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Subject: "Physics based solution to Three Circles Tangents problem"     Previous Topic | Next Topic
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mr_homm
Member since May-22-05
Jul-18-05, 10:58 AM (EST)
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"Physics based solution to Three Circles Tangents problem"
 
   Although there are already very nice proofs for the Three circles based on a 3D point of view, I would like to add one based on a physics point of view.

Name the three circles a, b, and c, so that the common tangents of a and b meet at C, those of a and c at B, and those of b and c at A. We assign each circle a mass (ma, mb, or mc) inversely proportional to its radius. If we use a positive mass for each circle, then the center of mass of each pair of circles lies at the intersection of their internal tangents, but if we make one mass positive and one negative, the center of mass will lie at the intersection of their external tangents. This is easy to prove using similar triangles and the definition of center of mass. (If you don't like the idea of negative mass, think of forces applied either pointing into the screen or out of it, and use center of force instead.)

Now it is a known fact that the coordinates of the center of mass of a combined system is just the weighted average of the coordinates of the centers of each system separately, and hence is collinear with them. If I choose systems {ma, -mb}, {mb,-mc}, and {ma, -mc}, then superimposing the first two systems gives the third, hence its center of mass is collinear with theirs. QED

By the way, this also shows why you can't prove the theorem for internal tangent pairs: if all the masses are positive, you can't combine two pairs of circles to get the remaining pair, because no cancellation is possible. The theorem does work again with one external tangent pair and two internal tangent pairs: {ma, mb} {mc, -mb} = {ma, mc}.

--Stuart Anderson


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alexbadmin
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1603 posts
Jul-18-05, 11:47 AM (EST)
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1. "RE: Physics based solution to Three Circles Tangents problem"
In response to message #0
 
   Yes, thank you.

In a similar vein, are you aware of the following problem:

All four sides of a quadrilateral touch a given sphere. Prove that the four tangency points are coplanar.


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mr_homm
Member since May-22-05
Jul-24-05, 06:41 AM (EST)
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2. "RE: Physics based solution to Three Circles Tangents problem"
In response to message #1
 
   >Yes, thank you.
>
>In a similar vein, are you aware of the following problem:
>
>All four sides of a quadrilateral touch a given sphere.
>Prove that the four tangency points are coplanar.

I haven't looked at this yet, but while taking a break from the 4 Pegs problem, I noticed that you can do a similar center of mass proof for the Soddy Circles and Eppstein Lines. Again, let each circle have a mass inversely proportional to its radius, and if there is a large circle to which the other three are internally tangent, let its mass be negative. If all 4 circles are externally tangent to one another, all masses are positive.

Then the center of mass of each two circle subsystem is at their point of tangency. The whole system is the sum of each complementary pair of two-circle subsystems, so by linearity, the center of mass of the whole system is on a line between the center of mass of the two subsystems. These lines are exactly Eppstein's lines, and since the center of mass of the whole system is on all three of them, they must be concurrent.

Note that he overall center of mass is always well defined even in the case where the masses are of mixed sign, because the masses of each of the three inner circles are greater than the mass of the outer circle, which is the only negative mass. Therefore, the total mass cannot be zero and the center of mass is well defined.

I had never given much thought to this center of mass trick, but now it has worked in two interesting cases, so I shall have to move it nearer the top of my bag of geometry tricks in the future.

Later, I'll look at the inscribed quadrilateral problem, but I wanted to post this now while it is fresh im my mind.

--Stuart Anderson


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alexbadmin
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1603 posts
Jul-24-05, 07:31 AM (EST)
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3. "RE: Physics based solution to Three Circles Tangents problem"
In response to message #2
 
   >I haven't looked at this yet, but while taking a break from
>the 4 Pegs problem, I noticed that you can do a similar
>center of mass proof for the Soddy Circles and Eppstein
>Lines.

This is the essence of Ceva's theorem. The ratio's on each side are exactly the ratio of the weights placed at the vertices.

https://www.cut-the-knot.org/Curriculum/Geometry/Barycentric.shtml

>Then the center of mass of each two circle subsystem is at
>their point of tangency. The whole system is the sum of
>each complementary pair of two-circle subsystems, so by
>linearity, the center of mass of the whole system is on a
>line between the center of mass of the two subsystems.
>These lines are exactly Eppstein's lines, and since the
>center of mass of the whole system is on all three of them,
>they must be concurrent.

The mechanics, unlike Ceva's theorem, works "on principle", thus avoiding the mecahnics of calculations.

