IMO ShortList 2002, geometry problem 1

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orl

3647 posts

orl

#1 h Sep 28, 2004, 12:45 PM • 2 Y

Y by Adventure10, Mango247

Let $B$ be a point on a circle $S_1$ , and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$ . Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$ . Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$ .

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orl

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orl

#2 h Sep 28, 2004, 12:46 PM • 2 Y

Y by Adventure10, Mango247

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

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grobber

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grobber

#3 h Sep 28, 2004, 7:09 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

I. for one, distinctly remember posting two or three of these on the forum, but what the heck!

They're nice problems; we can solve them all over again.

I remember giving an inversive proof, but I can't find it anymore, so here's another one:

Let $M$ be the circumcenter of $BCD$ . We have $\angle BMC=2\angle BDC$ , and we want to show that $\angle BMC=\angle BAC$ , so it's equivalent to showing that $\angle BAC=2\angle BDC$ . Let $C'$ be the intersection of $DC$ with $S_1$ .
The tangent through $C'$ to $S_1$ is parallel to $AC$ because $C'$ is the image $C$ through the homothety centered at $D$ which turns $S_2$ into $S_1$ . Now let $T$ be the intersection of the tangents to $S_1$ through $B,C'$ . We now have $\angle BAC=\pi-\angle BTC'=2\angle TC'B=2\angle BDC'=2\angle BDC$ , Q.E.D.

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darij grinberg

6555 posts

darij grinberg

#4 h Sep 28, 2004, 9:30 PM • 2 Y

Y by Adventure10, Mango247

The problem can be simplified (no arrangement conditions are necessary):

Let A, B and C be three non-collinear points, and $S_1$ and $S_2$ two circles such that the circle $S_1$ touches the line AB at the point B, the circle $S_2$ touches the line AC at the point C, and the circles $S_1$ and $S_2$ touch each other at a point D. Prove that the circumcenter of the triangle BCD lies on the circumcircle of the triangle ABC.

Here is my solution, using directed angles modulo 180:

If P is the circumcenter of triangle BCD, then we have to prove that this point P lies on the circumcircle of triangle ABC. In other words, we have to show that < BPC = < BAC. What we know is < BPC = 2 < BDC (since the point P is the circumcenter of triangle BCD).

If t is the common tangent to the circles $S_1$ and $S_2$ at their point of tangency S, then, since the tangents to a circle at the endpoints of a chord make equal angles with the chord, we have < (BD; t) = < (AB; BD) and < (CD; t) = < (AC; CD), so that

< BDC = < (BD; CD) = < (BD; t) - < (CD; t) = < (AB; BD) - < (AC; CD)
= < (AB; AC) + < (AC; BD) - (AC; CD) = < (AB; AC) + < (CD; BD)
= < BAC + < CDB = < BAC - < BDC.

Thus, 2 < BDC = < BAC, and hence < BPC = < BAC, as desired. $\blacksquare$

Darij

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orl

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orl

#5 h Sep 30, 2004, 12:49 PM • 2 Y

Y by Adventure10, Mango247

grobber wrote:

I. for one, distinctly remember posting two or three of these on the forum, but what the heck!

They're nice problems; we can solve them all over again.

Well, grobber, I told that you shall post the links for the ISL problems in the section where I and Peter VDD did it. I do not search the forum all the time. And from time to time these postings are not adapted to LaTeX. Rather than searching and adopting all threads I repost it.

The same for Russian Olympiad and Kvant problems. I really appreciate it that you post these nice problems but please collect the links an extra thread for convenience and to prevent people from reposting them. Thank you.

Other suggestions ?

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Ashegh

858 posts

Ashegh

#6 h Jun 17, 2006, 5:15 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Here is the first solution:

Name ,(the tangent $AB$ ),(the common tangent of the two circles from

$D$ ),and( the tangent $AC$ ),

$T_1,T_2,T_3$ , respectively.

Consider aline which is perpendicular, to $BD$ , and also bisects it.

The points which are on this line, have equal distance from $T_1,T_2$ ,

Consider aline which is perpendicular, to $CD$ , and also bisects it.

The points which are on this line , have equal distance from $T_2,T_3$ .

Then the intersection of these two lines, $O$ , have equal distance from

$T_1,T_3$ .

And we conclude that $O$ lies on the bisector of $\angle BAC$ .

But $O$ lies on line which is perpendicular to $BC$ , and also biscts it.

Then it lies on the circum circle of triangle $BAC$ .

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Ashegh

858 posts

Ashegh

#7 h Jun 17, 2006, 6:43 PM • 2 Y

Y by Adventure10, Mango247

The second solution:

Consider aline which is perpendicular, to $BD$ , and also bisects it.

Consider aline which is perpendicular, to $CD$ , and also bisects it.

And name their intersection, $O$ .

We should say that:

$180=\angle A+\angle BOC$ .

Ok………………………………………………………………………………………………….

$\angle BDC=X$ then : $\angle BOC=2\angle EOF=2(180-X)=360-2X$ .

we should know that if, $\angle BOC=360-2X+\angle A=180$ , holds or not.

Suppose it holds, then:

We draw a tangent at $S$ which is parallel to $AC$ . Then, $AC,A'M$ are parallel

And, $\angle A=\angle A'= \frac{%Error. "arcBDM" is a bad command. -%Error. "arcBM" is a bad command. }{2}$ .

$%Error. "arcBDM" is a bad command. =360-%Error. "arcBM" is a bad command.$

$\angle A=180-%Error. "arcBM" is a bad command. +180=360-%Error. "arcBM" is a bad command. =2X$

$X=180-\frac{%Error. "arcBM" is a bad command. }{2}$

and this fact is true, because of the quality of homotechy, $F,D,M$ are collinear.

(NOTE: $D$ is the inner homotechy center , and a tangent at $F$ , to circle $S'$ ,

is parallel to the tangent at $M$ , to the circle $S$ .

and finally: $\angle BDC=180-\angle BDM=180-\frac{%Error. "arcBM" is a bad command. }{2}$ .

And the second one is complete,too.

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Ashegh

858 posts

Ashegh

#8 h Jun 17, 2006, 6:46 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Now , I choose an $INVERSIVE$ proof for the third solution.

Really, and absolutely more nice than the two recent solution, I think.

Lets invert the shape round point $B$ ,and an arbitrary radius.

Now every thing which I will say , is about the inversive form of the shape.

It is clear that each point, for example $X$ , turns into its inversive form, $X'$ .

I should prove that, the center of circumcircle of triangle $BDC$ , lies on circle $CAB$ .

Or the other words, in the inversive shape, the inversion of the center of this circle,

Lies on the inversion of circle $CAB$ .

The inversion of circle $CAB$ , is $AC$ , and the inversion of circle $BDC$ , is $DC$ .

The inversion of the center of this circle is a point $B'$ ,with reflect point $B$ , at line $CD$ .

Then I should prove that:

$B'$ lies on $AC$

………………………………………………………………………………………………………………………...

and to do that, if $AC,BH$ meet at $O$ , and if $X=Y$ , the problem is solved.

$X=180-\angle BCP=\frac{%Error. "arcBCP" is a bad command. }{2}$

$Y=\angle PCA=\frac{%Error. "arcAP" is a bad command. }{2}$

we know that $P$ is the midpoint of the arc.because with center $C$ , $P,D$ are two homologues point.

And finally: $\overarc{BCP}=%Error. "arcAP" is a bad command.$ and every thing is ok.

It was really an enjoyable problem,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,.

I will think about the fourth solution.

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saeid

63 posts

saeid

#9 h Jun 17, 2006, 7:02 PM • 2 Y

Y by Adventure10, Mango247

You are so COOL ashegh.
Excelent Work

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Ashegh

858 posts

Ashegh

#10 h Jun 17, 2006, 7:25 PM • 2 Y

Y by Adventure10, Mango247

saeid wrote:

You are so COOL ashegh.
Excelent Work

ure wellcome my darling

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Altheman

6194 posts

Altheman

#11 h Jan 4, 2009, 9:51 AM • 1 Y

Y by Adventure10

Notation: (see diagram as well) Let $O$ be the circumcenter of $\triangle BCD$ . Let $BD$ intersect $S_2$ again at $D$ and let $CD$ intersect $S_1$ again at $F$ . Let $AC$ intersect $BF$ at $G$ . Let $\angle BDF=x$ .

(1) $\angle BOC=2x$ . Proof: $2x=2\angle BDF=2\pi-2\angle BDC=(\pi-2\angle BDO)+(\pi-2\angle CDO)$ . Since $BO=CO=DO$ $=\angle BOD+\angle DOC=\angle BOC$ .

