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cool_joh
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restoring axes
« on: Sep 12th, 2008, 5:27pm » |
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There is a parabola y = x2 drawn on the coordinate plane. The axes are deleted. Can you restore them with the help of compass and ruler?
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Eigenray
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Re: restoring axes
« Reply #1 on: Sep 24th, 2008, 12:51am » |
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1) Make two parallel lines intersecting the parabola and mark the points of intersection.
2) Construct the line joining the midpoints of the two line segments thus formed. This line is vertical!
Now we just need to take a (horizontal) line perpendicular to that one, intersecting the parabola, and the y-axis will be the perpendicular bisector of the resulting line segment. Now the x-axis is obvious; if we bisect the angle between the axes, we get the point (1,1), recovering the scale.
Rather nice. Now who wants to do the ellipse and hyperbola?
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| « Last Edit: Sep 24th, 2008, 1:02am by Eigenray » |
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Aryabhatta
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Re: restoring axes
« Reply #2 on: Sep 24th, 2008, 11:32am » |
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Interesting problem, strange that this seems to have been missed.
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Aryabhatta
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Re: restoring axes
« Reply #3 on: Sep 25th, 2008, 11:23am » |
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Here is an attempt for the ellipse (haven't tried proving it)
Take two sets of chords C1, C2 and D1, D2 with C1 and C2 parallel and D1 and D2 parallel, and C1 and D1 not parallel.
Join the midpoints of C1 and C2 to give a line c. Join the midpoints of D1 and D2 to give a line d.
The intersection of c and d will be the center of the ellipse!, say O.
Now just draw a circle through O which intersects the ellipse at four points. The chords of the circle (formed by the four points) which don't pass through O are either vertical or horizontal.
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Eigenray
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Re: restoring axes
« Reply #4 on: Sep 25th, 2008, 4:58pm » |
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To see why it's true for the ellipse reduce to the circle. The hyperbola can be reduced to the ellipse by going to C. But for the parabola I think we either need to use algebra or go projective.
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Barukh
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Re: restoring axes
« Reply #5 on: Sep 25th, 2008, 11:45pm » |
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on Sep 25th, 2008, 11:23am, Aryabhatta wrote:Here is an attempt for the ellipse (haven't tried proving it)
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If I'm not missing something, this is relevant.
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Hippo
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Re: restoring axes
« Reply #6 on: Sep 26th, 2008, 2:02pm » |
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on Sep 24th, 2008, 12:51am, Eigenray wrote:1) Make two parallel lines intersecting the parabola and mark the points of intersection.
2) Construct the line joining the midpoints of the two line segments thus formed. This line is vertical!
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OK, I found some time to check the "trivial" observation myself so for those slow in geometry as I am:
d(x1-x2) = x12-x22 = (x1-x2)(x1+x2) so avg(x1,x2)=d/2. The Barukh's links to projective geometry gives me new insights to geometry (or may be refreshes already forgotten?). Circle center search is obvious and the construction projected to nonparallel plane gives the same for elipse as mentioned above. I suppose midpoints of two parallel chords is the cruical part of all the constructions, but I cannot see geometric (projective) reasoning for parabola and I didn't start even algebraic reasoning for hyperbola. What the "going complex" mean?
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| « Last Edit: Sep 28th, 2008, 2:11pm by Hippo » |
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teekyman
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Re: restoring axes
« Reply #7 on: Sep 26th, 2008, 2:23pm » |
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on Sep 24th, 2008, 12:51am, Eigenray wrote:1) Make two parallel lines intersecting the parabola and mark the points of intersection.
2) Construct the line joining the midpoints of the two line segments thus formed. This line is vertical!
Now we just need to take a (horizontal) line perpendicular to that one, intersecting the parabola, and the y-axis will be the perpendicular bisector of the resulting line segment. Now the x-axis is obvious; if we bisect the angle between the axes, we get the point (1,1), recovering the scale.
Rather nice. Now who wants to do the ellipse and hyperbola? |
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Why was the line joining the midpoints of the two parallel lines vertical?
I can get this result with a little algebra, but how did you come up with that? Is there a geometric reason i'm not seeing?
