Let $H $ be the orthocenter of $ABC $ ,$AB\neq AC $ ,and let $F $ be a point on circumcircle of $ABC $ such that $\angle AFH=90^{\circ} $.$K $ is the symmetric point of $H $ wrt $B $.Let $P $ be a point such that $\angle PHB=\angle PBC=90^{\circ} $,and $Q $ is the foot of $B $ to $CP $.Prove that $HQ $ is tangent to tge circumcircle of $FHK $.
Problem
Source: Serbia 2018 TST day 2 p5
Tags: geometry, Inversion
29.05.2018 00:11
29.05.2018 03:15
Can you post Problem 3,4,6 please, or give me links to those problems.
29.05.2018 13:27
Let $AD, BE$ be the altitudes, $A'$ be the point diametrically opposite to $A$ w.r.t. $\odot(ABC)$. Notice that $BHCA'$ is parallelogram so let $M$ be the midpoint of $BC$, which is also the midpoint of $HA'$. First, note by Power of Point that $$HF\cdot HA' = 2\cdot HF\cdot HM = 2\cdot HB\cdot HE = HK\cdot HE$$so $A', F', K, E$ are concyclic. By Reim's theorem, it suffices to prove that $HQ\parallel A'E$. Let $X=\cdot(CEA')\cap BC$ and $T=BH\cap CP$. Since $$\frac{CA'}{CE} = \frac{BH}{CE} = \frac{BP}{BC}$$so $\text{Rt }\Delta BPC\sim\text{Rt }\Delta CA'E$ which implies $\angle BPC = \angle CA'E = \angle CXE$. But $\angle PBH=\angle ACB$ so $\Delta CEX\sim\Delta BPT$. Let $CV, XW$ be the altitudes of $\Delta CEX$. Since $EA'$ is diameter of $\odot(CEX)$, $VW\perp EA'$. Moreover $$\Delta CEX\cup V\cup W\sim\Delta BPT\cup Q\cup H\implies \measuredangle(VW, HQ) = \measuredangle(BP, CX) = 90^{\circ}$$hence $HQ\perp VW\perp EA'$ so $HQ\parallel EA'$ as desired.
29.05.2018 14:44
remark $BC$ is tangent to $(BHPQ)$ and $\angle HFB= 90^\circ -\angle C$ but $\angle HBC= 90^\circ -\angle C$ thus $ F $ is on $(BHP)$ then $HF$ is radical axis of $(AFH),(BHP)$ besides if $D,E$ are resectively the midpoint of $BC$ and the foot of $B$ on $AC$ then $DE$ is tangent to $AHF$ but $DB=DE $ and $DB$ tangent to $(HPB)$ so $D$ is on $HF$ ; also $DQ$ is tangent to $(HPB)$ since $DB=DQ$ hence $BHQF$ is harmonic so $\angle HBI=\angle QBF =\angle QHF$ where $I$ is the midpint of $HF$ moreover $B$ is midpoint of $HK$ then $FK \parallel BI \implies \angle HKF=\angle HB I$ therefore the result follows $\color{magenta} \square $ RH HAS
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08.06.2018 00:31
tenplusten wrote:
I would be very grateful if somebody could explained these transformations of points. I mean: why $P'\in EM $ as in fact $P'\in PH $, why $FHPQ'$ is cyclic and so on.
20.06.2018 11:29
$FHPQ'$ is cyclic because $\angle BFH = 90 - \angle C = \angle BPH$ (just angle chase) and considering diameter $BP$.
02.10.2019 00:16
18.02.2020 17:23
Nice and easy. Serbia 2018 TST D2 P5 wrote: Let $H $ be the orthocenter of $ABC $ ,$AB\neq AC $ ,and let $F $ be a point on circumcircle of $ABC $ such that $\angle AFH=90^{\circ} $.$K $ is the symmetric point of $H $ wrt $B $.Let $P $ be a point such that $\angle PHB=\angle PBC=90^{\circ} $,and $Q $ is the foot of $B $ to $CP $.Prove that $HQ $ is tangent to tge circumcircle of $FHK $. Perform a Negative Inversion $(\Psi)$ around $H$ which preserves $\odot(ABC)$. Let $DEF$ be the Orthic triangle of $\triangle ABC$ and $A'$ be the $A-\text{antipode}$ of $\odot(ABC)$. So, $\Psi:F\leftrightarrow A'$ and $\Psi:K\leftrightarrow E$. So, we just need to prove that $EA'\|QH$. Notice that $\angle HQB=\angle HPB=\angle EBC=\angle EQC\implies\angle HQE=90^\circ$. Let $EM\cap QH=T$ and $M$ be the midpoint of $BC$. So as $QM=ME\implies ME=MT$, also it's well known that $HM=MA'\implies HTA'E$ is a parallelogram $\implies A'E\|HQ$. So Inverting back we get that $\odot(FHK)$ is tangent to $HQ$. $\blacksquare$
10.12.2022 12:33
Let $M$ be midpoint of $BC$ and $N$ be midpoint of $HF$. It's well known that $F,H,M$ are collinear and pass through $A'$ the antipode of $A$. Claim $: PFQHB$ is cyclic. Proof $:$ Note that $\angle BPH = \angle HBC = \angle HAC = \angle BAA' = \angle BFA' = \angle BFH$ so $BPFH$ is cyclic and $\angle BQP = \angle BHP = \angle 90$ so $PFQHB$ is cyclic. Note that $\angle QHF = \angle QBF$ and we must prove $\angle QHF = \angle HKF$ so we must prove $BQ$ is tangent to $KBF$ and since $KB = BH$ we must prove $BQ$ is symmedian in $BFH$ and since $BFHQ$ is cyclic we can instead prove $FH$ is symmedian in $FBQ$. $MQ = MB$ and $\angle MBQ = \angle 90 - \angle QBP = \angle BPQ$ so $MB$ and $MQ$ are tangent to $FBQ$. we're Done.
27.07.2023 15:56