Angle chasing yet again

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TheDarkPrince

3042 posts

TheDarkPrince

#1 h Oct 28, 2018, 11:57 AM • 4 Y

Y by Maths_Guy, Adventure10, Mango247, Rounak_iitr

In a cyclic quadrilateral $ABCD$ with circumcenter $O$ , the diagonals $AC$ and $BD$ intersect at $X$ . Let the circumcircles of triangles $AXD$ and $BXC$ intersect at $Y$ . Let the circumcircles of triangles $AXB$ and $CXD$ intersect at $Z$ . If $O$ lies inside $ABCD$ and if the points $O,X,Y,Z$ are all distinct, prove that $O,X,Y,Z$ lie on a circle.

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leon.tyumen

183 posts

leon.tyumen

#2 h Oct 28, 2018, 12:57 PM • 3 Y

Y by Maths_Guy, Adventure10, Mango247

Since $OA=OC$ , $OB=OD$ , we can say that $O$ - "two bicyclist point" for circles $(BXC)$ and $(AXD)$ , then, $\angle OYX=90$ . Similary we can say, that $\angle OZX=90$ , and points $O, X, Y, Z$ lie on a circle. QED

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TheDarkPrince

3042 posts

TheDarkPrince

#4 h Oct 28, 2018, 1:06 PM • 4 Y

Y by Maths_Guy, dram, Adventure10, Mango247

Solution:

Claim. $A,D,O,Z$ are concyclic.
(Proof) $\angle AZX = \angle ABX = \angle ACD=\angle XZD$ . Thus $\angle AZD = 2\angle ABD = \angle AOD$ .

Main problem: As $AO = DO$ and $XZ$ is the angle bisector of $\angle AZD$ , from the claim $\angle OZX = 90^{\circ}$ . Similarly $\angle OYX = 90^{\circ}$ and we are done.

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mathetillica

333 posts

mathetillica

#5 h Oct 29, 2018, 1:12 PM • 2 Y

Y by Adventure10, Mango247

actually it must be $\Delta CXD$ in place of $\Delta BXD$

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AlastorMoody

2125 posts

AlastorMoody

#6 h Jan 10, 2019, 7:37 PM • 1 Y

Y by Adventure10

An Overkill

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415122

39 posts

415122

#7 h May 7, 2019, 1:05 PM • 1 Y

Y by Adventure10

Can anyone plz make figure?

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amar_04

1915 posts

amar_04

#9 h Jul 2, 2019, 2:06 PM • 1 Y

Y by Adventure10

leon.tyumen wrote:

Since $OA=OC$ , $OB=OD$ , we can say that $O$ - "two bicyclist point" for circles $(BXC)$ and $(AXD)$ , then, $\angle OYX=90$ . Similary we can say, that $\angle OZX=90$ , and points $O, X, Y, Z$ lie on a circle. QED

What is byclist point and why is $\angle OYX=90$ and $\angle OZX=90$ .

Also

TheDarkPrince wrote:

Solution:
Main problem: As $AO = DO$ and $XZ$ is the angle bisector of $\angle AZD$ , from the claim $\angle OZX = 90^{\circ}$ . Similarly $\angle OYX = 90^{\circ}$ and we are done.

How did he get $\angle OZX=90^\circ$ can anyone please explain.

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char2539

399 posts

char2539

#10 h Jul 2, 2019, 2:26 PM • 2 Y

Y by Adventure10, Mango247

Trivial.
Just note that $Y$ and $Z$ are the miquel points of $ABCD$ .Since $ABCD$ is cyclic we get $\angle OYX = \angle OZX=90^{\circ}$ .Hence the result follows.

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zuss77

520 posts

zuss77

#11 h Jul 2, 2019, 2:29 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

It Kazakhstan 2001.

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pkrmath2004

39 posts

pkrmath2004

#12 h Jul 14, 2019, 2:36 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

China Mathematical Olympiad 1992

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Pluto04

797 posts

Pluto04

#13 h Feb 12, 2020, 7:11 AM • 1 Y

Y by Adventure10

pkrmath2004 wrote:

China Mathematical Olympiad 1992

Yes indeed.
https://artofproblemsolving.com/community/c6h556271p3233203
For reference.

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lifeismathematics

1188 posts

lifeismathematics

#14 h Mar 24, 2023, 10:29 AM

Y by

very nice problem!

Claim 1:- points $O,Y,A,B$ are concyclic

Proof:-

we join $XY$ to get:

$\angle{AYB}=360^{\circ}-(\angle{XYA}+\angle{XYB})$

$=\angle{ADB}+\angle{ACB}$

$=2\angle{ADB}$

$=\angle{AOB}$

so we get that points $O,Y,A,B$ are concyclic $\blacksquare$

Claim 2:- Points $C,Z,O,B$ are concylic

Proof:- $\angle{BZC}=\angle{BZX}+\angle{CZX}$

$=\angle{XAB}+\angle{XDC}$

$=2\angle{CAB}$

$=\angle{BOC}$ , hence we get points $C,Z,O,B$ are concyclic $\blacksquare$

Claim 3:- $\angle{XYO}=90^{\circ}$

Proof:- $\angle{XYO}=\angle{XYA}-\angle{OYA}$

$=180^{\circ}-\frac{\angle{AOB}}{2}-\left(90^{\circ}-\frac{\angle{AOB}}{2}\right)$

$=90^{\circ}$ $\blacksquare$

similarly using Claim 2 we can prove that $\angle{XZO}=90^{\circ}$ hence we get that points $O,X,Y,Z$ lie on a circle $\blacksquare$

Attachments:

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Jishnu4414l

154 posts

Jishnu4414l

#15 h Sep 17, 2024, 2:11 PM

Y by

Cute problem!
Claim 1: Points $Z$ , $B$ , $C$ , $O$ are concyclic.
Proof: Notice that $\angle BZX=\angle XZC=180^{\circ}-\angle BAC$ . This gives $\angle BZC=2\angle BAC=\angle BOC$ , and we are done!
By symmetry, we have that points $B$ , $O$ , $Y$ , $D$ are concyclic.
Claim 2: $\angle OZX=90^{\circ}$
Proof: $\angle BZO=180^{\circ}-\angle OCB= 90^{\circ}+\angle BAC$ . This gives $\angle OZX=90^{\circ}$ , and we are done! By symmetry, $\angle OYX=90^{\circ}$ as well, giving that $O$ , $X$ , $Y$ , $Z$ lie on a circle.

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balllightning37

382 posts

balllightning37

#16 h Sep 17, 2024, 2:27 PM

Y by

Inner Miquel Points !?!??!!!
Invert about the circumcircle. It is well known that $X$ goes to the Miquel point of $ABCD$ . $Y$ and $Z$ then go to $AB\cap CD$ and $AD \cap BC$ , which are collinear with the Miquel point.

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