This was the most difficult problem at this year's German olympiad, only $16 \%$ of the marks were given to all students. I find that kind of weird, I did not qualify but did find this problem rather easy. #toodifficulttoqualifyinbavaria

lemma let $P,Q$ isogonal conjugates WRT the triangle $ABC $ , $\cal{C,C}$$_1,\cal{C}$$_2$ the circumcircles of $ABC,PBC,QBC$ resp. if $P',Q'$ are points of $\cal{C}$$_1,\cal{C}$$_2$ resp.such that $A-P-P'$ are collinear and $A-Q-Q'$ are collinear then $PQ\parallel P'Q'$ .we have previously prove it . back to the problem just take $ P=$ the circumcenter of $I_aBC$ which is the midpoint of arc $\overarc{BC}$ of $\cal{C}$ , $P'=A ,Q=$ the orthocenter of $I_aBC$ which is the symmetric of $I$ wrt the midpoint of $BC$ so $P,Q$ are isogonal wrt $I_aBC$ and $QBCM$ is cyclic then consider $Q'=M$ and applying the lemma yields the result. R HAS

I guess Barycentric coordinates work for this problem.

[asy][asy]size(7.5cm); pair A=(2,9),B=(0,0),C=(10,0),I,K,D,Dp,Ia,O,H,M,P; I=incenter(A,B,C);K=(B+C)/2;D=foot(I,B,C);Dp=2*K-D; O=extension(A,I,K,bisectorpoint(B,C));Ia=2*O-I; H=B+C-I;M=2*Dp-Ia;P=2*I-D; draw(circumcircle(A,B,C)^^circumcircle(Ia,B,C),green);draw(circle(I,abs(I-D)),orange); D(MP("A",A,N)--MP("B",B,S)--MP("D",D,S)--MP("K",K,S)--MP("D' ",Dp,SE)--MP("C",C,S)--A); D(A--MP("I",I,NE)--MP("O",O,SW)--MP("I_a",Ia,S)--MP("H",H,E)--MP("M",M,NE)--A,magenta); D(A--MP("P",P,E)--Dp,magenta);draw(P--I--D^^I--H--B--I--C--H--O^^B--Ia--C,magenta); dot(A^^B^^C^^I^^K^^D^^Dp^^O^^Ia^^H^^M^^P); [/asy][/asy] This is basically three well-known configurations tied together: Fact 5: In $\triangle ABC$, the midpoint of the arc $BC$ not containing $A$ is the center of $\odot (IBCI_a)$. Diameter of Incircle Lemma: In $\triangle ABC$, $A$, the antipode of the intouch point on $BC$ (wrt incircle) and the extouch point on $BC$ are collinear. Orthocenter-Parallelogram Lemma: (I made that name up) In $\triangle ABC$ with orthocenter $H$, and $A$'s antipode $A'$, $BHCA'$ is a parallelogram. In the main problem, let $O,H$ be the circumcenter and orthocenter of $I_aBC$ respectively. In $\triangle ABC$ let $D,D'$ be the in- and extouch point on $BC$ respectively, $K$ the midpoint of $BC$, $I$ the incenter. By Fact 5, $O$ is the midpoint of arc $BC$ and the midpoint of $II_a$. Also if $P$ is the antipode of $D$ in $\odot (I)$, we have $IK||PD'$, so by the Diameter of Incircle Lemma, $IK||AD$. Finally by the Orthocenter-Parallelogram Lemma applied to $I_aBC$, $H,K,I$ are collinear, so $IH||AD$. Now that all the pieces are together, we simply note that $$IH||AD\implies \frac{I_aI}{I_aA}=\frac{I_aH}{I_aD}\implies \frac{2I_aO}{I_aA}=\frac{I_aH}{\frac12 I_aM}\implies \frac{I_aO}{I_aA}=\frac{I_aH}{I_aM}\implies AM||OH$$And this gets the job done. $\blacksquare$

What program do you use to draw ? It's so nice.

That is Asymptote

The result follows from a result I found sometime ago. Let $ABC$ be a triangle. A triangle $DBC$ is completed which is similar to the orthic triangle $H_aH_bH_c$ and oriented similarly. Prove that the reflection of $AD$ in $BC$ is parallel to the Euler line of $ABC.$ Proof : Let the reflection of $ A$ in $BC$ be $A'.$ Let the orthocenter and circumcenter of $A'BC$ be $H,O$ resp. Then the circles $(BHC), (BOC)$ are conjugates under the $\sqrt{bc}$ inversion and flip in $A'BC$. So the reflection of $A$ in $BC$, i.e.$ A'$, is the image of $O$, and the point $D$ is the image of $H$. So we know that $OH$ and $AD$ are parallel, and this completes the proof.

Hmmm.. this problem was taken from Romania JBMO TST 2011(i found this on my RMC 2011 booklet)How this problem is so beautiful if it is taken from PREVIOUS TST?

