Problems

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Solution:

1. If \(\vec{PQ} = \vec{SR}\) we have a parallelogram. \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\) then we have a rhombus.
2. If the four points lie in a plane then \((\vec{RS} \times \vec{RP}) \cdot \vec{RQ} =0\), so \begin{align*} && 0 &=\left ( \begin{pmatrix}-r\\ s-1 \\ 0 \end{pmatrix} \times \begin{pmatrix}p-r\\ -1 \\ -1 \end{pmatrix}\right) \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ && &= \begin{pmatrix}1-s \\ -r \\r+(p-r)(1-s) \end{pmatrix} \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ &&&= (1-s)(1-r)-r(q-1)-r-(p-r)(1-s) \\ &&&=(1-s)(1-r-p+r)-rq \\ \Rightarrow && rq &= (1-s)(1-p) \end{align*}
   1. \(\,\) \begin{align*} && \vec{PQ} &= \vec{SR} \\ \Leftrightarrow && \begin{pmatrix}1-p\\q \\ 0 \end{pmatrix} &= \begin{pmatrix}r\\1-s \\ 0 \end{pmatrix} \\ \Leftrightarrow && 1-p = r & \quad ; \quad q = 1-s\\ \Leftrightarrow && 1= r+p & \quad ; \quad 1 = q+s\\ \end{align*} The centroid is \(\frac14 (p+1+r, q+s+1, 2)\) which is clearly \(\frac12(1,1,1)\) iff those equations are true.
   2. \(\,\) \begin{align*} && |\vec{PQ}| &= |\vec{PS}| \\ \Leftrightarrow && (1-p)^2+q^2+ 0^2 &= p^2+s^2+1)\\ \Leftrightarrow && 1-2p+p^2+q^2 &= p^2 + s^2 + 1 \\ \Leftrightarrow && -2p+q^2 &= s^2 \end{align*} From the previous equations we have \(r = 1-p\), and \(-2p+(1-s)^2 = s^2 \Rightarrow -2p + 1 -2s = 0 \Rightarrow s = \frac12 - p\) and \(q = \frac12 + p\) \begin{align*} && \cos PQR &= \frac{\vec{QP}\cdot \vec{QR}}{|\vec{QP}||\vec{QR}|} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -q \\ 0 \end{pmatrix} \cdot \begin{pmatrix}r-1\\ 1-q \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+q^2}\sqrt{(r-1)^2+(1-q)^2+1^2}} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -\frac12-p \\ 0 \end{pmatrix} \cdot \begin{pmatrix}-p\\ \frac12-p \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+(-\frac12-p)^2}\sqrt{p^2+(\frac12-p)^2+1^2}} \\ &&&= \frac{ p-p^2-\frac14+p^2}{\sqrt{p^2-2p+1+\frac14+p+p^2}\sqrt{p^2+\frac14-p+p^2+1}} \\ &&&= \frac{4p-1}{\sqrt{8p^2-4p+5}\sqrt{8p^2-4p+5}}\\ &&&= \frac{4p-1}{8p^2-4p+5}\\ \end{align*} For \(PQRS\) to be a square \(\cos PQR = 0\), ie \(p = \frac14\) and so \((p,q,r,s) = (\frac14, \frac34, \frac34, \frac14)\) and \(|PQ| = \sqrt{(1-p)^2+q^2} = \sqrt{\left ( \frac34 \right)^2 + \left ( \frac34 \right)^2 } = \frac{3\sqrt{2}}4\), notice that \(\left ( \frac{21}{20} \right)^2 = \frac{441}{400} < \frac{9}{8}\) (\(441 < 450\)) therefore the sides are at least as long as \(\frac{21}{20}\)

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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:

