Here's another circle problem which might interest you.

It takes some thought but is not that hard.

Perform the following steps:

1. Draw a circle whose radius = 1.

2. Draw a chord of the circle parallel to the X-axis, attempting to
divide the semi-circle's area in half.

3. Derive your formula to determine the value of k which the distance
from the X-axis to the chord that divides the semi-circle's area in half.

I first worked on this problem using BASIC on an Apple IIc. I generated the value of SIN using the first three terms of the series.

Let see your results... I hope mine is correct.

Earl Seely
Milton, DE
http://c:%5Chalf%20circle.jpg




http://c:%5Chalf%20circle.jpg

There are many ways to tackle this, but we can do this:

The area of half the semicircle will be Pi/4.

So, we can use the formula for the segment of a circle and solve for theta: \frac{1}{2}r^{2}({\theta}-sin{\theta})

\frac{1}{2}({\theta}-sin{\theta})=\frac{\pi}{4}

We find {\theta}=2.3098814600101

Now, sub this into the formual for the middle ordinate(the distance form the chord to the top of the circle): R(1-cos(\frac{\theta}{2}))

(1-cos(\frac{2.3098814600101}{2})\approx{.596}

But that is the middle ordinate length. The distance from the center to that point will be 1-.596=.404

The distance is about .40 units from the center. Is that what you got?

We can even use calculus to check it:

2\int_{0}^{.40397275329946}\sqrt{1-x^{2}}dx=\frac{\pi}{4}

My k value = 0.403977.

I worked with 1/8th of the circle whose area would be PI/8

Then I used angle A, the angle between the X-axis and the chord
dividing the semi-circle's area in the following formula:

A/2 + 1/2 SIN(A)*COS(A) = PI/8

where A/2 gave the first area swept by angle A and

1/2 SIN(A)*COS(A) is the area of the triangle.

PI/8 is 1/8th the area of a circle with radius = 1.

A/2 - PI()/8 + 1/2 SIN(A)*COS(A) = 0

Then I used Excel to home in on the value for angle A, from

which k = sin A.

That's exactly what I got as well. We must've done something right.

Hey Eeseely:

We can also use Newton's method to home in on the value.

I done that and I get .404

Rounded, of course.