Problems

2023 Paper 3 Q1

D: 1500.0 B: 1500.0

conic sections parabola tangent to circle distance from point to line quadratic equation discriminant triangle

The distinct points $P(2ap,\, ap^2)$ and $Q(2aq,\, aq^2)$ lie on the curve $x^2 = 4ay$, where $a > 0$.

Given that \[(p+q)^2 = p^2q^2 + 6pq + 5\,,\tag{$*$}\] show that the line through $P$ and $Q$ is a tangent to the circle with centre $(0,\, 3a)$ and radius $2a$.
Show that, for any given value of $p$ with $p^2 \neq 1$, there are two distinct real values of $q$ that satisfy equation $(*)$. Let these values be $q_1$ and $q_2$. Find expressions, in terms of $p$, for $q_1 + q_2$ and $q_1 q_2$.
Show that, for any given value of $p$ with $p^2 \neq 1$, there is a triangle with one vertex at $P$ such that all three vertices lie on the curve $x^2 = 4ay$ and all three sides are tangents to the circle with centre $(0,\, 3a)$ and radius $2a$.

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2022 Paper 2 Q5

D: 1500.0 B: 1500.0

inequality coordinate geometry triangle perpendicular distance optimisation area algebraic identity cauchy-schwarz inequality

Given that $a > b > c > 0$ are constants, and that $x$, $y$, $z$ are non-negative variables, show that \[ax + by + cz \leqslant a(x + y + z).\]

In the acute-angled triangle $ABC$, $a$, $b$ and $c$ are the lengths of sides $BC$, $CA$ and $AB$, respectively, with $a > b > c$. $P$ is a point inside, or on the sides of, the triangle, and $x$, $y$ and $z$ are the perpendicular distances from $P$ to $BC$, $CA$ and $AB$, respectively. The area of the triangle is $\Delta$.

1. Find $\Delta$ in terms of $a$, $b$, $c$, $x$, $y$ and $z$.
2. Find both the minimum value of the sum of the perpendicular distances from $P$ to the three sides of the triangle and the values of $x$, $y$ and $z$ which give this minimum sum, expressing your answers in terms of some or all of $a$, $b$, $c$ and $\Delta$.
1. Show that, for all real $a$, $b$, $c$, $x$, $y$ and $z$, \[(a^2+b^2+c^2)(x^2+y^2+z^2) = (bx-ay)^2 + (cy-bz)^2 + (az-cx)^2 + (ax+by+cz)^2.\]
2. Find both the minimum value of the sum of the squares of the perpendicular distances from $P$ to the three sides of the triangle and the values of $x$, $y$ and $z$ which give this minimum sum, expressing your answers in terms of some or all of $a$, $b$, $c$ and $\Delta$.
Find both the maximum value of the sum of the squares of the perpendicular distances from $P$ to the three sides of the triangle and the values of $x$, $y$ and $z$ which give this maximum sum, expressing your answers in terms of some or all of $a$, $b$, $c$ and $\Delta$.

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2019 Paper 1 Q1

D: 1500.0 B: 1500.0

differentiation optimisation trigonometric triangle area perimeter stationary points coordinate geometry

A straight line passes through the fixed point $(1 , k)$ and has gradient $- \tan \theta$, where $k > 0$ and $0 < \theta < \frac{1}{2}\pi$. Find, in terms of $\theta$ and $k$, the coordinates of the points $X$ and $Y$ where the line meets the $x$-axis and the $y$-axis respectively.

Find an expression for the area $A$ of triangle $OXY$ in terms of $k$ and $\theta$. (The point $O$ is the origin.) You are given that, as $\theta$ varies, $A$ has a minimum value. Find an expression in terms of $k$ for this minimum value.
Show that the length $L$ of the perimeter of triangle $OXY$ is given by $$L = 1 + \tan \theta + \sec \theta + k(1 + \cot \theta + \cosec \theta).$$ You are given that, as $\theta$ varies, $L$ has a minimum value. Show that this minimum value occurs when $\theta = \alpha$ where $$\frac{1 - \cos \alpha}{1 - \sin \alpha} = k.$$ Find and simplify an expression for the minimum value of $L$ in terms of $\alpha$.

