| terveloc |  18:41:23, 17 Nov 07 |
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| I got 1 + sqrt(2).
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| Jackson | 16:43:32, 09 Oct 07 |
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| ooh, bad Lawrence, didnt turn it into simplest radical form.... tsk tsk tsk... it should be (square root 2 +1) Just like sean bow.
isnt that simpler?
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| Lawrence Lee | 13:17:49, 09 Oct 07 |
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| Okay, this is pretty easy. You have a right triangle formed by the center of the inner circle, the center of any of the outer circles, and the point where two outer circles meet. Once you draw that triangle the solution becomes pretty straightforward as you realize that the hypotoneus of this triangle equals the radius of the inner circle plus the radius of the outer circle - r+1.
Therefore r+1 = square root of 2 x r and if you solve for r then r = 1 / (square root of 2 - 1)
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| Mabus | 17:13:57, 07 Oct 07 |
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| Try this: (2router)^2+(2router)^2 = (2router + 2rinner)^2
Pythagorean theorem
(the centers of the outer circles form a square = 2 triangles. )
<=> result of seanbow ...
<=> router = 1/(0.414213562...)
<=> router = 2.414213565 --> compare to drawing... works
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| seanbow | 13:58:34, 07 Oct 07 |
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| Or more generally:
Click to enlarge |
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| seanbow | 13:47:17, 07 Oct 07 |
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| Hm. I got 1 + sqrt(2) for the answer.
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| Gonzo | 10:48:23, 07 Oct 07 |
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| are you sure the answer cant be aproximated?
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