Problems

1. The sequence of numbers \(u_0, u_1, \ldots \) is given by \(u_0=u\) and, for \(n\ge 0\), \begin{equation} u_{n+1} =4u_n(1- u_n)\,. \tag{\(*\)} \end{equation} In the case \(u= \sin^2\theta\) for some given angle \(\theta\), write down and simplify expressions for \(u_1\) and \(u_2\) in terms of \(\theta\). Conjecture an expression for \(u_n\) and prove your conjecture.
2. The sequence of numbers \(v_0, v_1, \ldots\) is given by $v_0= v \text{ and, for }n\ge 0$, \[ v_{n+1} = -pv_n^2 +qv_n +r\,, \] where \(p\), \(q\) and \(r\) are given numbers, with \(p\ne0\). Show that a substitution of the form \(v_n =\alpha u_n +\beta\), where \(\alpha\) and \(\beta\) are suitably chosen, results in the sequence \((*)\) provided that \[ 4pr = 8 +2q -q^2 \,. \] Hence obtain the sequence satisfying \(v_0=1\) and, for \(n\ge0\), \(v_{n+1} = -v_n^2 +2 v_n +2 \,\).

In the triangle \(ABC\), the base \(AB\) is of length 1 unit and the angles at~\(A\) and~\(B\) are \(\alpha\) and~\(\beta\) respectively, where \(0<\alpha\le\beta\). The points \(P\) and~\(Q\) lie on the sides \(AC\) and \(BC\) respectively, with \(AP=PQ=QB=x\). The line \(PQ\) makes an angle of~\(\theta\) with the line through~\(P\) parallel to~\(AB\).

1. Show that \(x\cos\theta = 1- x\cos\alpha - x\cos\beta\), and obtain an expression for \(x\sin\theta\) in terms of \(x\), \(\alpha\) and~\(\beta\). Hence show that \begin{equation} \label{eq:2*} \bigl(1+2\cos(\alpha+\beta)\bigr)x^2 - 2(\cos\alpha + \cos\beta)x + 1 = 0\,. \tag{\(*\)} \end{equation} Show that \((*)\) is also satisfied if \(P\) and \(Q\) lie on \(AC\) produced and \(BC\) produced, respectively. [By definition, \(P\) lies on \(AC\) produced if \(P\) lies on the line through \(A\) and~\(C\) and the points are in the order \(A\), \(C\), \(P\)\,.]
2. State the condition on \(\alpha\) and \(\beta\) for \((*)\) to be linear in \(x\). If this condition does not hold (but the condition \(0<\alpha \le \beta\) still holds), show that \((*)\) has distinct real roots.
3. Find the possible values of~\(x\) in the two cases (a) \(\alpha = \beta = 45^\circ\) and (b) \(\alpha = 30^\circ\), \(\beta = 90^\circ\), and illustrate each case with a sketch.

A particle \(P\) of mass \(m\) is connected by two light inextensible strings to two fixed points \(A\) and \(B\), with \(A\) vertically above \(B\). The string \(AP\) has length \(x\). The particle is rotating about the vertical through \(A\) and \(B\) with angular velocity \(\omega\), and both strings are taut. Angles \(PAB\) and \(PBA\) are \(\alpha\) and \(\beta\), respectively. Find the tensions \(T_A\) and \(T_B\) in the strings \(AP\) and \(BP\) (respectively), and hence show that \(\omega^2 x\cos\alpha \ge g\). Consider now the case that \(\omega^2 x\cos\alpha = g\). Given that \(AB=h\) and \(BP=d\), where \(h>d\), show that \(h\cos\alpha \ge \sqrt{h^2-d^2}\). Show further that \[ mg < T_A \le \frac{mgh}{\sqrt{h^2-d^2}\,}\,. \] Describe the geometry of the strings when \(T_A\) attains its upper bound.

