Problems

Consider a fixed square \(ABCD\) and a variable point \(P\) in the plane of the square. We write the perpendicular distance from \(P\) to \(AB\) as \(p\), from \(P\) to \(BC\) as \(q\), from \(P\) to \(CD\) as \(r\) and from \(P\) to \(DA\) as \(s\). (Remember that distance is never negative, so \(p,q,r,s\geqslant 0\).) If \(pr=qs\), show that the only possible positions of \(P\) lie on two straight lines and a circle and that every point on these two lines and a circle is indeed a possible position of \(P\).

Suppose that \[{\rm f}''(x)+{\rm f}(-x)=x+3\cos 2x\] and \({\rm f}(0)=1\), \({\rm f}'(0)=-1\). If \({\rm g}(x)={\rm f}(x)+{\rm f}(-x)\), find \({\rm g}(0)\) and show that \({\rm g}'(0)=0\). Show that \[{\rm g}''(x)+{\rm g}(x)=6\cos 2x,\] and hence find \({\rm g}(x)\). Similarly, if \({\rm h}(x)={\rm f}(x)-{\rm f}(-x)\), find \({\rm h}(x)\) and show that \[{\rm f}(x)=2\cos x -\cos2x-x.\]

A child's toy consists of a solid cone of height \(\lambda a\) and a solid hemisphere of radius \(a\), made out of the same uniform material and fastened together so that their plane faces coincide. (Thus the diameter of the hemisphere is equal to that of the base of the cone.) Show that if \(\lambda < \sqrt{3}\) the toy will always move to an upright position if placed with the surface of the hemisphere on a horizontal table, but that if \(\lambda > \sqrt{3}\) the toy may overbalance. Show, however, that if the toy is placed with the surface of the cone touching the table it will remain there whatever the value of \(\lambda\). [The centre of gravity of a uniform solid cone of height \(h\) is a height \(h/4\) above its base. The centre of gravity of a uniform solid hemisphere of radius \(a\) is at distance \(3a/8\) from the centre of its base.]

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Solution:

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By symmetry the centre of mass will lie on the main axis. Taking the plane faces as \(x = 0\) we have the following centers of mass: \begin{align*} && \text{COM} && \text{Mass} \\ \text{Hemisphere} && -\frac{3a}{8} && \frac{2\pi a^3}{3} \\ \text{Cone} && \frac{\lambda a}{4} && \frac{\lambda \pi a^3}{3} \\ \text{Toy} && \bar{x} && \frac{(\lambda + 2)\pi a^3}{3} \\ \end{align*} Therefore, \begin{align*} && \frac{(\lambda + 2)\pi a^3}{3} \cdot \bar{x} &= -\frac{3a}{8} \cdot \frac{2\pi a^3}{3} + \frac{\lambda a}{4} \cdot \frac{\lambda \pi a^3}{3} \\ \Rightarrow && (\lambda + 2) \bar{x} &= \frac{(\lambda^2 -3)a}{4} \end{align*} Therefore the centre of mass will be inside the hemisphere (and it will always move to an upright position) iff \(\bar{x} < 0 \Leftrightarrow \lambda < \sqrt{3}\).

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For the toy to topple from this position, \(\bar{x}\) must be longer than it would need to be to form a right-angled triangle with the vertical at the plane face. The angle at this point will be \(\theta\), so we need: \(\bar{x} > a\tan \theta = a \frac{a}{\lambda a} = \frac{a}{\lambda}\) \begin{align*} && \bar{x} &> \frac{a}{\lambda} \\ \Leftrightarrow && \frac{(3-\lambda^2)a}{4(\lambda + 2)} &> \frac{a}{\lambda} \\ \Leftrightarrow && {(3-\lambda^2)\lambda} &> {4(\lambda + 2)} \\ \Leftrightarrow && -\lambda^3-\lambda -8 &> 0 \\ \end{align*} Contradiction! Therefore it can never topple when laid on its side.

