Multiple choice

A circle is inscribed into a rhombus ABCD with one angle $\displaystyle 60^{\circ} .$ The distance from the centre of the circle to the nearest vertex is equal to $1$. If P is any point of the circle, then $\displaystyle \left | PA \right |^{2}+\left | PB \right |^{2}+\left | PC \right |^{2}+\left | PD \right |^{2}$ is equal to

  1. $12$
  2. $11$
  3. $9$

  4. none
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B Correct answer
Explanation

For a rhombus with a 60-degree angle, the distance from the center to the vertex is related to the geometry of the inscribed circle. Using coordinates or geometric properties, the sum of squared distances from any point on the circle to the vertices of the rhombus is constant and equals 11.

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