INMO 2018 -- Problem #3

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INMO 2018 -- Problem #3

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Source: INMO 2018

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3465 posts

#1 h Jan 21, 2018, 11:53 AM • 4 Y

Y by samrocksnature, Adventure10, Mango247, Rounak_iitr

Let $\Gamma_1$ and $\Gamma_2$ be two circles with respective centres $O_1$ and $O_2$ intersecting in two distinct points $A$ and $B$ such that $\angle{O_1AO_2}$ is an obtuse angle. Let the circumcircle of $\Delta{O_1AO_2}$ intersect $\Gamma_1$ and $\Gamma_2$ respectively in points $C (\neq A)$ and $D (\neq A)$ . Let the line $CB$ intersect $\Gamma_2$ in $E$ ; let the line $DB$ intersect $\Gamma_1$ in $F$ . Prove that, the points $C, D, E, F$ are concyclic.

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Kayak

#2 h Jan 21, 2018, 11:55 AM • 4 Y

Y by lneis1, samrocksnature, Adventure10, Mango247

After Inverting at $B$ , the problem becomes trivial. (B becomes ortho centre of A*C*D* )

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#3 h Jan 21, 2018, 12:08 PM • 5 Y

Y by Aryan-23, samrocksnature, Adventure10, Mango247, Rounak_iitr

$BE$ is a diameter of $T_2$ .
$BF$ is a diameter of $T_1$ .

And you obtain $FEA$ is a straight line parrallel to $O_1O_2$ .

And rest is angle chasing.

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Vrangr

#4 h Jan 21, 2018, 12:23 PM • 4 Y

Y by AG234, govind7701, samrocksnature, Adventure10

$2\angle CFE = 2\angle BFC = \angle BO_1C = \angle O_1CO_2 - \angle O_1BO_2 = \angle O_1DO_2 - \angle O_1BO_2$ $=\angle CO_2B = \angle CO_2B = 2\angle CFB = 2\angle CFD \ \blacksquare$

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#5 h Jan 21, 2018, 12:30 PM • 5 Y

Y by DynamoBlaze, Arhaan, samrocksnature, Adventure10, Siddharthmaybe

Let $O_1B$ meet $\Gamma_2$ at $D'$ , we have
$\angle BDO_2=\angle DBO_2=180-\angle O_1BO_2=180-\angle O_1AO_2$ .
Hence $AO_1O_2D'$ is cyclic implying $D'=D$ giving $O_1,B,D$ are collinear. Similarly, $O_2,B,C$ are collinear.

This gives $BE$ and $BF$ are diameters of $\Gamma_2$ and $\Gamma_1$ respectively, giving $\angle FCE=\angle EDF=90$ .

Hence, alternate interior angles are equal and we are done!

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2527 posts

#6 h Jan 21, 2018, 12:32 PM • 2 Y

Y by samrocksnature, Adventure10

rd1452002 wrote:

$2\angle CFE = 2\angle BFC = \angle BO_1C = \angle O_1CO_2 - \angle O_1BO_2 = \angle O_1DO_2 - \angle O_1BO_2$ $\angle CO_2B = \angle CO_2B = 2\angle CFB = 2\angle CFD \ \blacksquare$

My solution

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1590 posts

#7 h Jan 21, 2018, 12:34 PM • 3 Y

Y by The-X-squared-factor, samrocksnature, Adventure10

Drunken_Master wrote:

Let $O_1B$ meet $\Gamma_2$ at $D'$ , we have
$\angle BDO_2=\angle DBO_2=180-\angle O_1BO_2=180-\angle O_1AO_2$ .
Hence $AO_1O_2D'$ is cyclic implying $D'=D$ giving $O_1,B,D$ are collinear. Similarly, $O_2,B,C$ are collinear.

This gives $BE$ and $BF$ are diameters of $\Gamma_2$ and $\Gamma_1$ respectively, giving $\angle FCE=\angle EDF=90$ .

Hence, alternate interior angles are equal and we are done!

Same solution. I wonder why did they give only one geo this year.

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328 posts

#8 h Jan 21, 2018, 12:38 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

AnArtist wrote:

Drunken_Master wrote:

Let $O_1B$ meet $\Gamma_2$ at $D'$ , we have
$\angle BDO_2=\angle DBO_2=180-\angle O_1BO_2=180-\angle O_1AO_2$ .
Hence $AO_1O_2D'$ is cyclic implying $D'=D$ giving $O_1,B,D$ are collinear. Similarly, $O_2,B,C$ are collinear.

This gives $BE$ and $BF$ are diameters of $\Gamma_2$ and $\Gamma_1$ respectively, giving $\angle FCE=\angle EDF=90$ .

Hence, alternate interior angles are equal and we are done!

Same solution. I wonder why did they give only one geo this year.

Usually they give 2 to 2.5 geo

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1600 posts

Vrangr

#10 h Jan 21, 2018, 1:07 PM • 3 Y

Y by AlastorMoody, samrocksnature, Adventure10

I prepared for 2-3 geometry questions because I am weak in geometry. Only 1 came :/
I should spent that time doing combi or algebra.

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449 posts

arulxz

#11 h Jan 21, 2018, 2:39 PM • 3 Y

Y by mueller.25, samrocksnature, Adventure10

Prove by angle chasing that $B$ is the incenter of $\triangle ACD$ .

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3042 posts

#12 h Jan 21, 2018, 3:09 PM • 5 Y

Y by Drunken_Master, Maths_Guy, FadingMoonlight, samrocksnature, Adventure10

Found a nice solution in the exam using phantom points!

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#13 h Jan 21, 2018, 3:51 PM • 7 Y

Y by Maths_Guy, hydrohelium, Spiralflux789, FadingMoonlight, samrocksnature, Adventure10, Mango247

My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$ .

As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$ . Similarly, $C'\equiv C$ .
Proved.

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#14 h Jan 23, 2018, 1:09 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

We can also use the concept of projective geometry to show O1,B,Dand F collinear.

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#15 h Jan 23, 2018, 10:40 AM • 5 Y

Y by JayantJha, samrocksnature, Adventure10, Mango247, Rounak_iitr

Claim 1 : $C , B , O_2$ and $D , B , O_1$ are co-linear.
Proof:
From the cyclic $\square O_1 A O_2 C$ we get,
$\angle O_1 C O_2 = 180^o - \angle O_1 A O_2$

$\triangle O_1 A O_2 \cong \triangle O_1 B O_2 \Rightarrow \angle O_1 A O_2 = \angle O_1 B O_2$

Let the line $O_2 B$ intersects $\tau_1$ at $K$
$\therefore \angle O_1 B K = 180^o - \angle O_1 B O_2 = 180^o - \angle O_1 A O_2$

$O_1 B = O_1 K$
$\therefore \angle O_1 K B = \angle O_1 B K = 180^o - \angle O_1 A O_2 = \angle O_1 C O_2$
Therefore, $K$ coincides with $C$

Thus , $C, B , O_2$ are co-linear. (Proved)
Similarly, $D, B , O_1$ are co-linear.

Claim 2 : $F , A , E$ are co-linear.
Proof : Trivial.

Now we complete our proof,
Let $\angle A E O_2 = \angle F E C = \alpha$
$\therefore E A O_2 = \alpha \Rightarrow \angle A O_2 B = 2\alpha$
$\Rightarrow \angle O_1 O_2 B = \angle O_1 O_2 C = \angle O_1 D C = \angle F D C = \alpha$
$\therefore \angle F E C = \angle F D C$
So, $C , D , E , F$ are concyclic [Q.E.D.]

Attachments:

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chandramauli_2000

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chandramauli_2000

#16 h Jan 28, 2018, 4:25 PM • 2 Y

Y by samrocksnature, Adventure10

all the ques pls?

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1590 posts

#17 h Jan 28, 2018, 4:27 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

@chandramauli see here

https://artofproblemsolving.com/community/c596374_2018_india_national_olympiad

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#18 h May 21, 2018, 12:45 PM • 2 Y

Y by samrocksnature, Adventure10

For those who didn't get Kayak's slick solution, he says: Invert about $B$ . Then $A \to A^*$ , $B \to B^*$ and $C \to C^*$ , while $\Gamma_1$ goes to the straight line $A^*C^*$ and $\Gamma_2$ goes to the straight line $A^*D^*$ . (Because $\angle O_1AO_2$ is obtuse, $C$ and $D$ , and hence $C^*$ and $D^*$ , are different from B.)