Kezer wrote: This was the most difficult problem at this year's German olympiad, only $16 \%$ of the marks were given to all students. I find that kind of weird, I did not qualify but did find this problem rather easy. #toodifficulttoqualifyinbavaria How many rounds are there? Which one was it? I guess half-final? After reading a status of Kezer it must have been a final.

^There are four rounds and this is indeed the final round.

Kezer wrote: Let $I_a$ be the $A$-excenter of a scalene triangle $ABC$. And let $M$ be the point symmetric to $I_a$ about line $BC$. Prove that line $AM$ is parallel to the line through the circumcenter and the orthocenter of triangle $I_aCB$. Let $H$ be the orthocenter and $O$ be the circumcenter of triangle $I_aCB$. Note that $\angle BAC=180^{\circ}-2\angle BI_aC$ hence $B,O,C,A$ are concyclic. Also $\angle AOC=\angle ABC=180^{\circ}-2\angle I_aBC$ hence $A$ lies on $I_aO$. Invert about circle with center $I_a$ and radius $\sqrt{I_aB \cdot I_aC}$ followed by reflection in bisector of angle $BI_aC$. Then $H \mapsto A$ and $O \mapsto M$ hence $HO \parallel AM$ as desired.

Dear Mathlinkers, this problem is in relation with the proof concerning the incenter of the Fuhrmann's triangle. http://jl.ayme.pagesperso-orange.fr/Docs/L'orthocentre%20du%20triangle%20de%20Fuhrmann.pdf p. 11 Ayme J.-L., Revistaoim (Spain) 23 (2006) ; http://www.oei.es/historico/oim/revistaoim/numero23.htm which nead a simple adaptation... Sincerely Jean-Louis

$r$ - radius of excircle, $O$ - center of circle circumscribed on triangle $BCI_a$, $H$ - orthocenter of this triangle. Easy to proof, that $ABOC$ is a quadrilateral inscribed in a circle, and by definition of $O$ we have $BO=CO$, so points $A,O,I_a$ lie on one line in this order. By law of sines $I_aO=\frac{a}{2\cos(A/2)}$. Easy trigonometry (from definition) gives $AI_a=\frac{r}{\sin(A/2)}$. $I_aM=2r, I_aH=\sqrt{4I_aO^2-a^2}=a\tan(A/2)$. $\frac{I_aO}{AI_a}=\frac{a}{r}\cdot\tan(A/2)=\frac{I_aH}{I_aM}$ hence by Tales we have $OH||AM$

Overkill: Note that $A$ lies on the neuberg cubic of $\triangle I_aBC$ so that's it.

Fumiko wrote: Overkill: Note that $A$ lies on the neuberg cubic of $\triangle I_aBC$ Why ?

TinaSprout wrote: Fumiko wrote: Overkill: Note that $A$ lies on the neuberg cubic of $\triangle I_aBC$ Why ? It is well known that if $P$ lies on the neuberg cubic of $\triangle ABC$ then $A$ lies on the neuberg cubic of $\triangle BPC$ .

Looking at the problem from the perspective of triangle $I_ABC$ leads to an almost immediate solution: it's easy to show that $\sqrt{bc}$ inversion swaps $O$ and $M$, and $H$ and $A$. Then $I_AH \cdot I_AA = I_AO \cdot I_AM$, so we are done by similar triangles.

Here’s my solution [asy][asy]size(10cm); pair A=(2,9),B=(0,0),C=(10,0),I,K,D,Dp,Ia,O,H,M,P; I=incenter(A,B,C);K=(B+C)/2;D=foot(I,B,C);Dp=2*K-D; O=extension(A,I,K,bisectorpoint(B,C));Ia=2*O-I; H=B+C-I;M=2*Dp-Ia;P=2*I-D; draw(circumcircle(A,B,C),orange); D(MP("A",A,N)--MP("B",B,S)--MP("D",D,S)--MP("K",K,S)--MP("D' ",Dp,SE)--MP("C",C,S)--A); D(A--MP("I",I,NE)--MP("O",O,SW)--MP("I_a",Ia,S)--MP("H",H,E)--MP("M",M,NE)--A,heavygreen); D(A--MP("P",P,E)--Dp,heavygreen);draw(P--I--D^^I--H--B--I--C--H--O^^B--Ia--C,red); dot(A^^B^^C^^I^^K^^D^^Dp^^O^^Ia^^H^^M^^P); [/asy][/asy] It is well known that $IKH \parallel AD’$. So, by comparing ratios, we have $\frac{I_aI}{2I_aA} = \frac{I_aO}{I_aA} = \frac{I_aH}{2I_aD’} = \frac{I_aH}{I_aM}$. Thus $\boxed{AM \parallel OH}$. $\blacksquare$.

