2025 AMC 10A Problem 6

Call the triangle $ABC$. Each angle is $60^\circ$, so the trisectors cut it into three $20^\circ$ parts and the middle wedge at $A$ is bounded by the ray making $20^\circ$ with $AB$ and the ray making $20^\circ$ with $AC$ (equivalently, $40^\circ$ with $AB$).

Six rays in all bound the three middle wedges, and the hexagon is the region common to the wedges, so each of its six sides lies on one of these rays and each of its vertices is a crossing of two rays drawn from two different corners. Take the side $AB$ and look at the two crossings it governs.

The two rays making $20^\circ$ with $AB$. One comes from $A$, one from $B$; call their crossing $P$. In triangle $ABP$ the angles at $A$ and $B$ are both $20^\circ$, so
$$\angle APB = 180^\circ - 20^\circ - 20^\circ = 140^\circ.$$
$P$ lies on the perpendicular bisector of $AB$, between $AB$ and the centre, and it is the hexagon vertex nearest to $AB$; the two sides meeting there are the two rays, so the interior angle at $P$ is $140^\circ$.

The two rays making $40^\circ$ with $AB$. Again one from each of $A$ and $B$; call their crossing $Q$. In triangle $ABQ$,
$$\angle AQB = 180^\circ - 40^\circ - 40^\circ = 100^\circ,$$
and $Q$ lies on the same perpendicular bisector, beyond the centre, with interior angle $100^\circ$.

Each of the three sides of $ABC$ supplies one such $P$ and one such $Q$, which accounts for six vertices, and the angles total $3(140^\circ) + 3(100^\circ) = 720^\circ$, exactly the interior angle sum of a hexagon. So these are all of them, and they alternate $140^\circ, 100^\circ, 140^\circ, 100^\circ, 140^\circ, 100^\circ$ around the figure.

The answer is $\boxed{\textbf{(C)}\ 100}$.

Nothing here needs lengths: each hexagon vertex is the apex of a triangle whose other two vertices are corners of $ABC$ and whose base angles are known multiples of $20^\circ$, so a single application of the angle sum settles each one. The check against $720^\circ$ is what licenses the claim that no other crossings are vertices, and it is worth running whenever a region is described as the intersection of several wedges.

Exploit the symmetry first. Rotating by $120^\circ$ about the centre maps the configuration to itself, so the hexagon's angles repeat in a pattern $\alpha, \beta, \alpha, \beta, \alpha, \beta$ with $3\alpha + 3\beta = 720^\circ$, that is $\alpha + \beta = 240^\circ$. The easy vertex to compute is the one nearest a side, giving $\alpha = 140^\circ$, so $\beta = 100^\circ$ and the smallest angle is $100^\circ$.

• Assuming the hexagon is regular because the configuration is symmetric, and answering 120, choice (E).
• Pairing a $20^\circ$ ray from $A$ with a $40^\circ$ ray from $B$, a crossing that lies outside the hexagon, and computing $180^\circ - 20^\circ - 40^\circ = 120^\circ$.
• Reporting the larger of the two alternating angles, or assuming the smallest angle must be the one adjacent to the triangle's side.