= Solution
An abstract <root system>, in the reduced crystallographic convention, is a finite subset $\Phi$ of a finite-dimensional real <Euclidean space> $E$ satisfying the following conditions:
* $0\notin\Phi$ and $\Phi$ spans $E$.
* For each $\alpha\in\Phi$, $\Phi\cap\mathbb R\alpha=\{\alpha,-\alpha\}$; this is the reduced condition.
* The <root reflection> $s_\alpha(v)=v-2(v,\alpha)\alpha/(\alpha,\alpha)$ preserves $\Phi$.
* Every <Cartan integer> $\langle\beta,\alpha^\vee\rangle=2(\beta,\alpha)/(\alpha,\alpha)$ is an integer.
The <root system> is irreducible if it is not a union of two nonempty mutually orthogonal <root systems>. The integrality condition is essential for the finite angle list below; arbitrary finite reflection systems need not satisfy it.
\b[The possible angles are $0$, $\pi/6$, $\pi/4$, $\pi/3$, $\pi/2$, $2\pi/3$, $3\pi/4$, $5\pi/6$, and $\pi$.] For <roots> $\alpha,\beta$, put
$$
a=\frac{2(\beta,\alpha)}{(\alpha,\alpha)},\qquad b=\frac{2(\alpha,\beta)}{(\beta,\beta)}.
$$
Both are <integers>, they have the same sign when nonzero, and
$$
ab=4\cos^2\theta.
$$
If the <roots> are linearly independent, strict <Cauchy-Schwarz inequality> gives $0\le ab<4$. Hence $ab\in\{0,1,2,3\}$. These values give, respectively,
$$
\begin{array}{c|c|c}
ab&\theta&\text{ratio of squared lengths, larger to smaller}\\\hline
0&\pi/2&\text{unrestricted by this calculation}\\
1&\pi/3,\ 2\pi/3&1\\
2&\pi/4,\ 3\pi/4&2\\
3&\pi/6,\ 5\pi/6&3
\end{array}
$$
For the last column, the possible absolute pairs $(|a|,|b|)$ are $(1,1)$, $(1,2)$, and $(1,3)$, up to interchange, and $a/b=(\beta,\beta)/(\alpha,\alpha)$. Parallel <roots> are equal or opposite by reducedness, giving $0$ or $\pi$. Thus the list is exhaustive, and the rank-two <root systems> realize each nonparallel possibility.
Now suppose two <simple roots> $\alpha,\beta$ have angle $5\pi/6$. Distinct <simple roots> have nonpositive inner product. Here is a proof from their defining sign property: if $(\delta,\epsilon)>0$, then $s_\delta(\epsilon)=\epsilon-n\delta$ with a positive <Cartan integer> $n$. This would be a <root> whose coefficients in the <root basis> include both a positive and a negative coefficient, which is impossible. Therefore, after normalizing the <simple roots> to unit length, any nonzero off-diagonal inner product is at most $-1/2$, by the angle list.
Let $u=\alpha/\|\alpha\|$, $v=\beta/\|\beta\|$, so $(u,v)=-\sqrt3/2$. If there were a third <simple root> with unit vector $w$, write $r=(u,w)\le0$ and $s=(v,w)\le0$. Independence of the <simple roots> makes their <Gram matrix> positive definite, so
$$
0<\det\begin{pmatrix}1&-\sqrt3/2&r\\-\sqrt3/2&1&s\\r&s&1\end{pmatrix}
=\frac14-r^2-s^2-\sqrt3rs.
$$
If either $r$ or $s$ were nonzero, its absolute value would be at least $1/2$. Since $rs\ge0$, the displayed determinant would then be nonpositive. Thus $r=s=0$. Every other <simple root> is orthogonal to both $\alpha$ and $\beta$: a <triple bond isolates a G2 component>.
Indeed, the <root reflections> in $\alpha,\beta$ preserve their plane and fix its orthogonal complement; the <root reflections> in the other <simple roots> preserve that complement and fix the plane. The given facts that these reflections generate the <Weyl group> and that every <root> is <Weyl group>-conjugate to a <simple root> imply that every <root> lies entirely in one of those two orthogonal subspaces. If there were any other <simple root>, this would make $\Phi$ reducible. Therefore irreducibility forces rank two.
Up to interchanging the <simple roots>, take $\alpha$ short and $\beta$ long. The squared-length ratio is $3$. Choose a scale and orthonormal coordinates so that
$$
\alpha=(1,0),\qquad\beta=\left(-\frac32,\frac{\sqrt3}{2}\right).
$$
The <Cartan integers> are $\langle\beta,\alpha^\vee\rangle=-3$ and $\langle\alpha,\beta^\vee\rangle=-1$. In coefficient coordinates $m\alpha+n\beta$, the two <root reflections> are
$$
s_\alpha(m,n)=(-m+3n,n),\qquad s_\beta(m,n)=(m,m-n).
$$
Their orbits of $\alpha$ and $\beta$ are, respectively,
$$
\{\pm\alpha,\pm(\alpha+\beta),\pm(2\alpha+\beta)\},\qquad
\{\pm\beta,\pm(3\alpha+\beta),\pm(3\alpha+2\beta)\}.
$$
These sets are closed under both displayed reflections; for example $s_\beta(\alpha)=\alpha+\beta$, $s_\alpha(\alpha+\beta)=2\alpha+\beta$, $s_\alpha(\beta)=3\alpha+\beta$, and $s_\beta(3\alpha+\beta)=3\alpha+2\beta$. The assumption that every <root> is conjugate to a <simple root> now determines all twelve <roots>, with no further ones possible.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-102-g2-roots.png]
{title=The twelve G2 roots, with the six positive roots labelled and the simple roots emphasized}
For completeness, this twelve-element set really is a <root system>. It is finite, spans the plane, and is reduced. The six <short roots> have length $1$ and directions spaced by $\pi/3$; the six <long roots> have length $\sqrt3$ and interleaved directions, also spaced by $\pi/3$. Their mutual <Cartan integers> are integral: within either length class they are $0$, $\pm1$, or $\pm2$ as applicable, and between the two length classes they are $0$, $\pm1$, or $\pm3$. Closure under all <root reflections> follows by conjugating the two simple reflections, since every listed <root> lies in a simple-root orbit. The nonorthogonal <simple roots> make it irreducible. We have therefore proved existence as well as uniqueness, up to isometry and overall scale:
$$
\boxed{\text{The unique irreducible type is the }G_2\text{ root system}.}
$$
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