Timeline for A generalization of Harcourt's theorem
Current License: CC BY-SA 4.0
12 events
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| Feb 11, 2021 at 1:40 | history | edited | Pedja | CC BY-SA 4.0 |
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| Feb 9, 2021 at 12:32 | comment | added | bathalf15320 | This is just a small suggestion which applies to all the versions of Harcourt's theorem. A back of the envelope calculation shows that the rhs of the equation at the end of your claim is, as a function of a point $(x,y)$, a quadratic form whose contours are circles with centre at the incentre. This reduces the proof to showing that the formula holds for one of the points of contact. It also allows extensions to results about other circles (i.e., other than the incircle). | |
| Feb 9, 2021 at 8:07 | history | edited | Pedja | CC BY-SA 4.0 |
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| Feb 6, 2021 at 11:43 | vote | accept | Pedja | ||
| Feb 6, 2021 at 10:18 | answer | added | Fedor Petrov | timeline score: 4 | |
| Feb 6, 2021 at 10:07 | comment | added | Pedja | @FedorPetrov I have formulated the claim in this way because Harcourt's theorem is about area of a triangle. | |
| Feb 6, 2021 at 9:48 | comment | added | Fedor Petrov | Then the formulation $r=(\sum n_id_i)/(\sum d_i)$ looks more natural for me. | |
| Feb 6, 2021 at 9:40 | comment | added | Pedja | @FedorPetrov Of course. | |
| Feb 6, 2021 at 9:28 | comment | added | Fedor Petrov | $K/s$ is just the inradius, right? | |
| Feb 6, 2021 at 9:12 | history | edited | Pedja | CC BY-SA 4.0 |
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| Feb 6, 2021 at 8:51 | history | edited | Pedja | CC BY-SA 4.0 |
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| Feb 6, 2021 at 8:22 | history | asked | Pedja | CC BY-SA 4.0 |