If \(\Gamma\) is a circle with centre \(C\), and \(A, B\) are two points in the same plane as \(\Gamma\) (but not on \(\Gamma\)), show that \(AX + BX\) attains its minimum as \(X\) varies on \(\Gamma\) at a point at which \(AX\) and \(BX\) are equally inclined to \(CX\). Hence or otherwise show that if \(A, B, C\) are three fixed points and \(P\) is a point (in the plane of \(A, B\) and \(C\)) at which \(AP + BP + CP\) is a minimum, then either \(P\) is one of the points \(A, B\), or \(C\), or else $$\widehat{BPC} = \widehat{CPA} = \widehat{APB} = \frac{2\pi}{3}.$$
Discover a general formula of which \begin{align} 1^3 + 3^3 + 5^3 &= 9 \times 17,\\ 1^3 + 3^3 + 5^3 + 7^3 &= 16 \times 31, \end{align} are particular cases. Prove the formula.
(i) Prove that, if \(a\), \(b\), \(c\) are in arithmetical progression, so are $$b^2 + bc + c^2, \quad c^2 + ca + a^2, \quad a^2 + ab + b^2.$$ Investigate whether the converse is true. (ii) Either find integers \(x\), \(y\) satisfying $$x^2 - 7y^2 = 10,$$ or prove that no such integers exist.
Express in partial fractions $$\frac{(x+1)(x+2)\cdots(x+n+1)-(n+1)!}{x(x+1)(x+2)\cdots(x+n)}.$$ Hence, or otherwise, prove that $$\frac{c_1}{1} - \frac{c_2}{2} + \frac{c_3}{3} - \ldots + (-1)^{n-1} \frac{c_n}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n},$$ where \(c_r = n!/r!(n-r)!\).
Solution: Notice that \(\deg\) of the numerator is equal to the degree of the denominator, so: \begin{align*} && \frac{(x+1)(x+2)\cdots(x+n+1)-(n+1)!}{x(x+1)(x+2)\cdots(x+n)} &= 1 + \frac{A_0}{x} + \frac{A_1}{x+1} + \cdots + \frac{A_n}{x+n} \\ \\ \Rightarrow && (x+1)\cdots(x+n+1) - (n+1)! &= x(x+1)\cdots(x+n) + A_0(x+1)\cdots(x+n) + \cdots + A_nx(x+1)\cdots(x+n-1) \\ x = 0: && 0 &= A_0 \\ x = -k: && 0 - (n+1)! &= A_k (-k)\cdots(-1) (1) \cdots (n-k) \\ \Rightarrow && 0-(n+1)! &= A_k (-1)^k k! (n-k)! \\ \Rightarrow && A_k &= (-1)^{k+1}\frac{n!}{k!(n-k)!} (n+1) \\ \Rightarrow && \frac{(x+1)(x+2)\cdots(x+n+1)-(n+1)!}{x(x+1)(x+2)\cdots(x+n)} &= 1 + (n+1) \sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{x+k} \end{align*} This expression is true if \(x \neq 0, -1, -2, \cdots, -n\). Notice however the LHS and RHS are both well-defined when \(x = 0\). Therefore we plug in \(0\) to see: \begin{align*} &&1 + (n+1) \sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k} &= \frac{(n+1)!\left ( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} + \frac{1}{n+1} \right)}{n!} \\ &&&= (n+1) \left ( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} + \frac{1}{n+1} \right) \\ \Rightarrow && \frac{1}{n+1} + \sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k} &= \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} + \frac{1}{n+1} \\ \Rightarrow && (-1)^{k+1} \binom{n}{k}\frac{1}{k} &= \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} \end{align*}
A square \(ABCD\) of side \(5a\) is divided into 25 squares each of side \(a\). In how many different ways can \(A\), \(C\) be joined by a path of length \(10a\) along sides of squares?
Points \(A_1\), \(A_2\), \(\ldots\), \(A_n\) (where \(n \geq 3\)) are equally spaced round the circumference of a circle. Their distances from a line drawn through the centre are \(d_1\), \(d_2\), \(\ldots\), \(d_n\). Prove that $$d_1^2 + d_2^2 + \ldots + d_n^2$$ is the same for every direction of the line.
Find, in terms of \(h\), \(k\), \(\sin 2\theta\) and \(\cos 2\theta\), the co-ordinates of the mirror-image of the point \((h, k)\) in the line \(x \cos \theta - y \sin \theta = 0\). \(M_1(P)\) and \(M_2(P)\) are respectively the mirror-images of \(P\) in two lines \(l_1\) and \(l_2\) which intersect at an angle \(\alpha\). Prove that \(M_1\{M_2(M_1(P))\}\) is the same as \(M_2\{M_1(P)\}\) for all positions of \(P\) if and only if \(\alpha = \frac{1}{3}\pi\).
A point \(P\) is given and two lines \(l\), \(m\) whose point of intersection \(Q\) is off the paper. You are given an ungraduated ruler and (if you wish, but preferably not) a pair of compasses. State and justify a construction for the line through \(P\) which would pass through \(Q\).
Prove that the feet of the normals from the point \((h, k)\) to the rectangular hyperbola \(xy = c^2\) are its intersections with the hyperbola $$x^2 - y^2 - hx + ky = 0.$$ Prove that, if \(hk = 4c^2\), the same four points are the feet of the normals to the second hyperbola from the point \((-\frac{1}{2}h, -\frac{1}{2}k)\).
The point \((at^2, at^3)\) on the curve \(ay^2 = x^3\) will be called the point \(t\). Prove that, if the points \(t_1\), \(t_2\), \(t_3\) are collinear, then $$\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} = 0.$$ Perpendicular lines through the origin \(O\) meet the curve at \(P\), \(Q\). \(PQ\) meets the curve again at \(R\). \(S\) is the point of contact of the tangent to the curve from \(R\). Prove that \(OP\), \(OQ\) bisect the angles between \(Ox\) and \(OS\).