I'll post just my idea, because i don't have time to write full proof. *It can be easily proven that H' lies on the circumcircle of ABC. *Now, AH'BC is a cyclic quadrilateral with perpendicular diagonals. *In fact there are 8 points that lie on a same circle with center S (the midpoint of OC'): -the 4 midpoints of the sides of the quadrilateral AH'BC and -the 4 feet of the perpendiculars from C' to the sides of the quadrilateral AH'BC. *The Brahmagupta theorem may also be helpful Cheers, Steff

Quote: Let $ABC$ be an acute triangle with orthocentre $H$, and let $M$ be the midpoint of $AC$. The point $C_1$ on $AB$ is such that $CC_1$ is an altitude of the triangle $ABC$. Let $H_1$ be the reflection of $H$ in $AB$. The orthogonal projections of $C_1$ onto the lines $AH_1$, $AC$ and $BC$ are $P$, $Q$ and $R$, respectively. Let $M_1$ be the point such that the circumcentre of triangle $PQR$ is the midpoint of the segment $MM_1$. Prove that $M_1$ lies on the segment $BH_1$. Let $N,$ $M'_1$ and $L$ respectively be the midpoints of $CB,$ $BH_1,$ $AH_1$. Then it is an obvious fact that $MNM'_1L$ is a rectangle. Therefore, it is inscribed in a circle with centered at $I$. Now, let $P'$ be the intersection of $C_1N$ with $AH_1$. Then it is followed that $\angle H_1C_1P=$ $\angle NC_1C=$ $\angle NCC_1\equiv$ $\angle BCH_1$ $=\angle C_1AH_1$, which implies $NC_1\perp$ $AH_1$. As the result, $P'$ must $\equiv$ $P$ $\Longrightarrow$ $P\in (I)$. The same argument holds for $Q$ and $R$. Thus, $(PQR)\equiv$ $(I)$ $\Longrightarrow$ $M'_1\equiv M_1$. The result of the problem is lead as follow. Our proof is completed then. $\square$

Does anybody know the official solution? )

Here's my solution. Recall that a podar triangle corresponding to a given point $P$ inside a triangle $ABC$ is the triangle formed by the feet of the orthogonal projections of $P$ onto the sides of the triangle $ABC$. It can also be easily proved that the projections of two isogonal points in triangle $ABC$ are concycic, which means the podar triangles corresponding to isogonal points have the same circumcircle and its center is the midpoint of the segment determined by those two points. Returning to our problem, the isogonal of $C_1$ with respect to the triangle formed by intersecting $AH_1$ with $BC$ is exactly $O$, the circumcenter of the triangle $ABC$. Therefore, the circle $PQR$ passes through $M$, thus we would have to prove that the point $M_1$ is also on the circle and $MM_1$ is a diameter for $(PQR)$. This is easy, since the circumcenter of $(PQR)$ is the midpoint of $OC_1$ which is the midpoint of $MM_1$ (where $M_1$ is the midpoint of $BH_1$ ).

http://mathproblems123.wordpress.com/2010/05/23/bmo-2010-geometry-problem/

This was nice except that it is pretty obvious if one has done the famous USA TST 2000 kite problem. Here is a brief sketch of my idea:- Let $S$ be the last perpendicular we can drop from $C_1$. The it is evident that $PQRS$ is a cyclic quad and then by our Lemma we have that the midpoints $M,M_1$ of $AC,BH_1$ are on this circle and that the same holds for the other midpoints. But then $MM_1$ is a diameter and we conclude.

My Solution. let $L$ be the midpoint of $BH_1$. since $H_1$ is the reflection of $H$ in $AB$ we have, $\measuredangle BH_1C=\measuredangle H_1HB=90^{o}-\measuredangle HBA=\measuredangle BAC$ therefore the quadrilateral $ACBH_1$ is Cyclic. and, $\measuredangle H_1C_1L=90^o-\frac{1}{2}\measuredangle HBH_1=H_1HB$, hence, $LC_1$ and $BH$ are parallel, and thus $LC_1$ is perpendicular to $AC$ and we know that $C_1Q$ is perpendicular to $AC$. hence, the points $L,C_1,Q$ are collinear. this mean that $Q$ lie on the Circle $(\Omega)$ (with diametre $ML$). Now, we have, $\overrightarrow{LP} \cdot \overrightarrow{MP}=(\overrightarrow{LC_1}+\overrightarrow{C_1P})(\overrightarrow{MA}+\overrightarrow{AP})=\overrightarrow{C_1P}\cdot\overrightarrow{MA} - \overrightarrow{LC_1} \cdot \overrightarrow{PA} = \sin\measuredangle H_1AC \cdot (MA \cdot C_1P-LC_1 \cdot PA)=X$ and we have, $\frac{LC_1}{C_1P}=\frac{\frac{1}{2}H_1B}{H_1C_1 \cdot \sin \beta}=\frac{1}{2}\frac{\frac{1}{\cos \measuredangle BAC}}{\sin \beta}=\frac{1}{2}\frac{AC}{AC_1 \cdot \sin(90^{o}-\measuredangle H_1AB)}=\frac{MA}{PA}.$ Or, $X=0.$ Therefore $P$ lie on the Circle $(\Omega).$ analogously, we prove that $R$ lie on the Circle $(\Omega)$ (By Symmetric!), and we deduce that $L=M_1$, and the result will be Clear.

