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2021 Paper 3 Q7
D: 1500.0 B: 1500.0
complex numbers unit circle cotangent modulus and argument perpendicularity orthocentre geometric transformations

  1. Let \[ z = \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} - e^{i\phi}}, \] where \(\theta\) and \(\phi\) are real, and \(\theta - \phi \neq 2n\pi\) for any integer \(n\). Show that \[ z = i\cot\!\bigl(\tfrac{1}{2}(\phi - \theta)\bigr) \] and give expressions for the modulus and argument of \(z\).
  2. The distinct points \(A\) and \(B\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\) and \(B\) are represented by the complex numbers \(a\) and \(b\), and \(O\) is at the origin. The point \(X\) is represented by the complex number \(x\), where \(x = a + b\) and \(a + b \neq 0\). Show that \(OX\) is perpendicular to \(AB\). If the distinct points \(A\), \(B\) and \(C\) in the complex plane, which are represented by the complex numbers \(a\), \(b\) and \(c\), lie on a circle with radius \(1\) and centre \(O\), and \(h = a + b + c\) represents the point \(H\), then \(H\) is said to be the orthocentre of the triangle \(ABC\).
  3. The distinct points \(A\), \(B\) and \(C\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\), \(B\) and \(C\) are represented by the complex numbers \(a\), \(b\) and \(c\), and \(O\) is at the origin. Show that, if the point \(H\), represented by the complex number \(h\), is the orthocentre of the triangle \(ABC\), then either \(h = a\) or \(AH\) is perpendicular to \(BC\).
  4. The distinct points \(A\), \(B\), \(C\) and \(D\) (in that order, anticlockwise) all lie on a circle with radius \(1\) and centre \(O\). The points \(P\), \(Q\), \(R\) and \(S\) are the orthocentres of the triangles \(ABC\), \(BCD\), \(CDA\) and \(DAB\), respectively. By considering the midpoint of \(AQ\), show that there is a single transformation which maps the quadrilateral \(ABCD\) onto the quadrilateral \(QRSP\) and describe this transformation fully.

Solution:

  1. \(\,\) \begin{align*} && z &= \frac{e^{i \theta} + e^{i \phi}}{e^{i \theta} - e^{i \phi}} \\ &&&= \frac{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})}{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})} \\ &&&= \frac{(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})/2}{i(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})/2i} \\ &&&= \frac{\cos \frac12(\theta-\phi)}{i \sin \frac12(\theta-\phi)} \\ &&&= -i \cot \tfrac12(\theta-\phi) \end{align*} Therefore \(|z| = \cot \tfrac12(\theta-\phi)\) and \(\arg z = \frac{\pi}{2}\)
  2. Since \(a,b\) lie on the unit circle, wlog \(a = e^{i \theta}, b = e^{i\phi}\). Not that the line \(OX\) has vector \(a+b\) and \(AB\) has vector \(b-a\) and not their ratio has argument \(\frac{\pi}{2}\) and hence they are perpendicular.
  3. \(AH\) has vector \(h - a = (a+b+c) - a = b+c\) which we've already established is perpendicular to \(c-b\) which is the vector for \(BC\) (unless \(b+c = 0\) in which case \(h = a\)).
  4. \(p = a +b+c, q = b+c+d\) etc. The midpoint of \(AQ = \frac12(a+b+c+d)\) which is the same as the midpoint of \(BR\), \(CS\) and \(DP\). Therefore we could say the transformation is reflection in the point \(\frac12(a+b+c+d)\)

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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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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 3 Q6
D: 1700.0 B: 1488.0
vectors 3D geometry plane equation centroid orthocentre circumcentre Euler line geometric proof perpendicular scalar product triangle geometry

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

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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 \(OQ.\)

  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.{]}

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