Timeline for A certain circle formed by perpendiculars
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 4 at 6:29 | comment | added | N M | You're right, I was looking at a more general situation. I interpreted the question to mean that CJ:JI:IA = AG:GH:HB = BK:KL:LC, but not CJ=IA etc. With this setup the six points are still concyclic, but the center is not necessarily the circumcenter. | |
| Sep 4 at 5:56 | comment | added | Toni Mhax | @NM, maybe you have another general form for the circle, but in the case i understood in the figure $OPRM$, $QRNO$, $PQMN$ are rectangles, as in the comment, this can be seen by Thales theorem (in basic triangles). I took $LC=KB=rCB$ and $CJ=AI=rAC$ and $AG=HB=rAB$ for the same ratio $r$... | |
| Sep 4 at 5:16 | comment | added | terceira | While it is true that the diagonals of the hexagon need not be concurrent, it is in fact true that the centre of the new circle coincides with the circumcentre. | |
| Sep 4 at 4:07 | comment | added | N M | In general the diagonals of the hexagon do not pass through the center of the circle, so I don't think either of the claims in the previous comment is correct. The circle is a Tucker circle of the triangle formed by either triple of corresponding perpendiculars. If you'd like I can write up a short proof as a proper answer. | |
| Sep 3 at 6:09 | comment | added | Toni Mhax | This is easy to prove when you notice the rectangles that the figure gives (using Thales theorem for example). The circumcircle of $ABC$ and this circle are concentric. For $PL=RK, RH=NG, KM=LO$, etc | |
| Sep 2 at 22:22 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tags
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| Sep 2 at 21:41 | history | asked | Benjamin L. Warren | CC BY-SA 4.0 |