On a triangle $ABC$, points $D$, $E$, $F$ are given on the segments $BC$, $AC$, $AB$ respectively such that $DE \parallel AB$, $DF \parallel AC$ and $\frac{BD}{DC}=\frac{AB^2}{AC^2}$ holds. Let the circumcircle of $AEF$ meet $AD$ at $R$ and the line that is tangent to the circumcircle of $ABC$ at $A$ at $S$ again. Let the line $EF$ intersect $BC$ at $L$ and $SR$ at $T$. Prove that $SR$ bisects $AB$ if and only if $BS$ bisects $TL$.
Problem
Source: 2023 Turkey NMO 2nd Round P6
Tags: geometry, circumcircle, symmedian
VicKmath7
23.12.2023 00:02
We will show that if $SR$ bisects $AB$, then $BS$ bisects $TL$. It's easy to see by angle chasing that $R$ is midpoint of the $A$-symmedian chord and that $BFRD$, $CERD$ and $BFEC$ are cyclic. Let $AD \cap (ABC)=D'$ and $M$ be the midpoint of $AB$; then $MR$ is midline in $\triangle ABD'$, so $\angle AES=\angle ARS=\angle AD'B=\angle ACB$, thus $SE \parallel BC$.
Let $X=(AEF) \cap (ABC)$ be the Miquel point of $BFEC$. Obviously $AX, EF, BC$ concur at $L$. Furthermore, $\angle AXS=\angle AES=\angle ACB$, which means that $S, X, B$ are collinear. Moreover, $\angle SAX=\angle ABX=\angle ALE$ as $LBFX$ is cyclic, so $AS \parallel EF$ and $EASF$ is a cyclic isosceles trapezoid. We have $\angle SFX=\angle SAX=\angle ABX$, so $SF$ touches $(FXB)$ and thus $SF^2=SX \cdot SB=SA^2$. Hence, $SF=SA=AE$, which implies $\beta=2\gamma$.
It's easy to show that $\angle BRF=\angle FER+\angle RCB$, so $(BRC)$ and $(AEF)$ touch at $R$, which means that $FE, BC$ and the common tangent concur at $L$. Moreover, $RD$ is the angle bisector of $\angle BRC$, so $LR=LD$. We have that $DL=DE$ since $\beta=2\gamma$. In addition $SE=2AE \cos \gamma=2DF \cos \gamma=DE=DL$ by LoS in $\triangle DEF$, which yields that $LDES$ is a rhombus as $DL \parallel ES$. Hence, $LS=LR$, which combined with the fact that $LR$ touches $(AEF)$ yields that $ARXS$ is harmonic. Finally, projecting $(SR;XA)=-1$ through $S$ onto $EF$ yields that $BS$ bisects $TL$, as desired.
$\textbf{Remark.}$ I think that the second direction doesn't contain anything new, so I won't bother typing the same things in reverse order. Also, I proved that $E, M, X$ are collinear, but I think I didn't use it; quite rich configuration - I am not even sure if I have included something I didn't actually use.
xeroxia
25.12.2023 03:00
Same solution but I'm not good at projective geometry. So here is my very long solution. Let the line $SR$ intersect the line $AB$ at $M$, and the line $BS$ intersect the line $TL$ at $N$. From similarity, we have $AF = AB \cdot \frac{DC}{BD + DC}$ and $$AF \cdot AB = AB^2 \cdot \frac{DC}{BD + DC} \qquad (1)$$From similarity, we have $AE = AC \cdot \frac{BD}{BD + DC}$ and $$AE \cdot AC = AC^2 \cdot \frac{BD}{BD + DC} \qquad (2)$$If we take the ratio of $(1)$ and $(2)$, we get $$\frac{AF \cdot AB}{AE \cdot AC} = \frac{AB^2 \cdot DC}{AC^2 \cdot BD} = \frac{BD \cdot DC}{DC \cdot BD} = 1 \qquad (3)$$In this case, $BFEC$ is a cyclic quadrilateral. Since $\angle BAS = \angle ACB = \angle