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?

  • $\begingroup$ Fix your title please $\endgroup$ – Alexander Gruber Nov 18 '13 at 6:56
  • $\begingroup$ goo.gl/8WZI2X $\endgroup$ – lab bhattacharjee Nov 18 '13 at 7:01
  • $\begingroup$ answer given is 11 but he is getting 11/3 $\endgroup$ – maths lover Nov 18 '13 at 7:09
  • $\begingroup$ The solution on that link is wrong. The answer I got is $8+2\sqrt{3}$. $\endgroup$ – Tigran Hakobyan Nov 18 '13 at 7:18
  • $\begingroup$ @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 $\sqrt{3}$. $\endgroup$ – achille hui Nov 18 '13 at 7:19

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