IMO ShortList 2002, geometry problem 3

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orl

3647 posts

orl

#1 h Sep 28, 2004, 12:49 PM • 9 Y

Y by Davi-8191, mathematicsy, Adventure10, jhu08, megarnie, ImSh95, Mango247, Rounak_iitr, ItsBesi

The circle $S$ has centre $O$ , and $BC$ is a diameter of $S$ . Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$ . Let $D$ be the midpoint of the arc $AB$ which does not contain $C$ . The line through $O$ parallel to $DA$ meets the line $AC$ at $I$ . The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$ . Prove that $I$ is the incentre of the triangle $CEF.$

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orl

3647 posts

orl

#2 h Sep 28, 2004, 12:51 PM • 5 Y

Y by jhu08, phongzel34-vietnam, ImSh95, 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

7849 posts

grobber

#3 h Oct 2, 2004, 6:38 AM • 6 Y

Y by Adventure10, jhu08, phongzel34-vietnam, ImSh95, Mango247, and 1 other user

Triangles $AOE,AOF$ are equilateral because $AE=OE=OA,\ AF=OF=OA$ .

The condition $\angle AOB<120^{\circ}\iff \angle AOC>60^{\circ}$ simply means that $F$ (assume $F$ is on the same side of $OA$ as $C$ ) lies on the arc $AC$ which doesn't contain $E$ , so $CA$ is the internal bisector of $\angle ECF$ because $A$ is the midpoint of the arc $EF$ (without the condition, $AC$ would be the external bisector). This means that $J$ lies on the bisector of $\angle ECF$ .

On the other hand, we have $AD\|OJ$ and $OD\|AJ$ because they're both $\perp AB$ , so $J$ is the reflection of $D$ in the midpoint of $OA$ . This means $\angle EJF=\angle EDF=120^{\circ}$ , and since $\angle ECF=\frac{\angle EOF}2=60^{\circ}$ and $J$ lies on the internal bisector of $\angle ECF$ , it means that $J$ is the incenter (the only other position on $AC$ from which $EF$ is seen under an angle of $120^{\circ}$ is $A$ , and it's clear that $J\ne A$ ).

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Mathias_DK

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Mathias_DK

#4 h Jun 5, 2009, 3:39 PM • 6 Y

Y by jhu08, phongzel34-vietnam, ImSh95, Adventure10, Mango247, and 1 other user

orl wrote:

The circle $S$ has centre $O$ , and $BC$ is a diameter of $S$ . Let $A$ be a point of $S$ such that $\angle AOB < 120{{}^\circ}$ . Let $D$ be the midpoint of the arc $AB$ which does not contain $C$ . The line through $O$ parallel to $DA$ meets the line $AC$ at $I$ . The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$ . Prove that $I$ is the incentre of the triangle $CEF.$

Consider the complex plane with origin $O$ . Wlog assume that $S$ is the unitcircle. For uppercase letter points, let the lowercase letter denote their corresponding complex number. (Except for $I$ , which will has the complex number $z_I$ )

$D$ is the midpoint of the arc $AB$ , ie. $\angle BOD = \angle DOA \iff \frac {b}{d} = \frac {d}{a} \iff a = d^2$ . (Since $|a| = |b| = |d| = 1|$ ) I will prove that $z_I = d^2 - d$ . Obviously $OI \parallel AD$ since $z_I - o = a - d = d^2 - d$ .

$\frac {z_I - a}{c - a} = \frac {d}{1 + d^2} \in \mathbb{R}$ since $\frac {d}{1 + d^2} = \frac {\frac {1}{d}}{\frac {1}{d^2} + 1} = \overline{\frac {d}{1 + d^2}}$ so $I \in AC$ , which concludes: $z_I = d^2 - d$ , since the point is unique.

