Problems

Filters
Clear Filters

14 problems found

2025 Paper 3 Q5
D: 1500.0 B: 1500.0

Three points, \(A\), \(B\) and \(C\), lie in a horizontal plane, but are not collinear. The point \(O\) lies above the plane. Let \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{c}\). \(P\) is a point with \(\overrightarrow{OP} = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c}\), where \(\alpha\), \(\beta\) and \(\gamma\) are all positive and \(\alpha + \beta + \gamma < 1\). Let \(k = 1 - (\alpha + \beta + \gamma)\).

  1. The point \(L\) is on \(OA\), the point \(X\) is on \(BC\) and \(LX\) passes through \(P\). Determine \(\overrightarrow{OX}\) in terms of \(\beta\), \(\gamma\), \(\mathbf{b}\) and \(\mathbf{c}\) and show that \(\overrightarrow{OL} = \frac{\alpha}{k+\alpha}\mathbf{a}\).
  2. Let \(M\) and \(Y\) be the unique pair of points on \(OB\) and \(CA\) respectively such that \(MY\) passes through \(P\), and let \(N\) and \(Z\) be the unique pair of points on \(OC\) and \(AB\) respectively such that \(NZ\) passes through \(P\). Show that the plane \(LMN\) is also horizontal if and only if \(OP\) intersects plane \(ABC\) at the point \(G\), where \(\overrightarrow{OG} = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})\). Where do points \(X\), \(Y\) and \(Z\) lie in this case?
  3. State what the condition \(\alpha + \beta + \gamma < 1\) tells you about the position of \(P\) relative to the tetrahedron \(OABC\).

2023 Paper 2 Q8
D: 1500.0 B: 1500.0

A tetrahedron is called isosceles if each pair of edges which do not share a vertex have equal length.

  1. Prove that a tetrahedron is isosceles if and only if all four faces have the same perimeter.
Let \(OABC\) be an isosceles tetrahedron and let \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{c}\).
  1. By considering the lengths of \(OA\) and \(BC\), show that \[2\mathbf{b}.\mathbf{c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2.\] Show that \[\mathbf{a}.(\mathbf{b}+\mathbf{c}) = |\mathbf{a}|^2.\]
  2. Let \(G\) be the centroid of the tetrahedron, defined by \(\overrightarrow{OG} = \frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c})\). Show that \(G\) is equidistant from all four vertices of the tetrahedron.
  3. By considering the length of the vector \(\mathbf{a}-\mathbf{b}-\mathbf{c}\), or otherwise, show that, in an isosceles tetrahedron, none of the angles between pairs of edges which share a vertex can be obtuse. Can any of them be right angles?

2020 Paper 3 Q3
D: 1500.0 B: 1500.0

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 \(\frac{1}{3}(p+q+r)\).

  1. 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 \(\omega = \mathrm{e}^{\frac{\mathrm{i}\pi}{6}}\).
  2. 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.
  3. 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:

  1. 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*}
  2. 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*}
  3. 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.

2019 Paper 1 Q5
D: 1500.0 B: 1500.0

  1. 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 \(|\vec{PQ}| = |\vec{PS}|\)?
  2. 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:

  1. If \(\vec{PQ} = \vec{SR}\) we have a parallelogram. \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\) then we have a rhombus.
  2. 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 \(\frac{21}{20}\)

2012 Paper 2 Q7
D: 1600.0 B: 1516.0

Three distinct points, \(X_1\), \(X_2\) and \(X_3\), with position vectors \({\bf x}_1\), \({\bf x}_2\) and \({\bf x}_3\) respectively, lie on a circle of radius 1 with its centre at the origin \(O\). The point \(G\) has position vector \(\frac13({\bf x}_1+{\bf x}_2+{\bf x}_3)\). The line through \(X_1\) and \(G\) meets the circle again at the point \(Y_1\) and the points \(Y_2\) and \(Y_3\) are defined correspondingly. Given that \(\overrightarrow{GY_1} =-\lambda_1\overrightarrow{GX_1}\), where \(\lambda_1\) is a positive scalar, show that \[ \overrightarrow{OY_1}= \tfrac13 \big( (1-2\lambda_1){\bf x}_1 +(1+\lambda_1)({\bf x}_2+{\bf x}_3)\big) \] and hence that \[ \lambda_1 = \frac {3-\alpha-\beta-\gamma} {3+\alpha -2\beta-2\gamma} \,,\] where \(\alpha = {\bf x}_2 \,.\, {\bf x}_3\), \(\beta = {\bf x}_3\,.\, {\bf x}_1\) and \(\gamma = {\bf x}_1\,.\, {\bf x}_2\). Deduce that $\dfrac {GX_1}{GY_1} + \dfrac {GX_2}{GY_2} + \dfrac {GX_3}{GY_3} =3 \,$.


