Let $\Delta ABC$ be an acute triangle and $H$ its orthocenter. $B_1$ and $C_1$ are the feet of heights from $B$ and $C$, $M$ is the midpoint of $AH$. Point $K$ is on the segment $B_1C_1$, but isn't on line $AH$. Line $AK$ intersects the lines $MB_1$ and $MC_1$ in $E$ and $F$, the lines $BE$ and $CF$ intersect at $N$. Prove that $K$ is the orthocenter of $\Delta NBC$.
Problem
Source: Moldova TST 2020
Tags: geometry, orthocenter
mira74
08.03.2020 01:17
It suffices to show that $BE$ and $CK$ are perpendicular. Drop the condition "not on $AH$". We show the following problem, which is clearly equivalent:
Let $ABC$ be a triangle with orthocenter $H$, let $BH$ and $CH$ meet opposite sides at $B_1$ and $C_1$, and let $M$ be a the midpoint of $AH$. Let $K$ be a variable point on $B_1C_1$, let $T$ be the second intersection of $CK$ with $(BC)$, let $AK$ meet $B_1M$ at $E$ and let $BT$ meet $B_1M$ at $E'$. Then, $E=E'$.
We proceed with mOviNG pOinTs. Note that the maps $$K \rightarrow T \rightarrow E'$$tethered onto $$B_1C_1 \rightarrow (BC) \rightarrow B_1M$$and $$K \rightarrow E$$tethered onto $$B_1C_1 \rightarrow B_1M$$are projective, so it suffices to check $3$ cases. $K=B_1,C_1$ are trivial. If $D=AH \cap B_1C_1$, $DBC$ has orthocenter $M$ by Brokard on $AB_1HC_1$, so $CD \perp BM$, as desired.
prodiguy
08.03.2020 01:26
Notice that in the similarity $\Delta AB_1B \sim \Delta AC_1C$ the points $E, K$ fit into isogonal conjugates, so $\angle ABE = \angle KCC_1,$ which means $CK \perp DE$ and we are done.
AlastorMoody
31.03.2020 03:37
Moldova TST 2020 P4 wrote:
Let $\Delta ABC$ be an acute triangle and $H$ its orthocenter. $B_1$ and $C_1$ are the feet of heights from $B$ and $C$, $M$ is the midpoint of $AH$. Point $K$ is on the segment $B_1C_1$, but isn't on line $AH$. Line $AK$ intersects the lines $MB_1$ and $MC_1$ in $E$ and $F$, the lines $BE$ and $CF$ intersect at $N$. Prove that $K$ is the orthocenter of $\Delta NBC$.
Solution: Restate as the following:
Restated Problem wrote:
Let $\Delta ABC$ be an arbitrary triangle with $H$ as orthocenter and $\Delta DEF$ as its orthic triangle. Let $K$ be a point on $DE$. Let $N$ be the orthocenter WRT $\Delta KBC$. Let $G$ be midpoint of $AH$. Define $L$ as $AK \cap BN$, $M$ as $AK$ $\cap$ $CN$. Show that $EL$ and $FM$ intersect on $AH$ at $G$
Proof: Let $\mathcal{H}$ be the circum-rectangular hyperbola WRT $\Delta ABC$ and passing through $K$. Since, $N$ is orthocenter of $\Delta KBC$ $\implies$ $N$ $\in$ $\mathcal{H}$. Also, $NK|| AH$. Let $AH$ $\cap$ $EF$ $=$ $P$. Let $PK$ intersect $\mathcal{H}$ at $T$. Let $Q$ be the orthocenter WRT $\Delta AKH$. We'll first show $DK \perp NQ$
\begin{align*} -1= (B,C ;D, EF \cap BC ) \stackrel{E}{=} (H , A ;D,P) \end{align*}From Pascal on $QAAKHH$ and $CAABHH$ $\implies$ $-1= (A,H ;K, PK \cap \mathcal{H})$ $\implies$ $DK$ is tangent to $\mathcal{H}$. Since, $\angle NKQ$ $=$ $90^{\circ}$ $\implies$ $DK \perp NQ$. Let $NC \cap HK=Z$
By Pascal on $KHBNCA$ $\implies$ $L,Z,E$ are collinear.
\begin{align*} -1 = (A,H ;K, PK \cap \mathcal{H} ) \stackrel{N}{=} (A,H ;\infty_{AH}, NT \cap AH ) \end{align*}Thus, $NT \cap AH$ is midpoint of $AH$ $\implies$ $G$ lies on $NT$. Apply Pascal on $KAHBNT$ $\implies$ $G$ $\in$ $LE$ and Applying Pascal on $HAKTNC$ $\implies$ $G$ $\in$ $MF$ $\qquad \blacksquare$
Mahdi_Mashayekhi
04.11.2022 20:30
Let's Have a simple solution for this problem... Let $AK$ meet $ABC$ at $P$. Claim $: B_1KPC$ is cyclic. Proof $:$ Note that $\angle CPK = \angle CPA = \angle 180 - \angle CBA = \angle 180 - \angle AB_1C_1 = \angle 180 - \angle CB_1K$. Claim $: BB_1PE$ is cyclic. Proof $:$ Note that $\angle BPE = \angle 180 - \angle APB = \angle 180 - \angle ACB = \angle 180 - \angle AC_1B_1 = \angle 180 - \angle AHB_1 = \angle 180 - \angle MHB_1 = \angle 80 - \angle MB_1H = \angle BB_1E$. Now Note that $\angle EBB_1 = \angle KPB_1 = \angle KCB_1$ so $\angle EBA = \angle KCC_1$ so of $BE$ meets $CK$ at $T$ then $BC_1TC$ is cyclic so $\angle BTC = \angle 90$ so $BE \perp CK$. Now with repeating same approach for $CF$ and $BK$ which is proving $BC_1KP$ and $CC_1FP$ are cyclic we can also prove that $BK \perp CF$ so $K$ is orthocenter of $BNC$.
AliAlavi
26.12.2022 14:56
Mahdi_Mashayekhi wrote: Let's Have a simple solution for this problem... Let $AK$ meet $ABC$ at $P$. Claim $: B_1KPC$ is cyclic. Proof $:$ Note that $\angle CPK = \angle CPA = \angle 180 - \angle CBA = \angle 180 - \angle AB_1C_1 = \angle 180 - \angle CB_1K$. Claim $: BB_1PE$ is cyclic. Proof $:$ Note that $\angle BPE = \angle 180 - \angle APB = \angle 180 - \angle ACB = \angle 180 - \angle AC_1B_1 = \angle 180 - \angle AHB_1 = \angle 180 - \angle MHB_1 = \angle 80 - \angle MB_1H = \angle BB_1E$. Now Note that $\angle EBB_1 = \angle KPB_1 = \angle KCB_1$ so $\angle EBA = \angle KCC_1$ so of $BE$ meets $CK$ at $T$ then $BC_1TC$ is cyclic so $\angle BTC = \angle 90$ so $BE \perp CK$. Now with repeating same approach for $CF$ and $BK$ which is proving $BC_1KP$ and $CC_1FP$ are cyclic we can also prove that $BK \perp CF$ so $K$ is orthocenter of $BNC$. Very nice solution ...