2010 Paper 2 Q6
Year: 2010
Paper: 2
Question Number: 6
Course: LFM Pure
Section: Introduction to trig
Problem
- Find the length of \(PD\).
- Show that the cosine of the angle between adjacent faces of the tetrahedron is \(1/3\).
- Find the radius of the largest sphere that can fit inside the tetrahedron.
Solution
- \(D\) must be above the centre (of any kind) of the equilateral triangle \(ABC\). Therefore it is a distance \(\frac23 \frac{\sqrt{3}}2 = \frac{\sqrt{3}}3\) from \(A\). \(D\) is \(1\) from \(A\), therefore by Pythagoras \(PD = \sqrt{1-\frac13} = \sqrt{\frac23}\)
- We can place \(D\) at \(\langle 0,0,\sqrt{\frac23}\rangle\) and \(A'\) (the midpoint of \(BC\)) at \(\langle-\frac{\sqrt{3}}{6},0,0 \rangle\) and we find: \begin{align*} && \cos \theta &= \frac{(\mathbf{a}'-\mathbf{d})\cdot \mathbf{a}'}{|\mathbf{a}'-\mathbf{d}|| \mathbf{a}'|} \\ &&&= \frac{|\mathbf{a}'|}{|\mathbf{a}'-\mathbf{d}|} \\ &&&= \frac{\frac{\sqrt{3}}{6}}{\sqrt{\frac23+\frac{3}{36}}} = \frac13 \end{align*}
- We have And therefore we must have \(\tan \frac{\cos^{-1} \frac13}{2} = \frac{r}{\frac{\sqrt{3}}{6}}\) therefore \begin{align*} && r &= \frac{\sqrt{3}}{6} \tan \left (\frac{\cos^{-1} \frac13}{2} \right) \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{1-\cos(\cos^{-1}\frac13)}{1+\cos(\cos^{-1}\frac13)}} \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{\frac23}{\frac43}} \\ &&&= \frac{\sqrt{6}}{12} \end{align*}
Examiner's report
— 2010 STEP 2, Question 6There were just under 1000 entries for paper II this year, almost exactly the same number as last year. Of this number, more than 60 scored over 90% while, at the other end of the scale, almost 200 failed to score more than 40 marks. In hindsight, many of the pure maths questions were a little too accessible and lacked a sufficiently tough 'difficulty gradient', so that scores were slightly higher than anticipated. This was reflected in the grade boundaries for the "1" and the "2" (around ten marks higher than is generally planned) in particular. Next year's questions may be expected to be a little bit more demanding, but only in the sense that the final 5 or 6 marks on each question should have rather more bite to them: it should certainly not be the case that all questions are tougher to get into at the outset. Most candidates attempted the requisite number of questions (six), although many of the weaker brethren made seven or eight attempts, most of which were feeble at best and they generally only picked up a maximum of 5 or 6 marks per question. It is a truth universally acknowledged that practice maketh if not perfect then at least a whole lot better prepared, and choosing to waste time on a couple of extra questions is not a good strategy on the STEPs. The major down-side of the present modular examination system is that students are not naturally prepared to approach the subject holistically; ally this to the current practice of setting highly-structured, fully-guided questions requiring no imagination, insight, depth or planning from A-level candidates in a system that fails almost nobody and rewards even the most modestly able with high grades in a manner reminiscent of a dentist giving lollipops to kids who have done little more than been brave and seen the course through, it is even more important to ensure a full and thorough preparation for these papers. The 20% of the entry who seem to be either unprepared for the rigours of a STEP, or unwittingly possessed of only a smattering of basic advanced-level skills, seems to be remarkably steady year-on-year, even in a year when their more suitably prepared compatriots found the paper appreciably easier than usual. 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.
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show LaTeX source
Problem source
Each edge of the tetrahedron $ABCD$ has unit length. The face
$ABC$ is horizontal, and
$P$ is the point in $ABC$ that is vertically below $D$.
\begin{questionparts}
\item Find the length of $PD$.
\item Show that the cosine of the angle between adjacent faces of the
tetrahedron is
$1/3$.
\item Find the radius of the largest sphere that can fit inside
the tetrahedron.
