Problems

2024 Paper 2 Q3

D: 1500.0 B: 1500.0

coordinate geometry unit circle trigonometric identities half-angle formulae geometric transformations rotation tangent half-angle substitution

The unit circle is the circle with radius 1 and centre the origin, $O$. $N$ and $P$ are distinct points on the unit circle. $N$ has coordinates $(-1, 0)$, and $P$ has coordinates $(\cos\theta, \sin\theta)$, where $-\pi < \theta < \pi$. The line $NP$ intersects the $y$-axis at $Q$, which has coordinates $(0, q)$.

Show that $q = \tan\frac{1}{2}\theta$.
In this part, $q \neq 1$.
1. Let $\mathrm{f}_1(q) = \dfrac{1+q}{1-q}$. Show that $\mathrm{f}_1(q) = \tan\frac{1}{2}\!\left(\theta + \frac{1}{2}\pi\right)$.
2. Let $Q_1$ be the point with coordinates $(0, \mathrm{f}_1(q))$ and $P_1$ be the point of intersection (other than $N$) of the line $NQ_1$ and the unit circle. Describe geometrically the relationship between $P$ and $P_1$.
1. $P_2$ is the image of $P$ under an anti-clockwise rotation about $O$ through angle $\frac{1}{3}\pi$. The line $NP_2$ intersects the $y$-axis at the point $Q_2$ with co-ordinates $(0, \mathrm{f}_2(q))$. Find $\mathrm{f}_2(q)$ in terms of $q$, for $q \neq \sqrt{3}$.
2. In this part, $q \neq -1$. Let $\mathrm{f}_3(q) = \dfrac{1-q}{1+q}$, let $Q_3$ be the point with coordinates $(0, \mathrm{f}_3(q))$ and let $P_3$ be the point of intersection (other than $N$) of the line $NQ_3$ and the unit circle. Describe geometrically the relationship between $P$ and $P_3$.
3. In this part, $0 < q < 1$. Let $\mathrm{f}_4(q) = \mathrm{f}_2^{-1}\!\Big(\mathrm{f}_3\!\big(\mathrm{f}_2(q)\big)\Big)$, let $Q_4$ be the point with coordinates $(0, \mathrm{f}_4(q))$ and let $P_4$ be the point of intersection (other than $N$) of the line $NQ_4$ and the unit circle. Describe geometrically the relationship between $P$ and $P_4$.

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2022 Paper 3 Q7

D: 1500.0 B: 1500.0

vector product cross product unit vector rotation geometric transformation vector triple product reflection

Let $\mathbf{n}$ be a vector of unit length in three dimensions. For each vector $\mathbf{r}$, $\mathrm{f}(\mathbf{r})$ is defined by \[ \mathrm{f}(\mathbf{r}) = \mathbf{n} \times \mathbf{r}\,. \]

Given that \[ \mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \quad \text{and} \quad \mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \] show that the $x$-component of $\mathrm{f}(\mathrm{f}(\mathbf{r}))$ is $-x(b^2+c^2)+aby+acz$. Show further that \[ \mathrm{f}(\mathrm{f}(\mathbf{r})) = (\mathbf{n}.\mathbf{r})\mathbf{n} - \mathbf{r}\,. \] Explain, by means of a diagram, how $\mathrm{f}(\mathrm{f}(\mathbf{r}))$ is related to $\mathbf{n}$ and $\mathbf{r}$.
Let $R$ be the point with position vector $\mathbf{r}$ and $P$ be the point with position vector $\mathrm{g}(\mathbf{r})$, where $\mathrm{g}$ is defined by \[ \mathrm{g}(\mathbf{s}) = \mathbf{s} + \sin\theta\,\mathrm{f}(\mathbf{s}) + (1-\cos\theta)\,\mathrm{f}(\mathrm{f}(\mathbf{s}))\,. \] By considering $\mathrm{g}(\mathbf{n})$ and $\mathrm{g}(\mathbf{r})$ when $\mathbf{r}$ is perpendicular to $\mathbf{n}$, state, with justification, the geometric transformation which maps $R$ onto $P$.
Let $R$ be the point with position vector $\mathbf{r}$ and $Q$ be the point with position vector $\mathrm{h}(\mathbf{r})$, where $\mathrm{h}$ is defined by \[ \mathrm{h}(\mathbf{s}) = -\mathbf{s} - 2\,\mathrm{f}(\mathrm{f}(\mathbf{s}))\,. \] State, with justification, the geometric transformation which maps $R$ onto $Q$.

