Problems

1. Show that, if \(a\) and \(b\) are complex numbers, with \(b \neq 0\), and \(s\) is a positive real number, then the points in the Argand diagram representing the complex numbers \(a + sbi\), \(a - sbi\) and \(a + b\) form an isosceles triangle. Given three points which form an isosceles triangle in the Argand diagram, explain with the aid of a diagram how to determine the values of \(a\), \(b\) and \(s\) so that the vertices of the triangle represent complex numbers \(a + sbi\), \(a - sbi\) and \(a + b\).
2. Show that, if the roots of the equation \(z^3 + pz + q = 0\), where \(p\) and \(q\) are complex numbers, are represented in the Argand diagram by the vertices of an isosceles triangle, then there is a non-zero real number \(s\) such that \[\frac{p^3}{q^2} = \frac{27(3s^2 - 1)^3}{4(9s^2 + 1)^2}\,.\]
3. Sketch the graph \(y = \dfrac{(3x-1)^3}{(9x+1)^2}\), identifying any stationary points.
4. Show that if the roots of the equation \(z^3 + pz + q = 0\) are represented in the Argand diagram by the vertices of an isosceles triangle then \(\dfrac{p^3}{q^2}\) is a real number and \(\dfrac{p^3}{q^2} > -\dfrac{27}{4}\).

A thin uniform wire is bent into the shape of an isosceles triangle \(ABC\), where \(AB\) and \(AC\) are of equal length and the angle at \(A\) is \(2\theta\). The triangle \(ABC\) hangs on a small rough horizontal peg with the side \(BC\) resting on the peg. The coefficient of friction between the wire and the peg is \(\mu\). The plane containing \(ABC\) is vertical. Show that the triangle can rest in equilibrium with the peg in contact with any point on \(BC\) provided \[ \mu \ge 2\tan\theta(1+\sin\theta) \,. \]

Show Solution

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Solution:

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Clearly the centre of mass will lie on the perpendicular from \(A\). We can also consider each side's wire as equivalent to a point mass at the centre of the side with mass proportional to the length of the side. Recalling that \(b = c\) (the triangle is isoceles we must have (for the \(y\)-coordinate \begin{align*} && a \cdot 0 + b \cdot \frac12 b \cos \theta + c \cdot \frac12 c \cos \theta &= (a+b+c) \overline{y} \\ \Rightarrow && b^2 \cos \theta &= (2b + 2b\sin \theta) \overline{y} \\ \Rightarrow && \overline{y} &= \frac{b \cos \theta}{2(1+\sin \theta)} \end{align*}

+ - ↺

\begin{align*} \text{N2}(\nearrow): && R - mg \cos \phi &= 0 \\ \text{N2}(\nwarrow): && F -mg \sin \phi &= 0 \\ \Rightarrow && F &\leq \mu R \\ \Rightarrow && \sin \phi &\leq \mu \cos \phi \\ \Rightarrow && \tan \phi &\leq \mu \end{align*} When the peg is at \(C\) \begin{align*} \tan \phi &= \frac{CM}{MG} \\ &= \frac{b\sin \theta}{\frac{b \cos \theta}{2(1+\sin \theta)}} \\ &= 2 \tan \theta(1+\sin \theta) \end{align*} Therefore \(2 \tan \theta(1+\sin \theta) \leq \mu\) as required.

In the triangle \(ABC\), the base \(AB\) is of length 1 unit and the angles at~\(A\) and~\(B\) are \(\alpha\) and~\(\beta\) respectively, where \(0<\alpha\le\beta\). The points \(P\) and~\(Q\) lie on the sides \(AC\) and \(BC\) respectively, with \(AP=PQ=QB=x\). The line \(PQ\) makes an angle of~\(\theta\) with the line through~\(P\) parallel to~\(AB\).

1. Show that \(x\cos\theta = 1- x\cos\alpha - x\cos\beta\), and obtain an expression for \(x\sin\theta\) in terms of \(x\), \(\alpha\) and~\(\beta\). Hence show that \begin{equation} \label{eq:2*} \bigl(1+2\cos(\alpha+\beta)\bigr)x^2 - 2(\cos\alpha + \cos\beta)x + 1 = 0\,. \tag{\(*\)} \end{equation} Show that \((*)\) is also satisfied if \(P\) and \(Q\) lie on \(AC\) produced and \(BC\) produced, respectively. [By definition, \(P\) lies on \(AC\) produced if \(P\) lies on the line through \(A\) and~\(C\) and the points are in the order \(A\), \(C\), \(P\)\,.]
2. State the condition on \(\alpha\) and \(\beta\) for \((*)\) to be linear in \(x\). If this condition does not hold (but the condition \(0<\alpha \le \beta\) still holds), show that \((*)\) has distinct real roots.
3. Find the possible values of~\(x\) in the two cases (a) \(\alpha = \beta = 45^\circ\) and (b) \(\alpha = 30^\circ\), \(\beta = 90^\circ\), and illustrate each case with a sketch.

The triangle \(OAB\) is isosceles, with \(OA = OB\) and angle \(AOB = 2 \alpha\) where \(0< \alpha < {\pi \over 2}\,\). The semi-circle \(\mathrm{C}_0\) has its centre at the midpoint of the base \(AB\) of the triangle, and the sides \(OA\) and \(OB\) of the triangle are both tangent to the semi-circle. \(\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3, \ldots\) are circles such that \(\mathrm{C}_n\) is tangent to \(\mathrm{C}_{n-1}\) and to sides \(OA\) and \(OB\) of the triangle. Let \(r_n\) be the radius of \(\mathrm{C}_n\,\). Show that \[ \frac{r_{n+1}}{r_n} = \frac{1-\sin\alpha}{1+\sin\alpha}\;. \] Let \(S\) be the total area of the semi-circle \(\mathrm{C}_0\) and the circles \(\mathrm{C}_1\), \(\mathrm{C}_2\), \(\mathrm{C}_3\), \(\ldots\;\). Show that \[ S = {1 + \sin^2 \alpha \over 4 \sin \alpha} \, \pi r_0^2 \;. \] Show that there are values of \(\alpha\) for which \(S\) is more than four fifths of the area of triangle~\(OAB\).