Problems

In this question, if \(O\), \(C\) and \(D\) are non-collinear points in three dimensional space, we will call the non-zero vector \(\mathbf{v}\) a \emph{bisecting vector} for angle \(COD\) if \(\mathbf{v}\) lies in the plane \(COD\), the angle between \(\mathbf{v}\) and \(\overrightarrow{OC}\) is equal to the angle between \(\mathbf{v}\) and \(\overrightarrow{OD}\), and both angles are less than \(90^\circ\).

1. Let \(O\), \(X\) and \(Y\) be non-collinear points in three-dimensional space, and define \(\mathbf{x} = \overrightarrow{OX}\) and \(\mathbf{y} = \overrightarrow{OY}\). Let \(\mathbf{b} = |\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}\).
   1. Show that \(\mathbf{b}\) is a bisecting vector for angle \(XOY\). Explain, using a diagram, why any other bisecting vector for angle \(XOY\) is a positive multiple of \(\mathbf{b}\).
   2. Find the value of \(\lambda\) such that the point \(B\), defined by \(\overrightarrow{OB} = \lambda\mathbf{b}\), lies on the line \(XY\). Find also the ratio in which the point \(B\) divides \(XY\).
   3. Show, in the case when \(OB\) is perpendicular to \(XY\), that the triangle \(XOY\) is isosceles.
2. Let \(O\), \(P\), \(Q\) and \(R\) be points in three-dimensional space, no three of which are collinear. A bisecting vector is chosen for each of the angles \(POQ\), \(QOR\) and \(ROP\). Show that the three angles between them are either all acute, all obtuse or all right angles.

1. The distinct points \(A\), \(Q\) and \(C\) lie on a straight line in the Argand diagram, and represent the distinct complex numbers \(a\), \(q\) and \(c\), respectively. Show that \(\dfrac {q-a}{c-a}\) is real and hence that \((c-a)(q^*-a^*) = (c^*-a^*)(q-a)\,\). Given that \(aa^* = cc^* = 1\), show further that \[ q+ ac q^* = a+c \,. \]
2. The distinct points \(A\), \(B\), \(C\) and \(D\) lie, in anticlockwise order, on the circle of unit radius with centre at the origin (so that, for example, \(aa^* =1\)). The lines \(AC\) and \(BD\) meet at \(Q\). Show that \[ (ac-bd)q^* = (a+c)-(b+d) \,, \] where \(b\) and \(d\) are complex numbers represented by the points \(B\) and \(D\) respectively, and show further that \[ (ac-bd) (q+q^*) = (a-b)(1+cd) +(c-d)(1+ab) \,. \]
3. The lines \(AB\) and \(CD\) meet at \(P\), which represents the complex number \(p\). Given that \(p\) is real, show that \(p(1+ab)=a+b\,\). Given further that \(ac-bd \ne 0\,\), show that \[ p(q+q^*) = 2 \,. \]

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 \(T\).

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Solution:

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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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1. Let \(\mathbf{x}_1\) and \(\mathbf{x}_2\) be the position vectors of the centres of \(C_1\) and \(C_2\), respectively. Show that the position vector of \(P\) is \[ \frac{r_1 \mathbf{x}_2- r_2 \mathbf{x}_1}{r_1-r_2} \,, \] where \(r_1\) and \(r_2\) are the radii of \(C_1\) and \(C_2\), respectively.
2. The circle \(C_3\) does not overlap either \(C_1\) or \(C_2\) and its radius, \(r_3\), satisfies \(r_1 \ne r_3 \ne r_2\). The focus of \(C_1\) and \(C_3\) is \(Q\), and the focus of \(C_2\) and \(C_3\) is \(R\). Show that \(P\), \(Q\) and \(R\) lie on the same straight line.
3. Find a condition on \(r_1\), \(r_2\) and \(r_3\) for \(Q\) to lie half-way between \(P\) and \(R\).

The points \(A\) and \(B\) have position vectors \(\bf a \) and \(\bf b\) with respect to an origin \(O\), and \(O\), \(A\)~and~\(B\) are non-collinear. The point \(C\), with position vector \(\bf c\), is the reflection of \(B\) in the line through \(O\) and \(A\). Show that \(\bf c\) can be written in the form \[ \bf c = \lambda \bf a -\bf b \] where \(\displaystyle \lambda = \frac{2\,{\bf a .b}}{{\bf a.a}}\). The point \(D\), with position vector \(\bf d\), is the reflection of \(C\) in the line through \(O\) and \(B\). Show that \(\bf d\) can be written in the form \[ \bf d = \mu\bf b - \lambda \bf a \] for some scalar \(\mu\) to be determined. Given that \(A\), \(B\) and \(D\) are collinear, find the relationship between \(\lambda\) and \(\mu\). In the case \(\lambda = -\frac12\), determine the cosine of \(\angle AOB\) and describe the relative positions of \(A\), \(B\) and \(D\).

Relative to a fixed origin \(O\), the points \(A\) and \(B\) have position vectors \(\bf{a}\) and \(\bf{b}\), respectively. (The points \(O\), \(A\) and \(B\) are not collinear.) The point \(C\) has position vector \(\bf c\) given by \[ {\bf c} =\alpha {\bf a}+ \beta {\bf b}\,, \] where \(\alpha\) and \(\beta\) are positive constants with \(\alpha+\beta<1\,\). The lines \(OA\) and \(BC\) meet at the point \(P\) with position vector \(\bf p\) and the lines \(OB\) and \(AC\) meet at the point \(Q\) with position vector \(\bf q\). Show that \[ {\bf p} =\frac{\alpha {\bf a} }{1-\beta}\,, \] and write down \(\bf q\) in terms of \(\alpha,\ \beta\) and \(\bf {b}\). Show further that the point \(R\) with position vector \(\bf r\) given by \[ {\bf r} =\frac{\alpha {\bf a} + \beta {\bf b}}{\alpha + \beta}\,, \] lies on the lines \(OC\) and \(AB\). The lines \(OB\) and \(PR\) intersect at the point \(S\). Prove that $ \dfrac{OQ}{BQ} = \dfrac{OS}{BS}\,$.

