CGMO5: Carlos Shine's Fact 5

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6870 posts

#1 h Aug 13, 2012, 10:43 PM • 18 Y

Y by Amir Hossein, Trivial, Durjoy1729, HamstPan38825, donotoven, HWenslawski, Jc426, jhu08, leozitz, mathematicsy, megarnie, rama1728, Adventure10, Mango247, Rounak_iitr, ItsBesi, ehuseyinyigit, and 1 other user

As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$ . Prove that $\angle ODB = \angle OEC$ .
$[asy]import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xmax=4.8,ymin=-3.69,ymax=3.71; pen zzttqq=rgb(0.6,0.2,0), wwwwqq=rgb(0.4,0.4,0), qqwuqq=rgb(0,0.39,0); pair A=(-2,2.5), B=(-3,-1.5), C=(2,-1.5), I=(-1.27,-0.15), D=(-2.58,0.18), O=(-0.5,-2.92); D(A--B--C--cycle,zzttqq); D(arc(D,0.25,-104.04,-56.12)--(-2.58,0.18)--cycle,qqwuqq); D(arc((-0.31,0.81),0.25,-92.92,-45)--(-0.31,0.81)--cycle,qqwuqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,1.35),linewidth(1.2)+dotted+wwwwqq); D(CR(O,2.87),linetype("2 2")+blue); D(D--O); D((-0.31,0.81)--O); D(A); D(B); D(C); D(I); D(D); D((-0.31,0.81)); D(O); MP( "A", A, dir(110)); MP("B", B, dir(140)); D("C", C, dir(20)); D("D", D, dir(150)); D("E", (-0.31, 0.81), dir(60)); D("O", O, dir(290)); D("I", I, dir(100)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

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v_Enhance

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v_Enhance

#2 h Aug 13, 2012, 10:45 PM • 21 Y

Y by Binomial-theorem, Sx763_, Kryptogram, Heisenberg09, JasperL, Durjoy1729, mathleticguyyy, Ultroid999OCPN, myh2910, A-Thought-Of-God, HamstPan38825, donotoven, Jc426, FIREDRAGONMATH16, jhu08, rayfish, Adventure10, Mango247, and 3 other users

Solution

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dinoboy

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dinoboy

#3 h Aug 13, 2012, 11:54 PM • 3 Y

Y by donotoven, jhu08, Adventure10

Let $\Gamma$ be the circumcenter of $\triangle ABC$ . Then I claim that $O = \Gamma \cap AI$ (intersection that's not $A$ obviously)
Let $F = \Gamma \cap AI$ .
As $F$ lies on the angle bisector of $\angle BAC$ , it follows $F$ is the midpoint of the minor arc $BC$ so $BF = FC$ .
Now we seek to show $FB=FI$ .
Note that $\angle BFI = \angle BFA = \angle BCA$ .
However, $\angle FBC = \angle FCB = \frac{180 - \angle BFC}{2} = \frac{\angle BAC}{2}$
Notice $\angle IBF = \frac{\angle BAC + \angle ABC}{2}$
We quickly derive then that $\angle BIF = \frac{\angle BAC + \angle ABC}{2} = \angle IBF$ . Thus $FB=FI=DC$ , hence $F$ is the circumcenter of $\triangle BIC$ so it follows $F=O$ lies on $AI$ .
But then $\angle DAO = \angle EAO$ and then since $AD = AE$ , we have $\triangle ADO \cong \triangle AEO \implies \angle ADO = \angle AEO \implies \angle OBD = \angle OEC$ as desired.

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subham1729

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subham1729

#4 h Aug 14, 2012, 12:11 AM • 7 Y

Y by donotoven, jhu08, Adventure10, Mango247, and 3 other users

Let $\angle {ODB}=\theta,\angle {OEC}=\alpha.$

So $\frac {OE}{OD}=\frac {Cos (A/2+\theta)}{Cos (B/2+\alpha)}$

Also $\frac {OE}{OD}=\frac{Cos(\theta)}{Cos(\alpha)}$ ,Combining done.

