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Length of chord from area of segment, and solving x - sinx = c
By Brad Rodgers (P1930) on Thursday, June 29, 2000 - 05:23 pm:

In the figure below, where r is the radius of the circle, b is 1/2 the base of the isosceles triangle and A is the area in green, how does one find what the base of the triangle is?



I have used integrals to get far enough to realize that the equation set up will be


A =  1
2
 pr2 - r2sin-1 æ
è

Ö
r2-b2
r
 ö
ø		 + b
Ö
 
r2-b2
 
[The three terms give the area of a semi-circle, the area of the sector not shown on the diagram, and the area of the triangle. - The Editor]

But how do you solve this?
By David Loeffler (P865) on Monday, July 3, 2000 - 10:40 pm:

Sorry if I'm being thick, but what are you looking for? Am I right in thinking that you have A and r, and you want to find b in terms of these?
If your setup is what I think you mean, I think you can dodge around the integrals by using area of sector=½r2q where q=angle subtended by sector. Thus we get A=½r2(q-sinq) where b=2r sin(½q) and we can substitute for theta to get something looking a bit like the equation you had above. But I haven't a clue how to invert this and recover b either.

David Loeffler

By Brad Rodgers (P1930) on Tuesday, July 4, 2000 - 02:09 am:

I am trying to solve for b, which is what you were looking for as well. A and r are constants. I am still not sure how to solve for b either though. But, I have little experience in trigonometry.

Anyways, it is interesting that when we define the angle opposite to the base of the triangle to be B, we can find the area to be

A=B/360×pi×r2 -b×sqrt(r2-b2)

As this is the same area defined in the previous formula, we can substitute and reach the conclusion that

sin((1/2 -B/360)pi)=sqrt(r2-1/4×L2)/r

Where L is the base of an iscosceles triangle, r the other sides and B is the angle opposite to L.

Sorry if this last part has been useless. Does anyone know how to solve for b in my first equation given A and r?

Thanks,

Brad

By Michael Doré (P904) on Tuesday, July 4, 2000 - 05:30 am:

Brad - if you go back to David's equation:

q - sinq = 2A/r2

where b = r sin(½q) then that's as simple as it's going to get. The equation x - sin x = C cannot generally be solved for x by analytical methods. I'm not sure how to prove this, but am pretty confident it's true.

If you're happy with an approximation then when A is small compared to R (in other words the proportion of the circle we're dealing with is small) we can approximate sin x by x - x3/3!. Now substituting this into:

q - sin q = 2A/r2

q3/6 = 2A/r2
q = cube root(12A/r2)

If q is close to p/2 then you can make a different approximation.

Don't know if that's any use...

Michael

By Michael Doré (P904) on Tuesday, July 4, 2000 - 11:53 am:

And then if you want an approximation for b:

b = r sin(½q)

which is approximately:

b = ½rq = cube root(3/2 × Ar)

Sorry I can't be any more helpful,

Michael

By Brad Rodgers (P1930) on Wednesday, July 5, 2000 - 10:49 pm:

Thanks, has anyone ever compiled a list of x's that equal

x-sin(x)=C

for given values of C, sort of like antiderivatives. I am sure that there is someway to find a general formula for this theorum. But perhaps not as some equations can't be solved. But anyways, the method you gave seems to work well enough for all practical purposes.

Thanks,

Brad

By Michael Doré (P904) on Wednesday, July 5, 2000 - 11:17 pm:

Brad - I'm fairly sure there is no analytical (closed form) solution for x - sin x = C. (Unless obviously if C takes a special value, like pi/2-1 for instance).

The problem is that sin x = [eix - e-ix]/2i. Now I know that ex - x = C cannot be solved analytically, so I doubt replacing the ex by sin x can.

I wouldn't say my method is OK for practical purposes. It is only really of use when the triangle is very narrow. I can do a similar one for when the base of the triangle is close to the diameter of the circle, and I'm sure other methods could be thought up in special cases.

As for compiling a table - I don't know if it has been done but it would be very easy to do using an Excel spreadsheet. Alternatively if you have a calculator to hand you can use iteration:

1) Choose a value of C, which you want to solve x for.
2) Put the calculator in radians mode.
3) Put in any number between 0 and 1 (say 0.5).
4) Now take the sin of this number.
5) Add C on.
6) Press equals, and go to 4) applying the steps to the new number.

I'm assuming your calculator is the old fashioned (and I think better) type in which you press the operator after inputting the number (rather than the new ones in which you type in the operation algebraically). If not that makes it more cumbersome.

I think this is going to always converge. After a lot of trials, the value of x will be the number in the display BEFORE C is added on. By the way, anyone know how many solutions to x - sin x = C there are on the complex plane?

Alternatively if you're into computer programming you could easily do the whole thing automatically.

Yours,

Michael

By Neil Morrison (P1462) on Thursday, July 6, 2000 - 08:43 pm:

I would to suggest iteration (graphic calculators do the donkey work) as the simplest way of solving the equation x - sin x = C.

Of course, for given x, x - 1 <= C <= x + 1.
Then suppose C = 4.7
we know already that x is between 3.7 and 5.7. So your iteration is simplified because you don't have to guess what to make x as the first guess.

Neil M