The vertices \(A\), \(B\), \(C\) and \(D\) of a square have coordinates \((0,0)\), \((a,0)\), \((a,a)\) and \((0,a)\), respectively. The points \(P\) and \(Q\) have coordinates \((an,0)\) and \((0,am)\) respectively, where \(0 < m < n < 1\). The line \(CP\) produced meets \(DA\) produced at \(R\) and the line \(CQ\) produced meets \(BA\) produced at \(S\). The line \(PQ\) produced meets the line \(RS\) produced at \(T\). Show that \(TA\) is perpendicular to \(AC\). Explain how, given a square of area \(a^2\), a square of area \(2a^2\) may be constructed using only a straight-edge. [Note: a straight-edge is a ruler with no markings on it; no measurements (and no use of compasses) are allowed in the construction.]

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Solution:

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Note that \(CP\) has equation \(\frac{y-0}{x-an} = \frac{a-0}{a-an} = \frac{1}{1-n} \Rightarrow y = \frac{x-an}{1-n}\) Therefore \(R = (0, -\frac{an}{1-n})\) Note that \(CQ\) has equation \(\frac{y-am}{x} = \frac{a-am}{a} = 1-m \Rightarrow y = (1-m)x + am\) Therefore \(S = (-\frac{am}{1-m}, 0)\) \(PQ\) has equation \(\frac{y}{x-an} = \frac{am-0}{0-an} \Rightarrow y = -\frac{m}{n}x +am\) \(SR\) has equation \(\frac{y}{x+\frac{am}{1-m}} = \frac{-\frac{an}{1-n}}{\frac{am}{1-m}} = -\frac{n(1-m)}{m(1-n)} \Rightarrow y =-\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n}\) So \(PQ \cap SR\) has \begin{align*} && -\frac{m}{n}x +am &= -\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n} \\ && x \left (\frac{n(1-m)}{m(1-n)} - \frac{m}{n} \right) &= -am - \frac{an}{1-n} \\ \Rightarrow && x \left ( \frac{n^2(1-m)-m^2(1-n)}{nm(1-n)} \right) &= -\frac{a(m(1-n)+n)}{1-n} \\ \Rightarrow && x \left ( \frac{(m-n)(mn-n-m)}{mn(1-n)} \right) &= \frac{a(mn-m-n)}{1-n} \\ \Rightarrow && x &= \frac{amn}{m-n} \\ && y &= -\frac{amn}{m-n} \end{align*} Therefore clearly \(TA\) is perpendicular to \(AC\) since they are the lines \(y = -x\) and \(y = x\) Given this method we can construct the perpendicular to the diagonal through the vertex. Doing this at \(A\) we can construct \(C'\) the reflection of \(C\) in \(AB\). We can do the same to find the reflection of \(A\) and so we have a square with sidelengths \(\sqrt{2}a\) and hence area \(2a^2\)

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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 \(ABDC\)?

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.

1. The point \(X\) has position vector \(s \mathbf{b}+t\mathbf{c}\). Describe the locus of \(X\) when \(s+t=1\).
2. 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 \(BD:DC=\gamma:\beta\).
3. The line \(BP\) cuts the line \(CA\) at \(E\), and the line \(CP\) cuts the line \(AB\) at \(F\). Show that \[ \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}=1\,. \]

Show that the equation \[ ax^{2}+ay^{2}+2gx+2fy+c=0 \] where \(a>0\) and \(f^{2}+g^{2}>ac\) represents a circle in Cartesian coordinates and find its centre. The smooth and level parade ground of the First Ruritanian Infantry Division is ornamented by two tall vertical flagpoles of heights \(h_{1}\) and \(h_{2}\) a distance \(d\) apart. As part of an initiative test a soldier has to march in such a way that he keeps the angles of elevation of the tops of the two flagpoles equal to one another. Show that if the two flagpoles are of different heights he will march in a circle. What happens if the two flagpoles have the same height? To celebrate the King's birthday a third flagpole is added. Soldiers are then assigned to each of the three different pairs of flagpoles and are told to march in such a way that they always keep the tops of their two assigned flagpoles at equal angles of elevation to one another. Show that, if the three flagpoles have different heights \(h_{1},h_{2}\) and \(h_{3}\) and the circles in which the soldiers march have centres of \((x_{ij},y_{ij})\) (for the flagpoles of height \(h_{i}\) and \(h_{j}\)) relative to Cartesian coordinates fixed in the parade ground, then the \(x_{ij}\) satisfy \[ h_{3}^{2}\left(h_{1}^{2}-h_{2}^{2}\right)x_{12}+h_{1}^{2}\left(h_{2}^{2}-h_{3}^{2}\right)x_{23}+h_{2}^{2}\left(h_{3}^{2}-h_{1}^{2}\right)x_{31}=0, \] and the same equation connects the \(y_{ij}\). Deduce that the three centres lie in a straight line.

The points \(P\) and \(R\) lie on the sides \(AB\) and \(AD,\) respectively, of the parallelogram \(ABCD.\) The point \(Q\) is the fourth vertex of the parallelogram \(APQR.\) Prove that \(BR,CQ\) and \(DP\) meet in a point.