We claim $A'B'OH$ is a parallelogram. Spiral sim at $A$ sending $A'O$ to $BH$ tells us that $A'O=BH\times\frac{AO}{AB}$ and spiral sim at $B$ sending $B'H$ to $AO$ gives $B'H=AO\times\frac{BH}{AB}$ clearly the two expressions are equal hence $A'O=B'H$ similarly we get opposite pair of equal sides yay.

Construction of $O,H,D,E$ motivate to use complex W.L.O.G. $O=0$ and$(ABC)$ as a unit circle. We have $h=a+b+c$ dan $|a|=|b|=|c|=|d|=|e|=1$ (imply $\overline{a}=1/a$ by $a\overline{a}=|a|^2$ property). Because $BE\;\bot\;AC$ we have $be+ac=0\iff e = -\frac{ac}{b}$ and $d=-\frac{bc}{a}$. Solve the system equations for $a'$: \[|a'-a| = |a'-h| = |a'-e|,\]we have \begin{align*} a' &= \frac{\left |\begin{matrix} h &h\overline{h} & 1\\ a & a\overline{a} & 1 \\ e&e\overline{e} & 1 \end{matrix} \right |}{\left |\begin{matrix} h &\overline{h} & 1\\ a & \overline{a} & 1 \\ e&\overline{e} & 1 \end{matrix} \right |} = \frac{\left |\begin{matrix} a+b+c &(a+b+c)\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right )&1 \\ a & 1 & 1\\ -\frac{ac}{b} & 1 & 1 \end{matrix} \right |}{\left |\begin{matrix} a+b+c & \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & 1\\ a & \frac{1}{a} & 1 \\ -\frac{ac}{b} & -\frac{b}{ac} & 1 \end{matrix} \right |}\\ &= \frac{-\frac{(a+b)(a+c)(b+c)^2}{b^2c}}{\frac{(b-a)(b+a)(b+c)^2}{ab^2c}} = -\frac{(a+c)a}{b-a}. \end{align*}Analogously, $b' = -\frac{(b+c)b}{a-b}.$ Therefore, the midpoint of $A'B'$ is \[m = \frac{a'+b'}{2} = \frac{1}{2}\left (\frac{-a^2-ac +b^2+bc}{b-a} \right )=\frac{(b-a)(a+b+c)}{2(b-a)}=\frac{a+b+c}{2},\]which is also the midpoint of $OH$.

Here is an alternate but similar way to prove the parallelogram claim: Note that $AE$ is the radical axis of $\odot(AHE)$ and $\odot(ABC)$, so $OA' \perp AE$. Also: \begin{align*} \measuredangle(B'H, AE) &= \measuredangle(B'H, HB) + \measuredangle(BE, EA) \\ &\stackrel{\triangle HB'B \sim \triangle AOB}{=} \measuredangle(OA, AB) + \measuredangle(BC, CA) \\ &\stackrel{AO \text{ and } AD \text{ are isogonal w.r.t. } \angle BAC}{=} \measuredangle(CA, AD) + \measuredangle(BC, CA) = \measuredangle(BC, AD) = 90^{\circ}. \end{align*}So, $B'H \perp AE$ as well. Therefore, $B'H \parallel OA'$. Similarly, $A'H \parallel OB'$, proving the claim.

From PoP on B' we can chase the length of B'O using trigonometry, same with A'H etc. Hence we can trigon bash this problem

