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.)
4 answers
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.
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
1 answer