= Solution
<Sperner's lemma> says that if a <triangle> is triangulated, its vertices are labelled $1,2,3$, and label $i$ is forbidden on the edge opposite vertex $i$, then an odd number of small triangles have all three labels.
To prove it, count incidences with edges whose endpoint labels are $1$ and $2$. Along the outer $12$-edge, the labels begin at $1$ and end at $2$, so the number of $12$ transitions is odd; no other boundary edge contributes. An interior edge contributes twice. A small triangle contributes an odd number precisely when its three labels are $1,2,3$: a triangle using only $1,2$ contributes two, and every other non-tricoloured triangle contributes zero. The number of tricoloured triangles is therefore odd.
Now suppose the closed sets $A_i$ in the question existed. Let $d_i(x)$ be the <distance from a point to a closed set> $A_i$. Empty triple intersection makes $d_1+d_2+d_3>0$, so
$$
r(x)=\frac{d_1(x)v_1+d_2(x)v_2+d_3(x)v_3}
{d_1(x)+d_2(x)+d_3(x)}
$$
is a <continuous function> from the large triangle to itself, where $v_i$ is the vertex opposite $\alpha_i$. Since the $A_i$ cover the triangle, at least one $d_i(x)$ vanishes, so $r(x)$ always lies on the boundary. On $\alpha_i\subseteq A_i$, its $i$th barycentric coordinate vanishes, hence $r(\alpha_i)\subseteq\alpha_i$. Thus $r$ is face-preserving on the boundary; its boundary restriction is homotopic there to the identity by the straight-line homotopy. This would give a map from the triangle into its boundary whose boundary degree is one, contradicting the <no-retraction theorem>, the standard topological consequence of Sperner's lemma. \b[Therefore the three closed sets must have a common point.]
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