Edit: I can now invent geometric reasons knowing the result, but I dont know how I could come up with it from scratch, so I'm curious to know how you thought of it.
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| « Last Edit: Sep 26th, 2008, 2:29pm by teekyman » |
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Hippo
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Re: restoring axes
« Reply #8 on: Sep 26th, 2008, 2:45pm » |
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May be he generalises the circle center search construction?
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| « Last Edit: Sep 26th, 2008, 2:46pm by Hippo » |
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cool_joh
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Re: restoring axes
« Reply #9 on: Sep 26th, 2008, 6:58pm » |
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in the first place, is it possible to make two (perfectly) parallel lines with the help of a compass and a ruler? 
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Obob
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Re: restoring axes
« Reply #10 on: Sep 26th, 2008, 9:13pm » |
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on Sep 26th, 2008, 6:58pm, cool_joh wrote:| in the first place, is it possible to make two (perfectly) parallel lines with the help of a compass and a ruler? |
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Sure. Draw one line first. Then draw two circles on this line which intersect one another in two points. Connect those two points of intersection to get a line perpendicular to the first line. Now repeat the construction to get a line perpendicular to the second; it is parallel to the first.
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Eigenray
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Re: restoring axes
« Reply #11 on: Sep 27th, 2008, 1:25am » |
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on Sep 26th, 2008, 2:02pm, Hippo wrote:| OK, I found some time to check the "trivial" observation myself so for those slow in geometry as I am |
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I didn't mean to imply that it was obvious. In fact I found it surprising, hence the exclamation point.
Since a linear transformation takes parallel lines to parallel lines, and midpoints to midpoints, if we can find the center of a circle just using the notions of parallel and midpoint (not perpendicular), then the same construction gives the center of an ellipse.
Given a conic section Ax2+Bxy+Cy2+Dx+Ey+F=0, any linear transformation multiplies the discriminant B2-4AC by the square of the determinant. So if we want a linear transformation taking a hyperbola to a circle, we have to allow complex coefficients. For example, (x,y) -> (x,iy) maps x2-y2=1 to x2+y2=1. (I suppose the simple geometric argument for the circle only works for 'real' lines, so maybe you still need to use algebra.)
But there is no linear transformation taking a parabola to a circle. For that I guess we would need to projectivize if we wanted a uniform answer. If we define the center of a parabola to be the point at infinity in the direction the parabola opens up, then we have the statement that for any conic section, the midpoints of parallel secants are all collinear with the center.
We can also prove it directly for a hyperbola: WLOG, y=1/x. If we intersect with the line y=mx+b, we get mx2+bx-1=0, so the sum of the x's will be -b/m, and the sum of the y's will be b, so the midpoint is (-b/m,b). Thus the line between the center (0,0) and the midpoint is y= -mx, which depends only on m.
on Sep 26th, 2008, 2:23pm, 1337b4k4 wrote:| Edit: I can now invent geometric reasons knowing the result, but I dont know how I could come up with it from scratch, so I'm curious to know how you thought of it. |
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What are these geometric reasons? I probably thought of it only because I must have seen the construction for an ellipse before.
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| « Last Edit: Sep 27th, 2008, 1:28am by Eigenray » |
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rmsgrey
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Re: restoring axes
« Reply #12 on: Sep 27th, 2008, 6:25am » |
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on Sep 26th, 2008, 6:58pm, cool_joh wrote:| in the first place, is it possible to make two (perfectly) parallel lines with the help of a compass and a ruler? |
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An alternative to Obob's construction requiring fewer arcs:
Pick a point on the line, P, and draw a circle, C, around it. Pick one of the two intersection points of the line with C, Q, and draw a second circle, D, of the same radius. Pick one of the two intersection points of C and D, R and draw circle E, with the same radius. D and E intersect at P and a new point, S, and the line RS is parallel to the original line PQ.
A shorthand description would be to make a rhombus out of two equilateral triangles (you could also construct two equilateral triangles of the same side-length on the same side of the line - the two vertices not on the line would be on a line parallel)
Obob's construction, with a slight modification, offers a way to construct the parallel through a given point - when constructing the second perpendicular, construct the one through the desired point (by drawing a circle centred on the chosen point and using its points of intersection with the first perpendicular as the centres of the two circles)
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