1. Let \(A = (0,0,0)\) and then \(B = 22b \mathbf{i}, D = 2d\mathbf{j}, C = 2b\mathbf{i}+2d\mathbf{j}\) and \(V = b \mathbf{i} + d\mathbf{j} + h\mathbf{k}\) We also have \begin{align*} && \tan \alpha &= \frac{h}{d}\\ && \tan \beta &= \frac{d}{b} \\ && \vec{AV} \times \vec{VB} &= \begin{pmatrix} b \\ d \\ h \end{pmatrix} \times \begin{pmatrix} -b \\ d \\ h \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ -2bh \\ 2db \end{pmatrix} \\ &&&= 2b \begin{pmatrix} 0 \\ -d \tan \alpha \\ d \end{pmatrix} \\ &&&= k \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \end{align*} similarly for the vector perpendicular to the other face it must be \(\begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix}\) Looking at the angle between these perpendicular (to find the angles between the faces we see: \begin{align*} \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \cdot \begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix} &= \cos \alpha \cos \beta \end{align*} But this is also \(\pi -\) the angle between the planes, ie \(\cos \theta = \cos \alpha \cos \beta\)
2. \(\,\) \begin{align*} && \cot^2 \phi &= \frac{b^2+d^2}{h^2} \\ && \cot^2 \alpha &= \frac{d^2}{h^2} \\ && \cot^2 \beta &= \frac{b^2}{h^2} \\ \Rightarrow && cot^2 \phi &= \cot^2 \beta+\cot^2 \alpha \end{align*} \begin{align*} && \cos^2 \phi &= \frac{b^2+d^2}{b^2+d^2+h^2} \\ && \cos^2 \alpha &= \frac{d^2}{d^2+h^2} \\ && \cos^2 \beta &= \frac{b^2}{b^2+h^2} \\ && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &= \frac{\frac{d^2}{d^2+h^2}+\frac{b^2}{b^2+h^2}-2\cdot \frac{d^2}{d^2+h^2} \cdot \frac{b^2}{b^2+h^2}}{1 - \frac{d^2}{d^2+h^2} \cdot\frac{b^2}{b^2+h^2}} \\ &&&= \frac{d^2(b^2+h^2)+b^2(d^2+h^2)-2d^2b^2}{(d^2+h^2)(b^2+h^2)-d^2b^2} \\ &&&= \frac{h^2(b^2+d^2)}{h^2(b^2+d^2+h^2)} \\ &&&= \frac{b^2+d^2}{b^2+d^2+h^2} \\ &&&= \cos^2\phi \end{align*} Also notice that \begin{align*} && \cos^2 \alpha + \cos^2 \beta &\underbrace{\geq}_{AM-GM} 2 \cos \alpha \cos \beta \\ &&&= 2 \cos \theta \\ \Rightarrow && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &\geq \frac{2 \cos \theta - 2\cos^2 \theta}{1-\cos^2 \theta} \\ &&&= \frac{2\cos \theta}{1+\cos \theta} = \cos \theta \frac{2}{1+\cos \theta} \\ &&&> \cos^2 \theta \\ \Rightarrow && \phi &< \theta \end{align*}

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Solution: The line through \(A\) perpendicular to \(BC\) is \(\mathbf{a} + \lambda\mathbf{u}\). The line through \(B\) perpendicular to \(CA\) is \(\mathbf{b} + \mu \mathbf{v}\). They intersect when \(\mathbf{a} + \lambda\mathbf{u} = \mathbf{b} + \mu \mathbf{v}\). Since \(\mathbf{v}\) is perpendicular to \(CA\), we must have \begin{align*} && \mathbf{a} + \lambda\mathbf{u} &= \mathbf{b} + \mu \mathbf{v} \\ \Rightarrow && \mathbf{a}\cdot(\mathbf{c}-\mathbf{a}) + \lambda\mathbf{u}\cdot(\mathbf{c}-\mathbf{a}) &= \mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) + \mu \mathbf{v}\cdot(\mathbf{c}-\mathbf{a}) \\ \\ \Rightarrow && \lambda &= \frac{\mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) -\mathbf{a}\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u}\cdot(\mathbf{c}-\mathbf{a})} \\ &&&= \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \end{align*} Therefore the point is \(\mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u}\). The line \(CP\) is \(\mathbf{c} + \nu \left (\mathbf{p} - \mathbf{c} \right)\), to check this is perpendicular with \(AB\) we should dot \(\mathbf{p}-\mathbf{c}\) with \(\mathbf{a}-\mathbf{b}\), ie \begin{align*} && (\mathbf{p}-\mathbf{c}) \cdot (\mathbf{a}-\mathbf{b}) &= \left ( \mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} - \mathbf{c}\right) \cdot ( \mathbf{a}-\mathbf{b}) \\ &&&= \left ( \mathbf{a}- \mathbf{c} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} \right) \cdot ( \mathbf{a}-\mathbf{c}+(\mathbf{c}-\mathbf{b})) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})}\mathbf{u} \cdot (\mathbf{a}-\mathbf{c}) + \\ &&&\quad (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) + \lambda \underbrace{\mathbf{u} \cdot (\mathbf{c}-\mathbf{b})}_{=0} \\ &&&=(\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) -(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})+ (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}+\mathbf{b}-\mathbf{a}+\mathbf{c}-\mathbf{b}) \\ &&&= 0 \end{align*} as required.