$\odot (O_{1}AO_{2}) \to \odot (O_{1}^*A^*O_{2}^*)$ , which is also the circumcircle of $\Delta A^*C^*D^*$ .

Now note:
1. $O_1^*$ is the reflection of $B$ in $A^*C^*$ ,
2. $O_2^*$ is the reflection of $B$ in $A^*D^*$ .

This forces $B$ to be the orthocenter of $A^*C^*D^*$ because there is only one point other than $A^*$ whose reflections in $A^*C^*$ and $A^*D^*$ respectively lie on $\odot (A^*C^*D^*)$ (why?) $-$ and the orthocenter clearly satisfies this condition. Hence, $B^*$ is the orthocenter of $\Delta A^*C^*D^*$ .

Since $E^*$ and $F^*$ are the feet of altitudes through $D^*$ and $C^*$ in $\Delta A^*C^*D^*$ , it's trivial to note that $C^*D^*E^*F^*$ is cyclic.

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Synthetic_Potato

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Synthetic_Potato

#19 h May 21, 2018, 1:58 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Extention: Let $AB$ meet the circle $\odot (O_1AO_2)$ again at $O$ . Prove that $O$ is the circumcenter of $BCD$ .
(Note: This is easier than the original problem

)

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1261 posts

#20 h May 21, 2018, 4:42 PM • 2 Y

Y by samrocksnature, Adventure10

Synthetic_Potato wrote:

Extention: Let $AB$ meet the circle $\odot (O_1AO_2)$ again at $O$ . Prove that $O$ is the circumcenter of $BCD$ .
(Note: This is easier than the original problem

)

Just note that $B$ is the incentre of $ACD$

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XbenX

#21 h Jul 19, 2018, 8:57 AM • 4 Y

Y by Pigeonhole_Biswaranjan_, samrocksnature, Adventure10, Mango247

A different approach:
After some easy angle chasing we get that $C,B,O_2,E$ and $D,B,O_1,F$ are collinar in that order .
By Power of point $B$ on $(CDO_2O_1) \Longleftrightarrow BD \cdot BO_1=BC \cdot BO_2$ and since $BO_1$ and $BO_2$ are the radiuses of their respective circles , we get that $BD \cdot BF=BC \cdot BE$ wich completes the proof.

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#22 h Dec 2, 2018, 6:07 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

TheDarkPrince wrote:

My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$ .

As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$ . Similarly, $C'\equiv C$ .
Proved.

I think
$\angle AO_1O_2=\angle BAO_2$ which can never be equal to $\angle BAO_2$ so your claim is wrong

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3042 posts

#23 h Dec 21, 2018, 2:36 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

hydrohelium wrote:

TheDarkPrince wrote:

My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$ .

As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$ . Similarly, $C'\equiv C$ .
Proved.

I think
$\angle AO_1O_2=\angle BAO_2$ which can never be equal to $\angle BAO_2$ so your claim is wrong

How is $\angle AO_1O_2 = \angle BAO_2$

I think my solution works fine.

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MathPassionForever

1663 posts

MathPassionForever

#24 h Dec 21, 2018, 2:48 PM • 3 Y

Y by AlastorMoody, samrocksnature, Adventure10

TheDarkPrince wrote:

hydrohelium wrote:

TheDarkPrince wrote:

My solution (something like this): Let $O_1B \cap \Gamma_2 = D'$ and $O_2B \cap \Gamma_1 = C'$ .

As $\triangle O_1D'A$ is not isosceles, $\angle AO_1O_2 = \angle O_2O_1B = \angle O_2O_1D'$ and $\angle O_2AD'=\angle O_2D'A$ gives $O_1AO_2D'$ is cyclic. So, $D'\equiv D$ . Similarly, $C'\equiv C$ .
Proved.

