Circles And Chords

page revision: 0, last edited: 15 Sep 2008 08:59

Circles And Chords

page revision: 0, last edited: 15 Sep 2008 08:59

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I'd like to upload a photo, but I can't. The farthest point is easily determined via the triangle inequality. For the closest point, take an arbitrary point on the circle, and join it to the centre. As two radii, the distance of the 'closer' point on the secant is the same as that of the arbitrary point's to the centre. The triangle formed by the arbitrary point and the 'closer' point along with the centre is therefore isosceles. Join the arbitrary point on the circle with the given interior point, and the angle formed will be smaller than the angle between the 'closer' point and the centre. The solution follows from here by part 45.

ReplyOptionsHmm, I don't think you gave a valid solution. First it is unclear: Which angle is smaller than which? What is "angle between" two points? Second, the "given point" is not required to be interior - it could be exterior to the circle. Third, I doubt you can get away with arguments based on comparison of angles: At this point in the text, nothing is known about angles relative to their position to a circle, so it would be hard to justify any claim, even when it is valid. Fourth, "easily determined" is not an argument: "the farthest point is easily determined via triangle inequality" is merely a rephrasing of the hint given to the problem.

I believe, if you make an accurate argument (based on this hint) for the farthest point, you'll see that a similar argument works for the closest point as well. Good luck!

Hey, thanks. I was intentionally being a bit informal; didn't want to spoil it. This is the argument I gave in my notebook:

Let a circle with centre O be given, and an arbitrary point A in its interior region also be given. Draw a secant through O and A, and mark its intersections with the circle as Q and P. Choose an arbitrary point M along the circle not equal to P or Q, and draw a line from it through A up to its intersection with the circle at N.

1. (part 48) NA < OA + ON, and ON = PO as radii of the same circle. Therefore, NA < OA + PO, and thus, because PO + OA are summands of the segment PA, and N is arbitrary, P is the farthest point from A.

2. Draw a line from M to O, then OM = OQ as radii of the same circle, so the triangle OQM is isosceles, and thus (part 35) the angles at the base are congruent (i.e. OQM = OMQ). But then in the triangle AQM, the angle AMQ < (AQM = OQM) because AMQ is a part of angle OMQ. Therefore, (part 45) AM > AQ as a side opposite to a greater angle in a triangle. Because M was chosen to be arbitrary, Q is the closest point on the circle to A.

It doesn't sound like my argument for point 2 was what you had in mind. Seemed a bit easy for an asterisked question. I'll see if I can approach the problem from another perspective. If it's possible to boost my karma, then I would appreciate it if you would do that so that I can post images here. Thanks.

1. What does N have to do with anything? You need to show that AM < AP.

2. Your argument to show that AM > AQ works, and it is also possible to show that AM<AP by the same argument, based on angles. [Why do you use different arguments for proving similar statements?]

Yet, what if A is outside the circle? (Your argument based on angles should also work, but might appear different due to a new disposition of the points involved).

1. AM < AP follows the same argument. Introducing N was unnecessary, yes.

2. I see it. The solution for AM < AP using the triangle inequality involved drawing 1 less auxiliary line, and was seen at a glance.

3. I incorrectly self-imposed this condition. As Polya said, the first step to problem-solving is to understand the problem. The solution for A outside the disk can use a similar angle-based argument—it's slightly different, so I'll take some time to post it for other readers later.

OK! So, the good news is that there are two fruitful approaches now: one based on comparing angles, the other based on the triangle inequality. Thanks!

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