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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: The tangent at \((at^2, 2at)\) can be found \begin{align*} && \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\ &&&= \frac{2a}{2at} = \frac1t \\ \Rightarrow && \frac{y-2at}{x-at^2} &= \frac1t \\ \Rightarrow && ty -x &= at^2 \\ \\ PT: && py - x &= ap^2 \\ QT: && qy - x &= aq^2 \\ \Rightarrow && (p-q)y &= a(p^2-q^2) \\ \Rightarrow && y &= a(p+q) \\ && x &= aq(p+q) - aq^2 \\ &&&= apq \end{align*} By the cosine rule: \begin{align*} && TP^2 &= FT^2 + FP^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ && (apq - ap^2)^2 + (a(p+q)-2ap)^2 &= (a-apq)^2+(a(p+q))^2 + \\ &&&\quad + (a-ap^2) + (2ap)^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ \Rightarrow && a^2p^2(q-p)^2 + a^2(q-p)^2 &= a^2(1-pq)^2+a^2(p+q)^2 + \\ &&&\quad + a^2(1-p^2)^2+4a^2p^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ && a^2(p^2+1)(q-p)^2 &= a^2(1+p^2)(1+q^2) + a^2(1+p^2)^2 + \\ &&&\quad - 2 \cdot a^2(1+p^2)\sqrt{(1+p^2)(1+q^2)} \cos \phi \\ \Rightarrow && \cos \phi &= \frac{a^2(1+p^2)(2+q^2+p^2-(q-p)^2)}{2 a^2 (1+p^2)\sqrt{(1+p^2)(1+q^2)}} \\ &&&= \frac{1+pq}{\sqrt{(1+p^2)(1+q^2)}} \end{align*} As required. Notice that by symmetry, \(\cos \angle TFQ = \frac{1+qp}{\sqrt{(1+q^2)(1+p^2)}} = \cos \phi\). Therefore they have the same angle and \(FT\) bisects \(PFQ\)

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Solution: \begin{align*} && {\bf r} &=\left( \e^{t}\cos t \right) {\bf i}+ \left(\e^t \sin t\right) {\bf j} \\ \Rightarrow && \dot{\bf r} &= \left( \e^{t}\cos t -\e^t \sin t\right) {\bf i}+ \left(\e^t \sin t+\e^t \cos t\right) {\bf j} \\ \Rightarrow && \mathbf{r}\cdot\dot{ \mathbf{r}} &= e^{2t}(\cos^2 t - \sin t \cos t) + e^{2t}(\sin^2 t+ \sin t \cos t) \\ &&&= e^{2t} (\cos^2 t + \sin ^2 t)\\ &&&= e^{2t} \\ \\ && | {\bf r}| &= e^{t} \\ && |{\bf \dot{r}}| &= e^t \sqrt{(\cos t - \sin t)^2 + (\sin t + \cos t)^2} \\ &&&= e^t \sqrt{2 \cos^2 t + 2 \sin^2 t} \\ &&&= \sqrt{2} e^t \\ \\ \Rightarrow && \frac{\mathbf{r}\cdot\dot{ \mathbf{r}}}{ |{\bf {r}}| |{\bf \dot{r}}|} &= \frac{e^{2t}}{\sqrt{2}e^te^t} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} Therefore the angle between the velocity and displacement is \(\frac{\pi}{4}\). \begin{align*} && \ddot{\bf{r}} &= \left( \e^{t}(\cos t - \sin t) - \e^t (\sin t + \cos t)\right) {\bf i}+ \left(\e^t (\sin t + \cos t) + \e^t(\cos t - \sin t)\right) {\bf j} \\ &&&= \left ( -2\e^{t} \sin t \right) {\bf i}+ \left ( 2\e^{t} \cos t \right) {\bf j} \\ \Rightarrow && {\bf r} \cdot \ddot{\bf{r}} &= 2e^{2t} \left ( -\sin t \cos t + \sin t \cos t \right) \\ &&&= 0 \end{align*} Therefore the acceleration is perpendicular.

