$\Bbb R^n$中的$n+1$条射线的夹角全都相等

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(Spectrum series) Five hundred mathematical challenges - Mathematical Association of America (1995)

Problem 498.

If the sum of the face angles at each vertex of a tetrahedron is 180°, prove that the tetrahedron is isosceles, i.e., the opposite edges are equal in pairs.

Problem 499.

Prove that every positive integer which is not a member of the infinite set below is equal to the sum of two or more distinct members of the set:$$\begin{array}{c}\left\{3,-2,2^{2} 3,-2^{3}, \ldots, 2^{2 k} 3,-2^{2 k+1}, \ldots\right\} \\ \quad=\{3,-2,12,-8,48,-32,192, \ldots\}\end{array}$$

Problem 500.

One is given $n + 1$ rays emanating from a common vertex in $n$-dimensional Euclidean space $\Bbb R^n$. If all the angles between pairs of the rays are congruent, determine the common angle.

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Problem 500.

For $E^{3}$, consider the 4 rays from the center of a regular tetrahedron to the vertices. By symmetry all the angles between the pairs of rays are equal. For $E^{n}$, we consider the $n+1$ rays from the center of a regular simplex ($n$-dimensional tetrahedron). Let $\overrightarrow{V}_{1}, \overrightarrow{V}_{2}, \ldots, \overrightarrow{V}_{n+1}$, denote vectors from the center to the vertices. By symmetry (show!) the circumcenter and centroid of the simplex are the same point. Thus $\raise.5ex{\large|}\overrightarrow{V_i}\raise.5ex{\large|}=R, i=1,2, \ldots, n+1$ and $$ \overrightarrow{V_{1}}+\overrightarrow{V_{2}}+\cdots+\overrightarrow{V}_{n+1}=\overrightarrow{0} $$ Then $$ \left(\overrightarrow{V}_{1}+\overrightarrow{V}_{2}+\cdots+\overrightarrow{V}_{n+1}\right) \cdot\left(\overrightarrow{V}_{1}+\overrightarrow{V}_{2}+\cdots+\overrightarrow{V}_{n+1}\right)=0 $$ or $$ (n+1) R^{2}+2\left(\begin{array}{c} n+1 \\ 2 \end{array}\right) \overrightarrow{V}_{i} \cdot \overrightarrow{V}_{j}=0 \text {. } $$ Finally, if $\theta$ is the common angle between any vector pair $\raise.5ex\Bigl(\overrightarrow{V_i}, \overrightarrow{V_j}\raise.5ex\Bigr)$, then $\cos \theta=-\frac{1}{n}$ and $\theta=\arccos \left(-\frac{1}{n}\right)$.

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Problem 498.

Cut the tetrahedron $A B C D$ along $A B, A C,$ and $A D$ and lay out flat as in Figure 245. Since $A B A, A C A$ and $A D A$ are all straight, $B, D$ and $C$ are the midpoints of the larger outer triangle. Then $\overline{A B}=\overline{C D}, \overline{B C}=\overline{D A}$ and $\overline{C A}=\overline{B D}$.

Problem 499.

Let the $i$th term of the given sequence $$ \{3,-2,12,-8,48,-32,192, \ldots\} $$ be $a_{i}$. We have $$ 1=a_{1}+a_{2}, \quad 2=a_{2}+a_{3}+a_{4}, \quad 3=a_{1} . $$ Suppose $n \geq 4$ and all integers $1,2, \ldots, n-1$ can be written as the sum of distinct $a_i$'s. There are four cases:

1. $n=4 k$ where $1 \leq k< n$. If $k=\sum_{i \in S} a_{i}$, where $S \subset\{1,2,3, \ldots\}$, then $n=\sum_{i\in S} a_{i+2}$ is the required representation;
2. $n=4 k+1$ where $1 \leq k< n$. If $k=\sum_{i \in S} a_{i}$, then $n=a_{1}+a_{2}+\sum_{i \in S} a_{i+2}$;
3. $n=4 k-2$ where $2 \leq k< n$. If $k=\sum_{i \in S} a_{i}$, then $n=a_{2}+\sum_{i \in S} a_{i+2}$;
4. $n=4 k+3$ where $1 \leq k< n$. If $k=\sum_{i \in S} a_{i}$, then $n=a_{1}+\sum_{i \in S} a_{i+2}$.