It is easy to show that $E$ and $F$ are constructed by rotating $A$ $60 ^\circ$ around $O$ positivily and negatively respectively. Let $\omega = e^{i\frac {\pi}{6}}$ , then $e = \omega^2d^2 = u^2, u = \omega d$ and $f = \overline{\omega}^2d^2 = v^2, v = \omega^5 d$ . Since $c = w^2, w = - i$ , and $u,v,w$ the incenter of $\triangle CEF$ is $- uv - vw - wu$ . (Wellknown. See the training materials from imomath.com)

And $- uv - vw - wu = - \omega^6d^2 + i(\omega d + \omega^5 d) = d^2 + i^2d = d^2 - d = z_I$ . So $I$ is the incenter of $\triangle CEF$ .

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Heebeen, Yang

81 posts

Heebeen, Yang

#5 h Jul 20, 2009, 4:20 AM • 4 Y

Y by phongzel34-vietnam, ImSh95, Vladimir_Djurica, Adventure10

enough to show $AD=AF=AI$ and observe just B D A I C O,
easily find $OA=AI$ by elementary angle chasing.
so we are done

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Zhero

2043 posts

Zhero

#6 h Mar 28, 2010, 5:12 PM • 4 Y

Y by phongzel34-vietnam, ImSh95, Adventure10, Mango247

grobber wrote:

Triangles $AOE,AOF$ are equilateral because $AE = OE = OA,\ AF = OF = OA$ .

The condition $\angle AOB < 120^{\circ}\iff \angle AOC > 60^{\circ}$ simply means that $F$ (assume $F$ is on the same side of $OA$ as $C$ ) lies on the arc $AC$ which doesn't contain $E$ , so $CA$ is the internal bisector of $\angle ECF$ because $A$ is the midpoint of the arc $EF$ (without the condition, $AC$ would be the external bisector). This means that $J$ lies on the bisector of $\angle ECF$ .

My solution was the same up to that point (though I think the $J$ should have been $I$ throughout the proof

.) Here's how I finished it:

We have that $m \angle OAC = m \angle ACO = \frac{m \angle BOA}{2} = m \angle DOA$ , so $DO || AI$ . Hence, $DOIA$ is a parallelogram, so $DO = AI$ . On the other hand, $\triangle AOF$ is equilateral, so we have that $AO = AI = AF$ , that is, $AI = AF$ , so $\triangle AIF$ is isosceles.

Let $m \angle AFE = \alpha$ and let $m \angle EFI = \beta$ . $m \angle AIF = \alpha + \beta$ since $\triangle AIF$ is isosceles, so $m \angle FIC = 180 - \alpha - \beta$ . Also, $\alpha = m \angle EFA = m \angle AEF = m\angle ACF$ , so $m \angle IFE = 180 - (180 - \alpha - \beta + \alpha) = \beta$ . Since $m \angle EFI = m \angle IFC$ , $I$ lies on the bisector of $\angle EFC$ . But $I$ also lies on the bisector of $\angle ECF$ , so $I$ is the incenter of $\triangle ECF$ .

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math154

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math154

#7 h Jul 31, 2010, 12:51 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

My solution is slightly different... unfortunately I missed the fact that $DO\|AI$ ...

WLOG let $E$ be on minor arc $BD$ . Let $OI$ hit the circle at $X$ and $Y$ so that $Y$ is on minor arc $FC$ . First we note that $AE=AF$ , so $I$ is on the angle bisector of $\angle{ECF}$ . Since $AO=AE=AF$ , it suffices to show that $AO=AI$ , which would imply that $A$ , the midpoint of minor arc $EF$ , is the circumcenter of $\triangle{EIF}$ and $I$ is thus the incenter of $\triangle{ECF}$ (since the incenter is on this circle, and we know that $I$ is on the angle bisector of $\angle{ECF}$ ). Using the facts that $AD=AB$ and $AD$ is parallel to $XY$ , we have (interpret all arcs as those not containing any point in the interior of minor arc $CX$ )
$\begin{align*} 2\angle{AOI}=2\widehat{AY}=\widehat{AY}+\widehat{DX}&=\widehat{AY}+\widehat{DB}+\widehat{BX}\\ &=\widehat{AY}+\widehat{DA}+\widehat{CY}=\widehat{DY}+\widehat{CY}=\widehat{AX}+\widehat{CY}=2\angle{AIO}, \end{align*}$ as desired.

