The triangle $ABC$ has $O$ as its circumcenter, and $H$ as its orthocenter. The line $AH$ and $BH$ intersect the circumcircle of $ABC$ for the second time at points $D$ and $E$, respectively. Let $A'$ and $B'$ be the circumcenters of triangle $AHE$ and $BHD$ respectively. If $A', B', O, H$ are not collinear, prove that $OH$ intersects the midpoint of segment $A'B'$.
Problem
Source: Indonesian National Mathematical Olympiad 2024, Problem 3
Tags: geometry, circumcircle, Indonesia, orthocenter, Indonesia MO, Inamo
28.08.2024 10:41
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.
28.08.2024 11:03
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 $OH$.
28.08.2024 11:11
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.
28.08.2024 11:49
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
28.08.2024 14:39
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.
28.08.2024 14:50
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 $HO$.
28.08.2024 17:47
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 )
31.08.2024 20:07
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
02.09.2024 10:57
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.
02.09.2024 11:50
Proof without word. (with the knowledge that $\triangle AHE$ and $\triangle{BDH}$ are isosceles)
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02.09.2024 12:19
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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05.09.2024 01:00
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)
11.09.2024 10:44
This is my problem. One of two official solutions has been posted, using complex. Another solution similar to @above (use directed angle)