Let $ABC$ be a scalene triangle with $I$ be its incenter. The incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. $Y$, $Z$ are the midpoints of $DF$, $DE$ respectively, and $S$, $V$ are the intersections of lines $YZ$ and $BC$, $AD$, respectively. $T$ is the second intersection of $\odot(ABC)$ and $AS$. $K$ is the foot from $I$ to $AT$. Prove that $KV$ is parallel to $DT$. Proposed by ltf0501
Problem
Source: 2022 IRN TWN Friendly Math Competition P3
Tags: geometry, incenter, Taiwan, Iran, parallel, incircle
laikhanhhoang_3011
21.06.2022 19:01
We change to a new problem: Quote: Let $ABC$ be a scalene triangle with $O$ is the circumcenter. $D$ is the intersection of tangent from $B,C$ to $(O)$, $H,G$ respectively.$M,N$ is midpoint of $AC,AB$. $MN$ cuts tangent at $A$ of $(O)$ and $AD$ at $T,E$. $(DHG)$ cut $DT$ at $F$. $OL \perp DT, L \in DT$. Prove that $AF//EL.$ We can easily prove $GNMH$ is cyclic by angle chasing. So $TA^2=TM.TN=TG.TH=TF.TD \Rightarrow \measuredangle TFA = \measuredangle TAD$ Let $K$ be $A-Dumpty$ point of $ABC$, $P=AD \cap BC$. We all know that $\overline{A,K,D}, \, TA=TK, \, K \in (BOC).$ We'll prove $TLKE$ is cyclic $\Leftrightarrow \measuredangle KLD=180^\circ - \measuredangle DEM = 180^\circ - \measuredangle DPC= \measuredangle DPB \Leftrightarrow DB^2=DB.DK$, which is true according to shooting lemma. So $\measuredangle TFA = \measuredangle TAK =\measuredangle TKA =\measuredangle KLT \Rightarrow AF // KL. $ (Q.E.D) Diagram:
trinhquockhanh
23.06.2022 05:31
so you used inversion to turn it into a new problem?
laikhanhhoang_3011
23.06.2022 05:34
no, $DEF$ in the original post is $ABC$ in my problem
trinhquockhanh
23.06.2022 05:46
laikhanhhoang_3011 wrote: no, $DEF$ in the original post is $ABC$ in my problem oh, thanks
Mahdi_Mashayekhi
24.12.2022 20:40
Nice Claim $: ATD$ is tangent to $BC$. Proof $:$ It's well known that $YZ$ is Radical Axis of $BIC$ and $DEF$ so $SD^2 = SB.SC = ST.SA$ Now we have that $\angle ATD = \angle ADC$ and we want to prove $\angle ADC = \angle AKV$ or instead $\angle VDS = \angle VKS$. Let $D'$ be reflection of $D$ across $VS$. Note that Now we want to prove $VD'KS$ is cyclic. Let $AD$ meet $EF$ at $P$. we have $\angle APD' = \angle AVS = \angle 180 - \angle DVS = \angle 180 - \angle D'VS$ so we can instead prove that $APD'K$ is cyclic. Claim $: AKFIE$ is cyclic. Proof $:$ $\angle IFA = \angle IEA = \angle IKA = \angle 90$. Claim $: FKE$ and $BTC$ are similar. Proof $:$ Note that $\angle KEF = \angle KAF = \angle TAF = \angle TAB = \angle TCB$ and $\angle KFE = \angle 180 - \angle KAE = \angle 180 - \angle TAC = \angle TBC$. Note that since $DD' \perp EF$ we have $\frac{FD'}{ED'} = \frac{BD}{CD}$ so $\angle BTD = \angle FKD'$. Also Note that $\angle BTD = \angle BTA - \angle DTA = \angle 180 - \angle C - \angle CDA = \angle 180 - \angle C - \angle B - \angle DAB = \angle EAP$ so $EAP = \angle FKD'$. Now we have $\angle EPA = \angle 180 - \angle EAP - \angle AEP = \angle 180 - \angle FKD' - \angle AEF = \angle FKA - \angle FKD' = \angle D'KA$ so $KAPD'$ is cyclic as wanted. we're Done.