AFE$, it follows that $EF \parallel AS$. Since $SFEA$ is a cyclic quadrilateral, it is also an isosceles trapezoid. Therefore, $\triangle FSE \cong \triangle EAF \sim \triangle BAC$ and since $\angle FST = \angle FAR = \angle BAD$, the line $AD$ divides $BC$ in the same ratio as $ST$ divides $FE$. That is, $$\frac{FT}{FE} = \frac{BD}{BC} = \frac{BF}{BA} \qquad (4)$$From the similarities due to parallel lines, we have $$\frac{LF}{LE + LF} = \frac{FD}{EC + FD} = \frac{FD}{EC + AE} = \frac{FD}{AD} = \frac{BF}{BA} \qquad (5)$$$$\frac{NF}{AS} = \frac{BF}{BA} \qquad (6)$$$$\frac{FT}{AS} = \frac{FM}{MA} \qquad (7)$$ Let $LN= x$, $TN=y$, $NF = z$, $TE=w$. Then $FT=y-z$, $FE = y - z + w$, $LF = x + z$, and $LE = x + y + w$. Now let's first show that $AM = BM \Longrightarrow LN = NT$. Then show $LN = NT \Longrightarrow AM = BM$. If $AM = BM$, by rearranging equation $(7)$ we get $$\frac{FT}{AS} = \frac{FM}{MA} = \frac{\frac{BA}{2} - BF}{\frac{BA}{2}} = 1 - \frac{2 \cdot BF}{BA} = 1 - 2 \cdot \frac{NF}{AS} \Longrightarrow 2\cdot NF + FT = AS$$In this case, $AS = 2z + y-z = y + z$. If we write equalities $(4)$, $(5)$, $(6)$, we get $$\frac{y-z}{y-z + w} = \frac{x+z}{2x + y + z + w} = \frac{z}{y+z}$$If we cross-multiply the $1st$ and $3rd$ ratios, we get $$y^2 - z^2 = yz - z^2 + wz \Longrightarrow wz = y^2 - yz = y(y-z) \qquad (8)$$If we cross-multiply the $2nd$ and $3rd$ ratios, we get $$xy + xz + yz + z^2 = 2xz + yz + z^2 + wz \Longrightarrow wz = xy - xz = x(y-z) \qquad (9)$$Since $y=TN > FN = z$, by combining $(8)$ and $(9)$ we get $x=y$, that is $LN=TN$. $\blacksquare$ Now let's assume $LN=NT$. We will have $x=y$. Equalities $(4)$, $(5)$, $(6)$ can be written as $$\frac{y-z}{y-z + w} = \frac{y+z}{2y + y + z + w} = \frac{z}{AS}$$Using the property of proportion, if we subtract the $1st$ ratio from the $2nd$ ratio, we get $\frac{2z}{2y + 2z} = \frac{z}{y+z}$ which implies $AS = y + z$. If we rewrite $(6)$ and $(7)$, we have $\frac{NF}{AS} = \frac{z}{y+z} = \frac{BF}{BA}$ and $\frac{FT}{AS} = \frac{y - z}{y+z} = \frac{FM}{MA}$ If we add twice the $1st$ equality to the $2nd$ equality, we get $\frac{2z + y-z}{y+z} = 1 = \frac{2\cdot BF}{BA} + \frac{FM}{MA}$. Now let $BF=p$, $FM=q$, and $MA=r$. We will have $BA=p+q+r$. Substituting, we get $$1 = \frac{2p}{p+q+r} + \frac{q}{r} = \frac{2pr + pq + q^2 + qr}{pr + qr + r^2}$$After simplifying, we get $pq + pr + q^2 - r^2 = p(q+r) + (q-r)(q+r) = (p+q - r)(q+r) = 0 \Longrightarrow p+q = r $ which means $BM = MA$. $\blacksquare$ Note: In both conditions, it can be shown that $AS = AE = SF$. This would require a bit of angle calculation that $\angle ABC = 2 \angle ACB$. From the property of medians (Simedian) ($AB^2:AC^2 = BD:DC$ ratio), it can be seen that $AD$ is a median in triangle $ABC$. That is, using a drawing tool, you can draw a triangle with $\angle B = 2\angle C$, mark the point $D$ as the symmetric of the median to $BC$ with respect to the angle bisector of $BC$. Following the directions in the problem, you will complete the drawing.
