kitchen table math, the sequel: help desk - Dr. Chung

## Tuesday, June 21, 2011

### help desk - Dr. Chung

I think the solution manual may be wrong on this one.

I'm a huge fan of John Chung's SAT Math, btw. This is the first time I've encountered an answer that struck me as wrong.

source: Dr. John Chung's SAT Math

Daniel Ethier said...

I think the answer should be 16 darts. What does he say?

Catherine Johnson said...

He says 16!

I'm going to re-do my solution & see why I don't get 16 ---

thanks!

Catherine Johnson said...

'careless error'

btw: as much as I love Chung, it's not an easy book to use if you're self-teaching

I often find the solutions hard to follow.

In this case, I misunderstood Chung's solution. The solution I **thought** he was using was in fact wrong...but getting balled up like that meant that I didn't manage to pick up on my 'careless error'...

(I'm using 'careless error' to mean that I was approaching the problem correctly but missed a step & then missed the fact that I'd missed a step.)

Thanks, Daniel!

Rivka said...

I got 16 as well.

What method did he use? Mine worked, but took a long time. I found the area of the inner circle (9pi) first. The middle circle is 36pi - 9pi (because it has a 9pi-sized "hole"), or 27pi. The outer circle is 81pi - 36pi (area of the other two rings, which have already been accounted for), or 45pi. Then I added the area of the shaded parts together, 9pi + 45pi = 54pi, and divided by the total area (81pi) to determine what percentage of the darts fall in the shaded range.

As I say, that worked but I'm wondering if there's a faster way.

Anonymous said...

I think the key here is that the area of circles is a square function (doesn't that sound funny!!!)

So, inner bit is 1, next bit is 4-1, outer bit is 9-4.

5+1=6. ( inner plus outer).
Middle is 3.

6:3 = 24:16.

-Mark Roulo

Michael Weiss said...

Mark,
Your last step has me baffled.

Don't you mean: inner + outer = 6 (by which I assume you mean, the shaded area is 6 times the center circle); whole dartboard is 9 (i.e. 9 times the center circle); 6:9 = 16:24?

My approach, by the way, was to ignore factors of pi (since I knew we were interested in a ratio of areas, the pi's would all cancel out anyway). Outermost circle area is 81, middle circle area is 36, innermost circle area is 9; so shaded area is 81-36+9 = 54. As a fraction of total area, it's 54/81 = 2/3. (That's the steps where the pi's would have canceled if I had been keeping track of them.) So 2/3 of the darts would be expected to hit the shaded area.

Anonymous said...

Michael,

Yes (embarrased) :-)

-Mark

Catherine Johnson said...

I'll try to type up his solution later.

When I read the solution I thought -- incorrectly -- that Chung was saying that the ratio of the radii to each other was equal to the ratio of the areas.

kcab said...

I wonder if it's similar to how I approached it:

shaded area = 9^2 - (6^2-3^2)

which is

notice powers of 3 s.t. total is 3 times as large as non-shaded, therefore shaded area is 2/3 of total,
or:

factor out 3^3 this becomes:

3^3(3-1)

which gets to the same ratio of 2:3 => 16:24

Stuart Buck said...

I just do it this way:
Area of outer shaded area: 9 pi^2 minus 6 pi^2.
Area of inner shaded area: 3 pi^2.
Total shaded area: (9 - 6 + 3) pi^2, or 6 pi^2.

Thus, ratio of shaded area to entire circle is 6 pi^2 over 9 pi^2. The pi^2 terms cancel out, leaving 6/9, or 2/3.

2/3 of 24 is 16.

jd2718 said...

If the inner circle has area A units, and the rings all have width equal to the radius of the inner circle, then the areas will be 1A, 3A, 5A, 7A, ... and a the probability of landing in a shaded region as a function of the number of rings (call it P(R)) will be
P(1) = 1
P(2) = 1/4
P(3) = 2/3
P(4) = 3/8
P(5) = 3/5
P(6) = 5/12
P(7) = 4/7
and there are all sorts of neat, accessible patterns to play with and explore.

Jonathan