Timeline for An open triangle problem in plane geometry
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Sep 6 at 9:27 | comment | added | terceira | There is a simple method which allows on to reduce this problem to a pair of quartics in two variables which Mathematics can solve explicitly, albeit with long and intricate formulae. One can assume that the vertices are $(0,0)$, $(1,0)$ and $(p,q)$ and that $P$ is $s A+t B+(1-s-t) C)$. The equality of the square distances to the corresponding points then provides the equations for $s$ and $t$. | |
| Dec 1, 2021 at 3:22 | vote | accept | Đào Thanh Oai | ||
| Nov 29, 2021 at 16:52 | answer | added | dan_fulea | timeline score: 5 | |
| Nov 15, 2021 at 10:08 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
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| Nov 15, 2021 at 9:55 | comment | added | Peter Taylor | Working numerically with the reference triangle $A = (\tfrac{31}{3}, \sqrt{62 + \tfrac29})$, $B = (0,0)$, $C=(6,0)$ I find $P \approx (5.253623554420555, 1.5549453416812575)$ which gives index $\approx 1.5549453416812575$ for the ETC 6-9-13 lookup table. The nearest indices bracketing this are $X(5324): 1.55475\ldots$ and $X(3368): 1.55504\ldots$ so the point does not appear to be in ETC. | |
| Nov 15, 2021 at 7:41 | answer | added | Toni Mhax | timeline score: 0 | |
| Nov 15, 2021 at 2:16 | comment | added | Đào Thanh Oai | @LSpice Thanks you very much, I edited again by your comment. | |
| Nov 15, 2021 at 2:15 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
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| Nov 15, 2021 at 2:12 | comment | added | LSpice | Despite saying that your question is about existence, your post still asks "how can one construct …", so that it is unclear what it means to say that the answer is negative. | |
| Nov 15, 2021 at 2:12 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
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| S Nov 15, 2021 at 2:07 | history | suggested | Toni Mhax | CC BY-SA 4.0 |
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| Nov 15, 2021 at 0:03 | review | Suggested edits | |||
| S Nov 15, 2021 at 2:07 | |||||
| Nov 14, 2021 at 17:17 | review | Close votes | |||
| Nov 20, 2021 at 1:03 | |||||
| Nov 14, 2021 at 16:07 | comment | added | zeb | Let $M$ be the midpoint of $BC$. The locus of points $P$ inside the angle $\angle BAC$ which satisfy $|PE| = |PF|$ meets the line $AM$ in at least three points: $A$, $M$, and the $180^\circ$ rotation of $A$ around $M$, so this locus definitely can't be a conic. | |
| Nov 14, 2021 at 15:57 | answer | added | hordubal | timeline score: 0 | |
| Nov 14, 2021 at 13:44 | comment | added | Wojowu | Existence should be trivial by continuity. For any $0<c<|BC|/2$ there is a point $P$ such that $|PE|=|PF|$. The length of $PD$ chances continuously from the length of the median from $A$ to $0$. So for some $c$ they all coincide. | |
| Nov 14, 2021 at 13:29 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
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| Nov 14, 2021 at 13:28 | comment | added | Đào Thanh Oai | @Wojowu Yes, I edited question to the conjecture that exist the point P. Now need a proof that. | |
| Nov 14, 2021 at 13:23 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
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| Nov 14, 2021 at 13:13 | comment | added | Wojowu | What exactly do you mean with "construct point"? Is proving it exists enough? | |
| Nov 14, 2021 at 11:50 | history | asked | Đào Thanh Oai | CC BY-SA 4.0 |