Re: Math Challenge I could not solve Message #10 Posted by PGILLET on 7 Jan 2013, 2:28 a.m., in response to message #7 by Thomas Klemm
That's how I saw the problem:
In the lens-shaped piece you can see an isosceles triangle with sides (1,1,r) and angles (a,a,pi-2*a).
Divided in two right-angled triangles, it leads to 1*cos(pi-2*a)=d and r*cos(a)=1-d => d=1-r*r/2.
The grazed surface can then be expressed as
S=2*(integ(sqrt(1-x*x),x,1-r*r/2,1)+integ(sqrt(r*r-x*x),x,r*r/2,r)).
But we know that integ(sqrt(r*r-x*x),x)=x/2*sqrt(r*r-x*x)+r*r/2*arcsin(x/r).
So developing and simplifying S naturally leads to
S=arccos(1-r*r/2)+r*r*arccos(r/2)-r/2*sqrt(4-r*r).
(It's just another way to the same formula as in the http://mathworld.wolfram.com/GoatProblem.html web site).
Solving S=pi/2 on r also :) leads to r=1.15872847...
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