Attachments:

$\phantom{I thought it would be both interesting and useful to post the grading scheme }$

Attachments:
BMO_2010_Grading_scheme.pdf (250kb)

It's quite trivial that $AH'BC$ is a cyclic quadrilateral. Let $J,K,L$ be the midpoint of segments $AH',H'B,BC$. $JKLM $ is the varignon quadrilateral of $AH'BC$. So $JKLM$ is a parallelogram. Also, as $AB \perp CH'$, $JKLM$ is a rectangle. Let us now drop perpendicular from $J,K,L,M $ to the opposite sides are $J',K',L',M'$. All these perpendiculars go through $C'$. Hence we get that $ J,K,L,M,J',K',L',M'$ is cyclic and we are done with the problem.

Let $AH'$ meet $BC$ at $B'$. See that $PQR$ is the pedal triangle of $C_1$ WRT $AB'C$. Now, observe that $O$, the circumcenter of $ABC$ is the isogonal conjugate of $C_1$ WRT $AB'C$. Let $M'$ be the mid point of $BH_1$. Now, since $AH_1BC$ is a cyclic quadrilateral with perpendicular diagonals, $R-C_1-M'$. So, $OM\parallel CM'$. Also, $C_1M' = \frac{H_1B}2 = \frac{HB}2 = OM$. So, $OMM'C_1$ is a parallelogram, which means $MM', OC_1$ share the same midpoint. Also, the circumcenter of $PQR$ is the midpoint of $OC_1$. So, $M_1=M'$ which implies $M_1\in \overline{H_1B}$. $\blacksquare$.

Special case of shortlist 2011 G3

By chasing angles: $H_1 \in (ABC)$ Notice that: $(PQR)$ is eight point circle. $\implies M \in (PQR)$ $M_1$ it is diametrically opposite to $M$ w.r.t $(PQR)$. Let $M'$ be the midpoint of $BH_1$ by eight point circle $MM'$ is diameter of $(PQR)$ i.e $M'$ it is diametrically opposite to $M$ w.r,t $(PQR)$ $\implies M'=M_1$.$\blacksquare$

Note $O , O_1$ be circumcenter of $(ABC)$ and $(PQR)$ ,respectively. Just an easy angle-chasing yields $H_1 \in (ABC)$ Hence $(PQR)$ is eight point circle $ \implies O_1$ is midpoint of $(C_1O)$ Let us continue with easy complex bash.Set $(ABC)$ as unit circle.Then $$h=a+b+c \wedge \bar{h} = \frac{ab+bc+ca}{abc}$$$$h_1=\frac{(a-b)\bar{h}+\bar{a}b-a\bar{b}}{\bar{a}-\bar{b}}=-\frac{ab}{c} \wedge \bar{h_1}=-\frac{c}{ab}$$$$m_1=2o_1-m=c_1-\frac{a+c}{2}=\frac{b(c-a)}{2c} \wedge \bar{m_1}=\frac{a-c}{2ab}$$It remains to show that $$ B,M_1,H_1 \text{ are collinear } \iff \begin {vmatrix} b & \frac{1}{b} & 1 \\ \frac{b(c-a)}{2c} & \frac{a-c}{2ab} & 1 \\ -\frac{ab}{c} & -\frac{c}{ab} & 1 \end {vmatrix} =0 \iff \begin {vmatrix} 1 & 1 & 1 \\ \frac{c-a}{2c} & \frac{a-c}{2a} & 1 \\ -\frac{a}{c} & -\frac{c}{a} & 1 \end {vmatrix} =0 \iff 0=0$$ so we are done

Really easy with eight-point circle properties: Obviously $AH_1BC$ is cyclic and notice that $(PQR)$ is its eight-point circle, so $M \in (PQR)$ which means that $M_1$ is diametrically opposite to $M$, but this implies that $M_1$ is the midpoint of $H_1B$ (because of the eight-point circle) and we're done.