An icosahedron - from the other side!
An icosahedron is rotated several times about different axes passing through its center.
Is it possible that none of the points on the surface of the icosahedron are in the same place as they were from before?
Note : The surface is considered the union of all faces, edges and vertices.
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8 solutions
You should also show that every 3 by 3 transformation matrix has a real eigenvalue since square matrices of even dimensions need not have a real eigenvalue.
But what if no points lie along the surface, and also this axis? For example, we can have this axis go through two vertices. Would these vertices be considered on the surface?
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Yes. The surface is considered to be the union of all faces, edges and vertices.
I got this wrong by mis-reading the question it's not "none of the VERTEX points ..." are in the same place but non of ANY arbitrary SURFACE point.
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Thanks for your observation. I made a clarifying note at the bottom.
I dont get it
I was wondering that is it applicable to all 3D figures?
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Yes it is! :) I just picked the icosahedron at random.
What if you rotate it once? Will there not be any two points on the surface that will change their position?
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Yes the two points that lie on the axis of the rotation (where that axis intersects the the surface of the shape).
By this can we say there are only 2 point which does so
Could one say that the problem is about the rotation axis of the problem cutting/going through (?) the surface twice, therefore not being able to change those two points? [Only talking about one rotation, because I still haven't got an intuitive grasp of combining two or more rotations - is that legit? Is "several" in math talk an arbitrary number n>0 or any number? When would you use "several"?]
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Yes in this context several means and arbitrary number >0. As an aid to visualising the truth that any number of arbitrary rotations in 3D is equivalent to one rotation. Consider a specific direction to be a reference direction say the X-axis. After a number of rotation the original X-axis will be point in a new 3D direction. This new direction will make an angle with the original X-axis. The whole transformation is the same as rotation about a unique axis that is simultaneously perpendicular to the original X-axis and the new 3D direction.
How do you know that the 'final' rotation has an axis?
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Hi Stewart... Not quite sure what you mean... I seems all rotations must have an axis by definition...
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I disagree. But with this view, my question still applies, albeit rephrased: How do you know that, when you combine the two rotations, the resulting transformation is a rotation by your definition?
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@Stewart Gordon – Once you have accepted that the first transformation is a rotation, then all subsequent ones are simply rotations about a new axis. Perhaps if you imagine a series of rotations of a sphere with a polar axis you can see that the polar axis ends up somewhere and that you could have moved the pole to that resulting place by one rotation.
@Stewart Gordon – I like to think of it as being able to represent a rotation by a rotation matrix, and two rotation matrices can be multiplied together to form a third rotation matrix which will have a new axis of rotation.
Alternatively, does it make more sense after reading some of the other comments/solutions to this question?
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@Geoff Pilling – All you've done is repeat the same thing that led to my original question.
In my mind, the basic definition of a rotation is a direct isometry that has at least one fixed point. By this definition, the question is one of whether every rotation has an axis. On the other hand, if you define a rotation as having an axis, then the question is of whether the transformation resulting from combining two rotations that have a fixed point in common is necessarily a rotation. Whichever definition you use, your solution is incomplete without a proof that the answer to the respective question is "yes", or at least a reference to a known theorem stating this.
the problem is not clearly explained. if you mean Before the last rotation the answer is obviously never. the intersection points between the rotating axis and the surface are always invariant . But if you mean Before as the initial position the answer is yes,
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Not sure I agree that there is any ending orientation (maintaining the same center point which is implied in the problem) where the positions of all the surface points are all different.
We all have realized this, but the proof is absolutely not complete. How do you know that if you rotate it with one axis, then another with a different axis, the new position of the points can be reached by a rotation with the axis passing through the Origo?
Here, as an alternative to Geoff's nice solution, is one that does not rely on the properties of SO(3).
Proof by contradiction. Suppose there is such a rotation R with no invariant point.
Imagine a sphere with the same center as the icosahedron and rotating with it. There is a one to one correspondence between the points on the sphere and the points on the icosahedron. (Found, for example, by drawing a line through the common center and the surfaces of the icosahedron and the sphere).
We will apply the hairy ball theorem , which states that there is no way to continuously comb a hairy ball without a hair sticking out. Technically, it says that there is no nonvanishing continuous tangent vector field on the 2-sphere (which is the surface of the 3-D sphere).