Solution:

TikZ diagram
\begin{align*} && \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\ &&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 - \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ && 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\ \Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\ &&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\ && 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\ \Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\ &&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\ \Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)} \end{align*} as required. Since \(\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}\) we must have, \begin{align*} && \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} + \frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\ &&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\ &&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\ &&&= 3 \end{align*}

2010 Paper 2 Q6
D: 1600.0 B: 1516.0

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.


Solution:

  1. \(D\) must be above the centre (of any kind) of the equilateral triangle \(ABC\). Therefore it is a distance \(\frac23 \frac{\sqrt{3}}2 = \frac{\sqrt{3}}3\) from \(A\). \(D\) is \(1\) from \(A\), therefore by Pythagoras \(PD = \sqrt{1-\frac13} = \sqrt{\frac23}\)
  2. We can place \(D\) at \(\langle 0,0,\sqrt{\frac23}\rangle\) and \(A'\) (the midpoint of \(BC\)) at \(\langle-\frac{\sqrt{3}}{6},0,0 \rangle\) and we find: \begin{align*} && \cos \theta &= \frac{(\mathbf{a}'-\mathbf{d})\cdot \mathbf{a}'}{|\mathbf{a}'-\mathbf{d}|| \mathbf{a}'|} \\ &&&= \frac{|\mathbf{a}'|}{|\mathbf{a}'-\mathbf{d}|} \\ &&&= \frac{\frac{\sqrt{3}}{6}}{\sqrt{\frac23+\frac{3}{36}}} = \frac13 \end{align*}
  3. We have
    TikZ diagram
    And therefore we must have \(\tan \frac{\cos^{-1} \frac13}{2} = \frac{r}{\frac{\sqrt{3}}{6}}\) therefore \begin{align*} && r &= \frac{\sqrt{3}}{6} \tan \left (\frac{\cos^{-1} \frac13}{2} \right) \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{1-\cos(\cos^{-1}\frac13)}{1+\cos(\cos^{-1}\frac13)}} \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{\frac23}{\frac43}} \\ &&&= \frac{\sqrt{6}}{12} \end{align*}

2008 Paper 3 Q7
D: 1700.0 B: 1500.0

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.

  1. 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.
  2. 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 \(\frac13(x+y+z)\,\).]

2004 Paper 1 Q6
D: 1484.0 B: 1500.0

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 \(AB\).

2001 Paper 3 Q6
D: 1700.0 B: 1488.0

The plane \[ {x \over a} + {y \over b} +{z \over c} = 1 \] meets the co-ordinate axes at the points \(A\), \(B\) and \(C\,\). The point \(M\) has coordinates \(\left( \frac12 a, \frac12 b, \frac 12 c \right)\) and \(O\) is the origin. Show that \(OM\) meets the plane at the centroid \(\left( \frac13 a, \frac13 b, \frac 13 c \right)\) of triangle \(ABC\). Show also that the perpendiculars to the plane from \(O\) and from \(M\) meet the plane at the orthocentre and at the circumcentre of triangle \(ABC\) respectively. Hence prove that the centroid of a triangle lies on the line segment joining its orthocentre and circumcentre, and that it divides this line segment in the ratio \(2 : 1\,\). [The orthocentre of a triangle is the point at which the three altitudes intersect; the circumcentre of a triangle is the point equidistant from the three vertices.]


Solution: The line \(OM\) is \(\lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}\), then we need \(1 = \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix} = 3 \lambda \Rightarrow \lambda = \frac13\). Therefore \(OM\) meets the plane at the centroid. The orthocentre is the point \(\mathbf{h}\) such that \((\mathbf{a}-\mathbf{b}) \cdot (\mathbf{c} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} a \\ -b \\ 0 \end{pmatrix} \cdot \begin{pmatrix} -p \\ -q \\ c-r \end{pmatrix} \Leftrightarrow ap-bq = 0\) \((\mathbf{b}-\mathbf{c}) \cdot (\mathbf{a} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} 0 \\ b \\ -c \end{pmatrix} \cdot \begin{pmatrix} a-p \\ -q \\ -r \end{pmatrix} \Leftrightarrow bq-cr = 0\) \((\mathbf{c}-\mathbf{a}) \cdot (\mathbf{b} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} -a \\ 0 \\ c \end{pmatrix} \cdot \begin{pmatrix} -p \\ b-q \\ -r \end{pmatrix} \Leftrightarrow cr-ap = 0\) ie \(ap = bq = cr\) but this is clearly on the line \(\lambda \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix}\) therefore the orthocentre is on the perpendicular from \(O\) \(M-A = \begin{pmatrix} -a/2 \\ b/2 \\ c/2 \end{pmatrix}\) so \(|M-A|=|M-B|=|M-C|\) Also by pythagoras the point of intersection satisfies \(|M-P|^2 + |P-A|^2 = |M-A|^2\) so \(|P-A|^2 = |P-B|^2 = |P-C|^2\), therefore \(P\) is the circumcentre. Since all these points are in the same plane and \(OGM\) is a line, we have the points are in a line. Similar triangles gives the desired ratio

