For reasons which are easy to explain, $OS\to OR$ is a projective transformation on the pencil through $O$, so it's determined by three particular images. In order to do this, just take $M=B$, then $N=C$, and then $MN\|BC$ (in the latter case the conclusion holds because $SR$ is the perpendicular bisector of $OD$, where $D$ is the midpt of $BC$). Since the conclusion holds for these three, it always holds. However, I'm getting kind of sick of projective geometry, so I'll try (I can't promise anything ) to find another clean, nice proof.

oh grobber I always have trouble with your projective solution would u mind explaining better and more.

I'll try, but I'm not that good at explaining stuff . Since $MN$ passes through a fixed point, the map $N\to M$ is projective (it clearly conserves cross-ratio). This means that $BN\to CM$ is also projective (from the pencil through $C$ to that through $B$). In turn, this means that the map $S\to R$ is projective (from the medial line of $ABC$ corresponding to $CA$ to that corresponding to $AB$). Of course, this implies what I said in the first sentence of my previous post: $OS\to OR$ is projective (a map from the pencil through $O$ to itself).

sam-n wrote: Let $ABC$ be a triangle, and $O$ the center of its circumcircle. Let a line through the point $O$ intersect the lines $AB$ and $AC$ at the points $M$ and $N$, respectively. Denote by $S$ and $R$ the midpoints of the segments $BN$ and $CM$, respectively. Prove that $\measuredangle ROS=\measuredangle BAC$. Here is my solution: We will work with directed angles modulo 180° and we will prove that < ROS = < CAB. Let B' and C' be the points diametrically opposite to the points B and C on the circumcircle of triangle ABC. We will denote the point of intersection of two lines g and h as $g\cap h$. Then, $BB^{\prime }\cap C^{\prime }C=O$, since the segments BB' and C'C are diameters of the circumcircle of triangle ABC and therefore meet at the center O of this circumcircle. Let the line B'N meet the circumcircle of triangle ABC at the point $\Phi$. Then, $B^{\prime }\Phi \cap CA=N$. Now, by the Pascal theorem, applied to the cyclic hexagon $BB^{\prime }\Phi C^{\prime }CA$, the points of intersection $BB^{\prime }\cap C^{\prime }C$, $B^{\prime }\Phi \cap CA$ and $\Phi C^{\prime }\cap AB$ are collinear. Since $BB^{\prime }\cap C^{\prime }C=O$ and $B^{\prime }\Phi \cap CA=N$, we thus see that the points O, N and $\Phi C^{\prime }\cap AB$ are collinear. In other words, the point $\Phi C^{\prime }\cap AB$ lies on the line ON. But since this point also lies on the line AB, it must be the point of intersection of the lines AB and ON. In other words, the point $\Phi C^{\prime }\cap AB$ must coincide with the point M. This yields that the point M lies on the line $\Phi C^{\prime }$. Equivalently, the line C'M passes through the point $\Phi$. Since the segment BB' is a diameter of the circumcircle of triangle ABC, while the point $\Phi$ lies on this circumcircle, we have $\measuredangle B\Phi B^{\prime }=90^{\circ }$. But since the line B'N passes through the point $\Phi$, this can be rewritten as $\measuredangle B\Phi N=90^{\circ }$. Thus, the point $\Phi$ lies on the circle with diameter BN. Hence, the point $\Phi$ is a common point of the circumcircle of triangle ABC and the circle with diameter BN. The other common point is, of course, the point B. Now, the line joining the two common points of two circles is always perpendicular to the line joining the centers of these circles. What are the centers of the circumcircle and of the circle with diameter BN ? Well, the center of the circumcircle is the point O, and the center of the circle with diameter BN is the midpoint S of the segment BN. Hence, with $\Phi$ and B being the common points of these two circles, we see that the line $\Phi B$ is perpendicular to the line OS. Thus, $\measuredangle \left( \Phi B;\;OS\right) =90^{\circ }$. Similarly, the line $\Phi C$ is perpendicular to the line OR, and we can conclude $\measuredangle \left( OR;\;\Phi C\right) =90^{\circ }$. Altogether, $\measuredangle ROS=\measuredangle \left( OR;\;OS\right)$ $=\measuredangle \left( OR;\;\Phi C\right)+\measuredangle \left( \Phi C;\;\Phi B\right)+\measuredangle \left( \Phi B;\;OS\right)$ $=90^{\circ }+\measuredangle \left( \Phi C;\;\Phi B\right)+90^{\circ }=180^{\circ }+\measuredangle \left( \Phi C;\;\Phi B\right)$. Since our angles are directed angles modulo 180°, we can write this as $\measuredangle ROS=\measuredangle \left( \Phi C;\;\Phi B\right)$. Now, $\measuredangle \left( \Phi C;\;\Phi B\right) =\measuredangle C\Phi B$. Since the point $\Phi$ lies on the circumcircle of triangle ABC, we have $\measuredangle C\Phi B=\measuredangle CAB$. Therefore, < ROS = < CAB, and the problem is solved. Darij Grinberg