(2) $BF\parallel CE$ . Proof: Since $S_1$ and $S_2$ are tangent at $D$ , there is a homothety that maps $S_1$ to $S_2$ centered at $D$ . It is easy to see that $BF$ goes to $CE$ . Since a homothety maps a line to another parallel line, we get the conclusion.

(3) $\angle ACE=\pi-x$ . Proof: $\angle ACE=\angle ACD+\angle DCE=\angle CED+\angle DCE$ by the tangent secant angle theorem in $S_2$ . $=\pi-\angle CDE=\pi=\angle BDF=\pi-x$ .

(4) $\angle BAC=\pi-2x$ . Proof: $\angle BAC=\angle AGF-\angle ABG=\angle ACE-\angle BDF=\pi-2x$ by the tangent secant angle theorem in $S_1$ , the parallel lines in (2), and (3).

(5) $ABOC$ is cyclic. Proof: $\angle BAC+\angle BOC=\pi-2x+2x=\pi$ by (1) and (4) so $ABOC$ is cyclic.

In other words, $O$ lies on the circumcircle of $\triangle ABC$ , as desired.

Attachments:

circumcenter on circumcircle.pdf (14kb)

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Heebeen, Yang

81 posts

Heebeen, Yang

#12 h Jul 20, 2009, 1:13 AM • 1 Y

Y by Adventure10

Another solution.
Define $M$ be $CD \cap S_{1}$ , and $AC$ meet $S_{1}$ at $E, F$ , $AF$ and $BM$ meet at $X$ .
$A,D,O$ (circumcenter of $\triangle BCD$ ) $B$ are cyclic $<=> \angle ABO=\angle OCE <=> \frac{\pi}{2}=\angle BXC+\angle OBC <=> X B C D cyclic.$
And using homothety, we easily know $\angle XCD=\angle XBD$ so $A D O B$ cyclic.

Sorry for my poor english...

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serialk11r

1449 posts

serialk11r

#13 h Dec 22, 2009, 7:22 PM • 1 Y

Y by Adventure10

Somewhat wierd solution, but I thought it was interesting.
Let $O_1$ be the center of the circle containing $B$ , and $O_2$ be the center of the circle containing $C$ . Let $A'$ be the intersection between $O_1B$ and $O_2C$ . $A'BCA$ is cyclic, so it suffices to show the circumcenter lies on the circumcircle of $A'BC$ .
The perpendicular bisectors of $BD$ and $CD$ are the angle bisectors of $\angle{O_2O_1A'}$ and $\angle{O_1O_2A'}$ , so the incenter of $O_1O_2A'$ is the circumcenter of $BCD$ . Let this point be $I$ . $A'I$ is thus the angle bisector of $\angle{O_1A'O_2}$ , and $BIC$ is isoceles.
If $A'I$ is not the perpendicular bisector of $BC$ , we are done. If $A'I$ is the perpendicular bisector of $BC$ , then $B,C,D$ are the points of tangency of the incircle to the sides since $A'B=A'C$ , and so $\angle{A'BI}=\angle{A'CI}=90$ , and so we are done.

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AK1024

228 posts

AK1024

#14 h Dec 28, 2010, 8:25 PM • 2 Y

Y by Adventure10, Mango247

I think all the solutions above just complicate the problem...

Diagram

Let $O$ be the circumcentre of $\triangle BDC$ and let $P$ be the intersection of the common tangent to $S_1$ and $S_2$ (passing through $D$ ) and $AC$ . Let the midpoints of $BD,CD$ be $M,N$ respectively. Let $PD$ and $BA$ intersect at $K$ . Obviously $D$ is the reflection of $B$ through the line $OK$ .

Then $CD$ is the polar of $P$ wrt $S_2$ so $O,P,N$ are collinear (all lie on the perpendicular bisector of $CD$ ). So by trig ceva on $\triangle ODC$ obviously $\angle ODP=\angle PCO$ (lol user pco

!). Now $\angle ACO=\angle PCO=\angle ODP=180^{\circ}- \angle KDO=180^{\circ}-\angle KBO=180^{\circ}-\angle ABO$ , so $ABOC$ is cyclic.

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StefanS

149 posts

StefanS

#15 h Jun 28, 2011, 6:21 PM • 3 Y

Y by Nafis_Noor, Adventure10, Mango247

The only solution without defining new lines or points.

