Prove that by the end of a party, attended by \(n \geq 2\) people, there are two people who have made the same number of new acquaintances. Now suppose that of all the pairs of people, precisely one pair have made the same number of new acquaintances. Show that there are at most two possibilities for this number. (It is not necessary to calculate the possibilities explicitly in terms of \(n\).)
Two triangles in a plane (\(ABC\), \(A'B'C'\)) are in perspective from a point \(O\) (i.e. \(AA'\), \(BB'\), \(CC'\) meet at \(O\)). It may be assumed that all the points are distinct. (i) Prove that the points of intersection of corresponding sides lie on a straight line. (ii) Suppose now that \(BC'\) meets \(B'C\) in \(P'\), \(CA'\) meets \(C'A\) in \(Q'\), and \(AB'\) meets \(A'B\) in \(R'\). Prove that \(Q'R'\), \(BC\), \(B'C'\) meet at a point, and hence using the converse of (i) (which may be assumed), prove that \(ABC\), \(P'Q'R'\) are in perspective.
If two variables \(x\) and \(z\) are related by \[z = x + \lambda g(z)\] where \(\lambda\) is a constant, then any smooth function \(F(z)\) satisfies Lagrange's Identity \[F(z) \equiv F(x) + \lambda g(x)\frac{dF(x)}{dx} + \sum_{n=2}^{\infty} \frac{\lambda^n}{n!}\frac{d^{n-1}}{dx^{n-1}}\left(\{g(x)\}^n\frac{dF(x)}{dx}\right),\] which you may use without proof. (i) By using Lagrange's identity, or otherwise, show that one root of the equation \[4z = 2 + z^3\] is given by \[z = \sum_{n=0}^{\infty} \frac{(3n)!}{(2n+1)!n!}\frac{1}{2^{4n+1}}\] (ii) By considering \(z = x + \lambda(z^2 - 1)\), or otherwise, prove the identity \[0 \equiv x^3 + \sum_{n=2}^{\infty} \frac{x^n}{n!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\] [You may assume all series converge.]
The triangle \(ABC\) is isosceles and has a right angle at \(B\). The sides \(AB\), \(BC\), \(AC\) are of unit length. Points \(X\), \(Y\), \(Z\) are selected at random on \(AB\), \(BC\), \(AC\), respectively. Let \(x\), \(y\) denote the distances of \(X\), \(Y\) from \(B\). Show that for fixed \(x\) and \(y\) the probability that \(ZXY\) and \(ZYX\) are both acute is \[\frac{x^2 + y^2}{x + y}.\] Hence show that the probability that both \(ZXY\) and \(ZYX\) are acute is \(\frac{2}{3} - \frac{1}{4} = \frac{5}{12}\).
A shooting gallery has two targets. A marksman has probability \(p\), \(q\) of hitting his aim when aiming for the first, second target respectively \((0 < p + q < 2)\). He never hits the target not aimed for, and each shot is independent of the others. He decides which target to aim for as follows: initially he aims for the first target; thereafter if his previous shot hit its mark, he fires at the same target, but if his previous shot missed, he aims at the other target. Obtain an expression for the probability that his \(n\)th shot hits its mark, and show that as \(n\) tends to infinity, this approaches \(\frac{p+q-2pq}{2-p-q}\).
The trace of a square matrix is defined to be the sum of its diagonal elements. If \(A\) and \(B\) are both two by two matrices, show that \[\text{trace}(AB) = \text{trace}(BA)\] If the elements of the two by two matrix \(A\) are functions of \(t\), \(\frac{dA}{dt}\) denotes the matrix whose elements are the derivatives of the corresponding elements of \(A\). If \(\Delta\) equals the determinant of the matrix \(A\), which may be assumed to be non-zero, show that \[\frac{1}{\Delta}\frac{d\Delta}{dt} = \text{trace}\left(A^{-1}\frac{dA}{dt}\right),\] where \(A^{-1}\) is the matrix inverse of \(A\). If, additionally, \(A\) satisfies the differential equation \[\frac{dA}{dt} = AB - BA,\] where the elements of \(B\) depend on \(t\), show that both \(\Delta\) and trace \((A^2)\) are independent of \(t\).
