Question:

Step 1: Determine the coordinates of the vertices of the rhombus and the radius of the inscribed circle. Let the center of the inscribed circle be the origin $O(0,0)$. The diagonals of a rhombus bisect each other perpendicularly at its center. We can align these diagonals with the coordinate axes. For a rhombus with one angle $60^\circ$, the other angles are $120^\circ$. Let the side length of the rhombus be $s$. The lengths of the half-diagonals are $s/2$ and $s\sqrt{3}/2$. The distance from the center to the nearest vertex is given as $1$. This corresponds to the shorter half-diagonal. So, $\frac{s}{2} = 1 \implies s=2$. The distances from the center to the vertices are $1$ and $\sqrt{3}$. Let the vertices be $A(\sqrt{3}, 0)$, $B(0, 1)$, $C(-\sqrt{3}, 0)$, and $D(0, -1)$. The radius $r$ of the inscribed circle is half the height of the rhombus. The height $h = s \sin(60^\circ) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$. Thus, the radius $r = \frac{h}{2} = \frac{\sqrt{3}}{2}$. Alternatively, using the specific form provided in the original solution: $r = \sqrt{3} \sin 30^\circ = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$. Since the circle is centered at the origin, its equation is $x^2 + y^2 = r^2$. Step 2: Express the sum of squares of distances from point $P$ to the vertices. Let $P(x,y)$ be any point on the circle. The coordinates of the vertices are $A(\sqrt{3}, 0)$, $B(0, 1)$, $C(-\sqrt{3}, 0)$, and $D(0, -1)$. We need to calculate $|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2$. The square of the distance from $P(x,y)$ to a vertex $V(x_v, y_v)$ is $(x-x_v)^2 + (y-y_v)^2$. $$|PA|^2 = (x - \sqrt{3})^2 + (y - 0)^2 = (x - \sqrt{3})^2 + y^2$$ $$|PB|^2 = (x - 0)^2 + (y - 1)^2 = x^2 + (y - 1)^2$$ $$|PC|^2 = (x - (-\sqrt{3}))^2 + (y - 0)^2 = (x + \sqrt{3})^2 + y^2$$ $$|PD|^2 = (x - 0)^2 + (y - (-1))^2 = x^2 + (y + 1)^2$$ Now, sum these four expressions: $$|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2 = [(x - \sqrt{3})^2 + y^2] + [x^2 + (y - 1)^2] + [(x + \sqrt{3})^2 + y^2] + [x^2 + (y + 1)^2]$$ Step 3: Simplify the sum of squares. Expand each term in the sum: $$ (x^2 - 2\sqrt{3}x + 3) + y^2 + x^2 + (y^2 - 2y + 1) + (x^2 + 2\sqrt{3}x + 3) + y^2 + x^2 + (y^2 + 2y + 1) $$ Group the like terms: $$ (x^2 + x^2 + x^2 + x^2) + (y^2 + y^2 + y^2 + y^2) + (-2\sqrt{3}x + 2\sqrt{3}x) + (-2y + 2y) + (3 + 1 + 3 + 1) $$ Simplify the grouped terms: $$ 4x^2 + 4y^2 + 0x + 0y + 8 $$ The sum simplifies to: $$ |PA|^2 + |PB|^2 + |PC|^2 + |PD|^2 = 4x^2 + 4y^2 + 8 = 4(x^2 + y^2) + 8 $$ Step 4: Substitute the circle equation to find the final value. Since $P(x,y)$ is any point on the inscribed circle, its coordinates satisfy the circle's equation $x^2 + y^2 = r^2$. From Step 1, we found the radius $r = \frac{\sqrt{3}}{2}$. Therefore, $x^2 + y^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$. Substitute this value into the simplified sum from Step 3: $$ |PA|^2 + |PB|^2 + |PC|^2 + |PD|^2 = 4\left(\frac{3}{4}\right) + 8 $$ $$ = 3 + 8 $$ $$ = 11 $$ The value of $|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2$ is $11$. The final answer is $\boxed{\text{11}}$.