In the game of ``Colonel Blotto'' there are two players, Adam and Betty. First Adam chooses three non-negative integers \(a_{1},a_{2}\) and \(a_{3},\) such that \(a_{1}+a_{2}+a_{3}=9,\) and then Betty chooses non-negative integers \(b_{1},b_{2}\) and \(b_{3}\), such that \(b_{1}+b_{2}+b_{3}=9.\) If \(a_{1} > b_{1}\) then Adam scores one point; if \(a_{1} < b_{1}\) then Betty scores one point; and if \(a_{1}=b_{1}\) no points are scored. Similarly for \(a_{2},b_{2}\) and \(a_{3},b_{3}.\) The winner is the player who scores the greater number of points: if the socres are equal then the game is drawn. Show that, if Betty knows the numbers \(a_{1},a_{2}\) and \(a_{3},\) she can always choose her numbers so that she wins. Show that Adam can choose \(a_{1},a_{2}\) and \(a_{3}\) in such a way that he will never win no matter what Betty does. Now suppose that Adam is allowed to write down two triples of numbers and that Adam wins unless Betty can find one triple that beats both of Adam's choices (knowing what they are). Confirm that Adam wins by writing down \((5,3,1)\) and \((3,1,5).\)
Two non-parallel lines in 3-dimensional space are given by \(\mathbf{r}=\mathbf{p}_{1}+t_{1}\mathbf{m}_{1}\) and \(\mathbf{r}=\mathbf{p}_{2}+t_{2}\mathbf{m}_{2}\) respectively, where \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are unit vectors. Explain by means of a sketch why the shortest distance between the two lines is \[ \frac{\left|(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}{\left|(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}. \]
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In this question, \(\mathbf{A,\mathbf{B\) }}and \(\mathbf{X\) are non-zero \(2\times2\) real matrices.} Are the following assertions true or false? You must provide a proof or a counterexample in each case.
Solution:
The integers \(a,b\) and \(c\) satisfy \[ 2a^{2}+b^{2}=5c^{2}. \] By considering the possible values of \(a\pmod5\) and \(b\pmod5\), show that \(a\) and \(b\) must both be divisible by \(5\). By considering how many times \(a,b\) and \(c\) can be divided by \(5\), show that the only solution is \(a=b=c=0.\)
Solution: \begin{array}{c|ccccc} a & 0 & 1 & 2 & 3 & 4 \\ a^2 & 0 & 1 & 4 & 4 & 1 \end{array} Therefore \(a^2 \in \{0,1,4\}\) and so we can have \begin{array} $2a^2+b^2 & 0 & 1 & 4 \\ \hline 0 & 0 & 1 & 4 \\ 1 & 2 & 3 & 1 \\ 4 & 3 & 4 & 2 \end{array} Therefore the only solution must have \(5 \mid a,b\), but then we can write them has \(5a'\) and \(5b'\) so the equation becomes \(2\cdot25 a'^2 + 25b'^2 = 5c^2\) ie \(5 \mid c^2 \Rightarrow 5 \mid c\). But that means we can always divide \((a,b,c)\) by \(5\), which is clearly a contradiction if we consider the lowest power of \(5\) dividing \(a,b,c\) for any solution.
Suppose that \(a_{i}>0\) for all \(i>0\). Show that \[ a_{1}a_{2}\leqslant\left(\frac{a_{1}+a_{2}}{2}\right)^{2}. \] Prove by induction that for all positive integers \(m\) \[ a_{1}\cdots a_{2^{m}}\leqslant\left(\frac{a_{1}+\cdots+a_{2^{m}}}{2^{m}}\right)^{2^{m}}.\tag{\ensuremath{*}} \] If \(n<2^{m}\), put \(b_{1}=a_{2},\) \(b_{2}=a_{2},\cdots,b_{n}=a_{n}\) and \(b_{n+1}=\cdots=b_{2^{m}}=A\), where \[ A=\frac{a_{1}+\cdots+a_{n}}{n}. \] By applying \((*)\) to the \(b_{i},\) show that \[ a_{1}\cdots a_{n}A^{(2^{m}-n)}\leqslant A^{2^{m}} \] (notice that \(b_{1}+\cdots+b_{n}=nA).\) Deduce the (arithmetic mean)/(geometric mean) inequality \[ \left(a_{1}\cdots a_{n}\right)^{1/n}\leqslant\frac{a_{1}+\cdots+a_{n}}{n}. \]
\textit{In this question, the argument of a complex number is chosen to satisfy \(0\leqslant\arg z<2\pi.\)} Let \(z\) be a complex number whose imaginary part is positive. What can you say about \(\arg z\)? The complex numbers \(z_{1},z_{2}\) and \(z_{3}\) all have positive imaginary part and \(\arg z_{1}<\arg z_{2}<\arg z_{3}.\) Draw a diagram that shows why \[ \arg z_{1}<\arg(z_{1}+z_{2}+z_{3})<\arg z_{3}. \] Prove that \(\arg(z_{1}z_{2}z_{3})\) is never equal to \(\arg(z_{1}+z_{2}+z_{3}).\)
Verify that if \[ \mathbf{P}=\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\qquad\mbox{ and }\qquad\mathbf{A}=\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix} \] then \(\mathbf{PAP}\) is a diagonal matrix. Put $\mathbf{x}=\begin{pmatrix}x\\ y \end{pmatrix}\( and \)\mathbf{x}_{1}=\begin{pmatrix}x_{1}\\ y_{1} \end{pmatrix}.$ By writing \[ \mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a} \] for a suitable vector \(\mathbf{a},\) show that the equation \[ \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11=0, \] where $\mathbf{b}=\begin{pmatrix}18\\ 6 \end{pmatrix}\( and \) \mathbf{x}^{\mathrm{T}} \( is the transpose of \)\mathbf{x},$ becomes \[ 3x_{1}^{2}-y_{1}^{2}=c \] for some constant \(c\) (which you should find).
Solution: \begin{align*} \mathbf{PAP} &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix}\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix} \\ &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}15 & -10\\ 30 & 5 \end{pmatrix} \\ &= \begin{pmatrix}75 & 0\\ 0 & -25 \end{pmatrix} \end{align*} Which is diagonal as required. Letting \(\mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a}\) \begin{align*} && \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11&=0 \\ \Leftrightarrow && (\mathbf{P}\mathbf{x}_{1}+\mathbf{a})^T\mathbf{A}(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 + \mathbf{x}_{1}^T\mathbf{PAa} + \mathbf{a}^T\mathbf{AP}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \end{align*} It would be nice if we picked \(\mathbf{a}\) such that \(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T = 0\), if \(\mathbf{a} = \begin{pmatrix} a_1 \\a_2 \end{pmatrix}\) then this equation becomes: \begin{align*} && 2\begin{pmatrix}-a_1 + 8a_2 & 8a_1+11a_2 \end{pmatrix} + \begin{pmatrix}18 & 6 \end{pmatrix} &= 0 \\ \Rightarrow && a_1 = 1, a_2 = -1 \end{align*} So our equation is now \begin{align*} && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1-6 +12 - 11 &= 0 \\ \Leftrightarrow && 25(3x_1^2 - y_1^2) &= 5 \\ \Leftrightarrow && 3x_1^2 - y_1^2 &= \frac{1}{5} \end{align*}