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Cartesian coordinates for a regular n-dimensional simplex in Rⁿ

One way to write down a regular n-simplex in Rⁿ is to choose two points to be the first two vertices, choose a third point to make an equilateral triangle, choose a fourth point to make a regular tetrahedron, and so on. Each step requires satisfying equations that ensure that each newly chosen vertex, together with the previously chosen vertices, forms a regular simplex. There are several sets of equations that can be written down and used for this purpose. These include the equality of all the distances between vertices; the equality of all the distances from vertices to the center of the simplex; the fact that the angle subtended through the new vertex by any two previously chosen vertices is $\pi /3$; and the fact that the angle subtended through the center of the simplex by any two vertices is ${\displaystyle \arccos(-1/n)}$.

It is also possible to directly write down a particular regular n-simplex in Rⁿ which can then be translated, rotated, and scaled as desired. One way to do this is as follows. Denote the basis vectors of Rⁿ by e₁ through e_n. Begin with the standard (n − 1)-simplex which is the convex hull of the basis vectors. By adding an additional vertex, these become a face of a regular n-simplex. The additional vertex must lie on the line perpendicular to the barycenter of the standard simplex, so it has the form (α/n, ..., α/n) for some real number α. Since the squared distance between two basis vectors is 2, in order for the additional vertex to form a regular n-simplex, the squared distance between it and any of the basis vectors must also be 2. This yields a quadratic equation for α. Solving this equation shows that there are two choices for the additional vertex:

${\displaystyle {\frac {1}{n}}\left(1\pm {\sqrt {n+1}}\right)\cdot (1,\dots ,1).}$

Either of these, together with the standard basis vectors, yields a regular n-simplex.

The above regular n-simplex is not centered on the origin. It can be translated to the origin by subtracting the mean of its vertices. By rescaling, it can be given unit side length. This results in the simplex whose vertices are:

${\displaystyle {\frac {1}{\sqrt {2}}}\mathbf {e} _{i}-{\frac {1}{n{\sqrt {2}}}}{\bigg (}1\pm {\frac {1}{\sqrt {n+1}}}{\bigg )}\cdot (1,\dots ,1),}$

for $1\leq i\leq n$, and

${\displaystyle \pm {\frac {1}{\sqrt {2(n+1)}}}\cdot (1,\dots ,1).}$

Note that there are two sets of vertices described here. One set uses $+$ in each calculation. The other set uses $-$ in each calculation.

This simplex is inscribed in a hypersphere of radius ${\displaystyle {\sqrt {n/(2(n+1))}}}$.

A different rescaling produces a simplex that is inscribed in a unit hypersphere. When this is done, its vertices are

${\displaystyle {\sqrt {1+n^{-1}}}\cdot \mathbf {e} _{i}-n^{-3/2}({\sqrt {n+1}}\pm 1)\cdot (1,\dots ,1),}$

where $1\leq i\leq n$, and

${\displaystyle \pm n^{-1/2}\cdot (1,\dots ,1).}$

The side length of this simplex is ${\textstyle {\sqrt {2(n+1)/n}}}$.

