Problems

2019 Paper 1 Q5

D: 1500.0 B: 1500.0

vectors quadrilateral parallelogram rhombus cube coplanarity centroid geometric proof scalar product optimisation

The four points $P$, $Q$, $R$ and $S$ are the vertices of a plane quadrilateral. What is the geometrical shape of $PQRS$ if $\vec{PQ} = \vec{SR}$? What is the geometrical shape of $PQRS$ if $\vec{PQ} = \vec{SR}$ and $|\vec{PQ}| = |\vec{PS}|$?
A cube with edges of unit length has opposite vertices at $(0,0,0)$ and $(1,1,1)$. The points $$P(p,0,0), \quad Q(1,q,0), \quad R(r,1,1) \quad \text{and} \quad S(0,s,1)$$ lie on edges of the cube. Given that the four points lie in the same plane, show that $$rq = (1-s)(1-p).$$
1. Show that $\vec{PQ} = \vec{SR}$ if and only if the centroid of the quadrilateral $PQRS$ is at the centre of the cube. Note: the centroid of the quadrilateral $PQRS$ is the point with position vector $$\frac{1}{4}(\vec{OP} + \vec{OQ} + \vec{OR} + \vec{OS}),$$ where $O$ is the origin.
2. Given that $\vec{PQ} = \vec{SR}$ and $|\vec{PQ}| = |\vec{PS}|$, express $q$, $r$ and $s$ in terms of $p$. Show that $$\cos PQR = \frac{4p-1}{5-4p+8p^2}.$$ Write down the values of $p$, $q$, $r$ and $s$ if $PQRS$ is a square, and show that the length of each side of this square is greater than $\frac{21}{20}$.

Solution:

If $\vec{PQ} = \vec{SR}$ we have a parallelogram. $\vec{PQ} = \vec{SR}$ and $|\vec{PQ}| = |\vec{PS}|$ then we have a rhombus.
If the four points lie in a plane then $(\vec{RS} \times \vec{RP}) \cdot \vec{RQ} =0$, so \begin{align*} && 0 &=\left ( \begin{pmatrix}-r\\ s-1 \\ 0 \end{pmatrix} \times \begin{pmatrix}p-r\\ -1 \\ -1 \end{pmatrix}\right) \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ && &= \begin{pmatrix}1-s \\ -r \\r+(p-r)(1-s) \end{pmatrix} \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ &&&= (1-s)(1-r)-r(q-1)-r-(p-r)(1-s) \\ &&&=(1-s)(1-r-p+r)-rq \\ \Rightarrow && rq &= (1-s)(1-p) \end{align*}
1. $\,$ \begin{align*} && \vec{PQ} &= \vec{SR} \\ \Leftrightarrow && \begin{pmatrix}1-p\\q \\ 0 \end{pmatrix} &= \begin{pmatrix}r\\1-s \\ 0 \end{pmatrix} \\ \Leftrightarrow && 1-p = r & \quad ; \quad q = 1-s\\ \Leftrightarrow && 1= r+p & \quad ; \quad 1 = q+s\\ \end{align*} The centroid is $\frac14 (p+1+r, q+s+1, 2)$ which is clearly $\frac12(1,1,1)$ iff those equations are true.
2. $\,$ \begin{align*} && |\vec{PQ}| &= |\vec{PS}| \\ \Leftrightarrow && (1-p)^2+q^2+ 0^2 &= p^2+s^2+1)\\ \Leftrightarrow && 1-2p+p^2+q^2 &= p^2 + s^2 + 1 \\ \Leftrightarrow && -2p+q^2 &= s^2 \end{align*} From the previous equations we have $r = 1-p$, and $-2p+(1-s)^2 = s^2 \Rightarrow -2p + 1 -2s = 0 \Rightarrow s = \frac12 - p$ and $q = \frac12 + p$ \begin{align*} && \cos PQR &= \frac{\vec{QP}\cdot \vec{QR}}{|\vec{QP}||\vec{QR}|} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -q \\ 0 \end{pmatrix} \cdot \begin{pmatrix}r-1\\ 1-q \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+q^2}\sqrt{(r-1)^2+(1-q)^2+1^2}} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -\frac12-p \\ 0 \end{pmatrix} \cdot \begin{pmatrix}-p\\ \frac12-p \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+(-\frac12-p)^2}\sqrt{p^2+(\frac12-p)^2+1^2}} \\ &&&= \frac{ p-p^2-\frac14+p^2}{\sqrt{p^2-2p+1+\frac14+p+p^2}\sqrt{p^2+\frac14-p+p^2+1}} \\ &&&= \frac{4p-1}{\sqrt{8p^2-4p+5}\sqrt{8p^2-4p+5}}\\ &&&= \frac{4p-1}{8p^2-4p+5}\\ \end{align*} For $PQRS$ to be a square $\cos PQR = 0$, ie $p = \frac14$ and so $(p,q,r,s) = (\frac14, \frac34, \frac34, \frac14)$ and $|PQ| = \sqrt{(1-p)^2+q^2} = \sqrt{\left ( \frac34 \right)^2 + \left ( \frac34 \right)^2 } = \frac{3\sqrt{2}}4$, notice that $\left ( \frac{21}{20} \right)^2 = \frac{441}{400} < \frac{9}{8}$ ($441 < 450$) therefore the sides are at least as long as $\frac{21}{20}$