First notice that triangle $AHE$ and $BHD$ are isosceles triangle and thus $A'$ and $B'$ lies on $AC$ and $BC$ respectively. Let $P, Q$ be midpoint of $AH$ and $BH$ respectively. Let $S_a = \frac{-a^2 + b^2 + c^2}{2}$, define $S_b$ and $S_c$ similarly. We are going to use Barycentric Coordinates. Let ABC be reference triangle and $A(1, 0, 0); B(0, 1, 0); C(0, 0, 1)$. Well known that $H(S_bS_c : S_cS_a : S_aS_b)$ and $O(a^2S_a : b^2S_b : c^2 S_c)$. Let $T = S_bS_c + S_cS_a + S_aS_b = \frac{1}{2} (a^2S_a + b^2S_b + c^2 S_c)$. We can compute that $Q(S_bS_c : S_cS_a + T : S_aS_b)$ and similarly $P(S_bS_c + T : S_c S_a : S_aS_b)$. Next, we can compute $A'$ by using the fact that $PA' // BC$. Hence, $P$, $A$, and $R$ are collinear where $R(0 : 1 : -1)$ is point of infinity of line $BC$. \[ \begin{vmatrix} x & 0 & z \\ 0 & 1 & -1\\ S_bS_c + T & S_cS_a & S_aS_b \end{vmatrix} = 0 \] Expanding, we can get $A'(c^2S_c + b^2S_b : 0 : a^2S_a)$. Similarly, $B'(0 : a^2S_a + c^2S_c : b^2S_b)$. Let $N$ be the midpoint of $A'B'$, we can see that $N(c^2S_c + b^2S_b : a^2S_a + c^2S_c : a^2S_a + b^2S_b)$. Now, we just need to prove that $O, N, H$ are collinear (and in fact $N$ is the midpoint of $OH$), that is: \[ \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ b^2S_b + c^2S_c & c^2S_c + a^2S_a & a^2S_a + b^2S_b\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ a^2S_a + b^2S_b + c^2S_c & c^2S_c + a^2S_a + b^2S_b & a^2S_a + b^2S_b + c^2S_c\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ S_aS_b + S_aS_c & S_aS_b + S_bS_c & S_aS_c + S_bS_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ T & T & T \end{vmatrix} = 0 \]which is evident from the last determinant.

Notice that the segment $AE$ is the reflection of $AH$ with respect to the line $AC$. Thus, if we let $X$ be the point on $AC$ such that $AH \bot HX$, the point $A'$ is the midpoint of $AX$. The construction of $X$ also means $HX$ is parallel to $BC$. Similarly, $B'$ is the midpoint of $BY$, where $Y$ is the point on $BC$ such that $HY$ is parallel to $AC$. Thus, the midpoint of $A'B'$ is just the centroid of $4$ points: $X$, $Y$, $A$, and $B$. Now notice that $XCYH$ is a parallelogram, so the midpoint of $XY$ is equal to the midpoint of $CH$. Then the midpoint of $A'B'$ is the centroid of $4$ points $A$, $B$, $C$, and $H$. Using complex coordinate or some other way, it is now easy to see that this centroid is actually the midpoint of $HO$.

godjuansan wrote: First notice that triangle $AHE$ and $BHD$ are isosceles triangle and thus $A'$ and $B'$ lies on $AC$ and $BC$ respectively. Let $P, Q$ be midpoint of $AH$ and $BH$ respectively. Let $S_a = \frac{-a^2 + b^2 + c^2}{2}$, define $S_b$ and $S_c$ similarly. We are going to use Barycentric Coordinates. Let ABC be reference triangle and $A(1, 0, 0); B(0, 1, 0); C(0, 0, 1)$. Well known that $H(S_bS_c : S_cS_a : S_aS_b)$ and $O(a^2S_a : b^2S_b : c^2 S_c)$. Let $T = S_bS_c + S_cS_a + S_aS_b = \frac{1}{2} (a^2S_a + b^2S_b + c^2 S_c)$. We can compute that $Q(S_bS_c : S_cS_a + T : S_aS_b)$ and similarly $P(S_bS_c + T : S_c S_a : S_aS_b)$. Next, we can compute $A'$ by using the fact that $PA' // BC$. Hence, $P$, $A$, and $R$ are collinear where $R(0 : 1 : -1)$ is point of infinity of line $BC$. \[ \begin{vmatrix} x & 0 & z \\ 0 & 1 & -1\\ S_bS_c + T & S_cS_a & S_aS_b \end{vmatrix} = 0 \] Expanding, we can get $A'(c^2S_c + b^2S_b : 0 : a^2S_a)$. Similarly, $B'(0 : a^2S_a + c^2S_c : b^2S_b)$. Let $N$ be the midpoint of $A'B'$, we can see that $N(c^2S_c + b^2S_b : a^2S_a + c^2S_c : a^2S_a + b^2S_b)$. Now, we just need to prove that $O, N, H$ are collinear (and in fact $N$ is the midpoint of $OH$), that is: \[ \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ b^2S_b + c^2S_c & c^2S_c + a^2S_a & a^2S_a + b^2S_b\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ a^2S_a + b^2S_b + c^2S_c & c^2S_c + a^2S_a + b^2S_b & a^2S_a + b^2S_b + c^2S_c\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ S_aS_b + S_aS_c & S_aS_b + S_bS_c & S_aS_c + S_bS_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ T & T & T \end{vmatrix} = 0 \]which is evident from the last determinant. name a more iconic duo than godjuansan and barycentric coordinates, i'll wait. (bro's only on aops now for posting bary solutions )