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Solution:

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\begin{align*} \text{N2}(\uparrow): && T_A \cos \alpha - T_B \cos\alpha - mg &= 0 \\ \Rightarrow && T_A \cos \alpha - T_B \cos\beta &= mg \\ \text{N2}(\leftarrow, \text{radially}): && T_A \sin \alpha + T_B \sin \beta &= m x \sin \alpha \omega^2 \\ \Rightarrow && T_A(\cos \alpha \sin \beta+\sin \alpha \cos \beta) &= mg \sin \beta + mx \sin \alpha \omega^2 \cos \beta \\ \Rightarrow && T_A &=\frac{mg\sin \beta + m x \sin \alpha \omega^2 \cos \beta }{\sin(\alpha + \beta)} \\ \Rightarrow && T_B(\sin \beta \cos \alpha- \cos \beta \sin \alpha)&= mx \sin \alpha \omega^2 \cos \alpha -mg \sin \alpha \\ \Rightarrow && T_B &= \frac{m x \sin \alpha \omega^2 \cos \alpha - mg \sin \alpha}{\sin(\beta - \alpha)} \\ &&&= \frac{m \sin \alpha(\omega^2 \cos\alpha - g)}{\sin (\beta - \alpha)} \end{align*} Since \(T_B \geq 0 \Rightarrow \omega^2 \cos\alpha - g \geq 0\) as required.

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\(\sqrt{h^2-d^2}\) is the length of the final side on the dashed right angle triangle with hypotenuse \(AB\). \(h \cos \alpha\) will be clearly longer as the angle \(\alpha\) will be smaller and so \(\cos \alpha\) will be larger. When \(\omega^2 x \cos \alpha = g\) we must have \(T_B = 0\). \(T_A\cos \alpha = mg \Rightarrow T_A > mg\) since \(\alpha \neq 0\). \(T_A = \frac{mg}{\cos \alpha} \leq \frac{mgh}{\sqrt{h^2-d^2}}\) \(T_A\) will attain it's upper bound when \(\angle APB\) is a right angle.

Two particles, \(A\) and \(B\), are projected simultaneously towards each other from two points which are a distance \(d\) apart in a horizontal plane. Particle \(A\) has mass \(m\) and is projected at speed \(u\) at angle \(\alpha\) above the horizontal. Particle \(B\) has mass \(M\) and is projected at speed \(v\) at angle \(\beta\) above the horizontal. The trajectories of the two particles lie in the same vertical plane. The particles collide directly when each is at its point of greatest height above the plane. Given that both \(A\) and \(B\) return to their starting points, and that momentum is conserved in the collision, show that \[ m\cot \alpha = M \cot \beta\,. \] Show further that the collision occurs at a point which is a horizontal distance \(b\) from the point of projection of \(A\) where \[ b= \frac{Md}{m+M}\, , \] and find, in terms of \(b\) and \(\alpha\), the height above the horizontal plane at which the collision occurs.

An equilateral triangle, comprising three light rods each of length \(\sqrt3a\), has a particle of mass \(m\) attached to each of its vertices. The triangle is suspended horizontally from a point vertically above its centre by three identical springs, so that the springs and rods form a tetrahedron. Each spring has natural length \(a\) and modulus of elasticity \(kmg\), and is light. Show that when the springs make an angle \(\theta\) with the horizontal the tension in each spring is \[ \frac{ kmg(1-\cos\theta)}{\cos\theta}\,. \] Given that the triangle is in equilibrium when \(\theta = \frac16 \pi\), show that \(k=4\sqrt3 +6\). The triangle is released from rest from the position at which \(\theta=\frac13\pi\). Show that when it passes through the equilibrium position its speed \(V\) satisfies \[ V^2 = \frac{4ag}3(6+\sqrt3)\,. \]