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2019 Paper 2 Q7

D: 1500.0 B: 1500.0

vectors scalar product unit vectors geometric proof equilateral triangle regular tetrahedron quadrilateral dot product identity symmetry

The points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$, respectively. Each of these vectors is a unit vector (so $\mathbf{a} \cdot \mathbf{a} = 1$, for example) and $$\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}.$$ Show that $\mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}$. What can be said about the triangle ABC? You should justify your answer.
The four distinct points $A_i$ ($i = 1, 2, 3, 4$) have unit position vectors $\mathbf{a}_i$ and $$\sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}.$$ Show that $\mathbf{a}_1 \cdot \mathbf{a}_2 = \mathbf{a}_3 \cdot \mathbf{a}_4$.
1. Given that the four points lie in a plane, determine the shape of the quadrilateral with vertices $A_1$, $A_2$, $A_3$ and $A_4$.
2. Given instead that the four points are the vertices of a regular tetrahedron, find the length of the sides of this tetrahedron.

Solution:

Given $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$, we can form the following results: \begin{align*} && \begin{cases} \mathbf{a} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{b} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{c} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= 0 \\ \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= 0 \\ \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} +\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -\frac12 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{b} = -\frac12 \\ \mathbf{a} \cdot \mathbf{c} = -\frac12 \\ \mathbf{b} \cdot \mathbf{c} = -\frac12 \\ \end{cases} \end{align*} The triangle must be equilateral since the angles between each vertex are the same.
We have $\displaystyle \sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}$ so $\displaystyle \mathbf{a}_i \cdot \sum_{i=1}^{4} \mathbf{a}_i = 0$ or for each $i$, $\displaystyle \sum_{j \neq i} \mathbf{a}_i \cdot \mathbf{a}_j = -1$. \begin{align*} && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_1 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ && \text{adding the first two, subtracting the last two} \\ \Rightarrow && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 +\cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ \Rightarrow && 2 (\mathbf{a}_1 \cdot \mathbf{a}_2) - 2(\mathbf{a}_3 \cdot \mathbf{a}_4) = 0 \end{align*} Rather than adding the first two and last two, we could have done any pair, resulting in the relations: \begin{align*} \mathbf{a}_1 \cdot \mathbf{a}_2 &= \mathbf{a}_3 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 &= \mathbf{a}_2 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_4 &= \mathbf{a}_2 \cdot \mathbf{a}_3 \end{align*}
1. The shape must be a parallelogram (from the angle requirement, but also cyclic quadrilateral (since all vectors are unit length), therefore it must be a rectangle
2. Given it's a regular tetrahedron, $\mathbf{a}_i \cdot \mathbf{a}_j$ must be the same for all $i \neq j$, ie $-\frac13$. We are interested in $|\mathbf{a}_i - \mathbf{a}_j|$ so consider, \begin{align*} |\mathbf{a}_i - \mathbf{a}_j|^2 &= (\mathbf{a}_i - \mathbf{a}_j) \cdot (\mathbf{a}_i - \mathbf{a}_j) \\ &= \mathbf{a}_i \cdot \mathbf{a}_i - 2 \mathbf{a}_i \cdot \mathbf{a}_j + \mathbf{a}_j \cdot \mathbf{a}_j \\ &= 1 - \frac23 + 1 \\ &= \frac43 \end{align*} Therefore the unit side lengths are $\frac{2}{\sqrt{3}}$

