Romania Junior TST 2022 Day 2 P1

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Source: Romania Junior TST 2022

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oVlad

#1 h May 16, 2022, 4:43 PM • 2 Y

Y by ImSh95, Mango247

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 $ABC.$

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#2 h May 16, 2022, 4:53 PM • 1 Y

Y by ImSh95

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 $|s|<|g|$ .

This post has been edited 1 time. Last edited by AndreiVila, May 16, 2022, 4:54 PM

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#3 h Aug 1, 2024, 2:57 AM

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$\angle{BCB'}=\angle{ACA'}=90^\circ\Longrightarrow \angle{ACB'}=\angle{BCA'}\Longrightarrow ABA'B'\hspace{1mm}\text{trapezoid} \Longrightarrow A'B'||AB||MN$ . 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.

https://i.imgur.com/hwgsAS8.jpeg

This post has been edited 6 times. Last edited by Anancibedih, Sep 17, 2024, 6:17 PM

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