Let $AA_1$ and $BB_1$ be the altitudes of an acute-angled, non-isosceles triangle $ABC$. Also, let $A_0$ and $B_0$ be the midpoints of its sides $BC$ and $CA$, respectively. The line $A_1B_1$ intersects the line $A_0B_0$ at a point $C'$. Prove that the line $CC'$ is perpendicular to the Euler line of the triangle $ABC$ (this is the line that joins the orthocenter and the circumcenter of the triangle $ABC$).
Problem
Source: Spanish Communities
Tags: Euler, geometry, circumcircle, LaTeX, parallelogram, projective geometry, perpendicular bisector
nsato
04.04.2006 12:12
As usual, let $H$ denote the orthocentre of triangle $ABC$. Let $X = A_0 B_1 \cap A_1 B_0$. We claim that $X$ lies on the Euler line of triangle $ABC$.
Let $P$ and $Q$ be the midpoints of $AH$ and $BH$, respectively. Then $P$, $Q$, $A_0$, $B_0$, $A_1$, and $B_1$ all lie on the nine-point circle of triangle $ABC$, centred at $N$, the nine-point centre of triangle $ABC$. Furthermore, $H = PA_1 \cap AB_1$, $X = A_0 B_1 \cap A_1 B_0$, and $N = PA_0 \cap QB_0$. Therefore, by Pascal's theorem, $H$, $X$, and $N$ are collinear. Since $H$ and $N$ lie on the Euler line, so does $X$.
Then, $C' = A_0 B_0 \cap A_1 B_1$, $C = A_0 A_1 \cap B_0 B_1$, and $X = A_0 B_1 \cap A_1 B_0$. Therefore, by results on poles and polars, $NX \perp CC'$. But $X$ and $N$ lie on the Euler line, so the result follows.
grobber
04.04.2006 12:48
By Pappus's Theorem applied to the lines $(A,B_1,B_0)$ and $(B,A_1,A_0)$ we conclude that $Q=A_1B_0\cap A_0B_1$ lies on the Euler line of $ABC$. Since $CC'$ is the polar of $Q$ wrt the nine-point circle, it follows that it must be perpendicular to the diameter of this circle which passes through $Q$, which, by the observation before, is precisely the Euler line.
xirti
30.04.2006 22:36
I'll state a more general statement (which appears in a ISL problem, from 1995 or smth... ): Given a quadrilateral $ABCD$ inscribed in a circle of center $O$, denote by $E,F$ intersections of extensions of opposite sides. We know that the circles circumscribed about $EAB,ECD,FBC,FAD$ have a common point $M$ (Miquel point). Then $OM\perp EF$. For a proof see we can show that the quadrilateral formed by circumcenters of $ABCD$,$ECD$,$EAB$ and the point $E$ is a parallelogram (*) because each pair of opposite edges is parallel with $DC$ and $AB$. Now, the line joining cicumcenters of $EAB$ and $ECD$ is the perpendicular bisector of the common chord $EM$ of those circles. Because that parallelogram (*), we know that this line pass trough midpoint of $EO$. Hence triangle $EMO$ is rectangular. And we are done. It is easy to prove that $CA_0HB_0$ is inscriptible and the circumcenter is the midpoint $T$ of $CH$. It is also easy to prove that $CA_1OB_1$ is inscriptible and the circumcenter is the midpoint $S$ of $CO$. It is obvious that $ST\parallel OH$ Aplying this statement for this problem we get that $OH\parallel ST\perp CC'$
xirti
01.05.2006 10:22
Sorry for makin lots of mistakes in my previous post. I will correct it. We need only that miquel point $M$ of the inscriptible quadrilateral $A_0A_1B_1A_0$ (the circumcircle in the nine point circle) lies on the line joining the points of intersetions of extensions of edges: $CC'$. As I proved in the previous post $OH$ is parallel with the line joining centers of the triangles $CA_0B_0$ and $CA_1B_1$. This line is perpendicular to the radical axis: $CM=CC'$. And we are done.
Yosh...
27.05.2007 20:26
Why if $X$ lies on euler line,we get $NX\perp CC'$ ? Thx...
nsato
13.11.2012 18:02
Let $\omega_1$ denote the nine-point circle of triangle $ABC$. Let $\omega_2$ denote the circle with diameter $AO$; note that $A_0$ and $B_0$ lie on $\omega_2$. Let $\omega_3$ denote the circle with diameter $AH$; note that $A_1$ and $B_1$ lie on $\omega_3$. The radical axis of $\omega_1$ and $\omega_2$ is $A_0 B_0$. The radical axis of $\omega_1$ and $\omega_3$ is $A_1 B_1$. Hence, $C' = A_0 B_0 \cap A_1 B_1$ is the radical center of $\omega_1$, $\omega_2$, and $\omega_3$. Let $D$ be the projection of $C$ onto $HO$. Since $\angle CDO = \angle CDH = 90^\circ$, $D$ lies on both $\omega_2$ and $\omega_3$. Hence, $CD$ is the radical axis of $\omega_2$ and $\omega_3$, which means $C'$ lies on $CD$. It follows that $CC'$ and $HO$ are perpendicular. [asy][asy] unitsize(0.6 cm); pair A0, B0, C0, D, H, O; pair[] A, B, C; C[0] = (5,12); A[0] = (0,0); B[0] = (14,0); A0 = (B[0] + C[0])/2; B0 = (A[0] + C[0])/2; A[1] = (A[0] + reflect(B[0],C[0])*(A[0]))/2; B[1] = (B[0] + reflect(C[0],A[0])*(B[0]))/2; C[1] = extension(A[1],B[1],A0,B0); H = orthocenter(A[0],B[0],C[0]); O = circumcenter(A[0],B[0],C[0]); D = extension(C[0],C[1],O,H); draw(circumcircle(C[0],A0,B0),red); draw(circumcircle(A0,B0,(A[0] + B[0])/2),red); draw(circumcircle(C[0],A[1],B[1]),red); draw(A[0]--B[0]--C[0]--cycle); draw(A[0]--A[1]); draw(B[0]--B[1]); draw(C[0]--D); draw(A0--B0); draw(A[1]--B[1]); draw(O--H); label("$A$", A[0], SW); label("$A_0$", A0, E); label("$A_1$", A[1], NE); label("$B$", B[0], SE); label("$B_0$", B0, W); label("$B_1$", B[1], W); label("$C$", C[0], N); label("$C'$", C[1], NW); label("$D$", D, S); dot("$H$", H, S); dot("$O$", O, SE); label("$\omega_1$", circumcenter(A0,B0,(A[0] + B[0])/2) + abs(A0 - circumcenter(A0,B0,(A[0] + B[0])/2))*dir(-50), dir(-50)); label("$\omega_2$", (C[0] + O)/2 + abs(C[0] - O)/2*dir(30), dir(30)); label("$\omega_3$", (C[0] + H)/2 + abs(C[0] - H)/2*dir(150), dir(150)); [/asy][/asy]
jayme
14.11.2012 07:28
Dear Mathlinkers, you can see at http://perso.orange.fr/jl.ayme vol. 2 Schroeter p. 5 Sincertely Jean-Louis