Messing around in higher dimensions 1

Geometry Level 5

The four dimensional analogue of a tetrahedron is called a 5-cell (also a pentachoron, or pentatope). This question is the four dimensional analogue of its inspiration .

Given five parallel hyperplanes in four dimensions, with each pair of neighbouring hyperplanes separated by a distance of 1 1 unit, is it possible to place a regular 5-cell so that each of its five vertices lies in a different one of these hyperplanes?

If the answer is "yes", enter the square of the edge-length of such a 5-cell. If the answer is "no", enter 0 0 .


The answer is 20.

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3 solutions

Michael Mendrin
Mar 6, 2020

For the figure above, the following are the coordinates of the vertices of the pentachoron, where all the edge lengths are 20 \sqrt{20} .

( 0 , 0 , 2 , 0 ) \left( 0,0,2,0 \right)
( 0 , 4 , 4 , 0 ) \left( 0,4,4,0 \right)
( 0 , 1 , 5 , 10 ) \left( 0,1,5,\sqrt { 10 } \right)
( 0 , 3 , 1 , 10 ) \left( 0,3,1,\sqrt { 10 } \right)
( 5 2 , 2 , 3 , 5 2 ) \left( \dfrac { 5 }{ \sqrt { 2 } } ,2,3,\sqrt { \dfrac { 5 }{ 2 } } \right)


The last vertex at the bottom of this list is at the center of this figure.

Parallel planes are shown passing through the 5 5 vertices having a spacing of 1 1

Great solution - I like the way you have shown the result, too. Can I ask how you found this particular shape? I'm interested because the method I used led to lots of 19 \sqrt{19} cropping up, whereas your vertex coordinates are much neater.

Chris Lewis - 1 year, 3 months ago

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I started off with the x-y plane, orthogonal to the 5 planes, and then the ( x , y ) (x,y) of the vertices are assigned to ( 0 , 2 ) (0,2) , ( 4.4 ) (4.4) , 1 , 5 ) 1,5) , ( 3 , 1 ) (3,1) and ( 2 , 3 ) (2,3) per figure. Then since ( 0 , 0 , 2 , 0 ) (0,0,2,0) , ( 0 , 4 , 4 , 0 ) (0,4,4,0) have a distance of 20 \sqrt{20} , it was just a matter of assigning z = 10 z=\sqrt{10} to the next pair to get the edges of 20 \sqrt{20} . and finally a little algebra and solving for ( u , z ) (u,z) takes care of the last one to ensure a pentachoron of edge length 20 \sqrt{20} . A total of 10 10 distances have to be worked out to make sure it all works out.

This is the same approach I used in Inspiration

Michael Mendrin - 1 year, 3 months ago
Chris Lewis
Mar 9, 2020

This is solvable using coordinate geometry. Let the five parallel planes be perpendicular to the w w -axis, and let the five vertices of the 5-cell be A , B , C , D , E A,B,C,D,E , with coordinates as below:

Point w w x x y y z z
A A 0 0 0 0 0 0 0 0
B B 1 1 B x B_x 0 0 0 0
C C 2 2 C x C_x C y C_y 0 0
D D 3 3 D x D_x D y D_y D z D_z
E E 4 4 E x E_x E y E_y E z E_z

Rather than make all the points arbitrary, we have put A A at the origin, then assigned the other vertices to the parallel hyperplanes with their w w -coordinates. We can further orient the 5-cell so that B B lies in the w x wx -plane, and so on. There's no loss of generality in doing this, but it reduces the number of unknowns to the 9 9 shown above, together with the edge-length s s .

We have ten unknowns; the ten equations we need are simply that the distance between any two vertices must be s s , that is, A B = A C = A D = = C D = s AB=AC=AD=\cdots=CD=s

This is a system of quadratic equations; with a bit of manipulation, it's possible to show that one solution is

A A 0 0 0 0 0 0 0 0
B B 1 1 19 \sqrt{19} 0 0 0 0
C C 2 2 8 19 \frac{8}{\sqrt{19}} 4 15 19 \frac{4\sqrt{15}}{\sqrt{19}} 0 0
D D 3 3 7 19 \frac{7}{\sqrt{19}} 5 57 \frac{\sqrt{5}}{\sqrt{57}} 5 3 \frac{5}{\sqrt3}
E E 4 4 6 19 \frac{6}{\sqrt{19}} 5 228 -\frac{\sqrt{5}}{\sqrt{228}} 5 12 -\frac{5}{\sqrt{12}}

Other solutions are possible in terms of the vertex positions, but these just amount to reflections and rotations; s = 20 s=\sqrt{20} is the only valid edge-length, so the required answer is s 2 = 20 s^2=\boxed{20} .

