Fairly standard configuration involving tangents

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Fairly standard configuration involving tangents

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Source: MEMO 2017 T6

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35 posts

abeker

#1 h Aug 25, 2017, 4:33 PM • 1 Y

Y by Adventure10

Let $ABC$ be an acute-angled triangle with $AB \neq AC$ , circumcentre $O$ and circumcircle $\Gamma$ . Let the tangents to $\Gamma$ at $B$ and $C$ meet each other at $D$ , and let the line $AO$ intersect $BC$ at $E$ . Denote the midpoint of $BC$ by $M$ and let $AM$ meet $\Gamma$ again at $N \neq A$ . Finally, let $F \neq A$ be a point on $\Gamma$ such that $A, M, E$ and $F$ are concyclic. Prove that $FN$ bisects the segment $MD$ .

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63 posts

#3 h Aug 25, 2017, 8:37 PM • 1 Y

Y by Adventure10

Let the perpendicular from $A$ to $BC$ meet $\Gamma$ at $H$ . Let $P \equiv AD \cap \Gamma$ and let $A’ \equiv AO \cap \Gamma$ .

Since $\angle EMF = \angle EAF = \frac{ \overarc{CF} + \overarc{A’C}}{2} = \frac{ \overarc{CF} + \overarc{BH}}{2}$ we get that $F$ , $M$ and $H$ are collinear.

On the other hand $\angle FEC = \angle MAF= \frac{ \overarc{CF} + \overarc{CN}}{2} = \frac{ \overarc{CF} + \overarc{BP}}{2}$ , thus $P$ , $E$ and $F$ are collinear.

$\angle BFM= \angle CFA’$ , thus $FA’$ is the symmedian of $\angle BFC$ in triangle $BFC$ , thus $F$ , $A’$ and $D$ are collinear. On the other hand $\angle A’FH= \angle A’AH= \angle A’OM$ , furthermore $FOMA’$ is cyclic. Notice that if we prove that $\tfrac{MF}{FD} = \tfrac{NA’}{NH}$ we are done by the general angle bisector theorem in triangle $MFD$ . Also $\tfrac{MF}{FD} = \tfrac{OA’}{OD}$ . Thus if we prove that $\tfrac{OA’}{NA’} = \tfrac{OD}{NH}$ we are done.

$\frac{OA’}{NA’} = \frac{ \cos \angle NAA’}{ \sin 2 \angle NAA’} = \frac{1}{2 \sin \angle NAA’} = \frac{1}{2 \sin \angle ADO} = \frac{ AO }{AA’ \sin \angle ADO} = \frac{ OD \sin \angle DAO}{ A’P \sin \angle DAO} = \frac{OD}{A’P} = \frac{OD}{NH}$

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241 posts

#5 h Aug 25, 2017, 9:38 PM • 4 Y

Y by tenplusten, mijail, Adventure10, Mango247

Let $G=AD\cap \Gamma$ , $A'=AO\cap \Gamma$ , $F'=DA'\cap \Gamma$ , $P=AF'\cap GA'$

1) Prove that $GMEA'$ is cyclic.
So from $DG$ is the G-symmedian of $\triangle BGC$ $\Longrightarrow$ $\measuredangle MEA$ $=$ $\measuredangle CAE$ $+$ $\measuredangle ECA$ $=$ $\measuredangle BGA$ $+$ $\measuredangle CGA'$ $=$ $\measuredangle CGM$ $+$ $\measuredangle CGA'$ $=$ $\measuredangle MGA'$ , hence $GMEA'$ is cyclic.

2) Prove that $P\in BC$ and $E\in F'G$ .
By pole and polars' theorem we get $AA'\cap GF'$ and $AF'\cap GA'$ lie in $BC$ , hence $P\in BC$ and $E\in F'G$ .

3) Prove that $F'=F$
Since $GMEA'$ is cyclic and $AF'$ , $ME$ , $GA'$ are concurrent in $P$ we get $PF'.PA=PA'.PG=PE.PM$ $\Longrightarrow$ $AF'EM$ is cyclic, hence $F'=F$ .

4) Prove that $GMFD$ is a harmonic quadrilateral.
It's so easy prove that $OF^2=OG^2=OA^2=OM.OD$ $\Longrightarrow$ $\measuredangle MFG$ $=$ $\measuredangle OAM$ $=$ $\measuredangle MDG$ , hence $GMFD$ is cyclic and $OG$ and $OF$ are tangent to $\odot (GMDF)$ and $O$ , $M$ , $D$ are collinear, hence $GMFD$ is a harmonic quadrilateral.

