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The "aha!" problems 5. Four-point square

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#11 plasmid


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Posted 07 June 2014 - 03:52 AM

If lines AC and BD are perpendicular and of different length, then line D-D' will not have two different perpendiculars with one going through A and one going through C... both perpendiculars will emanate from the same point. So in the end it wouldn't really describe a square so much as a point whose "edges" if extended would touch each of the four points.

But I do think your solution is far more Aha-y.
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#12 bonanova



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Posted 07 June 2014 - 06:58 AM

Good point. (no pun intended.)

That would be a degenerate case - a square with side zero.


And it wouldn't help to draw BD first.

You'd get AC' collinear with C, with the same result.

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