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Source: Korea National Olympiad 2020 P6

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#1 h Nov 25, 2020, 5:11 AM

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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.

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#2 h Nov 25, 2020, 3:35 PM

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cute

:-D

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#3 h Nov 25, 2020, 8:10 PM • 2 Y

Y by parmenides51, Aritra12

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.

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#4 h Nov 26, 2020, 3:58 AM • 3 Y

Y by gnoka, mathroyal, Brian_Xu

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

This post has been edited 1 time. Last edited by chrono223, Nov 26, 2020, 9:36 AM
Reason: added more words

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gnoka

#5 h Dec 1, 2020, 10:24 AM • 1 Y

Y by Brian_Xu

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?

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