Finally, since $GMFD$ is a harmonic quadrilateral we get $FG$ is F-symmedian of $\triangle FMD$ , but $\measuredangle MFG$ $=$ $\measuredangle MAE$ $=$ $\measuredangle DFN$ $\Longrightarrow$ $FN$ is F-median of $\triangle FMD$ , hence $FN$ bisects the segment $MD$ .

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1980 posts

#6 h Aug 25, 2017, 10:12 PM • 1 Y

Y by Adventure10

abeker wrote:

Let $ABC$ be an acute-angled triangle with $AB \neq AC$ , circumcentre $O$ and circumcircle $\Gamma$ . Let the tangents to $\Gamma$ at $B$ and $C$ meet each other at $D$ , and let the line $AO$ intersect $BC$ at $E$ . Denote the midpoint of $BC$ by $M$ and let $AM$ meet $\Gamma$ again at $N \neq A$ . Finally, let $F \neq A$ be a point on $\Gamma$ such that $A, M, E$ and $F$ are concyclic. Prove that $FN$ bisects the segment $MD$ .

Let $A_0$ lie on $\odot(ABC)$ such that $\overline{AA_0} \perp \overline{BC}$ and $M_0$ be symmetric to $M$ in point $O$ . Let $\Gamma$ denote inversion about circle $\odot(BC)$ followed by point-symmetry at $M$ . Let $K$ be the midpoint of $\overline{MD}$ . Notice that

$AA_0MM_0$ is an isosceles trapezoid, hence it is cyclic;
$\measuredangle AFM=\measuredangle AEM=\measuredangle AFA_0$ so $A_0$ lies on $\overline{MF}$ ;
$\Gamma$ switches each of the pairs $\{A, N\}, \{A_0, F\}, \{M_0, K\}$ .

Consequently, we see $K, N,$ and $F$ are collinear. $\blacksquare$

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464 posts

#7 h Nov 30, 2017, 3:10 PM • 1 Y

Y by Adventure10

hehe here is solution of LBT : by symmetry problem : triangle ABC is in (O). AO cut (O) = X, M is midpoint of BC,XM cut (O)=Y, Tagent of (O) at B and C meet each other at D,AD cut (O)=P. Prove YP bisects MD

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464 posts

#8 h Nov 30, 2017, 3:15 PM • 1 Y

Y by Adventure10

H is othorcentre of ABC , X,Y,M,H collinear , AH cut BC at G,cut (O) at S. BH ,CH cut AC,AB at M,N. YA,MN,BC concurrent at T (radical ). Y(AGCB)=-1 ==> Y,G,P collinear , A(YSBC)=-1 , Y,S,D collinear. HS//MD , GH=GS... .WED

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1740 posts

#9 h Apr 13, 2020, 6:18 AM • 1 Y

Y by Mango247

Solved with jclash and nukelauncher.

$[asy] size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,O,Ap,EE,M,D,NN,F,X,Fp,HA,K,SS; A=dir(120); B=dir(215); C=dir(325); O=(0,0); Ap=2O-A; EE=extension(A,O,B,C); M=(B+C)/2; D=-1/M; NN=2*foot(O,A,M)-A; F=2*foot(O,D,Ap)-Ap; X=(D+M)/2; Fp=reflect(O,D)*F; HA=-B*C/A; K=extension(A,D,X,Fp); SS=foot(A,B,C); draw(X--Fp--D,gray); draw(X--F--D,gray); draw(O--D,Dotted); draw(circumcircle(A,EE,F),dashed); draw(A--M,dashed); draw(circumcircle(A,B,C)); draw(Ap--A--HA); draw(A--B--C--A--D); draw(B--D--C); dot("$A$",A,NW); dot("$B$",B,B); dot("$C$",C,C); dot("$O$",O,NE); dot("$A'$",Ap,dir(165)); dot("$E$",EE,dir(240)); dot("$M$",M,NE); dot("$D$",D,S); dot("$F$",F,E); dot("$X$",X,SE); dot("$F'$",Fp,W); dot("$H_A$",HA,NE); dot("$K$",K,SW); dot("$S$",SS,NE); [/asy]$

Let $A'$ be the antipode of $A$ , let $S$ be the foot from $A$ to $\overline{BC}$ , let $H_A=\overline{AS}\cap(ABC)$ , let $K=\overline{AD}\cap(ABC)$ , let $F'=\overline{DH_A}\cap(ABC)$ .

