Timeline for What is the shape of the $n$-gon which gives the maximum of a function?
Current License: CC BY-SA 3.0
27 events
| when toggle format | what | by | license | comment | |
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| May 22, 2015 at 14:44 | history | edited | user9072 |
edited tags
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| Jul 31, 2013 at 14:47 | comment | added | Anton Petrunin | By the way, in Euclidean geometry $AB$ is the most standard way to denote the distance from $A$ to $B$. It was no need to edit the question. | |
| Jul 22, 2013 at 14:38 | vote | accept | mathlove | ||
| Jul 20, 2013 at 22:06 | answer | added | Sergei Ivanov | timeline score: 13 | |
| Jul 19, 2013 at 14:38 | comment | added | mathlove | $W$ is min only if the pentagon is a regular pentagon, so the proof is completed. | |
| Jul 19, 2013 at 8:30 | comment | added | mathlove | wrong: $$|P_2P_1|^2+|P_2P_3|^2\le|Q_2P_1|^2+|Q_2P_3|^2$$ right: $$|P_2P_1|^2+|P_2P_3|^2=|Q_2P_1|^2+|Q_2P_3|^2$$ | |
| Jul 19, 2013 at 6:48 | comment | added | mathlove | By using this transforming operation infinitely, you get the pentagon whose edges have the same length. Without loss of generality suppose that the pentagon whose edges have the same length $1$. Then, $A_5=2-\frac25W$. Here, $W=cosA+cosB+cosC+cosD+cosE$. | |
| Jul 19, 2013 at 6:47 | comment | added | mathlove | $$|P_2P_1|^2+|P_2P_3|^2\le|Q_2P_1|^2+|Q_2P_3|^2$$ and $$|P_2P_5|^2+|P_2P_4|^2\le|Q_2P_5|^2+|Q_2P_4|^2$$ Therefore, the $A_5$ for the pentagon $P_1Q_2P_3P_4P_5$ is larger than, or equals to the $A_5$ for the pentagon $P_1P_2P_3P_4P_5$. Note that the lengths of the other line segment is constant. | |
| Jul 19, 2013 at 6:47 | comment | added | mathlove | The point $Q_2$ is outside of the circle $C_{45}$, the center is the midpoint $M_{45}$ of the edge $P_4P_5$ and the radius is the line segment $M_{45}P_2$. This means the following two. | |
| Jul 19, 2013 at 6:46 | comment | added | mathlove | Next, draw the perpendicular bisector $L_{13}$ of the line segment $P_1P_3$. Final step is to determin an intersection $Q_2$ where the line $L_{13}$ crosses the circle $C_{13}$. This is the transforming operation. | |
| Jul 19, 2013 at 6:46 | comment | added | mathlove | First, find the two adjacent edges which has the largest of $||$$P_iP_{i+1}$$|$$-$$|$$P_{i+1}P_{i+2}$$|$$|$, say $P_1P_2$ $\ge$ $P_2P_3$. Next, draw the circle $C_{13}$, the center is the midpoint $M_{13}$ of the line segment $P_1P_3$ and the radius is the line segment $M_{13}P_2$. | |
| Jul 19, 2013 at 6:44 | comment | added | mathlove | I think I can prove that $A_5$ reaches the max only if the pentagon is a regular pentagon. Please judge whether my proof is correct or not. I use an transforming operation through which the value of $A_5$ doesn't change or increases. Here is the operation which consists of the following steps. | |
| Jul 17, 2013 at 15:05 | comment | added | mathlove | I got that $A_5\lt3$ for any pentagon by using the result that $A_4\le1$ for any quadrilateral and that there is no pentagon whose $A_5$ is 3. | |
| Jul 16, 2013 at 16:52 | comment | added | Joseph O'Rourke | Yes, it may be that the max is not uniquely achieved. | |
| Jul 16, 2013 at 15:50 | comment | added | mathlove | @Joseph O'Rourke: I think so. Though I've tried to prove that if the pentagon is not a regular pentagon, $A_5$ is not max, I'm facing difficulty. | |
| Jul 14, 2013 at 21:46 | comment | added | Joseph O'Rourke | Essentially the ratio is sum of diagonals$^2$ over sum of edges$^2$. For $n=5$, I suspect the max is achieved by the regular pentagon, $(3+\sqrt{5})/2 \approx 2.62$. | |
| Jul 14, 2013 at 17:24 | comment | added | mathlove | @Ricardo Andrade: Thank you very much. | |
| Jul 14, 2013 at 16:52 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
edited title
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| Jul 14, 2013 at 16:48 | comment | added | Ricardo Andrade | I edited the question to make the notation more explicit and standard. | |
| S Jul 14, 2013 at 16:46 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
I built on a suggested edit to define some notation in the question; tried to revert some of the changes to keep the question closer to the original; corrected grammar
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| S Jul 14, 2013 at 16:46 | history | suggested | N. Owad | CC BY-SA 3.0 |
I hopefully defined the terms the author of the question did not and cleaned up the notation, in such a way that there is no more confusion.
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| Jul 14, 2013 at 16:18 | review | Suggested edits | |||
| S Jul 14, 2013 at 16:46 | |||||
| Jul 14, 2013 at 15:51 | review | Close votes | |||
| Jul 14, 2013 at 17:59 | |||||
| Jul 14, 2013 at 15:38 | comment | added | mathlove | @Joseph O'Rourke: Thank you very much for pointing it out. In this problem, ${P_{i}P_{j}}^2$ means the square of the Euclidean length of the segment from $P_i$ to $P_j$. | |
| Jul 14, 2013 at 15:32 | comment | added | Steven Landsburg | Voting to close pending clarification of what $P_iP_j^2$ means. | |
| Jul 14, 2013 at 14:58 | comment | added | Joseph O'Rourke | What is the meaning of the notation $P_i P_j^2$? Do you mean $|P_i P_j|^2$, the square of the Euclidean length of the segment from $P_i$ to $P_j$? Or are you implying some type of multiplication between $P_i$ and $P_j$? | |
| Jul 14, 2013 at 14:29 | history | asked | mathlove | CC BY-SA 3.0 |