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Let $\mathbf{D}$ be the unit disk, is $$\inf_{\begin{array}{c} v_{1},v_{2},v_{3},v_{4}\in\mathbf{D},\\ v_{0}\in\mbox{convexhull}\left(v_{1},v_{2},v_{3},v_{4}\right) \end{array}}\max_{0\le i,j,k\le4}\frac{\mbox{perimeter}\left(\triangle v_{i}v_{j}v_{k}\right)}{\mbox{area}\left(\triangle v_{i}v_{j}v_{k}\right)} $$no less than $$\min_{v_{0},v_{1},v_{2},v_{3},v_{4}\in\partial\mathbf{D}}\max_{0\le i,j,k\le4}\frac{\mbox{perimeter}\left(\triangle v_{i}v_{j}v_{k}\right)}{\mbox{area}\left(\triangle v_{i}v_{j}v_{k}\right)}? $$

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Thanks in advance for any correct answer or any help that can lead to a correct answer. –  userior Feb 24 '13 at 7:19
    
or is there any counterexample? Thanks. –  userior Feb 24 '13 at 7:41
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If it's not too much trouble, could you please explain where you ran into this problem and whether or not there are partial results that you have already found? –  Yemon Choi Feb 24 '13 at 9:44
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1. perimeter/area = 2/inradius; so there is an equivalent formulation with "max min inradius($v_iv_jv_k$)". 2. Both quantities are constants. Wouldn't you rather know the value of those constants? –  Günter Rote Feb 24 '13 at 19:18
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@partial results: A natural conjecture is that the second quantity is minimized when the 5 points form a regular pentagon. Have you checked this? What is the value for a regular pentagon? Is the first quantity perhaps minimized for a square plus center? What is the value for this configuration? What is the best value that you know for each of these quantities? –  Günter Rote Feb 24 '13 at 19:25
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I think it is not hard to see that the second quantity is minimized by the regular pentagon. First, using Günter's observation, the problem is equivalent to find

$$\max_{v_{0},v_{1},v_{2},v_{3},v_{4}\in\partial\mathbf{D}} \min_{0\le i,j,k\le4} \mbox{inradius}(v_i,v_j,v_k).$$

If we don't have a regular pentagon, then there are 3 points who are contained in an arc whose length is strictly less than $4\pi/5$, suppose they are $v_1,v_2,v_3$ in this order. The $\mbox{inradius}(v_1,v_2,v_3)$ is maximized if $v_2$ is halfway between $v_1$ and $v_3$. So this quantity will be less than the inradius of three consecutive vertices of the pentagon, which is less than the inradius of three non-consecutive vertices of the pentagon.

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domotorp , thanks for your answer, but the main difficulty in the question is how to show that the quantity for the case that one point inside a quadrilateral is greater than that for a regular pentagon. Do you know how to show that? Thanks! –  userior Feb 25 '13 at 23:48
    
I think it would be the simplest to find the extreme configuration for that problem too but that seems harder, but probably not much. Maybe you can show by using Lagrange multipliers where the extreme value is. –  domotorp Feb 26 '13 at 17:03
    
@userior, why don't you calculate the value for the square plus center and compare? Then at least we have a conjecture, and we only need to concentrate on one problem. –  Günter Rote Feb 27 '13 at 16:32
    
@Günter, I calculated both the value for the square plus its center and the square plus an optimal point inside the square, and the values are both greater than the extremal value in the case when all the five points are on the boundary. But do you know if the square plus an point inside realizes the the minimum over the case of one point inside a quadrilateral, or how to find the optimal quadrilateral? Thanks! –  userior Mar 30 '13 at 2:53
    
@domotorp, Lagrange multipliers normally requires the function to be minimized to be differentiable, but the function here to be minimized is not differentiabe at many places. –  userior Mar 30 '13 at 3:18
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