K T
Mar 13, 2020

We work in a 4-dimensional euclidian space with a standard orthonormal standard basis.

We could rephrase the problem as: Given a 5-cell with vertices v 1 , . . . , v 5 \vec{v_1}, ..., \vec{v_5} ,  could we find a vector p = ( p 1 , . . p 4 ) \vec{p}=(p_1,..p_4) such that the dot products p v 1 , p v 2 , p v 3 , p v 4 and p v 5 \vec{p} \cdot \vec{v_1}, \vec{p} \cdot \vec{v_2}, \vec{p} \cdot \vec{v_3}, \vec{p} \cdot \vec{v_4} \text { and }\vec{p} \cdot \vec{v_5} form an arithmetic series?

We are going to project the vertices onto p \vec{p} . Note that all points that lie on a hyperplane perpendicular to a p \vec{p} are projected onto the same point. We want these projection points to be at regular distances. Once we have found an example, we will scale the 5-cell so that the distance between adjacent projection points is 1.

Projections of the vertices v i \vec{v_i} are given by ( p v i ) p p 2 \frac{(\vec{p} \cdot\vec{v_i})\vec{p}}{|p|^2} . The distance between the two projections then is p ( v 1 v 2 ) p \frac{p \cdot (\vec{v_1}-\vec{v_2})}{|p|} .

We will use the simplest 5-cell vertex coordinates: ( 2 , 0 , 0 , 0 ) , ( 0 , 2 , 0 , 0 ) , ( 0 , 0 , 2 , 0 ) , ( 0 , 0 , 0 , 2 ) , ( φ , φ , φ , φ ) (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(φ,φ,φ,φ) , so that the respective dot products with p \vec{p} are 2 p 1 , 2 p 2 , 2 p 3 , 2 p 4 , φ ( p 1 + p 2 + p 3 + p 4 ) 2p_1, 2p_2, 2p_3, 2p_4, φ(p_1+p_2+p_3+p_4)

If these form an arithmetic series, we can set d = p 2 p 1 d=p_2-p_1 and the dot products become 2 p 1 , 2 p 1 + 2 d , 2 p 1 + 4 d , 2 p 1 + 6 d , φ ( 4 p 1 + 6 d ) 2p_1, 2p_1+2d, 2p_1+4d, 2p_1+6d, φ(4p_1+6d) .

The last dot product has to be equal to 2 p 1 + 8 d 2p_1+8d , so we infer

φ ( 4 p 1 + 6 d ) = 2 p 1 + 8 d φ(4p_1+6d)=2p_1+8d

Working this out via ( 4 φ 2 ) p 1 = d ( 8 6 φ ) (4φ-2) p_1 =d(8-6φ) and d p 1 = 2 φ 1 4 3 φ = 2 5 5 3 5 \frac{d}{p_1}=\frac{2φ-1}{ 4-3φ} = \frac{2\sqrt{5}}{ 5-3\sqrt{5}} we arrive at

d p 1 = s = 5 + 3 2 \frac{d}{p_1}= s=  -\frac{ \sqrt{5}+3}{ 2}

Now the vector p \vec{p} can be written as p = p 1 ( 1 , 1 + s , 1 + 2 s , 1 + 3 s ) \vec{p}=p_1(1, 1+s, 1+2s, 1+3s)

and hence we find that p 2 = p 1 2 ( 4 + 12 s + 14 s 2 ) = ( 35 + 15 5 ) p 1 2 |p|^2= p_1^2(4+ 12s + 14s^2)= (35+15\sqrt{5})p_1^2

The distance between the two projections of v 1 \vec{v_1} and v 2 \vec{v_2} then is the projection of the difference vector v 2 v 1 = ( 2 , 2 , 0 , 0 ) \vec{v_2}-\vec{v_1}=(-2,2,0,0) which has length 2 p 1 + 2 p 2 p \frac{-2p1+2p2}{|p|} = 2 d p = 2 s p 1 p = \frac{2d}{|p|} =  \frac{2sp_1}{|p|} = 5 + 3 35 + 15 5 = \frac{\sqrt{5}+3}{ \sqrt{35+15\sqrt{5} }} = 1 5 10 = \frac{1}{5}\sqrt{10} . This is also the distance between the five parallel planes containing the vertices.

In order to get the planes a unit distance apart we now have to scale up the 5-cell by the reciprocal of this, which is a factor 1 2 10 \frac{1}{2}\sqrt{10} . The squared side lengths were 2 2 + 2 2 = 8 2^2+2^2=8 before scaling. These now become 8 × 10 4 = 20 8×\frac{10}{4}=\boxed{20} .

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