If we take a regular polygon and any line through its centre, then the sum of the squares of the distances from the vertices of the polygon to the line is independent of the orientation of the line or polygon. For example, in the following two diagrams, the sum of the squares of the lengths of the 5 blue line segments is the same.

Such a result is amenable to a proof via complex numbers. Without losing generality the real number line may be our line of interest and the points of the n-sided polygon may be described by the complex numbers for , where is the circumradius of the polygone and is an arbitrary real-numbered phase. Then the squared distance from a point to the real line is where . Summing this over gives our sum of squared distances as

Each of these sums is geometric in nature so can be simplified. For example,

Similarly, , being the conjugate of (2), and so from (1) our required sum of squared distances is , which is independent of proving the orientation independence.

More generally,

This enables us to generalise the above result to the following.

Given a regular n-sided polygon and line through its circumcentre, the sum of the mth power signed distances from the vertices of the polygon to the line is independent of the orientation when n > m.

By *signed distances*, we mean that points on different sides of the line will have distances of opposite sign.

To prove this, we define as before and this time our desired sum is

By (3), unless is an integer. For this can only occur in the case (if is even). Hence (4) becomes

Finally, what if the line does not pass through the centre of the polygon but is instead at distance from the centre?

This corresponds to replacing with in the above calculations and we find

Once again we find the sum is independent of the orientation of the line or polygon. In the particular case of this sum is , which also may be obtained by an application of the parallel axis theorem.

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