I think
$\angle AO_1O_2=\angle BAO_2$ which can never be equal to $\angle BAO_2$ so your claim is wrong

How is $\angle AO_1O_2 = \angle BAO_2$

I think my solution works fine.

People actually look up posts that old?

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#25 h Jan 9, 2019, 5:38 PM • 2 Y

Y by samrocksnature, Adventure10

Too easy for INMO, anyways here is an alternate proof.

Angle chase to get $C,B,O_{2},E$ are collinear. Similarly $D,B,O_{1},F$ are collinear.
Then notice $\angle FCE = \angle EDF = 90^{\circ}$ (Angle in a semicircle). By inscribed angles theorem, $C,D,E$ and $F$ are collinear. $\blacksquare$

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#26 h Jan 9, 2019, 7:25 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

vlohani wrote:

For those who didn't get Kayak's slick solution, he says: Invert about $B$ . Then $A \to A^*$ , $B \to B^*$ and $C \to C^*$ , while $\Gamma_1$ goes to the straight line $A^*C^*$ and $\Gamma_2$ goes to the straight line $A^*D^*$ . (Because $\angle O_1AO_2$ is obtuse, $C$ and $D$ , and hence $C^*$ and $D^*$ , are different from B.)
$\odot (O_{1}AO_{2}) \to \odot (O_{1}^*A^*O_{2}^*)$ , which is also the circumcircle of $\Delta A^*C^*D^*$ .

Now note:
1. $O_1^*$ is the reflection of $B$ in $A^*C^*$ ,
2. $O_2^*$ is the reflection of $B$ in $A^*D^*$ .

This forces $B$ to be the orthocenter of $A^*C^*D^*$ because there is only one point other than $A^*$ whose reflections in $A^*C^*$ and $A^*D^*$ respectively lie on $\odot (A^*C^*D^*)$ (why?) $-$ and the orthocenter clearly satisfies this condition. Hence, $B^*$ is the orthocenter of $\Delta A^*C^*D^*$ .

Since $E^*$ and $F^*$ are the feet of altitudes through $D^*$ and $C^*$ in $\Delta A^*C^*D^*$ , it's trivial to note that $C^*D^*E^*F^*$ is cyclic.

So how does O1*B perpendicular to A*C* imply that O1* is reflection of B in A*C*??

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2125 posts

#27 h Feb 21, 2019, 11:54 AM • 2 Y

Y by samrocksnature, Adventure10

$\angle ACO_2=\angle AO_1O_2=\frac{1}{2} \angle AO_1B=\angle ACB$ $\implies$ $BC$ passes through $O_2$ , similarly, $BD$ passes through $O_1$ , hence, $\angle FCE$ $=$ $90^{\circ}$ $=$ $\angle EDF$ which implies the conclusion

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#28 h Mar 24, 2020, 8:10 PM • 2 Y

Y by samrocksnature, Mango247

What about radical axis

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#29 h Apr 1, 2020, 3:59 AM • 2 Y

Y by Gerninza, samrocksnature

$\textcolor{red}{\text{Claim:}}$
$C,B,O_2$ are coliner. Similarly $O_1,B,D$ are similar .

Let's join $B,O_2$ and $B,0_1$ .Note that $AB \perp O_1O_2$ since it's radial axis of $\Gamma_1,\Gamma_2$ . Note that $\triangle AO_2B$ is isosceles .And hence $\angle AO_2O_1=\angle BO_2O_1=a$ and by similar argument we have $\angle BO_1O_2=\angle AO_1O_2=b$ and hence $\angle O_1AO_2=\angle O_1BO_2=180-(a+b)$ .

Suppose $O_2B$ met $\Gamma_1$ at $C'$ .
So $\angle O_1O_2C'=a$ . Again $\angle O_1C'B=\angle C'O_1B$ . Which means $A,O_1,C',D$ are concyclic . $\Rightarrow C'=C$ .

Similarly we can show that $O_1,B,D$ are coliner.

$\textcolor{red}{\text{Claim:}}$ $E,A,F$ are Coliner and $FE$ parallel to $O_1O_2$ .

In $\Gamma_2$ we get $\angle AO_2E =180-2b$ which gives $\angle AEB=\angle O_1O_2B =b$ .