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AK1024

228 posts

AK1024

#8 h Dec 28, 2010, 9:51 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Hmm.. I did exactly as Zhero did up until his angle-chase. Note that $AF=AO=AI=AE$ so $FEOI$ is cyclic with centre $A$ . Then $2\angle FEI=\angle FAI=\angle FAC=\angle FEC$ so $IE$ bisects $\angle FEC$ . This was much easier than G1!?

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sunken rock

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sunken rock

#9 h Jan 8, 2011, 3:27 PM • 5 Y

Y by Sx763_, Lifefunction, phongzel34-vietnam, ImSh95, Adventure10

Since $DO\perp AB$ , we conclude $DO\parallel AC$ and, with $AD\parallel OI$ , $ADOI$ is a parallelogram, i.e. $AI=OD$ , so $AE=AI=AF$ and $I$ is, indeed, the incenter of $\triangle CEF$ .

Best regards,
sunken rock

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Bertus

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Bertus

#10 h Aug 13, 2011, 9:38 PM • 5 Y

Y by HolyMath, phongzel34-vietnam, ImSh95, Adventure10, Mango247

My solution :
//cdn.artofproblemsolving.com/images/41ca7a0fc26d765bf55afd37d4f1c41cf9b81914.png

Let's show that $I$ is the incenter of the triangle $\triangle{CEF}$ :
First, it's easy to check that $OA=OE=AE=OF=AF$ and hence the quarilateral $AEOF$ is a rhombus, and we have $A$ is the middle of the arc $EF$ and since $A$ lies in the perpendicular bisector of $EF$ , it follow from a well-known fact that $CI$ is the bisector of $\angle{ECF}$ .
Otherwise, from the condition $OI\parallel AD$ , we have : $\angle{AIO}=\pi-\angle{CIO}=\pi-\angle{CAD}=\angle{DBC}$ . Now let $A'$ be the symetric of $A$ wrt $O$ , then $\angle{AOI}=\angle{DAO}=\angle{DAA'}=\frac{\pi}{2}-\angle{ACD}=\frac{\pi}{2}-\angle{DCB}=\angle{DBC}=\angle{AIO}$ since $CD$ is the bisector of $\angle{ACB}$ .
Hence wde get : $AO=AI$ which means that points $E$ , $I$ , $O$ and $F$ are concylic ( In particular in circle with center $A$ ), then :
$\angle{FEI}=\frac{1}{2}\angle{FAI}=\frac{1}{2}\angle{FEC}$ . And so $EI$ is the bisector of $\angle{FEC}$ and so we are done !

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capu

55 posts

capu

#11 h Feb 2, 2013, 12:14 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Very easy solution:

Since <BOD = <BCA then DO||AC and therefore AIOD is a parallelogram. Therefore AI = OD = OA = OE = OF = AF = AO = AE and therefore A is the centre of the circumcircle of quadrilateral FOIE. Therefore <FIE = 180º - <FAE / 2 = 120º since < FCE = 60º and < FAE = 120º (since FAEO is a 60º-rhombus) then if I' is the incentre of CFE, then C, I, I' are collinear since < FCI = arc AF = arc AE = < ACE and FIEI' are concyclic since < FIE = 120º = 90 + <FCE / 2 = < FI'E then I = I'. Done.

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exmath89

2572 posts

exmath89

#12 h Jul 9, 2013, 12:21 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Solution