I solved it with complex number ,realy short solution and always IQ free : assume that the circumcircle has radius 1, we denote A,B,C by a,b,c and also let MN intersect circumcircle at T and T' we denote X by x ( x is complex number) so T' is -t and u can prove that if we have four points on unit circle ,$t_{1},t_{2},t_{3},t_{4}$ then the intersection point of $t_{1}t_{2}$ and $t_{3}t_{4}$ is $\frac{\overline{t_{1}}+\overline{t_{2}}-\overline{t_{3}}-\overline{t_{4}}}{\overline{t_{1}t_{2}}-\overline{t_{3}t_{4}}}$ in this way u can find M and N then r=(m+c)/2 and s=(b+n)/2 now u should prove that $arg (\frac{r}{s})=arg (\frac{b-a}{c-a})$ (excuse me for bad explanation)

darij grinberg wrote: Let the line B'N meet the circumcircle of triangle ABC at the point $\Phi$. Then, $B^{\prime }\Phi \cap CA=N$. Now, by the Pascal theorem, applied to the cyclic hexagon $BB^{\prime }\Phi C^{\prime }CA$, the points of intersection $BB^{\prime }\cap C^{\prime }C$, $B^{\prime }\Phi \cap CA$ and $\Phi C^{\prime }\cap AB$ are collinear. Since $BB^{\prime }\cap C^{\prime }C=O$ and $B^{\prime }\Phi \cap CA=N$, we thus see that the points O, N and $\Phi C^{\prime }\cap AB$ are collinear. In other words, the point $\Phi C^{\prime }\cap AB$ lies on the line ON. But since this point also lies on the line AB, it must be the point of intersection of the lines AB and ON. In other words, the point $\Phi C^{\prime }\cap AB$ must coincide with the point M. This yields that the point M lies on the line $\Phi C^{\prime }$. You have just opened my eyes! I couldn't prove this fact (actually, I never used Pascal's theorem before). But your further arguments are too long. We may observe that OS is centerline (I hope it is right translation) in BB'N and OR is centerline in CC'M --> OS||FB' and OR||FC' , so <ROS = <B'FC'= <CAB (though this conclusion requires arguments why <ROS <> 180-<CAB). Great work, Darij!

Oh well, it was quite straightforward for me . I am used to apply Pascal, Brianchon, Desargues, Pappos and similar facts every day, even if the problem has nothing to do with projective geometry (at first sight!). Myth wrote: But your further arguments are too long. We may observe that OS is centerline (I hope it is right translation) in BB'N Indeed... I was stupid. I also don't know how is "centerline" in English; I usually say "midparallel" but I think it is hardly correct. Myth wrote: and OR is centerline in CC'M --> OS||FB' and OR||FC' , so <ROS = <B'FC'= <CAB (though this conclusion requires arguments why <ROS <> 180-<CAB). Actually, < ROS = 180° - < CAB happens if the triangle ABC is obtuse-angled; however, in this case, one of the points M and N lies on the extension of the corresponding side of triangle ABC, so one could argue whether this case is allowed... But in my opinion, such case distinctions are ugly, and when one uses directed angles modulo 180° one needn't care for them. Darij

