2012 Paper 2 Q6
Year: 2012
Paper: 2
Question Number: 6
Course: LFM Pure
Section: Introduction to trig
Problem
Solution
Examiner's report
— 2012 STEP 2, Question 6There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1528.8
Banger Comparisons: 4
Show LaTeX source
Problem source
A cyclic quadrilateral $ABCD$ has sides $AB$, $BC$, $CD$ and $DA$ of lengths $a$, $b$, $c$ and $d$, respectively. The area of the quadrilateral is $Q$, and angle $DAB$ is $\theta$. Find an expression for $\cos\theta$ in terms of $a$, $b$, $c$ and $d$, and an expression for $\sin\theta$ in terms of $a$, $b$, $c$, $d$ and $Q$.
Hence show that
\[
16Q^2 = 4(ad+bc)^2 - (a^2+d^2-b^2-c^2)^2
\,,
\]
and deduce that
\[
Q^2 = (s-a)(s-b)(s-c)(s-d)\,,
\]
where $s= \frac12(a+b+c+d)$.
Deduce a formula for the area of a triangle with sides of length $a$, $b$ and $c$.Solution source
\begin{center}
\begin{tikzpicture}
\def\a{2};
\coordinate (A) at ({\a*cos(0)},{\a*sin(0)});
\coordinate (B) at ({\a*cos(100)},{\a*sin(100)});
\coordinate (C) at ({\a*cos(200)},{\a*sin(200)});
\coordinate (D) at ({\a*cos(250)},{\a*sin(250)});
\draw[dashed] (0,0) circle (2);
\filldraw (A) circle (1.5pt) node[right] {$A$};
\filldraw (B) circle (1.5pt) node[above] {$B$};
\filldraw (C) circle (1.5pt) node[left] {$C$};
\filldraw (D) circle (1.5pt) node[below] {$D$};
\draw (A) -- (B) -- (C) -- (D) -- cycle;
\draw (A) -- (B) node[pos=0.5, above] {$a$};
\draw (B) -- (C) node[pos=0.5, above] {$b$};
\draw (C) -- (D) node[pos=0.5, below] {$c$};
\draw (D) -- (A) node[pos=0.5, below] {$d$};
\pic [draw, angle radius=.4cm, angle eccentricity=1.5, "$\theta$"] {angle = B--A--D};
\pic [draw, angle radius=.4cm, angle eccentricity=1.75, "$\pi-\theta$"] {angle = D--C--B};
\end{tikzpicture}
\end{center}
\begin{align*}
&& BD^2 &= a^2+d^2 - 2ad \cos \theta \\
&& BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\
\Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\
\Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\
\Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\
\\
&& Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\
&&&= \frac12 (ad+bc) \sin \theta \\
\Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\
\\
&& 1 &= \sin^2 \theta + \cos^2 \theta \\
&&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\
\Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\
\Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\
\Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\
&&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\
&&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\
\Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d)
\end{align*}
Since all triangles are cyclic, we can place $D$ at the same point as $A$ to obtain \textit{Heron's formula}
$A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac12(a+b+c)$
This was the second least popular of the pure maths questions, partly (it seems) because many candidates did not know what was meant by the term cyclic quadrilateral. The other immediate hurdle was that candidates needed to know that "opposite angles of a cyclic quad. are supplementary". We know this because of the large number of "attempts" that got no further than an initial diagram and a bit of working. Thus, the question was even less popular than the raw figures show. In reality, the question involved little more than some GCSE-level trigonometry, the difference of two squares factorisation and the result sin² + cos² = 1. Those who overcame the initial hurdles scored highly.