Problems

The triangle \(ABC\) has side lengths \(\left| BC \right| = a\), \(\left| CA \right| = b\) and \(\left| AB \right| = c\). Equilateral triangles \(BXC\), \(CYA\) and \(AZB\) are erected on the sides of the triangle \(ABC\), with \(X\) on the other side of \(BC\) from \(A\), and similarly for \(Y\) and \(Z\). Points \(L\), \(M\) and \(N\) are the centres of rotational symmetry of triangles \(BXC\), \(CY\!A\) and \(AZB\) respectively.

1. Show that \(| CM| = \dfrac {\ b} {\sqrt3} \,\) and write down the corresponding expression for \(| CL|\).
2. Use the cosine rule to show that \[ 6 \left| LM \right|^2 = a^2+b^2+c^2 + 4\sqrt3 \, \Delta \,, \] where \(\Delta\) is the area of triangle \(ABC\). Deduce that \(LMN\) is an equilateral triangle. Show further that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \,. \]
3. Show that the conditions \[ (a -b)^2 = -2ab \big( 1 -\cos(C-60^\circ)\big) \,\] and \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \] are equivalent. Deduce that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \(ABC\) is equilateral.

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Solution:

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1. Consider the equilateral triangle \(CYA\), notice that \(YM\) is a vertical line of symmetry, and \(\angle ACM = 30^\circ\) therefore \(\frac{AC/2}{CM} = \cos 30^\circ \Rightarrow |CM| = \frac{b}{2} \cdot \frac{2}{\sqrt{3}} = \frac{b}{\sqrt{3}}\). Similarly \(|CL| = \frac{a}{\sqrt{3}}\)
2. \(\,\) \begin{align*} && |LM|^2 &= |CM|^2 + |CL|^2 - 2 \cdot |CM| \cdot |CL| \cdot \cos \angle MCL \\ &&&= \frac{b^2}{3} + \frac{a^2}{3} - 2 \frac{ab}{3} \cos \left (\angle CMA + \angle CAB + \angle BCL \right) \\ &&&= \frac13 \left (b^2 + a^2 - 2ab \cos \left ( \frac{\pi}{3} + \angle CAB \right) \right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \cos \left ( \angle CAB \right) + \sqrt{3}ab \sin \angle CAB \right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \cos \angle CAB + 2\sqrt{3} \Delta\right) \\ &&&= \frac13 \left (b^2 + a^2 - ab \left (\frac{a^2+b^2-c^2}{2ab} \right) + 2\sqrt{3} \Delta\right) \\ &&&= \frac13 \left ( \frac12(a^2+b^2+c^2) + 2\sqrt{3}\Delta \right) \\ \Rightarrow && 6|LM|^2 &= a^2 + b^2 + c^2 + 4\sqrt{3} \Delta \end{align*} However, nothing in our reasoning here was special about \(LM\), therefore \(LN\) and \(MN\) also equal this value, and we find that the triangle is equilateral. The area of equilateral triangle [LMN] is \(\frac{\sqrt{3}}4 |LM|^2\), ie \begin{align*} &&& \text{areas are equal} \\ \Leftrightarrow && \Delta &= \frac{\sqrt{3}}4 |LM|^2 \\ &&&= \frac{\sqrt{3}}4 \frac{a^2+b^2+c^2+4\sqrt{3}\Delta}{6} \\ &&&= \frac{\sqrt{3}}{24} (a^2+b^2+c^2) + \frac12 \Delta \\ \Leftrightarrow && \Delta &= \frac{\sqrt{3}}{12}(a^2+b^2+c^2)\\ \Leftrightarrow && 4\sqrt{3}\Delta &=a^2+b^2+c^2\\ \end{align*}
3. \(\,\) \begin{align*} && (a-b)^2 &= -2ab(1 - \cos(C - 60^{\circ})) \\ \Leftrightarrow && a^2+b^2 - 2ab &=-2ab + 2ab \cos(C - 60^{\circ}) \\ \Leftrightarrow && a^2+b^2 &= ab \cos C+\sqrt{3}ab\sin C \\ \Leftrightarrow && a^2+b^2 &= ab \frac{a^2+b^2-c^2}{2ab} + 2\sqrt{3} \Delta \\ \Leftrightarrow && a^2+b^2+c^2 &= 4\sqrt{3}\Delta \end{align*} Since the LHS is non-positive, and the RHS is positive, the only way they can be equal is if they are both \(0\), ie \(a=b\) and \(C = 60^{\circ}\) ie \(ABC\) is equilateral.

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

Given that \(\alpha = \e^{\mathrm{i} \pi/3}\) , prove that \(1 + \alpha^2 = \alpha\). A triangle in the Argand plane has vertices \(A\), \(B\), and \(C\) represented by the complex numbers \(p\), \(q\alpha^2\) and \(- r\alpha\) respectively, where \(p\), \(q\) and \(r\) are positive real numbers. Sketch the triangle~\(ABC\). Three equilateral triangles \(ABL\), \(BCM\) and \(CAN\) (each lettered clockwise) are erected on sides \(AB\), \(BC\) and \(CA\) respectively. Show that the complex number representing \(N\) is \mbox{\(( 1 - \alpha) p- \alpha^2 r\)} and find similar expressions for the complex numbers representing \(L\) and \(M\). Show that lines \(LC\), \(MA\) and \(NB\) all meet at the origin, and that these three line segments have the common length \(p+q+r\).