Timeline for A necessary and sufficient condition for three diagonals of a hexagon to be concurrent
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 29 at 7:18 | answer | added | quinta | timeline score: 1 | |
| Sep 24 at 18:46 | answer | added | Toni Mhax | timeline score: 1 | |
| Sep 24 at 17:44 | comment | added | Michael Hardy | @GerryMyerson : On some past occasions on MathOverflow I have called such segments diagonals in a way where it would have been obvious what I meant and nobody objected. | |
| Sep 24 at 11:36 | comment | added | Đào Thanh Oai | @ToniMhax Can you help me show detail? | |
| Sep 24 at 6:08 | comment | added | Toni Mhax | Hello, first prove your identity for a triangle ($C\in (BD) $ etc) which reduces to Ceva theorem by the sine rule and the identity is true. Then for these concurrent diagonals at $P$ stretch point $C$ along $(FP)$ and apply the sine rule in new figure again to prove that the ratio is constant (here the changes are for angles $\beta$, $\epsilon$, $\gamma$ and $\delta$). Doing that for $A$ and $E$ proves the direct direction. The other direction is not true as @FedorPetrov noticed you may move the points to obtain the ratio 1 in this identity despite that the diagonals are not concurrent. | |
| Sep 24 at 3:54 | comment | added | Gerry Myerson | @Michael, proofwiki.org/wiki/Definition:Polygon/Chord says "A chord of a polygon $P$ is a straight line connecting two non-adjacent vertices of $P$." | |
| Sep 24 at 0:52 | comment | added | Michael Hardy | @GerryMyerson : Maybe. A chord of a circle is uncontroversially defined. One can say that a convex function is defined as one for which all chords of the graph are above (or at least not in any part below) the graph. But then maybe a "chord" of a polygon would connect points on the edge that are not vertices, so that doesn't fit. I have thought of them as "diagonals", but here that word is used somewhat differently. | |
| Sep 23 at 2:03 | comment | added | Đào Thanh Oai | @FedorPetrov You can click here geogebra.org/m/hadhw364 | |
| Sep 22 at 23:40 | comment | added | Gerry Myerson | @Michael, chord? | |
| Sep 22 at 14:06 | comment | added | Michael Hardy | Is there some word other than "diagonal" that refers to things like the line segment from $A$ to $C$ in this illustration? | |
| Sep 22 at 14:03 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 12 characters in body
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| Sep 22 at 10:13 | comment | added | Fedor Petrov | How do you check this with geogebra? | |
| Sep 22 at 10:06 | comment | added | Đào Thanh Oai | I checked by geogebra, I think $AD, BE, CF$ concurrent if and only if the above trigonometric formula is equal to 1. | |
| Sep 22 at 9:37 | comment | added | Fedor Petrov | Does not look like a necessary condition. This may be rewritten in terms of nine segments lengths on the diagonals (using Sine law), the expression is fractional linear with respect to each of six segments which are adjacent to the vertices of the hexagon. It can be easily equal to 1,no? | |
| Sep 22 at 9:24 | history | asked | Đào Thanh Oai | CC BY-SA 4.0 |