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2016 Paper 1 Q6

D: 1500.0 B: 1484.7

vectors position vectors quadrilateral parallel lines intersection ratio geometric proof collinearity

The sides $OA$ and $CB$ of the quadrilateral $OABC$ are parallel. The point $X$ lies on $OA$, between $O$ and $A$. The position vectors of $A$, $B$, $C$ and $X$ relative to the origin $O$ are $\bf a$, $\bf b$, $\bf c$ and $\bf x$, respectively. Explain why $\bf c$ and $\bf x$ can be written in the form \[ {\bf c} = k {\bf a} + {\bf b} \text{ and } {\bf x} = m {\bf a}\,, \] where $k$ and $m$ are scalars, and state the range of values that each of $k$ and $m$ can take. The lines $OB$ and $AC$ intersect at $D$, the lines $XD$ and $BC$ intersect at $Y$ and the lines $OY$ and $AB$ intersect at $Z$. Show that the position vector of $Z$ relative to $O$ can be written as \[ \frac{ {\bf b} + mk {\bf a}}{mk+1}\,. \] The lines $DZ$ and $OA$ intersect at $T$. Show that \[ OT \times OA = OX\times TA \text{ and } \frac 1 {OT} = \frac 1 {OX} + \frac 1 {OA} \,, \] where, for example, $OT$ denotes the length of the line joining $O$ and $T$.

Solution:

Notice that $\mathbf{x} = m\mathbf{a}$ since $OX$ is parallel to $OA$ and $0 < m < 1$ since $X$ lies between them. $\overline{OC} = \overline{OB} + \overline{BC} = \mathbf{b} + k\mathbf{a}$ since $BC$ is parallel to $OA$, $k$ can take any value. The line $OB$ is $\lambda \mathbf{b}$, the line $AC$ is $\mathbf{a} + \mu (\mathbf{c}-\mathbf{a}) = \mu \mathbf{b} +(1+ \mu(k-1)) \mathbf{a}$ Therefore they meet when $\mu = \lambda$ and $(1+\mu(k-1)) = 0$, ie $\mu = \frac{1}{1-k}$ so $D$ is $\frac{1}{1-k} \mathbf{b}$ The line $XD$ is $m\mathbf{a} + \nu ( \frac{1}{1-k} \mathbf{b} - m \mathbf{a}) $ and $BC$ is $\mathbf{b} + \eta \mathbf{a}$ so they meet when $\nu = 1-k$ and $\eta = m-(1-k)m = km$. Therefore $Y = \mathbf{b} + km \mathbf{a}$ Therefore the line $OY$ is $\alpha(\mathbf{b} + km \mathbf{a})$ and AB is $\mathbf{a} + \beta(\mathbf{b} - \mathbf{a})$ so they intersect when $\alpha = \beta$ and $\alpha km = (1-\alpha) \Rightarrow \alpha = \frac{1}{1+km}$. Therefore $Z = \mathbf{a} + \frac{1}{1+km} (\mathbf{b} - \mathbf{a}) = \frac{\mathbf{b}+km\mathbf{a}}{1+km}$ The lines $DZ$ and $OA$ are $\frac{1}{1-k} \mathbf{b} + \gamma \left ( \frac{1}{1-k} \mathbf{b} - \frac{\mathbf{b}+km\mathbf{a}}{1+km} \right)$ and $\delta \mathbf{a}$. Therefore they intersect when $\frac{1}{1-k} + \gamma \left (\frac{1}{1-k} - \frac{1}{1+km} \right) = 0 \Rightarrow \gamma = \frac{(1-k)(1+km)}{(k-1)k(m+1)} = -\frac{1+km}{k(m+1)}$ and $\delta = -\gamma \frac{km}{1+km} = \frac{m}{m+1}$. Therefore $OT = \frac{m}{m+1} |\mathbf{a}|, OA = |\mathbf{a}|, OX = m|\mathbf{a}|, TA = \frac{1}{m+1}|\mathbf{a}|$, Therefore $OT \times OA = OX \times TA$. Also $\frac{1}{OX} + \frac{1}{OA} = \frac{1}{m|\mathbf{a}|} + \frac{1}{|\mathbf{a}|} = \frac{m+1}{m|\mathbf{a}|} = \frac{1}{OT}$