Let the centers of $~$ $S_1 \wedge S_2$ $~$ be $~$ $O_1 \wedge O_2$ $~$ respectively, and $~$ $O$ $~$ be the circumcenter of $~$ $\triangle{BDC}.$

Solution

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Quidditch

818 posts

Quidditch

#36 h May 31, 2022, 11:46 AM

Y by

Let $O,O_1,O_2$ be center of circles $(BCD),S_1,S_2$ , respectively. Note that $O_1-D-O_2$ are collinear.

Since $O_2C=O_2D$ and $OC=OD$ , we have $\triangle O_2CO\cong \triangle O_2DO$ .
$\begin{align*} \angle O_2DO &= \angle O_2CO \\ &= \angle O_2CA+\angle ACO \\ &= 90^{\circ}+\angle ACO \end{align*}$ Similarly, $O_1B=O_1D$ and $OB=OD$ so $\triangle O_1DO\cong \triangle O_1BO$ .
$\begin{align*} \angle O_1DO &= \angle O_1BO \\ &= \angle ABO-\angle ABO_1 \\ &= \angle ABO-90^{\circ} \end{align*}$
Hence, we have
$\begin{align*} 180^{\circ} &=\angle O_2DO+\angle O_1DO \\ &= (90^{\circ}+\angle ACO)+(\angle ABO-90^{\circ}) \\ &= \angle ACO + \angle ABO. \end{align*}$ Thus, $A,B,C,O$ concyclic, as desired.

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JAnatolGT_00

559 posts

JAnatolGT_00

#37 h May 31, 2022, 2:27 PM

Y by

Let $O_1,O_2,O$ be centers of $S_1,S_2,\odot (BCD).$ Clearly $D\in O_1O_2,O_1B\perp AB,O_2C\perp AC,$ so $\measuredangle BAC=\angle (BO_1,CO_2)=\measuredangle BO_1D+\measuredangle DO_2C=2\measuredangle BDC=\measuredangle BOC\implies O\in \odot (ABC).$

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ETS1331

107 posts

ETS1331

#38 h Jul 2, 2022, 10:05 PM

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$[asy] size(10cm); import geometry; draw(circle((-1,0),1)); draw(circle((2,0),2)); circle c1 = circle((point)(-1,0),1); circle c2 = circle((point)(2,0),2); point B = 2*dir(65)+2, C = dir(85)-1, D = (0,0); point E = intersectionpoint(tangent(c1,D),tangent(c1,C)); point F = intersectionpoint(tangent(c2,D),tangent(c2,B)); point A = intersectionpoint(tangent(c2,B),tangent(c1,C)); dot("$B$", B, dir(60)); dot("$C$", C, dir(60)); dot("$D$", D, dir(-15)); dot("$E$", E, dir(55)); dot("$F$", F, dir(60)); dot("$A$", A, dir(60)); draw(C--A--F--D); draw(C--D--B); [/asy]$
Let $\ell$ be the common internal tangent to both circles, and let $E = AC \cap \ell$ and $F = AB \cap \ell$ . Let $O$ be the circumcenter of $BCD$ . To prove that $O \in (ABC)$ , we only need to show that $\measuredangle COB = \measuredangle CAB$ (all angles directed). However, $\measuredangle COB = 2 \measuredangle CDB$ (circumcenter angles). Furthermore, we have: $\begin{align*} \measuredangle CAB &= \measuredangle EAF \\ &= -\measuredangle AFE - \measuredangle FEA \\ &= - \measuredangle BFD - \measuredangle DEC \\ &= \measuredangle FDB + \measuredangle DBF + \measuredangle ECD + \measuredangle CDE \\ &= 2(\measuredangle CDE + \measuredangle FDB) \\ &= 2\measuredangle CDB \end{align*}$ which finishes.

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asdf334

7585 posts

asdf334

#39 h Jul 11, 2022, 8:49 PM

Y by

Does this work? (I think the part where $\measuredangle BO_1D=2\measuredangle BDO_1$ might be incorrect.)