Denote $O,H$ respectively as circumcenter and orthocenter of the triangle $BCI_A$. Let $a,b,c$ be complex numbers lying on unit circle such that $a^2,b^2,c^2$ are vertices of the triangle and $-(ab+bc+ca)$ is the incenter. Then $$e_A=ab-bc+ca,\ o=-bc,\ h=b^2+c^2+ab+bc+ca.$$Last equation can be obtained without complex bashing by observing that $BICH$ is a parallelogram. Reflection formula yields $$m=b^2+c^2-b^2c^2\cdot \overline{e_A}=b^2+c^2+\frac{a-b-c}{a}\cdot bc.$$Finally $$a^2-m=a^2-b^2-c^2+\frac{-a+b+c}{a}\cdot bc=\frac{(a-b)(a-c)(a+b+c)}{a}$$and $$h-o=(b+c+a)(b+c).$$If $a+b+c=0$ we have $a^2-m=h-o$ implying $AM\parallel HO$. If $a+b+c=0\neq 0$ $$\frac{a^2-m}{h-o}=\frac{(a-b)(a-c)}{a(b+c)}=\overline{\left(\frac{a^2-m}{h-o}\right)}.\square$$#1710

Here are two solutions which I found just now by just gazing at the diagram in #21. Kezer wrote: Let $I_a$ be the $A$-excenter of a scalene triangle $ABC$. And let $M$ be the point symmetric to $I_a$ about line $BC$. Prove that line $AM$ is parallel to the line through the circumcenter and the orthocenter of triangle $I_aCB$. [asy][asy]size(10cm); pair A=(2,9),B=(0,0),C=(10,0),I,K,D,Dp,Ia,O,H,M,P; I=incenter(A,B,C);K=(B+C)/2;D=foot(I,B,C);Dp=2*K-D; O=extension(A,I,K,bisectorpoint(B,C));Ia=2*O-I; H=B+C-I;M=2*Dp-Ia;P=2*I-D; draw(circumcircle(A,B,C),orange); D(MP("A",A,N)--MP("B",B,S)--MP("D",D,S)--MP("K",K,S)--MP("D' ",Dp,SE)--MP("C",C,S)--A); D(A--MP("I",I,NE)--MP("O",O,SW)--MP("I_a",Ia,S)--MP("H",H,E)--MP("M",M,NE)--A,heavygreen); D(A--MP("P",P,E)--Dp,heavygreen);draw(P--I--D^^I--H--B--I--C--H--O^^B--Ia--C,red); dot(A^^B^^C^^I^^K^^D^^Dp^^O^^Ia^^H^^M^^P); [/asy][/asy] \begin{align*} \textbf{First Solution:-} \end{align*} $\measuredangle CI_aH=\measuredangle AI_aB$ and $\measuredangle HCI_a=\measuredangle BAI_a\implies\Delta I_aHV\stackrel{+}{\sim}\Delta I_aBA\implies I_aH\cdot I_aA=I_aB\cdot I_aC$. So, $\sqrt{I_aB\cdot I_aC}$ Inversion around $I_a$ followed by a reflection around the angle bisector of $\angle BI_aC$ swaps $\{O,M\}$ and $\{H,A\}$. So, $\overline{OH}\|\overline{AM}$. $\blacksquare$ \begin{align*} \textbf{Second Solution:-} \end{align*} Notice that $\{A,M\}$ are pairs of Isogonal Conjugates WRT $\Delta BI_aC$ also $A$ lies on the Neuberg Cubic of $\Delta BI_aC$. So, $\overline{AM}\|\overline{OH}$ as Neuberg Cubic has its pivot at $\infty_{\overline{OH}}$. $\blacksquare$

We use complex numbers and set $(ABC)$ as the unit circle. Let the coordinates of $A$, $B$ and $C$ be $a^2$, $b^2$ and $c^2$ so that the incenter $J$ is given by $j=-ab-bc-ca$. It is well-known that the excenter $I_a$ is given by $j_a=ab+ca-bc$ so that the coordinates of $O$, the circumcenter of $I_ABC$ is given by $o=\frac12(j+j_a)=bc$ as it is the midpoint of $II_A$ by the incenter excenter lemma. Then we calculate \[m=\frac{(b-c)(\overline{ab+ca-bc})+\overline{b}c-b\overline{c}}{\overline{b}-\overline{c}}=\frac{(b-c)(\frac{1}{ab}+\frac{1}{ca}-\frac{1}{bc})+\frac{c}{b}-\frac{b}{c}}{\frac{1}{b}-\frac{1}{c}}\]As $OH$ is the Euler-line of $I_ABC$, we know that its centroid $s=\frac{j_a+b+c}{3}=\frac{ab+ca-bc+b+c}{3}$ also lies on that line. Therefore, it suffices to show that the points $m-a^2$ and $s-o$ are collinear with the orgin, meaning the ratio \[\left(\frac{m-a^2}{s-o}\right)\]should be real. From here on it is just straight calculation.