Dear Mathlinkers, an outline of my synthetic proof... 1. H appears as the Miquel’s point 2. AA’ goes through G and circularly 3. according to a Mannheim’s circle, we are done. Sincerely Jean-Louis

[asy][asy] /* First diagram for Rioplatense Olympiad 2013 Problem 6, free script by liberator, 23 August 2014 */ unitsize(4cm); pointpen=black; pen dsp = rgb(0.4,0.6,0.8); pen rcp = rgb(0.9,0,0); pen gcp = rgb(0,0.6,0); /* Initialize objects */ pair A = dir(110); pair B = dir(200); pair C = dir(340); pair A1 = midpoint(B--C); pair B1 = midpoint(C--A); pair C1 = midpoint(A--B); pair D = foot(A,B,C); pair E = foot(B,C,A); pair F = foot(C,A,B); pair G = centroid(A,B,C); pair H = orthocenter(A,B,C); pair Ap = intersectionpoint(circumcircle(B,H,C), A--A1); pair Bp = intersectionpoint(circumcircle(C,H,A), B--B1); pair Cp = intersectionpoint(circumcircle(A,H,B), C--C1); pair Ha = rotate(180,A1)*A; /* Draw objects */ draw(A--B--C--cycle, dsp+linewidth(1)); draw(A--Ha, dsp); draw(B--Ha, dsp); draw(C--Ha, dsp); draw(C--F, dsp); draw(circumcircle(B,H,C), rcp); draw(circumcircle(A,H,Ap), rcp); draw(circumcircle(Ap, Bp, Cp), gcp+linewidth(1)); /* Place dots on and label each point */ Drawing("A", A, dir(A)); Drawing("B", B, dir(B)); Drawing("C", C, dir(C)); Drawing("A'", Ap, 1.5*dir(70)); Drawing("B'", Bp, dir(-90)); Drawing("C'", Cp, dir(-150)); Drawing("D", D, dir(-90)); Drawing("E", E, dir(20)); Drawing("F", F, dir(-160)); Drawing("G", G, dir(-20)); Drawing("H", H, dir(90)); Drawing("H_a", Ha, dir(Ha)); [/asy][/asy] Let $\triangle DEF$ be the orthic triangle, and let $H_a$ be the antipode of $H$ w.r.t $(BHC)$. Note that $ABH_aC$ is a parallelogram. $AF \parallel H_aC$, so by Reim's theorem, $A,A',H_a$ are collinear: it follows that $AA'$ is a median; similarly, $BB', CC'$ are medians. [asy][asy] /* Second diagram for Rioplatense Olympiad 2013 Problem 6, free script by liberator, 23 August 2014 */ unitsize(4cm); pointpen=black; pen dsp = rgb(0.4,0.6,0.8); pen rcp = rgb(0.9,0,0); pen gcp = rgb(0,0.6,0); /* Initialize objects */ pair A = dir(110); pair B = dir(200); pair C = dir(340); pair A1 = midpoint(B--C); pair B1 = midpoint(C--A); pair C1 = midpoint(A--B); pair D = foot(A,B,C); pair E = foot(B,C,A); pair F = foot(C,A,B); pair G = centroid(A,B,C); pair H = orthocenter(A,B,C); pair Ap = intersectionpoint(circumcircle(B,H,C), A--A1); pair Bp = intersectionpoint(circumcircle(C,H,A), B--B1); pair Cp = intersectionpoint(circumcircle(A,H,B), C--C1); pair K = intersectionpoint(circumcircle(A,E,F), Line(Bp, H, -0.5, 2013)); /* Draw objects */ draw(A--B--C--cycle, dsp+linewidth(1)); draw(A--G, dsp); draw(B--G, dsp); draw(C--Cp, dsp); draw(Bp--K, dsp); draw(A--K, dsp); draw(circumcircle(A,E,F), rcp); draw(circumcircle(B,F,D), rcp); draw(circumcircle(C,D,E), rcp); draw(circumcircle(Ap, Bp, Cp), gcp+linewidth(1)); /* Place dots on and label each point */ Drawing("A", A, dir(A)); Drawing("B", B, dir(B)); Drawing("C", C, dir(C)); Drawing("A'", Ap, 1.5*dir(70)); Drawing("B'", Bp, dir(-110)); Drawing("C'", Cp, dir(-150)); Drawing("D", D, 1.6*dir(-90)); Drawing("E", E, 2*dir(60)); Drawing("F", F, 2.5*dir(160)); Drawing("G", G, dir(-90)); Drawing("H", H, 1.6*dir(90)); Drawing("K", K, dir(K)); [/asy][/asy] Let $K \equiv B'H \cap (AEF)$. By Reim's theorem on $(BDF), (AEF)$, $AK \parallel BB'$; by a converse of Reim's theorem, it follows that $A'HB'G$ is cyclic. Similarly, $B'C'HG$ is cyclic, so $A'GB'C'$ is cyclic, as required.