Solution: $y = (-\tan \theta)(x-1)+k$ so when $x = 0$, $y = k + \tan \theta$, so $Y = (0, k+\tan \theta)$. When $y = 0$, $x = 1 + \frac{k}{\tan \theta}$

$A = \frac12 (k+\tan \theta)\left ( 1 + \frac{k}{\tan \theta} \right) = k + \frac12 \left (\tan \theta + \frac{k^2}{\tan \theta} \right)$ Notice that $x + \frac{k^2}{x} \geq 2 k$ by AM-GM, so the minimum is $k + \frac12 \cdot 2k = 2k$
$\,$ \begin{align*} L &= k + \tan \theta + 1 + k \cot \theta + \sqrt{(k + \tan \theta)^2 + \left (1 + \frac{k}{\tan \theta} \right)^2} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{k^2 + 2 k \tan \theta +\tan^2 \theta + 1 + 2k \cot \theta + k^2\cot^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{\sec^2 \theta+ 2k \sec\theta\cosec \theta + k^2\cosec^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta +\sec \theta + k\cosec \theta\\ &= 1 + \tan \theta + \sec \theta + k (1 + \cot \theta + \cosec \theta) \end{align*} \begin{align*} && \frac{\d L}{\d \theta} &= \sec^2 \theta + \tan \theta \sec \theta + k(-\cosec^2 \theta - \cot \theta \cosec \theta ) \\ \Rightarrow && 0 &=\sec^2 \alpha+ \tan \theta \sec \alpha+ k(-\cosec^2 \alpha- \cot \alpha\cosec \alpha) \\ \Rightarrow && k &= \frac{\sec^2 \alpha+ \tan \alpha\sec \alpha}{\cosec^2 \alpha+ \cot \alpha\cosec \alpha} \\ &&&= \frac{\sin^2 \alpha(1 + \sin \alpha)}{\cos^2 \alpha (1+ \cos \alpha)} \\ &&&= \frac{(1-\cos^2 \alpha)(1 + \sin \alpha)}{(1-\sin^2 \alpha )(1+ \cos \alpha)} \\ &&&= \frac{1-\cos \alpha}{1-\sin \alpha} \\ \Rightarrow && L &= 1 + \tan \alpha + \sec \alpha + \frac{1-\cos \alpha}{1-\sin \alpha} \left (1 + \cot \alpha + \cosec \alpha \right) \\ &&&= \frac{1+\tan \alpha + \sec \alpha -\sin \alpha-\sin \alpha \tan \alpha-\tan \alpha}{1-\sin \alpha} + \\ &&&\quad \quad \frac{1+\cot \alpha + \cosec \alpha-\cos \alpha-\cos \alpha \cot \alpha -\cot \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\sec \alpha(1-\sin^2 \alpha)-\sin \alpha + \cosec \alpha(1-\cos^2 \alpha)-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\cos\alpha-\sin \alpha + \sin\alpha-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2}{1-\sin \alpha} \end{align*}

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2018 Paper 2 Q7

D: 1600.0 B: 1500.0

vectors geometric proof triangle position vectors ratio parallel lines intersection point cevians

The points $O$, $A$ and $B$ are the vertices of an acute-angled triangle. The points $M$ and $N$ lie on the sides $OA$ and $OB$ respectively, and the lines $AN$ and $BM$ intersect at $Q$. The position vector of $A$ with respect to $O$ is $\bf a$, and the position vectors of the other points are labelled similarly. Given that $\vert MQ \vert = \mu \vert QB\vert $, and that $\vert NQ \vert = \nu \vert QA\vert $, where $\mu$ and $\nu$ are positive and $\mu \nu <1$, show that \[ {\bf m} = \frac {(1+\mu)\nu}{1+\nu} \, {\bf a} \,. \] The point $L$ lies on the side $OB$, and $\vert OL \vert = \lambda \vert OB \vert \,$. Given that $ML$ is parallel to $AN$, express $\lambda$ in terms of $\mu$ and $\nu$. What is the geometrical significance of the condition $\mu\nu<1\,$?

