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A Property of Rhombi

Let's do this. Select a few points on a straight line. On each of the segments defined by the successive points erect an isosceles triangle, and let all the triangles have equal sides. Complete the space between two successive triangles to a rhombus. This produces some shapes that may be called uneven pyramids that combine into an outline of a mountain ridge. The space between any two successive shapes can be completed to a rhombus, and the process can continue until there emerges a single uneven pyramid. This shape has an interesting property. Note that it has three extreme points: two on the base line and one at the top. The fact is that the three points form an isosceles triangle. More accurately, the top point is equidistant from the two points on the base line.


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To see why this is so, reflect the whole structure in the base line. The combined shape consists entirely of rhombi with equal sides. Its four broken sides are all equal. The opposite two (there are two pairs of them) are equal because their constituent segments correspond to each other via a chain of rhombi, so that they are equal and parallel. The broken sides that meet on the base line are reflections of each other and are therefore equal. We see then that all four broken sides are equal. This means that the four extreme points (top, bottom, and the two on the base line) form a rhombus one of whose diagonals is horizontal, the other is vertical. A swarm of rhombi begot another rhombus! Its top half is an isosceles triangle, as required.

This property of rhombi furnishes a proof of the Circle-Stacking Theorem, where the rhombi are formed by touching circles with sides equal to twice their common radius.

Copyright © 1996-2009 Alexander Bogomolny

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