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Apr 6th, 2002, 10:32 PM
#1
Thread Starter
Frenzied Member
Geometry
Let's say you have a square whose sides are 8 units long. You also have a circle that is congruent to the square on all four sides. We have ANOTHER circle congruent to the circle at one point and to the square at two points. What is the radius of the smaller circle?
Code:
+------------+
| / \ |
| / \ |
|/ \|
| |
|\ /|
| \ / |
|()\ / |
+------------+
We need to find the radius of circle ().
MicroBasic
Dragon Shadow Trainer
There is no good or evil in the world...only programmers and fools .
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Apr 7th, 2002, 07:07 AM
#2
I think...
Radius of big circle = 4
Distance centre to vertex = 4*sqrt(2)
let radius of smaller circle be R
=> 4*sqrt(2) = 4 + R + R*sqrt(2)
=> 4*(sqrt(2)-1) = R*(sqrt2+1)
=> 4 *(sqrt(2)-1)/(sqrt(2)+1) = R
=> R = 0.68629... (I think, but i only used a calc. with no bodmas, so it is unreliable ) It may be wrong, so you probably should check.
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Apr 7th, 2002, 05:02 PM
#3
Hyperactive Member
Verified.
The exact figure is 4 (sqrt(2) -1)^2.
There are 10 types of people in the world - those that understand binary, and those that don't.
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Apr 8th, 2002, 05:13 AM
#4
WOW!
Cool, i'd never thought before, but:
(sqrt(2) - 1)/(sqrt(2) + 1) = (sqrt(2) - 1)^2
i.e. (sqrt(2) - 1)^-1 = (sqrt(2) + 1)
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Apr 8th, 2002, 12:10 PM
#5
Hyperactive Member
Hmm, a useful case of rationalising the denominator. Worth remembering.
There are 10 types of people in the world - those that understand binary, and those that don't.
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