>I had never given much thought to this center of mass trick,
>but now it has worked in two interesting cases, so I shall
>have to move it nearer the top of my bag of geometry tricks
>in the future.
>
>Later, I'll look at the inscribed quadrilateral problem, but
>I wanted to post this now while it is fresh im my mind.

Yes, thank you. The quadrilateral problem is of the sorts for which the mechanical solution is almost transparent, but for which a geometric solution is less than trivial.

Into each of the points place a weight inversely proportional to the length of the tangent from the point to the sphere. Then each point of tangency will be the center of mass of two adjacent vertices. The center of mass of the four points will lie on the line joining two opposite tangency points. Taken in two ways, the center of mass of the system lies on two segments that must then intersect making the four tangency points coplanar.


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mr_homm
Member since May-22-05
Jul-27-05, 11:21 AM (EST)
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4. "RE: Physics based solution to Three Circles Tangents problem"
In response to message #3
 
   >
>Yes, thank you. The quadrilateral problem is of the sorts
>for which the mechanical solution is almost transparent, but
>for which a geometric solution is less than trivial.
>
>Into each of the points place a weight inversely
>proportional to the length of the tangent from the point to
>the sphere. Then each point of tangency will be the center
>of mass of two adjacent vertices. The center of mass of the
>four points will lie on the line joining two opposite
>tangency points. Taken in two ways, the center of mass of
>the system lies on two segments that must then intersect
>making the four tangency points coplanar.

That is a very neat proof. It'seems that this sort of center of mass trick will work with problems involving related proportions, because the mass ratios determine the distance ratios.

In fact, looking at Ceva's and Menelaus' theorems together gives a neat unified proof based on center of mass very much like the one I gave above for the three circles' tangents. Start with the mechanical proof of Ceva's Theorem as given at
www.cut-the-knot.org/Generalization/CevaPlus.shtml#ceva
but allow the masses to be negative to get center of mass points D, E, and F to lie on the extensions of the sides of the triangle. Then both Ceva's amd Menelaus' Theorems follow from the two basic properties of center of mass:
Collinearity: Using COM{} to mean "center of mass of," COM{X}, COM{Y}, COM{X,Y} are collinear, where {X} and {Y} are systems of masses, and {X,Y} is the combined system.
Scaling: COM{aX} = COM{X}, where aX is the mass system X with all masses scaled up by a factor of a.

There is a trick though: you cannot assume that there is a single global system of 3 masses that (pairwise) produce the points D, E, and F as centers of mass. If you do assume that, you can only get Ceva's theorem. Instead, assume that there is a mass at each point A, B, and C, but form a separate system of two masses for each side the triangle ABC, choosing the signs of the masses in each system independently. You can then assume COM{A,B} = F, COM{B,C} = D, and COM{C,A} = E.

If the signs for a mass are chosen differently in the two systems it lies in, then let's say there is a sign mismatch. If there are 2 mismatches, say at A and B, then reversing the signs of both masses in system {A,B} removes both mismatches, and does not change the position of F, by the rescaling property. If there are no mismatches, then you can choose all signs as positive, and you get a global system {A,B,C} of which {A,B}, {B,C}, and {C,A} are subsystems. Ceva's theorem then follows from COM{A,B,C} = COM{A,{B,C}} = COM{B,{C,A}} = COM{C,{A,B}} and the collinearity property.

If there are 3 mismatches, reversing the signs of both masses in one system reduces them to 1 mismatch. With 1 mismatch, you cannot form a global system with {A,B}, {B,C}, and {C,A} as subsystems, so Ceva's theorem does not follow. Instead, you can now sum the two systems that share the mismatch to get the third system. For instance, if the mismatch is at B, them {A,B} + {B,C} = {A,C}, since the masses at B cancel out in the superposition. Now the collinearity property gives Menelaus' Theorem.

From this point of view, the difference between Ceva's and Menelaus' Theorems is a combinatorial one, a matter of whether the signs allow one pattern or another to be made with the systems.

--Stuart Anderson


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alexbadmin
Charter Member
1603 posts
Jul-28-05, 11:45 AM (EST)
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5. "RE: Physics based solution to Three Circles Tangents problem"
In response to message #4
 
   There is a 50 years old Russian book on the theory and applications of the center of mass. I've been charmed by it from my school days and, from the inception of the site, thought writing a few pages to formalize the concept. With so many examples the idea becomes almost irresistable.


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