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1993 Paper 1 Q8
D: 1500.0 B: 1500.0
coordinate geometry sphere circle intersection 3D geometry geometric proof trigonometry cosine rule

  1. Prove that the intersection of the surface of a sphere with a plane is always a circle, a point or the empty set. Prove that the intersection of the surfaces of two spheres with distinct centres is always a circle, a point or the empty set. {[}If you use coordinate geometry, a careful choice of origin and axes may help.{]}
  2. The parish council of Little Fitton have just bought a modern sculpture entitled `Truth, Love and Justice pouring forth their blessings on Little Fitton.' It consists of three vertical poles \(AD,BE\) and \(CF\) of heights 2 metres, 3 metres and 4 metres respectively. Show that \(\angle DEF=\cos^{-1}\frac{1}{5}.\) Vandals now shift the pole \(AD\) so that \(A\) is unchanged and the pole is still straight but \(D\) is vertically above \(AB\) with \(\angle BAD=\frac{1}{4}\pi\) (in radians). Find the new angle \(\angle DEF\) in radians correct to four figures.

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1993 Paper 1 Q10
D: 1500.0 B: 1500.0
energy elastic spring Hooke's law equilibrium moments rigid body rod trigonometry modulus of elasticity optimisation

A small lamp of mass \(m\) is at the end \(A\) of a light rod \(AB\) of length \(2a\) attached at \(B\) to a vertical wall in such a way that the rod can rotate freely about \(B\) in a vertical plane perpendicular to the wall. A spring \(CD\) of natural length \(a\) and modulus of elasticity \(\lambda\) is joined to the rod at its mid-point \(C\) and to the wall at a point \(D\) a distance \(a\) vertically above \(B\). The arrangement is sketched below. \noindent

\psset{xunit=0.8cm,yunit=0.8cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-1.55,-0.7)(4.5,5.27) \psline(0,5)(0,-1.16) \psline(0,0)(3.11,3.89) \pscoil[coilheight=1,coilwidth=0.2,coilarm=0.05](0,2.53)(1.47,1.84) \rput[tl](-0.56,0.41){\(B\)} \rput[tl](1.59,1.89){\(C\)} \rput[tl](3.41,4.28){\(A\)} \rput[tl](-0.56,2.84){\(D\)} \parametricplot{0.7583777142101807}{3.8999703677999737}{1*0.16*cos(t)+0*0.16*sin(t)+3.22|0*0.16*cos(t)+1*0.16*sin(t)+3.77} \psline(3.1,3.66)(3.33,3.88) \begin{scriptsize} \psdots[dotsize=5pt 0](0,0) \psdots[dotstyle=*](3.11,3.89) \psdots[dotstyle=*](0,2.53) \end{scriptsize} \end{pspicture*} \par
Show that if \(\lambda>4mg\) the lamp can hang in equilibrium away from the wall and calculate the angle \(\angle DBA\).

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1993 Paper 1 Q12
D: 1500.0 B: 1484.0
projectiles clay pigeon trigonometry simultaneous equations inequality optimisation vertical launch inclined firing

In a clay pigeon shoot the target is launched vertically from ground level with speed \(v\). At a time \(T\) later the competitor fires a rifle inclined at angle \(\alpha\) to the horizontal. The competitor is also at ground level and is a distance \(l\) from the launcher. The speed of the bullet leaving the rifle is \(u\). Show that, if the competitor scores a hit, then \[ l\sin\alpha-\left(vT-\tfrac{1}{2}gT^{2}\right)\cos\alpha=\frac{v-gT}{u}l. \] Suppose now that \(T=0\). Show that if the competitor can hit the target before it hits the ground then \(v < u\) and \[ \frac{2v\sqrt{u^{2}-v^{2}}}{g}>l. \]

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1992 Paper 1 Q6
D: 1500.0 B: 1519.4
small angle approximation linear approximation trigonometry parallax error analysis differentiation distant point telescopes

