What is the probability that a randomly drawn chord will be longer than the radius of the circle?

Prove it.

My brother thinks this may be to solution to the geometical problem which you presented. He has asked that I post it on your web site to entertain comments.

Please share my solution on the site. I'd be

interested to know what they think.

Basically, the probability is SQRT(3)/2.

The reason is that the length of chords vary from

0 (really a line tangent to the circle) to 2r, the diameter. If you start at the tangent line and draw chords perpendicular to a radius line, the chords will be less than zero until you hit some point on the radius, between the edge of the circle and the center. After that, your chords will be greater than the radius as you approach the center where the length of the chord equals the diameter.

The proportion of the distance from the center of

the circle to the point where a chord drawn perpendicular to the radius line is the same length as the radius, divided by the length of the radius, is the probability that a chord drawn

perpendicular to the radius line is longer than

the radius.

Since an infinite number of radii can be drawn

for a circle, we can assume an infinite number of chords that are perpendicular to each radius line. We also know that you cannot draw a chord in a circle that is not perpendicular to a radius line. For each radius line the proportions will stay the same, so the overall probability will

likewise remain the same.

Thus our problem is to find the point on a single

radius line where a chord drawn perpendicular to the radius is equal to the radius.

1. We draw a chord equal to the radius of the

circle and draw two radii, one from each end of the chord to the center of the circle. This forms an equilateral triangle, all sides being equal. So the angles at each corner of the triangle are 60 degrees, again by definition.

2. We now draw a radius from the center of the

circle perpendicular to the chord. This line will bisect the chord and divide our equilateral triangle into two right triangles.

3. Each right triangle is a 30-60-90 degree right triangle with hypotenuse = r and the short leg = r/2. By the definition of a 30-60-90 triangle, the long leg equals r * SQRT(3)/2.

4. The proportion between the distance from the

center of the circle to the point where the radius line bisects the chord and the radius of the circle is our probability. So we divide ...

r * SQRT(3)/2 SQRT(3)

--------------- = -------- = SQRT(3)/2

r 2

QED