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Source: HSGS TST 2014-2015, Viet Nam

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3212 posts

#1 h Oct 18, 2014, 10:22 AM • 3 Y

Y by buratinogigle, Adventure10, Mango247

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 $AK=AL$ .

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2312 posts

#2 h Dec 4, 2014, 10:54 AM • 5 Y

Y by shinichiman, Pirkuliyev Rovsen, buratinogigle, Adventure10, and 1 other user

My solution:

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

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#3 h Dec 22, 2014, 9:25 PM • 4 Y

Y by wiseman, buratinogigle, Adventure10, Mango247

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 $AD$ $\Longrightarrow$ $AK=AL.$

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#4 h Dec 27, 2014, 7:17 AM • 5 Y

Y by shinichiman, Arab, AlastorMoody, Adventure10, Mango247

I proposed this problem for HSGS TST 2014, here is my solution

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 $\dfrac{KS}{KC}=\dfrac{SP}{BC}=\dfrac{SP}{MP}.\dfrac{MP}{BC}=\dfrac{AB}{AN}.\dfrac{BP-BM}{BC}=\dfrac{AB}{AN}.\dfrac{BN-CN}{BC}$ . We will prove that $\dfrac{KS}{KC}=\dfrac{DE}{DC}$ by the way to show $\dfrac{AB}{AN}.\dfrac{BN-CN}{BC}=\dfrac{DE}{DC}$ , it is equivalent to

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

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

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

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

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

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

The last equality is true from $AD$ is bisector of $ABC$ . Thus, $\dfrac{KS}{KC}=\dfrac{DE}{DC}$ deduced $KD\parallel AE$ . Similarly, $LD\parallel AF$ . From this we have

$\dfrac{DK}{DL}=\dfrac{DK}{SE}.\dfrac{SE}{TF}.\dfrac{TF}{DL}=\dfrac{CD}{CE}.\dfrac{AE}{AF}.\dfrac{FB}{BD}=\dfrac{CD}{BD}.\dfrac{FB}{CE}.\dfrac{AE}{AF}\quad (1)$ .

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

From $(1),(2)$ we have $\dfrac{DK}{DL}=1$ 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 $AK=AL$ .

Attachments:

Figure2580.pdf (11kb)

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