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Topic: Three Collinear Intersections (Read 2720 times) 

william wu
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Three Collinear Intersections
« on: Apr 2^{nd}, 2003, 3:56am » 
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Take any three circles of differing radius in the plane. For each pair of circles, draw two tangent lines that touch both circles and cross each other outside the convex hull of the circles. You now have three intersections, one for each pair of circles. Prove that these three points lie on a line. Note: A difficult geometry puzzle, but with a simple solution. [ Wording last edited on 7:02 AM 4/2/2003 by William Wu ]

« Last Edit: Apr 2^{nd}, 2003, 7:07am by william wu » 
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Boody
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Re: Three Collinear Intersections
« Reply #1 on: Apr 2^{nd}, 2003, 6:50am » 
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on Apr 2^{nd}, 2003, 3:56am, william wu wrote:...Draw two tangents touching the circles and find the intersection of the two tangents... 
 Just idea of reformulation: Draw the two tangents touching the circles and forming a cone with their intersection as summit (with an intersection which is not between the two circles). But cones means 3D : perhaps this can introduce a misunderstood. In fact this riddle seems work also in 3D with 3 spheres and the 3 cones we can do with. This ridlle is the projection in the plan formed by the centre of the 3 spheres.


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Rujith de Silva
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Re: Three Collinear Intersections
« Reply #3 on: Apr 9^{th}, 2003, 11:35am » 
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Here's a proof that I saw ages ago (sorry, can't remember where): Wrap spheres around the circles, with the same center and radii. For each pair of spheres, wrap a cone around them. This gives three cones. The tangent lines around each pair of circles lie on the corresponding cone, and intersect at the vertices of the cone. Now place a plane tangent to the three spheres. This plane will obviously be tangent to the three cones. In particular, the vertices of the cones will lie on the plane. Now place another plane, also touching the three spheres, from the other side. The vertices of the cones lie on this plane as well. But the two planes intersect in a straight line. Therefore the intersections of the tangent lines lie on this straight line. Hope the above description is clear!

« Last Edit: Apr 10^{th}, 2003, 6:23am by Rujith de Silva » 
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SWF
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Re: Three Collinear Intersections
« Reply #4 on: Apr 13^{th}, 2003, 5:12pm » 
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That is a clever proof, Rujith de Silva. Here is how I would do it in a more mundane way: Number the circles, 1, 2, 3, with radii r_{i}, say r_{1}>r_{2}>r_{3}. If A, B, and C are the vectors from center of circle 1 to the 3 intersection points, and V_{ij} is the vector from center of circle i to center of circle j, then: A= r_{1}/(r_{1}r_{2})*V_{12} B= r_{1}/(r_{1}r_{3})*V_{13} which can be derived by looking at the similar trianges formed from tangent lines, intersections points, and lines through centers of a pair of circles. The vector C is found in a similar way, but is a more complicated expression because it needs to be expressed as a vector relative to center of circle 1: C=V_{12}+r_{2}/(r_{2}r_{3})*(V_{13} V_{12}) With some algebra, it is seen that if x=r_{2}(r_{1}r_{3})/r_{1}/(r_{2}r_{ 3}), C=x*B+(1x)*A This is the vector formula for a line through intersection points A and B, so three intersection points lie on the same line.


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Icarus
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Re: Three Collinear Intersections
« Reply #5 on: Apr 13^{th}, 2003, 7:01pm » 
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Rujith's Proof is one I have seen before, and is a very slick and geometrically appealing proof. Unfortunately, as it stands, it does not apply in every case. If the smallest circle is located between the other two, then you cannot get a plane tangent to all three cones. This difficulty can be avoided by viewing the original plane situation to be the projection onto a skew plane of the spheres and cones, with the smallest sphere not lying directly between the other two.


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Rujith de Silva
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Re: Three Collinear Intersections
« Reply #6 on: Apr 14^{th}, 2003, 6:01am » 
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on Apr 13^{th}, 2003, 7:01pm, Icarus wrote:Unfortunately, as it stands, it does not apply in every case. If the smallest circle is located between the other two, then you cannot get a plane tangent to all three cones. 
 Hi Icarus, do you mean if the spheres' centers are collinear? If they are merely NEARLY collinear, and the smallest sphere is between the other two, then I think tangent planes can still be constructed, with two spheres on one side of the plane, and the other sphere on the other side. Please let me know if that is not correct.


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Icarus
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Re: Three Collinear Intersections
« Reply #7 on: Apr 14^{th}, 2003, 6:54pm » 
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Actually the problem is more general than I stated. If the smallest sphere lies anywhere within the cone formed by the other two, then at most one plane tangent to that cone can also touch the smallest sphere. Since two planes are necessary for the proof, this situation cannot be allowed: O ^{O} O or this one O O O without special treatment. However, if you treat these as perspective views, you have the freedom to move the small sphere forward or backward until it is no longer inside the cone of the other two. So the proof still works, it just needs some tweaking in this special case.


