overkill

Radical Axis easily kills this problem. It shouldn't be P5

the intended solution (most probably) From the length conditions, we reflect $H$ across $D$, we call this point $K$. Since $AH=HK$, $OH \perp AH$. From our length conditions, we realise that $S$ lies on $(BDHF)$ and $T$ lies on $(CDHE)$. We also let the antipodes of $B$ and $C$ on the circumcircle of $(ABC)$ be $B’$ and $C’$ respectively. It is also known that $S,H,B’$ and $T,H,C’$ are collinear. From there, we realise that $BHOC$ is a trapezoid and we do angle chasing to prove $\angle HSD = \angle HST$. Proof: $B,H,O,T$ are concyclic. Solution: $$\angle OTB = \angle OBT = \angle B’BT = \angle B’C’T = \angle B’C’H = \angle HBC = 180 - \angle OHB$$ From here: $$\angle HSD = \angle HBC = \angle B’BT = \angle B’ST = \angle HST$$ which proves $S,H,T$ concyclic.

This problem was proposed by me. Here is an alternative solution. Let $SH$ and $TH$ meet the circumcircle of $ABC$ again at $P$ and $Q$, respectively. Also, let $AH$ meet the circumcircle of $ABC$ again at $X$. Note that $H$ is the midpoint of $AX$ and $D$ is the midpoint of $HX$. Note that $SH$ passes through the midpoint of $CA$, denoted by $M$. Hence, $AP=CH=CX=2HM=HP$, and, also, $AQ=HQ$. This implies $PQ$ bisect $AH$. By Butterfly theorem, $ST$ also passes through the midpoint of $HX$ as desired.

If you have ever seen this problem which has exactly same config (just different goal), this problem is very cute. Fortunately i think very few contestants is already done Balkan MO before the test

demmy wrote: If you have ever seen this problem which has exactly same config (just different goal), this problem is very cute. Fortunately i think very few contestants is already done Balkan MO before the test FYI, I was inspired by the P3 of IMO2015. lol

Let line $AD$ intersect $(ABC)$ at $X$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. It is a well known property of the HM configuration that the points $M$, $H$, and $T$ and the points $N$, $H$, and $S$ are collinear. Notice that $AMHN\sim ABXC$ so $AMHN$ is cyclic. The rest is just an angle-chase: $$\angle SDB=\angle SHB=90^{\circ}-\angle ANH=\angle AMH-90^{\circ}=\angle THC=\angle TDC$$

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My Opinion on Day 1