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2017 Paper 2 Q8

D: 1600.0 B: 1500.0

vectors geometric proof perpendicular triangle altitudes orthocentre scalar product line equations

All vectors in this question lie in the same plane. The vertices of the non-right-angled triangle $ABC$ have position vectors $\bf a$, $\bf b$ and $\bf c$, respectively. The non-zero vectors $\bf u$ and $\bf v$ are perpendicular to $BC$ and $CA$, respectively. Write down the vector equation of the line through $A$ perpendicular to $BC$, in terms of $\bf u$, $\bf a$ and a parameter $\lambda $. The line through $A$ perpendicular to $BC$ intersects the line through $B$ perpendicular to $CA$ at $P$. Find the position vector of $P$ in terms of $\bf a$, $\bf b$, $\bf c$ and $\bf u$. Hence show that the line $CP$ is perpendicular to the line $AB$.

Solution: The line through $A$ perpendicular to $BC$ is $\mathbf{a} + \lambda\mathbf{u}$. The line through $B$ perpendicular to $CA$ is $\mathbf{b} + \mu \mathbf{v}$. They intersect when $\mathbf{a} + \lambda\mathbf{u} = \mathbf{b} + \mu \mathbf{v}$. Since $\mathbf{v}$ is perpendicular to $CA$, we must have \begin{align*} && \mathbf{a} + \lambda\mathbf{u} &= \mathbf{b} + \mu \mathbf{v} \\ \Rightarrow && \mathbf{a}\cdot(\mathbf{c}-\mathbf{a}) + \lambda\mathbf{u}\cdot(\mathbf{c}-\mathbf{a}) &= \mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) + \mu \mathbf{v}\cdot(\mathbf{c}-\mathbf{a}) \\ \\ \Rightarrow && \lambda &= \frac{\mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) -\mathbf{a}\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u}\cdot(\mathbf{c}-\mathbf{a})} \\ &&&= \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \end{align*} Therefore the point is $\mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u}$. The line $CP$ is $\mathbf{c} + \nu \left (\mathbf{p} - \mathbf{c} \right)$, to check this is perpendicular with $AB$ we should dot $\mathbf{p}-\mathbf{c}$ with $\mathbf{a}-\mathbf{b}$, ie \begin{align*} && (\mathbf{p}-\mathbf{c}) \cdot (\mathbf{a}-\mathbf{b}) &= \left ( \mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} - \mathbf{c}\right) \cdot ( \mathbf{a}-\mathbf{b}) \\ &&&= \left ( \mathbf{a}- \mathbf{c} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} \right) \cdot ( \mathbf{a}-\mathbf{c}+(\mathbf{c}-\mathbf{b})) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})}\mathbf{u} \cdot (\mathbf{a}-\mathbf{c}) + \\ &&&\quad (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) + \lambda \underbrace{\mathbf{u} \cdot (\mathbf{c}-\mathbf{b})}_{=0} \\ &&&=(\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) -(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})+ (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}+\mathbf{b}-\mathbf{a}+\mathbf{c}-\mathbf{b}) \\ &&&= 0 \end{align*} as required.

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2014 Paper 1 Q7

D: 1516.0 B: 1500.0

vectors geometric proof triangle ratio intersection of lines position vectors section formula

In the triangle $OAB$, the point $D$ divides the side $BO$ in the ratio $r:1$ (so that $BD = rDO$), and the point $E$ divides the side $OA$ in the ratio $s:1$ (so that $OE =s EA$), where $r$ and $s$ are both positive.

The lines $AD$ and $BE$ intersect at $G$. Show that \[ \mathbf{g}= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b} \,, \] where $\mathbf{a}, \mathbf{b}$ and $\mathbf{g}$ are the position vectors with respect to $O$ of $A$, $B$ and $G$, respectively.
The line through $G$ and $O$ meets $AB$ at $F$. Given that $F$ divides $AB$ in the ratio $t:1$, find an expression for $t$ in terms of $r$ and $s$.

