# A zigzag circuit in the triangle

Let $\overset { \triangle }{ ABC }$ be a triangle and O its incenter. Starting from a generic point P$\in$AC, we reflect it across the bisector AO obtaining ${P}_{1}$. Reflecting it across the bisector BO we obtain ${P}_{2}$. Reflection of ${P}_{2}$ across CO gives ${P}_{3}$. If ${P}_{3}=P$ we have finished. If not, we continue reflecting it across AO obtaining ${P}_{4}$. After reflection across BO we arrive at ${P}_{5}$ and finally with a reflection across CO to ${P}_{6}$. But we claim that

Proposition 1:  ${P}_{6}=P$, that is the circuit closes after at most two turns, and it closes at once with ${P}_{3}=P$  iff P is a point of tangence of the inscribed cercle..

Analitic proof:
let’s call AP=x. Then $A{P}_{1}=x$, ${P}_{2}B={P}_{1}B=c-x$; $C{P}_{3}=C{P}_{2}=BC-{P}_{2}B= a-(c-x)=a-c+x$; $A{P}_{4}=A{P}_{3}=AC-C{P}_{3}=b-(a-c+x)=b-a+c-x$; $B{P}_{5}=B{P}_{4}=AB-A{P}_{4}=c-(b-a+c-x)=a-b+x$; $C{P}_{6}=C{P}_{5}=BC-B{P}_{5}=a-(a-b+x)=b-x \Rightarrow {P}_{6}=P$ closing the circuit. Q.E.D

From the proof we also have that the circut closed with ${P}_{3}=P$ exactly when $C{P}_{3}$+AP=c that is (a-c+x)+x = b that gives x=(b-a+c)/2. But one recognizes that this is the distance from A of the point of tangence with AC of the inscribed circle. Then the circuit closes at once iff the starting point is a point of tangence of the inscribed triangle.

Synthetic proof: by construction ${P}_{1}$ is symmetric of P with axes AO and then OP=$O{P}_{1}$. Similarly O${P}_{1}$=O${P}_{2}$, $O{P}_{2}=O{P}_{3}$, $O{P}_{3}=O{P}_{4}$, $O{P}_{4}=O{P}_{5}$, $O{P}_{5}=OP$ that is P,${P}_{1}$,${P}_{2}$,${P}_{3}$,${P}_{4}$,${P}_{5}$ all belong to the same circle of center O. On this circle the two chords $P{P}_{1}$ and ${P}_{3}{P}_{4}$ are parallel being perpendicular to AO and similarly ${P}_{4}{P}_{5}$//${P}_{1}{P}_{2}$ and ${P}_{5}{P}_{6}$//${P}_{2}{P}_{3}$. If we sign on a circle the points P, ${P}_{1}$, ${P}_{2}$, ${P}_{3}$ and draw ${P}_{3}{P}_{4}$ parallel to $P{P}_{1}$ meeting the circle in ${P}_{4}$, then the parallel ${P}_{4}{P}_{5}$ to ${P}_{1}{P}_{2}$ and then the parallel ${P}_{5}{P}_{6}$ to ${P}_{2}{P}_{3}$ we must have ${P}_{6}$=P. In fact arcs ${P}_{3}P$ and ${P}_{1}{P}_{4}$ must be equal being intercepted by parallel chords. Similarly arc(${P}_{1}{P}_{4}$)=arc(${P}_{5}{P}_{2}$)=arc(${P}_{3}{P}_{6}$). But then arc(${P}_{3}{P}_{6}$)=arc(${P}_{3}P$) implies that ${P}_{6}$=P. Thus the circuit closes at the second round. Q.E.D.

From the proof we have also that  the chords ${P}_{1}{P}_{4}$, ${P}_{2}{P}_{5}$, ${P}_{3}P$ are all equal. This is confirmed also by the preceeding calculations that give easily their common value (b-a+c-2x). O, incenter of ABC, is equidistant from AB,AC,BC ad hence from the chords ${P}_{1}{P}_{4}$,${P}_{2}{P}_{5}$,${P}_{3}P$P and this confirms again that they all have the same length. But these distances are given joining O with the middle points M,N,S of the chords ${P}_{1}{P}_{4}$,${P}_{2}{P}_{5}$,${P}_{3}$P that must then coincide with the points of tangence of the inscribed triangle.
But vaying P these three points M,N,S are fixed and fixed are the lenghts MS,MN,NS. But these are the semisum of P${P}_{1}$ and ${P}_{3}{P}_{4}$, ${P}_{1}{P}_{2}$ and ${P}_{4}{P}_{5}$, ${P}_{2}{P}_{3}$ and ${P}_{5}{P}_{6}$ (consider the isosceles trapezioes $P{P}_{1}{P}_{3}{P}_{4}$…). Then we have:
Proposition 2: The length of the double circuit is constant varying P and is two times the length of the triangle with vertices the points of tangence of the inscribed circle.

Moreover:
Proposition 3$\overset { \triangle }{ {P}_{1}{P}_{3}{P}_{5} \simeq }\overset { \triangle }{ {P}_{2}{P}_{4}{P}_{6} }$ and the congruence is given by rotation around the incenter O of angle $\widehat { {P}_{1}O{P}_{4}}$. .
Proof: $\widehat { {P}_{1}O{P}_{4}}=\widehat { {P}_{5}O{P}_{2}}=\widehat { {P}_{3}O{P}_{6}}$ because all subtend equal chords. Then the rotation of such angle around O sends ${P}_{1}\rightarrow {P}_{4},{P}_{5}\rightarrow {P}_{2},{P}_{3}\rightarrow {P}_{6}$ and then overlaps exactly the two triangles. Q.E.D.

Elementary proof of Pascal’s theorem:
The three points of intersecions of the three couples of opposite sides of an exagon inscribed in a conic are collinear.
Proof: With a proiective (with conservation of incidence, collinearity…) we transform the conic in a circle. With another such transformation we throw two of those three point to infinite, that is  we make ${P}_{1}{P}_{2}$//${P}_{4}{P}_{5}$ and ${P}_{3}{P}_{4}$//$P{P}_{1}$. arc(${P}_{1}{P}_{4}$)=arc(${P}_{5}{P}_{2}$)=arc(${P}_{3}{P}_{6})\Rightarrow arc({P}_{1}{P}_{4})=arc({P}_{3}{P}_{6})\Rightarrow {P}_{3}{P}_{4}//{P}_{1}{P}_{6}$. This means that ${P}_{3}{P}_{4}\cap{P}_{1}{P}_{6}$ is at infinite too, that is the three point of intersections of opposite sides are collineated. Transforming back to the original conics the same is true there because proiectivities preserve incidence, collineation and so on.

Comment: magazine Archimede, quoted in references, refused to publish this article with the following reasons: ” very interesting property of triangles but too difficult…, most italian math teachers are not graduated in mathematics…, too inaccurate…many references to little-known results, properties and theorems… ”

When I answered that I could make corrections and improvements they answered: ” no absolutely, better to devote yourself to a more simple and contextualized teaching…!!!”

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