2025 AMC 10A Problem 6

Each $60^\circ$ vertex angle splits into three $20^\circ$ pieces; consider triangles formed by trisectors from two vertices on one side

Use the triangle angle sum at the intersections with base-angle pairs $40^\circ,40^\circ$ and $20^\circ,20^\circ$

Label the equilateral triangle $ABC.$ Each $60^\circ$ angle splits into three $20^\circ$ pieces. Take the outermost trisectors from $A$ and $B$: they meet at base angles $\tfrac23 \cdot 60^\circ = 40^\circ,$ so the hexagon vertex there has angle $180^\circ - 2\cdot 40^\circ = 100^\circ.$ The innermost trisectors from $A$ and $B$ meet at base angles $20^\circ,$ giving apex $180^\circ - 2\cdot 20^\circ = 140^\circ,$ and by vertical angles that’s the opposite hexagon angle. So the six angles alternate $100^\circ$ and $140^\circ.$ The smallest is $100^\circ.$ Therefore, the answer is C.