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sunken rock

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sunken rock

#5 h Aug 15, 2012, 4:34 PM • 5 Y

Y by AlastorMoody, A-Thought-Of-God, donotoven, jhu08, Adventure10

Clearly, $I, O$ belong to angle bisector of $\angle BAC$ , while $D,E$ are symmetrical about it, done.

Best regards,
sunken rock

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Orin

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Orin

#6 h Aug 15, 2012, 6:11 PM • 3 Y

Y by donotoven, jhu08, Adventure10

Let $AI$ intersect $\odot (ABC)$ again at $O'$ .
Since $AI$ is the internal bisector of $\angle A,O'B=O'C=O'I \Rightarrow O'$ is the circumcircle of $\triangle BCI \Rightarrow O'=O$ .
$A,I,O$ are collinear. $AD=AE$ and $\angle IAD=\angle IAE \Rightarrow D,E$ are reflections of each other in $AO$ .
Hence $\angle ADO=\angle AEO \Rightarrow \angle ODB=\angle OEC$

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ACCCGS8

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ACCCGS8

#7 h Aug 17, 2012, 11:11 AM • 4 Y

Y by donotoven, jhu08, Adventure10, Mango247

Let $\frac{A}{2} = \alpha$ , $\frac{B}{2} = \beta$ and $\frac{C}{2} = \gamma$ .
$\angle BIC = \angle ABI + \angle ACI + \angle BAC = 2\alpha + \beta + \gamma = 90^\circ + \alpha$ so $\angle BOC = \angle BOI + \angle COI = 2(\angle ICB + \angle IBC) = 2(180^\circ - \angle BIC) = 180^\circ - 2\alpha$ so $ABOC$ is cyclic.
Thus $O$ is the intersection point (closer to $I$ than $A$ ) of the perpendicular bisector of $BC$ with the circumcircle of $ABC$ . Thus $A$ , $I$ , $O$ are collinear.
Then by SAS, $DAO$ and $EAO$ are congruent so $\angle ADO = \angle AEO$ so $\angle ODB = \angle OEC$ .

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Konigsberg

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Konigsberg

#8 h Aug 13, 2014, 10:43 AM • 4 Y

Y by Devastator, donotoven, jhu08, Adventure10

No pencil, no paper.

We would first show that proving AIO collinear is sufficient. Since ADIE is a kite, I is on the perpendicular bisector of DE, so O lies on the perpendicular bisector of DE. Then DO=OE, so ADOE is a kite: <ADO=<AEO so <ODB=<OEC.

Now we would show that AIO is collinear. <AIB=180-A/2-B/2. <BCI=C/2, so <BOI=C. Since BO=OI, <BIO=90-C/2. Thus <AIB+<BIO=180-A/2-B/2+90-C/2=180. Hence AIO is collinear and we are done.

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v_11

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v_11

#9 h Feb 14, 2018, 12:10 PM • 6 Y

Y by Drunken_Master, Devastator, donotoven, jhu08, Adventure10, Mango247

Well known that $A,I,O$ are collinear. Now $\triangle ADO \cong \triangle AEO \implies \angle ODB = \angle OEC$ as desired. $\blacksquare$

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jayme

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jayme

#10 h Feb 14, 2018, 12:34 PM • 4 Y

Y by donotoven, jhu08, Adventure10, Mango247

Dear Mathlinkers,

also at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32163

Sincerely
Jean-Louis

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Devastator

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Devastator

#11 h Jun 6, 2018, 6:43 AM • 4 Y

Y by donotoven, jhu08, Adventure10, Mango247

This is trivial if you know Chicken Feet Theorem beforehand

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PhysicsMonster_01

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PhysicsMonster_01

#13 h Sep 16, 2019, 1:47 PM • 3 Y

Y by donotoven, jhu08, Adventure10

@above what is that?

Solution

Question: Can we use the incenter-excenter lemma directly on contests?