The plot of `Rhode Island Red and the Henhouse of Doom' calls for the heroine to cling on to the circumference of a fairground wheel of radius \(a\) rotating with constant angular velocity \(\omega\) about its horizontal axis and then let go. Let \(\omega_{0}\) be the largest value of \(\omega\) for which it is not possible for her subsequent path to carry her higher than the top of the wheel. Find \(\omega_{0}\) in terms of \(a\) and \(g\). If \(\omega>\omega_{0}\) show that the greatest height above the top of the wheel to which she can rise is \[\frac{a}{2}\left(\frac{\omega}{\omega_{0}} -\frac{\omega_{0}}{\omega}\right)^{\!\!2}.\]

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Solution:

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\begin{align*} \uparrow: && v &= u + at \\ \Rightarrow && T &= \frac{a \omega \sin \theta}{g} \\ && s &= ut + \frac12 gt^2 \\ \Rightarrow && s &= a\omega \sin \theta \cdot \frac{a \omega \sin \theta}{g} - \frac12 g \left ( \frac{a \omega \sin \theta}{g} \right) ^2 \\ &&&= \frac1{2g} a^2 \omega^2 \sin^2 \theta \\ s < \text{distance to top}: && \frac1{2g} a^2 \omega^2 \sin^2 \theta &< a(1- \cos \theta) \\ \Rightarrow && \omega^2 &< \frac{2g}{a} \frac{1-\cos \theta}{\sin^2 \theta} \\ &&&= \frac{2g}{a} \frac{2 \sin^2 \tfrac12 \theta}{4 \sin^2 \tfrac12 \theta \cos^2 \tfrac12 \theta} \\ &&&= \frac{g}{a} \sec^2 \tfrac12 \theta \\ &&&\leq \frac{g}{a} \tag{since it holds for all \(\theta\) it holds for min \(\theta\)}\\ \Rightarrow && \omega_0 &= \sqrt{\frac{g}{a}} \\ \\ && \text{max height} &= \frac1{2g} a^2 \omega^2 \sin^2 \theta - a(1-\cos \theta) \\ &&&= \frac1{2g} a^2 \omega^2 (1-\cos^2 \theta) - a(1-\cos \theta) \\ &&&= \frac{a}{2} \left (- \frac{\omega^2}{\omega_0^2} \cos^2 \theta + 2 \cos \theta + \frac{\omega^2}{\omega_0^2}-2 \right) \\ &&&= \frac{a}{2} \left (-\left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta \right)^2 + \frac{\omega^2}{\omega_0^2}-2+\frac{\omega_0^2}{\omega^2} \right) \\ &&&= \frac{a}{2} \left ( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right)^2 - \frac{a}{2} \left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta \right)^2 \end{align*} If \(\omega > \omega_0\) we can find a \(\theta\) such that the second bracket is \(0\), hence the maximium height is as desired.

A particle hangs in equilibrium from the ceiling of a stationary lift, to which it is attached by an elastic string of natural length \(l\) extended to a length \(l+a\). The lift now descends with constant acceleration \(f\) such that \(0 < f < g/2\). Show that the extension \(y\) of the string from its equilibrium length satisfies the differential equation $$ {{\rm d}^2 y \over {\rm d} t^2} +{g \over a}\, y = g-f. $$ Hence show that the string never becomes slack and the amplitude of the oscillation of the particle is \(af/g\). After a time \(T\) the lift stops accelerating and moves with constant velocity. Show that the string never becomes slack and the amplitude of the oscillation is now \[\frac{2af}{g}|\sin {\textstyle \frac{1}{2}}\omega T|,\] where \(\omega^{2}=g/a\).

1. Let \(X_{1}\), \(X_{2}\), \dots, \(X_{n}\) be independent random variables each of which is uniformly distributed on \([0,1]\). Let \(Y\) be the largest of \(X_{1}\), \(X_{2}\), \dots, \(X_{n}\). By using the fact that \(Y<\lambda\) if and only if \(X_{j}<\lambda\) for \(1\leqslant j\leqslant n\), find the probability density function of \(Y\). Show that the variance of \(Y\) is \[\frac{n}{(n+2)(n+1)^{2}}.\]
2. The probability that a neon light switched on at time \(0\) will have failed by a time \(t>0\) is \(1-\mathrm{e}^{-t/\lambda}\) where \(\lambda>0\). I switch on \(n\) independent neon lights at time zero. Show that the expected time until the first failure is \(\lambda/n\).

By considering the coefficients of \(t^{n}\) in the equation \[(1+t)^{n}(1+t)^{n}=(1+t)^{2n},\] or otherwise, show that \[\binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\cdots +\binom{n}{r}\binom{n}{n-r}+\cdots+\binom{n}{n}\binom{n}{0} =\binom{2n}{n}.\] The large American city of Triposville is laid out in a square grid with equally spaced streets running east-west and avenues running north-south. My friend is staying at a hotel \(n\) avenues west and \(n\) streets north of my hotel. Both hotels are at intersections. We set out from our own hotels at the same time. We walk at the same speed, taking 1 minute to go from one intersection to the next. Every time I reach an intersection I go north with probability \(1/2\) or west with probability \(1/2\). Every time my friend reaches an intersection she goes south with probability \(1/2\) or east with probability \(1/2\). Our choices are independent of each other and of our previous decisions. Indicate by a sketch or by a brief description the set of points where we could meet. Find the probability that we meet. Suppose that I oversleep and leave my hotel \(2k\) minutes later than my friend leaves hers, where \(k\) is an integer and \(0\leqslant 2k\leqslant n\). Find the probability that we meet. Have you any comment? If \(n=1\) and I leave my hotel \(1\) minute later than my friend leaves hers, what is the probability that we meet and why?

