Timeline for Does every ellipse inside a tetrahedron inside a ball fit in a triangle inside the ball?
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| Oct 9, 2012 at 1:01 | comment | added | Marcos Cossarini | Good job! I don't understand the projection through $C$ from $\omega$ to $\Omega$ (in the comment between brackets in the last paragraph), but anyway, from the equality of ratios, it is easy to see that the fourth point of intersection $C′$ should satisfy $(C'A,C'C;C'X,C'Y)=(C'M,C'C,C'X,C'Y)$, so it should be on the line $AM$. In case anyone is wondering, the expression "cross ratio of the four points $A$, $C$, $X$, $Y$ with respect to $\omega$" is justified because given four points E,F,G,H and a number $\lambda$, the locus of the points T such that $(TE:TF:TG:TH)=\lambda$ is a conic section. | |
| Oct 8, 2012 at 8:36 | comment | added | Gjergji Zaimi | This is a very pretty argument. | |
| Oct 8, 2012 at 3:46 | history | answered | zeb | CC BY-SA 3.0 |