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The Spaniard

Another Maths Problem

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If you have not met this one before, in my opinion the solution is very neat and impressive.

Consider three circles drawn in a plane. For clarity imagine that the circles are of different sizes, and not overlapping.

Then there are three ways of choosing two of the circles, drawing the two external common tangents to the chosen pair, and plotting the meeting point of these tangents.

Thus there are three such meeting points in the plane.

Prove that these three points are collinear.

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Hint: Think in three dimensions, reasoning about the three spheres for which the three co-planar circles are great circles.

For each of the three possible pairs of circles think of the unique and well defiined 3D cone which touches both corresponding 3D spheres all the way around.

The intersection of each such 3D cone with the plane containing the three circles is precisely the pair of external common tangents of the two relevant circles.

Hence the three points of interest in the plane are the apex points of these three cones.

Why are these three apexes collinear?

No algebra or calculus is required.

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Why in the world should we need to go into a higher dimension to solve this?  Did we need to start thinking about cuboids and pyramids to solve the square-in-triangle one?

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14 minutes ago, Qetesuesi said:

Why in the world should we need to go into a higher dimension to solve this?  Did we need to start thinking about cuboids and pyramids to solve the square-in-triangle one?

To grasp the description, I think, instead of talking about tangents it's easier to visualise balls in a cone.

Passed it to someone at work and he's solved it algebraically, although it's left me with the feeling that an easier explanation is possible.

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52 minutes ago, Riedquat said:

To grasp the description, I think, instead of talking about tangents it's easier to visualise balls in a cone.

Passed it to someone at work and he's solved it algebraically, although it's left me with the feeling that an easier explanation is possible.

Care to post that proof here?  I think it would involve handling at least five independent variables (degrees of freedom).

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38 minutes ago, Qetesuesi said:

Care to post that proof here?  I think it would involve handling at least five independent variables (degrees of freedom).

He's saying it isn't completely rigorous... I'm not sure I can figure out what he's scribbled enough to translate it into something postable.

The gist of the principle is (I think):

Treat one of the crossing points as the origin, and two of the circles (A and B ) lying on the x-axis. Take circle B and circle C (i.e. the one not on the x-axis). The intersection of the tangents of that one forms a line between that point and the origin. Circle A is part of an infinite set of circles that include your original chosen one that fits in the tangents, with a radius proportional to the distance from the origin. Move it out from the origin and its intersecting tangents with C will draw a line. That line can be calculated as being a line, and the B to C point lies on because that's also part of that series. That last part might be a bit iffy because it fails the non overlapping condition.

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I get the set-up in Monge's theorem, but I can't make the final leap of logic.

Actually, thinking about it now (after a nights sleep) - maybe I can. In 3 dimensions the points must intersect in a plane. Now when you view that plane 'end on' the points must be on a line. Hmmm, that's about as far as I can get. They definitely have to be on a straight line when viewed from a particular angle.

 

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49 minutes ago, frozen_out said:

I get the set-up in Monge's theorem, but I can't make the final leap of logic.

Actually, thinking about it now (after a nights sleep) - maybe I can. In 3 dimensions the points must intersect in a plane. Now when you view that plane 'end on' the points must be on a line. Hmmm, that's about as far as I can get. They definitely have to be on a straight line when viewed from a particular angle.

It is possible to have two planes that will touch all three spheres (if you only had two spheres you could wobble the plane from side to side, until it's at the right angle to touch a third), and it will touch each one at a tangent. You could draw the lines between the two spheres on both planes, and they'd meet where the planes meet.

Any two planes in 3 dimensional space will intersect along a line, as long as they're not parallel (imagine two sheets of card).

In three dimensions the spheres can be imagined sitting in cones, which just touch the planes. Therefore any opposite lines along those cones are also tangents to the spheres that would meet along the line where the planes intersect (the points of the cones are there). So you could take a cut through this 3D example to create the original 2D one.

I agree with the OP, it's a rather nice simple way of proving it without having to use any maths at all.

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I must admit it took me a long time to understand the sphere explanation of Monge's theorem.

I had proved it algebraically in the mean time. 

Define 3 circles with centres (x1..3, y1..3) with radius r1..3.

The locations of the intersections of the tangents can be calculated as follows, where (xa,ya) represents the apex formed by the tangents of circles 1 and 2.

xa = x1 + r1/(r1-r2)*(x2-x1); ya = y1 + r1/(r1-r2)*(y2-y1)

The apices B and C (corresponding to circles 1 and 3, and 2 and 3) can be computed in the same way.

You can now compute the slopes of the lines Gab, Gac and Gbc, which will be equal if the points are colinear.

Gab = (xa-xb)/(ya-yb)

Gab = Gac = Gbc = (x1(r2-r3) - x2(r1-r3) - x3(r2-r1))/(y1(r2-r3) - y2(r1-r3) - y3(r2-r1))

 

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11 minutes ago, ChumpusRex said:

I must admit it took me a long time to understand the sphere explanation of Monge's theorem.

I had proved it algebraically in the mean time. 

Define 3 circles with centres (x1..3, y1..3) with radius r1..3.

The locations of the intersections of the tangents can be calculated as follows, where (xa,ya) represents the apex formed by the tangents of circles 1 and 2.

xa = x1 + r1/(r1-r2)*(x2-x1); ya = y1 + r1/(r1-r2)*(y2-y1)

The apices B and C (corresponding to circles 1 and 3, and 2 and 3) can be computed in the same way.

You can now compute the slopes of the lines Gab, Gac and Gbc, which will be equal if the points are colinear.

Gab = (xa-xb)/(ya-yb)

Gab = Gac = Gbc = (x1(r2-r3) - x2(r1-r3) - x3(r2-r1))/(y1(r2-r3) - y2(r1-r3) - y3(r2-r1))

 

V = I x R

I get thirty-seven grand a year - and a half-decent quality desk - for understanding that.

You must earn a fortune knowing all those fancy equations Chumps.

 

XYY

 

                                                                                                               

The dog's kennel is not the place to keep a sausage - Danish proverb

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13 minutes ago, happy_renting said:

Picture 3 non-intesecting circles, like a uselessly sh*t Venn diagram.

You obviously did the same crap maths course I did. Very 70s.:o

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On 20/12/2016 at 0:20 PM, The Spaniard said:

If you have not met this one before, in my opinion the solution is very neat and impressive.

Consider three circles drawn in a plane. For clarity imagine that the circles are of different sizes, and not overlapping.

Then there are three ways of choosing two of the circles, drawing the two external common tangents to the chosen pair, and plotting the meeting point of these tangents.

Thus there are three such meeting points in the plane.

Prove that these three points are collinear.

Are we talking Boeing or Airbus here?

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32 minutes ago, The XYY Man said:

 

V = I x R

I get thirty-seven grand a year - and a half-decent quality desk - for understanding that.

You must earn a fortune knowing all those fancy equations Chumps.

 

XYY

 

 

                                                                                                               

 

The dog's kennel is not the place to keep a sausage - Danish proverb

 

Physics teachers get more than I thought.

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When looking at the diagram and mentally moving the circles it is indeed "Self evident" as John Edson Sweet said. Why do we need the wordy explanations?For those with limited spatial imagination?

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