Thứ Sáu, 9 tháng 1, 2015

Solution for HSGS TST 2014 Round 2 Day 1

Problem. Let $(O)$ be a circumcircle of triangle $ABC$. $M,N$ are point on arc $BC$ not containing $A$ such that $MN \parallel BC$ (ray $AM$ is between ray $AB$ and ray $AC$). Let $P,Q$ are points on ray $BM,CN$ respectively such that $BP=BN=CM=CQ$. $AM,AN$ intersect $PQ$ at $S,T$, respectively. $BT,CS$ intersect $CQ,BP$ at $L,K$ respectively. Prove that $AK=AL$.

Solution 1 (By Telv Cohl). WLOG $ BN>BM, CM>CN $ .
Let $ M'=AM \cap BC, N'=AN \cap BC $ .
Let $ X $ be the intersection of $ BC $ and the bisector of $ \angle BAC $ .

Since $ \angle SMP=\angle BNA, \angle MPS=\angle MBC=\angle BAN $ ,
so we get $ \triangle MPS \sim \triangle NAB \sim \triangle CAM' $ .
Similarly, we can prove $ \triangle NQT \sim \triangle MAC \sim \triangle BAN' $ .

From Ptolemy theorem we get $ AC \cdot BN+AB \cdot CN=BC \cdot AN $ ... $ (1) $

Since $ PM=BP-BM=BN-CN $ ,
so combine with $ (1) $ we get $ AB \cdot PM=(AB+AC) \cdot BN-BC \cdot AN $ .

$ \Longleftrightarrow AB+AC=BC \cdot \frac{AN}{BN}+PM \cdot \frac{AB}{BN} $ .

$ \Longleftrightarrow (AB+AC) \cdot BN=BC \cdot AN+AN \cdot PS $ . ... $ (2) $

Since $ CX=\frac{AC}{AB+AC} \cdot BC , CM'=\frac{CA}{NA} \cdot BN $ ,

so from $ (2) $ we get $ \frac {CX}{CM'}=\frac{BC}{BC+PS}=\frac{CK}{CS} $ . ie. $ KX \parallel SM' $

Similarly, we can prove $ LX \parallel SM' $ ,

so we get $ \frac{KX}{LX}=\frac{SM' \cdot \frac{CX}{CM'}}{TN' \cdot \frac {BX}{BN'}}=\frac{AM' \cdot \frac{AC}{CM'}}{AN' \cdot \frac{AB}{BN'}}=\frac{AC \cdot \frac{AB}{BN}}{AB \cdot \frac{AC}{CM}}=1 $ . ie. $ XK=XL $

Since $ \angle AXK=180^{\circ} - \angle MAX=180^{\circ}-\angle XAN=\angle AXL $ ,
so we get $ \triangle AXK \cong \triangle AXL $ and $ AK=AL $ .

Q.E.D

Solution 2 (By Luis González). Let $D$ be the midpoint of the arc $BC$ of $(O).$ $AD,AM$ cut $BC$ at $E,F,$ resp. $\widehat{BPD}=\widehat{BND}=\widehat{CBD}$ $\Longrightarrow$ $\odot(PDB)$ is tangent to $BC.$  If $U \equiv AC \cap PS,$ then $SPU \parallel BC$ and $PMB$ is antiparallel to $AU$ WRT $SA,SU.$ Thus if the isogonal of $SC$ WRT $\angle ASU$ cuts $AU$ at $J,$ from Steiner theorem we get

$\frac{KM}{KP}=\frac{JU}{JA}=\frac{SU^2}{SA^2} \cdot \frac{CA}{CU}=\frac{FC^2}{FA^2} \cdot \frac{FA}{FS}=\frac{FC^2}{FA \cdot FS}.$ But $\frac{FS}{BP}=\frac{FS}{CM}=\frac{FM}{BM}$

$\Longrightarrow \frac{KM}{KP}=\frac{FC^2 \cdot BM}{FA \cdot FM \cdot CM}=\frac{FC^2 \cdot BM}{FB \cdot FC \cdot CM}=\frac{FC}{FB} \cdot \frac{BM}{CM}=\frac{AC}{AB}=\frac{EC}{EB}$

$\Longrightarrow \frac{KM \cdot KB}{KB \cdot KP}=\frac{EB \cdot EC}{EB^2}.$

Thus, ratio of the powers of $K,E$ WRT $(O),\odot(PDB)$ are equal $\Longrightarrow$ $K,E$ lie on a circle coaxal with $(O)$ and $\odot(PDB),$ i.e. $K \in \odot(BED)$ and similarly $L \in \odot(CED).$ Further, $\odot(BED) \cup K$ and $\odot(CED) \cup L$ are symmetric about $AD$ because $\widehat{DBE}=\widehat{DCE}$ and $\widehat{EDK}=\widehat{EDL}=\pi-\widehat{CBM},$ i.e. $K,L$ are symmetric about $AD$ $\Longrightarrow$ $AK=AL.$

Solution 3 (By Tran Quang Hung). Let $AM,AN$ cut $BC$ at $E,F$. We have $\angle SMP=\angle BMA=\angle BNA$ and $\angle MSP=\angle AMN=\angle ABN$. From this $\triangle MSP\sim\triangle NBA$. Bisector of angle $A$ cuts $BC$ at $D$. We have $\frac{KS}{KC}=\frac{SP}{BC}=\frac{SP}{MP}.\frac{MP}{BC}=\frac{AB}{AN}.\frac{BP-BM}{BC}=\frac{AB}{AN}.\frac{BN-CN}{BC}$. We will prove that $\frac{KS}{KC}=\frac{DE}{DC}$ by the way to show $\frac{AB}{AN}.\frac{BN-CN}{BC}=\frac{DE}{DC}$, it is equivalent to

$\frac{AB.BN-AB.CN}{AN.BC}=\frac{DE}{DC}$

$\frac{AB.BN-AN.BC+AC.BN}{AN.BC}=\frac{DE}{DC}$ (Apply Ptolemy's theorem)

$\frac{(AB+AC)BN}{AN.BC}=\frac{DE}{DC}+1$

$\frac{(AB+AC)BN}{AN.BC}=\frac{CE}{CD}$

$\frac{BN}{AN}CE=\frac{CD.BC}{AB+AC}$ (Because of$\triangle ANB\sim\triangle ACE$)

$\frac{AC}{BC}=\frac{CD}{AB+AC}$.

The last equality is true from $AD$ is bisector of $ABC$. Thus, $\frac{KS}{KC}=\frac{DE}{DC}$ deduced $KD\parallel AE$. Similarly, $LD\parallel AF$. From this we have
$\frac{DK}{DL}=\frac{DK}{SE}.\frac{SE}{TF}.\frac{TF}{DL}=\frac{CD}{CE}.\frac{AE}{AF}.\frac{FB}{BD}=\frac{CD}{BD}.\frac{FB}{CE}.\frac{AE}{AF}\quad (1)$.

Note that, $\triangle ACE\sim\triangle ANB$ và $\triangle ABF\sim\triangle AMC$ therefore $\frac{CE}{BF}=\frac{CE}{BN}.\frac{CM}{BF}=\frac{AE}{AB}.\frac{AC}{AF}=\frac{DC}{DB}.\frac{AE}{AF}\quad (2)$.

From $(1),(2)$ we have $\frac{DK}{DL}=1$ hay $DK=DL$. Easily seen $\angle ADK=180^\circ-\angle DAE=180^\circ-\angle DAF=\angle ADL$. From this $\triangle ADK=\triangle ADL\ (c.g.c)$ deduced $AK=AL$.

Reference.


 

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