So $AE$ parallel to $O_1O_2$ by similar argument $AF$ parallel to $O_1O_2$ .

Our claim is proved.

Note that $C,O_1,O_2,D$ are concyclic so $CB.BO_2=DB.BO_1 \cdots (*)$ .

Since $\triangle BO_1O_2 \sim BEF$ hence $\frac{BO_1}{BF}=\frac{BO_2}{BE}$ .

From $(*)$ we get $CB.BE =DB .BF$ so we are done .

This post has been edited 1 time. Last edited by ftheftics, Apr 1, 2020, 4:01 AM
Reason: Jjsj

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#30 h Apr 22, 2021, 10:13 AM • 1 Y

Y by samrocksnature

XbenX wrote:

A different approach:
After some easy angle chasing we get that $C,B,O_2,E$ and $D,B,O_1,F$ are collinar in that order .
By Power of point $B$ on $(CDO_2O_1) \Longleftrightarrow BD \cdot BO_1=BC \cdot BO_2$ and since $BO_1$ and $BO_2$ are the radiuses of their respective circles , we get that $BD \cdot BF=BC \cdot BE$ wich completes the proof.

First you need to prove that $(CDO_2O_1)$ is cyclic..

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#31 h May 5, 2021, 5:12 AM • 2 Y

Y by samrocksnature, kamatadu

Wait what? i just realised this problem is a part of https://artofproblemsolving.com/community/q1h1952359p21527833
$C-B-O_2$ are collinear, so is $D-B-O_1$
$FB$ and $BE$ is diameter so $\angle FCE = 90^\circ = \angle EDF$ and we are done.

edit-

arulxz wrote:

Prove by angle chasing that $B$ is the incenter of $\triangle ACD$ .

yes indeed, this was the brazil problem lol

This post has been edited 1 time. Last edited by MrOreoJuice, May 5, 2021, 7:19 AM

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#33 h Feb 6, 2022, 10:37 AM • 1 Y

Y by tuymurod2005

$\angle CAB =\angle DAB,\angle FAB = \angle EAB \implies FB \cdot BD = CB \cdot BE$

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51 posts

#34 h Feb 23, 2022, 10:43 AM • 1 Y

Y by Rounak_iitr

$\measuredangle DO_1O_2 =\measuredangle DAO_2 =\measuredangle O_2DA =\measuredangle O_2O1A =\measuredangle BO_1O_2$ $\implies D,B,O_1$ are collinear, hece $BE$ is the diameter of $\Gamma_1$ similiarly we get $BF$ as the diameter of $\Gamma_2$ ,
$\implies \measuredangle FDE = \measuredangle FCE = 90^{\circ}$ , Hence $C,D,E,F$ are concyclic

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Mahdi_Mashayekhi

697 posts

Mahdi_Mashayekhi

#35 h Mar 6, 2022, 6:07 AM

Y by

Claim: $C,B,O_2$ and $D,B,O_1$ are collinear.
Proof : $\angle ADO_1 = \angle AO_2O_1 = \angle AO_2B/2 = \angle ADB$ so $D,B,O_1$ are collinear. we prove the other one with same approach.

Now we have $\angle ECF = \angle BCF = \angle 90 = \angle BDE = \angle FDE$ so $CDEF$ is cyclic.

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87 posts

#36 h Apr 9, 2022, 7:08 PM

Y by

Just invert about $B$ , using the inversion distance formula we find out that $O_1'$ and $O_2'$ are the reflection of $B$ over $A'C'$ and $A'D'$ respectively, then note that since both $O_1'$ and $O_2'$ lie on $(A'C'D')$ , $B'$ is the orthocenter of $\triangle{A'C'D'}$ and therefore, $C,B,O_2,E$ and $D,B,O_1,F$ are in fact collinear, the conclusion now follows immediately.

