Let $X$ be a variable point on the side $BC$ of a triangle $ABC$. Let $B'$ and $C'$ be points on the rays $[XB$ and $[XC$, respectively, satisfying $B'X=BC=C'X$. The line passing through $X$ and parallel to $AB'$ cuts the line $AC$ at $Y$ and the line passing through $X$ and parallel to $AC'$ cuts the line $AB$ at $Z$. Prove that all lines $YZ$ pass through a fixed point as $X$ varies on the line segment $BC$.
Problem
Source: Turkey EGMO TST 2016 P3
Tags: geometry
tastymath75025
28.05.2016 00:52
funny problem
Denote by $M$ the midpoint of segment $BC$, and let $\ell$ be the line through $A$ parallel to $BC$.
Denote $Y' = XY \cap \ell, Z' = XZ \cap \ell$. Then by Pappus's Theorem on lines $(BXC), (Z'AY')$, we have $Z = AB \cap XZ', Y = AC \cap XY', P = CZ' \cap BY'$ are collinear.
it suffices to show that point $P$ is fixed, Notice that $P$ is the center of negative homothety between segments $Y'Z', BC$. By parallelogram properties, $AZ' = XC' = BC, AY' = XB' = BC$. Thus $A$ is the midpoint of $Y'Z'$ so it corresponds to $M$ when the homothety is applied. In other words, $P\in AM$.
But the homothety has a fixed ratio $\frac{Y'Z'}{BC} = \frac{2BC}{BC}=2$. Thus point $P$ divides $AM$ in the ratio $2:1$ so $P$ is the centroid $G$ of $\triangle ABC$, which is fixed as desired.
badumtsss
30.01.2022 17:39
This problem can be easily solved using barycentric coordinates. Let $X= (0:m:n)$, then we get that $$B'=(0:2m+n:-m) \;\text{and}\;C'=(0:-n:2n+m)$$which means $$Y=(m:0:m+n) \; \text{and} \; Z=(n:m+n:0)$$. Then the equation of line $YZ$ is, $$mn(y+z-2x)+m^2(z-x)+n^2(y-x)=0$$where $P=(x:y:z)$ is a point on $XY$. One can easily deduce that the centroid $G=(1:1:1)$ satisfies the equation, so $G$ is our desired point.
Hakurei_Reimu
30.01.2022 19:46
Let's solve this problem with vector. Let $BX=aBC$ $\Rightarrow \overrightarrow{AZ}=\dfrac{1}{a+1}\overrightarrow{AB}$, $\overrightarrow{AY}=\dfrac{1}{2-a}\overrightarrow{AC}$. Notice that $\dfrac{a+1}{3}+\dfrac{2-a}{3}=1$. $\Rightarrow \dfrac{a+1}{3}\overrightarrow{AZ}+\dfrac{2-a}{3}\overrightarrow{AY}$ $=\dfrac13(\overrightarrow{AB}+\overrightarrow{AC})$ $=\overrightarrow{AG}$, $G$ is the centroid of $\triangle ABC$, which is fixed.
sevket12
14.08.2022 16:51
Lets solve this problem with moving point. We can easily check that $X \rightarrow Y$ is projective so $Y$ $\rightarrow$ $Z$ is projective and when $X=\infty_{BC}$ then $Y=Z=A$ so by steiner conic we can get that $YZ$ passes through fixed point
cursed_tangent1434
03.12.2024 12:39
Nice and different problem. Maybe slightly too easy for the Problem 3 position. Denote by $E$ and $F$ the points such that $AXC'E$ and $AXB'F$ are parallelograms respectively. Denote by $G$ the centroid of $\triangle ABC$. The following is the key claim. Claim : Points $E , X , Y$ and $F , X , Z$ as well as points $E , G , B$ and $F , G , C$ are collinear. Proof : We will show one set of collinearities, the other is entirely similar. Note that since $AXC'E$ is a parallelogram, $E$ lies on the line through $A$ parallel to $\overline{BC}$. Further, $XC'=AE$. But then, \[AE=XC'=BC\]so $AEBC$ is also a parallelogram. It is well known that diagonals of a parallelogram bisect each other, which implies that $\overline{BE}$ is in fact the $B-$median of $\triangle ABC$. Thus points $B , G$ and $E$ must indeed be collinear. Similarly, \[AE=XC'=XB'\]so $AEXB'$ is also a parallelogram. Thus, $E$ lies on the line through $X$ parallel to $\overline{AB'}$, which implies that points $E$ , $X$ and $Y$ are also collinear, as claimed. To finish, simply note that applying Pappus theorem on lines $FAE$ and $BXC$ we have that $Y = \overline{AC} \cap \overline{EX}$ , $Z = \overline{AB} \cap \overline{FX}$ and $G = \overline{CF} \cap \overline{BE}$. Thus, the line $\overline{YZ}$ indeed passes through a fixed point $G$ - the centroid of $\triangle ABC$, as point $X$ varies on the line segment $BC$.