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Topic: Circles Within a Circle (Read 889 times) |
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ThudnBlunder
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Circles Within a Circle
« on: Jun 13th, 2003, 9:06pm » |
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Imagine a circle, within which are three smaller circles that fit perfectly and all touch each other.
There will be a bit in the middle that has three curved sides.
What percentage of the the total area is this central bit?
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| « Last Edit: Jun 15th, 2003, 9:38am by ThudnBlunder » |
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Chronos
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Re: Circles Within a Circle
« Reply #1 on: Jun 13th, 2003, 11:17pm » |
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I think you need to specify that the three inner circles are all the same size. Without that constraint, the central region can be arbitrarily small.
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ThudnBlunder
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Re: Circles Within a Circle
« Reply #2 on: Jun 14th, 2003, 2:22am » |
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Quote:| I think you need to specify that the three inner circles are all the same size. |
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HINT: I believe I did.
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| « Last Edit: Jun 14th, 2003, 2:50am by ThudnBlunder » |
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redPEPPER
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Re: Circles Within a Circle
« Reply #3 on: Jun 14th, 2003, 5:50am » |
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How so? Is it by saying they "fit perfectly"? Maybe you can define that a little better. I assumed that meant they all touch the outer circle. But it's very possible to draw three circles that touch each other and the outer circle without these circles being the same size.
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LZJ
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Re: Circles Within a Circle
« Reply #4 on: Jun 14th, 2003, 5:54am » |
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I got (6*sqrt3 - 3*pi) / [2*pi * (7 + 4*sqrt3)], which comes up to about 1.10557 %.
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ThudnBlunder
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Re: Circles Within a Circle
« Reply #5 on: Jun 14th, 2003, 10:42am » |
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Quote:| How so? Is it by saying they "fit perfectly"? |
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I don't mean that.
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redPEPPER
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Re: Circles Within a Circle
« Reply #6 on: Jun 15th, 2003, 6:41am » |
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Could you clarify? I must be missing something. The way I understand it, each of these graphs complies to the riddle: three circles that touch each other and that "fit perfectly" (i.e. that also touch the outer circle). But clearly the red area is not the same on every graph.
I assume you mean A. Can you explain how the other three don't comply with the riddle?
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LZJ
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Re: Circles Within a Circle
« Reply #7 on: Jun 15th, 2003, 6:59am » |
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Perhaps the 3 circles of equal size comes from THUDandBLUNDER saying "There will be a bit in the middle".
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ThudnBlunder
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Re: Circles Within a Circle
« Reply #8 on: Jun 15th, 2003, 8:10am » |
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Quote:| Perhaps the 3 circles of equal size comes from THUDandBLUNDER saying "There will be a bit in the middle". |
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Yeah, that's right, LZJ!
Quote:| Can you explain how the other three don't comply with the riddle? |
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If the circles were not of equal diameter, the "bit in the middle" would not be in the middle, would it?  
MIDDLE: A point or area equidistant from all sides of something.
(American Heritage Dictionary)
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| « Last Edit: Jun 16th, 2003, 12:33am by ThudnBlunder » |
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redPEPPER
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Re: Circles Within a Circle
« Reply #9 on: Jun 15th, 2003, 10:01am » |
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The riddle doesn't say that the shape is formed so that the bit is in the middle. It says that, if we "imagine a circle, within which are three smaller circles that fit perfectly and all touch each other", there will be a bit in the middle. I formed the shape in four different ways, and found a bit in the middle of the three circles, in compliance with a broader meaning of the word "middle" which, until proven otherwise, is as acceptable as the other meaning:
Middle : being at neither extreme : INTERMEDIATE
(Merriam-Webster dictionary)
Oh, the importance of a properly worded riddle
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| « Last Edit: Jun 15th, 2003, 10:10am by redPEPPER » |
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Sir Col
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Re: Circles Within a Circle
« Reply #10 on: Jun 15th, 2003, 12:46pm » |
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This thread has been most amusing! 
My solution...
If we join the three centres of the internal circles, each with radius r, we form an equilateral triangle. The height, h=sqr(3)r. It can be seen that the centre of the large circle is also the centre of the triangle, and the distance from the centre of an equilateral triangle to its apex is (2/3)h. Hence the radius of large circle, R=r+(2/3)h=(r/3)(3+2sqr(3)).
As the triangle is equilateral, 3 sectors of 60o are contained inside it, which is equivalent to 1/2 of one small circle. That is, the area of small circles inside the triangle is pi*r2/2. Area of triangle=sqr(3)r2. Therefore area in middle is r2(sqr(3)–pi/2).