Let the center of $S_1$ be $O_1$ and let the center of $S_2$ be $O_2$ . Also let the circumcenter of $BCD$ be $O$ . Then let $BO_1$ and $CO_2$ intersect at $A'$ . Then we have
$\begin{align*} \measuredangle BAC&=\measuredangle BA'C \\ &=\measuredangle A'O_1O_2 + \measuredangle O_1O_2A' \\ &=\measuredangle BO_1D+\measuredangle DO_2C \\ &=2(\measuredangle BDO_1+\measuredangle O_2DC) \\ &=2\measuredangle BDC \\ &=\measuredangle BOC \end{align*}$ and we are done. $\blacksquare$

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asdf334

7585 posts

asdf334

#40 h Jul 11, 2022, 9:04 PM • 1 Y

Y by Mango247

OKAY so there is apparently config issues. aint no way im actually gonna fix the proof

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fuzimiao2013

3302 posts

fuzimiao2013

#41 h Jul 19, 2022, 4:20 PM • 1 Y

Y by Mango247

Let $O_1$ denote the center of $S_1$ and $O_2$ be defined similarly. Let $O$ be the circumcenter of $BCD$ . The key claim is that

Claim: $\measuredangle OBO_1 = \measuredangle O_1DO$ .
Proof: We know that $\measuredangle O_1BD = \measuredangle BDO_1$ and $\measuredangle OBD =\measuredangle BDO$ by isosceles triangles. Subtracting gives the result.

It is sufficient to prove that $\color{blue} OBAC$ is cyclic. The rest is straightforward angle chasing.
$\begin{align*}\measuredangle OBA &= \measuredangle O_1DO + 90^{\circ} \\ &= \measuredangle O_2DO + 90^{\circ} \\ &= \measuredangle CDO + \measuredangle O_2DC + 90^{\circ} \\ &= \measuredangle OCD + \measuredangle DCO_2 + 90^{\circ} \\ &= \measuredangle OCO_2 + 90^{\circ} \\ &= \measuredangle OCA,\end{align*}$ and we're done. $\square$

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ike.chen

1162 posts

ike.chen

#42 h Jul 30, 2022, 3:26 PM

Y by

Let $I$ be the incenter of $ABC$ , segment $AC$ meet $S_1$ again at $Y$ and $Z$ so that $AY < AZ$ , the common tangent to $S_1$ and $S_2$ meet $AC$ at $M$ , segment $BC$ meet $S_1$ again at $X$ , the midpoint of arc $YZ$ not containing $D$ be $N$ , the reflection of $C$ over $M$ be $E$ , and $B_1$ be the point on $S_1$ such that $BB_1 \parallel YZ$ . The parallel condition implies $N$ is also the midpoint of arc $BB_1$ .

Because $ME = MC = MD$ follows from equal tangents, we have $\angle CDE = 90^{\circ}.$ Now, observe $MC^2 = |Pow_{S_2}(M)| = MD^2 = |Pow_{S_1}(M)| = MY \cdot MZ$ so $-1 = (C, E; Y, Z)$ . Thus, the Right Angle and Bisectors Lemma implies $DE$ bisects $\angle YDZ$ , which means $D, E, N$ are collinear.

Now, we know $\angle BDB_1 = 180^{\circ} - \angle BNB_1 = 180^{\circ} - \angle BNX - \angle XNB_1$ $= 180^{\circ} - \angle ABX - \angle XBB_1 = 180^{\circ} - \angle ABC - \angle ACB = \angle BAC.$ This gives $\angle BDC = \angle BDN + \angle NDC = \frac{\angle BDB_1}{2} + \angle EDC$ $= \frac{\angle A}{2} + 90^{\circ} = \angle BIC$ which finishes via the Incenter-Excenter Lemma. $\blacksquare$

Remark: This solution is a bit overkill.

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lrjr24

967 posts

lrjr24

#43 h Aug 8, 2022, 6:40 PM

Y by

Let $O_1$ be the center of $S_1$ , $O_2$ be the center of $S_2$ , and $O$ be the circumcenter of $BCD$ . We have
$\begin{align*} \measuredangle BOC &= 2 \measuredangle BDC \\ &=( \measuredangle BCD + \measuredangle DBC ) \\ &=2( \measuredangle BCA + \measuredangle ACD + \measuredangle DBA + \measuredangle ABC ) \\ &= 2(\measuredangle BAC + \measuredangle DBA + \measuredangle ACD) \\ &= 2 \measuredangle BAC + 2 \measuredangle DB O_1 + \measuredangle CO_2D \\ &= 2 \measuredangle BAC + \measuredangle DO_1B + \measuredangle CO_2D \\ &= 2\measuredangle BAC + \measuredangle CBO_1 + \measuredangle O_2CB \\ &= 2 \measuredangle BAC + \measuredangle CBA - 90 + \measuredangle ACB +90 \\ &= \measuredangle BAC \\ \end{align*}$ This means $B,A,O,C$ are concyclic as desired so we are done.

