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A cone is cut by a plane, making a closed curve. Two spheres, each inscribed into the cone, are tangent to the plane, at points $A$ and $B$ respectively.

Find the point $C$ on the cut line (i.e. the curve formed by the intersection of the cone and plane) such that $CA + CB$ is at an extrema.

--Vladimir Arnold

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My solution is below. Is it correct? Is the proof rigorous, easy to follow, and well written? How could it be improved?


Solution: Let the line $AB$ intersect the cone at points $M, N$. The points $M, N$ are the desired points. That is, $CA + CB$ is at an extrema if and only if $C = M$ or $C = N$.

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Proof: Consider the family of confocal ellipses with $R,S$ as foci. For any point $G$, the quantity $GR + GS$ is fixed for all points on a particular ellipse of this family. Thus, given an arbitrary smooth closed convex curve $\Gamma$, the points $G$ on $\Gamma$ where $GR + GS$ is an extremum are precisely the points where an ellipse of this family is tangent to $\Gamma$.

Therefore, in the given problem, $C$ is an extrema if and only if it is a point where the cut line is tangent to some ellipse with foci $A, B$.

The cut line is clearly an ellipse with major axis along the line $AB$.

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Therefore, our problem becomes:

Given a fixed ellipse $\Gamma$ whose major axes lies on line $AB$, and the family of ellipses with foci at $A, B$, at what points on $\Gamma$ will $\Gamma$ be tangent to some ellipse of this family?

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Every ellipse intersects its major axis with a tangent perpendicular to this axis. Therefore, the ellipses which intersect $M$ or $N$ will be tangent at those points.

Furthermore, there can be no other points of tangency, since as we move around an ellipse, the slope of the tangent varies strictly monotonically (this last point seems admittedly like hand waving.)


Update

Reviewing Dandelin spheres, it's clear that not only are the foci on the line $AB$, the foci are the points $A,B$! This means $MA = BN$ and my solution is wrong. Rather, the value $CA + CB$ is constant.

If so, my question becomes: Where is the "proof" wrong?

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  • $\begingroup$ Your proof is "wrong" because the tangent ellipse is the same as the curve. But I'd also say that the problem is misleading. $\endgroup$ Commented Feb 26 at 10:53

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