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2009 Paper 1 Q4

D: 1500.0 B: 1500.0

trigonometry cosine rule triangle trig identity algebraic manipulation ratio

The sides of a triangle have lengths $p-q$, $p$ and $p+q$, where $p>q> 0\,$. The largest and smallest angles of the triangle are $\alpha$ and $\beta$, respectively. Show by means of the cosine rule that \[ 4(1-\cos\alpha)(1-\cos\beta) = \cos\alpha + \cos\beta \,. \] In the case $\alpha = 2\beta$, show that $\cos\beta=\frac34$ and hence find the ratio of the lengths of the sides of the triangle.

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2009 Paper 1 Q8

D: 1500.0 B: 1484.0

circle tangent incircle triangle trigonometric identity tan 2α double angle formula distance from point to line coordinate geometry

The equation of the circle $C$ is \[ (x-2t)^2 +(y-t)^2 =t^2, \] where $t$ is a positive number. Show that $C$ touches the line $y=0\,$. Let $\alpha$ be the acute angle between the $x$-axis and the line joining the origin to the centre of $C$. Show that $\tan2\alpha=\frac43$ and deduce that $C$ touches the line $3y=4x\,$.
Find the equation of the incircle of the triangle formed by the lines $y=0$, $3y=4x$ and $4y+3x=15\,$. Note: The incircle of a triangle is the circle, lying totally inside the triangle, that touches all three sides.

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2007 Paper 2 Q8

D: 1600.0 B: 1529.3

vectors locus ratio collinearity Cevians Menelaus-Ceva geometric proof position vectors triangle

The points $B$ and $C$ have position vectors $\mathbf{b}$ and $\mathbf{c}$, respectively, relative to the origin $A$, and $A$, $B$ and $C$ are not collinear.

The point $X$ has position vector $s \mathbf{b}+t\mathbf{c}$. Describe the locus of $X$ when $s+t=1$.
The point $P$ has position vector $\beta \mathbf{b}+\gamma\mathbf{c}$, where $\beta$ and $\gamma$ are non-zero, and $\beta+\gamma\ne1$. The line $AP$ cuts the line $BC$ at $D$. Show that $BD:DC=\gamma:\beta$.
The line $BP$ cuts the line $CA$ at $E$, and the line $CP$ cuts the line $AB$ at $F$. Show that \[ \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}=1\,. \]

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2004 Paper 1 Q6

D: 1484.0 B: 1500.0

coordinate geometry centroid medians perpendicularity orthocentre triangle concurrent lines midpoint

The three points $A$, $B$ and $C$ have coordinates $\l p_1 \, , \; q_1 \r$, $\l p_2 \, , \; q_2 \r$ and $\l p_3 \, , \; q_3 \r\,$, respectively. Find the point of intersection of the line joining $A$ to the midpoint of $BC$, and the line joining~$B$ to the midpoint of $AC$. Verify that this point lies on the line joining $C$ to the midpoint of~$AB$. The point $H$ has coordinates $\l p_1 + p_2 + p_3 \, , \; q_1 + q_2 + q_3 \r\,$. Show that if the line $AH$ intersects the line $BC$ at right angles, then $p_2^2 + q_2^2 = p_3^2 + q_3^2\,$, and write down a similar result if the line $BH$ intersects the line $AC$ at right angles. Deduce that if $AH$ is perpendicular to $BC$ and also $BH$ is perpendicular to $AC$, then $CH$ is perpendicular to $AB$.

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2001 Paper 1 Q1

D: 1516.0 B: 1500.0

coordinate geometry inequality geometric proof triangle square pigeonhole principle optimisation extremal principle

The points $A$, $B$ and $C$ lie on the sides of a square of side 1 cm and no two points lie on the same side. Show that the length of at least one side of the triangle $ABC$ must be less than or equal to $(\sqrt6 -\sqrt2)$ cm.

