Given $\vartriangle ABC$ with circumcenter $O$. Let $D$ be a point satisfying $\angle ABD = \angle DCA$ and $M$ be the midpoint of $AD$. Suppose that $BM,CM$ intersect circle $(O)$ at another points $E, F$, respectively. Let $P$ be a point on $EF$ so that $AP$ is tangent to circle $(O)$. Prove that $A, P,M,O$ are concyclic.
Problem
Source: https://artofproblemsolving.com/community/c6h1740077p11309077
Tags: geometry, circumcircle, Concyclic, equal angles
parmenides51
21.03.2020 12:41
Solution :
Lemma :
$\triangle ABC.$ $D,E$ lie on $(ABC)$ such that $DBEC$ is a harmonic quadrilateral. An arbitrary circle through $A, D$ intersect $AB, AC$ again at $N, M.$ $EM$ intersect $(ABC)$ again at $K.$ then $BK$ bisects $MN.$
Solution of lemma :
Let $MD$ cuts $(ABC)$ again at $L,$ then by Reim theorem $MN \parallel BL$
So $-1=(D,B;E,C)\overset{M}{=}(L,BM \cap (ABC);K,A)=B(L,M;K,N)$ so $BK$ bisects $MN$
Back to the main problem :
Construct parallelogram $AIDJ$ with $I, J \in AC, AB$ resp. Let $K$ be the reflection of $D$ wrt $IJ,$ then $AIJK$ is a isosceles trapezoid
Since $\triangle DJB \sim \triangle DIC,$ $\frac{JB}{IC}=\frac{DJ}{DI}=\frac{KJ}{JI} \Rightarrow$ $K$ is the center of the spiral similarity that map $JI$ to $BC \Rightarrow K \in (O)$
Let $S = (AD \cap (O)) \neq A,$ $SI$ meets $(O)$ again at $E',$ then by lemma $BE'$ passes through the midpoint of $IJ$ which is $M \Rightarrow E' \equiv E$
Anagously $CF$ passes through $M$
We have $\frac{AI}{AJ}=\frac{DJ}{DI}=\frac{JB}{IC} \Rightarrow JA.JB=IA.IC \Rightarrow \mathcal{P}_{I/(O)}=\mathcal{P}_{J/(O)} \Rightarrow OI=OJ,$ combine with $MI=MJ \Rightarrow OM \perp IJ$
It suffices to prove $IJ$ passes through $P$
Let $KM$ meets $(O)$ again at $L$
Since inversion preserve cross ratio, $(L,E;A,F)=(K,B;S,C)=-1,$ so $PL$ is tangent to $(O) \Rightarrow A, P, L, O$ are concyclic
Also $S(L,I;M,J)=(L,E;A,F)=-1,$ combine with $MI=MJ :,$ $SL \parallel AK$
$\angle AML = 2m(\overarc{AL})=\angle AOL$ so $M$ lies on $APLO$
Since $PA=PL,$ $MP$ is angle bisector of $\angle AML,$ so $P$ lies on $IJ.$
Q.E.D
Dianademon
09.02.2021 11:09
parmenides51 wrote:
Solution :
Lemma :
$\triangle ABC.$ $D,E$ lie on $(ABC)$ such that $DBEC$ is a harmonic quadrilateral. An arbitrary circle through $A, D$ intersect $AB, AC$ again at $N, M.$ $EM$ intersect $(ABC)$ again at $K.$ then $BK$ bisects $MN.$
Solution of lemma :
Let $MD$ cuts $(ABC)$ again at $L,$ then by Reim theorem $MN \parallel BL$
So $-1=(D,B;E,C)\overset{M}{=}(L,BM \cap (ABC);K,A)=B(L,M;K,N)$ so $BK$ bisects $MN$
Back to the main problem :
Construct parallelogram $AIDJ$ with $I, J \in AC, AB$ resp. Let $K$ be the reflection of $D$ wrt $IJ,$ then $AIJK$ is a isosceles trapezoid
Since $\triangle DJB \sim \triangle DIC,$ $\frac{JB}{IC}=\frac{DJ}{DI}=\frac{KJ}{JI} \Rightarrow$ $K$ is the center of the spiral similarity that map $JI$ to $BC \Rightarrow K \in (O)$
Let $S = (AD \cap (O)) \neq A,$ $SI$ meets $(O)$ again at $E',$ then by lemma $BE'$ passes through the midpoint of $IJ$ which is $M \Rightarrow E' \equiv E$
Anagously $CF$ passes through $M$
We have $\frac{AI}{AJ}=\frac{DJ}{DI}=\frac{JB}{IC} \Rightarrow JA.JB=IA.IC \Rightarrow \mathcal{P}_{I/(O)}=\mathcal{P}_{J/(O)} \Rightarrow OI=OJ,$ combine with $MI=MJ \Rightarrow OM \perp IJ$
It suffices to prove $IJ$ passes through $P$
Let $KM$ meets $(O)$ again at $L$
Since inversion preserve cross ratio, $(L,E;A,F)=(K,B;S,C)=-1,$ so $PL$ is tangent to $(O) \Rightarrow A, P, L, O$ are concyclic
Also $S(L,I;M,J)=(L,E;A,F)=-1,$ combine with $MI=MJ :,$ $SL \parallel AK$
$\angle AML = 2m(\overarc{AL})=\angle AOL$ so $M$ lies on $APLO$
Since $PA=PL,$ $MP$ is angle bisector of $\angle AML,$ so $P$ lies on $IJ.$
