Year: 2023
Paper: 2
Question Number: 8
Course: UFM Pure
Section: Vectors
No solution available for this problem.
Many candidates were able to express their reasoning clearly and presented good solutions to the questions that they attempted. There were excellent solutions seen for all of the questions. An area where candidates struggled in several questions was in the direction of the logic that was required in a solution. Some candidates failed to appreciate that separate arguments may be needed for the "if" and "only if" parts of a question and, in some cases, candidates produced correct arguments, but for the wrong direction. In several questions it was clear that candidates who used sketches or diagrams generally performed much better that those who did not. Sketches often also helped to make the solution clearer and easier to understand. Several questions on the STEP papers ask candidates to show a given result. Candidates should be aware that there is a need to present sufficient detail in their solutions so that it is clear that the reasoning is well understood.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A tetrahedron is called isosceles if each pair of edges which do not share a vertex have equal length.
\begin{questionparts}
\item Prove that a tetrahedron is isosceles if and only if all four faces have the same perimeter.
\end{questionparts}
Let $OABC$ be an isosceles tetrahedron and let $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$ and $\overrightarrow{OC} = \mathbf{c}$.
\begin{questionparts}
\setcounter{enumi}{1}
\item By considering the lengths of $OA$ and $BC$, show that
\[2\mathbf{b}.\mathbf{c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2.\]
Show that
\[\mathbf{a}.(\mathbf{b}+\mathbf{c}) = |\mathbf{a}|^2.\]
\item Let $G$ be the centroid of the tetrahedron, defined by $\overrightarrow{OG} = \frac{1}{4}(\mathbf{a}+\mathbf{b}+\mathbf{c})$.
Show that $G$ is equidistant from all four vertices of the tetrahedron.
\item By considering the length of the vector $\mathbf{a}-\mathbf{b}-\mathbf{c}$, or otherwise, show that, in an isosceles tetrahedron, none of the angles between pairs of edges which share a vertex can be obtuse. Can any of them be right angles?
\end{questionparts}
In part (i), most candidates answered the "only if" direction of the argument successfully, often with a diagram. Many candidates did not realise they needed to prove "if" separately, but those that did usually answered this well. Most students wrote a list of simultaneous equations in the edge lengths to solve. Appeals to symmetry were accepted without much detail needed. Many candidates did not attempt the later parts of the question. Part (ii) was generally answered well. Many students attempted the cosine rule, which required some detail relating it to the given problem. A surprising number of candidates would set the direction vectors of each edge equal, rather than just their lengths. Some ended up confusing scalars and vectors due to poor notation. In part (iii) many candidates found a.g etc. rather than |AG|. This could be made to work but needed to be made relevant to the question to earn marks. Some candidates thought a, b, c were the components of a vector and attempted to use Pythagoras, which got no credit. Several candidates found |AG|² etc. in a non-symmetric form and attempted to appeal to symmetry, which was not accepted. Part (iv) proved to be quite tricky for most candidates. Many ignored the questions prompting entirely or failed to relate it to a relevant geometrical idea. Few candidates used cosine successfully and attained the final two marks. Several candidates stated that right angles were possible at the end.