Let $K$ be midpoint of $[BC]$, let $P$ be midpoint of $[AH]$ and let $X$ be center of $\odot(BHC)$. Let $(P)$ be circle on diameter $[AH]$ and $(X) \equiv \odot(BHC)$. $A'H$ is radical axis of circles $(P), (X),$ perpendikular to their center line $PX$. $[AP] = [KX]$ and $AP \parallel KX$ $\implies$ $AKXP$ is parallelogram $\implies$ $(AK \equiv AG) \parallel PX$ $\implies$ $(AK \equiv AG) \perp A'H$. $A' \in (P)$ $\implies$ $AA' \perp A'H$ as well $\implies$ $A' \in (AK \equiv AG)$ $\implies$ $GA' \perp A'H$ $\implies$ $A'$ is on circle with diameter $[GH]$. Similarly, $B', C'$ are on this circle.

Its not hard to see the circle is the circle with diametre $HG$.

How did you get that IDMaster,

See this $AA'$,$BB'$,$CC'$ are medians so it trivially follows that $A'$,$B'$,$C'$ lie on the circle with diameter $HG$.

Let $M$ and $N$ be midpoints of $AC$ and $DF$. Then (since $M$ is circumcenter of $ACDF$) is $MN\perp DF$. Now realize, that $B'$ is center of spiral similarity moving $AD$ to $CF$ and therefore also moving them to $MN$. Therefore it's also center of spiral similarity moving $MA$ to $ND$. Therefore $\angle B'FB=90-\angle B'FC=90-\angle B'NM=\angle B'ND=\angle B'MA$, so $FB'MA$ is cyclic and analogically $B'DCM$ is cyclic. Since $B'M$ is their common chord and $B$ has the same power to both of them (since $BA\cdot BF=BC\cdot BD$), we get that $B'$ lies on $BM$. Then $\angle HB'G=180-\angle HB'B=180-90=90$ or $\angle HB'G=\angle HB'B=90$, so anyway $B'$ lies on circle with diameter $HG$. Analogically for $A'$ and $C'$ and we're done. Q.E.D.

This is in fact a well known result, for example see Roger Johnson's book. The points $A',B',C'$ are known as the vertices of the $D-$ triangle of $ABC$.

WizardMath wrote: This is in fact a well known result, for example see Roger Johnson's book. The points $A',B',C'$ are known as the vertices of the $D-$ triangle of $ABC$. Also the isgonal conjugate of Artzt parabola's foci.

Some more beautiful properties: $A'$ is on the McCay's circle as well as on the $A$-Apollonius circle, and is the focus of the $A-$Brocard parabola. It is the point of intersection of the circles through $A,B$ and $A,C$ tangent to $BC$. And it is the reflection of the endpoint of the $A-$symmedian chord in the circumcircle of $ABC$ in $BC$. It is on the circumcircle of $BHC$, $(AH)$, and the isogonal conjugate of the point of the second brocard triangle corresponding to $A$. It also forms an isosceles trapezoid with the midpoint of $AH$, midpoint of $BC$ and reflection of O in BC.

WizardMath wrote: Some more beautiful properties: $A'$ is on the McCay's circle as well as on the $A$-Apollonius circle, and is the focus of the $A-$Brocard parabola. It is the point of intersection of the circles through $A,B$ and $A,C$ tangent to $BC$. And it is the reflection of the endpoint of the $A-$symmedian chord in the circumcircle of $ABC$ in $BC$. It is on the circumcircle of $BHC$, $(AH)$, and the isogonal conjugate of the point of the second brocard triangle corresponding to $A$. It also forms an isosceles trapezoid with the midpoint of $AH$, midpoint of $BC$ and reflection of O in BC. The same as me（but Wizard said much more）.In my memory,these property were spotted(maybe firstly) by an American student...

Hey guys! The main part of the problem is to see that A' lies on the A-median and so on. Let us consider D, E and F the feet of A, B and C altitudes. It is quite intuitive to consider the intersection L of the lines EF and BC, since it may lie on the line HA' by simple radical center. By construction: (B, C; L, D)=-1. Projecting by H onto (AFHE), we have: (E, F; A', A)=-1. Now is just happiness: Projecting our last result by A at the line BC, we see that E and F go to C and B, respectively, but as the center of (AFHE) lies on AH, A goes to the point of infinity. Then, A' must go to M! Nice problem!

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%201.pdf p. 53... Sincerely Jean-Louis

This basically reduces to proving that the three HM points and the centroid are concyclic. In fact, I will prove an even stronger claim which include the orthocenter in this concyclicity. Because $\angle HA'G = 90^{\circ}$ and $\angle HB'G = 90^{\circ},$ we know that $HA'GB'$ is cyclic. Furthermore, using a similar argument, we find that $\angle HC'G=90^{\circ}$ and $\angle HB'G=90^{\circ}$, which shows that $HC'B'G$ is cyclic. Since these two cyclic quads share at least three points, we know that they are in fact the same circle, and so we have shown the desired (stronger) claim. This problem is also killed by reim's theorem, as presented in @post#3's blog.