A regular hexagon is inscribed in a unit circle. If the hexagon is rotated counter-clockwise about its center by an angle of $150^\circ$, how many of its original 6 vertices will coincide with the vertices of the unrotated hexagon?

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### 1. Strategy & Theoretical Principles A regular hexagon has rotational symmetry of order 6. Its rotational symmetry angles are integer multiples of the vertex separation angle: $$\theta_k = k \times \left(\frac{360^\circ}{6}\right) = k \times 60^\circ, \quad k \in \mathbb{Z}$$ ### 2. Analytical Formulation & Calculation The vertices will coincide with the original positions if and only if the rotation angle is an exact integer multiple of $60^\circ$. The given rotation angle is $\theta = 150^\circ$. Dividing by the fundamental symmetry angle: $$\frac{150^\circ}{60^\circ} = 2.5$$ Since $2.5$ is not an integer, no vertex falls onto any existing vertex position. Therefore, the number of vertices that coincide with their original positions is **0**. ### 3. Distractor Forensics - For $\theta = 120^\circ$ or $180^\circ$, all 6 vertices coincide. For non-integer multiples of $60^\circ$, exactly 0 vertices coincide. ### 4. Exam Trap & Standard Reference - **Trap Alert:** For regular $n$-gons, either all vertices coincide (if $\theta = k \cdot 360/n$) or none coincide. - **Standard Reference:** GATE General Aptitude Official Syllabus - Spatial Reasoning.

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