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419 posts

#37 h Dec 20, 2022, 7:50 PM

Y by

Proof
Remark
I hope you enjoyed reading my proof

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4191 posts

#38 h May 4, 2023, 3:38 AM

Y by

Let $\angle O_1BC=\angle O_1CB=\alpha$ , $\angle O_1AB=\angle O_1BA=\beta$ , and $\angle O_2AB=\angle O_2BA=\gamma.$

Main Claim: $C,B,O_2$ are collinear. We wish to show that $\alpha+\beta+\gamma=180$ . Note that $\angle AO_2O_1=90-\gamma$ since $AB\perp O_1O_2$ . Thus, from cyclic $AO_1CO_2$ , we have $\angle ACO_1=90-\gamma$ . Since $O_1C=O_1A$ , $\angle AO_1C=2\gamma$ . However, we have $\angle AO_1B=180-2\beta$ , and $\angle BO_1C=180-2\alpha$ . Hence, $2\gamma=180-2\beta+180-2\alpha\rightarrow \alpha+\beta+\gamma=180,$ as desired.

Hence, $BE$ is a diameter of $\Gamma_2$ , so $\angle BDE=90$ . Similarly, $\angle BCF=90$ . Hence, $C,D,E,F$ are concyclic and we are done.

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481 posts

#39 h May 9, 2023, 3:31 PM • 2 Y

Y by aansc1729, HoripodoKrishno

Bro what? Why Invert this :maybe:

:maybe:

?

Anyways here is the sketch (typing full solns is too stressful due to my 50 WPM :stretcher:

:stretcher:

).

https://i.imgur.com/RZLbbWr.png

You get that $E$ and $F$ are the $B-$ antipodes in $\Gamma_1$ and $\Gamma_2$ respectively from some angle chasing.

Then a bit of homothety shows $EF\parallel O_1O_2$ and $\overline{A-E-F}$ and so you get $\measuredangle ECF=\measuredangle ECB=90^\circ=\measuredangle BDF=\measuredangle EDF$ which finishes.

This post has been edited 1 time. Last edited by kamatadu, May 9, 2023, 3:33 PM

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#40 h Aug 13, 2023, 8:07 PM

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To be fair, inversion (especially about $A$ ) doesn't help a ton, but I'll just stick with it

In the inverted picture, $\omega_1^*$ and $\omega_2^*$ become two lines that intersect at $B$ , and $(O_1AO_2)$ becomes the line $\overline{CD}$ , on which the reflections of $A$ over $\omega_1^*$ and $\omega_2^*$ lie. Now, observe

$O_1$ and $O_2$ lie on $(ABD)$ and $(ACD)$ , the two images of the lines, respectively, by the converse of Fact 5;
$A, E, F$ are collinear as $\angle BAE = \angle BAF = 90^\circ$ ;
$CDFE$ is cyclic by a quick angle chase.

Hence done!

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1703 posts

#41 h Dec 6, 2023, 4:09 AM

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We invert about $A$ with radius $1$ . We can see that $F^*C^*B^*$ are collinear, $E^*D^*B^*$ are collinear, $O_1^*C^*D^*O_2^*$ are collinear, $F^*O_1^*B^*D^*$ are concyclic, and $E^*O_2^*B^*C^*$ are concyclic.

\begin{lemma}
$O_1^*$ is the reflection of $A$ over $F^*B^*$ .
\end{lemma}
\begin{proof}
EGMO lemma 8.10
\end{proof}

This we have $\angle AF^*C^* = \angle C^*F^*O_1^* = \angle C^*D^*B^*$ $\blacksquare$

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cursed_tangent1434

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cursed_tangent1434

#42 h Apr 9, 2024, 12:17 AM

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Me when I wanted a medium geo and tried this only for it to turn out trivial.

The following observation is key.

Claim : The points $D$ , $B$ and $O_1$ are collinear. Similarly, the points $C$ , $B$ and $O_2$ are collinear.
Proof : Simply note that
$\measuredangle AO_1D = \measuredangle AO_2D = 2\measuredangle ADO_2 = 2\measuredangle AO_1O_2 = \measuredangle AO_1B$ from which it is clear that $D-B-O_1$ as claimed. Similarly, we can also show $C-B-O_2$ .

Now, note that this means $BF$ and $BE$ are diameters of $\Gamma_1$ and $\Gamma_2$ respectively. This then gives us,
$\measuredangle FCE = \measuredangle FCB = 90^\circ = \measuredangle BDE = \measuredangle FDE$ from which it is clear that the points $C, D, E, F$ are concyclic as required.

