Problems

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Solution:

1. Given \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\), we can form the following results: \begin{align*} && \begin{cases} \mathbf{a} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{b} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{c} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= 0 \\ \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= 0 \\ \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} +\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -\frac12 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{b} = -\frac12 \\ \mathbf{a} \cdot \mathbf{c} = -\frac12 \\ \mathbf{b} \cdot \mathbf{c} = -\frac12 \\ \end{cases} \end{align*} The triangle must be equilateral since the angles between each vertex are the same.
2. We have \(\displaystyle \sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}\) so \(\displaystyle \mathbf{a}_i \cdot \sum_{i=1}^{4} \mathbf{a}_i = 0\) or for each \(i\), \(\displaystyle \sum_{j \neq i} \mathbf{a}_i \cdot \mathbf{a}_j = -1\). \begin{align*} && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_1 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ && \text{adding the first two, subtracting the last two} \\ \Rightarrow && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 +\cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ \Rightarrow && 2 (\mathbf{a}_1 \cdot \mathbf{a}_2) - 2(\mathbf{a}_3 \cdot \mathbf{a}_4) = 0 \end{align*} Rather than adding the first two and last two, we could have done any pair, resulting in the relations: \begin{align*} \mathbf{a}_1 \cdot \mathbf{a}_2 &= \mathbf{a}_3 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 &= \mathbf{a}_2 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_4 &= \mathbf{a}_2 \cdot \mathbf{a}_3 \end{align*}
   1. The shape must be a parallelogram (from the angle requirement, but also cyclic quadrilateral (since all vectors are unit length), therefore it must be a rectangle
   2. Given it's a regular tetrahedron, \(\mathbf{a}_i \cdot \mathbf{a}_j\) must be the same for all \(i \neq j\), ie \(-\frac13\). We are interested in \(|\mathbf{a}_i - \mathbf{a}_j|\) so consider, \begin{align*} |\mathbf{a}_i - \mathbf{a}_j|^2 &= (\mathbf{a}_i - \mathbf{a}_j) \cdot (\mathbf{a}_i - \mathbf{a}_j) \\ &= \mathbf{a}_i \cdot \mathbf{a}_i - 2 \mathbf{a}_i \cdot \mathbf{a}_j + \mathbf{a}_j \cdot \mathbf{a}_j \\ &= 1 - \frac23 + 1 \\ &= \frac43 \end{align*} Therefore the unit side lengths are \(\frac{2}{\sqrt{3}}\)

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Solution: Note that \({\bf p}_i \cdot {\bf p}_i = 1\) and \({\bf p}_i \cdot {\bf p}_j\) are all equal when \(i \neq j\) by symmetry and commutativity. \begin{align*} && 0 &= {\bf p}_i \cdot \left ( {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \\ &&&= 1 + \sum_{j \neq i} {\bf p}_i \cdot {\bf p}_j \\ &&&= 1 + 3 {\bf p}_i \cdot {\bf p}_j \\ \Rightarrow && {\bf p}_i \cdot {\bf p}_j &= -\frac13 \end{align*}

