Parallel lines in an acute triangle

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Parallel lines in an acute triangle

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Source: VNTST 2025 P2

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#1 h Mar 25, 2025, 9:01 AM • 3 Y

Y by MS_asdfgzxcvb, NO_SQUARES, lolsamo

Let $ABC$ be an acute, non-isosceles triangle with orthocenter $H$ . Let $D, E, F$ be the reflections of $H$ over $BC, CA, AB$ , respectively, and let $A', B', C'$ be the reflections of $A, B, C$ over $BC, CA, AB$ , respectively. Let $S$ be the circumcenter of triangle $A'B'C'$ , and let $H'$ be the orthocenter of triangle $DEF$ . Define $J$ as the center of the circle passing through the three projections of $H$ onto the lines $B'C', C'A', A'B'$ . Prove that $HJ$ is parallel to $H'S$ .

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#2 h Mar 25, 2025, 9:03 AM • 2 Y

Y by MS_asdfgzxcvb, NO_SQUARES

My proof. Let $(ABC)$ be the unit circle, and denote $D(d^2), E(e^2), F(f^2)$ , so that $A(-ef), B(-fd), C(-de)$ .
Thus, the orthocenter $H$ of $ABC$ is
$h = a + b + c = -(ef + fd + de),$ and the orthocenter $H'$ of $DEF$ is
$h' = d^2 + e^2 + f^2.$ Since $A'$ is the reflection of $A$ over $BC$ , we obtain
$a' = b + c - bc\bar{a} = -d(-d + e + f).$ Similarly,
$b' = -e(d - e + f), \quad c' = -f(d + e - f).$ From this, the circumcenter $S$ of triangle $A'B'C'$ is
$s = -\frac{\begin{vmatrix} {a'\bar {a'}} & a' & 1 \\ {b'\bar {b'}} & b' & 1 \\ {c'\bar {c'}} & c' & 1 \\ \end{vmatrix}}{\begin{vmatrix} a' & \bar {a'} & 1 \\ b' & \bar {b'} & 1 \\ c' & \bar {c'} & 1 \\ \end{vmatrix}} = \frac{-2def(d^{2} + e^{2} + f^{2})}{ef(e+f)+fd(f+d)+de(d+e)-def}.$ Next, the projections $P, Q, R$ of $H$ onto $B'C', C'A', A'B'$ are given by
$p = \frac{d(e^3+f^3)-d^2(2e^2+2f^2+3ef)+3e^2f^2}{2(-ef+fd+de)},$ $q = \frac{e(f^3+d^3)-e^2(2f^2+2d^2+3fd)+3f^2d^2}{2(ef-fd+de)},$ $r = \frac{f(d^3+e^3)-f^2(2d^2+2e^2+3de)+3d^2e^2}{2(ef+fd-de)}.$ From this, the circumcenter $J$ of triangle $PQR$ is
$j = -\frac{\begin{vmatrix} {p\bar {p}} & p & 1 \\ {q\bar {q}} & q & 1 \\ {r\bar {r}} & r & 1 \\ \end{vmatrix}}{\begin{vmatrix} p & \bar {p} & 1 \\ q & \bar {q} & 1 \\ r & \bar {r} & 1 \\ \end{vmatrix}} = \frac{d^{2} + e^{2} + f^{2} - 4(ef + fd + de)}{4}.$ Finally, it is straightforward to verify that
$\frac{j - h}{s - h'} = \frac{def - ef(e+f) - fd(f+d) - de(d+e)}{4(def+ef(e+f)+fd(f+d)+de(d+e))} = \overline{\left(\frac{j - h}{s - h'}\right)},$ which implies that $HJ \parallel H'S$ . This completes the proof.

This post has been edited 1 time. Last edited by buratinogigle, Mar 25, 2025, 10:43 AM
Reason: Typo

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#3 h Mar 26, 2025, 2:08 PM • 5 Y

Y by Funcshun840, ErTeeEs06, AlephG_64, ehuseyinyigit, MS_asdfgzxcvb

Firstly, notice the midpoint of $OS$ is the Kosnita point of triangle $ABC$ , which is the schiffler point of triangle $DEF$ , so it lies of the euler line of triangle $DEF$ , so $O,S,H'$ are collinear.
Now let $P,Q,R$ be the centers of $B'C'H$ and so on. It is easy to see that triangle $PQR$ is just a homothety of $ABC$ at $O$ with ratio 2. (Let $P'$ be reflection of $O$ over $A$ , then easy lengths with feet dropped gives $P'B'=P'H=P'C'$ .)

This means that $O$ is the center of triangle $PQR$ .
Let $(PQR)$ meet $(A'B'C')$ at $X,Y$ , then by RMMSL 2023 G3, $XY$ is the perpendicular bisector of the line connecting $H$ to it's isogonal conjugate in triangle $A'B'C'$ , so $J$ is the foot from $H$ onto $XY$ , so $HJ$ and $OS$ are both perpendicular to $XY$ , as desired.

This post has been edited 1 time. Last edited by lolsamo, Mar 26, 2025, 2:09 PM

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#4 h Friday at 3:49 PM • 2 Y

Y by MS_asdfgzxcvb, lolsamo

Here is my solution for this problem
Solution
Let $O, G, N, K$ be centroid, NPC center, Kosnita point of $\triangle ABC;$ $\triangle N_aN_bN_c$ be pedal triangle of $N$ WRT $\triangle ABC,$ $M$ be midpoint of $BC;$ $AH$ intersects $BC$ at $H_a$ . We have $\overline{NN_a} = \dfrac{1}{2}(\overline{OM} + \overline{HH_a}) = \dfrac{1}{4}(\overline{AH} + \overline{HD}) = \dfrac{1}{4}\overline{AD} = \dfrac{1}{4}\overline{HA'}$ Combine with $\overline{GN} = \dfrac{1}{4}\overline{GH},$ we have $\overline{GN_a} = \dfrac{1}{4}\overline{GA'}$ . Similarly, we have $G$ is center of the homothety with ratio $k = 4$ that transforms $\triangle N_aN_bN_c$ into $\triangle A'B'C'$ . So if we let $P$ be center of $(N_aN_bN_c)$ then $\overline{GP} = \dfrac{1}{4}\overline{GS}$ . From this, we have $\dfrac{\overline{OG}}{\overline{ON}} \cdot \dfrac{\overline{KN}}{\overline{KP}} \cdot \dfrac{\overline{SP}}{\overline{SG}} = \dfrac{2}{3} \cdot 2 \cdot \dfrac{3}{4} = 1$ or $O, K, S$ are collinear. Note that $K$ is Schiffler point of $\triangle DEF$ then $K \in OH',$ hence $S \in OH'$ . Let $N'$ be isogonal conjugate of $N$ WRT $\triangle N_aN_bN_c$ . From property 2 in here, we have $NN' \parallel OK$ . But $HJ$ passes through isogonal conjugate of $H$ WRT $\triangle A'B'C'$ and $\triangle A'B'C' \cup H$ is image of $\triangle N_aN_bN_c \cup N$ through the homothety with center $G,$ ratio $k = 4$ then $HJ \parallel NN'$ . Therefore, $HJ \parallel OK$ of $HJ \parallel H'S$ .

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