Timeline for Showing that a $1:\sqrt{2}:2$ triangle can be inscribed in a regular heptagon in this way
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Oct 21 at 9:48 | history | notice added | Shaun♦ | Contest Question | |
| S Oct 21 at 9:48 | history | locked | Shaun♦ | ||
| Oct 21 at 8:34 | comment | added | Dan | @dezdichado For synthetic solutions for angles involving $\frac{\pi}{7}$, see here. | |
| Oct 21 at 4:19 | comment | added | Dan | This is question 4/2/37 in the on-going USA Mathematical Talent Search contest. The rules state that participants "may not use 'live' help. For example, they may not ask for help on online forums...". | |
| Oct 20 at 16:28 | history | edited | svj | CC BY-SA 4.0 | added 107 characters in body |
| Oct 20 at 15:00 | vote | accept | svj | ||
| Oct 20 at 16:13 | |||||
| Oct 19 at 19:23 | history | edited | svj | CC BY-SA 4.0 | added 120 characters in body |
| Oct 19 at 19:22 | comment | added | svj | i was thinking something with heptagonal triangles. | |
| Oct 19 at 18:58 | comment | added | dezdichado | I am not sure I ever seen a nice, synthetic solution for angles involving $\frac{\pi}{7}.$ Would be pleasantly surprised if someone comes up with it. | |
| Oct 19 at 18:46 | history | edited | svj | CC BY-SA 4.0 | added 134 characters in body |
| Oct 19 at 15:29 | history | edited | svj | CC BY-SA 4.0 | added 124 characters in body |
| Oct 19 at 15:27 | history | undeleted | svj | ||
| Oct 19 at 15:11 | history | deleted | svj | via Vote | |
| Oct 19 at 12:48 | history | edited | svj | CC BY-SA 4.0 | added 88 characters in body |
| Oct 19 at 12:48 | comment | added | svj | I know, but I am interested in a purely geometric solution. | |
| Oct 19 at 12:47 | history | undeleted | svj | ||
| Oct 19 at 12:19 | history | deleted | svj | via Vote | |
| Oct 19 at 8:49 | comment | added | user398505 | You could perhaps use that $\cos(2 \pi/7), \cos(4 \pi/7), \cos(6 \pi/7)$ are the roots of the polynomial $p(x)=8x^3+4x^2-4x-1$. Then use Cardano's formula to compute exactly these cosines and the corresponding sines. Then compute all the points exactly. | |
| Oct 19 at 6:56 | history | edited | Blue | CC BY-SA 4.0 | More-informative title; incorporated key fact into the description of inscription |
| Oct 19 at 6:09 | review | Close votes | |||
| Oct 19 at 7:12 | |||||
| Oct 19 at 4:46 | history | edited | svj | CC BY-SA 4.0 | added 64 characters in body |
| Oct 19 at 4:45 | comment | added | svj | sory, shoul have specified that in this specific configuration, $BF \perp FE$. | |
| Oct 19 at 4:44 | comment | added | insipidintegrator | I don't understand why this is special. I could choose any point X on the side HI and then claim that the triangle ABX can be "inscribed" in a regular heptagon. | |
| Oct 19 at 4:42 | history | edited | insipidintegrator | CC BY-SA 4.0 | added 2 characters in body |
| S Oct 19 at 4:37 | review | First questions | |||
| Oct 19 at 4:59 | |||||
| S Oct 19 at 4:37 | history | asked | svj | CC BY-SA 4.0 |