Problems

Let \(f(x, y, z) \equiv x^2 + y^2 + z^2 - xy - yz - zx\). Show that \[f(x, y, z) = (x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z),\] where \(\omega\) is a complex cube root of 1. Prove that if \((y - z)^n + (z - x)^n + (x - y)^n\), (\(n \geq 2\)), has \(f(x, y, z)\) as a factor then 3 does not divide \(n\); and that if the expression has \(\{f(x, y, z)\}^2\) as a factor then \(n\) is of the form \(3m + 1\).

A triangle is called chromatic if all its sides are the same colour. Each pair of \(n\) distinct points \(P_1\), \ldots, \(P_n\) in space is connected with either a red line or a black line. Prove that if \(n = 6\) there must be at least 2 chromatic triangles of the form \(P_iP_jP_k\). Deduce that if \(n = 7\) there are at least 3 such chromatic triangles.

A solid cone is described by the following equations (in cylindrical polar coordinates \((r, \phi, z)\)): \begin{align} r &\leq -z\tan\alpha \quad (\alpha < \pi/2)\\ z &\leq 0\\ z &\geq mr\sin\phi-a \quad (m < \cot\alpha) \end{align} \(\alpha\), \(m\) and \(a\) are constants with \(a > 0\). Sketch the cone. The cone is placed on its side on a plane, with its vertex at a point \(O\). The cone is in contact with the plane along the line segment \(OP\), which is initially of length \(a\sec\alpha\). The cone now rolls on the plane, the vertex remaining at \(O\). Obtain the polar equation of the locus of the point \(P\) in terms of the distance \(R\) from \(O\) to \(P\), and the angle between the line \(OP\) and its initial direction. What conditions are required on \(\alpha\) to ensure that the curve is closed?

Let \(P\) be a point on the circumcircle of the triangle \(ABC\), and let \(L\), \(M\) and \(N\) be the feet of the perpendiculars from \(P\) to \(BC\), \(CA\) and \(AB\). Prove that \(L\), \(M\) and \(N\) are collinear. The altitudes of the triangle that pass through \(A\), \(B\) and \(C\) meet the circumcircle again at \(D\), \(E\) and \(F\) respectively. Let \(PD\) meet \(BC\) in \(D'\), \(PE\) meet \(CA\) in \(E'\) and \(PF\) meet \(AB\) in \(F'\). Prove that \(D'\), \(E'\), \(F'\) and \(H\), where \(H\) is the orthocentre of the triangle, lie on a line parallel to the line \(LMN\).

Let \(A'\) be a point in the plane of a triangle \(BCD\). Let \(BC\) and \(A'D\) meet at \(X\), and \(A'B\) meet at \(Y\), and let \(l\) denote the line \(XY\). Let \(l\) meet \(BD\) and \(A'C\) at \(P\) and \(Q\) respectively. Prove that \(P\) and \(Q\) divide \(XY\) internally and externally in the same ratio (i.e. that \(\frac{XP}{PY} = -\frac{XQ}{QY}\)). Suppose that \(ABCD\) is a tetrahedron and that \(A'\) is a general point in the plane of the face \(BCD\). Show that it is possible to find points \(B'\), \(C'\) and \(D'\) so that the two tetrahedra \(ABCD\) and \(A'B'C'D'\) have the property that the plane of any face of either tetrahedron contains precisely one vertex of the other. Are \(B'\), \(C'\) and \(D'\) uniquely determined?

(i) Let \(f(x) = e^{-1/x^2}\) for \(x \neq 0\), and \(f(0) = 0\). Prove that \(f^{(n)}(x)\) exists for all \(x\) and for all \(n\); calculate \(f^{(n)}(0)\). Comment upon Maclaurin's theorem applied to \(f(x)\). (ii) Let \(g(x) = x^2\sin \frac{1}{x}\) for \(x \neq 0\), \(g(0) = 0\). Determine, for every \(x\), the value of \(g'(x)\). What can you say about \(g''(x)\)?

It is agreed, in private, by two union leaders that ultimately the pay, \(x\), of a xerographer should be three quarters of the pay, \(y\), of a yogi. Publicly the xerographers' leader claims that, on the grounds of comparability, \(x\) should be governed by the equation \[\dot{x}+2x = 3y+t.\] The chief yogi, being more pessimistic about inflation, makes a claim for \[\dot{y}+2y = 3x+e^{2t}.\] The claims of these powerful unions are met in full. Has the gentleman's agreement been breached?

An impatient motorist, travelling from home to office, has to cross \(n\) sets of traffic lights which break his route into \(n + 1\) stretches of road. The chance that he is delayed at any set of lights is \(\phi\) (\(0 < \phi < 1\)), independently of what happens at any other set. His chance of being involved in an accident on the first stretch of road is \(p = 1 - q\). If he is delayed at any set of lights his chance of not being involved in an accident on the subsequent stretch of road becomes a fraction \(\theta\) (\(0 < \theta < 1\)) of what it was on the preceding stretch; if he is not delayed, it stays unchanged. Show that his chance of reaching the office without an accident is \[P_n(q) = q^{n+1} \prod_{r=1}^{n} \{(1 - \phi) +\phi\theta^r\}.\]

The random variables \(X_1, X_2, \ldots, X_n\) are independent and have identical probability distributions. The function \(\phi\) of \(n\) arguments is such that \(\phi(X_1, X_2, \ldots, X_n)\) has expectation \(\mu\) and variance \(\sigma^2\). Furthermore, \(\phi\) is not symmetric, so that there is at least one pair of suffixes \((i, j)\) such that with positive probability \[\phi(X_1, \ldots, X_i, \ldots, X_j, \ldots, X_n) \neq \phi(X_1, \ldots, X_j, \ldots, X_i, \ldots, X_n).\] The symmetrisation \(\psi\) of \(\phi\) is defined by \[\psi(X_1, \ldots, X_n) = \frac{1}{n!}\sum\phi(X_{i_1}, \ldots, X_{i_n})\] where the summation is over all \(n!\) permutations \((i_1, i_2, \ldots, i_n)\) of \((1, 2, \ldots, n)\). Prove that \(\psi(X_1, X_2, \ldots, X_n)\) has expectation \(\mu\) but variance less than \(\sigma^2\). [A simpler version, using exactly the same strategy of proof, has \(n = 2\).]

A massless hoop, of radius \(a\), stands vertically on a rough plane. A weight is attached to the rim of the hoop so that the radius to the weight makes an angle \(\theta_0\) (\(0 \leq \theta_0 < \pi\)) to the upward vertical. In the subsequent motion the hoop remains vertical and rolling occurs without slipping until the vertical reaction at the point of contact with the plane is zero. Show that this occurs when \(\theta = \theta_1\) where \(\frac{1}{2}\pi \leq \theta_1 < \pi\). At the moment when the vertical reaction is zero, the plane is removed. Show that the velocity of the weight when it reaches the former level of the plane is \[2\sqrt{2ag}\cos\left(\frac{\theta_1}{2}\right).\]