A thin circular path with diameter \(AB\) is laid on horizontal ground. A vertical flagpole is erected with its base at a point \(D\) on the diameter \(AB\). The angles of elevation of the top of the flagpole from \(A\) and \(B\) are \(\alpha\) and \(\beta\) respectively (both are acute). The point \(C\) lies on the circular path with \(DC\) perpendicular to \(AB\) and the angle of elevation of the top of the flagpole from \(C\) is \(\phi\). Show that \(\cot\alpha\cot \beta = \cot^2\phi\). Show that, for any \(p\) and \(q\), \[ \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) = \tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q) .\] Deduce that, if \(p\) and \(q\) are positive and \( p+q \le \tfrac12 \pi\), then \[ \cot p\cot q\, \ge \cot^2 \tfrac12(p+q) \, \] and hence show that \(\phi \le \tfrac12(\alpha+\beta)\) when \( \alpha +\beta \le \tfrac12 \pi\,\).

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Solution:

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\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

A cyclic quadrilateral \(ABCD\) has sides \(AB\), \(BC\), \(CD\) and \(DA\) of lengths \(a\), \(b\), \(c\) and \(d\), respectively. The area of the quadrilateral is \(Q\), and angle \(DAB\) is \(\theta\). Find an expression for \(\cos\theta\) in terms of \(a\), \(b\), \(c\) and \(d\), and an expression for \(\sin\theta\) in terms of \(a\), \(b\), \(c\), \(d\) and \(Q\). Hence show that \[ 16Q^2 = 4(ad+bc)^2 - (a^2+d^2-b^2-c^2)^2 \,, \] and deduce that \[ Q^2 = (s-a)(s-b)(s-c)(s-d)\,, \] where \(s= \frac12(a+b+c+d)\). Deduce a formula for the area of a triangle with sides of length \(a\), \(b\) and \(c\).

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Solution:

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\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)

Prove the identity \[ 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta)= \sin 3\theta\, . \tag{\(*\)}\]

1. By differentiating \((*)\), or otherwise, show that \[ \cot \tfrac19\pi - \cot \tfrac29\pi + \cot \tfrac49\pi = \sqrt3\,. \]
2. By setting \(\theta = \frac16\pi-\phi\) in \((*)\), or otherwise, obtain a similar identity for \(\cos3\theta\) and deduce that \[ \cot \theta \cot (\tfrac13\pi-\theta) \cot (\tfrac13\pi+\theta) =\cot3\theta\,. \] Show that \[ \cosec \tfrac19\pi -\cosec \tfrac59\pi +\cosec \tfrac79\pi = 2\sqrt3\,. \]

A particle is projected at an angle \(\theta\) above the horizontal from a point on a horizontal plane. The particle just passes over two walls that are at horizontal distances \(d_1\) and \(d_2\) from the point of projection and are of heights \(d_2\) and \(d_1\), respectively. Show that \[ \tan\theta = \frac{d_1^2+d_\subone d_\subtwo +d_2^2}{d_\subone d_\subtwo}\,. \] Find (and simplify) an expression in terms of \(d_1\) and \(d_2\) only for the range of the particle.

A thin non-uniform bar \(AB\) of length \(7d\) has centre of mass at a point \(G\), where \(AG=3d\). A light inextensible string has one end attached to \(A\) and the other end attached to \(B\). The string is hung over a smooth peg \(P\) and the bar hangs freely in equilibrium with \(B\) lower than~\(A\). Show that \[ 3\sin\alpha = 4\sin\beta\,, \] where \(\alpha\) and \(\beta\) are the angles \(PAB\) and \(PBA\), respectively. Given that \(\cos\beta=\frac45\) and that \(\alpha\) is acute, find in terms of \(d\) the length of the string and show that the angle of inclination of the bar to the horizontal is \(\arctan \frac17\,\).

