Consider a triangle whose sides are segments of $\color{red}{\text{line}}$, $\color{blue}{\text{line}}$, $\color{green}{\text{line}}$ falling in the circum-circle $c$. Let $\color{red}{\text{P}}$,$\color{green}{\text{P}}$, $\color{blue}{\text{P}}$ be the poles (with respect to $c$) of the corresponding sides of the triangle.

Now, take a point $P$ different from the poles. Connect the poles with $P$. The connecting lines will intersect the corresponding edges or the elongations of theses edges mentioned above (perhaps in the $\infty$). (Corresponding means: $\color{red}{\text{ red broken line}}$ with $\color{red}{\text{ red edge line }}$, etc.

Then connect the vertices of the triangle with the the opposite intersection points mentioned above as shown in the figure below (white lines). The white lines will meet in one point. (Perhaps in the infinity; then the white lines are parallel.)

I call this point the $P$-pole point of the triangle with respect to its circum-circle and point $P$. I cannot prove that the pole point always exists. (It exists even if the white lines are parallel.) Any help, please? Any known results?

enter image description here

The same statement can be told easier in the language of hyperbolic geometry: Take an ideal triangle and a point $P$ not on the sides. Drop perpendiculars from $P$ to the sides of the triangle. Consider the intersection points. Then connect these intersection points with the opposite vertices with suitable parallels. These parallels will meet in one point, the "pole point of the ideal triangle-with respect to $P$. (See the figure below.)

enter image description here

To be honest I don't have a clue as to how to prove the statements given above. I found the "pole point" in the clear blue.

  • $\begingroup$ That “pole point” depends not only on the ideal triangle, but also on the point $P$ from which you dropped the perpendiculars. The term you chose suggests otherwise, so I'd make this clear. $\endgroup$ – MvG May 13 '15 at 13:15
  • $\begingroup$ Right, thank you for warning. So I edit. $\endgroup$ – zoli May 13 '15 at 13:28
  • $\begingroup$ I think I have an olympiad-like proof of your first statement (using a lot of Ceva and and a bit of projections), but I guess it might not satisfy you, since you also formulated the statement with hyperbolic geometry (I am not familiar at all with hyperbolic geometry). $\endgroup$ – Segipp May 13 '15 at 13:32

Here I am going to prove the first formulation of the statement in completely non-analytic way. In this proof, by "sides" of a triangle I will mean lines containing these sides.

I will first prove a lemma: let $A_1A_2A_3$ is any triangle and let $B_1,B_2,B_3$ be points on sides of this triangle opposite to $A_1,A_2,A_3$ respectively, such that $A_1B_1,A_2B_2,A_3B_3$ are concurrent at point $P$. Also let $C_1,C_2,C_3$ lie on sides of triangle $B_1B_2B_3$ such that $A_1C_1,A_2C_2,A_3C_3$ are concurrent at point $Q$. Then lines $B_1C_1,B_2C_2,B_3C_3$ are concurrent.


(NOTE: we don't know yet if red point is a point of concurrence)

Proof of lemma: Let $X_i$ be the intersection of line $A_iB_i$ with side of triangle $B_1B_2B_3$, and let $Y_i$ be the intersection of line $A_iC_i$ with side of triangle $A_1A_2A_3$. Also let's denote by $[XYZ]$, for collinear points $X,Y,Z$, value $\frac{XY}{YZ}$ if $Y$ is between $X,Z$ and $-\frac{XY}{YZ}$ otherwise. We can formulate Ceva's theorem as "$AK,BL,CM$ are concurrent cevians of $\triangle ABC$ iff $[BKC]\cdot[CLA]\cdot[AMB]=1$.

By projective invariance of cross-ratio we have $\frac{[A_1B_3A_2]}{[A_1Y_3A_2]}=\frac{[B_2X_3B_1]}{[B_2C_3B_1]},\frac{[A_2B_1A_3]}{[A_2Y_1A_3]}=\frac{[B_3X_1B_2]}{[B_3C_1B_2]},\frac{[A_3B_2A_1]}{[A_3Y_2A_1]}=\frac{[B_1X_2B_3]}{[B_1C_2B_3]}$ so by multiplying we get $\frac{[A_1B_3A_2]}{[A_1Y_3A_2]}\frac{[A_2B_1A_3]}{[A_2Y_1A_3]}\frac{[A_3B_2A_1]}{[A_3Y_2A_1]}=\frac{[B_2X_3B_1]}{[B_2C_3B_1]}\frac{[B_3X_1B_2]}{[B_3C_1B_2]}\frac{[B_1X_2B_3]}{[B_1C_2B_3]}$. Now by Ceva's theorem, numerators of both sides and denominator of LHS are all equal to $1$, so denominator of RHS is equal to $1$, so again by Ceva $A_1Y_1,A_2Y_2,A_3Y_3$ are concurrent and lemma follows.