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2020 Paper 3 Q3

D: 1500.0 B: 1500.0

complex numbers rotation centroid equilateral triangle parallelogram roots of unity geometric transformations

Given distinct points $A$ and $B$ in the complex plane, the point $G_{AB}$ is defined to be the centroid of the triangle $ABK$, where the point $K$ is the image of $B$ under rotation about $A$ through a clockwise angle of $\frac{1}{3}\pi$. Note: if the points $P$, $Q$ and $R$ are represented in the complex plane by $p$, $q$ and $r$, the centroid of triangle $PQR$ is defined to be the point represented by $\frac{1}{3}(p+q+r)$.

If $A$, $B$ and $G_{AB}$ are represented in the complex plane by $a$, $b$ and $g_{ab}$, show that \[ g_{ab} = \frac{1}{\sqrt{3}}(\omega a + \omega^* b), \] where $\omega = \mathrm{e}^{\frac{\mathrm{i}\pi}{6}}$.
The quadrilateral $Q_1$ has vertices $A$, $B$, $C$ and $D$, in that order, and the quadrilateral $Q_2$ has vertices $G_{AB}$, $G_{BC}$, $G_{CD}$ and $G_{DA}$, in that order. Using the result in part (i), show that $Q_1$ is a parallelogram if and only if $Q_2$ is a parallelogram.
The triangle $T_1$ has vertices $A$, $B$ and $C$ and the triangle $T_2$ has vertices $G_{AB}$, $G_{BC}$ and $G_{CA}$. Using the result in part (i), show that $T_2$ is always an equilateral triangle.

Solution:

Note that the vector $\overrightarrow{AB}$ is $b-a$, and if we rotate this by $\frac13\pi$ we get $e^{-i\pi/3}(b-a)$ after rotating it. Therefore the point $K$ is represented by $a + e^{-i\pi/3}(b-a)$ and so $G_{AB}$ is \begin{align*} && g_{ab} &= \tfrac13(a + b + a + e^{-i\pi/3}(b-a)) \\ &&&= \tfrac13((1+ e^{-i\pi/3})b+(2-e^{-i\pi/3})a)\\ &&&= \tfrac13((1+\tfrac12 - \tfrac{\sqrt3}{2}i)b + ((2-\tfrac12+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac13((\tfrac32 - \tfrac{\sqrt3}{2}i)b + ((\tfrac32+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac1{\sqrt3}((\tfrac{\sqrt3}2 - \tfrac{1}{2}i)b + ((\tfrac{\sqrt3}2+\tfrac{1}{2}i)a) \\ &&&= \frac{1}{\sqrt3}(\omega^* b + \omega a) \end{align*}
First note that $Q_1$ is a parallelogram iff $c - a = (b-a) + (d-a)$ ie $a + c = b+d$ (indeed this is true for all quadrilaterals), so. \begin{align*} && Q_1 &\text{ is a parallelogram} \\ \Longleftrightarrow && a + b &= c + d \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega - \omega^*)(a + c) &= \frac{1}{\sqrt{3}}(\omega -\omega^*)(b + d) \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega a + \omega^*b)+\frac{1}{\sqrt{3}}(\omega c + \omega^*d) &=\frac{1}{\sqrt{3}}(\omega b + \omega^*c)+\frac{1}{\sqrt{3}}(\omega d + \omega^*a) \\ \Longleftrightarrow && g_{ab}+g_{cd} &=g_{bc}+g_{da} \\ \Longleftrightarrow && Q_2 &\text{ is a parallelogram} \\ \end{align*}
We consider $\frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}}$ so \begin{align*} && \frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}} &= \frac{(\omega a + \omega^*b)-(\omega b + \omega^* c)}{(\omega c + \omega^*a)-(\omega b + \omega^* c)} \\ &&&= \frac{\omega a- \omega^* c -(\omega- \omega^*)b }{\omega^*a-\omega b -(\omega^* -\omega )c} \\ &&&= \frac{\omega^2 a- c -(\omega^2- 1)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- \omega^4 c -(1- \omega^4)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- (1-\omega^2) c -\omega^2b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&= \omega^2 \end{align*} Therefore the triangle is equilateral.

