Trapezoids, and blogs

Atrios remarked on the (slow, in-progress) death of the open internet, as in blogs etc. And Jonathan’s offhand remark about his blog’s heyday is another anecdatum pointing in the same direction. Anyhow, that’s not what this post is about; it’s about the second of two geometry puzzles from JD2718:

Are there four segments from which it is possible to construct a quadrilateral, but from which it is not possible to construct a trapezoid?

The answer, I suppose, is “it depends whether you think a parallelogram is a trapezoid”: if the opposite pairs of edges are equal in length, you’ll always have a parallelogram. But if there are a pair of opposite unequal sides, you can always do something: if we have edges a, b, c, d with a > c, then we can draw the figure thusly

with x = \dfrac{a - c + \frac{b^2 - d^2}{a - c}}{2} and y =  \dfrac{a - c - \frac{b^2 - d^2}{a - c}}{2}. (These formulas allow x and y to be negative, in case b and d differ quite a lot, pushing the top edge off to one side or the other. Hopefully I have not missed any subtleties about different drawings!)

When the edge lengths are obliging, one can think in terms of hinges:

Since the right vertex of the blue edge begins below the left vertex but ends above it, at some point they must be level, i.e., the blue edge must be parallel to the black edge.

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2 Responses to Trapezoids, and blogs

  1. jd2718 says:

    I never remarked on how elegantly simple this proof is. Until now.

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