Question

A circle is inscribed into a rhombus abcd with one angle 60. The distance from the center of the circle to the nearest vertex is equal ti 1 if p is any point of the circle then |pa|^2+|pb|^2+|pc|^2+|pd|^2 is equal to

Answer 1

Thank you for asking this question. Here is your answer:

If ∠DAB=∠DCB=60°

This means that DAB  and DCB are equilateral

so ∠ADB=60°

∠POB=θP=(√3/2 cosθ,√3/2sinθ)

A=(0,3–√)B=(1,0)C=(0,−3–√)D=(−1,0)

(Let O the center of the circle

and the radius of the circle is r=AOsin60°

=√3/2 and the distance

AO=√3

If there is any confusion please leave a comment below.