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2020 Paper 3 Q4

D: 1500.0 B: 1500.0

3x3 matrices reflection rotation plane geometry dot product unit vector linear transformations

The plane $\Pi$ has equation $\mathbf{r} \cdot \mathbf{n} = 0$ where $\mathbf{n}$ is a unit vector. Let $P$ be a point with position vector $\mathbf{x}$ which does not lie on the plane $\Pi$. Show that the point $Q$ with position vector $\mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\mathbf{n}$ lies on $\Pi$ and that $PQ$ is perpendicular to $\Pi$.

Let transformation $T$ be a reflection in the plane $ax+by+cz=0$, where $a^2+b^2+c^2=1$. Show that the image of $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ under $T$ is $\begin{pmatrix} b^2+c^2-a^2 \\ -2ab \\ -2ac \end{pmatrix}$, and find the images of $\mathbf{j}$ and $\mathbf{k}$ under $T$. Write down the matrix $\mathbf{M}$ which represents transformation $T$.
The matrix \[ \begin{pmatrix} 0.64 & 0.48 & 0.6 \\ 0.48 & 0.36 & -0.8 \\ 0.6 & -0.8 & 0 \end{pmatrix} \] represents a reflection in a plane. Find the cartesian equation of the plane.
The matrix $\mathbf{N}$ represents a rotation through angle $\pi$ about the line through the origin parallel to $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$, where $a^2+b^2+c^2=1$. Find the matrix $\mathbf{N}$.
Identify the single transformation which is represented by the matrix $\mathbf{NM}$.

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2017 Paper 3 Q2

D: 1700.0 B: 1500.1

complex numbers rotation transformation composition of transformations geometric proof exponential form commutativity

The transformation $R$ in the complex plane is a rotation (anticlockwise) by an angle $\theta$ about the point represented by the complex number $a$. The transformation $S$ in the complex plane is a rotation (anticlockwise) by an angle $\phi$ about the point represented by the complex number $b$.

The point $P$ is represented by the complex number~$z$. Show that the image of $P$ under $R$ is represented by \[ \e^{{\mathrm i} \theta}z + a(1-\e^{{\rm i} \theta})\,. \]
Show that the transformation $SR$ (equivalent to $R$ followed by $S$) is a rotation about the point represented by $c$, where \[ %\textstyle c\,\sin \tfrac12 (\theta+\phi) = a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi \,, \] provided $\theta+\phi \ne 2n\pi$ for any integer $n$. What is the transformation $SR$ if $\theta +\phi = 2\pi$?
Under what circumstances is $RS =SR$?

Solution:

We can map $a \mapsto 0$, rotate the whole plane, then shift the plane back to $a$, so $z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})$
$z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})$ \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle $\phi + \theta$ and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If $\phi + \theta = 2\pi$, then $z \mapsto z + (b-a)(1 - e^{i\phi})$ which is a translation.
If $\phi + \theta \neq 2 \pi$ then $RS = ST$ if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If $\phi + \theta \neq 2 \pi$ then $RS = ST$ if $b = a$ or $e^{i\phi} = e^{i \theta}$ ie rotation through the same angle.

