Azerbaijan NMO 2015

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Source: Azerbaijan NMO 2015

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IstekOlympiadTeam

542 posts

IstekOlympiadTeam

#1 h Oct 21, 2015, 2:40 PM • 4 Y

Y by ahmedosama, opptoinfinity, Adventure10, Rounak_iitr

In the convex quadrilateral $ABCD$ angle $\angle{BAD}=90$ , $\angle{BAC}=2\cdot\angle{BDC}$ and $\angle{DBA}+\angle{DCB}=180$ . Then find the angle $\angle{DBA}$

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Imgoodinmaths_dude

805 posts

Imgoodinmaths_dude

#2 h Feb 7, 2016, 8:39 AM • 2 Y

Y by Adventure10, Mango247

Can it be done without trig?

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77 posts

izat

#3 h Feb 20, 2016, 6:16 AM • 3 Y

Y by Adventure10, Mango247, ClassyPeach

Let $R$ be reflection of $B$ with respect to $AD$ . Since $BR$ is perpendicular to $AD$ , we get $B, A , R$ are collinear. Thus, $180-\angle DCB=\angle DBA = \angle DBR = \angle DRB$ , so $D,C,B,R$ are concyclic. $\angle CAB = 2\angle CDB = 2\angle CRB$ , therefore $\angle ACR= \angle CRA$ , it means $CA=AR=AB$ and $\angle BCR=90$ .Finally, $\angle RBD= \angle RCD = \angle BCD - 90$ , however, $\angle RBD + \angle BCD= 180$ ,so $\angle ABD=45$

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#4 h Mar 11, 2016, 7:08 PM • 2 Y

Y by Adventure10, Mango247

45 сршйфвш

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#5 h Mar 12, 2016, 1:40 AM • 2 Y

Y by Adventure10, Mango247

My solution is quite simple
B,C,D are on circumference of circle A. Therefore triangle ABD is right-angle isosceles triangle. Angle ABD = 45.

This post has been edited 2 times. Last edited by Davidng, Mar 12, 2016, 1:45 AM
Reason: latex???

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507 posts

#6 h Sep 21, 2016, 10:06 AM • 2 Y

Y by Adventure10, Mango247

using trig it is easy

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#7 h Sep 21, 2016, 10:15 AM • 2 Y

Y by Adventure10, Mango247

Quote:

B,C,D are on circumference of circle A

how

This post has been edited 2 times. Last edited by fighter, Sep 21, 2016, 10:16 AM

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AlexLewandowski

157 posts

AlexLewandowski

#8 h Jan 22, 2017, 5:12 PM • 2 Y

Y by Adventure10, Mango247

Extend DC to P such that BP is perpendicular to CD. Join AP to intersect BC at Q.
$ABPD$ is cyclic.
So, $ABD$ = $APD$
or, $BCP$ = $APD$
Thus, $QC$ = $QP$ .
Similar angle chasing gives,
$CBP$ = $APB$
Hence, $QB$ = $QP$ .
Thus, QC = QP = QB,
or, Q is the circumcenter of BPC. But, Q lies in the bisector of BAC. Thus, AP is the perpendicular bisector of BC. So, $\Delta BCP$ is isosceles.
Hence, $ABD$ = $BCP$ = $45$ °

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472 posts

Ari004

#9 h Jun 30, 2018, 12:13 PM • 2 Y

Y by Adventure10, Mango247

Let $B'$ be the reflection of $B$ in $A$ .From the above angle conditions, $\angle AB'S+\angle BCD=180^\circ$ .So, $B'BCD$ is cyclic.Now, $\angle BDC=\angle BB'C$ .Let $C'$ be the reflection of $C$ in $A$ ,hence $\angle BB'C=\angle B'BC'$ ,by angle chase, $\angle BC'C=\angle B'CA$ ,so $\bigtriangleup B'AC$ is $A$ isosceles.So, $AB=AC=AB'$ ,hence $A$ is the circumcenter of $\bigtriangleup BB'C$ ,so $\angle BCB'=90^\circ$ ,hence $\angle BDB'=90^\circ$ .Now, $\bigtriangleup BDB'$ is right angled and $DA \perp BB'$ ,by angle chase we have $\angle ABD=45^\circ$ .

This post has been edited 1 time. Last edited by Ari004, Jun 30, 2018, 12:14 PM

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#10 h Jul 31, 2018, 7:09 PM • 2 Y

Y by Adventure10, Mango247

fighter wrote:

Quote:

B,C,D are on circumference of circle A

how

If you reflect $B$ and $C$ across $AD$ you have cyclic quadrilaterals $BDC'B'$ and $BCDB'$ as $\angle B'BD+\angle B'C'D=\angle ABD+\angle BCD=180^\circ$ . So $BCDC'B'$ is a cyclic pentagon. Furthermore, $A$ lies on the perpendicular bisector of $BB'$ , so the circumcircle of $BDB'$ lies on $AD$ . As $\angle B'AB=180^\circ$ , $A$ is in fact the circumcenter.

So $B,C,D$ are on the circumference of circle $A$

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374 posts

#11 h Dec 29, 2023, 1:03 PM

Y by

Very easy ... 45°

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30628 posts

#12 h May 12, 2024, 11:02 AM

Y by

an auxiliary point solution

This post has been edited 2 times. Last edited by parmenides51, May 18, 2024, 8:49 AM

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