Constant points $B$ and $C$ lie on the circle $\omega$. The point middle of $BC$ is named $M$ by us. Assume that $A$ is a variable point on the $\omega$ and $H$ is the orthocenter of the triangle $ABC$. From the point $H$ we drop a perpendicular line to $MH$ to intersect the lines $AB$ and $AC$ at $X$ and $Y$ respectively. Prove that with the movement of $A$ on the $\omega$, the orthocenter of the triangle $AXY$ also moves on a circle.
Problem
Source: Iran MO Third Round G2
Tags: geometry, dilation, homothety, orthocenter
09.09.2022 16:52
Let $A'$ be the anti-pod of $A$ wrt $(ABC)$ and consider a point $F$ st $AHMF$ is parallelogram , which is a fix point. Now since $HBCA'$ is parallelogram and $A'XBH$ is cyclic , we have $\angle HXA' = \angle HBA' = \angle BAC$ , and similar for $Y,C$ which means $HM$ is the perpendicular bisector of $XY$. Now one can easily deduce that $XY = HM . C_1$ where $C_1$ is constant since $|sin \angle BAC|$ and $|cos \angle BAC|$ are constant. But in every triangle $\triangle ABC$ with orthocenter $H$ , we have $AH = 2R.cos \angle A$ So $AH'=HM.C_2$where $H'$ is the orthocenter of $\triangle AXY$ and $C_2$ is constant(this is followed by the fact that: $2R_{(AXY)}= \frac {XY}{sin \angle BAC}$) So since $F \in AH'$ and $AF=HM$ , we'd have $AF=FH'.C_3$ where $C_3$ is constant and we're done since $F$ would be a homothety center of $(BAC)$ and a new circle which is the desired circle of the problem.
15.09.2022 17:22
25.09.2022 13:20
16.11.2022 03:41
We prove that the ortocenter of $\triangle AXY$, $H'$ is the image of a fixed point in a homothety centered at $A$, solving the problem. First, see that $AH$ is fixed since $$AH=\lvert h-a\rvert=\lvert b+c\rvert,$$which is constant. Now let $D$ be the intersection between $AH'$ and the perpendicular bisector of $BC$. Since $$AH\parallel DM\text{ and }AD\perp XY\perp HM,$$we have $AHMD$ a parallelogram, and thus $DM=AH$, which is fixed, and thus $D$ is a fixed point. Now let $A'$ be the antipode of $A$ with respect to $(ABC)$. Since $H, M$ and $A'$ are collinear, we angle chase: $$\angle A'XY=\angle A'BH=\angle A'CH=\angle A'YX,$$and therefore $A'XY$ is isosceles with $H$ being $A'$'s projection onto $XY$, and thus its midpoint. Futhermore, since $\angle A'BH=\angle A$, the angles of $\triangle A'XY$ are fixed and so are the angles of $\triangle MXY$, since $M$ is the midpoint of $A'H$. Now, if $O$ is $(AXY)$'s center, we get that the angles of $\triangle OXY$ are fixed and $$\frac{AD}{AH'}=\text{constant}\iff\frac{HM}{OH}=\text{constant},$$which is true since $O,H$ and $M$ are collinear (since $H$ is the midpoint of $XY$) and the angles of triangles $HMX$ and $HXO$ are fixed. Thus, there is a homothety with fixed ratio centered at $A$ that takes $D\to H'$, and since $A$ moves around a circle so do $H'.$ $\blacksquare$
16.12.2022 01:12
this problem reminded me of that other one that i already posted a while ago