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2015 Paper 1 Q6

D: 1484.0 B: 1500.0

vectors geometric proof quadrilateral midpoints square right angle optimisation position vectors

The vertices of a plane quadrilateral are labelled $A$, $B$, $A'$ and $B'$, in clockwise order. A point $O$ lies in the same plane and within the quadrilateral. The angles $AOB$ and $A'OB'$ are right angles, and $OA=OB$ and $OA'=OB'$. Use position vectors relative to $O$ to show that the midpoints of $AB$, $BA'$, $A'B'$ and $B'A$ are the vertices of a square. Given that the lengths of $OA$ and $OA'$ are fixed (and the conditions of the first paragraph still hold), find the value of angle $BOA'$ for which the area of the square is greatest.

Solution: Let $O$ be the origin, and let $\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}'$ be the four points. The conditions give us \begin{align*} && \mathbf{a} \cdot \mathbf{b} &= 0 \\ && |\mathbf{a}| &= |\mathbf{b}| \\ && \mathbf{a}' \cdot \mathbf{b}' &= 0 \\ && |\mathbf{a}'| &= |\mathbf{b}'| \\ \end{align*} So \begin{align*} \text{midpoint }AB \text{ to midpoint } BA' &= (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}'))\cdot (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}')) \\ &= \frac12(\mathbf{a}-\mathbf{a}')\cdot \frac12(\mathbf{a} - \mathbf{a}') \\ \text{midpoint }BA' \text{ to midpoint } A'B' &= (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}')) \cdot (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}'))\\ &= \frac12(\mathbf{b}-\mathbf{b}')\cdot \frac12(\mathbf{b} - \mathbf{b}') \\ &= \frac14 (|\mathbf{b}|^2 + |\mathbf{b}'|^2 - 2\mathbf{b}\cdot\mathbf{b}')\\ &= \frac14(|\mathbf{a}|^2 + |\mathbf{a}'|^2 - 2\mathbf{b}\cdot\mathbf{b}') \\ \text{midpoint }A'B' \text{ to midpoint } B'A &= (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a})) \cdot (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a}))\\ &= \frac12(\mathbf{a}'-\mathbf{a})\cdot \frac12(\mathbf{a}' - \mathbf{a}) \\ \text{midpoint }B'A \text{ to midpoint } AB &= (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b})) \cdot (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b}))\\ &= \frac12(\mathbf{b}'-\mathbf{b})\cdot \frac12(\mathbf{b}' - \mathbf{b}) \\ \end{align*} So it's sufficient to prove $\mathbf{a}\cdot \mathbf{a}' = \mathbf{b}\cdot \mathbf{b}'$ but this is clear from looking at a diagram for 1 second. Given the length of the square is what it is, we want to minimise $\mathbf{b}\cdot \mathbf{b}'$ which is when they are vertically opposite each other, ie $\angle BOA' = 90^\circ$

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2014 Paper 2 Q7

D: 1600.0 B: 1486.9

modulus function curve sketching graphical solution piecewise linear number of solutions inequality quadrilateral absolute value equations

