# Sector of circle radius and areas

• Feb 17th 2013, 11:19 AM
Mouhaha
Sector of circle radius and areas
Hi everybody,

I could not solve the following problem.
Look at the attachment.

It is not homework.

I need it to solve another problem.

I did not succeed so I hope that you will.
• Feb 17th 2013, 07:47 PM
Mouhaha
Re: Sector of circle radius and areas
It seems that it is very hard problem.
So for now I abandon.
• Feb 17th 2013, 08:20 PM
Re: Sector of circle radius and areas
(given) Area ABCD = Area CDEF - Area OEF = Area OCD - Area OEF - Area OEF = Area OCD - 2 Area OEF = (y^2)(a/2) - 2(x^2)(a/2)
(derived) Area ABCD = Area OAB - Area OCD = (r^2)(a/2) - (y^2)(a/2)

Hence
(y^2)(a/2) - 2(x^2)(a/2) = (r^2)(a/2) - (y^2)(a/2)
2(y^2 - x^2) = r^2
y^2 - x^2 = (r^2)/2

If x and y are integers, then the difference of squares is an integer, so r^2/2 is an integer, say n.

EDIT:
(r^2)/2 = (r/sqrt(2))^2
Hence
(r/sqrt(2))^2 + x^2 = y^2 and hence (r/sqrt(2),x,y) would be a Pythagorean triplet.

that's all I have so far, but I hope that helps a little. my intuition says it's impossible, so I would continue down this line of reasoning to find a contradiction.
• Feb 18th 2013, 06:46 AM
Mouhaha
Re: Sector of circle radius and areas
Thank you very much.
You gave me little hope.
If we can not find x and y we could at least bound it.
Here is my idea :
The area CDEF is always equal to area ABCD + area OEF.
We have to find DF such as DF = OF+DB
The value of computed DF then will give us an upper bound for y-x
Am I right?
• Feb 18th 2013, 01:07 PM
Mouhaha
Re: Sector of circle radius and areas
Another idea comes to my mind.
We could find :
- an upper bound (or lower?) to y-x by assuming that the area OEF goes to zero
- and lower bound (or upper?) to y-x by assuming that the area ABCD goes to zero.

So lower bound <(y-x)< upper bound.

Is it easy to compute?
• Feb 19th 2013, 12:05 AM
Re: Sector of circle radius and areas
I do like that idea of boundedness and considering the difference to find either a contradiction or a bound.

Suppose $y-x = M$ for some positive integer M.
then $y^2 - x^2 = (M+x)^2 - x^2 = M^2 + 2Mx = \frac{r^2}{2}$

then $x=\frac{r^2-2M^2}{4M}$ and $y = \frac{r^2+2M^2}{4M}$

and since x and y are integers $4M$ must divide both $r^2-2M^2$ and $r^2+2M^2$

Recall that if (ab) divides c, then a divides c, so 4 must divide both $r^2-2M^2$ and $r^2+2M^2$

edit://

so $r^2-2M^2 = 4k$ for some integer k and $r^2+2M^2=4m$ for some integer m.
If k = m then M = 0. If m < k then $r^2+2M^2 < r^2 - 2M^2$, a contradiction.
If k < m $r^2 - 2M^2 < r^2 + 2M^2$

so now I would investigate the case of k<m, but I might just be complicating things or made a mistake. interesting problem though. I also considered looking at the problem when a = 360 degrees.