Explain briefly, by means of a diagram, or otherwise, why \[ \mathrm{f}(\theta+\delta\theta)\approx\mathrm{f}(\theta)+\mathrm{f}'(\theta)\delta\theta, \] when \(\delta\theta\) is small. Two powerful telescopes are placed at points \(A\) and \(B\) which are a distance \(a\) apart. A very distant point \(C\) is such that \(AC\) makes an angle \(\theta\) with \(AB\) and \(BC\) makes an angle \(\theta+\phi\) with \(AB\) produced. (A sketch of the arrangement is given in the diagram.) \noindent

\psset{xunit=0.8cm,yunit=0.8cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-4.18,-0.94)(4.4,5.22) \psline(-4,0)(4,0) \psline(-2,0)(2,5) \psline(2,5)(1,0) \rput[tl](-2.3,-0.14){\(A\)} \rput[tl](1.08,-0.14){\(B\)} \rput[tl](-1.6,0.46){\(\theta\)} \rput[tl](1.24,0.52){\(\theta+\phi\)} \rput[tl](2.14,5.1){\(C\)} \end{pspicture*} \par
If the perpendicular distance \(h\) of \(C\) from \(AB\) is very large compared with \(a\) show that \(h\) is approximately \((a\sin^{2}\theta)/\phi\) and find the approximate value of \(AC\) in terms of \(a,\theta\) and \(\phi.\) It is easy to show (but you are not asked to show it) that errors in measuring \(\phi\) are much more important than errors in measuring \(\theta.\) If we make an error of \(\delta\phi\) in measuring \(\phi\) (but measure \(\theta\) correctly) what is the approximate error in our estimate of \(AC\) and, roughly, in what proportion is it reduced by doubling the distance between \(A\) and \(B\)?

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1991 Paper 1 Q1
D: 1484.0 B: 1513.2
trigonometry trig identity proof sin²θ identity cubic equation angle sum constraint double angle formula

If \(\theta+\phi+\psi=\tfrac{1}{2}\pi,\) show that \[ \sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi=1. \] By taking \(\theta=\phi=\tfrac{1}{5}\pi\) in this equation, or otherwise, show that \(\sin\tfrac{1}{10}\pi\) satisfies the equation \[ 8x^{3}+8x^{2}-1=0. \]

Solution: \begin{align*} S &= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi \\ &= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}(\tfrac\pi2-\theta-\phi)+2\sin\theta\sin\phi\sin(\tfrac\pi2-\theta-\phi) \\ &= \sin^{2}\theta+\sin^{2}\phi+\cos^{2}(\theta+\phi)+2\sin\theta\sin\phi\cos(\theta+\phi) \\ &= \sin^{2}\theta+\sin^{2}\phi+\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right)^2+2\sin\theta\sin\phi\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right) \\ &= \sin^{2}\theta+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi-\sin^2 \theta \sin^2 \phi \\ &= \sin^{2}\theta(1-\sin^2 \phi)+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\ &= \sin^{2}\theta\cos^2 \phi+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\ &= \sin^{2}\phi+\cos^2 \phi \\ &= 1 \end{align*} Suppose \(\theta = \phi = \tfrac15 \pi, \psi = \tfrac1{10}\pi\). Also let \(s = \sin \tfrac1{10}\) \begin{align*} 1 &= 2\sin^2 \tfrac15 \pi + \sin^2 \tfrac1{10} \pi + 2 \sin^2\tfrac15 \pi \sin \tfrac1{10} \pi \\ &= 8\sin^2 \tfrac1{10} \pi \cos^2 \tfrac1{10} \pi + \sin^2 \tfrac1{10} \pi + 8 \sin^2 \tfrac1{10} \pi \cos^2 \tfrac1{10} \pi \sin \tfrac1{10} \pi \\ &= 8\sin^2 \tfrac1{10} \pi(1- \sin^2 \tfrac1{10} \pi) + \sin^2 \tfrac1{10} \pi + 8 \sin^2 \tfrac1{10} \pi (1-\sin^2 \tfrac1{10} \pi) \sin \tfrac1{10} \pi \\ &= 8s^2(1-s^2)+s^2 + 8s^2(1-s^2)s \\ &= -8 s^5 - 8 s^4 + 8 s^3 + 9 s^2 \end{align*} Therefore \(s\) is a root of \(8s^5+8s^4-8s^3-9s^2+1 = 0\), but notice that \begin{align*} 8s^5+8s^4-8s^3-9s^2+1 &= (s-1)(8 s^4 + 16 s^3 + 8 s^2 - s - 1 ) \\ &= (s-1)(s+1)(8s^3+8s^2-1) \end{align*} Therefore since \(\sin \tfrac{1}{10} \pi \neq \pm 1\) it must be a root of \(8x^3+8x^2-1=0\)