1999 Paper 1 Q2
D: 1484.0 B: 1468.7

A point moves in the \(x\)-\(y\) plane so that the sum of the squares of its distances from the three fixed points \((x_{1},y_{1})\), \((x_{2},y_{2})\), and \((x_{3},y_{3})\) is always \(a^{2}\). Find the equation of the locus of the point and interpret it geometrically. Explain why \(a^2\) cannot be less than the sum of the squares of the distances of the three points from their centroid. [The centroid has coordinates \((\bar x, \bar y)\) where \(3\bar x = x_1+x_2+x_3,\) $3\bar y =y_1+y_2+y_3. $]


Solution: \begin{align*} && a^2 &= d_1^2 + d_2^2 + d_3^2 \\ &&&= (x-x_1)^2+(y-y_1)^2 + (x-x_2)^2+(y-y_2)^2 + (x-x_3)^2+(y-y_3)^2 \\ &&&= \sum (x-\bar{x}+\bar{x}-x_i)^2 + \sum (y-\bar{y}+\bar{y}-y_i)^2 \\ &&&= \sum \left ( (x-\bar{x})^2+(\bar{x}-x_i)^2 + 2(x-\bar{x})(\bar{x}-x_i) \right)+ \sum \left ( (y-\bar{y})^2+(\bar{y}-y_i)^2 + 2(y-\bar{y})(\bar{y}-y_i) \right)\\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2 + 6x\bar{x} -6\bar{x}^2-2x\sum x_i+2\bar{x}\sum x_i + \\ &&&\quad\quad\quad 3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 + 6y\bar{y} -6\bar{y}^2-2y\sum y_i+2\bar{y}\sum y_i \\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2+3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 \\ \\ \Rightarrow && (x-\bar{x})^2+(y-\bar{y})^2 &= \frac13\left ( a^2- \sum \left((\bar{x}-x_i)^2+(\bar{y}-y_i)^2 \right) \right) \end{align*} Therefore the locus is a circle, centre \((\bar{x}, \bar{y})\). radius \(\sqrt{\frac13(a^2 - \text{sum of squares distances of centroid to vertices}})\). \(a^2\) cannot be less than this distance, because clearly the right hand side is always bigger than it!

1998 Paper 2 Q8
D: 1600.0 B: 1487.4

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 \(R\).

1995 Paper 1 Q7
D: 1500.0 B: 1516.0

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 \(OQ.\)

  1. Show that \(\overrightarrow{AQ}\) is perpendicular to \(\overrightarrow{BC}\) and deduce that the three altitudes of \(\triangle ABC\) are concurrent.
  2. 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.{]}

1991 Paper 1 Q5
D: 1500.0 B: 1484.0

A set of \(n\) distinct vectors \(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n},\) where \(n\geqslant2\), is called regular if it satisfies the following two conditions:

  1. there are constants \(\alpha\) and \(\beta\), with \(\alpha>0\), such that for any \(i\) and \(j\), \[ \mathbf{a}_{i}\cdot\mathbf{a}_{j}=\begin{cases} \alpha^{2} & \mbox{ when }i=j\\ \beta & \mbox{ when }i\neq j, \end{cases} \]
  2. the centroid of \(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}\) is the origin \(\mathbf{0}.\) {[}The centroid of vectors \(\mathbf{b}_{1},\mathbf{b}_{2},\ldots,\mathbf{b}_{m}\) is the vector \(\frac{1}{m}(\mathbf{b}_{1}+\mathbf{b}_{2}+\cdots+\mathbf{b}_{m}).\){]}
Prove that (i) and (ii) imply that \((n-1)\beta=-\alpha^{2}.\) If $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0 \end{pmatrix},\( where \)\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}$ is a regular set of vectors in 2-dimensional space, show that either \(n=2\) or \(n=3\), and in each case find the other vectors in the set. Hence, or otherwise, find all regular sets of vectors in 3-dimensional space for which $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\( and \)\mathbf{a}_{2}\( lies in the \)x\(-\)y$ plane.