Area of large circle, A=(pi*r2/3)(6+4sqr(3)).
So ratio of middle bit to total circle=(r2(sqr(3)–pi/2))/((pi*r2/3)(6+4sqr(3))).
Which simplifies(?) to (12–6sqr(3)+pi(3–2sqr(3)))/(4pi), and is about 1.19% (different to LZJs?)
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| « Last Edit: Jun 15th, 2003, 12:50pm by Sir Col » |
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ThudnBlunder
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Re: Circles Within a Circle
« Reply #11 on: Jun 15th, 2003, 2:58pm » |
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Quote:Middle : being at neither extreme : INTERMEDIATE
(Merriam-Webster dictionary) |
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But, whereas your definition leads nowhere, my definition leads to an answer which cheekily asserts
its own existence.
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| « Last Edit: Jun 16th, 2003, 12:34am by ThudnBlunder » |
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redPEPPER
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Re: Circles Within a Circle
« Reply #12 on: Jun 16th, 2003, 3:59am » |
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Fair enough. Here's another one for you.
Draw two parallel lines. Draw two more parallel lines that intersect the first two. This will form a shape. Given that one side measures 2 meters, what is the area of the shape?
Answer:
If one side of the square is two meters, then obviously the square area is 4 square meters. Duh!
What? Couldn't calculate the area? You got a parallelogram and not a square? But I said "shape" as meaning "square".
Yes, a parallelogram is also a shape. But, whereas your definition leads nowhere, my definition leads to an answer which cheekily asserts its own existence.
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wowbagger
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Re: Circles Within a Circle
« Reply #13 on: Jun 16th, 2003, 5:50am » |
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LOL 
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ThudnBlunder
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Re: Circles Within a Circle
« Reply #14 on: Jun 16th, 2003, 10:02am » |
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Touché, redPEPPER, but I was thinking more in terms of the Sum/Product puzzle, where knowing that there exists a unique solution allows one to derive it. Or the Hole Drilled Through Sphere puzzle, where the same assumption allows one to solve it much more quickly.
In fact, your example is not analogous, as stated. If you gave me that puzzle and asked (as I did) for a numerical value, and not merely for 'the area of the shape', I would be forced to the conclusion that the shape is a square.
(Actually, the problem was intended to be part puzzle, part riddle - if you see what I mean.)
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| « Last Edit: Jun 17th, 2003, 12:28pm by ThudnBlunder » |
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LZJ
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Re: Circles Within a Circle
« Reply #15 on: Jun 16th, 2003, 10:12pm » |
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Amusing debate...
Anyway, Sir Col, shouldn't the area of the large circle be (pi*r2/3)(7+4sqr(3)), and not (pi*r2/3)(6+4sqr(3))? That's where the difference in our answer lies.
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Sir Col
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Re: Circles Within a Circle
« Reply #16 on: Jun 17th, 2003, 12:33am » |
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Thanks, LZJ, I just couldn't see where I'd gone wrong. In which case I get exactly the same way as you, the ratio is 3(2sqr(3)–pi)(7–4sqr(3))/(2pi).
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Chronos
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Re: Circles Within a Circle
« Reply #17 on: Jun 17th, 2003, 3:24pm » |
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OK, so we've specified that the concave triangle is in the middle. I would interpret that to mean, at its strictest, that the center of the circle is in the interior of that region. But that still leaves us wiggle room. I can't support the interpretation that "in the middle" means that the centroid of the concave triangle is the same as of the circle.
This is not analogous to the hole-in-the-sphere problem. There, the answer really is independent of the size of the sphere, although we probably don't know that other than inference from the fact that the problem asks for a unique solution. In this problem, though, the answer does depend on the relative sizes of the circles, so the implication that there is a unique answer is simply incorrect.
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Sir Col
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Re: Circles Within a Circle
« Reply #18 on: Jun 18th, 2003, 5:23pm » |
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If it's any consolation, T&B, I enjoyed the puzzle and used common sense to solve it. It seemed to me that it was an irresistibly interesting and challenging geometrical problem. If I'd missed some subtle trick that you had intended, then it wouldn't have bothered me in the slightest, as I would have just enjoyed solving the problem as I had interpreted it.
*adds it to his growing list of tricky problems for his classes*
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| « Last Edit: Jun 18th, 2003, 5:25pm by Sir Col » |
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