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        riddles >> medium >> Not Both Algebraic
        
(Message started by: ThudanBlunder on Nov 2^nd, 2008, 2:25pm)

Title: Not Both Algebraic
Post by ThudanBlunder on Nov 2^nd, 2008, 2:25pm

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.

Title: Re: Not Both Algebraic
Post by ThudanBlunder on Dec 19^th, 2008, 12:16am

HINT:[hide]Prove that both x and y are algebraic implies http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif is algebraic.[/hide]

Title: Re: Not Both Algebraic
Post by Eigenray on Dec 19^th, 2008, 12:38am

Let the circle have radius 1, so the sides of the rectangle are 2cos t, 2sin t. Then
[hideb]
A = sin t cos t/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif and B = (sin t + cos t)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif are both algebraic. But
1 = sin²t + cos²t = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifB)² - 2(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifA)
forces http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifto be algebraic, as a root of a non-zero polynomial with algebraic coefficients.[/hideb]