# (PA)^2 + (PB)^2 +(PC)^2 + (PD)^2 is equal to?

A circle is inscribed into the rhombus ABCD with one angle 60. The distance from the centre of the circle to the narest vertex is equal to 1. If P is any point of the circle ,then $$(PA)^2 + (PB)^2 +(PC)^2 + (PD)^2$$ is equal to?

• Fix your title please – Alexander Gruber Nov 18 '13 at 6:56
• goo.gl/8WZI2X – lab bhattacharjee Nov 18 '13 at 7:01
• answer given is 11 but he is getting 11/3 – maths lover Nov 18 '13 at 7:09
• The solution on that link is wrong. The answer I got is $8+2\sqrt{3}$. – Tigran Hakobyan Nov 18 '13 at 7:18
• @mathslover That answer is very close, it just make the mistake $\frac{1}{\sqrt{3}} > 1$. You can get the right answer by scaling the geometric figure by $\sqrt{3}$. – achille hui Nov 18 '13 at 7:19