For the quadrilateral $ABCD$, let $AC$ and $BD$ intersect at $E$, $AB$ and $CD$ intersect at $F$, and $AD$ and $BC$ intersect at $G$. Additionally, let $W, X, Y$, and $Z$ be the points of symmetry to $E$ with respect to $AB, BC, CD,$ and $DA$ respectively. Prove that one of the intersection points of $\odot(FWY)$ and $\odot(GXZ)$ lies on the line $FG$. Proposed by chengbilly
Problem
Source: 2024 Taiwan TST Round 1 Independent Study 2-G
Tags: geometry
MathLuis
07.03.2024 08:45
Let $EX \cap BC=X'$ and $EZ \cap DA=Z'$, let $H$ the projection from $E$ to $FG$ and $M$ midpoint of $EG$ and let $N$ the midpoint of $HE$. Main Claim: $MNX'Z'$ is cyclic Proof: Note that $GHX'EZ'$ is cyclic with diameter $GE$ and center $M$, also from Menelaus-Ceva config we have $-1=(D, C; GE \cap CD, F) \overset{G}{=} (Z', X'; E, H)$. Therefore let the tangents from $Z',X'$ to $(GE)$ meet $HE$ at $T$, now clearly $MX'TZ'$ is cyclic with diameter $MT$, however we also have $\angle MNE=90$ therefore $MNX'TZ'$ is cyclic as desired. Finishing: Scale with center $E$ and radius 2 to get $GHXZ$ cyclic, and now analogously we can get $FHWY$ cyclic also therefore $(FWY), (GXZ)$ meet at $H$ which lies on $FG$ as desired thus we are done
Try to reverse-enginer this problem as much as you can until you find something that forces the entire problem (the cross ratio thing, which ofc its forcing).
Hakurei_Reimu
07.03.2024 15:02
(This solution is quite not intuitive, but it's very clean) Let $EW\cap AB=P, EW\cap CD=Q, EY\cap AB=R, EY\cap CD=S$. Claim: The intersection of $PS$ and $QR$ lies on $FG$. proof: Let $PS\cap QR=T, FE\cap AD=U, FE\cap PS=V$. By La Hire's Theorem(or Brocard's Theorem), $E$ lies on the polar of $F$ wrt $(PRQS)$, $(ABCD)$. Therefore, $(A,D;U,G)=-1$ and $(P,S;V,T)=-1$. Notice that $FA=FP, FD=FS, FU=FV$, so we must have $FG=FT$. Back to the main problem. It's obvious that $E$ is the orthocenter of $\triangle FQR$ and $PRQS$ is cyclic. By the refletion property of orthocenter, $W,Y$ are lying on $(FQR)$. Let the second intersection of $(FQR)$ and $(FPS)$ be $H$. Now, by radical axis theorem on $(FQR), (FPS)$ and $(PRQS)$, we have $PS$, $QR$ and $FH$ are concurrent. With the claim, we get $FH=FG$. Note that $\angle FQE=\dfrac\pi2=\angle FHE$, which implies the second intersection of $(FWY)\ (=(FQR))$ and $FG$ is the projection of $E$ on $FG$. Similarly, we can get the second intersection of $(GXZ)$ and $FG$ is also the projection of $E$ on $FG$. Hence, one of the intersection of $(FWY)$ and $(GXZ)$ lies on $FG$, which is $H$. $\blacksquare$
VicKmath7
07.03.2024 15:41
We claim that both circles pass through the foot of the perpendicular from $E$ to $FG$, let it be $T$; we ignore $W, Y$ and we will focus on proving $GTXZ$ being cyclic.
Since $GZ=GX$, we need $FG$ to be the external angle bisector of $\angle XTZ$, and since $ET \perp FG$, we need $ET$ to be the internal angle bisector of $\angle XTZ$.
We claim $\triangle TXE \sim \triangle TEZ$, which will finish the problem. Add the midpoints $S, R$ of $EX, EZ$ and observe that $TSER$ is cyclic with diameter $GE$, so $\angle TRE=\angle TSX$. It suffices to show that $\triangle TRE \sim \triangle TSX$, so we need $\frac{RE} {RT}=\frac {SX} {ST} \iff RE \cdot TS=ES \cdot TR$, i.e. we want $TRES$ to be harmonic, which is true by projecting through $G$ onto $AB$ combined with Ceva-Menelaus.
sami1618
15.08.2024 23:40
We claim that $P$, the foot of $E$ onto $FG$, is the desired intersection point. Let $X'$, $Z'$, $G'$, $P'$ be the midpoints of $EX$, $EZ$, $EG$, and $EP$. Let $GE$ meet $DC$ at $E'$. It suffices to show that $Z'X'P'G'$ is cyclic. Notice that $P$, $X'$, and $Z'$ all lie on the circle with diameter $EG$. We claim that $X'P'E$ is similar to $XPZ'$ we have that $\angle X'EP'=\angle X'Z'P$ and $$\frac{P'E}{PZ'}\colon \frac{X'E}{X'Z'}=\frac{1}{2}\frac{PE}{PZ'}\colon\frac{X'E}{X'Z'}=\frac{1}{2}(EZ';PX')=\frac{1}{2}(E'D; FC)=\frac{1}{2}(1-(E'F;DC))=1$$The second to last equality following from the properties of the anharmonic group (we ignored directed lengths as the do not really matter). Similarly one can show that $EZ'P'$ is similar to $X'Z'P$. Thus $$\angle X'P'Z'=\angle X'P'E+\angle EP'Z'=2\angle X'PZ'=\angle X'G'Z'$$
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