Year: 2014
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Vectors
No solution available for this problem.
A 10% increase in the number of candidates and the popularity of all questions ensured that all questions had a good number of attempts, though the first two questions were very much the most popular. Every question received at least one absolutely correct solution. In most cases when candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five sixths gave in at least six attempts.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
The four distinct points $P_i$ ($i=1$, $2$, $3$, $4$) are the
vertices, labelled anticlockwise, of a cyclic quadrilateral. The
lines $P_1P_3$ and $P_2P_4$ intersect at $Q$.
\begin{questionparts}
\item
By considering the
triangles $P_1QP_4$ and $P_2QP_3$ show that
$(P_1Q)( QP_3) = (P_2Q) (QP_4)\,$.
\item Let $\+p_i$ be the position vector of the point $P_i$
($i=1$, $2$, $3$, $4$). Show that there
exist numbers $a_i$, not all zero, such
that
\begin{equation}
\sum\limits_{i=1}^4 a_i =0 \qquad\text{and}\qquad
\sum\limits_{i=1}^4 a_i \+p_i ={\bf 0} \,. \tag{$*$}
\end{equation}
\item Let $a_i$ ($i=1$,~$2$, $3$,~$4$) be any numbers, not all zero,
that satisfy~$(*)$. Show that $a_1+a_3\ne 0$ and that the lines
$P_1P_3$ and $P_2P_4$ intersect at the point with position vector
\[
\frac{a_1 \+p_1 + a_3 \+p_3}{a_1+a_3} \,.
\]
Deduce that $a_1a_3 (P_1P_3)^2 = a_2a_4 (P_2P_4)^2\,$.
\end{questionparts}
Roughly two fifths of the candidates attempted this with a mean score of just over three marks making it the least well attempted question on the paper. Most could do part (i), which is GCSE material, but frustratingly quite a few stated that the triangles were similar with no justification. Part (ii) was by far the most poorly attempted part with a lot of hand-waving arguments. Part (iii) was done well by virtue of only the best candidates making it past part (i) with 75% of solutions containing good proofs by contradiction for the first result and the last two parts were pretty well done.