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1991 Paper 2 Q4
D: 1600.0 B: 1484.0
trigonometry roots of unity quadratic equation cos nθ geometric series sum of cosines direct substitution pentagon 17-gon

Let \(y=\cos\phi+\cos2\phi\), where \(\phi=\dfrac{2\pi}{5}.\) Verify by direct substitution that \(y\) satisfies the quadratic equation \(2y^{2}=3y+2\) and deduce that the value of \(y\) is \(-\frac{1}{2}.\) Let \(\theta=\dfrac{2\pi}{17}.\) Show that \[ \sum_{k=0}^{16}\cos k\theta=0. \] If \(z=\cos\theta+\cos2\theta+\cos4\theta+\cos8\theta,\) show that the value of \(z\) is \(-(1-\sqrt{17})/4\).

Solution: Note that \(\cos 4 \phi = \cos \phi, \cos 3 \phi = \cos 2 \phi\) \begin{align*} && LHS & = 2y^2 \\ &&&= 2 \left ( \cos \phi + \cos 2 \phi \right)^2 \\ &&&= 2 \cos ^2 \phi + 2 \cos^2 2 \phi + 4 \cos \phi \cos 2 \phi \\ &&&= \cos 2 \phi+1+ \cos4 \phi+1+2 \left ( \cos \phi + \cos 3 \phi \right) \\ &&&= \cos 2 \phi + 2 + \cos \phi + 2 \cos \phi + 2 \cos 2 \phi \\ &&&= 3(\cos \phi + \cos 2 \phi) + 2 \\ &&&= 3 y + 2 \\ &&&= RHS \end{align*} Therefore \(y\) satisfies \(2y^2 = 3y+2\), which we can solve: \begin{align*} && 0 &= 2y^2-3y-2 \\ &&&= (2y+1)(y-2) \\ \Rightarrow && y &= -\frac12,2 \end{align*} Since \(\cos \phi \neq 1\), \(y \neq 2\), therefore \(y = -\frac12\). \begin{align*} && \sum_{k=0}^{16} \cos k \theta &= \sum_{k=0}^{17} \textrm{Re} \left ( e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \sum_{k=0}^{16}e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \frac{1-e^{17 \theta i}}{1-e^{i \theta}} \right ) \\ &&&= 0 \end{align*} Suppose \(z = \cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta\) \begin{align*} z^2 &= \left (\cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta \right)^2 \\ &= \cos^2 \theta + \cos^2 2 \theta + \cos^2 4 \theta + \cos^2 8 \theta \\ & \quad \quad 2( \cos \theta \cos 2 \theta + \cos \theta \cos 4 \theta + \cos \theta \cos 8 \theta + \\ & \quad \quad \quad \cos 2 \theta \cos 4 \theta + \cos 2 \theta \cos 8 \theta + \cos 4 \theta \cos 8 \theta) \\ &= \frac12 \left (\cos 2 \theta + 1+ \cos 4 \theta + 1 + \cos 8 \theta + 1 + \cos 16 \theta + 1 \right ) + \\ &\quad \quad ( \cos \theta + \cos 3 \theta + \cos 3 \theta + \cos 5 \theta + \cos 7 \theta + \cos 9 \theta + \\ & \quad \quad \quad \cos 2 \theta + \cos 6 \theta + \cos 6 \theta + \cos 10 \theta +\cos 4 \theta + \cos 12 \theta ) \\ &= \frac12 z + 2 + \\ & \quad \quad ( \cos 3 \theta + \cos 6 \theta - \cos 8 \theta - \cos 11 \theta \\ & \quad \quad \quad - \cos 13 \theta - \cos 14 \theta - \cos 15 \theta - \cos 16 \theta - 1) \\ &= \frac12 z + 1 - z \\ &= -\frac12 z +1 \end{align*} Therefore \(z\) satisfies \(z^2=-\frac12 z+1 \Rightarrow z = \frac{-\frac12 \pm \sqrt{\frac14+4}}{2} = \frac{-1 \pm \sqrt{17}}{4}\) Therefore \(z = \frac{\sqrt{17}-1}{4}\) since \(z > 0\)

