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Post by Admin on Oct 6, 2016 9:05:24 GMT
Gary I have amended and republished my solution to exercise 3 and my commentary. Most of the changes are to remove ambiguities and to make clarifications as a result of our discussions on this forum. If you think anything else needs changing let me know. Vasco |
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Gary
GaryVasco
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Post by Gary on Oct 6, 2016 16:29:15 GMT
Vasco,
I have been thinking about point 6 of the commentary. The plotting of $\widehat{G}$ and $\widehat{J}$ is based on two points of intersection. How can we say with certainty that the lines through v are tangent to G and J at the points of intersection of $\widehat{G}$ and $\widehat{J}$ with $\widehat{C}$? I think it is true based on my own solution, but $\widehat{C}$ is not a great circle, so I think it has to be proven. The only way I can see to prove it is by a conformal projection from $C$ in $\mathbb{C}$.
Gary
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Post by Admin on Oct 6, 2016 18:36:52 GMT
Gary
First of all:
Given the circle $\widehat{C}$ there is only one possible point $v$ for the apex of the cone, because the sloping sides of the cone are required to just touch (not intersect) $\Sigma$ (NB. not $\widehat{C}$).
Secondly:
We can draw a circle on $\Sigma$ which passes through the points of intersection of any line through $v$ with $\Sigma$.
Thirdly:
Since the points of intersection of this circle with $\widehat{C}$ lie on $\Sigma$, the line from $v$ through this point just touches $\Sigma$ at this point. It follows that this line is tangent to the circle we have just drawn (not $\widehat{C}$) and since this line on the cone is orthogonal to $\widehat{C}$, it follows that our circle is orthogonal to $\widehat{C}$.
Vasco
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Gary
GaryVasco
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Post by Gary on Oct 6, 2016 23:17:15 GMT
Gary First of all: Given the circle $\widehat{C}$ there is only one possible point $v$ for the apex of the cone, because the sloping sides of the cone are required to just touch (not intersect) $\Sigma$ (NB. not $\widehat{C}$). Secondly: We can draw a circle on $\Sigma$ which passes through the points of intersection of any line through $v$ with $\Sigma$. Thirdly: Since the points of intersection of this circle with $\widehat{C}$ lie on $\Sigma$, the line from $v$ through this point just touches $\Sigma$ at this point. It follows that this line is tangent to the circle we have just drawn (not $\widehat{C}$) and since this line on the cone is orthogonal to $\widehat{C}$, it follows that our circle is orthogonal to $\widehat{C}$. Vasco Vasco, I accept the first point, and it presents a problem for my proof because there is no obvious reason why my line should exactly hit a particular v. Perhaps I am missing something about your third point. It sounds like you are describing a line (call it "the intersect tangent") from v to one of the points (call it "the intersect point") of a second circle (say $\widehat{G}$) where it intersects $\widehat{C}$. The circle $\widehat{G}$ is drawn through two points where a line (call it "the projection from v") from v intersects $\Sigma$. I see that the intersect tangent just touches $\Sigma$ at the intersect point by the construction of the cone and $\widehat{C}$, and it seems intuitive that it would also be tangent to $\widehat{G}$, but is it proven for this context? I have reservations about it, but I think I will accept it and move on. Gary |
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Post by Admin on Oct 6, 2016 23:36:09 GMT
Gary
The crucial point is that the circle and the line are in the same plane - that defined by $v$ and the two points of intersection of the circle with $\widehat{C}$, and so if the circle touches the line then that is the definition of a tangent! As Needham would say: Done!
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 7, 2016 0:56:57 GMT
Vasco,
The doubt lingers that the line can only "touch" the circle (without intersecting it) at the same point that the circle intersects $\widehat{C}$. I suppose it is the positive curvature that makes that possible. Nevertheless, I have rewritten my proof with this point accepted and will post it in a minute or two.
Gary
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Post by Admin on Oct 7, 2016 5:43:09 GMT
Vasco, The doubt lingers that the line can only "touch" the circle (without intersecting it) at the same point that the circle intersects $\widehat{C}$. I suppose it is the positive curvature that makes that possible. Nevertheless, I have rewritten my proof with this point accepted and will post it in a minute or two. Gary Gary Draw a picture with pencil and paper and compasses and you will see how it works. Looking at your latest solution I see that you have already done this! Vasco |
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