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fclvbfm934

759 posts

fclvbfm934

#13 h May 15, 2014, 5:15 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15.46cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.300000000000004, xmax = 11.16000000000001, ymin = 1.480000000000002, ymax = 6.300000000000006; /* image dimensions */ /* draw figures */ draw(circle((2.180000000000002,1.960000000000002), 4.201190307520005)); draw((-2.020000000000002,1.860000000000002)--(6.380000000000007,2.060000000000002)); draw((1.057644715918773,6.008495846150148)--(2.180000000000002,1.960000000000002)); draw((-1.887277891922414,3.012259734989030)--(5.124922607841190,4.956236111161120)); draw((1.057644715918773,6.008495846150148)--(-1.194073337803743,4.463123870510994)); draw((1.057644715918773,6.008495846150148)--(6.380000000000007,2.060000000000002)); draw((-1.887277891922414,3.012259734989030)--(6.380000000000007,2.060000000000002)); draw((5.124922607841190,4.956236111161120)--(6.380000000000007,2.060000000000002)); draw((2.180000000000002,1.960000000000002)--(4.431718053722520,3.505371975639156)); draw((-1.194073337803743,4.463123870510994)--(2.180000000000002,1.960000000000002)); draw((1.057644715918773,6.008495846150148)--(-1.887277891922414,3.012259734989030)); draw((1.057644715918773,6.008495846150148)--(5.124922607841190,4.956236111161120)); draw((5.124922607841190,4.956236111161120)--(2.180000000000002,1.960000000000002)); draw((2.180000000000002,1.960000000000002)--(-1.887277891922414,3.012259734989030)); draw((4.431718053722520,3.505371975639156)--(5.124922607841190,4.956236111161120)); draw((4.431718053722520,3.505371975639156)--(-1.887277891922414,3.012259734989030)); /* dots and labels */ dot((2.180000000000002,1.960000000000002),dotstyle); label("$O$", (2.260000000000003,2.080000000000003), NE * labelscalefactor); dot((-2.020000000000002,1.860000000000002),dotstyle); label("$B$", (-1.940000000000002,1.980000000000003), NE * labelscalefactor); dot((6.380000000000007,2.060000000000002),dotstyle); label("$C$", (6.460000000000007,2.180000000000003), NE * labelscalefactor); dot((1.057644715918773,6.008495846150148),dotstyle); label("$A$", (1.140000000000002,6.060000000000006), NE * labelscalefactor); dot((-1.194073337803743,4.463123870510994),dotstyle); label("$D$", (-1.120000000000001,4.580000000000005), NE * labelscalefactor); dot((-1.887277891922414,3.012259734989030),dotstyle); label("$E$", (-1.800000000000002,3.140000000000004), NE * labelscalefactor); dot((5.124922607841190,4.956236111161120),dotstyle); label("$F$", (5.200000000000006,5.080000000000004), NE * labelscalefactor); dot((4.431718053722520,3.505371975639156),dotstyle); label("$I$", (4.520000000000006,3.620000000000004), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $\angle DOB = \theta$ . Notice that $\angle AIO = \angle IOC + \angle ICO = \theta + 90 - 1.5 \theta = \angle AOI$ , so $AO = AI = AF = AE$ . Therefore, $EFIO$ is cyclic, giving us $\angle EIF = \angle EOF = 120^{\circ}$ . Notice that $\angle ECF = 60^{\circ}$ , so $I$ must lie on the circumcircle formed by the incenter of $CEF$ and $E, F$ . But $\angle ACE = \angle ACF = 30^{\circ}$ , so $I$ lies on the angle bisector as well. Therefore, $I$ is the incenter of $CEF$

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Wolstenholme

543 posts

Wolstenholme

#14 h Aug 4, 2014, 4:41 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

This is easily done with complex numbers as well. WLOG $S$ is the unit circle. WLOG let $A, B, C, D, I, E, F$ have complex coordinates $a^2, -1, 1, d, x, e, f$ respectively.

Clearly $d = -ia$ . Since $OD \perp AB$ by definition and since $AI \perp AB$ since $BC$ is a diameter of $S$ we have that quadrilateral $ADOI$ is a parallelogram and so we have that $x = a^2 + ai$ .

Now, note that $E, F \in S$ and that the projections of either point onto $AO$ has complex coordinate $\frac{a^2}{2}$ . This implies that $e, f$ are the roots of the equation $z^2 - a^2z + a^4 = 0$ . Therefore by Vieta's formulas $e + f = a^2$ and $ef = a^4$ .

Now since the circumcircle of $CEF$ is $S$ it suffices to show that $x = -\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f})$ . Choosing the appropriate sign for the square roots given that $\angle{AOB} < 120$ we have that $\sqrt{ef} = -a^2$ and since $(\sqrt{e} + \sqrt{f})^2 = e + f + 2\sqrt{ef} = -a^2$ we have that $\sqrt{c}(\sqrt{e} + \sqrt{f}) = -ai$ so $-\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f}) = a^2 + ai$ as desired.

