Prove that if the normals at four points of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) are concurrent, and two of the points lie on the line \(\frac{lx}{a}+\frac{my}{b}+1=0\), the other two will lie on the line \(\frac{x}{al}+\frac{y}{bm}+1=0\). Prove also that if two lines drawn through the point \((\frac{2}{3}a, \frac{2}{3}b)\) meet the ellipse at four points, the normals at which are concurrent, one of the lines is \(\frac{x}{a}-\frac{y}{b}=2\), and find the equation of the other.
Prove that the locus of a point \(P\) from which the pair of tangents to a given hyperbola are harmonic conjugates to the pair of parallels through \(P\) to the asymptotes, is a similar and similarly situated hyperbola.
A conic is inscribed in a triangle \(ABC\), touching the sides at \(P, Q, R\). The lines \(QR, RP, PQ\) meet \(BC, CA, AB\) in \(L, M, N\) respectively. Prove that \(L, M, N\) are collinear. Prove also that if the line \(LMN\) passes through the centre of the conic, it will, for different conics, envelop a conic touching the sides of the triangle \(ABC\).
If \begin{align*} u_n &= 1 - \frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)(n-3)}{4!} - \dots, \\ v_n &= n - \frac{n(n-1)(n-2)}{3!} + \frac{n(n-1)(n-2)(n-3)(n-4)}{5!} - \dots, \end{align*} where \(n\) is a positive integer, shew that \[ u_n^2+v_n^2 = 2(u_n u_{n-1} + v_n v_{n-1}). \]
A bag contains six balls, each of which is known to be black or white, either colour being a priori equally likely. Two balls are drawn and found to be one black and one white: these are replaced and two others are drawn. Shew that the chance of their being both black is 19/75. NB: This question assumes the number of black balls 0-6 is equally likely with probability 1/7.
Solution: (Assuming each ball is equally likely to be black or white) \begin{array}{c|c|c} n\text{ black} & \mathbb{P}(n \text{ black}) & \mathbb{P}(BW|\text{number of black}=n) & \mathbb{P}(BB|\text{number of black}=n)\\ \hline 1 & \frac{6}{2^6} & \frac{5}{15} = \frac13 & 0 \\ 2 & \frac{15}{2^6} & \frac{2 \cdot 4}{15} = \frac{8}{15} & \frac{1}{15} \\ 3 & \frac{20}{2^6} & \frac{3 \cdot 3}{15} = \frac{3}{5} & \frac{3}{15} = \frac15 \\ 4 & \frac{15}{2^6} & \frac{4 \cdot 2}{15} = \frac{8}{15} & \frac{6}{15} = \frac{2}{5} \\ 5 & \frac{5}{2^6} &\frac{5}{15} = \frac13 & \frac{10}{15} = \frac{2}{3} \end{array} \begin{align*} && \mathbb{P}(BB | BW) &= \frac{\mathbb{P}(BB \text{ and } BW)}{\mathbb{P}(BW)} \\ &&&= \frac{15 \cdot \frac{8}{15} \cdot \frac{1}{15} + 20 \cdot \frac{9}{15} \cdot \frac{3}{15} + 15\cdot\frac{8}{15}\cdot \frac{6}{15} + 6\cdot\frac{5}{15} \cdot \frac{10}{15}}{6 \cdot \frac{5}{15} +15 \cdot \frac{8}{15} + 20 \cdot \frac{9}{15} + 15\cdot\frac{8}{15}+ 6\cdot\frac{5}{15}} \\ &&&= \frac{15 \cdot 8 + 20 \cdot 27 + 15 \cdot 48 + 6 \cdot 50}{15(30 + 120+180+120+30)} \\ &&&= \frac{120+540+720+300}{15 \cdot 480}\\ &&&= \frac{168}{15 \cdot 48} = \frac{84}{15 \cdot 24} = \frac{21}{15 \cdot 6}\\ &&&= \frac{7}{30} \end{align*}
Prove that \[ \sin(A+B) = \sin A \cos B + \cos A \sin B, \] taking \(A, B, A+B\) to be acute angles. If any two of \[ \sin(B+C)+\sin(C+A)+\sin(A+B), \] and the three similar expressions obtained by permuting \(A, B, C, D\) are equal, then all four are equal, provided that no two of \(A, B, C, D\) differ by a multiple of \(\pi\).
Obtain the formula \[ \cos n\theta = 2^{n-1}\cos^n\theta - \frac{n}{1!}2^{n-3}\cos^{n-2}\theta + \frac{n(n-3)}{2!}2^{n-5}\cos^{n-4}\theta - \dots, \] and give the expansion arranged in ascending powers of \(\cos\theta\). If \(p_p, p_q, p_r\) denote three of the distances \(p_1, p_2, \dots, p_n\) of a point \(P\) in the plane of a regular polygon from the vertices, prove that \[ \Sigma p_p^2 p_q^2 p_r^2 = n(n-2)(a^2+c^2)\{(n-1)(a^2+c^2)^2-6a^2c^2\}/6, \] where \(a\) is the radius of the circle round the polygon, \(c\) is the distance of \(P\) from the centre of the circle, and the summation extends over all possible groups of three vertices.
Prove that \[ \left(\frac{d}{dx}-\tan x\right)^n u_n = n! u_0, \] where \[ u_n = x^n \sec x. \]
Trace the curve \[ (x^2-y^2)^2-4y^2+y=0. \]
Prove that in general a system of coplanar forces acting on a rigid body can be reduced to a single force acting at an assigned point together with a couple. Forces \(P, 2Q, 3P, 4Q, 5P, 6Q\) act along the sides of a regular hexagon taken in order. Reduce this system to a single force acting at the centre of the hexagon together with a couple. Find the relation between \(P\) and \(Q\) if the single force acts through the point of intersection of the lines of action of (a) \(6Q\) and \(P\), (b) \(2Q\) and \(3P\).