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1990 Paper 1 Q1
D: 1500.0 B: 1500.0
trigonometry optimisation area semicircle angle in semicircle double angle formula quadrilateral maximisation

\(\,\)

TikZ diagram
In the above diagram, \(ABCD\) represents a semicircular window of fixed radius \(r\) and centre \(D\), and \(AXYC\) is a quadrilateral blind. If \(\angle XDY=\alpha\) is fixed and \(\angle ADX=\theta\) is variable, determine the value of \(\theta\) which gives the blind \(maximum\) area. If now \(\alpha\) is allowed to vary but \(r\) remains fixed, find the maximum possible area of the blind.

Solution: The area for \(\alpha\) fixed is \(\frac12 r^2 \sin \alpha + \frac12 r^2 \sin \theta + \frac12 r^2 \sin (\pi - \theta - \alpha)\) So we wish to maximise \(V = \sin \theta + \sin(\pi - \theta-\alpha)\) \begin{align*} && V &= \sin \theta + \sin(\pi - \theta-\alpha)\\ &&&= 2\sin \l \frac{\pi-\alpha}2\r\cos \l \frac{2\theta + \alpha - \pi}{2}\r \end{align*} The largest \(\cos\) can be is \(1\) when \(\displaystyle 2\theta + \alpha - \pi = 0 \Rightarrow \theta = \frac{\pi-\alpha}2\). (ie we split the remaining area exactly in half). We are now trying to maximise \(W = \sin \alpha + 2 \sin \frac{\pi - \alpha}2\) ie \begin{align*} && W &= \sin \alpha + 2 \cos \frac{\alpha}{2} \\ \Rightarrow && \frac{\d W}{\d \alpha} &= \cos \alpha-\sin \frac{\alpha}{2} \\ &&&= 1-2 \sin^2 \frac{\alpha}{2} - \sin \frac{\alpha}{2} \\ &&&= (1+\sin \frac{\alpha}{2})(1-2\sin \frac{\alpha}{2}) \end{align*} Therefore \(\frac{\alpha}{2} = -\frac{3\pi}{2}, \frac{\alpha}{2} = \frac{\pi}{6}, \frac{5\pi}{6} \Rightarrow \alpha = -3\pi, \frac{\pi}{3}, \frac{5\pi}{3}\). The only turning point in our range is \(\frac{\pi}{3}\). This is obvious a a maximum by symmetry or checking the end points, but we could also check the second derivative \(\frac{\d^2 W}{\d \alpha^2} = -\sin \frac{\pi}{3}-\cos \frac{\pi}{3} < 0\) so we have a maximum. Therefore the largest possible area is: \(\displaystyle \frac{3\sqrt{3}}{4}r^2\)

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1990 Paper 1 Q12
D: 1516.0 B: 1484.0
energy conservation elastic spring Hooke's law equilibrium potential energy modulus of elasticity trigonometry optimisation

\(\,\)