\(Q\) has position $\mathbf{r}' = \left( \e^{t-T}\cos (t-T) \right) {\bf i}+ \left(\e^{t-T} \sin (t-T)\right) {\bf j}\( for \)t > T$. \begin{align*} && {\bf r' \cdot r} &= e^{2t-T} \left (\cos t \cos (t-T) + \sin t \sin(t - T) \right) \\ &&&= e^{2t-T} \cos (t - (t-T)) \\ &&&= e^{2t-T} \cos T \\ \\ && |{\bf r'}- {\bf r} |^2 &= |{\bf r}|^2 + |{\bf r}'|^2 - 2 {\bf r' \cdot r} \\ &&&= e^{2t} + e^{2(t-T)} - 2e^{2t-T} \cos T \\ &&&= e^{2t} \left (1 - 2e^{-T} \cos T + e^{-2T} \right) \\ \Rightarrow && |{\bf r'}- {\bf r} | &= e^{t} \sqrt{1 - 2e^{-T} \cos T + e^{-2T} } \end{align*} as required

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Solution:

\(\mathbf{p} = \lambda \mathbf{a} +(1- \lambda) \mathbf{b}\) Applying the Sine rule, we can see that: \begin{align*} && \frac{OA}{\sin \angle APO} = \frac{AP}{\sin \angle AOP} \\ && \frac{OB}{\sin \angle BPO} = \frac{BP}{\sin \angle BOP} \\ \end{align*} But \(\angle AOP = \angle BOP = \frac12 \angle AOP\) (since \(OP\) bisects the angle) and \(\angle APO = 180^{\circ} -\angle BPO\), so their sines are also equal. Therefore \begin{align*} && \frac{a}{AP} &= \frac{b}{BP} \\ \Rightarrow && \frac{a}{b} &= \frac{1-\lambda}{\lambda} \\ \Rightarrow && \lambda &= \frac{b}{a+b} \end{align*} Note that \(\mathbf{p} = \frac{b\mathbf{a} + a \mathbf{b}}{a+b} = \mathbf{a} + \frac{a}{a+b}(\mathbf{b}-\mathbf{a})\) and \(\mathbf{q} = \mathbf{b} +\frac{a}{a+b}(\mathbf{a}-\mathbf{b}) = \frac{a\mathbf{a} +b \mathbf{b}}{a+b}\) Therefore \begin{align*} && OQ^2 &= \frac{1}{(a+b)^2} \left (a\mathbf{a} + b \mathbf{b} \right)\cdot \left (a\mathbf{a} + b \mathbf{b} \right) \\ &&&= \frac{a^4+b^4+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ &&OP^2 &= \frac{1}{(a+b)^2} \left (b\mathbf{a} + a \mathbf{b} \right)\cdot \left (b\mathbf{a} + a \mathbf{b} \right) \\ &&&= \frac{2a^2b^2+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ \\ && OQ^2 - OP^2 &= \frac{a^4+b^4-2a^2b^2}{(a+b)^2} \\ &&&= \frac{(a^2-b^2)^2}{(a+b)^2} \\ &&&= (a-b)^2 \end{align*}

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Solution:

1. Suppose the vertices are \((\pm1, 0,0), (0,\pm1,0), (0,0,\pm1)\), then clearly this is an octahedron. We can measure the angle between the faces, by looking at vectors in the same plane and also in two of the faces: \(\langle \frac12, \frac12, - 1\rangle\) and \(\langle \frac12, \frac12, 1\rangle\), then by considering the dot product: \begin{align*} && \cos \theta &= \frac{\frac14+\frac14-1}{\frac14+\frac14+1} \\ &&&= \frac{-2}{6} = -\frac13 \end{align*}
2. The volume of our octahedron is \(2 \cdot \frac13 \cdot \underbrace{\sqrt{2}^2}_{\text{base}} \cdot \underbrace{1}_{\text{height}} = \frac43\). The centre of two touching faces are \(\langle \frac13, \frac13, \frac13 \rangle\) and \(\langle \frac13, \frac13, -\frac13 \rangle\) and so the length of the side of the cube is \(\frac23\) and so the volume of the cube is \(\frac8{27}\). Therefore the ratio is \(\frac{2}{9}\)

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Solution: Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

1. By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
2. Since \begin{align*} && (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\ \Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\ &&3^6 \cdot 3^4 &\geq 3^{10} \end{align*} Therefore we must be in the equality case, ie \(p = 8\lambda, 2q = 4\lambda, 3r = \lambda\) as well as \(64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3\) so we have \[ (p,q,r) = \left (24, 6, 1 \right) \]

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Solution: Suppose \({\bf p} = \lambda {\bf a}\) and \({\bf p} + {\bf q} = {\bf a} + {\bf b}\) then \begin{align*} {\bf a} \cdot : && {\bf a} \cdot {\bf p} + {\bf a} \cdot {\bf p} &= {\bf a} \cdot {\bf a} + {\bf a} \cdot {\bf b} \\ && \lambda + 0 &= 1 + 3 = 4 \\ \Rightarrow && \mathbf{p} &= 4 \mathbf{a} \\ && \mathbf{q} &= \mathbf{b} - 3\mathbf{a} \\ \\ && \mathbf{P} &= 4p\mathbf{a} \\ && \mathbf{Q} &= q\mathbf{b} - 3q\mathbf{a} \\ \\ \mathbf{a} \cdot : && \mathbf{a} \cdot \mathbf{P} + \mathbf{a} \cdot \mathbf{Q} + \mathbf{a} \cdot \mathbf{R} &= \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} \\ && 4p &= 1+3-2 \\ \Rightarrow && p &= \tfrac12 \\ \\ && {\bf P} + {\bf Q} + {\bf R} &= {\bf a}+{\bf b}+{\bf c} \\ \mathbf{b} \cdot : && \mathbf{b} \cdot \mathbf{P} + \mathbf{b} \cdot \mathbf{Q} + \mathbf{b} \cdot \mathbf{R} &= \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} \\ && 12p + 25q - 9q &= 3+25+2 \\ && 6+16q &= 30 \\ \Rightarrow && q &= \tfrac{3}{2}\\ && \\ && \mathbf{P} &= 2\mathbf{a} \\ && \mathbf{Q} &= \tfrac32 \mathbf{b} - \tfrac92 \mathbf{a} \\ && \mathbf{R} &= \tfrac72\mathbf{a} -\tfrac12 \mathbf{b} + \mathbf{c} \end{align*}

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Solution:

Set \(A\) to be the origin, then \(B = \langle 2, 0, 0 \rangle, G = \langle 2, 1, \lambda \rangle, H = \langle 0, 1, \lambda \rangle\), in particular \begin{align*} && AG&= \langle 2, 1, \lambda \rangle \\ && BH &= \langle -2, 1, \lambda \rangle \\ \Rightarrow && \cos \alpha &= |\frac{-4+1+\lambda^2}{\sqrt{2^2+1^2+\lambda^2}\sqrt{(-2)^2+1^2+\lambda^2}}| \\ &&&= |\frac{-3+\lambda^2}{5+\lambda^2}| \end{align*} \begin{align*} && \text{Volume} &= 2\lambda \\ && \text{Surface area} &= 2\cdot2\lambda + 2\cdot\lambda + 2\cdot2 \\ \Rightarrow && R&= \frac{\lambda}{3\lambda + 2} < \frac{1}{3} \\ && \frac14 &\leq R \\ \Rightarrow && 3\lambda +2 &\leq 4\lambda \\ \Rightarrow &&2 & \leq \lambda \end{align*} Then \(\frac{\lambda^2-3}{5+\lambda^2}\) is increasing as \(\lambda\) increases, in particularly the smallest value is \(\frac{1}{9}\).