A circle is inscribed in a rhombus $ABCD$ with one angle $60°$. The distance from the centre of the circle to the nearest vertex is equal to $1$. If $P$ is any point on the circle, then $|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2$ is equal to:
Step-by-Step Solution
Key Concept: For a rhombus with an inscribed circle, use the relationship between the inradius, side length, and angles to establish the coordinate system. Then apply the formula ∑|PA|² = ∑[(x-xᵢ)² + (y-yᵢ)²] = 4(x² + y² + d²) where d² is related to the vertices' positions, noting that for any point P on the circle, x² + y² = r².
Given $r = \sqrt{3}\sin 30° = \frac{\sqrt{3}}{2}$, compute $(PA)^2 + (PB)^2 + (PC)^2 + (PD)^2 = (x-\sqrt{3})^2 + y^2 + x^2 + (y-1)^2 + (x+\sqrt{3})^2 + y^2 + x^2 + (y+1)^2 = 4(x^2 + y^2 + 2) = 4\left(\frac{3}{4} + 2\right) = 11$.
Correct Answer: 2