Year: 2015
Paper: 2
Question Number: 3
Course: LFM Stats And Pure
Section: Combinatorics
As in previous years the Pure questions were the most popular of the paper with questions 1, 2 and 6 the most popular. The least popular questions on the paper were questions 8, 11 and 13 with fewer than 250 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1483.4
Banger Comparisons: 1
Three rods have lengths $a$, $b$ and $c$, where $a< b< c$. The three rods can be made into a triangle (possibly of zero area) if $a+b\ge c$.
Let $T_{n}$ be the number of triangles that can be made with three rods chosen from $n$ rods of lengths $1$, $2$, $3$, $\ldots$ , $n$ (where $n\ge3$). Show that $T_8-T_7 = 2+4+6$ and evaluate $T_8 -T_6$. Write down expressions for $T_{2m}-T_{2m-1}$ and $T_{2m} - T_{2m-2}$.
Prove by induction that $T_{2m}=\frac 16 m (m-1)(4m+1)\,$, and find the corresponding result for an odd number of rods.Every $T_7$ triangle is valid, so we are interested in new triangles which have $8$ has a longest side.
We can have:
\begin{array}{c|c|c}
\text{longest} & \text{middle} & \text{shortest} \\ \hline
8 & 7 & 1-6 \\
8 & 6 & 2-5 \\
8 & 5 & 3-4
\end{array}
which is $6+4+2$ extra triangles.
The new ones excluding all the sixes are:
\begin{array}{c|c|c}
\text{longest} & \text{middle} & \text{shortest} \\ \hline
8 & 7 & 1-6 \\
8 & 6 & 2-5 \\
8 & 5 & 3-4 \\
7 & 6 & 1-5 \\
7 & 5 & 2-4 \\
7 & 4 & 3 \\
\end{array}
Ie $2+4+6 + 1 + 3+5$
$T_{2m}-T_{2m-1} = 2 \frac{(m-1)m}{2} = m(m-1)$ and $T_{2m}-T_{2m-2} = \frac{(2m-2)(2m-1)}{2}$
$T_4 = 3$ ($1,2,3$, $1,3,4$, $2,3,4$)
and $\frac16 \cdot 2 \cdot 1 \cdot 9 = 3$ so the base case holds.
Suppose it's true for some $m = k$, then
\begin{align*}
&& T_{2(k+1)} &= T_{2k} + \frac{2m(2m+1)}{2} \\
&&&= \frac{m(m-1)(4m+1)}{6} + \frac{6m(2m+1)}{6}\\
&&&= \frac{m(4m^2-3m-1+12m+6)}{6} \\
&&&= \frac{m(4m^2+9m+5)}{6}\\
&&&= \frac{m(4m+5)(m+1)}{6}\\
&&&= \frac{(m+1-1)(4(m+1)+5)(m+1)}{6}\\
\end{align*}
as required, therefore it is true by induction.
For odd numbers, we can see that $T_{2m-1} = \frac{m(m-1)(4m+1)}{6} - m(m-1) = \frac{m(m-1)(4m-5)}{6}$
On the whole attempts at this question were good with a significant number of candidates obtaining full marks. In the first part of the question a number of candidates did not interpret the difference between successive terms of the sequence as triangles which included a particular length edge and chose to enumerate all possible cases – if this was carried out correctly it was still possible to achieve full marks on this section. Even when unsuccessful in this part of the question many candidates were able to write down correct expressions for the general cases. The proof by induction was generally well done, although a number of candidates failed to justify the first case fully (which can easily be done by enumerating all of the cases). The final part of the question (the corresponding result for an odd number of rods) was not attempted by all candidates. Of those that did, those who attempted to use induction rather than applying the result from earlier in the question struggled to reach the correct answer.