also measure 2r, i.e., 8 units. Using the Pythagorean theorem, we can find the lengths of Square EFGH's sides by setting up this equation:

[x^2] + [x^2] = [8^2]

This becomes

2(x^2) = 64, or

x^2 = 32.

But wait-- we don't actually need to know the lengths of the sides of Square EFGH: we only need to know the area, which we now know is 32.

At this point, we have all the information we need to solve the problem.

The combined area of Regions 1 through 4 is the difference of 64 and the area of Circle Q, already given as 16π. Let's use the following shorthand:

A(S) = Area of the bigger square
A(s) = Area of the smaller square
A(Q) = Area of Circle Q

So:

A(S) - A(Q) = 64 - 16π

This means the area of a single region, whether it be 1 or 2 or 3 or 4, must be one fourth the above, or 16 - 4π.

Meanwhile, the combined area of Regions 5, 6, 7, and 8 can be calculated thus:

A(Q) - A(s) = 16π - 32

The original problem asks for the combined area of Regions 1, 5, and 6, so we need to know the area of two out of four of the above-named regions (Regions 5 and 6 only), i.e., half of [16π - 32]. That would be 8π - 16.

At this point we're ready for the last step. More shorthand to help us out:

A(1) = Area of Region 1
A(5 + 6) = Combined Area of Regions 5 and 6
A(1 + 5 + 6) = Combined Area of Regions 1, 5, and 6

We know that

A(1) = 16 - 4π
A(5 + 6) = 8π - 16

All we need to do is add these together to find A(1 + 5 + 6).

A(1 + 5 + 6) =

[16 - 4π] + [8π - 16],

which becomes a simple

4π.

Quite Easily Done.