since cecil chose not to answer me, i now pose this question to the board
This is a question my calc teacher posed that no one has been able to solve. You are given a circle with two chords (not diameters). They intersect so that there are four segments (2 from each chord). The lengths of the 2 segments of the first chord are 18 and 6. The lengths of the 2 segments of the second chord are 9 and 12.
Find the length of the diameter.
the answer is the square root of the sum of each segment length squared = (9^2+12^2+18^2+6^2)^(1/2)
This can’t, in general, be true if I understand your description of the problem. I think you’re saying that you can take two intersecting chords of any circle, and the geometric mean of the four segments of those two chords yields the diameter.
I can easily picture a large circle, with two very short but intersecting chords. The geometric mean of the four segments would be a tiny number compared to the diameter.
It might happen to work out that the solution for this particular set of numbers works out that way, but in general it won’t be the way to solve it.
It’s not the geometric mean, it’s the square-root of the sum of the squares. At any rate, it isn’t generally true, although it is true if the chords happen to intersect at a 90 degree angle.
evanmsc, I think you messed up the problem statment. I’m getting that there is an infinite family of solutions with the segment lengths you gave. I can get a solution with any circle diameter of at least 24. I get the solution you posted two ways, for one of which the angle between the chords is 90 degrees.
2 things -
first - my understanding of the problem is that you are not given the angle between the segments, but that you could use 90 degrees in showing the truth of the idea in a specific case
second- how specifically do you go from the segments to the diameter (without me giving you the answer)… rather than from the answer to the measurement of the angle of intersection or anything else
This only works if the answer is independent of the angle, but for this problem, it isn’t.
I’m just about to go to bed, so I’m not going to make a long post right now. I started by drawing the perpendicular bisector to each chord. These intersect at different distances for different angles between the chords. I can show that the same radius circle centered at this intersection goes through both chord ends (hard to quickly describe without a picture).
This, in turn, would of course also yield a radius of the circle, and the diameter, too. Using a construction isn’t really a proof, but I’d WAG by knowing this construction,
it then probably wouldn’t be hard to prove the OP’s statement. (Seems Pythagorean…)
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ZenBeam is right of course. With the two chords, 9 * 12 has to equal 6 * 18, which means that if you take any circle with a chord of length 18+6, and from the interior point of that chord draw a line of length 9 to the circle (always possible–just start on the side of length 18 and swing it until it hits the circle), then that line extended will be a chord whose other segment must be 12.