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Consider two numbers $a>b>0$. Let $A_1,A_2,A_3$ be three convex sets in ${\mathbb R}^2$ such that $\mu(A_i)=a$, $\mu(A_i\cap A_j)=b$ ($\mu$ is the usual measure on ${\mathbb R}^2$). What is the minimal possible value of $\mu(A_1\cap A_2\cap A_3)$?

Surely, if we omit the convexity assumption, the answer is trivial. But it is not necessary realizable by convex sets. Consider, for instance, the case $a=2b$: the answer is nonzero!

It seems that the optimal construction is the following one. Take a triangle $ABC$, and cut three trapezoids $A_BA_CBC$, $B_AB_CCA$, $C_AC_BBA$ (here $A_BA_C\parallel BC$, $B_A,C_A\in BC$, similar relations for the others). E.g., for $a=2b$ the altitude of the trapezoid is $2/5$ of the altitude of triangle; hence the answer in this case seems to be $1/16$.

The generalizations are also interesting. E.g., what happens if we fix the areas but omit the relation that they are equal?

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Did you mean $\mu(A_1\cap A_2)\cap A_3$ instead of $\mu(A_1\cap A)2\cap A_3$? –  Joseph O'Rourke Apr 3 '12 at 15:34
    
@Ilya, Joseph: I hope I made it right(?) –  Anton Petrunin Apr 3 '12 at 15:54
    
@Anton: That must be what he meant. Thanks. –  Joseph O'Rourke Apr 3 '12 at 16:09
    
@Anton: Yes, thank you. –  Ilya Bogdanov Apr 3 '12 at 16:58

2 Answers 2

Let $b=ra.$ Then your construction gives $$\mu(A_1\cap A_2\cap A_3)=0,\frac{(5r-2)^2}{8r}a \text{ or }(2r-1)a$$according as $r\le\frac25,\frac25\le r \le \frac23$ or $\frac23 \le r.$

Thanks to the nice illustration I can say that as $r$ decreases from $r=\frac12$ to $r=\frac25,$ the small triangle shrinks to a point and beyond that there is an empty triangle in the middle. But as it increases the triangle of overlap grows and touches the sides at $r=\frac23$ and beyond that becomes a hexagon of overlap.

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Here is Ilya's construction:
Trapezoids

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Just wondering how have you produced this nice picture, technically? –  Seva Apr 4 '12 at 4:51
    
@Seva: I drew this one in Adobe Illustrator. –  Joseph O'Rourke Apr 4 '12 at 10:20

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