Problem. Let $ABC$ be a triangle isncribed $(O)$, incenter $I$, median $AM$. $E,F$ lie on $CA,AB$ su

Problem. Let $ABC$ be a triangle isncribed $(O)$, incenter $I$, median $AM$. $E,F$ lie on $CA,AB$ such that $ME\perp IC, MF\perp IB$. $(MEF)$ cuts $BC$ again at $D$. $S$ is midpoint uel of arc $BC$ contain $A$ of $(O)$. The line passing through $S$ parrallel uel to $OI$ cuts the line passing through $D$ perpendicular to $BC$ at $T$. $K,L$ are symmetric of $T$ through $E,F$. Prove that $CK,BL$ and $ST$ are concurrent on $(O)$. uel Solution (By Telv Cohl). Let $ O' $ be the reflection of $ O $ in $ I $ . Let $ B' $ be the reflection of $ B $ in $ F $ . Let $ C' $ be the reflection of $ C $ in $ E $ . Let $ X_1=TS \cap BL, X_2=TS \cap CK $ . Since $ B'B=BC=CC' $ , so we get $ S \in \odot (AB'C') uel $ (see hard problem (#post 11)) . Since $ IB, IC $ is the perpendicular bisector of $ MF, ME $, respectively , so we get $ I $ is the center of $ \odot (MEF) $ and $ DT $ is the reflection of $ MS $ in $ I $ , hence the reflection $ O' $ of $ O $ in $ I $ lie on $ DT $ . From hard problem (#post 9) we get $ OI \perp B'C' $ and the length uel of $ OI $ is equal to the radius of $ \odot (AB'C') $ , so combine with $ OO' \parallel ST $ we get $ T $ is the midpoint of arc $ B'C' $ in $ \odot (AB'C') $ and $ ST $ is the diameter of $ \odot (AB'C') $ . Easy to see $ TB' \parallel LB $ and $ TC' \parallel uel KC $ . Since $ \angle BX_1S=180^{\circ}- \angle STB'=180^{\circ}-\angle SCB $ , so we get $ X_1 \equiv TS \cap BL \in (O) $ . Similarly, we can prove $ X_2 \equiv TS \cap CK \in (O) $ , so we get $ X_1 \equiv X_2 $ and $ TS, BL, CK $ are concurrent on $ (O) $ . Q.E.D Solution (By Tran Quang Hung). $AI$ cuts $(O)$ again at $N$. $SN$ is diameter of $(O)$. $G$ is midpoint of $DM$ and $J$ is midpoint of $ST$. $IG$ and $JG$ are perpendicular to $DM$ so $I,J,G$ are collinear. This, $IJ\parallel SO$ and $JS\parallel IO$ deduce $IJSO$ is parallelogram. Therefore uel $ITJO$ is parallelogram, too, so $TI\parallel JO$. Easily seen $TN\parallel JO$. Therefore $T,I,N$ are collinear or $T$ lies on $AI$. Let $P$ is symmetric of $B$ through $F$. We have $BP=2BF=2BM=BC$. From $OBN$ and $SBC$ are similar. We have $\frac{BP}{BS}=\frac{BC}{BS}=\frac{BN}{BO}=\frac{NI}{NO}=\frac{NT}{NS}$. Easily seen $\angle PBS=\angle TNS$ thus $PBS$ and $TNS$ are similar. Deduce $\angle BPS=\angle NTS$ thus $ATPS$ is cyclic. $ST$ cuts $(O)$ again at $R$. We have angle charsing $\angle ABL=\angle TPF=\angle TSA=\angle ABR$. Thus $BL$ passe through $R$. Similarly $CK$ passes through $R$. We are done. Solution (By Luis González). Let $Y,Z$ be the reflections of $C,B$ on $E,F$ and let $R$ be the midpoint of the arc $BC$ of $(O).$ Clearly $BZ=BC=CY$ and $S$ is the center of the rotation taking uel $CY$ into $BZ,$ thus $A \in \odot(SYZ)$ $\Longrightarrow$ $AR$ cuts $\odot(SYZ)$ again at the midpoint $T'$ of its arc $YZ.$ Hence, the spiral similarity with center $S,$ taking $YZ$ into $CB,$ takes $T'$ into $R$ $\Longrightarrow$ $\triangle SRT' \sim \triangle SCY$ $\Longrightarrow$ $\frac{RS}{RT'}=\frac{CS}{CY}=\frac{CS}{BC}=\frac{OR}{RB}=\frac{OR}{RI}$ $\Longrightarrow$ $ST' \parallel OI$ $\Longrightarrow$ $I$ is midpoint of $RT'.$ But since $I$ is circumcenter of $\triangle MEF,$ then $M,D$ are symmetric across the perpendicular uel from $I$ to $DM$ $\Longrightarrow$ $(DT' \parallel MR) \perp BC$ $\Longrightarrow$ uel $T \equiv T'.$ Now if $ST$ cuts $(O)$ again at $J,$ we have $\angle ACJ=\angle AST=\angle AYT$ $\Longrightarrow$ $YT \parallel CJ,$ i.e. $J \in CK$ and similarly $J \in BL.$   Solution uel (By dibyo_99). Let $B', C'$ on $AB, AC$ such that $BB' = BC = CC'$. We know that $S$ is the center of the spiral similarity taking $BB'$ to $CC'$. Consequently, $ASB'C'$ is cyclic. Now, since $IE = IM$ and $IM = IF$, note that $I$ is the center of $(MEF)$. uel Therefore, $IM = ID$, implying that $I$ lies on the perpendicular bisector of $MD$. Since $STDM$ is a trapezium, the perpendicular bisector $l$ of $DM$ cuts $ST$ at its midpoint. Let $P = SO \cap TI$. Then, $l \parallel SO$, meaning $I$ is the midpoint uel of $TP$. Since $IO\parallel ST$, $O$ must be the midpoint of $SP$. Therefore, $P$ is the midpoint of arc $BC$ of $(ABC)$. Let $ST \cap (ABC) = X$. Note that $LBTB'$ is a parallelogram. Then, \[ \angle{ABX} = \angle{ASX} = \angle{AST} = \angle{AB'T} = 180^{\circ}-\angle{ABL} \]So, $L,B,X$ are collinear. Similarly, $C,K,X$ are also collinear. uel Therefore, $CK,BL,ST$ are all concurrent on $(O)$. Reference. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=619172
►  2014 (67) ►  tháng một (23) ►  tháng hai (17) ►  tháng ba (4) ►  tháng tư (2) ►  tháng năm (7) ►  tháng sáu (6) ►  uel tháng bảy (1) ►  uel tháng tám (3) ►  tháng chín (2) ►  tháng mười (1) ►  tháng mười hai (1) ▼  2015 (18) ▼  tháng một (10) Về bài hình học thi VMO năm 2015 ngày thứ nhất Solution for HSGS TST 2014 Round 1 Day 2 Solution for HSGS TST 2014 Round 2 Day 1 Solu