Timeline for Maximum area of the intersection of a parallelogram and a triangle
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2018 at 16:50 | comment | added | Wlodek Kuperberg | @DavidG.Stork: Actually, the example given by Noam D. Elkies (see above), shows that the base of the best isosceles triangle need not be parallel to any of the square's sides, and if you rotate it in any way, the area of the intersection will decrease. The reference given by yarchik mathematica.stackexchange.com/questions/156677/… indicates that the maximum should indeed be $2(\sqrt{2} -1)$, but the arguments you give don't look like a proof at all. | |
| Jan 30, 2018 at 1:06 | comment | added | David G. Stork | @WlodekKuperberg: "rotate the base..." (to be parallel to the edge of the square) and keep the opposite vertex in place. If possible, then translate the entire triangle to have its base overlap the edge of the square. In short: you don't want to place one triangle vertex on an edge and the other outside that edge. | |
| Jan 29, 2018 at 21:40 | comment | added | Wlodek Kuperberg | @DavidG.Stork "If the base of the triangle is not parallel to the base of the quadrilateral, rotate the base (keeping its area constant to see that the overlap area cannot decrease." This statement is very ambiguous. I am trying hard to figure out what it is supposed to mean. If your triangle is isosceles, with a very short base but a very long height, then it is certainly better to place the triangle diagonally across the square rather than with its base on the side of the square. | |
| Jan 28, 2018 at 2:51 | comment | added | David G. Stork | If the base of the triangle is not parallel to the base of the quadrilateral, rotate the base (keeping its area constant to see that the overlap area cannot decrease. As such, one can assume the base of the triangle is parallel to the base of the quadrilateral. | |
| Jan 27, 2018 at 22:56 | comment | added | Wlodek Kuperberg | @DavidG.Stork: Without loss of generality you may assume that the parallelogram is a square OR you may assume that the triangle is isosceles or even equilateral, but you CANNOT assume both, let alone that the base of the isosceles triangle is collinear with one side of the square. If you IMPOSE these, you lose generality, which destroys your "proof". | |
| Jan 25, 2018 at 19:56 | comment | added | David G. Stork | @WlodekKuperberg: Draw the figure imposing symmetry of the triangle (and of course square). The unit area constraint means that a single parameter (the width of the base of the triangle) determines everything about the problem. Just look at Joseph O'Rourke's figure! | |
| Jan 25, 2018 at 19:53 | comment | added | Wlodek Kuperberg | @DavidG.Stork "Write an equation for the overlap area..." can you be more specific about how you reduce the whole problem to just one parameter? It is clear that $x$ determines the triangle, but what determines the parallelogram? Why do you say "the square"? | |
| Sep 30, 2017 at 20:16 | comment | added | Wlodek Kuperberg | @Joseph O'Rourke: Nice graphics, Joe! | |
| Sep 29, 2017 at 6:34 | comment | added | Timothy Budd | @yarchik If you move the tip as well as the base in such a way that the intersections with the vertical sides are unchanged, then the overlapping area does not change. This continuously deforms the one triangle into the other. | |
| Sep 29, 2017 at 1:18 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
3D image added.
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| Sep 28, 2017 at 23:23 | comment | added | David G. Stork | @Joseph O'Rouke's answer is correct. You assume the triangle has base along the horizontal axis and is symmetric with respect to the vertical axis, and determined entirely by the location of the horizontal position $(x,0)$ of the corner. Write an equation for the overlap area, take the derivative with respect to $x$, set it to $0$ and solve for $x$. If $x=0$ is the center of the square, the solution is $\sqrt{2}/2$--the isosceles triangle. | |
| Sep 28, 2017 at 17:04 | comment | added | orangeskid | @Noam D. Elkies: your solution is the other, cut in half | |
| Sep 28, 2017 at 16:38 | comment | added | yarchik | @coudy But you have to move 2 vertices, not 1. How exactly do you move? | |
| Sep 28, 2017 at 14:32 | comment | added | coudy | @Elkies Is it curious? It is not hard to convince oneself that moving the upper edge of the triangle on a horizontal line does not change the overlaping area. | |
| Sep 28, 2017 at 2:21 | comment | added | Noam D. Elkies | Curiously the unit square has the same overlap $2(\sqrt2-1)$ with an isosceles right triangle of side $\sqrt{2}$ placed so that its right angle coincides with one of the square's. | |
| Sep 28, 2017 at 1:25 | comment | added | Wlodek Kuperberg | This looks like a good candidate for the answer in the plane, Joe. | |
| Sep 28, 2017 at 0:45 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Exact area.
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| Sep 28, 2017 at 0:32 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |