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What's 8 plus five (in digits only)?

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File: AnUnexpectedFraction

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----
My latest posts can be
found here:
  * ColinsBlog
----
Previous blog posts:
  * YouHaveToAdmireTheirOptimism
  * RepresentativesMatter
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  * APuzzleAboutPuzzles
  * HowNotToDoTwitter
  * Calculating52FactorialByHand
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  * TheOtherOtherRopeAroundTheEarth
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  * ThePointOfTheBanachTarskiTheorem
  * SieveOfEratosthenesInPython
  * FastPerrinTest
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  * FindingPerrinPseudoPrimes_Part2
  * FindingPerrinPseudoPrimes_Part1
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----
Additionally, some earlier writings:
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********
!! 2016/09/18 - An Unexpected Fraction

On 2016-09-09,
''' <a href="https://twitter.com/MEImaths">@MEIMaths</a>
tweeted an image and said:

[[[>
https://www.solipsys.co.uk/images/QuadrilateralProblem_Question.png
----
|>> Quadrilateral Problem <<|
]]]
********> width=10%
******** width=80%
[[[
An Unexpected Fraction?

Start with any convex quadrilateral.  Mark the
midpoint of each side, join these midpoints to
the vertex two places clockwise around the
quadrilateral.

What fraction of the original quadrilateral is
the new quadrilateral?
]]]
******** width=10%
********<

The way the question is phrased makes you feel two things.

Firstly, you have to feel that the answer is likely to be surprising
in some way, so that means it's likely to be a nice puzzle.

But secondly, it seems clear from the phrasing that the answer must
be the same regardless of the particular quadrilateral chosen.

Well, I've seen this before in the context of a square, and the answer
there is lovely - it's a fifth.  The proof in the case of a square is
fairly simple, and there are a number of ways of going about it.  So
there we are, it's sorted, the answer is a fifth.

Nice one.

[[[<
https://www.solipsys.co.uk/images/QuadrilateralProblem_Square.png
----
|>> Quadrilateral Problem _ The Square case <<|
]]]
But then Hugh Hunt sent me an email saying:

|>> [[[
"In the limiting case of _
an equilateral triangle _
the answer is a sixth.

"Am I doing something _
wrong?"
]]] <<|

[[[>
https://www.solipsys.co.uk/images/QuadrilateralProblem_Triangle.png
----
|>> Quadrilateral Problem _ Limiting Triangle <<|
]]]
That's a good question.  If we take two adjacent vertices of the square
and move them together, just before they meet we do still have a convex
quadrilateral, so because we expect the answer to be constant (because
of the context) we would expect the answer still to be a fifth, but the
answer is now pretty obviously a sixth.

So what's going on?

The short answer seems to be that
''' <a href="https://twitter.com/MEImaths">@MEIMaths</a>
mis-spoke themselves, and despite the wording, they didn't actually
mean */any/* quadrilateral. But what might they have meant, and what
answers might be possible?

So here are some interesting questions:

* Under what circumstances is the answer one fifth?

* Is the answer always between a sixth and a fifth?

* If not, what are the extreme values?

* Is every value in the range $[\frac{1}{6},\frac{1}{5}]$ achievable?

So even if it /was/ a mistake, there is interesting mathematics to be
found.  That, at least, is no surprise.

''' <img src="/cgi-bin/CountHits.py?AnUnexpectedFraction" alt="" />

----

!! References

Here's the original tweet:
  * https://twitter.com/MEImaths/status/774161733761966080

Adam Atkinson has created a GeoGebra twitchet you can play with:
  * http://www.ghira.mistral.co.uk/quadrilateral.ggb

----

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********<