SerdarBozdag
28.01.2024 00:25
$AD$ is symmedian. $AS\cap BC=P, M$ is the midpoint of $LT$, $N$ is the midpoint of $AB$. $K = (AFE) \cap (BAC)$. $AD \cap FE = J$. $AD \cap (ABC) = G$. $AFDE$ is parallelogram, so $J$ is the midpoint of $EF$. Because $AJ$ is the symmedian of $ABC$, $BCEF$ is cyclic. Thus, $\angle AFE = \angle ACB = \angle SAF \implies AS \parallel EF$. $-1 = (AS,AJ;AE,AF) = (S,R;E,F) $. In addition, $S$ and $A$ are reflections of each other across the perpendicular bisector of $EF$. This gives $JD = JA = JS$. Because $L$ is the midpoint of $PD$, $LF \cdot LE = LB \cdot LC = LD^2 = LS^2$ which implies $LS $ is tangent to $(AEF)$. Note that $L$ is the center of $(PSRD)$ with diameter $PD$. By radical axis theorem on $(AEF), (BFEC), (ABC)$, $K$ is on $AL$. $-1 = (SS,SR;SK,SA) = (L,T; SK \cap EF,\parallel_{LT}) \implies M$ is on $SK$. We know that $PR \perp AD$, so $R$ is the midpoint of $AG$ ( $PG$ is tangent to $(ABC)$ ). Now $$BS \textbf{ bisects } TL \iff \angle SKA =\angle C \iff \angle ARS = \angle C \iff \angle ARS = \angle AGB \iff \angle ARS = \angle ARN \iff SR \textbf{ bisects } AB$$
bin_sherlo
02.08.2024 16:46
Let $AS\cap BC=K, \ SB\cap (AEF)=U, \ SB\cap LT=N, \ SD\cap AB=M, \ AD\cap EF=Q$. Claim $1$: $B,C,E,F$ are cyclic and $AS\parallel EF$. Proof $1$: Since $AD$ is symedian in $ABC$ and median in $AEF,$ we get that $EF$ and $BC$ are antiparallel which gives that $(BCEF)$ is cyclic. $\angle AFE=\angle C=\angle FAS$ hence $\overline{ASK}\parallel \overline{EFL}$.$\square$ Claim $2$: $L$ is the miquel point of $SFDE$ and $L$ is the circumcenter of $(KDRS)$. Proof $2$: $\frac{LB}{LD}=\frac{LF}{LE}=\frac{LD}{LC}$ thus, $LD^2=LB.LC$ Also $(K,D;B,C)=-1$ hence $LK=LD$. Let $SE\cap DF=X$ and $SF\cap DE=Y$. We have \[\angle FYD=\angle FDE-\angle DFY=\angle A-180+2\angle B=\angle B-\angle C=\angle FLD\]Hence $(LFDY)$ is cyclic. \[\angle SXF=\angle ESF-\angle XFS=\angle A-180+2\angle B=\angle B-\angle C=\angle ELD\]$(LEDX)$ is cyclic. These two give that $L$ is the miquel point of $SDEF$. $\angle FSL=\angle DEL=\angle AFE=\angle C=\angle FES$ Thus, $LS$ is tangent to $(AEF)$. $LS^2=LE.LF=LB.LC=LD^2$ yields $LK=LD=LS$. $\angle SRA=\angle SEA=\angle FEA-\angle FES=\angle B-\angle C=\angle SKD$ gives that $S,K,D,R$ are cyclic and since $L$ is the circumcenter of $(KSD),$ we get $LK=LS=LR=LD$.$\square$ Claim $3$: $E,U,K$ are colliear. Proof $3$: Pascal at $EUSSAF$ gives the desired result.$\square$ Claim $4$: If $ST\cap BC=P,$ then $NL=NT\iff (K,B;L,P)=-1\iff (A,U;S,R)=-1\iff A,U,L$ are collinear. Proof $4$: By Menelaus at $TPL$, \[NL=NT\iff \frac{BL}{BP}=\frac{ST}{SP}\overset{TL\parallel SK}{=}\frac{KL}{KP}\iff -1=(K,B;L,P)=(SK,SB;SL,SP)=(A,U;S,R)\]\[(BC_{\infty},L;K,D)=-1=(A,U;S,R)=(AA,AU;AS,AR)=(BC_{\infty},AU\cap BC;K,D)\iff A,U,L \ \text{are collinear}\]Which gives that $NL=NT\iff A,U,L$ are collinear.$\square$ Claim $5$: $A,U,L$ are collinear $\iff \angle B=2\angle C\iff SE\parallel KC$. Proof $5$: When we invert from $A$ with radius $\sqrt{AF.AB}$, $U$ and $L$ swaps since $(AEF)$ and $BC$ swaps. Hence $U,L,F,B$ and $U,L,S,K$ are cyclic. (Inversion is unique hence we can use iff) \[\angle B-\angle C=\angle SKB=180-\angle LUS=\angle BUL=\angle BFL=\angle AFE=\angle C\]Thus, $A,U,L$ are collinear $\iff \angle B=2\angle C$ and $\angle SEA=\angle B-\angle C=\angle C=\angle BCA$ so $\angle B=2\angle C\iff SE\parallel KC$$\square$ Claim $6$: If $\angle B=2\angle C,$ then $MA=MB$. Proof $6$: $\angle ASE=\angle B-\angle C=\angle C=\angle SEA$ hence $AS=AE$. \[\angle MFQ=\angle AFE=\angle SRA=\angle MRQ\]yields $M,Q,F,R$ are cyclic. $\angle QMR=\angle QFR=\angle EFR=\angle ESR$ so $MQ\parallel SE\parallel \overline{BDC}$. Since $Q$ is the midpoint of $AD$, we get that $M$ is the midpoint of $AB$.$\square$ Claim $7$: If $MA=MB,$ then $\angle B=2\angle C$. Proof $7$: We have $MQ\parallel BC$ hence \[\angle AMQ=\angle B=\angle FRA=\angle FRQ\]which gives that $M,Q,R,F$ are cyclic. $\angle AFE=\angle MFQ=\angle MRQ=\angle ARS$ thus, $AS=AE\iff \angle C=\angle ASE=\angle SEA=\angle B-\angle C$ as desired.$\blacksquare$