With the rotation R , we now define a map T : S 2 → Tangent space of S 2 , where T ( x ) = the vector direction from x to R ( x ) , with length equal to the distance (along the surface of the sphere) from x to R ( x ) . Then, since R ( x ) = x so T ( x ) = 0 , which gives us a nonvanishing continuous tangent vector field. Hence, we have a contradiction.
Brouwer's fixed point theorem can only be applied to compact convex sets though, and the surface of a sphere isn't a convex set. Perhaps if u can show that the surface of a sphere has a bijective mapping to a compact convex set, this proof would work.
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No a sphere is not homeomorphic to a compact convex and there is no way to make this proof work : the antipodal map x ↦ − x is a continuous homeomorphism of the sphere without fixed point. Of course, the antipodal map is an element of O ( 3 ) ∖ S O ( 3 ) so you cannot get the antipodal map using only rotations.
Right. As Roland points out, Brouwer's fixed point theorem doesn't work (immediately).
Peter is actually invoking the hairy ball theorem , IE that there is no nonvanishing continuous tangent vector field on the 2-sphere (which is the surface of the 3-D sphere).
Thanks for pointing out this error in my 'solution', and for Calvin's suggestion that it could be rescued by the 'hairy ball' theorem. I'm kicking myself for not checking the conditions under which Brouwer would hold.
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Can you update your solution accordingly? It is a nice way to think about this problem.
In fact, there are connections between the Hairy Ball Theorem and the fact that rotation matrices can be decomposed to 1 × 1 and 2 × 2 blocks, which makes these arguments "similar at the base level".
Hey I loved the hairy ball theorem which last I had seen in minute physics,which is a youtube channel!!
How are you defining T ( x ) when R ( x ) is the point diametrically opposite x , so as to maintain continuity of T ?
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Yes Stewart, the pesky antipodal point is a problem for my solution. I'm in half a mind to delete it out of embarrassment. However it has given folk some fun, and maybe it can still be patched up.
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One possible fix is that by orientation, we cannot have an open set (on the surface of the sphere) get mapped to the antipodal point, so it should be possible to define T ( x ) as a limit of points around it.
E.g. Given a point x , take a sequence of neighborhoods N i that converge to x , and define T(x) = \lim_{n \rightarrow \infty} \lim_{y \in N_n
If you rotate you have an axis, and on the "poles" of that axis those 2 points don't change position (can we say a point rotates?), no matter the axis selected (vertex,edge, any point on the surface of the icosahedron)
It says several different axes. So you can't consider poles as never moving
ELI5 solution. Brower's fixed point theorem says that any continuous transformation in any object with no holes will leave at least one point unmoved.Since the object has no holes, and a rotation is a continuous transformation, so the theorem applies.
I don't know but for any 3d shape it's not possible
Because the two surface points will always be stationary(those on the axis of rotation)
But I think I didn't understood the actual problem, correct me if I am wrong
As it is mentioned that the icosahedron has been rotated around different axis which does not change the location and moreover as there is no change in the location but due to rotating it becomes smaller and smaller,the points will eventually overlap.
Becomes smaller and smaller? Not quite sure what you mean... Could you elaborate?
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I meant the area of the figure becomes smaller and smaller!!
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Why does the area of the figure become smaller and smaller?
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@Geoff Pilling – Because according to the question it is stated": The surface is considered the union of all faces, edges and vertices." And please do correct me if I am wrong!!!
The icosahedron is homeomorphic to a sphere, so we may consider rotations on the sphere instead. The sphere is a compact, convex set (in R^3, a Euclidean space) and the composition of rotations is a continuous function from the sphere to itself, so by the Brouwer Fixed Point Theorem, the sphere has a point fixed by the single composed rotation, and thus so does the icosahedron via the homeomorphism (so the answer is "No").
Imagine a sphere with its center matching that of the icosahedron and radius such that it forms a tangent at the point where this rotation axis meets the icosahedron (there will be two points of tangent on the two sides away from this axis?).
Now, as the icosahedron rotates along this axis, the sphere rotates along the same axis. The point where this sphere connects with the icosahedron will remain stationary. Because all points along the axis of a rotating sphere (including the two on its surface) remain stationary.
Since the icosahedron can be reoriented with a series of rotations, and the rotation matrices can be multiplied together to form a "new" single rotation, the points on the surface that lie upon the axis of this final rotation will not have changed location.