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2014 Paper 3 Q5

D: 1700.0 B: 1500.0

complex numbers parallelogram square geometric proof Argand diagram midpoint rotation quadrilateral

A quadrilateral drawn in the complex plane has vertices $A$, $B$, $C$ and $D$, labelled anticlockwise. These vertices are represented, respectively, by the complex numbers $a$, $b$, $c$ and $d$. Show that $ABCD$ is a parallelogram (defined as a quadrilateral in which opposite sides are parallel and equal in length) if and only if $a+c =b+d\,$. Show further that, in this case, $ABCD$ is a square if and only if ${\rm i}(a-c)=b-d$. Let $PQRS$ be a quadrilateral in the complex plane, with vertices labelled anticlockwise, the internal angles of which are all less than $180^\circ$. Squares with centres $X$, $Y$, $Z$ and $T$ are constructed externally to the quadrilateral on the sides $PQ$, $QR$, $RS$ and $SP$, respectively.

If $P$ and $Q$ are represented by the complex numbers $p$ and $q$, respectively, show that $X$ can be represented by \[ \tfrac 12 \big( p(1+{\rm i} ) + q (1-{\rm i})\big) \,. \]
Show that $XY\!ZT$ is a square if and only if $PQRS$ is a parallelogram.

Solution: The vector representing the side $AB$ is $b - a$ and the vector representing the side $DC$ is $c - d$. $ABCD$ is a parallelogram if and only if these opposite sides are parallel and equal in length, which is given by $b - a = c - d$, or equivalently $a + c = b + d$. Similarly, if $a + c = b + d$, then $c - b = d - a$, so the side $BC$ is parallel and equal in length to the side $AD$. Thus, $a + c = b + d$ is the necessary and sufficient condition for $ABCD$ to be a parallelogram. In a parallelogram, the shape is a square if and only if the diagonals are equal in length and perpendicular to each other. The diagonals are represented by the vectors $c - a$ and $d - b$. For these to be equal in length and perpendicular, one must be a $90^\circ$ rotation of the other. Since $A, B, C, D$ are labeled anticlockwise, a $90^\circ$ anticlockwise rotation of the vector $\vec{AC}$ (which is $c-a$) would point in the direction of $\vec{DB}$ (which is $b-d$ if we consider the relative orientation). Specifically: $i(c - a) = d - b \implies -i(a - c) = d - b \implies i(a - c) = b - d$. Thus, $ABCD$ is a square if and only if $i(a - c) = b - d$.

The midpoint of the side $PQ$ is $\frac{1}{2}(p + q)$. To find the centre $X$ of the square built externally on $PQ$, we start at the midpoint and move a distance equal to half the side length in a direction perpendicular to $PQ$. Since $P, Q, R, S$ are anticlockwise, the outward direction is a $90^\circ$ clockwise rotation of the vector $\vec{PQ}$. A clockwise rotation of $90^\circ$ corresponds to multiplication by $-i$. \[ x = \frac{p+q}{2} + (-i)\left(\frac{q-p}{2}\right) = \frac{p + q - iq + ip}{2} = \frac{1}{2} \big( p(1+i) + q(1-i) \big) \]
From part (i), we have the representations for the centres: \begin{align*} x &= \tfrac{1}{2}(p(1+i) + q(1-i)) \\ y &= \tfrac{1}{2}(q(1+i) + r(1-i)) \\ z &= \tfrac{1}{2}(r(1+i) + s(1-i)) \\ t &= \tfrac{1}{2}(s(1+i) + p(1-i)) \end{align*} As shown in the first part of the problem, $XYZT$ is a square if and only if: (1) $x+z = y+t$ (it is a parallelogram) (2) $i(x-z) = y-t$ (it is a square) First, examine condition (1): \begin{align*} x+z - (y+t) &= \tfrac{1}{2} \big[ (p+r)(1+i) + (q+s)(1-i) - (q+s)(1+i) - (r+p)(1-i) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(1+i - (1-i)) - (q+s)(1+i - (1-i)) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(2i) - (q+s)(2i) \big] \\ &= i(p+r - (q+s)) \end{align*} Thus, $x+z = y+t$ if and only if $p+r = q+s$, which is the condition that $PQRS$ is a parallelogram. Next, examine condition (2): \begin{align*} i(x-z) &= \tfrac{1}{2} i \big[ p(1+i) + q(1-i) - r(1+i) - s(1-i) \big] \\ &= \tfrac{1}{2} \big[ p(i-1) + q(i+1) - r(i-1) - s(i+1) \big] \\ y-t &= \tfrac{1}{2} \big[ q(1+i) + r(1-i) - s(1+i) - p(1-i) \big] \\ \text{So, } i(x-z) - (y-t) &= \tfrac{1}{2} \big[ p(i-1 + 1-i) + q(i+1 - 1-i) + r(-i+1 - 1+i) + s(-i-1 + 1+i) \big] \\ &= 0 \end{align*} Since $i(x-z) = y-t$ is an identity (always true for any $PQRS$), $XYZT$ is a square if and only if it is a parallelogram. As established above, this occurs if and only if $PQRS$ is a parallelogram.