This is literally the worst solution possible but I felt that I had to post it for its ridiculousness. In fact, you can easily prove that $ADOI$ is a parallelogram with solely complex numbers (it requires solving a simple linear system of two equations).

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v_Enhance

6882 posts

v_Enhance

#15 h Jul 9, 2015, 7:34 PM • 5 Y

Y by jam10307, vsathiam, ImSh95, Adventure10, Mango247

A bit too easy for an IMO #2?

By http://www.mit.edu/~evanchen/handouts/Fact5/Fact5.pdf as usual it suffices to show that $AJ = AE = AF$ , but obviously $AE = AF = AO$ so we simply wish to check that $\angle AOJ$ is isosceles which is angle chasing.

This post has been edited 1 time. Last edited by v_Enhance, Apr 25, 2016, 9:11 PM
Reason: missing space

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huashiliao2020

1292 posts

huashiliao2020

#61 h May 29, 2023, 5:33 PM • 1 Y

Y by ImSh95

orl wrote:

The circle $S$ has centre $O$ , and $BC$ is a diameter of $S$ . Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$ . Let $D$ be the midpoint of the arc $AB$ which does not contain $C$ . The line through $O$ parallel to $DA$ meets the line $AC$ at $I$ . The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$ . Prove that $I$ is the incentre of the triangle $CEF.$

First, note that AOE and AOF are equilateral (perp. bisectors from vertice). Trivially CI is an angle bisector since they both are inscribed angles of 60 degrees, and <ECF=60 degrees. Lemma: ADOI is a parallelogram. Proof: <DOB=2<DCB=<ACB, so by corresponding angles DO//AC, two pairs of parallel lines make a parallelogram. Now, we see that EF perp. AO, EF a chord, so EF is bisected by AO, but so is DI (diagonals are bisected in parallelogram), all concurrent at the midpoint of AO=midpoint of EF=midpoint of DI, so EDFI is a parallelogram. Now, we have 120=<EDF=<EIF, but there exists the unique point incenter iff in a triangle ABC incenter I we have <BIC=90+1/2<BAC, which is true, since 120=90+30, so I is the incenter of CEF, as desired. $\blacksquare$

Additional facts we could've derived: OI is the interior angle bisector of <FOG, since we know <EOF=<EIF, EOIF is cyclic => 1/2<FOG=<FEG=<FEI=<FOI=<FOH, as desired.

Edit: Because angle AOB never exceeds 120 degrees, it can be seen that E and F lie on opposite sides of AO

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pikapika007

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pikapika007

#62 h Jun 29, 2023, 11:17 PM • 1 Y

Y by ImSh95

lol this is so stupid

Note that $ADOI$ is a parallelogram since
$\angle BOD = \frac{\angle BOA}{2} = \angle BCA$ and $AFOE$ is a rhombus, so $AE = AF = FO = OD = AI$ and Fact 5 finishes. $\blacksquare$

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john0512

4191 posts

john0512

#63 h Jul 21, 2023, 10:47 PM

Y by

Claim: $A$ is the circumcenter of $\triangle EIF$ . Let $P$ be the intersection of $AD$ and $BC$ . Let $\angle ACD=\angle BCD=\alpha$ . Then, $\angle BAD=\alpha,$ so $\angle APB=90-3\alpha.$ Thus, $\angle IOC=90-3\alpha,$ and since $\angle AOB=4\alpha$ we have $\angle AOI=90-\alpha$ . However, $\angle AIO=90-3\alpha+2\alpha=90-\alpha$ as well, so $AO=AI$ . Clearly, $AO=AE=AF$ since $\triangle AEO$ and $\triangle AFO$ are equilateral, so $AE=AF=AI$ , thus $A$ is the circumcenter of $\triangle EFI$ .

