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Topic: Re: Geometry question. (Read 904 times) |
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flamingdragon
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Re: Geometry question.
« on: Dec 12th, 2006, 6:40pm » |
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Is there a point in the middle of the circle?
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Sameer
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Re: Geometry question.
« Reply #1 on: Dec 12th, 2006, 6:50pm » |
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can you edit the post for better title? locus means the set of all poiints P that will satisfy your conditions... so it will be some sort of curve equation
as for e.g.
consider a circle around origin... consider point x on (-r,0) co-ordinate...
Consider a chord a angle t w.r.t positive x axis intersecting circle at P(x,y)
Then height = y = lsint
distance of the projection of chort on x-axis from origin = (r-lcost)
Using pythagoras you get l = 2rcost
Also x^2 + y^2 = l^2 so the midpoint P1 would be l/2 in lenght giving x^2+y^2=l^2/4=r^2cos^t
where -pi/2 < t < pi/2 that would be the locus for one set of points.. i forget the name of the curve tho... now as you rotate x around the circle you will get this curve rotated by that much angle.... i hope i had my math right...
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| « Last Edit: Dec 12th, 2006, 7:05pm by Sameer » |
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towr
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Re: Geometry question. (about Locuses (word?))
« Reply #2 on: Dec 13th, 2006, 1:09am » |
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I'm not entirely familiar with the terminology; what is a chord, is it a line-piece between x and a point on the circle?
In that case, I think, you'd get a circle half the size of Y, with its center halfway between x and the center of Y. That also collapses all three seperate cases into one btw..
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Grimbal
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Re: Geometry question. (about Locuses (word?))
« Reply #3 on: Dec 13th, 2006, 3:17am » |
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Yep, I vote for the circle. But not necessarily half the size of Y. It has a diameter that goes from the point x to the center of Y.
The trick is to understand that x, the midpoint of the chord and the center of Y form a right angle.
And I think the plural of locus is loci, and it means "locations".
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| « Last Edit: Dec 14th, 2006, 5:51am by Grimbal » |
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towr
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Re: Geometry question. (about Locuses (word?))
« Reply #4 on: Dec 13th, 2006, 5:32am » |
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on Dec 13th, 2006, 3:17am, Grimbal wrote:| The trick is to understand that x, the midpoint of the cord and the center of Y form a right angle. |
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?!
I must be fundamentally mistaken about what a chord is then. I'd have though the chord could be tangent to the circle (If x is outside it). But if what you say is true that's impossible.
Maybe someone could draw it. Or maybe I could look it up in mathworld.. Whichever takes the least effort of my part..
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SMQ
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Re: Geometry question. (about Locuses (word?))
« Reply #5 on: Dec 13th, 2006, 5:38am » |
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My understanding is that a chord is a line segment connecting any two points on a circle; e.g. a diameter is a chord which additionally passes through the center of the circle.
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towr
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Re: Geometry question. (about Locuses (word?))
« Reply #6 on: Dec 13th, 2006, 5:57am » |
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on Dec 13th, 2006, 5:38am, SMQ wrote:| My understanding is that a chord is a line segment connecting any two points on a circle; e.g. a diameter is a chord which additionally passes through the center of the circle. |
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So we have a line through x that intersects the circle, and the linepiece in the circle is the chord?
Well, that completely changes things.
Except when x is on the circle, in which case you'll get a circle half the size that touches x and the center of Y.
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SMQ
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Re: Geometry question. (about Locuses (word?))
« Reply #7 on: Dec 13th, 2006, 6:18am » |
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on Dec 13th, 2006, 5:57am, towr wrote:| [...] when x is on the circle [...] you'll get a circle [...] that touches x and the center of Y. |
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Actually, I believe that's what you'll get for any X (for X outside Y you only get the portion of the circle within Y).
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| « Last Edit: Dec 13th, 2006, 6:19am by SMQ » |
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Sameer
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Re: Geometry question. (about Locuses (word?))
« Reply #8 on: Dec 13th, 2006, 10:42am » |
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on Dec 13th, 2006, 5:57am, towr wrote:
So we have a line through x that intersects the circle, and the linepiece in the circle is the chord?
Well, that completely changes things.
Except when x is on the circle, in which case you'll get a circle half the size that touches x and the center of Y. |
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How come I am getting the equation to be x^2+y^2=r^2cost This according to me is not a circle...
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towr
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Re: Geometry question. (about Locuses (word?))
