All edges in the <Type E(p,q,k) Coxeter graph> are unlabelled, so the factor $a^2$ in part c is one. Separate the arm of length $p$ from the central vertex. The remaining two arms form a type $A_{q+k+1}$ chain, while deleting the central vertex leaves the disjoint type $A_q$ and type $A_k$ chains. Using $\det G(A_r)=r+1$ in the formula from part c yields
$$
\begin{aligned}
\det G(E(p,q,k))
&=(p+1)(q+k+2)-p(q+1)(k+1)\\
&=(p+1)(q+1)(k+1)
\left(\frac1{p+1}+\frac1{q+1}+\frac1{k+1}-1\right).
\end{aligned}
$$
The associated <bilinear form> is degenerate exactly when
$$
\frac1{p+1}+\frac1{q+1}+\frac1{k+1}=1.
$$
The positive-integer solutions of $1/a+1/b+1/c=1$, up to permutation, are $(3,3,3)$, $(2,4,4)$, and $(2,3,6)$. Consequently the degenerate arm-length triples are the permutations of
$$
(2,2,2),\qquad(1,3,3),\qquad(1,2,5).
$$
For every other allowed $(p,q,k)$ the determinant is nonzero, so the form is <nondegenerate bilinear form>[nondegenerate].

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 111 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-111/2/d/solution