The triangle $ABC$ is given. On the arc $BC$ of its circumscribed circle, which does not contain point $A$, the variable point $X$ is selected, and on the rays $XB$ and $XC$, the variable points $Y$ and $Z$, respectively, so that $XA = XY = XZ$. Prove that the line $YZ$ passes through a fixed point. Proposed by A. Kuznetsov
Problem
Source: Tuymaada 2024 Senior league P6
Tags: geometry
09.07.2024 22:17
11.07.2024 23:46
Identical to Australia MO 2024 P2 https://artofproblemsolving.com/community/c6h3262920p29992384
12.07.2024 01:16
The problem is true even if the length condition is replaced with $AX=AY=AZ$?! With the misread: The fixed point is the orthocenter $H$ of $\triangle ABC$. Let $M$ and $N$ be the midpoints of $\overline{XY}$ and $\overline{XZ}$, respectively. Since $(AMXN)$ is cyclic, it follows from spiral similarity that $\triangle AMN \sim \triangle ABC$. Also, since line $MN$ is the Simson line of $A$ with respect to $\triangle BXC$, it passes through the foot $D$ from $A$ to line $BC$. Letting $R$ be the reflection of $H$ over $D$, we have $\angle RXN = \angle RAC = \angle MNX$, where the last step follows from our similarity. So, $\overline{RX} \parallel \overline{MN}$ which means that line $MN$ bisects $\overline{XH}$. Now, since a homothety of scale factor $2$ centered at $X$ sends line $MN$ to line $YZ$, it follows that $H$ always lies on line $YZ$.
01.08.2024 19:53
Note that $\angle ABY + \angle ACZ = 180$ so $ABY$ and $ACZ$ are concurrent on $YZ$ at point $P$. Note that $\angle PBA = \angle PYA = \angle ZYA = \angle \frac{ZXA}{2} = \angle \frac{B}{2}$ so $PB$ is angle bisector of $B$. similarly $PC$ is angle bisector of $C$ so $P$ is incenter of $ABC$ which is a fixed point.
01.08.2024 20:09
Mahdi_Mashayekhi wrote: Note that $\angle ABY + \angle ACZ = 180$ so $ABY$ and $ACZ$ are concurrent on $YZ$ at point $P$. Note that $\angle PBA = \angle PYA = \angle ZYA = \angle \frac{ZXA}{2} = \angle \frac{B}{2}$ so $PB$ is angle bisector of $B$. similarly $PC$ is angle bisector of $C$ so $P$ is incenter of $ABC$ which is a fixed point. HUH......................................................