A highly symmetric way to construct a regular n-simplex is to use a representation of the cyclic group Z_n+1 by orthogonal matrices. This is an n × n orthogonal matrix Q such that Qⁿ⁺¹ = I is the identity matrix, but no lower power of Q is. Applying powers of this matrix to an appropriate vector v will produce the vertices of a regular n-simplex. To carry this out, first observe that for any orthogonal matrix Q, there is a choice of basis in which Q is a block diagonal matrix

${\displaystyle Q=\operatorname {diag} (Q_{1},Q_{2},\dots ,Q_{k}),}$

where each Q_i is orthogonal and either 2 × 2 or 1 × 1. In order for Q to have order n + 1, all of these matrices must have order dividing n + 1. Therefore each Q_i is either a 1 × 1 matrix whose only entry is 1 or, if n is odd, −1; or it is a 2 × 2 matrix of the form

${\displaystyle {\begin{pmatrix}\cos {\frac {2\pi \omega _{i}}{n+1}}&-\sin {\frac {2\pi \omega _{i}}{n+1}}\\\sin {\frac {2\pi \omega _{i}}{n+1}}&\cos {\frac {2\pi \omega _{i}}{n+1}}\end{pmatrix}},}$

where each ω_i is an integer between zero and n inclusive. A sufficient condition for the orbit of a point to be a regular simplex is that the matrices Q_i form a basis for the non-trivial irreducible real representations of Z_n+1, and the vector being rotated is not stabilized by any of them.

In practical terms, for n even this means that every matrix Q_i is 2 × 2, there is an equality of sets

${\displaystyle \{\omega _{1},n+1-\omega _{1},\dots ,\omega _{n/2},n+1-\omega _{n/2}\}=\{1,\dots ,n\},}$

and, for every Q_i, the entries of v upon which Q_i acts are not both zero. For example, when n = 4, one possible matrix is

${\displaystyle {\begin{pmatrix}\cos(2\pi /5)&-\sin(2\pi /5)&0&0\\\sin(2\pi /5)&\cos(2\pi /5)&0&0\\0&0&\cos(4\pi /5)&-\sin(4\pi /5)\\0&0&\sin(4\pi /5)&\cos(4\pi /5)\end{pmatrix}}.}$

Applying this to the vector (1, 0, 1, 0) results in the simplex whose vertices are

${\displaystyle {\begin{pmatrix}1\\0\\1\\0\end{pmatrix}},{\begin{pmatrix}\cos(2\pi /5)\\\sin(2\pi /5)\\\cos(4\pi /5)\\\sin(4\pi /5)\end{pmatrix}},{\begin{pmatrix}\cos(4\pi /5)\\\sin(4\pi /5)\\\cos(8\pi /5)\\\sin(8\pi /5)\end{pmatrix}},{\begin{pmatrix}\cos(6\pi /5)\\\sin(6\pi /5)\\\cos(2\pi /5)\\\sin(2\pi /5)\end{pmatrix}},{\begin{pmatrix}\cos(8\pi /5)\\\sin(8\pi /5)\\\cos(6\pi /5)\\\sin(6\pi /5)\end{pmatrix}},}$

each of which has distance √5 from the others. When n is odd, the condition means that exactly one of the diagonal blocks is 1 × 1, equal to −1, and acts upon a non-zero entry of v; while the remaining diagonal blocks, say Q₁, ..., Q_{(n − 1) / 2}, are 2 × 2, there is an equality of sets

${\displaystyle \left\{\omega _{1},-\omega _{1},\dots ,\omega _{(n-1)/2},-\omega _{n-1)/2}\right\}=\left\{1,\dots ,(n-1)/2,(n+3)/2,\dots ,n\right\},}$

and each diagonal block acts upon a pair of entries of v which are not both zero. So, for example, when n = 3, the matrix can be

${\displaystyle {\begin{pmatrix}0&-1&0\\1&0&0\\0&0&-1\\\end{pmatrix}}.}$

For the vector (1, 0, 1/√2), the resulting simplex has vertices

${\displaystyle {\begin{pmatrix}1\\0\\1/\surd 2\end{pmatrix}},{\begin{pmatrix}0\\1\\-1/\surd 2\end{pmatrix}},{\begin{pmatrix}-1\\0\\1/\surd 2\end{pmatrix}},{\begin{pmatrix}0\\-1\\-1/\surd 2\end{pmatrix}},}$

each of which has distance 2 from the others.