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Limerent

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Limerent

#14 h Oct 3, 2020, 7:43 PM • 2 Y

Y by donotoven, jhu08

Isn't it called symmetry wrt. $AI$ ? :huh:

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Pleaseletmewin

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Pleaseletmewin

#15 h Nov 4, 2020, 7:23 PM • 2 Y

Y by donotoven, jhu08

By the Incenter Excenter Lemma, $A, I$ , and $O$ are collinear. Now, we note that $DI=EI$ and $IO=IO$ trivially. Furthermore, also note that $\angle DIO=180^\circ-\angle AID=180^\circ-(90^\circ-\angle DAI)=90+\angle DAI \ \ \text{and} \ \ \angle EIO=180^\circ-\angle AIE=180^\circ-(90^\circ-\angle EAI)=90^\circ+\angle EAI=90^\circ+\angle DAI.$ This implies that $\triangle DIO\cong\triangle EIO$ which in turn, implies $\angle ODI=\angle OEI$ . Finally, $\angle ODB=90^\circ-\angle ODI \ \ \text{and} \ \ \angle OEC=90^\circ-\angle OEI=90^\circ-\angle ODI$ as desired. $\blacksquare$

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franzliszt

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franzliszt

#22 h Feb 3, 2021, 4:09 PM • 2 Y

Y by donotoven, jhu08

Storage.

By fact 5, $A,I,O$ are on the $A$ angle-bisector. Note that $AD=AE,\measuredangle DAO=\measuredangle OAE,$ and $AO=AO$ . So by SAS Congruence, we are done.

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Aaronjudgeisgoat

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Aaronjudgeisgoat

#56 h Jun 6, 2024, 10:41 PM

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By the Incenter-Excenter Lemma, $A, B, O,$ and $C$ are concyclic, and $A, I,$ and $O$ are collinear. Futhermore, $\triangle DAO$ and $\triangle EAO$ are congruent by $SAS$ , since $AD=AE.$ This means that $\angle ADO=\angle AEO,$ so $\angle ODB= \angle OEC.$

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G0d_0f_D34th_h3r3

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G0d_0f_D34th_h3r3

#57 h Jun 9, 2024, 8:15 AM

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We know that $A, I$ and $O$ are collinear (By Incenter-Excenter Lemma).
In $\Delta ADO$ and $\Delta AEO$ ,
Since, $\overline{AD} = \overline{AE}$ (Tangent to the incircle) and $\angle DAO = \angle DAI = \angle EAI = \angle EAO$ .
So, $\Delta ADO \cong \Delta AEO$ (By SAS Congruence Criterion)
Hence, $\angle BDO = 180^{\circ} - \angle ODA = 180^{\circ} - \angle OEA = \angle CEO$ .

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Akacool

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Akacool

#58 h Jun 9, 2024, 9:17 AM

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$A$ , $I$ , and $O$ are collinear and $\angle DAO = \angle EAO$ , $AD = AE$ which means $\Delta ADO$ and $\Delta AEO$ are equal implying the result.

This post has been edited 2 times. Last edited by Akacool, Jun 9, 2024, 9:18 AM

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L13832

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L13832

#59 h Jul 13, 2024, 4:46 PM

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By \textbf{Incenter-Excenter Lemma} we have that $O$ lies on $AI$ . We also have $AD=AE$ and $\angle DAO=\angle EAO$ . So by SAS congruence criteria $\triangle ADO \cong \triangle AEO\Rightarrow \angle ADO=\angle AEO \Rightarrow \boxed{\angle BDO =\angle CEO}$ .

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aops-g5-gethsemanea2

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aops-g5-gethsemanea2

#60 h Jul 24, 2024, 3:53 AM

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Many people posted solutions to this problem, but I will post mine anyway. I wrote this while working on EGMO.

Solution

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dudade

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dudade

#61 h Aug 1, 2024, 7:33 PM

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By Incenter-Excenter Lemma, $O$ is the midpoint of arc $BC$ in $(ABC)$ . Since $AD = AE$ and $AI$ is the angle bisector of $\angle A$ , then $\triangle ADO \cong \triangle AEO$ . Therefore, $\angle ADO \cong \angle AEO$ , so $\angle ODB \cong \angle OEC$ , as desired.