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jelena_ivanchic

151 posts

jelena_ivanchic

#44 h Oct 22, 2022, 6:28 PM

Y by

solution(has config issues but can be fixed with directed angles).

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lambda5

28 posts

lambda5

#45 h Feb 20, 2023, 1:09 AM

Y by

Let $O$ be the circuncenter of $BCD$ . It would be sufficient to prove that
$360^{\circ}-2\angle CDB=\angle DOC=180^{\circ}- \angle BAC$
$\iff 180^{\circ}=2\angle CDB - \angle BAC$
Now invert at $C:$
We need
$180^{\circ}=2\angle C'B'D' - \angle A'B'C' \iff 90^{\circ} = \angle C'B'D' - \frac{\angle A'B'C'}{2}$
Let the line parallel to the tangent at $D'$ through $B'$ meet the circles $(B'D'), (B'A'C')$ at $E, F$ . Let M be the midpoint of the arc $A'C'$ not containing $B'$
$A'C' \parallel EF \Rightarrow \angle EB'D' + \angle C'B'M = \angle D'B'M = 90$
But $90^{\circ}=\angle D'B'M = \angle C'B'D' - \frac{\angle A'B'C'}{2}$

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huashiliao2020

1292 posts

huashiliao2020

#46 h May 29, 2023, 12:33 AM • 1 Y

Y by Amir Hossein

Nice angle chasing. We have <BFC=360-2<BDC=2(<BCD+<CBD)=2(<ACB+<ACD+<ABC-<ABD)=2(180-<BAC+<ACD-<ABD)=360-2<BAC+<CED-<BOD (tangent angle rules)=... I'm not sure how to continue, I ask someone to post a solution like this. Edit: I saw lrjr's but i didnt get exactly how he got the third to last line from the fourth

Anyways, second attempt: First, we have <FBD=<FDB, <ODB=<OBD. Then we have <FBO=<FDO. We have <FBA=90+<FBO=90+180-<EDF=180+90-<ECF=180-<FCA, so ABCF is cyclic, as desired. $\blacksquare$

Edit: After comparing with others, it seems like directed angles are necessary, but I don't want to do them right now.

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starchan

1605 posts

starchan

#47 h Aug 5, 2023, 7:31 AM

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only angle chasing!
solution

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bal09

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bal09

#49 h Jun 18, 2024, 3:21 AM

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I HAD POSTED SOLUTION IN PHOTO CHECK IT

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ezpotd

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ezpotd

#50 h Aug 25, 2024, 2:18 AM

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Draw in the circle centers $O_1, O_2$ , let $O_1DO_2$ meet $S_1$ again at $D'$ , we then see that $\angle CAB = \angle DO_2C - \angle D'O_1B$ (the condition on $C$ guarantees this is positive) by comparing arguments of lines. Now $\angle CDB = 180 - \frac 12 (180 - (180 - \angle D'O_1 B)) - \frac 12 (180 - \angle DO_2C) = 90 + \frac 12 (\angle DO_2C - \angle D'O_1B)$ . Since this is guarantee to be obtuse, we can conclude that the desired circumcenter $O_3$ is on the opposite side of $BC$ as $D$ , and $D$ is on the same side of $BC$ as $A$ , so we it remains to check $\angle CO_3B + \angle CAB = 180$ , which is trivial since $CO_3B = 2(180 - \angle CDB) = 180 - (\angle DO_2C - \angle D'O_1B)$ as desired.

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Avron

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Avron

#51 h Apr 8, 2025, 8:16 PM

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(We assume the circles are internally tangent, for the other config just change some angles). Let the line $DC$ intersect $S_1$ again at $C'$ . Consider the homothety taking $S_2$ to $S_1$ centered at $D$ . The line $AC$ goes to a line paralell to it through $C'$ tangent to $S_1$ . Let this line intersect $AB$ at $X$ . Now $\angle XC'D-\angle BC'D=\angle XC'B=\angle BDC'$ , thus
$\angle BAC = 360-((\angle ABD+\angle DBC')+\angle BC'D+\angle ACC')=360-((\angle BC'D+180-\angle XC'D)+\angle BC'D+180-\angle XC'D)=2(\angle XC'D-\angle BC'D)=2\angle BDC'$ so if $O$ is the circumcenter of $BCD$ then $\angle BOC=2\angle BDC'=2\angle BDC=\angle BAC$ and we're done.

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