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Suppose we have $n$ lines in general position in the plane. Prove that there are at least $n-2$ ''small'' triangles. Here a "small" triangle is a triangle that is not contained in any larger triangle.

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what does "general position" mean? i guess you are trying to exclude the possibility of parallel lines? –  Suvrit Nov 25 '10 at 21:40
    
Yes, and also no $3$ lines concurrent. –  Vinoth Nov 25 '10 at 22:01
    
You mean at least n-2 triangles? –  Gjergji Zaimi Nov 25 '10 at 22:23
    
Yes, sorry - I've edited it now. –  Vinoth Nov 25 '10 at 22:27

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up vote 7 down vote accepted

It is well-known problem, but quite now I am unable to find a link on AoPS. For any line $a$ take all $n-1$ points, in which it meets other lines, and for any two consecutive points $B=a\cap b$, $C=a\cap c$ consider the triangle, formed by lines $a$, $b$, $c$ and draw a flower inside this triangle near the midpoint of its side $BC$. Totally, we get $(n-2)n$ flowers. On the other hand, in any part, which is not a triangle, we have at most two flowers (because any two flowers in the same part must lie on neighbouring sides of this part). Since we have $n(n+1)/2+1$ parts (simple induction), and $2n$ of them are unbounded (common sense), we get at most $3T+2(n(n+1)/2+1-2n-T)$ flowers, hence $(n-2)n\leq n^2-3n+2+T$, $T\geq n-2$, where $T$ is the number of triangular parts.

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Sorry I don't quite understand why "in any part which is not a triangle we have at most two flowers" - surely in any part with $n$ sides we have $n$ flowers? If we have a part with $n$ sides, how could any of those sides fail to have a flower on it? –  Vinoth Nov 25 '10 at 23:01
    
No, say, in regular pentagon we have 0 flowers. All flowers are in other parts. (We draw exactly one flower for each segment, on one side, not on both.) –  Fedor Petrov Nov 25 '10 at 23:05
    
Thank you! Very nice solution. –  Vinoth Nov 25 '10 at 23:29

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