Consider $\triangle CEF$ . $I$ lies on $CA$ (since $A$ is the arc midpoint), as well as the circle centered at $A$ passing through $E$ and $F$ . Thus, $I$ is either the incenter or the $C$ -excenter.

Claim: $I$ is inside the circle $S$ . Note that $ADOI$ is a parallelogram since $\angle DOB=\angle ICO=2\alpha$ and we already know the other sides are parallel. Thus, $I$ is the reflection of $D$ over the midpoint of $AO$ . Clearly, $\angle DAO<60$ , thus $I$ lies inside the circle, hence done.

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cj13609517288

1926 posts

cj13609517288

#64 h Aug 27, 2023, 3:34 AM

Y by

Note that triangles $AFO$ and $AEO$ are equilateral triangles so we get some easy angles. In particular, $\angle FCA=\angle ECA$ .

Claim. $DAJO$ is a parallelogram.
Proof. $DA\parallel OJ$ by definition and
$\angle BOD=2\angle BCD=\angle BCA \Longrightarrow DO\parallel AJ.$
Let the midpoint of $AO$ be $M$ . Then a reflection about point $M$ takes $F$ to $E$ , $A$ to $O$ , and $J$ to $D$ . Therefore,
$\angle FJC=180^{\circ}-\angle FJA=180^{\circ}-\angle EDO =180^{\circ}-\left(90^{\circ}-\frac12\angle EOD\right) =90^{\circ}+\frac12\angle EOD.$ However,
$\angle FEC=\angle FAC=60^{\circ}+\angle OAC=60^{\circ}+\angle ACB=60^{\circ}+\frac12\angle AOB=60^{\circ}+\angle DOA=\angle DOE\;\blacksquare$

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eibc

600 posts

eibc

#65 h Aug 28, 2023, 1:53 AM

Y by

how did this make the shortlist

Note that $AEOF$ is a rhombus so
$\angle ACE = \angle AFE = \angle FEA = \angle FCA,$ and $I$ lies on the internal bisector of $\angle ECF$ . But because
$\angle DAC = 180^{\circ} - \angle DBC = 180^{\circ} - \angle ADO,$ we find that $ADOI$ is a parallelogram, and thus $AI = DO = FO = AF = AE$ . So, $A$ is the center of $(EFI)$ , and since $A$ is also the midpoint of arc $EF$ not containing $C$ , the conclusion follows by fact 5.

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asdf334

7585 posts

asdf334

#66 h Nov 10, 2023, 10:31 PM

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hello,

$AI=DO=AO=AE=AF$ so by Fact 5 done.

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lian_the_noob12

173 posts

lian_the_noob12

#67 h Jan 23, 2024, 5:27 PM

Y by

$\color{magenta}\boxed{\textbf{SOLUTION G3}}$

$\color{green} \textbf{Claim 1:}$ $CA$ is the angle bisector of $\angle ECF$
$\textbf{proof :}$ As $EF$ is the perpendicular bisector of $OA$ , $A$ is the midpoint of arc $EF$ Hence done $\square$

$\color{green} \textbf{Claim 2:}$ $\triangle AOF$ is equilateral
$\textbf{proof :}$ Let $OA \cap EF \equiv X$
$\triangle AXF$ and $\triangle OXF$ are congruent. Hence $AF=OF=OA \square$

$\color{green} \textbf{Claim 3:}$ $ADOI$ is a parallelogram
$\textbf{proof :}$ $\angle DAC = 180- \angle DBC = 180 - \angle ADO$ and we have $AD \parallel OI \square$

So, $AI = DO = FO = AF= AE$ And by Incenter-Excenter Lemma we are done $\blacksquare$

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lelouchvigeo

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lelouchvigeo

#68 h Jan 24, 2024, 5:28 PM

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It is easy to prove $CI$ is angle bisector. Just observe $AI=DO=AE$ and we are done.

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mahmudlusenan

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mahmudlusenan

#69 h Feb 7, 2024, 11:39 AM • 1 Y

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Prove that: $AI=AF=AE$

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Math4Life7

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Math4Life7

#70 h Mar 1, 2024, 2:50 AM

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We can see that $EO = EA$ since this is the perpendicular bisector. Also, we can see that $EO = OA$ . Thus, $\triangle OEA$ is equilateral. Similarly, $\triangle OAF$ is equilateral.