The function $\f$ is defined by $\f(x)= |x-a| + |x-b| $, where $a < b$. Sketch the graph of $\f(x)$, giving the gradient in each of the regions $x < a$, $a < x < b$ and $x > b$. Sketch on the same diagram the graph of $\g(x)$, where $\g(x)= |2x-a-b|$. What shape is the quadrilateral with vertices $(a,0)$, $(b,0)$, $(b,\f(b))$ and $(a, \f(a))$?
Show graphically that the equation \[ |x-a| + |x-b| = |x-c|\,, \] where $a < b$, has $0$, $1$ or $2$ solutions, stating the relationship of $c$ to $a$ and $b$ in each case.
For the equation \[ |x-a| + |x-b| = |x-c|+|x-d|\,, \] where $a < b$, $c < d$ and $d-c < b-a$, determine the number of solutions in the various cases that arise, stating the relationship between $a$, $b$, $c$ and $d$ in each case.

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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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2009 Paper 2 Q1

D: 1600.0 B: 1516.0

simultaneous equations symmetry curve sketching quadrilateral area algebraic manipulation coordinate geometry rectangular hyperbola

Two curves have equations $\; x^4+y^4=u\;$ and $\; xy = v\;$, where $u$ and $v$ are positive constants. State the equations of the lines of symmetry of each curve. The curves intersect at the distinct points $A$, $B$, $C$ and $D$ (taken anticlockwise from $A$). The coordinates of $A$ are $(\alpha,\beta)$, where $\alpha > \beta > 0$. Write down, in terms of $\alpha$ and $\beta$, the coordinates of $B$, $C$ and $D$. Show that the quadrilateral $ABCD$ is a rectangle and find its area in terms of $u$ and $v$ only. Verify that, for the case $u=81$ and $v=4$, the area is $14$.

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2009 Paper 2 Q8

D: 1600.0 B: 1484.8

vectors position vectors collinearity parallelogram geometric proof parametric line through points quadrilateral

The non-collinear points $A$, $B$ and $C$ have position vectors $\bf a$, $\bf b$ and $\bf c$, respectively. The points $P$ and $Q$ have position vectors $\bf p$ and $\bf q$, respectively, given by \[ {\bf p}= \lambda {\bf a} +(1-\lambda){\bf b} \text{ \ \ \ and \ \ \ } {\bf q}= \mu {\bf a} +(1-\mu){\bf c} \] where $0<\lambda<1$ and $\mu>1$. Draw a diagram showing $A$, $B$, $C$, $P$ and $Q$. Given that $CQ\times BP = AB\times AC$, find $\mu$ in terms of $\lambda$, and show that, for all values of $\lambda$, the the line $PQ$ passes through the fixed point $D$, with position vector ${\bf d}$ given by ${\bf d= -a +b +c}\,$. What can be said about the quadrilateral $ABDC$?

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1997 Paper 1 Q5

D: 1500.0 B: 1484.0

vectors geometric proof quadrilateral midpoints distance formula scalar product invariant

Four rigid rods $AB$, $BC$, $CD$ and $DA$ are freely jointed together to form a quadrilateral in the plane. Show that if $P$, $Q$, $R$, $S$ are the mid-points of the sides $AB$, $BC$, $CD$, $DA$, respectively, then \[|AB|^{2}+|CD|^{2}+2|PR|^{2}=|AD|^{2}+|BC|^{2}+2|QS|^{2}.\] Deduce that $|PR|^{2}-|QS|^{2}$ remains constant however the vertices move. (Here $|PR|$ denotes the length of $PR$.)

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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

$\,$

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.

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1988 Paper 1 Q8

D: 1500.0 B: 1468.0

vectors geometric proof scalar product perpendicular midpoint quadrilateral skew quadrilateral equal opposite sides

$ABCD$ is a skew (non-planar) quadrilateral, and its pairs of opposite sides are equal, i.e. $AB=CD$ and $BC=AD$. Prove that the line joining the midpoints of the diagonals $AC$ and $BD$ is perpendicular to each diagonal.

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