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Korea National MO 2020 P6 wrote: Let $ABCDE$ be a convex pentagon such that quadrilateral $ABDE$ is a parallelogram and quadrilateral $BCDE$ is inscribed in a circle. The circle with center $C$ and radius $CD$ intersects the line $BD, DE$ at points $F, G(\neq D)$, and points $A, F, G$ is on line l. Let $H$ be the intersection point of line $l$ and segment $BC$. Consider the set of circle $\Omega$ satisfying the following condition. Circle $\Omega$ passes through $A, H$ and intersects the sides $AB, AE$ at point other than $A$. Let $P, Q(\neq A)$ be the intersection point of circle $\Omega$ and sides $AB, AE$. Prove that $AP+AQ$ is constant. Let $\angle EAF=x$, $\angle BAF=y$ and $\angle FCE=z$. Note that, $$\frac{BF}{\sin \angle FCB}=\frac{FC}{\sin \angle DBC}=\frac{DC}{\sin \angle CED}=\frac{DE}{\sin \angle ECD} \,\, (*)$$Since $$\frac{BF}{DE}=\frac{BF}{BA}=\dfrac{\sin y}{\sin x}$$and $$\frac{\sin \angle FCB}{\sin \angle ECD}=\frac{\sin (x+y-z) }{\sin (2y-z)},$$hence $(*)$ easily rewrites as $$\sin y\sin(2y-z)-\sin x \sin (x+y-z)=0,$$which due to the trigonometric identity $$\sin A \sin B=\frac{\cos(A-B)-\cos(A+B)}{2}$$and the fact $(x+y-z)-x=(2y-z)-y$ rewrites as $$\cos(2x+y-z)=\cos(3y-z)$$ Hence $2x+y-z=2k\pi \pm (3y-z)$ with $k$ an integer. Let now $2x+y-z=P, \,\, 3y-z=Q$. We prove the following Claim. Claim: $0<P+Q<2\pi$. Proof: Note that $P+Q=2(x+2y-z)$. Since $z<\angle FCD=2y$, we infer that $x+2y-z>0$, hence $P+Q>0$. In addition, since $\angle CDG=\frac{\pi}{2}-\frac{DFG}{2}=\frac{\pi}{2}-\frac{x}{2}$ we have $$\angle ECD<\angle CDG=\frac{\pi}{2}-\frac{x}{2}$$Hence, $$z=2y-\angle ECD>2y+\frac{x}{2}-\frac{\pi}{2},$$therefore $$P+Q=2(x+2y-z)<\pi+x< \pi +\angle EAB<2\pi,$$as desired $\blacksquare$. To the problem. We seperate cases depending on $k$. Case 1: $k \geq 1$. Then, if $P+Q=2k\pi$, we have a contradiction since $2\pi>P+Q=2k\pi \geq 2\pi$. Hence, $P-Q=2k\pi \geq 2\pi$ which implies that $x-y \geq \pi$, which is again a contradiction since $$\pi>\angle EAB>x \geq y+\pi>\pi$$Case 2: $k=0$. Then, $P=\pm Q$. If $P=-Q$ then we have a contradiction using the Claim. Hence $P=Q$, that is $x=y$. Case 3: $k \leq -1$. Then, $P \pm Q \leq -2\pi$. Since $P+Q>0$, it must be the case $P-Q \leq -2\pi$, hence $y-x \geq \pi$, which is a contradiction since $y < \pi$. To conclude, $x=y$. Since $x=y$, we have $$\angle PQH=\angle PAH=y=x=\angle QPH,$$hence $PH=QH$. Apply Ptolemy's in the quadrilateral $APHQ$ to obtain $$(AP+AQ)PH=AH \cdot PQ \Rightarrow AP+AQ=\frac{AH \cdot PQ}{PH}=\frac{AH \sin \angle PHQ}{\sin \angle PQH}=\frac{AH \sin 2x}{\sin x}=2AH \cos x,$$which is obviously constant, hence we are done.

by tolemi enough to prove that $l$ is angle bisector of $\angle{BAE}$, which is equivalent to CB=CE by angle chasing, which is trivial by simson

chrono223 wrote: by tolemi enough to prove that $l$ is angle bisector of $\angle{BAE}$, which is equivalent to CB=CE by angle chasing, which is trivial by simson nice solution! I'll make a video for it.Can I post it on bilibli?