Timeline for Theorems in Euclidean geometry with attractive proofs using more advanced methods
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 20, 2020 at 6:11 | comment | added | user131781 | Every triangle is affinely equivalent to an equilateral one for which the result is trivial (in this case you get a circle, of course). All the claims of the result are affinely invariant. This method can be used to give the ellipse in an explicit form or a geometric construction of it. | |
| Feb 1, 2012 at 18:38 | comment | added | darij grinberg | ... touching the line $BC$ at $X$, and so on, and thus you obtain as a particular case the following fact: Let $X$, $Y$, $Z$ be points on the sidelines $BC$, $CA$, $AB$ of a triangle $ABC$. Then, there exists a conic touching $BC$, $CA$, $AB$ at $X$, $Y$, $Z$ if and only if either the lines $AX$, $BY$, $CZ$ concur or the points $X$, $Y$, $Z$ are collinear (in which case it is a degenerate conic). This uses Ceva and Menelaos. Not to say this is simpler than the affine-transformation proof, but in my eyes it shows that the problem has nothing to do with midpoints of edges. | |
| Feb 1, 2012 at 18:36 | comment | added | darij grinberg | Generalization: Let $ABC$ be a triangle. Let $X$ and $X'$ be points on the line $BC$. Let $Y$ and $Y'$ be points on the line $CA$. Let $Z$ and $Z'$ be points on the line $AB$. Then, the points $X$, $X'$, $Y$, $Y'$, $Z$, $Z'$ lie on one conic (possibly degenerate) if and only if $\frac{BX}{XC}\cdot\frac{BX'}{X'C}\cdot\frac{CY}{YA}\cdot\frac{CY'}{Y'A}\cdot\frac{AZ}{ZB}\cdot\frac{AZ'}{Z'B}=1$, where the segments are directed. This is easy to prove using Pascal and Menelaos. When two points like $X$ and $X'$ coincide, a conic passing through $X$ and $X'$ has to be understood as a conic ... | |
| Jul 12, 2010 at 20:27 | comment | added | Gjergji Zaimi | Yes, sorry, I meant fractional linear :) | |
| Jul 11, 2010 at 7:46 | comment | added | Victor Protsak | Gjergij, Steiner chain deserves its own entry, but the key issue is that the transformation be $\textit{conformal},$ so that it preserves circles, and not merely linear. | |
| Jul 11, 2010 at 5:12 | comment | added | Gjergji Zaimi | Linear transformations also give a nice proof of en.wikipedia.org/wiki/Steiner_chain | |
| Jul 11, 2010 at 4:07 | history | answered | Jack Huizenga | CC BY-SA 2.5 |