Timeline for Proofs without words
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2023 at 11:12 | comment | added | user376343 | This is a Dandelin's theorem. | |
| Dec 15, 2016 at 13:36 | comment | added | DanielWainfleet | There are more gems in Dandelin's figure but I need words : Let circles $k_1,\; k_2$ lie in planes $J_1.\; J_2 $ respectively. Let the ellipse be in plane $J_E$. Then $J_1\cap J_E ,\; J_2\cap J_E$ are the directrices.. Let line $l_P$ be the tangent to the ellipse at $P,$ with $l_P\subset J_E.$ Let $l_P$ meet $J_1,\;J_2$ at $Q_1\;,Q_2$ respectively. For $ i=1,2 , $ triangles $Q_iPP_i ,\; Q_iPF_i$ are congruent (as $Q_iP_i.\; Q_iF_i$ are tangent to sphere $G_i$), So angles $Q_iPP_i = Q_iPF_i$ . But $Q_1PQ_2$ and $P_1PP_2$ are lines. So angles $Q_2PF_2=Q_2PP_2=Q_1PP_1=Q_1PF_1.$ | |
| Jun 17, 2014 at 2:47 | history | edited | senshin | CC BY-SA 3.0 |
rehost to imgur to prevent linkrot
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| Mar 5, 2014 at 22:49 | comment | added | Tim Campion | I was confused by the perspective. In case anyone is having the same problem: the perspective is from a point that is (below the apex $S$ of the cone but) above the base of the cone (circle $k2$) and the bottom half-sphere ($G2$). I mistakenly thought we were looking up through $k2$ into the inside of the cone -- I think this is because in my browser, at least, the the circle $k2$ gets thicker when it passes behind the ellipse and should if anything get thinner. It's generally a nice drawing, though, and a nice proof. | |
| Nov 9, 2013 at 3:21 | comment | added | Włodzimierz Holsztyński | I've learned this and related proofs from Hilbert and Cohn-Vossen (but these proofs still originated mostly with Dadelin). | |
| Jul 29, 2013 at 13:14 | comment | added | Patrick Da Silva | @aorq : Riiiight. Thanks for the clarification! It is indeed very visual. | |
| Jul 29, 2013 at 8:53 | comment | added | aorq | @PatrickDaSilva: $PF1 = PP1$ because tangents to a circle/sphere have equal length. The total distance is thus equal to $PF1 + PF2 = PP1 + PP2 = P1P2$, which is constant. | |
| Jul 28, 2013 at 18:35 | comment | added | Patrick Da Silva | How does the picture explain the invariance of the total distance to two foci? I don't see it ; I haven't done geometry in a while though, I'm guessing it's some triviality... refresh my memory please? :) | |
| Feb 18, 2011 at 22:17 | comment | added | godelian | Indeed. And there are also similar visual proofs for the hyperbola and the parabola | |
| May 15, 2010 at 18:47 | comment | added | Andrés E. Caicedo | Yes, this one is beautiful. | |
| Dec 22, 2009 at 16:54 | history | answered | aorq | CC BY-SA 2.5 |