« Reply #9 on: Dec 13th, 2006, 11:20am » |
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on Dec 13th, 2006, 10:42am, Sameer wrote:| How come I am getting the equation to be x^2+y^2=r^2cost This according to me is not a circle... |
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I'm not sure how you got that
We have l = 2r cos(t),
taking half, and putting it back into the parametric equation (l cos(t) - r, l sin(t))
We get
(r cos2(t) - r, r cos(t) sin(t))
->
(r/2 [2cos2(t) - 1) - r/2, r/2 [2 cos(t) sin(t)])
using double angle formulas that's
(r/2 cos(2t) - r/2, r/2 sin(2t))
or
r/2( cos(2t), sin(2t)) + (-r/2, 0)
Which looks perfectly circular to me.
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| « Last Edit: Dec 13th, 2006, 11:22am by towr » |
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Barukh
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Re: Geometry question. (about Locuses (word?))  Locus.PNG
« Reply #10 on: Dec 14th, 2006, 4:10am » |
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koonie, consider the following purely geometric argument.
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Icarus
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Re: Geometry question. (about Locuses (word?))
« Reply #11 on: Dec 14th, 2006, 5:42am » |
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To be more explanatory (don't look until you've tried to figure it out yourself):
| hidden: | The perpendicular to a chord passing through the center of the circle always intersects the chord at its midpoint. Let Y be the circle, with center O, and let L be the circle having X and O as endpoints of a diameter. Because XO is a diameter, for any other point M on L, the angle XMO is right. Therefore if we extend XM to a chord of the original circle (assuming that XM intersects the original circle), we must have that M is the midpoint of that chord, or conversely, that the midpoint of the chord lies on L. We can even say this in the unique case when M = X if we consider the line XM to be the tangent to L at X. Every line passing through X intersects L in such a point M, so every chord of Y which passes through X must have its midpoint on L, and every point on L which lies within Y is the midpoint of such a chord.
I.e. Regardless of where X lies, the locus of points satisfying the condition is the portion of L that lies within Y. |
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This brings up the curious question in the case of X outside Y of the points of the circle L that are outside Y. By analogy, they should in some sense be the midpoint of a "chord" of Y passing though X. How can one define this "chord"?
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| « Last Edit: Dec 14th, 2006, 5:49am by Icarus » |
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SMQ
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Re: Geometry question. (about Locuses (word?))
« Reply #12 on: Dec 14th, 2006, 8:23am » |
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on Dec 14th, 2006, 5:42am, Icarus wrote:| By analogy, they should in some sense be the midpoint of a "chord" of Y passing though X. How can one define this "chord"? |
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Consider the parametric interpretation of the problem where a chord of the unit circle lies on a line passing through point X = (a,b). The equation for the line can be given in terms of a parameter t as (x,y) = (a+ut, b+vt) where (u,v) is a unit vector, i.e. u2+v2=1.
The endpoints of the chord are the intersections of this line with the unit circle: x2+y2=1. Substituting gives (u2+v2)t2 + 2(au+bv)t + (a2+b2-1) = 0, and since (u,v) is a unit vector this further reduces to t2 + 2(au + bv)t + (a2+b2-1) = 0.
Solving this equation can give complex results for t when (a,b) lies outside the unit circle, but this situation could be envisioned as an imaginary z axis extending above and below the x-y plane. Now chords outside of the unit circle are "diagonal" perpendicular to the x-y plane, but their midpoint is still in the x-y plane; specifically the point given by t = -(au+bv).
Alternatively by taking only the real portion of the resulting values of t, the entire chord can effectively be projected into the x-y plane. I'll try to attach a diagram of the resulting structure if I get a chance.
[edit]Er, strike that, the resulting chords are perpendicular, not diagonal, and so the projection into the x-y plane is just a point.[/edit]
--SMQ
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| « Last Edit: Dec 14th, 2006, 8:27am by SMQ » |
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Icarus
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Re: Geometry question. (about Locuses (word?))
« Reply #13 on: Dec 16th, 2006, 9:29am » |
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This is a way of seeing what it ought to be, but it doesn't solve the problem by itself. To solve the problem, you have to show that every such midpoint must lie on your circle (there are infinitely many of them, so just plotting a few doesn't cut it), and you also have to show that every point on the circle is the midpoint of chord through X.
If you cannot show these two things, then there is always the possibility that you failed to notice other midpoints that don't lie on the circle, or that some points on the circle were not the midpoints of chords.
Barukh's diagram shows the way to prove both.
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