Year: 2020
Paper: 3
Question Number: 3
Course: UFM Pure
Section: Complex numbers 2
In spite of the change to criteria for entering the paper, there was still a very healthy number of candidates, and the vast majority handled the protocols for the online testing very well. Just over half the candidates attempted exactly six questions, and whilst about 10% attempted a seventh, hardly any did more than seven. With 20% attempting five questions, and 10% attempting only four, overall, there were very few candidates not attempting the target number. There was a spread of popularity across the questions, with no question attracting more than 90% of candidates and only one less than 10%, but every question received a good number of attempts. Likewise, there was a spread of success on the questions, though every question attracted at least one perfect solution.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 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 $\frac{1}{3}\pi$.
\textbf{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 $\frac{1}{3}(p+q+r)$.
\begin{questionparts}
\item 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 $\omega = \mathrm{e}^{\frac{\mathrm{i}\pi}{6}}$.
\item 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.
\item 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.
\end{questionparts}\begin{questionparts}
\item 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*}
\item 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*}
\item 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.
\end{questionparts}
This was the second least popular pure question, but many candidates produced good solutions to it, including some very elegant ones, and the mean score was just shy of half marks. The most successful candidates were those confident in manipulating terms of the form e^(iθ). Candidates demonstrating a good knowledge of classical geometry also did well. Several candidates abandoned their attempts after part (i) or (ii). Part (i) was generally well answered. The most common mistakes were to rotate anticlockwise rather than clockwise, or to omit the "+a" from their expression for k. Some candidates (including many of those with the wrong angle) still achieved the required result with no or incorrect working; as the required result was in the question, candidates could not be rewarded without correct justification. In part (ii) most candidates attempted to show both implications at the same time, which tended to be more successful if candidates started with Q2 being a parallelogram. Despite the question stating the order of the vertices, some candidates used an incorrect direction for one of their lines. Most candidates used the fact that ABCD is a parallelogram ⇒ b − a = c − d (or an equivalent), though a few candidates tried to use the condition that pairs of opposite sides are parallel (not appreciating the fact that if one set of sides are parallel and equal in length then the shape must be a parallelogram). Some candidates did not appreciate that they had to state that ω − ω* ≠ 0 before being able to deduce b − a = c − d from the corresponding result for Q2. Instead, these candidates either simply cancelled ω − ω* without justification, or attempted (unsuccessfully) to argue that the coefficients of ω and ω* could be directly equated. In part (iii), some candidates stopped their attempts after finding an expression for g_BC − g_AB (or similar), but most of those who attempted this part went on to produce good solutions. There were some elegant and creative solutions to this part, but the most common approach was to try to show that G_CA G_AB is a rotation of G_BC G_AB by π/3 about G_AB. A common error here was instead to try to show that G_CA G_AB is a rotation of G_CA G_AB by −π/3 about G_AB (sometimes arising from an incorrect labelling of T2). Another error which arose in several attempts was to try and compute |g_AB − g_BC| by treating a, b and c as real numbers. Whilst this approach led to an expression which was totally symmetric in a, b and c, leading these candidates to 'conclude' that T2 was equilateral, very little credit could be gained for such an approach. Only one candidate attempted to prove part (iii) independently of part (i).