2014 Paper 1 Q7
Year: 2014
Paper: 1
Question Number: 7
Course: UFM Pure
Section: Vectors
Problem
- The lines \(AD\) and \(BE\) intersect at \(G\). Show that \[ \mathbf{g}= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b} \,, \] where \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{g}\) are the position vectors with respect to \(O\) of \(A\), \(B\) and \(G\), respectively.
- The line through \(G\) and \(O\) meets \(AB\) at \(F\). Given that \(F\) divides \(AB\) in the ratio \(t:1\), find an expression for \(t\) in terms of \(r\) and \(s\).
Solution
Examiner's report
— 2014 STEP 1, Question 7More than 1800 candidates sat this paper, which represents another increase in uptake for this STEP paper. The impression given, however, is that many of these extra candidates are just not sufficiently well prepared for questions which are not structured in the same way as are the A-level questions that they are, perhaps, more accustomed to seeing. Although STEP questions try to give all able candidates "a bit of an intro." into each question, they are not intended to be easy, and (at some point) imagination and real flair (as well as determination) are required if one is to score well on them. In general, it is simply not possible to get very far into a question without making some attempt to think about what is actually going on in the situation presented therein; and those students who expect to be told exactly what to do at each stage of a process are in for a shock. Too many candidates only attempt the first parts of many questions, restricting themselves to 3-6 marks on each, rather than trying to get to grips with substantial portions of work – the readiness to give up and try to find something else that is "easy pickings" seldom allows such candidates to acquire more than 40 marks (as was the case with almost half of this year's candidature, in fact). Poor preparation was strongly in evidence – curve-sketching skills were weak, inequalities very poorly handled, algebraic capabilities (especially in non-standard settings) were often pretty poor, and the ability to get to grips with extended bits of working lacking in the extreme; also, an unwillingness to be imaginative and creative, allied with a lack of thoroughness and attention to detail, made this a disappointing (and, possibly, very uncomfortable) experience for many of those students who took the paper. On the other side of the coin, there was a very pleasing number of candidates who produced exceptional pieces of work on 5 or 6 questions, and thus scored very highly indeed on the paper overall. Around 100 of them scored 90+ marks of the 120 available, and they should be very proud of their performance – it is a significant and noteworthy achievement.
Rating Information
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
In the triangle $OAB$, the point $D$ divides the side $BO$ in the ratio $r:1$ (so that $BD = rDO$), and the point $E$ divides the side $OA$ in the ratio $s:1$ (so that $OE =s EA$), where $r$ and $s$ are both positive.
\begin{questionparts}
\item The lines $AD$ and $BE$ intersect at $G$.
Show that
\[
\mathbf{g}= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b} \,,
\]
where $\mathbf{a}, \mathbf{b}$ and $\mathbf{g}$ are the position vectors with respect to $O$ of $A$, $B$ and $G$, respectively.
\item The line through $G$ and $O$ meets $AB$ at
$F$. Given that $F$ divides $AB$ in the ratio $t:1$, find an expression for $t$ in terms of $r$ and $s$.
\end{questionparts}Solution source
\begin{center}
\begin{tikzpicture}
\coordinate (O) at (0,0);
\coordinate (A) at (3,1);
\coordinate (B) at (1,4);
\coordinate (D) at ($(O)!0.3!(B)$);
\coordinate (E) at ($(O)!0.4!(A)$);
\draw (O) -- (A) -- (B) -- cycle;
\filldraw (O) circle (1pt) node[below] {$O$};
\filldraw (A) circle (1pt) node[right] {$A$};
\filldraw (B) circle (1pt) node[above] {$B$};
\filldraw (D) circle (1pt) node[left] {$D$};
\filldraw (E) circle (1pt) node[below] {$E$};
\draw (A) -- (D);
\draw (E) -- (B);
\end{tikzpicture}
\end{center}
Notice that $\mathbf{d} = \frac{1}{r+1} \mathbf{b}$ and $\mathbf{e} = \frac{s}{s+1}\mathbf{a}$. We must also have that the line $AD$ is $\mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right)$ and $BE$ is $\mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right)$ at their point of intersection we must have
\begin{align*}
&& \mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right) &= \mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\
[\mathbf{a}]: && 1 + \lambda &= -\frac{\mu s}{s+1} \\
[\mathbf{b}]: && -\frac{\lambda}{r+1} &= 1 + \mu \\
\Rightarrow && \lambda &= -\frac{1+s+\mu s}{s+1} \\
\Rightarrow && \mu &= \frac{1+s+\mu s}{(1+r)(1+s)} - 1 \\
\Rightarrow && (1+r+rs)\mu &= 1+s - 1 - r - s - rs \\
\Rightarrow && \mu &= -\frac{r+rs}{1+r+rs} \\
\Rightarrow && \mathbf{g} &= \mathbf{b} -\frac{r+rs}{1+r+rs}\left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\
&&&= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b}
\end{align*}
\item The line $OG$ is $\lambda \mathbf{g}$. The line $AB$ is $\mathbf{a} + \mu(\mathbf{b}-\mathbf{a})$, so we need
\begin{align*}
&& \lambda \mathbf{g} &= \mathbf{a} + \mu(\mathbf{b}-\mathbf{a}) \\
[\mathbf{a}]: && \lambda \frac{rs}{1+r+rs} &= 1-\mu \\
[\mathbf{b}]: && \lambda \frac{1}{1+r+rs} &= \mu \\
\Rightarrow && \lambda \frac{1+rs}{1+r+rs} &= 1 \\
\Rightarrow && \lambda &= \frac{1+r+rs}{1+rs} \\
\Rightarrow && \mu &= \frac{1}{1+rs}
\end{align*}
Therefore the line is divided in the ratio $rs : 1$, and therefore we have proven Ceva's Theorem.
Of the pure maths questions, this was by far the least popular, with attempts from only around a quarter of the candidature. Very few attempts proceeded beyond the opening stages. Surprisingly – despite the fact that the whole question can be done with little more than the knowledge of how to split a line segment in a given ratio, vector equations of lines joining two given points, and the finding of a point of intersection of two lines – most attempts began with incorrect statements of the position vectors d and e in terms of b and a (respectively). Most candidates wisely gave up at this point, though some persevered, but with little success. The mean score of under 3/20 on this question reflects the paucity and brevity of efforts. Geometers amongst the readership may notice that this result follows from a combination of Menelaus' Theorem and Ceva's Theorem. A handful of candidates noticed it too, and scored most of the marks for relatively little effort.