Question 1182792: For n>_3, consider 2n points spaced regularly on a circle with alternate points black and white and a point placed at the centre of the circle.
The points are labelled -n,-n+1,...,n-1,n so that:
(a) the sum of the labels on each diameter through three of the points is a constant s, and
(b) the sum of the labels on each black-white-black triple of consecutive points on the circle is also s.
1.Show that the label on the central point is 0 and s=0. &
2.Show that such a labelling exists if and only if n is even.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to approach this problem:
**1. Showing the central point is 0 and s = 0:**
* **Central Point Label:** Let 'c' be the label of the central point. Consider any diameter passing through two points on the circle, say labeled 'i' and 'j'. According to condition (a), i + j + c = s. Now, consider the diameter perpendicular to the first one. Let the points on this diameter be labeled 'k' and 'l'. Then k + l + c = s. Since the points are spaced regularly, if i and j are at opposite ends of a diameter, so are k and l. For every point labeled 'x', there will be another point labeled '-x' on the opposite end of the diameter. So, if i and j are at opposite ends of a diameter, and k and l are at opposite ends of a diameter, then i = -j and k = -l. Therefore, i + j = 0 and k + l = 0. Therefore, 0 + c = s and 0 + c = s. This implies that c = s.
Now, consider a black-white-black triple. Let the labels be i, j, and k. Then i + j + k = s. Since the points are spaced regularly, and the labeling goes from -n to n, we know that the labels of opposite points are additive inverses of each other. For any point labeled 'x', there is a point labeled '-x' on the opposite end of the diameter.
Consider the black-white-black triple i, j, k. Because the points are spaced regularly, the point opposite to j is -j. Because the sum of diametrically opposite points is 0 and the sum of consecutive points in a black-white-black triple is s, we can deduce that the point opposite to j is -j. Since the points are spaced regularly, the point opposite to j is -j. Then i + (-j) + k = 0 (because they are on a diameter).
We also know that i + j + k = s. Subtracting the two equations, we get 2j = s.
Since we can choose any j, consider the case where j = 0. Then s = 2 * 0 = 0.
Since c = s, and s = 0, therefore c = 0.
* **Value of s:** We've shown that the label of the central point (c) is equal to s. Consider any diameter. Let the labels of the two points on the circle be 'x' and 'y'. Then x + y + c = s. Since the points are spaced regularly, we know that x = -y. Thus, x + y = 0. Therefore, 0 + c = s, which means c = s. Because c = 0, s = 0.
**2. Showing existence of labeling if and only if n is even:**
* **If n is even:** We can label the points as follows: Start at an arbitrary point and label it -n. Move clockwise and label the next point -n+1. Continue until you reach n. This labeling satisfies both conditions. The sum of labels on a diameter is -i + i + 0 = 0 = s, and the sum of a black-white-black triple is -i + i + 0 = 0 = s.
* **If n is odd:** Suppose such a labeling exists. Consider the sum of the labels of all the points on the circle. Since the points are labeled from -n to n, the sum of all the labels is 0.
Now consider the sum of all black-white-black triples. There are 2n such triples. Each point on the circle belongs to exactly two black-white-black triples.
Let the labels of the points be l_1, l_2, ..., l_{2n}.
The sum of labels of all black-white-black triples is 2(l_1 + l_2 + ... + l_{2n}) = 2 * 0 = 0.
Since the sum of each black-white-black triple is s, the sum of all black-white-black triples is 2n * s.
Therefore 2ns = 0, which means s = 0. But we know that the sum of labels on each diameter is s. Since the labels on the diameter are additive inverses, the sum of labels on each diameter is 0.
However, if n is odd, we cannot pair up all the points on the circle to form diameters because there will be one point left unpaired. This implies that the labels can't be from -n to n.
Therefore, such a labeling is impossible when n is odd.
Therefore, such a labeling exists if and only if n is even.
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