Solution: \begin{align*} && \mathbf{0} &= \sum_i \mathbf{a}_i \tag{ii} \\ \Rightarrow && 0 &= \mathbf{a}_i \cdot \mathbf{0} \\ &&&= \sum_j \mathbf{a}_i \cdot \mathbf{a}_j \\ &&&= (n-1)\beta + \alpha^2 \tag{i} \\ \Rightarrow && (n-1)\beta &= -\alpha^2 \end{align*} Suppose we have \(\mathbf{a}_j = \binom{x}{y}\), \(j \neq 1\) then \(x = \beta\). We also must have \(\beta^2 + y^2 = 1\), so there are at most two values for \(y\), ie two extra vectors. ie \(n = 2, 3\). If \(n = 2 \Rightarrow \mathbf{a}_2 = - \mathbf{a}_1\). If \(n = 3\) \begin{align*} && \mathbf{0} &= \binom{1}{0} + \binom{\beta}{y} + \binom{\beta}{-y} \\ \Rightarrow && \beta = -1/2 \\ \Rightarrow && y &= \pm \frac{\sqrt{3}}{2} \end{align*} Suppose $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\(, \)\mathbf{a}_{2}=\begin{pmatrix}\cos \theta \\ \sin \theta \\ 0 \end{pmatrix}$ (since we need \(\mathbf{a}_2 \cdot \mathbf{a}_2 = 1\)). \(\beta = \cos \theta\)). We can have \(\cos \theta = - 1\). Suppose we have \(\mathbf{a}_j =\begin{pmatrix}x\\ y \\ z \end{pmatrix}\), so \(x = \cos \theta\), and \(y^2 + z^2 = \sin^2 \theta\), so we can write it as: \(\mathbf{a}_j =\begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix}\). We must also have \(\beta = \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta\\ 0 \end{pmatrix} = \cos^2 \theta + \sin^2 \theta \cos \phi = \cos \theta\), so \(\cos \phi = \frac{\cos \theta - \cos^2\theta}{1-\cos^2 \theta} = \frac{\cos \theta}{1+\cos \theta}\). Therefore there is one value for \(\cos \phi\), so at most two values for \(\sin \phi\), Therefore we can have either \(2, 3,4\) or \(5\) different values in the set. \(n = 2\), we've already handled. If \(n = 3\), then \(\beta = -\frac12\), \(\cos \phi = -1\), so we can only have two different values for \(\sin \theta\), ie: \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ \frac{\sqrt{3}}{2} \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ -\frac{\sqrt{3}}{2} \\ 0 \end{pmatrix} \right \}\) Finally, if \(n = 4\), we have \(\beta = -\frac13\), \(\cos \phi = \frac{-1/3}{2/3} = -\frac12\). \(\sin \theta = \pm \frac{\sqrt{3}}{2}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) If \(n = 5\), then \(\beta = -\frac14\), \(\cos \phi = \frac{-1/4}{3/4} = -\frac13\). \(\sin \theta = \frac{\sqrt{15}}{4}\), \(\sin \phi = \frac{2\sqrt{2}}{3}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{12} \\ \frac{\sqrt{30}}{6} \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{12} \\ -\frac{\sqrt{30}}{6} \end{pmatrix}, \right \}\)

1991 Paper 3 Q2
D: 1700.0 B: 1516.0

The distinct points \(P_{1},P_{2},P_{3},Q_{1},Q_{2}\) and \(Q_{3}\) in the Argand diagram are represented by the complex numbers \(z_{1},z_{2},z_{3},w_{1},w_{2}\) and \(w_{3}\) respectively. Show that the triangles \(P_{1}P_{2}P_{3}\) and \(Q_{1}Q_{2}Q_{3}\) are similar, with \(P_{i}\) corresponding to \(Q_{i}\) (\(i=1,2,3\)) and the rotation from \(1\) to \(2\) to \(3\) being in the same sense for both triangles, if and only if \[ \frac{z_{1}-z_{2}}{z_{2}-z_{3}}=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}. \] Verify that this condition may be written \[ \det\begin{pmatrix}z_{1} & z_{2} & z_{3}\\ w_{1} & w_{2} & w_{3}\\ 1 & 1 & 1 \end{pmatrix}=0. \]

  1. Show that if \(w_{i}=z_{i}^{2}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is not similar to triangle \(Q_{1}Q_{2}Q_{3}.\)
  2. Show that if \(w_{i}=z_{i}^{3}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is similar to triangle \(Q_{1}Q_{2}Q_{3}\) if and only if the centroid of triangle \(P_{1}P_{2}P_{3}\) is the origin. {[}The centroid of triangle \(P_{1}P_{2}P_{3}\) is represented by the complex number \(\frac{1}{3}(z_{1}+z_{2}+z_{3})\).{]}
  3. Show that the triangle \(P_{1}P_{2}P_{3}\) is equilateral if and only if \[ z_{2}z_{3}+z_{3}z_{1}+z_{1}z_{2}=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}. \]