A circle is inscribed (i.e. touches all four sides) into rhombus ABCD with one angle 60 degree. The distance from centre of circle to the nearest vertex is 1. If P is any point on the circle, then value of
$|PA|^2+|PB|^2+|PC|^2+|PD|^2$ will be?
Can something hint the starting approach for this question.
Hint:
If $\angle DAB=\angle DCB=60°$ than The triangles $DAB$ and $DCB$ are equilateral, so $\angle ADB=60°$. Let $O$ the center of the circle, than the radius of the circle is $r=AO\sin 60°=\frac{\sqrt{3}}{2}$ and the distance $AO=\sqrt{3}$.
Now you can use a coordinate sistem with center $O$, a point on the circle has coordinates $P=(r\cos \theta, r \sin \theta)$ and you can find the distances from the vertices of the rhombus.
So you can prove if the sum of the squares of these distances is constant and find its value.
$$\angle POB=\theta \qquad P=\left(\frac{\sqrt{3}}{2}\cos \theta,\frac{\sqrt{3}}{2}\sin \theta \right)$$
$$ A=\left(0,\sqrt{3} \right) \quad B=\left(1,0\right) \quad C=\left(0,-\sqrt{3} \right) \quad D=\left(-1,0 \right) $$