$ABCD$ is a parallelogram with $\angle BAD \neq 90$. Circle centered at $A$ radius $BA$ denoted as $\omega _1$ intersects the extended side of $AB,CB$ at points $E,F$ respectively. Suppose the circle centered at $D$ with radius $DA$, denoted as $\omega _2$, intersects $AD,CD$ at points $M,N$ respectively. Suppose $EN,FM$ intersects at $G$, $AG$ intersects $ME$ at point $T$. $MF$ intersects $\omega _1$ at $Q \neq F$, and $EN$ intersects $\omega _2$ at $P \neq N$. Prove that $G,P,T,Q$ concyclic.
Problem
Source: CSMO 2019 Grade 11 Problem 2
Tags: geometry
MarkBcc168
30.07.2019 09:51
Compare the configuration with 2014 Taiwan TST Round 1 P4.
NahTan123xyz
30.07.2019 17:39
Let $X\neq A$ be the second intersection of $\Gamma_1$ and $\Gamma_2$. Observe that $E,X,T,M$ are, respectively, the reflections of $A$ across point $B$, line $BD$, midpoint of $BD$, point $D$. Hence, $E,X,T,M$ are collinear. We claim that line $AGT$ is the internal angle-bisector of $\angle EAM$.
Let $a,b$ denote the radius of $\Gamma_1, \Gamma_2$ respectively.
Assume that $EN$ intersects $AM$ at $U$; $MF$ intersects $AE$ at $V$.
Since $\ \dfrac{AU}{UD}\ =\ \dfrac{AE}{DN}\ =\ \dfrac{2a}{b}$, and $\ AD\ =\ DM\ =\ b,$
we obtain that $$\dfrac{AU}{UM}\ =\ \dfrac{2a}{2a+2b}\ =\ \dfrac{a}{a+b}.$$Analogously, $$\dfrac{AV}{VB}\ =\ \dfrac{b}{a+b}.$$By Ceva Thm, we obtain that $$\dfrac{ET}{TM}\ =\ \dfrac{a}{b}\ =\ \dfrac{AE}{AM}.$$This concludes our claim.
Assume that $\angle ABD\ =\ \alpha, \angle ADB\ =\ \beta$.
We have $$\angle XTG\ =\ \dfrac{\pi-\alpha+\beta}{2}$$Also,
\begin{align*}
\angle XPE\ &=\ \angle XMN
\\ &=\ \beta+\angle DMN
\\ &=\ \beta+\dfrac{\pi-\alpha-\beta}{2}\ =\ \dfrac{\pi-\alpha+\beta}{2}
\end{align*}
Hence, $\square XTGP$ is concyclic. Analogously, $\square XTGQ$ is concyclic. In summary, the points $G,P,T,Q$ are concyclic as desired.
f6700417
27.10.2022 05:26
Is there a logical mistake ?
Assume that $\odot B \cap \odot D=K$. We have $\angle EKA=\angle MKA=\frac{\pi}{2}$.
Thus, $E, K, C, M$ are collinear.
We have $\triangle ABF \sim \triangle NDA$. Thus $\frac{AF}{NA}=\frac{AB}{ND}=\frac{FB}{BC}$.
Obviously $F, A, N$ are collinear, $EF//MN$.
Since $\triangle AFE \sim \triangle ANM$, we have $\frac{FA}{AN}=\frac{EF}{MN}=\frac{FG}{GM}$.
Hence, $AG//MN//EF$.
We have $P, K, T, G$ ;$Q, K, T, G$are concyclic. $\square$