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2008 Paper 1 Q7

D: 1484.0 B: 1500.0

trigonometry rotation equilateral triangle perpendicular distance coordinate geometry geometric proof compound angle formulae if and only if

The point $P$ has coordinates $(x,y)$ with respect to the origin $O$. By writing $x=r\cos\theta$ and $y=r\sin\theta$, or otherwise, show that, if the line $OP$ is rotated by $60^\circ$ clockwise about $O$, the new $y$-coordinate of $P$ is $\frac12(y-\sqrt3\,x)$. What is the new $y$-coordinate in the case of an anti-clockwise rotation by $60^\circ\,$? An equilateral triangle $OBC$ has vertices at $O$, $(1,0)$ and $(\frac12,\frac12 \sqrt3)$, respectively. The point $P$ has coordinates $(x,y)$. The perpendicular distance from $P$ to the line through $C$ and $O$ is $h_1$; the perpendicular distance from $P$ to the line through $O$ and $B$ is $h_2$; and the perpendicular distance from $P$ to the line through $B$ and $C$ is $h_3$. Show that $h_1=\frac12 \big\vert y-\sqrt3\,x\big\vert$ and find expressions for $h_2$ and $h_3$. Show that $h_1+h_2+h_3=\frac12 \sqrt3$ if and only if $P$ lies on or in the triangle $OBC$.

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2008 Paper 3 Q7

D: 1700.0 B: 1500.0

complex numbers Argand diagram equilateral triangle rotation geometric proof parallelogram centroid Napoleon's theorem

The points $A$, $B$ and $C$ in the Argand diagram are the vertices of an equilateral triangle described anticlockwise. Show that the complex numbers $a$, $b$ and $c$ representing $A$, $B$ and $C$ satisfy \[2c= (a+b) +\mathrm{i}\sqrt3(b-a).\] Find a similar relation in the case that $A$, $B$ and $C$ are the vertices of an equilateral triangle described clockwise.

The quadrilateral $DEFG$ lies in the Argand diagram. Show that points $P$, $Q$, $R$ and $S$ can be chosen so that $PDE$, $QEF$, $RFG$ and $SGD$ are equilateral triangles and $PQRS$ is a parallelogram.
The triangle $LMN$ lies in the Argand diagram. Show that the centroids $U$, $V$ and $W$ of the equilateral triangles drawn externally on the sides of $LMN$ are the vertices of an equilateral triangle. \noindent [{\bf Note:} The {\em centroid} of a triangle with vertices represented by the complex numbers $x$,~$y$ and~$z$ is the point represented by $\frac13(x+y+z)\,$.]

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2003 Paper 2 Q4

D: 1600.0 B: 1484.0

coordinate geometry circle area of segment integration inverse trigonometric functions translation rotation perpendicular distance

The line $y=d\,$, where $d>0\,$, intersects the circle $x^2+y^2=R^2$ at $G$ and $H$. Show that the area of the minor segment $GH$ is equal to \begin{equation} R^2\arccos \left({d \over R}\right) -d\sqrt{R^2 - d^2}\;. \tag {$*$} \end{equation} In the following cases, the given line intersects the given circle. Determine how, in each case, the expression $(*)$ should be modified to give the area of the minor segment.

Line: $y=c\,$; \ \ \ circle: $(x-a)^2+(y-b)^2=R^2\,$.
Line: $y=mx+c\, $; \ \ \ circle: $x^2+y^2=R^2\,$.
Line: $y=mx+c\,$; \ \ \ circle: $(x-a)^2+(y-b)^2=R^2\,$.

