This problem was proposed to 2022 IMO and sadly did not make it into ISL I'm really proud of the problem though! In case you find it too easy, here are something else that you can prove: Let $ABC$ be a scalene triangle with circumcenter $O$ and orthocenter $H$. Let $AYZ$ be another triangle sharing the vertex $A$ such that its circumcenter is $H$ and its orthocenter is $O$. Let the circumcircle of $AHZ$ intersect $AY$ again at $X$, and let the intersections of $XZ$ with $AB, AC$ be $C', B'$, respectively. Show that if $Z$ is on $BC$, then the circumcircles of $ABC, AB'C'$ and $AYZ$ have two common points. More properties in this diagram: if $Y'$ is on $AZ$ with $OY'$ parallel to $HY$, then $HY'M$ are collinear. Lastly, consider pairs $(A,Z),(O,H),(B,B'),(C,C'),(Y,Y').$ If $(P_1,P_2), (Q_1,Q_2)$ are two pairs among those, then $\Delta MP_1Q_1\sim \Delta MQ_2P_2$ and $\Delta MP_1Q_2\sim \Delta MQ_1P_2$.

These are some nice configurations

USJL wrote: Let $ABC$ be a scalene triangle with circumcenter $O$ and orthocenter $H$. Let $AYZ$ be another triangle sharing the vertex $A$ such that its circumcenter is $H$ and its orthocenter is $O$. Show that if $Z$ is on $BC$, then $A,H,O,Y$ are concyclic. Proposed by usjl Hello, would you be kind enough to provide a diagram please?, I ran some calculations to construct Z,Y and it seems like they shouldn't exist, don't know where I messed up but would like to try the problem. Hope you're well, and thanks in advance.

IMD2 wrote: USJL wrote: Let $ABC$ be a scalene triangle with circumcenter $O$ and orthocenter $H$. Let $AYZ$ be another triangle sharing the vertex $A$ such that its circumcenter is $H$ and its orthocenter is $O$. Show that if $Z$ is on $BC$, then $A,H,O,Y$ are concyclic. Proposed by usjl Hello, would you be kind enough to provide a diagram please?, I ran some calculations to construct Z,Y and it seems like they shouldn't exist, don't know where I messed up but would like to try the problem. Hope you're well, and thanks in advance. Hi, I am not sure how I should provide a diagram without spoiling the entire problem for you. Let me assure you that those configurations do exist though (and there are more than one possible configurations!).

@IMD2 Define the point where AH and BC intersect "K". It is possible if we draw the triangle ABC and H such that |AH|>|HK|. That way, we can find a point Z on BC such that |AH|=|HZ| for sure. Then choosing Y to be the orthocenter of the triangle AOZ. Since I'm new to the community I cant post pictures.

We find ∠AHY = 2∠AZY and ∠AOY = 180°-∠AZY Im guessing we should show ∠AZY=60° but I have no idea how Then we can say A,H,O,Y is cyclic

Define M as the midpoint of |BC| and L as the intersection of AH and BC. Let B, C, Z collinear respectively. Define N as the intersection of AY and BC Define T as the foot of the perpendicular H to AY A, H, O, Y is cyclic if and only if ∠AHY=∠AOY We find ∠AHY = 2∠AZY and ∠AOY = 180°-∠AZY so its equivalent to show that ∠AZY=60° ZO is perpendicular to AN and AL is perpendicular to ZN. So, ∠OZM = ∠NZL = ∠NAL = ∠YAH = ∠TAH = ∠TYH = ∠AYH = (180°- ∠AHY)/2 = 90°-∠AZY and ∠ZOM = 90° - ∠OZM = ∠AZY Thus, THY and MOZ are similar triangles with angles ∠AZY, 90°-∠AZY and 90° |HY|/|TH|=|OZ|/|OM| by this similarity Because of the properties of O and H, 2|TH|=|ZO| and |HY|=|AH|=2|OM| 2|TH|/|OM|=|OZ|/|OM|=|HY|/|TH|=2|OM|/|TH| Thus, |TH|/|OM|=|OM|/|TH| and we can conclude that |OM|=|TH| and since 2|TH|=|ZO| we find |ZO|=2|OM| and thus, <AZY= <ZOM = 60° Proof is complete if I'm not mistaken

Let $M$ be the midpoint of $BC$, then it is well-known that $\overrightarrow{AH} = 2\overrightarrow{OM}$, so $\text{proj}_{AH}(\overrightarrow{OZ})=\frac 12 \overrightarrow{AH}$. This is the only information we retain from $\Delta ABC$. Let $\angle AZY = z$ and the radius of $(AYZ)$ be $R$. Note that the angle between the lines $OZ$ and $AH$ is $z$, while $ZO = 2R \cos z$. Hence \[ 2R \cos z \cdot \cos z = \frac 12 R \]which gives $z = 60^\circ$ or $120^\circ$, either of which imply the desired claim.