Year: 2008
Paper: 1
Question Number: 7
Course: LFM Pure
Section: Trigonometry 2
No solution available for this problem.
There were significantly more candidates attempting this paper this year (an increase of nearly 25%), but many found it to be very difficult and only achieved low scores. The mean score was significantly lower than last year, although a similar number of candidates achieved very high marks. This may be, in part, due to the phenomenon of students in the Lower Sixth (Year 12) being entered for the examination before attempting papers II and III in the Upper Sixth. This is a questionable practice, as while students have enough technical knowledge to answer the STEP I questions at this stage, they often still lack the mathematical maturity to be able to apply their knowledge to these challenging problems. Again, a key difficulty experienced by most candidates was a lack of the algebraic skill required by the questions. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many students were simply unable to progress on some questions because they did not know how to handle the algebra. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was also pleasing that one of the applied questions, question 13, attracted a very large number of attempts. However, the examiners were again left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers and other available resources to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1500.0
Banger Comparisons: 0
The point $P$ has coordinates $(x,y)$ with respect to
the origin $O$. By writing $x=r\cos\theta$ and $y=r\sin\theta$,
or otherwise, show that, if the line $OP$ is rotated by $60^\circ$
clockwise about $O$, the new $y$-coordinate of $P$
is $\frac12(y-\sqrt3\,x)$.
What is the new $y$-coordinate in the case of an anti-clockwise
rotation by $60^\circ\,$?
An equilateral triangle $OBC$ has vertices at $O$, $(1,0)$ and
$(\frac12,\frac12 \sqrt3)$, respectively.
The point $P$ has coordinates
$(x,y)$.
The perpendicular distance from $P$ to the line
through $C$ and $O$ is $h_1$;
the perpendicular distance from $P$ to the line
through $O$ and $B$ is $h_2$;
and the perpendicular distance from $P$ to the line through $B$ and $C$ is
$h_3$.
Show that
$h_1=\frac12 \big\vert y-\sqrt3\,x\big\vert$ and find expressions
for $h_2$ and $h_3$.
Show that $h_1+h_2+h_3=\frac12 \sqrt3$ if and only if $P$ lies on or
in the triangle $OBC$.
This was a reasonably popular question, tackled by about half of the candidates. Most confidently showed that y = ½(y − √3 x) and went on to deduce the result for the clockwise rotation. A small number of candidates lost marks here because their presentation either failed to make clear which answer corresponded to which direction of rotation, or the directions were reversed. Several candidates would have been helped by including a sketch in their solution. About two-thirds of the candidates were unable to progress beyond this point. Of those who continued, the majority succeeded in finding h₁, either by a direct argument or, more usually, by using the earlier result as intended by the question. Despite the hint of h₁ being given with absolute value signs, a large number of candidates then claimed that h₂ = y rather than the correct |y|, suggesting that they do not understand what absolute values mean and when they should be used. Very few candidates correctly determined h₃, the most common incorrect answer being h₃ = ½|y + √3 x| to parallel the answer for h₁. Again, clear diagrams are essential if marks are to be gained for questions such as these. There was also evidence of confusion in the algebraic manipulation of absolute values, with some candidates confusing |a−b| with |a|−|b|, thereby giving answers such as h₃ = ½|y + √3 x| − ½√3. Only a handful of candidates made a significant attempt at the final part of the question, and of those who did, the main difficulty stemmed from not appreciating that to prove an "if and only if" statement, one has to prove the implication in both directions. The sample solutions use the triangle inequality; it could equally and straightforwardly be argued by considering all eight possible cases of where the point P might lie with respect to each of the three sides of the triangle.