Q.E.D
But BE,CF passes through M is condition
IndoMathXdZ
06.03.2021 20:27
What a rich configuration. First of all, we define the following set of points: Define $X = AB \cap CD, Y = AC \cap BD, H = XY \cap BC, I = AD \cap BC, J = AD \cap (ABC), K = AH \cap (ABC), G = KM \cap (ABC)$. Claim 01. $BXYC$ is cyclic. Proof. Notice that $\measuredangle XBY \equiv \measuredangle ABD = \measuredangle DCA \equiv \measuredangle XCY$. Claim 02. $DK \perp AH$. Proof. Well known. Claim 03. $MG = MJ$. Proof. From our previous claim, we have $\angle DKA = 90^{\circ}$ and $M$ being midpoint of $AD$. This forces $MA = MK$. Furthermore, by power of point, \[ MJ \cdot MA = \text{Pow}_M (ABC) = MK \cdot MG \Rightarrow MJ = MG \] Claim 04. $AMOG$ cyclic. Proof. Since $MG = MJ$, we conclude that $AK \parallel GJ$. Therefore, \[ \angle AMG = \angle KMD = 2 \angle KAD \equiv 2 \angle KAJ = 2 \angle AJG = 2 \angle ACG = \angle AOG \] Claim 05. $AEGF$ is a harmonic quadrilateral. Proof. We have \[ -1 = (C,B;I,H) \overset{A}{=} (C,B;J,K) \overset{M}{=} (F,E;A,G) \]and hence the conclusion. Claim 06. $APMO$ is cyclic. Proof. To finish this, just notice that since $AEGF$ is a harmonic quadrilateral, tangent of $A$ and $C$ on $(ABC)$, and $EF$ concur at a point, which forces $P$ lies on the tangent of $G$. Therefore, $PA$ and $PG$ tangent to $(ABC)$, and therefore $PAOG$ is cyclic. Since we already have $AMOG$ being cyclic. $A,P,M,O,G$ are cyclic.
Motivational Remark. Another fresh example of dissecting the figure before looking at the whole picture. The definition of $P$ could be first omitted, and we could focus more on the base configuration instead. The condition $\angle ABD = \angle DCA$ could be rephrased into $BXYC$ being cyclic, and $D = BY \cap XC$ quite naturally. From here, we already get a "nearly" complete quadrilateral. Therefore, we would want to complete the picture, by adding point $H$ and point $K$. This further implies $\angle DKA = 90^{\circ}$ from a well-known complete quadrilateral property. Now, the definition of points $E$ and $F$ is quite annoying, as they involve lines intersecting circle, where the line is somehow annoying to deal with. Therefore, this inspired the projective approach, which is motivated more after seeing $P = AA \cap EF$. Knowing that we need to have $\angle PAO -= 90^{\circ}$, we draw another tangent from $P$ to $(ABC)$, which is point $G$. Therefore, instead of proving $APMO$ cyclic, we could just prove $AMOG$ being cyclic, and $AEFG$ being harmonic quadrilateral. Now, I rephrase the question into:
Rephrased problem wrote:
Suppose $G = (AMO) \cap (ABC)$, prove that $AEFG$ is a harmonic quadrilateral.
Of course, I need to prove $(A,G;E,F) = -1$. Since $E = BM \cap (ABC)$ and $F = CM \cap (ABC)$, we project this wrt $M$ and get $(J,GM \cap (ABC); B,C) = -1$. Combining with the obvious harmonic bundle obtained from the complete quadrilateral configuration, it suffices to prove that $K,M,G$ are collinear. Therefore, we once again rephrase the question into an equivalent one:
Rephrased problem II wrote:
Suppose $G = (AMO) \cap (ABC)$, prove that $G,M,K$ are collinear.
I think it's easier to prove $A,M,O,G$ cyclic instead of $G,M,K$ collinear. So, I use "phantom" point here, and wanted to prove $AMOG$ is cyclic if $G,M,K$ are collinear. Now, a little angle chasing forces you to prove that the length of arc $AG$ and $KJ$ are the same, or in other words, $AK \parallel GJ$. However, we have $\angle DKA = 90^{\circ}$, which means $MA = MK$, and therefore by POP, $MG = MJ$, and we are done.
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