Let $K$, $L$ be midpoints of $AB$, $AC$. We have to prove $<KOR+<LOS=180$. Let the spiral similarity of center $O$ map$ L$ to $N$ and $S$ to $X_1$. Analogously define $X_2$ and let $MX_2$ and $NX_1$ meet at $X$. I claim $X=X_1=X_2$ which solves the problem since then $<LOS+<KOR=<NOX+<MOX=180$. Of course $NX_1=\frac{LS \cdot ON}{OL}=\frac{AM}{2 \sin {ANM}}=R_{AMN}$. Analogously we deduce $MX_2=R_{AMN}$ Also $<ONX=<OMX=<OLS=<OKR=90-A$. Now allpying sine theorem in the isosceles triangle $MNX$ we get $MX=MX_1$ and hence $X=X_1=X_2$ QED

darij grinberg wrote: Here is my solution: .. Let B' and C' be the points diametrically opposite to the points B and C on the circumcircle of triangle ABC. ... Nice problem, good solution and very good idea.

Suppose $CC'$ and $BB'$ are diameters of circumcircle of $ABC$ we have: $OS||B'N$ and $OR||C'M$ and from pascal theorem $B'N$ and $C'M$ intersct point $T$ on circle. Now $<B'TC'=<A$ So $<ROS=<A$

There is my solution: Let 0, a, b, c be the respective complex coordinates of the points O, A, B, C. (with |a|=|b|=|c|=1 Assume WLOG that the equation of the line MN is $z-\bar{z}=0$ How $M%Error. "inAB" is a bad command. $ and $M%Error. "inMN" is a bad command. $, so $m+\bar{m}ab=a+b$ and $m-\bar{m}=0$ ence: $m=\frac{a+b}{ab+1}$ and for analogie $n=\frac{a+c}{ac+1}$ Furthermore, we can find the complex coordinates of R and S: $2r=\frac{a+b+c+abc}{1+ab}$ and $2s=\frac{a+b+c+abc}{1+ac}$ So, <ROS=<BAC iff $arg(\frac{r}{s})=arg(\frac{b-a}{c-a}) \iff z=\frac{1+ab}{1+ac}\frac{c-a}{b-a} \in R_{+} \iff z=\bar{z}$ And by a simple calculus we may find $z=\bar{z}$

Very nice problem and very nice Darij's solution (with Pascal's theorem) ! Where found I the original solution from math-competition ?

http://www.mathlinks.ro/Forum/viewtopic.php?t=64807 will help

Are you a Chinese?My solution has been posted here,plz click this website: http://gjmath.cn/bbs/dispbbs.asp?boardID=16&ID=5345&page=1

See my solution to this problem on my Youtube channel here: https://www.youtube.com/watch?v=osoL5xXOUnU

We will use the following point labeling: Quote: Let $ABC$ be a triangle, and $O$ the center of its circumcircle. Let a line through the point $O$ intersect the lines $AB$ and $AC$ at the points $X$ and $Y$, respectively. Denote by $M$ and $N$ the midpoints of the segments $BY$ and $CX$, respectively. Prove that $\measuredangle MON=\measuredangle BAC$. We will use complex numbers with $(ABC)$ as the unit circle and $\ell$ as the real line (rotate appropriately). Set $O=0, A=a,B=b,C=c$ and let $\ell$ intersect $(ABC)$ at points $P=-1,Q=1$. By Lemma 8 in https://web.evanchen.cc/handouts/cmplx/en-cmplx.pdf, $X=\frac{a+b}{ab+1},Y=\frac{a+c}{ac+1}$ so we clearly have $M=m=\frac{b+y}2=\frac{a+b+c+abc}{2(ac+1)},N=n=\frac{c+x}2=\frac{a+b+c+abc}{2(ab+1)}$. To show that $\measuredangle BAC=\measuredangle MON$, we need to show $\text{arg~}\frac{a-b}{a-c}=\text{arg~}\frac{o-m}{o-n}$ or equivilently $\frac{o-m}{o-n}\div \frac{a-b}{a-c} \in\mathbb{R}$. Indeed, \begin{align*} \frac{o-m}{o-n}\div \frac{a-b}{a-c}&= \frac{0-\frac{a+b+c+abc}{2(ac+1)}}{0-\frac{a+b+c+abc}{2(ab+1)}}\div \frac{a-b}{a-c}\\ &= \frac{ab+1}{ac+1}\cdot \frac{a-c}{a-b}\\ &=\frac{\left(\frac{1}{ab}+\frac 11\right)\left( \frac 1a-\frac 1c\right)}{\left(\frac{1}{ac}+\frac 11\right)\left( \frac 1a-\frac 1b\right)} \end{align*} where the last equality follows from $$\frac{\left(\frac{1}{ab}+\frac 11\right)\left( \frac 1a-\frac 1c\right)}{\left(\frac{1}{ac}+\frac 11\right)\left( \frac 1a-\frac 1b\right)}=\frac{\left(\frac{ab+1}{ab}\right)\left( -\frac{a-c}{ac} \right)}{\left(\frac{ac+1}{ac}\right)\left( -\frac{a-b}{ab} \right)} =\frac{ab+1}{ac+1}\cdot \frac{a-c}{a-b}$$ so $\frac{o-m}{o-n}\div \frac{a-b}{a-c} \in\mathbb{R}\iff \measuredangle BAC=\measuredangle MON$ as desired.