Claim: $\overline{FF'}\parallel\overline{BC}$ .

Proof. Let $T=\overline{AF}\cap\overline{BC}$ . Then $TB\cdot TC=TA\cdot TF=TM\cdot TE$ , so $-1=(BC;ET)\stackrel A=(BC;A'F).$ But $-1=(BC;H_AF')$ , so the claim is true. $\blacksquare$

Claim: $S$ , $K$ , $X$ collinear.

Proof. Let $L$ be the midpoint of $\overline{AK}$ , let $Y$ be the point so that $ABCY$ is an isosceles trapezoid, and let $Z$ be the reflection of $A$ over $\overline{BC}$ . Note that $K$ , $M$ , $Y$ collinear, and since $BYCZ$ is a parallelogram, $K$ , $M$ , $Z$ collinear.

But by homothety $(A,2)$ , we have $\overline{SL}\parallel\overline{ZK}=\overline{KM}$ , so $\triangle ASL\sim\triangle DKM$ . In particular, $\frac{AK}{KD}=\frac{2AL}{KD}=\frac{2AS}{MD}=\frac{AS}{XD},$ so $S$ , $K$ , $X$ collinear, as claimed. $\blacksquare$

By reflection over $\overline{OD}$ , it suffices to prove $X$ , $K$ , $F'$ collinear. By Brokard's theorem, $\overline{AH_A}\cap\overline{KF'}$ lies on the polar of $\overline{AK}\cap\overline{H_AF'}=D$ , or $\overline{BC}$ . Thus $X$ , $K$ , $S$ , $F'$ are all collinear, and we are done.

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202 posts

#10 h Apr 13, 2020, 4:52 PM

Y by

I'm surprised nobody complex-bash it because it's fairly easy.

First let's make some synthetic observations.

Let $\{T\}=AF\cap BC$ and $A'$ the antipode of $A$ .By the power of point we have $TB\cdot TC=TA\cdot TF=TM\cdot TE$ .So there is a inversion of pole $T$ which swaps $\{B,C\}$ , $\{M,E\}$ , $\{T,\infty\}$ . By the famous fact that inversion preserves cross-ratio we have $-1=(B;C;M;\infty)=(C;B;E;T)\overset{A}{=}(C,B,A',F)$ so $FBA'C$ is harmonic which means $F,A',D$ are collinear.

Now we are ready to bash!!!

As usual let $(ABC)$ be the unit circle and we denote by $x$ the affixe of $X$ .Also set $OD$ to be the real axis,so $b=\overset{-}{c}$ , $d=\frac{2}{b+c}=\frac{1}{m}$ and $t=\frac{d+\frac{1}{d}}{2}=\frac{d^2+1}{2d}$ ,where $T$ is the midpoint of $MD$ .

We use the following lemma.
Lemma:Let P be a point on the unit circle and X be other point.Then the affixe of $P'$ , the second intersection of $PX$ with the unit circle is $\frac{p-x}{p\overset {-}{x}-1}$
Proof:Click to reveal hidden text
With the lemma we easily obtain $f=\frac{a+d}{ad+1}$ ,and $n=\frac{a-m}{\overset{-}{m}a-1}=\frac{ad-1}{a-d}$ .

Now,we have $\frac{f-n}{f-t}=\frac{\frac{a+d}{ad+1}-\frac{ad-1}{a-d}}{\frac{a+d}{ad+1}-\frac{d^2+1}{2d}}=\frac{\frac{a^2-d^2-a^2d^2+1}{(ad+1)(a-d)}}{\frac{2ad+2d^2-ad^3-ad-d^2-1}{2d(ad+1)}}=2d\cdot \frac{a^2(1-d^2)+1-d^2}{(a-d)(d^2-1+ad(1-d^2))}=2d\cdot \frac{a^2+1}{(a-d)(ad-1)}=2d\cdot \frac{\frac{1}{a^2}+1}{(\frac{1}{a}-d)(\frac{d}{a}-1)}=2d\cdot \frac{\overline{a^{2}}+1}{ \overline{(a-d)}\overline{(ad-1)}}=\overline{2d\cdot \frac{a^2+1}{(a-d)(ad-1)}}$ so $\frac{f-n}{f-t} \in \mathbb{R}$ which means that $FN$ bisects $MD$ . $\square$

This post has been edited 1 time. Last edited by Doru2718, Jul 23, 2021, 10:24 AM
Reason: .