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2001 Paper 2 Q7

D: 1600.0 B: 1470.8

complex numbers Argand diagram regular polygon rotation circumcentre geometric construction multiplication by cis

In an Argand diagram, $O$ is the origin and $P$ is the point $2+0\mathrm{i}$. The points $Q$, $R$ and $S$ are such that the lengths $OP$, $PQ$, $QR$ and $RS$ are all equal, and the angles $OPQ$, $PQR$ and $QRS$ are all equal to ${5{\pi}}/6$, so that the points $O$, $P$, $Q$, $R$ and $S$ are five vertices of a regular 12-sided polygon lying in the upper half of the Argand diagram. Show that $Q$ is the point $2 + \sqrt 3 + \mathrm{i}$ and find $S$. The point $C$ is the centre of the circle that passes through the points $O$, $P$ and $Q$. Show that, if the polygon is rotated anticlockwise about $O$ until $C$ first lies on the real axis, the new position of $S$ is $$ - \tfrac{1}{2} (3\sqrt 2+ \sqrt6)(\sqrt3-\mathrm{i})\;. $$

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2000 Paper 3 Q3

D: 1700.0 B: 1484.0

complex numbers Argand diagram equilateral triangle roots of unity Napoleon's theorem geometric proof rotation cube roots of unity

Given that $\alpha = \e^{\mathrm{i} \pi/3}$ , prove that $1 + \alpha^2 = \alpha$. A triangle in the Argand plane has vertices $A$, $B$, and $C$ represented by the complex numbers $p$, $q\alpha^2$ and $- r\alpha$ respectively, where $p$, $q$ and $r$ are positive real numbers. Sketch the triangle~$ABC$. Three equilateral triangles $ABL$, $BCM$ and $CAN$ (each lettered clockwise) are erected on sides $AB$, $BC$ and $CA$ respectively. Show that the complex number representing $N$ is \mbox{$( 1 - \alpha) p- \alpha^2 r$} and find similar expressions for the complex numbers representing $L$ and $M$. Show that lines $LC$, $MA$ and $NB$ all meet at the origin, and that these three line segments have the common length $p+q+r$.

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2000 Paper 3 Q11

D: 1700.0 B: 1484.0

moments of inertia rigid body rotation friction angular acceleration normal reaction energy conservation critical coefficient of friction

A thin beam is fixed at a height $2a$ above a horizontal plane. A uniform straight rod $ACB$ of length $9a$ and mass $m$ is supported by the beam at $C$. Initially, the rod is held so that it is horizontal and perpendicular to the beam. The distance $AC$ is $3a$, and the coefficient of friction between the beam and the rod is $\mu$. The rod is now released. Find the minimum value of $\mu$ for which $B$ strikes the horizontal plane before slipping takes place at $C$.

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1998 Paper 1 Q5

D: 1484.0 B: 1529.9

complex numbers Argand diagram regular hexagon rotation rationalising denominators real and imaginary parts

In the Argand diagram, the points $Q$ and $A$ represent the complex numbers $4+6i$ and $10+2i$. If $A$, $B$, $C$, $D$, $E$, $F$ are the vertices, taken in clockwise order, of a regular hexagon (regular six-sided polygon) with centre $Q$, find the complex number which represents $B$.
Let $a$, $b$ and $c$ be real numbers. Find a condition of the form $Aa+Bb+Cc=0$, where $A$, $B$ and $C$ are integers, which ensures that \[\frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i}\] is real.

