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.)
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.
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.