TikZ diagram
In the above diagram, \(ABC\) represents a light spring of natural length \(2l\) and modulus of elasticity \(\lambda,\) which is coiled round a smooth fixed horizontal rod. \(B\) is the midpoint of \(AC.\) The two ends of a light inelastic string of length \(2l\) are attached to the spring at \(A\) and \(C\). A particle of mass \(m\) is fixed to the string at \(D\), the midpoint of the string. The system can be in equilibrium with the angle \(CAD\) equal to \(\pi/6.\) Show that \[ mg=\lambda\left(\frac{2}{\sqrt{3}}-1\right). \] Write the length \(AC\) as \(2xl\), obtain an expression for the potential energy of the system as a function of \(x\). The particle is held at \(B\), and the spring is restored to its natural length \(2l.\) The particle is then released and falls vertically. Obtain an equation satisfied by \(x\) when the particle next comes to rest. Verify numerically that a possible solution for \(x\) is approximately \(0.66.\)

Solution:

TikZ diagram
\(|AB| = l \cos \tfrac{\pi}{6} = \frac{\sqrt{3}}{2}l\) therefore \(|AC| = \sqrt{3}l\) and the compression is \((2l - \sqrt{3}l)\) and so \(T_2 = \frac{\lambda}{2l} (2l - \sqrt{3}l) = \frac12\lambda(2- \sqrt{3})\) \begin{align*} \text{N2}(\rightarrow, A): && T_1 \cos \tfrac{\pi}{6} - T_2 &= 0 \\ \Rightarrow && T_1 &= \frac12 \frac{2\lambda(2-\sqrt{3})}{\sqrt{3}} \\ &&&= \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \\ \text{N2}(\uparrow, D): && 2T_1 \cos \frac{\pi}{3} - mg &= 0 \\ \Rightarrow && mg &= \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \end{align*} Suppose \(|AC| = 2xl\), then: \begin{array}{c|c} \text{energy} & \\ \hline \text{GPE} & -mg \sqrt{l^2 - x^2l^2} \\ \text{EPE} & \frac12 \frac{\lambda (2l - 2lx)^2}{2l} \\ \text{KE} & \frac12 m v^2 \end{array} Therefore \[ E = \frac12 mv^2 + \lambda l (1-x)^2-mgl \sqrt{1-x^2}\] Initially, \(E = 0 + 0 + 0 = 0\). When the particle first comes to rest: \begin{align*} \text{COE}: && 0 &= E \\ &&&= \lambda l^2 (1-x)^2 - mgl \sqrt{1-x^2} \\ &&&= \lambda l (1-x)^2 - l \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \sqrt{1-x^2} \\ \Rightarrow && (1-x)^2 &= \sqrt{1-x^2} \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \Rightarrow && (1-x)^2(1-x^2)^{-1/2} &= \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \Rightarrow && (1-2x+x^2)(1+\frac12 x^2+\cdots) &= \left ( \frac{2}{\sqrt{3}} - 1 \right) \\ \end{align*} If \(x = \frac23\) then \((1-x)^2(1-x^2)^{-1/2} = \frac19 \cdot \left ( \frac{5}{9} \right)^{-1/2} = \frac{\sqrt{5}}{15}\) If \(2\sqrt{3}-3 \approx \frac{\sqrt{5}}5\) we're done.

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1990 Paper 2 Q2
D: 1600.0 B: 1500.0
trigonometry product-to-sum formulas circle theorems Ptolemy's theorem inscribed angle theorem geometric proof sine rule cyclic quadrilateral

Prove that if \(A+B+C+D=\pi,\) then \[ \sin\left(A+B\right)\sin\left(A+D\right)-\sin B\sin D=\sin A\sin C. \] The points \(P,Q,R\) and \(S\) lie, in that order, on a circle of centre \(O\). Prove that \[ PQ\times RS+QR\times PS=PR\times QS. \]

Solution: \begin{align*} \sin(A+B)\sin(A+D) - \sin B \sin D &= \sin (A+B)\sin(\pi - B-C) - \sin B \sin (\pi - A - B - C) \\ &= \sin (A+B)\sin(B+C) - \sin B \sin(A+B+C) \\ &= \sin(A+B)\sin (B+C) - \sin B (\sin (A+B)\cos C +\cos(A+B) \sin C) \\ &= \sin(A+B)\cos B \sin C + \cos(A+B)\sin B \sin C \\ &= \sin A \sin C \cos^2 B + \cos A \sin B \cos B \sin C - \cos A \cos B \sin B \sin C + \sin A \sin^2 B \sin C \\ &= \sin A \sin C (\cos^2 B + \sin^2 B) \\ &= \sin A \sin C \end{align*}