Now notice that $J$ lies on the angle bisector of $\angle ECF$ . This is because $A$ is the midpoint of arc $EF$ .

Now since $OD$ hits $AB$ at a $90$ degree angle and same with $CA$ , we can see that $OD \parallel CA$ . Thus $J$ is the reflection of $D$ over the midpoint of $OA$ . This means that $\angle EJF = \angle EDF = 120$ . Thus we can conclude. $\blacksquare$

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blueberryfaygo_55

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blueberryfaygo_55

#71 h Apr 26, 2024, 1:06 AM

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Claim: $CI$ is the angle bisector of $\angle ECF$ .
Proof. In cyclic quadrilateral $EAFC$ , we have $\angle ACF = \angle AEF = \angle AFE = \angle ACE$ , and the result follows. $\blacksquare$

Claim: $ADOI$ is a parallelogram.
Proof. $D$ lies on the perpendicular bisector of $AB$ , so we have $OD \perp AB$ . However, $AC \perp AB$ since $BC$ is the diameter of $S$ . Thus, $OD \parallel AC$ , or $OD \parallel AI$ , and combining with the given $OI \parallel AD$ , the claim is shown. $\blacksquare$

Now, since $AO$ and $EF$ bisect one another, $AEOF$ is a rhombus, which implies that $AF = AO = OD = AI$ . Let $I'$ be the incenter of $\angle CEF$ . By the Incenter-Excenter Lemma in $\Delta CEF$ , $I'$ is the unique point on the angle bisector of $\angle ECF$ such that $AI' = AE = AF$ , but $AI = AE = AF$ and $CI$ is the angle bisector of $\angle ECF$ , so $I' = I$ , as desired. $\blacksquare$

This post has been edited 1 time. Last edited by blueberryfaygo_55, Apr 26, 2024, 1:36 AM

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EpicBird08

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EpicBird08

#72 h Jun 5, 2024, 1:06 AM

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Claim: $ADOI$ is a parallelogram.
Proof: We already know by definition that $OI \parallel DA;$ it suffices to show that $DO \parallel AI.$ But $\measuredangle DOA = 2\measuredangle DCA = \measuredangle BCA = \measuredangle OCA = \measuredangle CAO,$ which implies the claim.

Since $DO = AO,$ we get that $AO = AI.$ But it is clear (say by properties of regular hexagons) that $AE = AF = AO.$ Thus $AE = AF = AI.$ Combining this with the fact that $CI$ bisects $\angle ECF$ clearly, we conclude by the Incenter-Excenter Lemma that $I$ is the incenter of $CEF.$

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BestAOPS

707 posts

BestAOPS

#73 h Jul 16, 2024, 8:26 PM

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Simple angle chasing reveals that triangles $ODA$ and $AIO$ are congruent, and thus $AI = AO = r$ , where $r$ is the radius of the circle. Furthermore, $AF = FO = AO = AE$ , so $F,I,E$ lie on a circle with center $A$ . Lastly, notice that $A$ is the arc midpoint of arc $FE$ , so the incenter-excenter lemma applies.

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Maximilian113

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Maximilian113

#74 h Dec 2, 2024, 8:47 PM

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It is clear that $\triangle AEO, \triangle AFO$ are equilateral. Therefore, $AE=AF \implies \angle ECA = \angle FCA, \text{ and } OE=OA=OF,$ so the by Incenter-Excenter Lemma the desired conclusion follows. QED

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Avron

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Avron

#75 h Apr 10, 2025, 10:39 AM

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Let the radius of $S$ be $r$ . Note that $AE=OE=r=OF=AF$ so $A$ is the midpoint of arc $EF$ therefore it's enough to prove that $AI=AE=r$ . $DO$ is the perpendicular bisector of $AB$ so $DO \perp AB$ , also $AI\perp AB$ so $AI\parallel DO$ and $AIOD$ is a parallelogram $\Rightarrow AI=DO=r$ and we're done.

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