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SWF
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Re: Three Collinear Intersections
« Reply #8 on: Apr 14^{th}, 2003, 8:05pm » 
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Perhaps the mundane vector proof is not so bad afterall. The original question seems to leave out a few cases: it implies that the intersection points must lie outside the convex hull formed by the circles. Also, what if one of the small circles is completely inside a larger circle.


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Icarus
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Re: Three Collinear Intersections
« Reply #9 on: Apr 14^{th}, 2003, 8:43pm » 
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I think you are misreading the convex hull part of the problem statement. The only point of that is to specify which two bitangent lines are being considered. Given any two disjoint circles (with disjoint interiors as well  i.e. one circle is not inside the other), there are 4 lines tangent to both circles. By the "convex hull" bit, William is just throwing out the two lines that pass between the circles. As for having one circle inside another  in this case there can be no line tangent to both, so the theorem certainly does not apply. I was amazed at how simple your vector proof was. When I tried to do this, I only thought of 2D geometric approaches, and didn't get very far. I think it shows the great utility of vectors, that they provide the answer in only a few lines. I originally came across the problem, including the spheres and cones proof and the shortcomings of it, in a Martin Gardner book.


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James Fingas
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Re: Three Collinear Intersections
« Reply #10 on: Apr 17^{th}, 2003, 1:15pm » 
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There is a more general way to look at this problem. Consider this: 1) Shrink the radii of all three circles to 1/2 of their original values. Prove that the line of tangent intersections remains in the same place. 2) Draw two of the tangentpairs so they intersect between the circles, and one set so it intersects outside the convex hull of the circles (like they all did before). Prove that these intersection points are also in a line. 3) Consider this one a hint: Draw another circle, at the centroid of the triangle defined by the centers of the three original circles, and with the radius as the average of the three original radii. Prove that all pairwise outside tangent intersections still lie on the same line.


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Rujith de Silva
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Re: Three Collinear Intersections
« Reply #11 on: Apr 18^{th}, 2003, 7:04am » 
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on Apr 14^{th}, 2003, 6:54pm, Icarus wrote:However, if you treat these as perspective views, you have the freedom to move the small sphere forward or backward until it is no longer inside the cone of the other two. 
 That doesn't seem right  I don't think it's invariant under projection. Just to verify the scenario: suppose there are two disjoint differentsized circles, and you draw the common outside tangents. Construct spheres and cones as usual. Construct a projection plane going through the common centers. Then the projection of the spheres will be the circles, and the projection of the cones will be the tangent lines. With me so far? Okay, now move one circle/sphere a long distance in a line perpendicular to the plane. Its projection will remain the original circle. But the projection of the new wrapping cone will NOT be the original tangent lines. To see why, consider the arc joining the tangent points of the cone and the stationary circle. This changes as the other circle recedes into the distance.


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Icarus
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Re: Three Collinear Intersections
« Reply #12 on: Apr 18^{th}, 2003, 9:04pm » 
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But the cone remains tangent to both spheres, and its extremities always project into lines tangent to the circles. Since the circles do not change, the lines tangent to them cannot change either.


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Rujith de Silva
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Re: Three Collinear Intersections
« Reply #13 on: Apr 22^{nd}, 2003, 8:47am » 
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on Apr 18^{th}, 2003, 9:04pm, Icarus wrote:But the cone remains tangent to both spheres, and its extremities always project into lines tangent to the circles. 
 I believe that the cone's extremities DON'T project to lines that are tangent to the circles. Could you explain why they do? Otherwise, I'll attempt to explain why I believe they do not.


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James Fingas
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Re: Three Collinear Intersections
« Reply #14 on: Apr 22^{nd}, 2003, 11:32am » 
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Rujith, Icarus is right here. Here is a somewhat clumsy way to show it: Project both lines vertically to form two planes. The planes touch the cone in two lines. Draw these lines on the cone. The sphere touches the cone in a circle. Draw this circle on the cone. The lines each intersect the circle at one point. At that point, the surface of the cone and the surface of the sphere both point in the same direction (because they're tangent there), and that direction is horizontal (because the vertical plane is also tangent there). Draw the horizontal equator of the sphere. This is the circle which you project onto the plane. It consists of all points where the sphere surface faces horizontally. Therefore, the sphere equator, the circlecone intersection, and the planecone intersections all meet at these two points. The projection of these two points onto the plane are where the lines are tangent to the circle.


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Rujith de Silva
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Re: Three Collinear Intersections
« Reply #15 on: Apr 22^{nd}, 2003, 1:44pm » 
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Okay, I understand the explanation regarding the projection of the cone. Thanks, Icarus & James!


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