A rocket burns fuel at a rate equal to \(k\) times its instantaneous mass, the fuel being ejected at a fixed velocity \(u\) relative to the rocket. It is fired vertically upwards from the surface of the Earth. The gravitational force exerted by the Earth on the rocket varies as the inverse square of the distance, \(r\), of the rocket from the centre of the Earth. Show that it is necessary to have \(ku > g\), where \(g\) is the acceleration due to gravity at the Earth's surface, and that the time taken to reach a distance \(R\) from the centre of the Earth is \[\int_a^R \left\{\frac{1}{2(kur-ag)(r-a)}\right\}^{1/2}\,dr\] where \(a\) is the radius of the Earth. If \(ku\) is very much greater than \(g\), show that, on reaching \(r = R\), the rocket is lighter than it would be in the absence of the Earth's gravity by a factor \(e^{-\lambda}\) where \(\lambda^2 = ag^2/2ku^3\) and \(R = a(\sec\phi)^2\).
A particle of mass \(m\) moves horizontally in a long horizontal cylinder. The walls and one end of the cylinder are fixed, but the other end is a piston of mass \(\beta^2 m\) which can move freely. Collisions between the particle and the cylinder or the piston are perfectly elastic. Let \(V_n\) be the speed of the piston and \(U_n\) be the speed of the particle, both measured just before the \((n+1)\)th collision between the particle and piston. Show that \[2U_n = (\beta^2-1)V_n - (\beta^2+1)V_{n-1}\] and \[V_n - \frac{2(\beta^2-1)}{\beta^2+1}V_{n-1} + V_{n-2} = 0\] Verify that these equations have a solution with \[V_n = Ae^{ins} + Be^{-ins}\] where \(s\) is an angle between 0 and \(\pi\) to be determined, and \(A\) and \(B\) are complex constants. Hence show that if the piston is initially stationary, the particle will only collide with the piston \(N\) times, where \(N\) is the integral part of \[\frac{\pi}{4\cot^{-1}\beta} + \frac{1}{2}\]
A small bullet of mass \(m\) strikes the centre of one of the faces of a uniform cubical block of wood, of edge \(a\) and mass \(M\), at right angles. The block rests on a smooth horizontal plane. The bullet enters the wood at \(t = 0\), travelling with velocity \(v\), and it is then retarded by a force of magnitude \(ku\) where \(u\) is the velocity of the bullet relative to the block. Gravity and all other forces may be neglected, apart from the retarding forces between the bullet and the block and between the block and the plane. Show that, whilst in the block, the bullet has at time \(t\) travelled a distance \[\frac{v}{k}(1-e^{-kt}) + e^{-kt}\int_0^t G(s)e^{ks}\,ds\] through the block, where \(MdG/dt\) is the frictional force between the block and the plane, \(G(0) = 0\) and \(b = k(m+M)/mM\). Deduce that the bullet spends less time in the block than if the plane were smooth. Show that, if the bullet goes right through the block, it leaves with velocity \(v - ka/m\), and find the condition for it to do this if friction between block and plane can be neglected.
(i) Prove that \[\frac{d}{dt}\left(\frac{\mathbf{u}}{|\mathbf{u}|}\right) = \frac{1}{|\mathbf{u}|^3}\left(\mathbf{u} \times \frac{d\mathbf{u}}{dt}\right) \times \mathbf{u},\] where \(\mathbf{u}\) is any function of \(t\). (ii) A particle \(P\) of unit mass is acted on by a force of magnitude \(\mu/|r|^2\) directed towards a fixed point \(O\), where \(\mu\) is a constant and \(\mathbf{r} = \overrightarrow{OP}\). Its equation of motion is \[\frac{d\mathbf{v}}{dt} = -\frac{\mu}{|r|^3}\mathbf{r},\] where \(\mathbf{v} = d\mathbf{r}/dt\). By taking an appropriate product with this equation and integrating, prove that \(\mathbf{r} \times \mathbf{v}\) is a constant vector \(\mathbf{h}\) and deduce that, if \(\mathbf{h}\) is non-zero, the motion of \(P\) is confined to the plane through \(O\) perpendicular to \(\mathbf{h}\). Show that \[\mu\frac{d}{dt}\left(\frac{\mathbf{r}}{|r|}\right) = \frac{d\mathbf{v}}{dt} \times \mathbf{h}\] and hence by integration show also that \[\mu(\mathbf{a} \cdot \mathbf{r} + |r|) = |\mathbf{h}|^2\] for some constant vector \(\mathbf{a}\). Relate the magnitude and direction of \(\mathbf{a}\) to the geometry of the orbit of the particle.