1. Find all the values of \(\theta\), in the range \(0^\circ < \theta < 180^\circ\), for which \(\cos\theta=\sin 4\theta\). Hence show that \[ \sin 18^\circ = \frac14\left( \sqrt 5 -1\right). \]
2. Given that \[ 4\sin^2 x + 1 = 4\sin^2 2x \,, \] find all possible values of \(\sin x\,\), giving your answers in the form \(p+q\sqrt5\) where \(p\) and \(q\) are rational numbers.
3. Hence find two values of \(\alpha\) with \(0^\circ < \alpha < 90^\circ\) for which \[ \sin^23\alpha + \sin^25\alpha = \sin^2 6\alpha\,. \]

Three non-collinear points \(A\), \(B\) and \(C\) lie in a horizontal ceiling. A particle \(P\) of weight \(W\) is suspended from this ceiling by means of three light inextensible strings \(AP\), \(BP\) and \(CP\), as shown in the diagram. The point \(O\) lies vertically above \(P\) in the ceiling.

1. Show that the unit vector in the direction \(PB\) can be written in the form \[ -\frac13\, {\bf i} - \frac{\sqrt2\,}3\, {\bf j} + \frac{\sqrt2\, }{\sqrt3 \,} \,{\bf k} \,,\] where \(\bf i\,\), \(\, \bf j\) and \(\bf k\) are the usual mutually perpendicular unit vectors with \(\bf j\) parallel to \(OA\) and \(\bf k\) vertically upwards.
2. Find expressions in vector form for the forces acting on \(P\).
3. Show that \(U=\sqrt6 V\) and find \(T\), \(U\) and \(V\) in terms of \(W\).

A thin uniform circular disc of radius \(a\) and mass \(m\) is held in equilibrium in a horizontal plane a distance \(b\) below a horizontal ceiling, where \(b>2a\). It is held in this way by \(n\) light inextensible vertical strings, each of length \(b\); one end of each string is attached to the edge of the disc and the other end is attached to a point on the ceiling. The strings are equally spaced around the edge of the disc. One of the strings is attached to the point \(P\) on the disc which has coordinates \((a,0,-b)\) with respect to cartesian axes with origin on the ceiling directly above the centre of the disc. The disc is then rotated through an angle \(\theta\) (where \(\theta<\pi\)) about its vertical axis of symmetry and held at rest by a couple acting in the plane of the disc. Show that the string attached to~\(P\) now makes an angle \(\phi\) with the vertical, where \[ b\sin\phi = 2a \sin\tfrac12 \theta\,. \] Show further that the magnitude of the couple is \[ \frac {mga^2\sin\theta}{\sqrt{b^2-4a^2\sin^2 \frac12\theta \ } \ }\,. \] The disc is now released from rest. Show that its angular speed, \(\omega\), when the strings are vertical is given by \[ \frac{a^2\omega^2}{4g} = b-\sqrt{b^2 - 4a^2\sin^2 \tfrac12\theta \;}\,. \]

Show that \[ \sin(x+y) -\sin(x-y) = 2 \cos x \, \sin y \] and deduce that \[ \sin A - \sin B = 2 \cos \tfrac12 (A+B) \, \sin\tfrac12 (A-B) \,. \] Show also that \[ \cos A - \cos B = -2 \sin \tfrac12(A+B) \, \sin\tfrac12(A-B)\,. \] The points \(P\), \(Q\), \(R\) and \(S\) have coordinates \(\left(a\cos p,b\sin p\right)\), \(\left(a\cos q,b\sin q\right)\), \(\left(a\cos r,b\sin r\right)\) and \(\left(a\cos s,b\sin s\right)\) respectively, where \(0\le p < q < r < s <2\pi\), and \(a\) and \(b\) are positive. Given that neither of the lines \(PQ\) and \(SR\) is vertical, show that these lines are parallel if and only if \[ r+s-p-q = 2\pi\,. \]

Each edge of the tetrahedron \(ABCD\) has unit length. The face \(ABC\) is horizontal, and \(P\) is the point in \(ABC\) that is vertically below \(D\).

1. Find the length of \(PD\).
2. Show that the cosine of the angle between adjacent faces of the tetrahedron is \(1/3\).
3. Find the radius of the largest sphere that can fit inside the tetrahedron.