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Topic: Not Both Algebraic (Read 478 times) |
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ThudnBlunder
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Not Both Algebraic
« on: Nov 2nd, 2008, 2:25pm » |
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A rectangle R is inscribed in a circle C. Let x be the ratio of the area of C to the area of R and let y be the ratio of the circumference of C to the perimeter of R.
Prove that not both of x and y can be algebraic numbers.
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ThudnBlunder
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Re: Not Both Algebraic
« Reply #1 on: Dec 19th, 2008, 12:16am » |
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HINT:Prove that both x and y are algebraic implies  is algebraic.
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| « Last Edit: Dec 19th, 2008, 12:31am by ThudnBlunder » |
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Eigenray
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Re: Not Both Algebraic
« Reply #2 on: Dec 19th, 2008, 12:38am » |
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Let the circle have radius 1, so the sides of the rectangle are 2cos t, 2sin t. Then
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A = sin t cos t/ and B = (sin t + cos t)/ are both algebraic. But
1 = sin2t + cos2t = (B)2 - 2(A)
forces to be algebraic, as a root of a non-zero polynomial with algebraic coefficients. |
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