2007 Paper 3 Q6

Year: 2007
Paper: 3
Question Number: 6

Course: LFM Stats And Pure
Section: Complex Numbers (L8th)

Difficulty: 1700.0 Banger: 1472.0

Problem

The distinct points \(P\), \(Q\), \(R\) and \(S\) in the Argand diagram lie on a circle of radius \(a\) centred at the origin and are represented by the complex numbers \(p\), \(q\), \(r\) and \(s\), respectively. Show that \[ pq = -a^2 \frac {p-q}{p^*-q^*}\,. \] Deduce that, if the chords \(PQ\) and \(RS\) are perpendicular, then \(pq+rs=0\). The distinct points \(A_1\), \(A_2\), \(\ldots\), \(A_n\) (where \(n\ge3\)) lie on a circle. The points \hbox{\(B_1\), \(B_2\), \(\ldots\), \(B_{n}\)} lie on the same circle and are chosen so that the chords \(B_1B_2\), \(B_2B_3\), \(\ldots\), \(B_nB_{1}\) are perpendicular, respectively, to the chords \(A_1A_2\), \(A_2A_3\), \(\ldots\), \(A_nA_1\). Show that, for \(n=3\), there are only two choices of \(B_1\) for which this is possible. What is the corresponding result for \(n=4\)? State the corresponding results for values of \(n\) greater than 4.

No solution available for this problem.

Examiner's report
— 2007 STEP 3, Question 6
Below Average Least popular Pure question; very little success

This was the least popular Pure question and very little success was achieved by the few that attempted it. The first result was often obtained correctly by expressing each of the four complex numbers in modulus-exponential form, but then the perpendicularity was the stumbling block.

Source: Cambridge STEP 2007 Examiner's Report · 2007-full.pdf
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1472.0

Banger Comparisons: 2

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Problem source
The distinct points $P$, $Q$, $R$ and $S$ in the Argand diagram
lie on a circle of radius $a$ centred at the origin and are represented
by the complex numbers $p$, $q$, $r$ and $s$, respectively.
Show that 
\[
pq = -a^2 \frac {p-q}{p^*-q^*}\,.
\]
Deduce that, if the chords
$PQ$ and $RS$ are perpendicular, then $pq+rs=0$.
The distinct
points $A_1$, $A_2$, $\ldots$, $A_n$ (where $n\ge3$) lie on a circle. The 
points
\hbox{$B_1$, $B_2$, $\ldots$, $B_{n}$} lie on the same circle and are chosen 
so that the chords $B_1B_2$, $B_2B_3$,  $\ldots$, $B_nB_{1}$
are perpendicular, respectively, to the chords 
$A_1A_2$, $A_2A_3$,  $\ldots$, $A_nA_1$.
Show that, for  $n=3$, there are only two choices of $B_1$
for which this is possible. What is the corresponding result for $n=4$?
State the corresponding results  for values of $n$ greater than 4.