Let $M,N$ and $P$ be the midpoints of sides $BC,CA$ and $AB$ respectively, of the acute triangle $ABC.$ Let $A',B'$ and $C'$ be the antipodes of $A,B$ and $C$ in the circumcircle of triangle $ABC.$ On the open segments $MA',NB'$ and $PC'$ we consider points $X,Y$ and $Z$ respectively such that \[\frac{MX}{XA'}=\frac{NY}{YB'}=\frac{PZ}{ZC'}.\] Prove that the lines $AX,BY,$ and $CZ$ are concurrent at some point $S.$ Prove that $OS<OG$ where $O$ is the circumcenter and $G$ is the centroid of triangle $ABC.$
Problem
Source: Romania Junior TST 2022
Tags: romania, Romanian TST, geometry
AndreiVila
16.05.2022 19:53
Very stupid complex bash: Set $(ABC)=S^1$. The condition is equivalent to $$x=\lambda a' + (1-\lambda) m =-\lambda a + \frac{1-\lambda}{2}b + \frac{1-\lambda}{2}c$$and similarly for $y,z$, for some fixed $\lambda\in (0,1)$. Consider the constant $l=\frac{1+\lambda}{3+\lambda}$. Observe that $$la+(1-l)x=g\frac{3-3\lambda}{3+\lambda}=s,$$therefore $\{S\}=AX\cap BY\cap CZ$. Also, since $\lambda\in (0,1)$, the constant $\frac{3-3\lambda}{3+\lambda}\in (0,1)$, thus $|s|<|g|$.
Anancibedih
01.08.2024 05:57
$\angle{BCB'}=\angle{ACA'}=90^\circ\Longrightarrow \angle{ACB'}=\angle{BCA'}\Longrightarrow ABA'B'\hspace{1mm}\text{trapezoid} \Longrightarrow A'B'||AB||MN$. Hence $NMA'B'$ is trapezoid and $XY$ is parallel to the edges by the ratio $\frac{|NY|}{|NB'|}=\frac{|MX|}{|MA'|}$. Thus $XY||AB$. Symmetrically, $XZ||BC$ and $ZY||AC$. In such a case, $ZC,AX,BY$ is concurrent clearly satisfied. Now let's take any rectangle $ABCD$. Let $T$ be the intersection point of the diagonals. Let's take a triangle $RBC$ with side $BC$ as an side and outside $ABCD$. $lemma:$ A line passing through $A$ and a line passing through $D$ intersect on $ZK$, it is parallel to $BC$ with the line segment passing through two points where these lines intersect with the lines $BZ$ and $CZ$ respectively. $proof:$ Let's delete $ZC$. Let's take a $P$ on $[KZ]$. Let $DP\cap BZ=S$. Lemma $\Longleftrightarrow$ $Z,F,C$ are collinear. with $F$ be the intersection of the line passing through $S$ and parallel to $BC$, with $AP$. Let $SF\cap PZ=E,BC\cap PZ=Q$. We have to prove that $\frac{|EF|}{|QC|}=\frac{|ZE|}{|ZQ|}$. The second ratio is equal to $\frac{|SE|}{|BQ|}$ from triangle $BQZ$. Let $AD\cap ZP=K$. It is clear that $|BQ|=|DK|$. So $\frac{|SE|}{|KD|}$. And this ratio is equal to $\frac{|EP|}{|PK|}$ from the butterfly similarity and $\frac{|EF|}{|AK|}$ from the butterfly similarity too, hence $\frac{|EF|}{| QC|}$ is equal to. $\blacksquare$ If this lemma is used for rectangle $C'BCB'$ and trapezoid $C'B'NP$ $\Longrightarrow$ $S,O,G$ are collinear and $Z$ and $Y$ is between of the points for $S=O,S=G$ $((B',C'),(B,N))$. So $|OS|<|OG|$. We're done.