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#43 h Jun 4, 2024, 5:12 PM • 1 Y

Y by GeoKing

Let $O_2B$ intersect $\Gamma_1$ at $C'$ (phantom point), then,
$\angle O_1C'B = \angle O_1BC' = 180^o - \angle O_1BO_2$ As $O_1$ and $O_2$ are the centers of $\Gamma_1$ and $\Gamma_2$ , we have $O_1B=O_1A$ , $O_2B=O_2A$ , which implies that,
$\angle O_1BA = \angle O_1AB , \angle O_2BA = \angle O_2AB$ $\therefore \angle O_1BO_2 = \angle O_1AO_2$
Hence, $\angle O_1C'B = \angle O_1AO_2 = \angle O_1CB$ , $\Rightarrow C \equiv C'$ ,
$\therefore C,B,O_2,E$ are collinear,
Similarly, $D,B,O_1,F$ are collinear, $\Rightarrow BF$ and $BE$ are diameters of $\Gamma_1$ and $\Gamma_2$ respectively.
$\therefore \angle BDE= \angle FDE = \angle FCE = 90^o$ , $\Rightarrow CDEF$ is a cyclic quadrilateral.

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773 posts

#44 h Sep 20, 2024, 5:03 PM

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$\text{[math]}$

Attachments:

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#47 h Feb 19, 2025, 10:09 PM

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Claim: $FO_1BD$ is collinear.
Proof: Invert at $A.$ Then, we have that $(ABC)$ goes to a line $\ell_1,(ABD)$ goes to a line $\ell_2, O_1$ goes to the reflection of $A$ across $\ell_1, O_2$ goes to the reflection of $A$ across $\ell_2, C, D$ going to the intersections of $\ell_1, \ell_2$ with $O_1^*O_2^*.$ We also have that $E^*, F^*$ go to the intersections of $\ell_1, \ell_2$ with $(AB^*D^*)$ and $(AC^*D^*).$
Now, note that $B^*$ is the circumcenter of $AO_1O_2$ since $B^*$ is the intersection of $\ell_1, \ell_2.$ Thus, $\angle AB^*D^* = \frac12\angle AB^*O_2^* = \angle AO_1^*D,$ so $AO_1^*B^*D^*$ is cyclic. However, since $E^*$ also lies on $AB^*D^*,$ we have that $AO_1^*B^*D^*E^*$ is cyclic. Inverting back, we get our desired claim.

By Thales, we are done.