Now we want to find this kind of configuration in your problem. If we take colorful points $P$ from your problem as $A_1,A_2,A_3$ in the lemma and vertices of your triangle as $B_1,B_2,B_3$ then we have that $A_1B_1,A_2B_2,A_3B_3$ are concurrent by using certain degenerated form of Brianchon's theorem (note that the circle in question is tangent to the sides of triangle formed by colorful points $P$, because pole of a line intersecting a circle is intersection of tangents through intersection points). Now black point $P$ from your problem will work like point $Q$ from lemma, if we take $C_1,C_2,C_3$ to be colorful points on triangle's sides. By lemma, lines $B_1C_1,B_2C_2,B_3C_3$ will be concurrent - but these are precisely the white lines you ask about. So the pole indeed always exists (except for the cases when black $P$ is at one of triangle's sides).

  • $\begingroup$ Dear @Vojowu, thank you for the answer. Please, give me some time to study it. $\endgroup$ – zoli May 13 '15 at 16:19

Any two ideal triangles are congruent, or in other words, any non-degenerate triangle inscribed into the unit circle can be mapped to any other by a projective transformation which fixes the unit circle. So without loss of generality you can restrict your considerations to one special case.

I like coordinates, so I'd start with $\triangle A_1A_2A_3$, the matrix $M$ of its circumcircle, and point $P$ chosen as follows (in homogeneous coordinates):

$$P=\begin{pmatrix}x\\y\\z\end{pmatrix} \qquad M=\begin{pmatrix}1&&\\&1&\\&&-1\end{pmatrix}$$

\begin{align*} A_1&=\begin{pmatrix}1\\0\\1\end{pmatrix} & A_2&=\begin{pmatrix}0\\1\\1\end{pmatrix} & A_3&=\begin{pmatrix}0\\-1\\1\end{pmatrix} \\ a_1=A_2\vee A_3&=\begin{pmatrix}1\\0\\0\end{pmatrix} & a_2=A_3\vee A_1&=\begin{pmatrix}1\\-1\\-1\end{pmatrix} & a_3=A_1\vee A_2&=\begin{pmatrix}1\\1\\-1\end{pmatrix} \\ B_1=Ma_1&=\begin{pmatrix}1\\0\\0\end{pmatrix} & B_2=Ma_2&=\begin{pmatrix}1\\-1\\1\end{pmatrix} & B_3=Ma_3&=\begin{pmatrix}1\\1\\1\end{pmatrix} \\ b_1=P\vee B_1&=\begin{pmatrix}0\\z\\-y\end{pmatrix} & b_2=P\vee B_2&=\begin{pmatrix}y+z\\z-x\\-x-y\end{pmatrix} & b_3=P\vee B_3&=\begin{pmatrix}y-z\\z-x\\x-y\end{pmatrix} \\ C_1=a_1\wedge b_1&=\begin{pmatrix}0\\y\\z\end{pmatrix} & C_2=a_2\wedge b_2&=\begin{pmatrix}z+y\\x-z\\2z-x+y\end{pmatrix} & C_3=a_3\wedge b_3&=\begin{pmatrix}z-y\\z-x\\2z-x-y\end{pmatrix} \\ c_1=A_1\vee C_1&=\begin{pmatrix}y\\z\\-y\end{pmatrix} & c_2=A_2\vee C_2&=\begin{pmatrix}3z-2x+y\\y+z\\-y-z\end{pmatrix} & c_3=A_3\vee C_3&=\begin{pmatrix}3z-2x-y\\y-z\\y-z\end{pmatrix} \end{align*}

$A_i$ are the corners of the triangle. $a_i$ its edges, your solid colored lines. $B_i$ their poles, your $\color{red}P,\color{green}P,\color{blue}P$. $b_i$ the connection of these to $P$, i.e. the perpendiculars, which you drew using dashed lines. $C_i$ are the points where these intersect the edges. Finally, $c_i$ are the lines connecting these, which you drew in white.

These three lines $c_i$ are concurrent if their determinant is zero.

$$\begin{vmatrix} y&3z-2x+y&3z-2x-y\\ z&y+z&y-z\\ -y&-y-z&y-z \end{vmatrix}=0$$

The point of concurrency is

$$Q=c_1\wedge c_2= \begin{pmatrix}y^2- z^2 \\ 2xy - 2yz \\ y^2 + 2xz - 3z^2\end{pmatrix}$$

There are some degenerate situations, where that point becomes undefined, i.e. the null vector. Computing all solutions to this set of three quadratic homogeneous equations, you find the degenerate situations to be

$$P\in\left\{ \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\-1\\1\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix} \right\}=\{B_1,B_2,B_3\}$$

Those are the cases you explicitely excluded, so in all other situations, everything is fine.

  • $\begingroup$ Dear @MvG, thank you for the answer. Please, give me some time to study it. $\endgroup$ – zoli May 13 '15 at 16:20

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