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2000 Paper 2 Q3

D: 1600.0 B: 1484.0

small angle approximation cosine rule triangle linear approximation error bounds optimisation perturbation

The lengths of the sides $BC$, $CA$, $AB$ of the triangle $ABC$ are denoted by $a$, $b$, $c$, respectively. Given that $$ b = 8+{\epsilon}_1, \, c=3+{\epsilon}_2,\, A=\tfrac{1}{3}\pi + {\epsilon}_3, $$ where ${\epsilon}_1$, ${\epsilon}_2$, and $ {\epsilon}_3$ are small, show that $a \approx 7 + {\eta}$, where ${\eta}= {\left(13 \, {{\epsilon}_1}-2\,{\epsilon}_2 + 24{\sqrt 3} \;{{\epsilon}_3}\right)}/14$. Given now that $$ {\vert {\epsilon}_1} \vert \le 2 \times 10^{-3}, \ \ \ {\vert {\epsilon}_2} \vert \le 4\cdot 9\times 10^{-2}, \ \ \ {\vert {\epsilon}_3} \vert \le \sqrt3 \times 10^{-3}, $$ find the range of possible values of ${\eta}$.

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1998 Paper 2 Q8

D: 1600.0 B: 1487.4

vectors geometric proof triangle centroid midpoint concurrent lines plane intersection 3D geometry

Points $\mathbf{A},\mathbf{B},\mathbf{C}$ in three dimensions have coordinate vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, respectively. Show that the lines joining the vertices of the triangle $ABC$ to the mid-points of the opposite sides meet at a point $R$. $P$ is a point which is {\bf not} in the plane $ABC$. Lines are drawn through the mid-points of $BC$, $CA$ and $AB$ parallel to $PA$, $PB$ and $PC$ respectively. Write down the vector equations of the lines and show by inspection that these lines meet at a common point $Q$. Prove further that the line $PQ$ meets the plane $ABC$ at $R$.

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1995 Paper 1 Q7

D: 1500.0 B: 1516.0

vectors geometric proof triangle circumcentre centroid orthocentre nine-point circle perpendicular scalar product concurrency

Let $A,B,C$ be three non-collinear points in the plane. Explain briefly why it is possible to choose an origin equidistant from the three points. Let $O$ be such an origin, let $G$ be the centroid of the triangle $ABC,$ let $Q$ be a point such that $\overrightarrow{GQ}=2\overrightarrow{OG},$ and let $N$ be the midpoint of $OQ.$

Show that $\overrightarrow{AQ}$ is perpendicular to $\overrightarrow{BC}$ and deduce that the three altitudes of $\triangle ABC$ are concurrent.
Show that the midpoints of $AQ,BQ$ and $CQ$, and the midpoints of the sides of $\triangle ABC$ are all equidistant from $N$.

{[}The centroid of $\triangle ABC$ is the point $G$ such that $\overrightarrow{OG}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}).$ The altitudes of the triangle are the lines through the vertices perpendicular to the opposite sides.{]}

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1995 Paper 2 Q7

D: 1600.0 B: 1516.7

coordinate geometry triangle incircle Heron's formula area of triangle cosine rule semi-perimeter sphere proof

The diagram shows a circle, of radius $r$ and centre $I$, touching the three sides of a triangle $ABC$. We write $a$ for the length of $BC$ and $\alpha$ for the angle $\angle BAC$ and so on. Let $s=\frac{1}{2}\left(a+b+c\right)$ and let $\triangle$ be the area of the triangle.

By considering the area of the triangles $AIB,$ $BIC$ and $CIA$, or otherwise, show that $\Delta=rs$.
By using the formula $\Delta=\frac{1}{2}bc\sin\alpha$, show that \[ \Delta^{2}=\tfrac{1}{16}[4b^{2}c^{2}-\left(2bc\cos\alpha\right)^{2}]. \] Now use the formula $a^{2}=b^{2}+c^{2}-2bc\cos\alpha$ to show that \[ \Delta^{2}=\tfrac{1}{16}[(a^{2}-\left(b-c\right)^{2})(\left(b+c\right)^{2}-a^{2})] \] and deduce that \[ \Delta=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \]
A hole in the shape of the triangle $ABC$ is cut in the top of a level table. A sphere of radius $R$ rests in the hole. Find the height of the centre of the sphere above the level of the table top, expressing your answer in terms of $a,b,c,s$ and $R$.

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