Let $ABC$ be a triangle with incircle $\omega$. Let $A_0,B_0,C_0$ be points outside $\omega$. The tangents from $A_0$ to $\omega$ intersect $BC$ at $A_1,A_2$. Define $B_1,B_2$ and $C_1,C_2$ similarly. Is it true that $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic $\Gamma$ if and only if $\Delta ABC$ and $\Delta A_0B_0C_0$ are perspective from a point? I find this to be true in all cases I could possibly check using Geogebra. I suppose it is true. Yet I have no proof or reference on this fact. So something on this topic would be helpful. Thanks a lot.
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3$\begingroup$ I suspect this is more a problem of projective geometry than of Euclidean geometry, since everything (except possibly the fact that $\omega$ is an incircle, which I'm guessing we could replace with other conics) is in terms of incidence, tangency, etc. $\endgroup$– Todd TrimbleCommented Jun 26, 2014 at 22:27
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1$\begingroup$ I believe that the theorems of Pascal and Brianchon will be relevant here. $\endgroup$– Peter MuellerCommented Jun 27, 2014 at 10:07
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1 Answer
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As noted by Todd this is a problem in projective geometry of conics. So we can use the duality principle and reformulate the "only if" part by the following statement (the other part is equivalent):
Suppose $C$ and $D$ are two conics in the projective plane. For every point $A$ in $C$, we associate a line $T(A)$ in this way: Let $A_1,A_2$ are two other intersection points of tangents of $A$ to $D$ with the conic $C$. Define $T(A)$ to be the line $A_1A_2$. Then for every three points $X,Y,Z$ on $C$, $\Delta XYZ$ is perspective to the triangle with edges $T(X),T(Y), T(Z)$.
To see the equivalence with the original problem, consider $X,Y,Z$ to be the dual points of edges of $\Delta ABC$, $C$ the dual of $\omega$ and $D$ the dual of the conic through $A_1,A_2,B_1,B_2,C_1,C_2$.
(All conics here are nonsingular.)
We use two lemmas:
Lemma 1: Two triangles are perspective iff there exists a polarity which sends every vertex of one of them to the corresponding edge of the other.
Lemma 2: Let $T$ ba a correlation (projective isomorphism of the plane to its dual) and $C$ a conic. If for every points $x,y\in C$, $x \in T(y) \Longleftrightarrow y\in T(x)$, then $T$ is a polarity.
Proof: $T$ is equal to its dual on $C$.
Now for the proof of our statement it suffices to see that $T$ is an algebraic isomorphism between $C$ and dual of a conic in the plane. (The only thing that should be checked is injectivity of $T$ which is true in the generic case. More precisely when there not exists a quadrilateral inscribed in $C$ and circumscribed about $D$.) So $T$ can be extended to a correlation in the plane which obviously satisfies the condition of Lemma 2 for $C$. Therefore $T$ is a restriction of a polarity and Lemma 1 completes the proof.
answered Jun 27, 2014 at 13:53
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$\begingroup$ I have some questions. See I am not too comfortable with projective geometry, so it took me long to reply. What is a polarity? Do you mean a projective transformation? Also I don't understand your one line proof. How do you conclude $T$ is its dual on $C$? And how does that finish the lemma? Again you prove the statement in the last paragraph, which you state to be obvious, but I don't understand anything. See these things may be obvious for a geometer but I am not really experienced in Projective geometry. So a bit of explanation would help, thanks. $\endgroup$– shadow10Commented Jun 29, 2014 at 10:44
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1$\begingroup$ @shadow10 You can see the definition of these basic concepts(polarity, correlation, ...) in classical books on algebraic geometry, like "Hodge & Pedoe" or "Semple & Roth". A correlation is a projective transformation $T:\mathbb{P}(V)\to\mathbb{P}(V^\vee)$ or a morphism from the points of $\mathbb{P}(V)$ to hyperplanes in $\mathbb{P}(V)$. So there is a matrix $A$ for which $x\in T(y) \Longleftrightarrow xAy=0.$ The dual of this correlation is given by the transpose of $A$. And a correlation is called polarity if it's equal to it's dual, i.e. $A^t = \lambda A$. $\endgroup$ Commented Jun 29, 2014 at 11:37
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2$\begingroup$ @shadow10 By these definitions you can easily prove the lemmas. In the last paragraph you first need to show that $T$ gives a bijection between $C$ and lines tangent a conic $E$, which is the same as a conic in the dual plane. A bijection between two conics(in two planes) can be extended to a projective morphism iff it respects the cross ratios on two conics. The important point here is that the bijections given by rational functions (i.e. algebraic morphisms) respect the cross ratio, and this is the fact used in that paragraph. $\endgroup$ Commented Jun 29, 2014 at 11:44
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$\begingroup$ Thank you for these references, let me read a bit, and then get back to you. @Mostafa $\endgroup$– shadow10Commented Jun 29, 2014 at 11:47