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1998 Paper 3 Q5

D: 1700.0 B: 1516.0

Taylor series matrix exponential matrix rotation shear transformation commutativity geometric proof 2x2 matrix

The exponential of a square matrix ${\bf A}$ is defined to be $$ \exp ({\bf A}) = \sum_{r=0}^\infty {1\over r!} {\bf A}^r \,, $$ where ${\bf A}^0={\bf I}$ and $\bf I$ is the identity matrix. Let $$ {\bf M}=\left(\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array} \right) \,. $$ Show that ${\bf M}^2=-{\bf I}$ and hence express $\exp({\theta {\bf M}})$ as a single $2\times 2$ matrix, where $\theta$ is a real number. Explain the geometrical significance of $\exp({\theta {\bf M}})$. Let $$ {\bf N}=\left(\begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right) \,. $$ Express similarly $\exp({s{\bf N}})$, where $s$ is a real number, and explain the geometrical significance of $\exp({s{\bf N}})$. For which values of $\theta$ does $$ \exp({s{\bf N}})\; \exp({\theta {\bf M}})\, = \, \exp({\theta {\bf M}})\;\exp({s{\bf N}}) $$ for all $s$? Interpret this fact geometrically.

Solution: \begin{align*} \mathbf{M}^2 &= \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}^2 \\ &= \begin{pmatrix} 0 \cdot 0 + (-1) \cdot 1 & 0 \cdot (-1) + (-1) \cdot 0 \\ 1 \cdot 0 + 0 \cdot 1 & 1 \cdot (-1) + 0 \cdot 0 \end{pmatrix} \\ &= \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix} \\ &= - \mathbf{I} \end{align*} \begin{align*} \exp(\theta \mathbf{M}) &= \sum_{r=0}^\infty \frac1{r!} (\theta \mathbf{M})^r \\ &= \sum_{r=0}^\infty \frac{1}{r!} \theta^r \mathbf{M}^r \\ &= \cos \theta \mathbf{I} + \sin \theta \mathbf{M} \\ &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{align*} This is a rotation of $\theta$ degrees about the origin. \begin{align*} && \mathbf{N}^2 &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}^2 \\ && &= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \\ \Rightarrow && \exp(s\mathbf{N}) &= \sum_{r=0}^\infty \frac{1}{r!} (s\mathbf{N})^r \\ &&&= \mathbf{I} + s \mathbf{N} \\ &&&= \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix} \end{align*} This is a shear, leaving the $y$-axis invariant, sending $(1,1)$ to $(1+s, 1)$. Suppose those matrices commute, for all $s$, ie \begin{align*} && \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} \cos \theta - s \sin \theta & -\sin \theta + s \cos \theta \\ \sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & s \cos \theta - \sin \theta \\ \sin \theta & s \sin \theta + \cos \theta \end{pmatrix} \\ \Rightarrow && \sin \theta &= 0 \\ \Rightarrow && \theta &=n \pi, n \in \mathbb{Z} \end{align*} Clearly it doesn't matter when we do nothing. If we are rotating by $\pi$ then it also doesn't matter which order we do it in as the stretch happens in both directions equally.

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1994 Paper 3 Q6

D: 1700.0 B: 1501.5

complex numbers rotation Argand diagram geometric transformation composition of transformations unit circle necessary and sufficient condition

The four points $A,B,C,D$ in the Argand diagram (complex plane) correspond to the complex numbers $a,b,c,d$ respectively. The point $P_{1}$ is mapped to $P_{2}$ by rotating about $A$ through $\pi/2$ radians. Then $P_{2}$ is mapped to $P_{3}$ by rotating about $B$ through $\pi/2$ radians, $P_{3}$ is mapped to $P_{4}$ by rotating about $C$ through $\pi/2$ radians and $P_{4}$ is mapped to $P_{5}$ by rotating about $D$ through $\pi/2$ radians, each rotation being in the positive sense. If $z_{i}$ is the complex number corresponding to $P_{i},$ find $z_{5}$ in terms of $a,b,c,d$ and $z_{1}.$ Show that $P_{5}$ will coincide with $P_{1},$ irrespective of the choice of the latter if, and only if \[a-c=\mathrm{i}(b-d)\] and interpret this condition geometrically. The points $A,B$ and $C$ are now chosen to be distinct points on the unit circle and the angle of rotation is changed to $\theta,$ where $\theta\neq0,$ on each occasion. Find the necessary and sufficient condition on $\theta$ and the points $A,B$ and $C$ for $P_{4}$ always to coincide with $P_{1}.$

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