2025 AMC 10A Problem 6

1. Call the triangle $ABC$. Each corner is $60^\circ$, and trisecting cuts it into three $20^\circ$ slices. At corner $A$, the middle slice lies between two rays: one makes $20^\circ$ with side $AB$, the other makes $40^\circ$ with side $AB$.

1. The hexagon is the region inside all three middle slices, so its six sides lie along these six rays. Each corner of the hexagon is where a ray from one triangle corner crosses a ray from another. Draw a picture and focus on side $AB$; it helps find two hexagon corners.

The corner close to side $AB$.

1. Take the ray from $A$ making $20^\circ$ with $AB$, and the ray from $B$ making $20^\circ$ with $BA$. They meet at a point $P$ just above the middle of $AB$.

4. In triangle $ABP$, two angles are $20^\circ$, so the third is
$$\angle APB = 180^\circ - 20^\circ - 20^\circ = 140^\circ.$$
The hexagon sits in the angle directly opposite this one (on the far side of $P$ from $AB$), and opposite angles are equal. So this hexagon angle is $140^\circ$.

The corner far from side $AB$.

1. Now take the two rays that make $40^\circ$ with $AB$, one from $A$ and one from $B$. They meet at a point $Q$, also on the line down the middle of the triangle, but on the other side of the center.

6. In triangle $ABQ$,
$$\angle AQB = 180^\circ - 40^\circ - 40^\circ = 100^\circ,$$
and the hexagon lies inside this angle, so this hexagon angle is $100^\circ$.

All six corners.

1. Sides $BC$ and $CA$ give the same two kinds of corners, so there are three $140^\circ$ angles and three $100^\circ$ angles.

1. Check: a hexagon's angles add to $(6-2) \times 180^\circ = 720^\circ$, and $3 \times 140^\circ + 3 \times 100^\circ = 420^\circ + 300^\circ = 720^\circ$. Correct, so there are no other corners.

The smallest angle is $100^\circ$.

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

Each hexagon corner is the tip of a triangle whose base is a side of $ABC$ and whose base angles are known, so one use of "angles in a triangle add to $180^\circ$" finds it. Checking the total against $720^\circ$ confirms you found every corner. Next time a shape is cut out by several rays, look for small triangles built on sides you already know.

Use symmetry. Turning the picture one third of a full turn about the center gives the same picture, so the hexagon's angles repeat in a pattern: big, small, big, small, big, small. Call them $x$ and $y$. Then $3x + 3y = 720^\circ$, so $x + y = 240^\circ$. The corner nearest a side is easy (step 4): $140^\circ$. So the other angle is $240^\circ - 140^\circ = 100^\circ$, and that is the smallest.

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 $\boxed{\textbf{(C)}\ 100}$.