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william wu
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Three Collinear Intersections  
« on: Apr 2nd, 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 2nd, 2003, 7:07am by william wu » IP Logged


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Re: Three Collinear Intersections  
« Reply #1 on: Apr 2nd, 2003, 6:50am »
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on Apr 2nd, 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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Re: Three Collinear Intersections  
« Reply #2 on: Apr 2nd, 2003, 7:06am »
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thanks, i just edited the wording ... let me know if you think it could be better
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Re: Three Collinear Intersections  
« Reply #3 on: Apr 9th, 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 10th, 2003, 6:23am by Rujith de Silva » IP Logged
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Re: Three Collinear Intersections  
« Reply #4 on: Apr 13th, 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 ri, say r1>r2>r3.  If A, B, and C are the vectors from center of circle 1 to the 3 intersection points, and Vij is the vector from center of circle i to center of circle j, then:
A= r1/(r1-r2)*V12
B= r1/(r1-r3)*V13
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=V12+r2/(r2-r3)*(V13- V12)
With some algebra, it is seen that if x=r2(r1-r3)/r1/(r2-r 3),
C=x*B+(1-x)*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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Re: Three Collinear Intersections  
« Reply #5 on: Apr 13th, 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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Re: Three Collinear Intersections  
« Reply #6 on: Apr 14th, 2003, 6:01am »
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on Apr 13th, 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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Re: Three Collinear Intersections  
« Reply #7 on: Apr 14th, 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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Re: Three Collinear Intersections  
« Reply #8 on: Apr 14th, 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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Re: Three Collinear Intersections  
« Reply #9 on: Apr 14th, 2003, 8:43pm »
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I think you are mis-reading the convex hull part of the problem statement. The only point of that is to specify which two bi-tangent 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 short-comings of it, in a Martin Gardner book.
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Re: Three Collinear Intersections  
« Reply #10 on: Apr 17th, 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 tangent-pairs 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 pair-wise outside tangent intersections still lie on the same line.
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Re: Three Collinear Intersections  
« Reply #11 on: Apr 18th, 2003, 7:04am »
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on Apr 14th, 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 different-sized 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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Re: Three Collinear Intersections  
« Reply #12 on: Apr 18th, 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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Re: Three Collinear Intersections  
« Reply #13 on: Apr 22nd, 2003, 8:47am »
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on Apr 18th, 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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Re: Three Collinear Intersections  
« Reply #14 on: Apr 22nd, 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 circle-cone intersection, and the plane-cone 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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Re: Three Collinear Intersections  
« Reply #15 on: Apr 22nd, 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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