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Mathandski

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Mathandski

#62 h Aug 23, 2024, 9:29 PM

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ehuseyinyigit

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ehuseyinyigit

#63 h Aug 26, 2024, 2:46 PM

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By Incenter Lemma, $A$ , $I$ and $O$ are collinear and $O\in (ABC)$ . The result follows by $ADOE$ kite.

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Vedoral

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Vedoral

#64 h Dec 3, 2024, 2:24 PM

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Likeminded2017

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Likeminded2017

#65 h Dec 30, 2024, 5:08 PM

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By Fact 5 $A,I,O$ collinear. Observe by SAS $\triangle ADO \cong \triangle AEO.$ Then $\angle BDO=\angle DAO+\angle DOI=\angle EAO+\angle EOI=\angle EOC.$

This post has been edited 1 time. Last edited by Likeminded2017, Dec 30, 2024, 5:09 PM

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eg4334

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eg4334

#66 h Dec 31, 2024, 10:22 PM

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By Fact $5$ , $AIO$ is collinear. Also $AD=AE$ by tangents, so $\triangle DAO \cong EAO$ so the exterior $D$ and $E$ angles are equal which finishes.

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EaZ_Shadow

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EaZ_Shadow

#67 h Dec 31, 2024, 10:59 PM • 4 Y

Y by 1217662, 1218865, 1220118, Ad112358

O is the center of the circle that goes through B, C, I, and by the incenter excenter lemma, it also goes through $I_A$ . From there, A, O, I are collinear and the result follows because triangle ADO is congruent to triangle AEO.

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cappucher

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cappucher

#68 h Jan 7, 2025, 6:38 AM

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Because $I$ is the incenter of $\triangle{ABC}$ , we have that $ID = IE$ ; we also have that $\angle{BAI} = \angle{CAI}$ by properties of the incenter. From the incenter-excenter lemma, we have that $O$ lies on line $AI$ . Since $\angle{ADI} = \angle{AEI} = 90^{\circ}$ , we have that $\angle{DIO} = \angle{EIO}$ . Thus, by Side-Angle-Side congruence (because obviously $IO = IO$ ), we have that $\triangle{DIO} \cong \triangle{EIO}$ .

This means that $\angle{IDO} = \angle{IEO} \implies 90^{\circ} - \angle{IDO} = 90^{\circ} - \angle{IEO} \implies \angle{BDO} = \angle{OEC}$ , as desired.

This post has been edited 1 time. Last edited by cappucher, Jan 7, 2025, 6:39 AM

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blueprimes

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blueprimes

#69 h Mar 3, 2025, 11:17 PM

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Clearly $O$ is the image of the $A$ -excircle dilated by a factor of $\dfrac{1}{2}$ with respect to $I$ , thus $A, I, O$ are collinear. Now $\angle DIO = 180^\circ - \angle AID = 180^\circ - \angle AIE = \angle EIO$ and $DI = EI$ , so by SAS Congruence we obtain $\triangle DIO \cong \triangle EIO$ . Then
$\angle ODB = \angle IDB - \angle IDO = \angle IEC - \angle IEO = \angle OEC$ as needed.

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LeYohan

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LeYohan

#70 h Mar 30, 2025, 11:54 AM

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By the Incenter/Excenter lemma, $O$ is the intersection of $BI$ with $(ABC)$ .
Looking at triangles $DIO$ and $EIO$ , we notice that $\angle DIO = \angle EIO = 90 + \frac{\angle A}{2}$ .
By the $SAS$ criterion, $\triangle DIO \cong \triangle EIO \implies \angle AOD = \angle AOE$ , and because $\angle DAO = \angle EAO \implies \triangle ADO \cong \triangle AEO \implies \angle ADO = \angle AEO$ , so $\angle ODB = 180 - \angle ADO = \angle OEC$ , as desired. $\square$

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