TikZ diagram
Using the extended form of the sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) where \(R\) is the circumradius, we have \begin{align*} PR \times QS &= 2R \sin (A+D) \times 2R \sin (A+B) \\ &= 4R^2 \l \sin A \sin C + \sin B \sin D \r \\ &= 2R \sin A \times 2R \sin C + 2R \sin B 2R \sin D \\ &= PS \times QR + PQ \times RS \end{align*} as required.

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1989 Paper 2 Q1
D: 1600.0 B: 1516.0
trigonometry cos nθ triple angle formula cubic equation substitution solving cubics via trig Chebyshev polynomial

Prove that \(\cos3\theta=4\cos^{3}\theta-3\cos\theta\). Show how the cubic equation \[ 24x^{3}-72x^{2}+66x-19=0\tag{*} \] can be reduced to the form \[ 4z^{3}-3z=k \] by means of the substitution \(y=x+a\) and \(z=by\), for suitable values of the constants \(a\) and \(b\). Hence find the three roots of the equation \((*)\), to three significant figures. Show, by means of a counterexample, or otherwise, that not all cubic equations of the form \[ x^{3}+\alpha x^{2}+\beta x+\gamma=0 \] can be solved by this method.

Solution: \begin{align*} \cos 3\theta &= \cos 2\theta\cos\theta - \sin 2\theta \sin \theta \\ &= (2\cos^2\theta-1)\cos \theta - 2\cos \theta \sin^2 \theta \\ &= 2\cos^3\theta-\cos \theta - 2\cos \theta(1- \cos^2 \theta) \\ &= 4\cos^3 \theta - 3\cos \theta \end{align*} \begin{align*} 0 &= 24x^{3}-72x^{2}+66x-19 \\ &= 24(y+1)^3-72(y+1)^2+66(y+1)-19 \\ &= 24(y^3+3y^2+3y+1)-72(y^2+2y+1)+66(y+1)-19\\ &= 24y^3+(72-144+66)y+(24-72+66-19) \\ &= 24y^3-6y-1 \\ &= 24b'^3z^3 - 6b'z - 1 \\ &= \frac{2}{\sqrt{3}}(4 z^3 -3z) - 1 \\ \end{align*} Therefore if \(b = \sqrt{3}, a = 1\), we have: \(4z^3 - 3z = \frac{\sqrt{3}}{2}\) So if \(z = \cos \theta \Rightarrow \cos 3\theta = \frac{\sqrt{3}}2 \Rightarrow 3 \theta = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6} \Rightarrow \theta = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}\). Since \(\frac{\cos x}{\sqrt{3}} < 1\) we only need to approximate the first part to 2 significant figures. Therefore: \begin{align*} \sqrt{3} &\approx 1 + \frac{1}{1 + \frac{1}{2+\frac11}} = \frac{7}{4} = 1.75\\ \cos \tfrac{\pi}{18} &\approx \cos \frac{1}{6} \approx 1 - \frac{1}{2} \frac{1}{6^2} = \frac{71}{72} \approx 1 - \frac{1}{70} = 1 - 0.014 = 0.986 \\ \frac{\cos \tfrac{\pi}{18}}{\sqrt{3}} & \approx \frac{.986}{1.75} = 0.57 \\ \end{align*} Final answers: \(1.57, 0.803, 0.629\). We wouldn't be able to solve \(x^3 + 1= 0\) using this method, as we would have 2 non-real roots

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1988 Paper 1 Q6
D: 1500.0 B: 1516.0
complex numbers Argand diagram vectors perpendicular convex hexagon geometric proof translation trigonometry

The complex numbers \(z_{1},z_{2},\ldots,z_{6}\) are represented by six distinct points \(P_{1},P_{2},\ldots,P_{6}\) in the Argand diagram. Express the following statements in terms of complex numbers:

  1. \(\overrightarrow{P_{1}P_{2}}=\overrightarrow{P_{5}P_{4}}\) and \(\overrightarrow{P_{2}P_{3}}=\overrightarrow{P_{6}P_{5}}\,\);
  2. \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\).
If \((i)\) holds, show that \(\overrightarrow{P_{3}P_{4}}=\overrightarrow{P_{1}P_{6}}\,\). Suppose that the statements \((i)\) and \((ii)\) both hold, and that \(z_{1}=0,\) \(z_{2}=1,\) \(z_{3}=z,\) \(z_{5}=\mathrm{i}\) and \(z_{6}=w.\) Determine the conditions which \(\mathrm{Re}(z)\) and \(\mathrm{Re}(w)\) must satisfy in order that \(P_{1}P_{2}P_{3}P_{4}P_{5}P_{6}\) should form a convex hexagon. Find the distance between \(P_{3}\) and \(P_{6}\) when \(\tan(\angle P_{3}P_{2}P_{6})=-2/3.\)

Solution:

  1. \(\overrightarrow{P_{1}P_{2}}=\overrightarrow{P_{5}P_{4}}\) is equivalent to \(z_2 - z_1 = z_4 - z_5\). \(\overrightarrow{P_{2}P_{3}}=\overrightarrow{P_{6}P_{5}}\) is equivalent to \(z_3-z_2 = z_5 - z_6\).
  2. \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) is equivalent to \(\frac{z_4 - z_2}{z_6-z_3} \in i\mathbb{R}\)
If \(z_2 - z_1 =z_4 - z_5\) and \(z_3-z_2 = z_5 - z_6\) then adding we get \(z_3 - z_1 = z_4 - z_6\) or \(z_4 - z_3 = z_6-z_1\), which is equivalent to \(\overrightarrow{P_{3}P_{4}}=\overrightarrow{P_{1}P_{6}}\,\).
TikZ diagram
\(\textrm{Re}(z) > 1, \textrm{Re}(w) < 0, \textrm{Re}(z) +\textrm{Re}(w)=1\). (We only need one of the first two constraints, since the other is implied by the former). Since \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) we must have that \(\textrm{Im}(z) = \textrm{Im}(w)\). Combined with the vector logic we must have that \(\textrm{Im}(z) = \frac12\). Let \(z = a + \frac12i\) and \(w = (1-a) + \frac12i\). Since \(w - 1 = k(3-2i)(z-1)\) (the angle constraint) we must have that: \begin{align*} &&-a+\frac12i &= k(3-2i)((a-1) \frac12i) \\ &&&= k( 3 a - 2+(\frac72 - 2 a)i) \\ \Rightarrow && \frac{3a-2}{-a} &= \frac{\frac72-2a}{\frac12} \\ \Rightarrow && 3a-2&= 4a^2-7a \\ \Rightarrow && 0 &= 4a^2-10a+2 \\ \Rightarrow && a &= \frac{5 \pm \sqrt{17}}{4} \end{align*} Since \(a > 1, a = \frac{5 +\sqrt{17}}{4}\) and the distance is: \begin{align*} \left | z - w \right | &= | a+\frac12i - ((1-a) +\frac12i ) | \\ &= |2a-1| \\ &= \frac{5+\sqrt{17}}{2}-1 \\ &= \frac{3+\sqrt{17}}{2} \end{align*}

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1987 Paper 2 Q2
D: 1500.0 B: 1500.0
trigonometry trigonometric identity proof sum-product formulas factorization angle conditions

Show that if at least one of the four angles \(A\pm B\pm C\) is a multiple of \(\pi\), then \begin{alignat*}{1} \sin^{4}A+\sin^{4}B+\sin^{4}C & -2\sin^{2}B\sin^{2}C-2\sin^{2}C\sin^{2}A\\ & -2\sin^{2}A\sin^{2}B+4\sin^{2}A\sin^{2}B\sin^{2}C=0. \end{alignat*}

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