\end{questionparts}Solution source
\begin{questionparts}
\item $D$ must be above the centre (of any kind) of the equilateral triangle $ABC$. Therefore it is a distance $\frac23 \frac{\sqrt{3}}2 = \frac{\sqrt{3}}3$ from $A$. $D$ is $1$ from $A$, therefore by Pythagoras $PD = \sqrt{1-\frac13} = \sqrt{\frac23}$
\item We can place $D$ at $\langle 0,0,\sqrt{\frac23}\rangle$ and $A'$ (the midpoint of $BC$) at $\langle-\frac{\sqrt{3}}{6},0,0 \rangle$ and we find:
\begin{align*}
&& \cos \theta &= \frac{(\mathbf{a}'-\mathbf{d})\cdot \mathbf{a}'}{|\mathbf{a}'-\mathbf{d}|| \mathbf{a}'|} \\
&&&= \frac{|\mathbf{a}'|}{|\mathbf{a}'-\mathbf{d}|} \\
&&&= \frac{\frac{\sqrt{3}}{6}}{\sqrt{\frac23+\frac{3}{36}}} = \frac13
\end{align*}
\item We have \begin{tikzpicture}
% --- Definitions ---
\def\R{2} % Radius of the circle
% Calculate angle alpha = arccos(1/3)
\pgfmathsetmacro{\ang}{acos(1/3)}
% Calculate distance from Origin to Tangent points
% Geometry: tan(alpha/2) = R / distance
% We need L = R * cot(alpha/2)
% By half angle identities, if cos(a) = 1/3, then cot(a/2) = sqrt(2).
\pgfmathsetmacro{\dist}{\R * sqrt(2)}
% --- Coordinates ---
\coordinate (O) at (0,0); % Origin (Vertex)
\coordinate (T1) at (\dist, 0); % Tangent point on flat line
\coordinate (T2) at (\ang:\dist); % Tangent point on angled line
% Center of circle is at (T1 x, R) because T1 is on the x-axis
\coordinate (C) at (\dist, \R);
% --- Drawing ---
% 1. The Circle
\draw[thick, fill=blue!5] (C) circle (\R);
\fill[black] (C) circle (1.5pt) node[above right] {$C$};
% 2. The Lines (drawn slightly longer than the tangent distance)
\draw[thick] (O) -- ++(0: \dist + 1.5);
\draw[thick] (O) -- ++(\ang: \dist + 1.5);
% 3. The Radii
\draw[red, thick, dashed] (C) -- (T1) node[midway, right] {$r$};
\draw[red, thick, dashed] (C) -- (T2) node[midway, above left] {$r$};
% 4. Tangent Points
\fill[red] (T1) circle (1.5pt);
\fill[red] (T2) circle (1.5pt);
% 5. Right Angle Markers
% At T1 (flat)
\draw ($(T1)!0.25cm!(O)$) -- ++(0,0.25) -- ($(T1)!0.25cm!(C)$);
% At T2 (angled) - we rotate the marker to match the line slope
\draw[rotate around={\ang:(T2)}] ($(T2)!0.25cm!(O)$) -- ++(0,-0.25) -- ($(T2)!0.25cm!(C)$);
% 6. Label the main angle
\pic [draw, "$\tiny \cos^{-1}(1/3)$", angle radius=.5cm, angle eccentricity=1.5] {angle = T1--O--T2};
\draw (C) -- (O);
\end{tikzpicture}
And therefore we must have $\tan \frac{\cos^{-1} \frac13}{2} = \frac{r}{\frac{\sqrt{3}}{6}}$ therefore
\begin{align*}
&& r &= \frac{\sqrt{3}}{6} \tan \left (\frac{\cos^{-1} \frac13}{2} \right) \\
&&&= \frac{\sqrt{3}}6 \sqrt{\frac{1-\cos(\cos^{-1}\frac13)}{1+\cos(\cos^{-1}\frac13)}} \\
&&&= \frac{\sqrt{3}}6 \sqrt{\frac{\frac23}{\frac43}} \\
&&&= \frac{\sqrt{6}}{12}
\end{align*}
\end{questionparts}
Of the pure maths questions on the paper, only Q5 and this one attracted attempts from under half the candidature; this despite the fact that it is obviously (to the trained eye, at least) the easiest question on the paper. Parts (i) and (ii) require nothing more than GCSE trigonometry, and (iii) can be done in one line if one knows a little bit about geometric centres of 3-d shapes. Clearly 3-dimensional objects, and the associated trig., are sufficiently daunting to have put most folks off either completely or early on in the proceedings, and the average mark scored here was under 10.