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Let a convex quadrilateral ABCD with perimeter 1,d is the maximum of AB,AC,AD,BC,BD,CD,prove that d is not less than 1/3 we can prove that parallelogram ABCD with perimeter 1,than one of AC,BD is more than 1/3 but the general case is very difficult to solve.

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2 Answers 2

up vote 5 down vote accepted

The answer given by ε-δ (a kite inscribed in a Reuleaux triangle) can be found in

Ball, D. G. (1973), "A generalisation of π", Mathematical Gazette 57 (402): 298–303, doi:10.2307/3616052, JSTOR 3616052;

He doesn't give an explicit proof that this is optimal but says it can be done by "some tedious but not very difficult trigonometry. See also

Griffiths, David; Culpin, David (1975), "Pi-optimal polygons", Mathematical Gazette 59 (409): 165–175, doi:10.2307/3617699, JSTOR 3617699

for extensions to higher order polygons.

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Quadrilateral Image

It is not true.

Take $A=0$, $B=1$, $C=e^{\frac\pi6 i}$ $D=e^{\frac\pi3 i}$. Then $d=1$ and the perimeter is $2+\tfrac1{\cos\frac{\pi}{12}} >3$.

I am sure that $\frac1{2+\tfrac1{\cos\frac{\pi}{12}}}$ is the optimal bound.

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How is this convex? Gerhard "Ask Me About System Design" Paseman, 2012.01.09 –  Gerhard Paseman Jan 9 '12 at 17:02
    
it is convex, $A$, $B$ and $D$ are vertices of equilateral triangle and $C$ is on the extension of the bisector at $A$. –  ε-δ Jan 9 '12 at 17:07
    
OK. For some reason C kept appearing (to me) on the inside of triangle ABD. Gerhard "Not Enough Coffee This Morning" Paseman, 2012.01.09 –  Gerhard Paseman Jan 9 '12 at 18:15
    
$|e^{ix}|\equiv 1$, it can not be inside... –  ε-δ Jan 9 '12 at 18:42
    
(I fixed the image URL...) –  Joseph O'Rourke Jan 9 '12 at 19:46

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