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Topic: Reflective Circle (Read 1143 times) |
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ThudnBlunder
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Reflective Circle
« on: Dec 14th, 2004, 7:16am » |
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I think this ought to be in Hard, but hey, it's too mathy.
Given a circle C and two points A and B external to it, find the point(s) on the circle such that both AC and BC make equal angles with the tangent to the circle passing through point C.
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| « Last Edit: Dec 14th, 2004, 7:17am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Aryabhatta
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Re: Reflective Circle
« Reply #1 on: Dec 14th, 2004, 9:03am » |
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on Dec 14th, 2004, 7:16am, THUDandBLUNDER wrote:I think this ought to be in Hard, but hey, it's too mathy.
Given a circle C and two points A and B external to it, find the point(s) on the circle such that both AC and BC make equal angles with the tangent to the circle passing through point C. |
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Did you mean: Given a circle and two points A and B, find all points C on the circle such that AC and BC make equal angles with tangent to the circle at C? (You have named the circle C which makes it confusing to read).
One other question, do A and B have to be on the same side of the tangent? Also the angle AC makes with the tangent seems not clearly defined.. which one do we take? x or 180-x? My guess (based on the title of this thread) is you want the tangent to be the segement PCQ, with C in between P and Q and the corresponding angles are ACP and BCQ.
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JocK
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Re: Reflective Circle
« Reply #2 on: Dec 14th, 2004, 12:05pm » |
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If M is the centre of the circle, then the reflection point C on the circle satisfies: [angle] MCA = [angle] BCM.
You want this equation parametrised in some way? (Don't see the problem yet...) And yes, there is always two solutions, at most one corresponds to an "outside reflection".
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| « Last Edit: Dec 14th, 2004, 12:15pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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ThudnBlunder
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Re: Reflective Circle
« Reply #3 on: Dec 14th, 2004, 7:35pm » |
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on Dec 14th, 2004, 9:03am, Aryabhatta wrote:
Did you mean: Given a circle and two points A and B, find all points C on the circle such that AC and BC make equal angles with tangent to the circle at C? (You have named the circle C which makes it confusing to read).
One other question, do A and B have to be on the same side of the tangent? Also the angle AC makes with the tangent seems not clearly defined.. which one do we take? x or 180-x? My guess (based on the title of this thread) is you want the tangent to be the segement PCQ, with C in between P and Q and the corresponding angles are ACP and BCQ.
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Sorry about the confusing double use of 'C'.
Physically, the problem asks for a method to find the point(s) on the boundary of a circular pool table at which the cue ball must be aimed, such that it hits the target ball after one bounce off the cushion. (For example, can it be constructed using only ruler and compasses?) To put it yet another way, find an isosceles triangle whose sides pass through two given points inside a given circle.
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| « Last Edit: Dec 15th, 2004, 2:57am by ThudnBlunder » |
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