Sum of Squared Lengths in an Octagon inscribed in a Circle

Question

Regular octagon $ABCDEFGH$ is inscribed in a circle of radius 5. Let $P$ be a point on this circle. Compute $$PA^2 + PB^2 + PC^2 + PD^2 + PE^2 + PF^2 + PG^2 +PH^2.$$

Answer 1

The vertex of the regular Octagon in the complex plane are the solutions $z_i=5e^{\frac{2k\pi i}{8}},\;i=0\ldots 7$ of the equation

$z^8=5^8$

The point $P$ is the number $5e^{it}$

The requested sum is

$$\sum_{i=0}^7\,(5e^{it}-z_i)^2=\sum_{i=0}^7\,(5e^{it}-5e^{\frac{2k\pi i}{8}})^2=8\cdot (25+25)=400$$

Hope this can be useful

Answer 2

By the Pythagorean theorem, if you take two opposite points, such as $A$ and $E$, summing the squares of their distances from $P$ gives the square of the distance between the two, which is 10 ($PA^2+PE^2=100$). Therefore since 4 such pairs exist, our answer is 400.

Answer 3

Use the Thales theorem and Pythagoras theorem and we get $4\cdot (2r)^2=16r^2$

Answer 4

Hint: How much is $PA^2+PE^2$? ($\triangle AEP$ is right.) Then do $PB^2+PF^2$ etc. and finally add them all up.

Answer 5

If $A_1,A_2,\ldots,A_n$ are the vertices of a regular $n$-agon inscribed in a circle with radius $R$ (centered at the origin $O$), the sum $PA_1^2+\ldots+PA_n^2$ can be seen as the moment of inertia of a set of points with respect to $P$. Since $O$ is also the centroid / barycenter of such set of points, by the parallel axis theorem the previous sum only depends on $PO^2$:

$$ \sum_{k=1}^{n}PA_k^2 = n\left(OP^2+R^2\right).$$ If $P$ is a point on the circumcircle of our regular polygon, the RHS just equals $2n R^2$.