We had an age old solution via the method of moving points at the start of this post, so here's another! This was my first ever solution using the Method of Moving Points so YAY. We denote the line through $O$ by $\ell$ and its intersections with lines $\overline{AB}$ and $\overline{AC}$ by $X$ and $Y$. Let $m_B$ and $m_C$ denote the $B-$midline and $C-$midline of $\triangle ABC$ respectively. We will use moving points. The points $A$ , $B$ , $C$ and hence $O$ will be fixed. We let $M$ vary along $m_C$. First note that the map, $M\mapsto X \mapsto Y \mapsto N$ is projective since it is induced by the perspectivities $m_C \overset{C}{\to}\overline{AB}\overset{O}{\to} \overline{AC} \overset{B}{\to}m_B$. Further, the map $N\to M$ from $m_C$ to $m_B$ (rotation at $O$ by the fixed angle $\angle BAC$) is also projective. Now, it suffices to check that these maps coincide for three specific cases. $\bullet \hspace{2.5pt}$ If $M=m_C \cap \overline{BC}$ (midpoint of $BC$), then $X=B$ , $Y = \overline{BO} \cap \overline{AC}$, and $N= \overline{BO} \cap m_B$. Thus it is easy to see that, $\measuredangle (OM,ON) = \measuredangle BAC$. $\bullet \hspace{2.5pt}$ If $M=m_C \cap \overline{OC}$, then $X = \overline{OC} \cap \overline{BC}$, $Y=C$, and $N= \overline{m_B} \cap \overline{BC}$ (midpoint of $BC$). Thus it is easy to see that, $\measuredangle (OM,ON) = \measuredangle BAC$. $\bullet \hspace{2.5pt}$ If $M=m_C \cap \overline{AC}$ (midpoint of $AC$), then $X=A$ , $Y=A$ and $N=m_B \cap \overline{AB}$ (midpoint of $AB$). Then $AMON$ must be cyclic and it is easy to see that, $\measuredangle (OM,ON) = \measuredangle MAN = \measuredangle BAC$. Having checked it for three cases, the problem is solved.

A silly untethered MMP approach: It is well known that the angle $\angle BAC$ is equal to \[\frac{1}{2i}\log(AB,AC;AI,AJ)\]where $I,J$ are the two circle points (outline of proof: take conic/circle $(ABCIJ)$; for any point $P$ on this conic/circle, we have $(PB,PC;PI,PJ)=(AB,AC;AI,AJ)$ and $\angle BPC = \angle BAC$ by inscribed angle theorem). Note that $I$ and $J$ are fixed and complex (never overlaps with $B,C$ if they're always defined in the real plane). Thus we want to prove that the cross ratio of $(OR, OS; OI, OJ)$ is the same as the cross ratio of $(AB, AC; AI, AJ)$. Note that $B$ and $C$ are fixed, so $(AB,AC; AI, AJ)$ has degree $0$. It is well known that the cross ratio of four elements in a pencil has degree equal to the sum of the four individual elements minus the number of times that three of them overlap plus 2x the number of times that all four overlap. $R$ and $S$ move with degree $1$ (they never overlap) and $O$ is fixed. Thus $(OR, OS; OI, OJ)$ has degree $2$, so we only need to check three cases. This is trivial by letting the line pass through $B,C,A$.