Let A1,A2...A12 be twelve equally spaced points on a circle with radius 1. Find

24

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?

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.

you copied of AoPS

To find the sum of the squares of the distances between these twelve equally spaced points on a circle, we can make use of the properties of a regular dodecagon (a polygon with twelve sides of equal length).

Let's start by visualizing the dodecagon and labeling the points A1, A2, ..., A12 on the circle.

A12 A1 A2
*-------*-------*
/ \
A11 A3
| |
A10 A4
\ /
A9 *-------*-------* A5
| |
A8 A6
\ /
*-------*-------*
A7 A12 A1

Now, notice that each pair of adjacent points (A1, A2), (A2, A3), ..., (A11, A12) forms one side of a right-angled triangle, where the hypotenuse is the radius of the circle (which is 1 in this case) and the other two sides are the distances between the points.

Let's compute the square of each of these distances:

(A1 A2)^2 = (A2 A3)^2 = ... = (A11 A12)^2 = (A12 A1)^2

The sum S of all these squares can then be expressed as:

S = (A1 A2)^2 + (A2 A3)^2 + ... + (A11 A12)^2
= (A12 A1)^2 + (A1 A2)^2 + (A2 A3)^2 + ... + (A11 A12)^2
= 2(A1 A2)^2 + 2(A2 A3)^2 + ... + 2(A11 A12)^2
= 2[(A1 A2)^2 + (A2 A3)^2 + ... + (A11 A12)^2]

Note that the terms inside the brackets represent the sum of the squares of the sides of a regular dodecagon inscribed in a circle with radius 1.

For a regular dodecagon, the ratio of the length of a side to the radius of the circle is given by sin(π/12). Therefore, (A1 A2)^2 + (A2 A3)^2 + ... + (A11 A12)^2 is equal to the sum of squares of the sides of a regular dodecagon with a radius of 1, multiplied by 2.

Now, let's calculate this value using the formula for the sum of squares of the sides of a regular polygon:

S = 2 * [12 * sin²(π/12)]

Using a calculator, we find:

S = 2 * [12 * (0.13053)^2]
≈ 2 * (12 * 0.016999)
≈ 0.407976

So, the sum of the squares of the distances between the twelve equally spaced points on the circle is approximately 0.407976.