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#48 h Apr 26, 2025, 5:12 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.51570034117379, xmax = 40.82314408852699, ymin = -16.69519592790025, ymax = 21.93029312646862; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); pen ttffqq = rgb(0.2,1,0); pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffqqff = rgb(1,0,1); pen ffwwqq = rgb(1,0.4,0); pen xfqqff = rgb(0.4980392156862745,0,1); /* draw figures */ draw(circle((-3,-1), 9.501880502737109), linewidth(1) + blue); draw(circle((12.983545658162647,-1.0859342715935283), 10.371101277912132), linewidth(1) + red); draw(circle((4.978910084569212,-3.4354033592755004), 8.342313567590866), linewidth(1) + qqccqq); draw((-3,-1)--(12.983545658162647,-1.0859342715935283), linewidth(1) + red); draw((4.455727850687627,4.890488545509315)--(12.983545658162647,-1.0859342715935283), linewidth(1) + ttffqq); draw((4.455727850687627,4.890488545509315)--(-3,-1), linewidth(1) + ttffqq); draw((-0.10455216140722312,-10.049978723858084)--(21.50932605200054,4.81921690282699), linewidth(1) + blue); draw((9.085859717659737,-10.69675036732216)--(-10.445019518138734,4.904017061529285), linewidth(1) + red); draw((-10.445019518138734,4.904017061529285)--(21.50932605200054,4.81921690282699), linewidth(1) + ffdxqq); draw((-10.445019518138734,4.904017061529285)--(-0.10455216140722312,-10.049978723858084), linewidth(1) + ffdxqq); draw((-0.10455216140722312,-10.049978723858084)--(9.085859717659737,-10.69675036732216), linewidth(1) + ffdxqq); draw((9.085859717659737,-10.69675036732216)--(21.50932605200054,4.81921690282699), linewidth(1) + ffdxqq); draw((4.455727850687627,4.890488545509315)--(4.419722115642268,-6.935777692513815), linewidth(1) + ffqqff); draw((4.455727850687627,4.890488545509315)--(9.085859717659737,-10.69675036732216), linewidth(1) + ffwwqq); draw((4.455727850687627,4.890488545509315)--(-0.10455216140722312,-10.049978723858084), linewidth(1) + ffwwqq); draw(circle((5.532254937295013,4.899928348557101), 15.977274978600997), linewidth(1) + linetype("4 4") + ffqqff); draw((-3,-1)--(-0.10455216140722312,-10.049978723858084), linewidth(1) + xfqqff); draw((12.983545658162647,-1.0859342715935283)--(9.085859717659737,-10.69675036732216), linewidth(1) + xfqqff); /* dots and labels */ dot((-3,-1),dotstyle); label("$O_1$", (-4.514173220805075,-1.4528504714676924), NE * labelscalefactor); dot((4.455727850687627,4.890488545509315),dotstyle); label("$A$", (4.362760922855928,5.995113785652763), NE * labelscalefactor); dot((12.983545658162647,-1.0859342715935283),dotstyle); label("$O_2$", (13.889226833126274,-1.7126631781114292), NE * labelscalefactor); dot((4.419722115642268,-6.935777692513815),dotstyle); label("$B$", (4.2328545695340605,-9.030721081910016), NE * labelscalefactor); dot((-0.10455216140722312,-10.049978723858084),dotstyle); label("$C$", (-0.6169826211490247,-11.67215026612134), NE * labelscalefactor); dot((9.085859717659737,-10.69675036732216),dotstyle); label("$D$", (9.125993877991101,-12.105171443860902), NE * labelscalefactor); dot((21.50932605200054,4.81921690282699),dotstyle); label("$E$", (22.72285885901332,4.782654487981991), NE * labelscalefactor); dot((-10.445019518138734,4.904017061529285),dotstyle); label("$F$", (-11.745626889055746,4.955862959077816), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
A Great Angle Chasing Exercise :love:

:love:

$\color{red}\textbf{Claim:-}$ $C,D,E,F$ are concyclic.
$\color{green}\textbf{Proof:-}$ By Angle Chasing we get, $\angle O_1AC =\angle O_1CA = 180-2\angle AO_1C = 90-\angle CFA = 90+\angle ABC$ Also, $\angle O_2AD = \angle O_2DA =180-2\angle AO_2D = 90-\angle AED = 90+\angle ABD$ Since, $\angle AFC = \angle ABE$ and $\angle AED = \angle ABF$ $\implies C,B,O_2$ are collinear and $D,B,O_1$ are collinear points.
Therefore, $BE$ and $BF$ are the diameters of circumcircles of $\triangle ABE$ and $\triangle ABF \implies \angle FAB = \angle BAE = 90^o \implies F,A,E$ are collinear and parallel to $O_1O_2.$ Now By Angle Chasing we get, $\angle O_1AC = \angle O_1CA = \angle AO_2O_1 = \angle O_1DC$ And $\angle O_2AD = \angle O_2DA = \angle EFB = \angle AO_1O_2 = \angle ACO_2$ Therefore $B$ is the incentre of $\triangle ACD.$

Now finish this problem from this angle relation $\angle AEC = \angle FEC = \angle FDC \implies C,D,E,F$ are concyclic points.

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bjump

#49 h Apr 30, 2025, 4:01 PM

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Invert at $A$ . By Egmo Lemma 8.10 $B'D'$ is the perpendicular bisector of $AO_1'$ . Similar bisector statement holds for $AO_2'$ .
$2\angle AB'D' = 2\angle AO_1'O_2' = \angle AB'O_2'$ which gives $(AO_1'B'C')$ is a circle so $DO_1BE$ collinear, finishes by PoP.

This post has been edited 1 time. Last edited by bjump, Apr 30, 2025, 4:01 PM

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