By reflecting the orthocenter, we can clearly see that A' lies on AC and B' lies on BC. Notice that A' is the intersection of the perpendicular bisector of AH with line AC and B' is the intersection of the perpendicular bisector of BH with line BC, therefore we can simply use cartesian coordinates to prove that the midpoint of OH and midpoint of A'B' coincide

To generalize the problem, we use directed angle mod $180$. The main idea here is to prove that $A'OB'H$ is a parallelogram. To prove that, we can simply prove that $\triangle{A'OE}$ is congruent to $\triangle{B'OD}$. To do this, try to prove 2 of the 3 angles are the same and use the fact that $OD=OE$, then the conclusion follows from that.

Proof without word. (with the knowledge that $\triangle AHE$ and $\triangle{BDH}$ are isosceles)

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KHOMNYO2 wrote: Proof without word. (with the knowledge that $\triangle AHE$ and $\triangle{BDH}$ are isosceles) The diagram also holds for obtuse, for anyone wondering how does my food taste (even though it looked quite different). Even though, one diagram is enough but i had the urge to add its obtuse version too, lol

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Btw, here is my full solution. I solved this problem by doing the case when $\triangle{ABC}$ is acute, the obtuse one can be proved similarly. It is well-known that $\triangle{AHE}$ and $\triangle{BHD}$ are isosceles triangles and also $O$ and $H$ are isogonal conjugates w.r.t the angles of $\triangle{ABC}$. Because of this, clearly $A'$ and $B'$ lies on $AC$ and $BC$. Note that $\angle{CA'E}=2\cdot\angle{A'AE}=2\cdot\angle{CAE}=\angle{COE}$ and similarly $\angle{CB'D}=2\cdot\angle{B'BD}=2\cdot\angle{CBD}=\angle{COD}$ $\Longrightarrow$ $A'OCE$ and $B'OCD$ are cyclic quads. Now by considering the fact that $OD=OE$ and $\angle{A'EO}=\angle{A'CO}=\angle{ACO}=\angle{CAO}=\angle{BAH}=\angle{BAD}=\angle{BCD}=\angle{B'CD}=\angle{B'OD}$ and $\angle{A'OE}=\angle{A'CE}=\angle{ACE}=\angle{ABE}=\angle{ABH}=\angle{OBC}=\angle{OCB}=\angle{OCB'}=\angle{B'DO}$, thus $\triangle{A'OE}$ is congruent to $\triangle{B'OD}$. For the final act, we have $B'H=B'D=A'O$ and $A'H=A'E=B'O$, hence $A'OB'H$ is a parallelogram and so $OH$ bisects $A'B'$. $\blacksquare$ Actually, I have another solution that doesn't requires any additional cyclic quads, but it's written in Indonesian so I hope you guys can read it

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P3_OSN_2024___Day_1_compressed.pdf (66kb)

This is my problem. One of two official solutions has been posted, using complex. Another solution similar to @above (use directed angle)