Question

Let A1,A2...A12 be twelve equally spaced points on a circle with radius 1. Find
(A1 A2)^2 + (A1 A3)^2 + ... + A11 A12)^2.
(The sum includes the square of the distance between any pair of points, so the sum includes 12 choose 2 = 66 terms.)

Answer 1

24

Answer 2

What does the sums of squares of lengths make you think about? If you still don't know, what neat property is applied to all right triangles?

Answer 3

This is a solution (sorry if LaTeX or Asymptote programming doesn't appear on this database):
Since the sum involves all pairs of points, it doesn't matter what order the points are in around the circle. Without loss of generality, let them be in order $A_1, A_2, \ldots, A_{12}.$

The sum has twelve terms that are equal to $(A_1 A_2)^2$, and twelve terms that are equal to $(A_2 A_7)^2$. Since $\overline{A_1 A_7}$ is a diameter of the circle, $\angle A_1 A_2 A_7 = 90^\circ$. Then by Pythagoras on right triangle $A_1 A_2 A_7$,
\[(A_1 A_2)^2 + (A_2 A_7)^2 = (A_1 A_7)^2 = 2^2 = 4.\]
Therefore, the sum of the twelve terms that are equal to $(A_1 A_2)^2$ and the twelve terms that are equal to $(A_2 A_7)^2$ is $12 \cdot 4 = 48$.

[asy]
unitsize(2 cm);

for(int i = 1; i <= 12; ++i) {
dot(dir(360/12*i));
label("$A_{" + string(i) + "}$", dir(90 + 360/12*(i - 1)), dir(90 + 360/12*(i - 1)));
}

draw(Circle((0,0),1));
draw(dir(90)--dir(270)--dir(120)--cycle);
draw(dir(90)--dir(150)--dir(270));
draw(dir(90)--dir(180)--dir(270));
[/asy]

Similarly, the sum has twelve terms that are equal to $(A_1 A_3)^2$, and twelve terms that are equal to $(A_3 A_7)^2$, and their sum is also
\[12 [(A_1 A_3)^2 + (A_3 A_7)^2] = 12 \cdot (A_1 A_7)^2 = 48.\]

The sum has twelve terms that are equal to $(A_1 A_4)^2$. Note that $A_1A_4 = A_4A_7$, so the sum of the 12 terms that equal $(A_1A_4)^2$ is
\[12 \cdot (A_1 A_4)^2 = 6 [(A_1 A_4)^2 + (A_4 A_7)^2] = 6 \cdot (A_1 A_7)^2 = 24.\]

Finally, the sum has six terms that are equal to $(A_1 A_7)^2$, so the their sum is $6 \cdot (A_1 A_7)^2 = 6 \cdot 4 = 24$.

We have accounted for all $12 + 12 + 12 + 12 + 12 + 6 = 66$ terms in the sum, and the sum is equal to $48 + 48 + 24 + 24 = \boxed{144}$.

If you can't read it, basically the answer if 144.

Answer 4

you copied of AoPS