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Solution: \begin{align*} && (1+t)^{n}(1+t)^{n}&=(1+t)^{2n} \\ [t^n]: &&\sum_{k=0}^n \underbrace{\binom{n}{k}}_{t^k\text{ from left bracket}} \underbrace{\binom{n}{n-k}}_{t^{n-k}\text{ from right bracket}} &= \binom{2n}{n} \end{align*}

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From each point, we can get to the diagonal ahead of us, so each move only takes us one diagonal closer together. Therefore we can only meet on the diagonal. The number of routes we can meet at is \begin{align*} && R &= \sum_{k=0}^n \underbrace{\binom{n}{k}}_{\text{I go up } k}\underbrace{\binom{n}{n-k}}_{\text{she goes down }n-k} \\ &&&= \binom{2n}{n} \end{align*} Therefore the probability is \(\displaystyle \frac1{2^{2n}} \binom{2n}n\). If I leave \(2k\) minutes late, then we will be attempting meet on a diagonal which is \(2k\) closer to me. The probability this occurs is \begin{align*} && \frac{1}{2^{2n}}\sum_{j=0}^{n-k}\binom{n-k}{j}\binom{n+k}{n-j} &= \frac{1}{2^{2n}}\binom{2n}{n} \end{align*} (by considering the coefficient of \(t^n\) in \((1+t)^{n+k}(1+t)^{n-k} =(1+t)^{2n}\)) This probability is unchanged, because you can consider the two paths as one path by one random person, conditional on them meeting and the delay doesn't change anything. If \(n = 1\) and I leave late, the only way we meet is if we end up walking towards each other down the same street (not at an intersection). This means I need to walk towards the intersection she reaches after the first minute \(\frac12\) and she needs to walk towards me \(\frac12\) so we have probability \(\frac14\)

The random variable \(X\) is uniformly distributed on \([0,1]\). A new random variable \(Y\) is defined by the rule \[ Y=\begin{cases} 1/4 & \mbox{ if }X\leqslant1/4,\\ X & \mbox{ if }1/4\leqslant X\leqslant3/4\\ 3/4 & \mbox{ if }X\geqslant3/4. \end{cases} \] Find \({\mathrm E}(Y^{n})\) for all integers \(n\geqslant 1\). Show that \({\mathrm E}(Y)={\mathrm E}(X)\) and that \[{\mathrm E}(X^{2})-{\mathrm E}(Y^{2})=\frac{1}{24}.\] By using the fact that \(4^{n}=(3+1)^{n}\), or otherwise, show that \({\mathrm E}(X^{n}) > {\mathrm E}(Y^{n})\) for \(n\geqslant 2\). Suppose that \(Y_{1}\), \(Y_{2}\), \dots are independent random variables each having the same distribution as \(Y\). Find, to a good approximation, \(K\) such that \[{\rm P}(Y_{1}+Y_{2}+\cdots+Y_{240000} < K)=3/4.\]

Define \(\cosh x\) and \(\sinh x\) in terms of exponentials and prove, from your definitions, that \[ \cosh^{4}x-\sinh^{4}x=\cosh2x \] and \[ \cosh^{4}x+\sinh^{4}x=\tfrac{1}{4}\cosh4x+\tfrac{3}{4}. \] Find \(a_{0},a_{1},\ldots,a_{n}\) in terms of \(n\) such that \[ \cosh^{n}x=a_{0}+a_{1}\cosh x+a_{2}\cosh2x+\cdots+a_{n}\cosh nx. \] Hence, or otherwise, find expressions for \(\cosh^{2m}x-\sinh^{2m}x\) and \(\cosh^{2m}x+\sinh^{2m}x,\) in terms of \(\cosh kx,\) where \(k=0,\ldots,2m.\)

For all values of \(a\) and \(b,\) either solve the simultaneous equations \begin{alignat*}{1} x+y+az & =2\\ x+ay+z & =2\\ 2x+y+z & =2b \end{alignat*} or prove that they have no solution.