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Hiding a triangle (solution)

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for the problem,.


Suppose the origins equilateral triangle has sides of length 1. Then its three vertices are each 1 unit from the other two.

If we tried to cover it with two other triangles, then one of them would have to cover two vertices of the original triangle, because there are three vertices distributed between two covering triangles. So at least one of the covering triangles would have to contain points 1 unit apart, so it couldn’t be smaller than the original triangle.

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The Roots of Coincidence

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Today of all days I remembered this old question.

I first came across it in the 1970s but it seems to have been first asked more than 200 years before, and perhaps humorously. This is my attempt at a reconstruction.

Gentleman doth Enquire recently: whether it is Probable that: at any time which should be nominated: In the City of London there must bee,Two Persons, who have exactly the same numerosity of Haires on their Heads (supposing the Capital had sufficient Barbers to perform this Census &&).

I answer: For Certain. Since the Roy. Soc. of London have determined by what is called field-work, that the the count of Haires on any person's never doth, even if they be be-wigged, extend beyond ca. 250 Thousand.

Thus having thought to the vast Numbers of London-dwellers (which doth over-top 1 mill.) since the latter number be greater than the first, a co-incidence cannot be avoided.

Permalink 1 comment (latest comment by Simon Reed, Tuesday, 25 Oct 2016, 07:06)
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