1. \(\,\) \begin{align*} && (XP_i)^2 &= ({\bf p}_i- {\bf x)}\,.\,({\bf p}_i- {\bf x}) \\ &&&= {\bf p}_i \cdot {\bf p}_i - 2 {\bf p}_i \cdot {\bf x} + {\bf x} \cdot {\bf x} \\ &&&= 2 - 2 {\bf p}_i \cdot {\bf x} \\ \Rightarrow && \sum_i (XP_i)^2 &= \sum_i \left (2 - 2 {\bf p}_i \cdot {\bf x} \right) \\ &&&= 8 - 2 \sum_i {\bf p}_i \cdot {\bf x} \\ &&&= 8 - 2 \left ( {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \cdot {\bf x} \\ &&&= 8 \end{align*}
2. Notice we have \(1 = \left \|\begin{pmatrix} a \\0 \\b \end{pmatrix} \right \|= a^2 + b^2\) and \(-\frac13 = \begin{pmatrix} a \\0 \\b \end{pmatrix} \cdot \begin{pmatrix} 0 \\0 \\ 1 \end{pmatrix} = b\). So \(b = -1/3\) and \(a = \sqrt{1-b^2} = 2\sqrt{2}/3\). Suppose another of the vertices has coordinates \((u,v,w)\) we must have \begin{align*} && 1 &= u^2+v^2+w^2 \\ && -\frac13&=w \\ && -\frac13 &= \frac{2\sqrt{2}}3 u +\frac19 \\ \Rightarrow && u &= -\frac{\sqrt2}3 \\ \Rightarrow && 1 &= \frac19 + \frac29 + v^2 \\ \Rightarrow && v &= \pm \sqrt{\frac{2}{3}} \end{align*} So \(P_3, P_4 = (-\frac{\sqrt2}3, \pm \frac{\sqrt{6}}3, -\frac13)\)
3. \(\,\) \begin{align*} && \sum_{i=1}^4 (XP_i)^4 &= \sum_i \left (2 - 2 {\bf p}_i \cdot {\bf x} \right)^2 \\ &&&= 4 \sum_i \left (1 - {\bf p}_i \cdot {\bf x} \right)^2 \\ &&&= 4 \sum_i (1 - 2{\bf p}_i \cdot {\bf x} + ({\bf p}_i \cdot {\bf x})^2) \\ &&&= 16 + 4\sum_i ({\bf p}_i \cdot {\bf x})^2 \\ &&&=16+ 4\left ( z^2+\left (\frac{2\sqrt{2}}3x-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x-\frac{\sqrt{6}}3y-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x+\frac{\sqrt{6}}3y-\frac13z \right)^2 \right) \\ &&&= 16+4 \left ( \frac43z^2 + \left (\frac89 + \frac29+\frac29 \right)x^2+\left (\frac69 + \frac69 \right)y^2 + 0xz + 0yz + 0zx \right) \\ &&&= 16+ 4\cdot\frac43(x^2+y^2+z^2) \\ &&&=16+\frac{16}{3}=\frac{64}{3} \end{align*}

Note: It may be better to view the last part of this question in terms of linear transformations. There are two possible approaches. One is to show \(T:{\bf x} \mapsto \sum_i ({\bf p}_i \cdot x) {\bf p}_i\) is \(\frac43I\) (easy since it has three eigenvectors with the same eigenvalue which span \(\mathbb{R}^3\) and we are interested in the value \({\bf x} \cdot T\mathbf{x} = \frac43 \lVert {\bf x} \rVert^2\). The second is to consider \(\sum_I ({\bf p}_i \cdot {\bf x})^2 = {\bf x}^TM{\bf x}\) where \(M = \sum_i {\bf p}_i{\bf p}_i^T\) and note that this matrix is invariant under rotations.

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Solution:

1. \(D\) must be above the centre (of any kind) of the equilateral triangle \(ABC\). Therefore it is a distance \(\frac23 \frac{\sqrt{3}}2 = \frac{\sqrt{3}}3\) from \(A\). \(D\) is \(1\) from \(A\), therefore by Pythagoras \(PD = \sqrt{1-\frac13} = \sqrt{\frac23}\)
2. We can place \(D\) at \(\langle 0,0,\sqrt{\frac23}\rangle\) and \(A'\) (the midpoint of \(BC\)) at \(\langle-\frac{\sqrt{3}}{6},0,0 \rangle\) and we find: \begin{align*} && \cos \theta &= \frac{(\mathbf{a}'-\mathbf{d})\cdot \mathbf{a}'}{|\mathbf{a}'-\mathbf{d}|| \mathbf{a}'|} \\ &&&= \frac{|\mathbf{a}'|}{|\mathbf{a}'-\mathbf{d}|} \\ &&&= \frac{\frac{\sqrt{3}}{6}}{\sqrt{\frac23+\frac{3}{36}}} = \frac13 \end{align*}
3. We have
   

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   And therefore we must have \(\tan \frac{\cos^{-1} \frac13}{2} = \frac{r}{\frac{\sqrt{3}}{6}}\) therefore \begin{align*} && r &= \frac{\sqrt{3}}{6} \tan \left (\frac{\cos^{-1} \frac13}{2} \right) \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{1-\cos(\cos^{-1}\frac13)}{1+\cos(\cos^{-1}\frac13)}} \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{\frac23}{\frac43}} \\ &&&= \frac{\sqrt{6}}{12} \end{align*}