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256 posts

#11 h Aug 26, 2020, 3:45 PM • 1 Y

Y by mkomisarova

Let $A'$ is antipode of $A$ with respect to $\odot(ABC)$ . Assume that $FN \cap MD = G$ .

Claim 1: Quadrilateral $AOND$ is cyclic
Proof: Clearly quadrilateral $OBDC$ is cyclic. From Pop we deduce that:
$OM \cdot MD = BM \cdot MC = AM \cdot MN$ This proves that $OBDC$ is cyclic as desired.

Claim 2: Points $F, A',D$ are collinear
Proof: Consider $\sqrt{bc}$ inversion followed by reflection over angle bisector of $\angle BAC$ . The claim transforms into:

Quote:

Given triangle $\triangle ABC$ with orthocenter $H$ and Humpty point $H_A$ . Let $M$ be a intersection of $\odot(ABC)$ and $A$ symmedian. Assume that altitude from $A$ intersects $BC$ at point $D$ and $\odot(ABC)$ at point $E$ . Let $F$ be intersection of $ME$ and $BC$ . Prove that quadrilateral $AFDH_A$ is cyclic.

Clearly $-1 =(M,A;B,C) \overset {E}{=} (F,D;B,C)$ , therefore $F$ is $A$ Ex - point. Consequently points $F,H,H_A$ are collinear. But note that $\angle FH_AA = \angle FDA = 90$ , which implies that $AFDH_A$ is cyclic as desired.

Now we go back to the main problem. Note that:
$\angle ADO = \angle ODN = \angle OAN =\angle ONA = \angle NFA'= \angle NFD = x$ This implies that $GD$ is tangent to $\odot(FND)$ , therefore $GD^ 2 = GN \cdot GF$ .

Claim 3: Quadrilateral $OMA'F$ is cyclic
Proof: From PoP we deduce that:
$DA' \cdot DF = DB^ 2 = DM \cdot DO$ This implies that $OMA'F$ is cyclic as desired.

Claim 4: $\angle GMN = \angle MFN$
Sketch of Proof: Note that:
$\angle MFN = \angle MFA'- \angle NFD = \angle MOA'-x =\angle B - \angle C - x$ On another hand:
$\angle GMN = \angle DAA'= \angle A - 2 \angle BAD + \angle OAN = \angle B - \angle C -x$
Now we deduce that $MG$ is tangent to $\odot(MFN)$ , therefore:
$GM^ 2 = GF \cdot GN = GD^2$ This implies that $FN$ bisects $MD$ as desired.

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1071 posts

gghx

#12 h Jun 26, 2021, 2:34 AM

Y by

Interestingly the first solution that is neither a bash nor involves any projective geometry.

Easy to see $O,M,D$ are collinear. Also, $AOND$ is cyclic by radax.
Let $FE \cap \Gamma=K$ . By Reim, $KN//ME \implies BK=NC \implies AK$ is a symmedian so $AKD$ is collinear.

Let $A'$ be the $A$ antipode.
$\angle FKD = 180-\angle AKF = 180- \angle AA'F = 90+\angle FAA' = 90+\angle FAE = 90+\angle FME = \angle DMF$ so $DKMF$ is cyclic.
Now $\angle KFD = \angle KMD = \angle NMT = \angle OMA = 90-\angle AMB = 90-\angle ACK = 90-\angle AA'K = \angle KAA' = \angle KFA'$ , which means $FA'D$ collinear.

Let $T$ be the intersection of $FN$ with $MD$ .

Lastly, $\angle TDN = \angle MAO = \angle NAA' = \angle NFA' = \angle TFD$ and $\angle MFT = \angle MFE + \angle KFN = \angle NAA'+\angle KAN=\angle KAA'=\angle KFD=\angle KMD=\angle TMN$ . Thus, $TM^2=TN \times TF=TD^2$

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#13 h Nov 7, 2021, 7:45 PM

Y by

$N$ is humpty point in $\bigtriangleup FMD$ and done

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#14 h Jan 12, 2022, 1:37 PM

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The diametrical opposite of $A$ lies at $DF,FE,AD$ and $(ABC)$ intersect at one point.

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