20080130, 18:53  #1 
May 2004
New York City
5·7·11^{2} Posts 
Ratio of Areas
Show that the ratio of the areas of the circumscribed to
inscribed circles of a right triangle with integral sides can never be a Mersenne prime. 
20080203, 19:02  #2 
"William"
May 2003
New Haven
2^{6}×37 Posts 
The winky is because the ratio can never be an integer.
We can limit ourselves to right triangles whose sides are primitive Pythagorean triples because scaling the triangle larger will not change the ratio. Recall that all primitive Pythagorean triples can be generated from p and q relatively prime and opposite parity as p[sup]2[/sup]  q[sup]2[/sup] 2pq p[sup]2[/sup] + q[sup]2[/sup] The radius r of the inscribed circle can be found by taking the area of the three triangles using a side of the right triangle as base and r as height. r = ab / (a+b+c) The center of the circumscribed circle has to be halfway along each side, so it's radius R is R = c/2 The ratio R/r is c(a+b+c)/ab Substituting the expressions for primitive Pythagorean triples we get (p[sup]2[/sup] + q[sup]2[/sup])(2p[sup]2[/sup]+2pq)/((4(p[sup]2[/sup]  q[sup]2[/sup])pq) Which simplifies to (p[sup]2[/sup] + q[sup]2[/sup])/((2(pq)q) The numerator is odd because p and q are opposite parity. The denominator is even. Hence this ratio is never an integer. As a further exericise, show that if the ratio is a half integer, then it is half a Fermat Prime. William 
20080213, 03:20  #3  
May 2004
New York City
5·7·11^{2} Posts 
I just want to point out that a "simpler" expression for r is r = (a+bc)/2, and so R/r = c/(a+bc), and then if you know (or using p and q) that c must be odd and a,b must be opposite parity, then the denominator must be even. The rest is as you said. Quote:
a half integer, then q must = 1, p must = 2, so a,b,c = 3,4,5, so 2R/r = 5, which is lo and behold a Fermat Prime! (Thanks for understanding the winky.) Last fiddled with by davar55 on 20080213 at 03:36 

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