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Kingom

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Can a cube be inscribed in a cone so that 7 vertices of the cube lie on the surface of the cone?
 

seanieg89

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I am assuming that "cone" means a set like {(x,y,z): x >= 0, y^2+z^2=c^2x^2}, rather than the usual solid with a circular hat, otherwise the problem is trivial.

Under this assumption, the answer is no. There is probably a nicer way to do this, but I will sketch my argument below.


Pf: One face F of the cube must have all vertices on the cone, and so by considering the intersection of this face's plane with the cone we obtain a conic section circumscribing a square. As we are only considering a single branch of the "double-cone", the possibilities for the conic are as follows:

-Degenerate (A point, a line, or a planar "V" shape.)
-Ellipse (including circles for simplicity of nomenclature)
-Parabola
-Hyperbola.

In fact, out of the above list, only ellipses can possibly circumscribe a square. Ruling out the first two degenerate cases is trivial. To rule out the V, the parabola, and the hyperbola, there are multiple approaches. The argument sketched in the next paragraph is in theme with the rest of the proof, and is hopefully convincing to you.

If a square is circumscribed by any of the above planar curves, this implies that each vertex V of the square is simultaneously visible from some point P outside the curve (a sufficiently far point on the axis of symmetry) and hence outside the square. Other ways of stating this property is that a point light source at P will illuminate every V, or that each line segment PV only intersects the square at V. However, we can never illuminate all vertices of a square from a single exterior point light source. Indeed, such a light source must lie in a quadrant (the exterior region closes to a particular vertex), from which it is easy to show the opposite vertex will not be illuminated.

The upshot of all of this is that the conic must be an ellipse. This also implies that the plane through the face F' OPPOSITE F must also cut the cone in a similar but not congruent ellipse that must circumscribe three of the four vertices of the same size square that F circumscribes. The remainder of the proof is showing that this cannot occur.

Basic HS circle geometry shows:

i) If three vertices of a rectangle lie on a circle, then so does the fourth.
ii) Each rectangle is circumscribed by a circle of a unique radius.

Now choosing standard coordinates and applying the linear transformation (x,y)->(x/a,y/b) maps the circle to an ellipse and the square to a cyclic parallelogram, i.e. a rectangle.

Applying the same linear transformation to the intersection of the plane containing F' with the cone and using i), we obtain a congruent rectange circumscribed by a circle of different radius. This contradicts ii) and completes the proof.


Apologies for the haphazard writeup, hopefully the proof is convincing enough. I am happy to flesh out any part that is unclear.
 
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seanieg89

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This kind of question makes me wish that geometric transformations were discussed more in HS. The reduction of a geometric fact about conics to a geometric fact about circles using parallel projection is kinda cute.
 

Kingom

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I am assuming that "cone" means a set like {(x,y,z): x >= 0, y^2+z^2=c^2x^2}, rather than the usual solid with a circular hat, otherwise the problem is trivial.

Under this assumption, the answer is no. There is probably a nicer way to do this, but I will sketch my argument below.


Pf: One face F of the cube must have all vertices on the cone, and so by considering the intersection of this face's plane with the cone we obtain a conic section circumscribing a square. As we are only considering a single branch of the "double-cone", the possibilities for the conic are as follows:

-Degenerate (A point, a line, or a planar "V" shape.)
-Ellipse (including circles for simplicity of nomenclature)
-Parabola
-Hyperbola.

In fact, out of the above list, only ellipses can possibly circumscribe a square. Ruling out the first two degenerate cases is trivial. To rule out the V, the parabola, and the hyperbola, there are multiple approaches. The argument sketched in the next paragraph is in theme with the rest of the proof, and is hopefully convincing to you.

If a square is circumscribed by any of the above planar curves, this implies that each vertex V of the square is simultaneously visible from some point P outside the curve (a sufficiently far point on the axis of symmetry) and hence outside the square. Other ways of stating this property is that a point light source at P will illuminate every V, or that each line segment PV only intersects the square at V. However, we can never illuminate all vertices of a square from a single exterior point light source. Indeed, such a light source must lie in a quadrant (the exterior region closes to a particular vertex), from which it is easy to show the opposite vertex will not be illuminated.

The upshot of all of this is that the conic must be an ellipse. This also implies that the plane through the face F' OPPOSITE F must also cut the cone in a similar but not congruent ellipse that must circumscribe three of the four vertices of the same size square that F circumscribes. The remainder of the proof is showing that this cannot occur.

Basic HS circle geometry shows:

i) If three vertices of a rectangle lie on a circle, then so does the fourth.
ii) Each rectangle is circumscribed by a circle of a unique radius.

Now choosing standard coordinates and applying the linear transformation (x,y)->(x/a,y/b) maps the circle to an ellipse and the square to a cyclic parallelogram, i.e. a rectangle.

Applying the same linear transformation to the intersection of the plane containing F' with the cone and using i), we obtain a congruent rectange circumscribed by a circle of different radius. This contradicts ii) and completes the proof.


Apologies for the haphazard writeup, hopefully the proof is convincing enough. I am happy to flesh out any part that is unclear.
Well done! This is basically the solution I have as well.
 

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