IMO 2010 Problem 2

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3647 posts

orl

#1 h Jul 7, 2010, 5:33 PM • 14 Y

Y by BuiBaAnh, Davi-8191, a_friendwr_a, megarnie, Adventure10, Mango247, Rounak_iitr, Tastymooncake2, Funcshun840, cubres, MS_asdfgzxcvb, and 3 other users

Given a triangle $ABC$ , with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$ . Let $E$ be a point on the arc $BDC$ , and $F$ a point on the segment $BC$ , such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$ . If $G$ is the midpoint of $IF$ , prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$ .

Proposed by Tai Wai Ming and Wang Chongli, Hong Kong

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silouan

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silouan

#2 h Jul 7, 2010, 6:16 PM • 42 Y

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It suffices to prove that $\angle{IDG}=\angle{AEI}$ . Taking the excenter we have to prove that the triangles $AFI_a$ and $AIE$ are similar. But this is easy because it is enough to show that $\frac{AF}{AI_a}=\frac{AI}{AE}$ . But from the similarity of ABF,AEC we have that $\frac{AE}{AC}=\frac{AB}{AF}$ . So we have to prove that $AI\cdot AI_a=AB\cdot AC$ which is clearly true.

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kalantzis

25 posts

kalantzis

#3 h Jul 7, 2010, 7:22 PM • 12 Y

Y by futurestar, a_friendwr_a, megarnie, Adventure10, Mango247, Mad_SciSt, Tastymooncake2, farhad.fritl, cubres, and 3 other users

I have a different aproach..
Suppose EI cuts $\Gamma$ at K. Let the parallel from I to BC cut AF to P. Then $AKPI$ is cyclic (because AK is antiparallel to IP or by simple angle chase).
Now we prove that D,P,K are collinear:From the cyclic $AKPI$ $\widehat{AKP}= \widehat{PID}= \widehat{PIB} + \widehat{BID}= \widehat{B} + \widehat{A}/2$ . But $\widehat{AKD}= \widehat{ABD}= \widehat{ABC} + \widehat{CBD}= \widehat{B} + \widehat{A}/2$ .

If line DPK cuts BC at Q, it suffices to prove that $IQ \parallel AF$ since then, $PIQF$ will be parallelogram and G the intersection of DK and IE.
Since $\widehat{IAP}= \widehat{IKP}$ and we want to show $\widehat{IAP}= \widehat{DIQ}$ it is enough $DI^2= DQ \cdot DK$
But DI=DB (well-known fact) so we have to prove $DB^2=DQ \cdot DK$ which is obvious from similar triangles $DBQ,DKB$ since arcs BD and DC are equal.
QED
Image not found

Note: the condition $\angle BAF=\angle CAE <\frac{1}2\angle BAC$ it is not necessary.

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m.candales

186 posts

m.candales

#4 h Jul 7, 2010, 9:54 PM • 7 Y

Y by Illuzion, a_friendwr_a, megarnie, Adventure10, Mango247, Tastymooncake2, cubres

Let $L$ be the second intersection of $IE$ and $\Gamma$ . I will prove that $G, D, L$ are collinear

Let $M$ be a point on the prolongation of $AD$ ( $D$ is between $A$ and $M$ ) such that $\angle{IBM}=90$
$\angle{IBC}=\frac{1}{2}\angle{B}$ and $\angle{CBD}=\frac{1}{2}\angle{A}$ . Then $\angle{DBM}=\frac{1}{2}\angle{C}$ ,
$\angle{BIM}=\frac{1}{2}\angle{A}+\frac{1}{2}\angle{B}$ . Then $\angle{BMI}=\frac{1}{2}\angle{C}$ .
Then $BD=DM$ . Then $BD$ bisects $IM$ because $\angle{IBM}=90$ . Then $ID=DM$
(I believe all this results are well-known)

$\triangle{ABM}$ is similar to $\triangle{AIC}$ . Then $\frac{AM}{AC}=\frac{AB}{AI}$

$\triangle{AEC}$ is similar to $\triangle{ABF}$ . Then $\frac{AE}{AC}=\frac{AB}{AF}$

Then we have $\frac{AM}{AE}=\frac{AF}{AI}$ . But $\angle{FAM}=\angle{IAE}$ .
Then $\triangle{AFM}$ and $\triangle{AIE}$ are similar. Then $\angle{FMA}=\angle{IEA}=\angle{LDA}$
Then $LD$ and $FM$ are parallel. But $ID=DM$ then $LD$ bisects $FI$ . Then $G, D, L$ are collinear.

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m.candales

186 posts

m.candales

#5 h Jul 7, 2010, 10:21 PM • 5 Y

Y by a_friendwr_a, Adventure10, Mango247, Tastymooncake2, cubres

I just want to add that there is another way of continuing my solution after we got that $ID=DM$ which I believe is well-known.
This is the continuation:
Let $AI=i, AD=l, AB=c, AC=b$ . Then $AM=2l-i$
$\triangle{ABM}$ is similar to $\triangle{AIC}$ . Then $\frac{AM}{AB}=\frac{AC}{AI}$ . Then $\frac{2l-i}{c}=\frac{b}{i}$
Then $(2l-i)i = bc$ . Then $il-i^2=bc-il$ (*)

Let $N$ the intersection of $LD$ and $AF$ .
$\frac{AN}{AD}=\frac{AI}{AE}$ because $\triangle{AND}$ and $\triangle{AIE}$ are similar.
Then $AN=\frac{il}{AE}$
$\frac{AF}{AB}=\frac{AC}{AE}$ because $\triangle{ABF}$ and $\triangle{AEC}$ are similar. Then $AF=\frac{bc}{AE}$
Then $FN=AF-AN=\frac{bc-il}{AE}$

Let $G'$ the intersection of $LD$ and $AF$ . Then $\frac{IG'}{G'F}\frac{FN}{AN}\frac{AD}{ID}=1$ by Menelaus

Then $\frac{IG'}{G'F}=\frac{ID}{AD}\frac{AN}{FN}=\frac{il(l-i)}{(bc-il)l}=\frac{il-i^2}{bc-il}=1$ by (*)
Then $G'=G$ , and then $D, G, L$ are collinear.

This problem can also be solved automatically using complex numbers. The solution is long and painful to write, but I will try to post it soon

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NickNafplio

422 posts

NickNafplio

#6 h Jul 7, 2010, 11:59 PM • 4 Y

Y by Mathematicalx, a_friendwr_a, Adventure10, cubres

Another solution:

We need to prove that <GMI = <IEA. It is well known (and it can be proved easily) that the midpoint D of the arc BC is the center of the circumcircle U of the triangle BIC. Let T be the symmetric of I with respect to D, which is the intersection of the line AD and the circle U, then FT//MG and <FTI = <GMI, so we need <FTI = <IEA, or its enough to prove that triangles AFT and AIE are similliar, or equivalently AI/AE = AF/AT <=> AI*AT = AE*AF (1). From the similliar triangles ABF and AEC we have AE/AC = AB/AF <=> AE*AF = AB*AC. So we need AI*AT = AB*AC <=> AI/AB = AC/AT, which is true because the triangles ABI, ATC are similliar (<BAI = <TAC = A/2, <ABI = <ABC/2 = <ADC/2 = <ATC)!

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armpist

527 posts

armpist

#7 h Jul 8, 2010, 12:25 AM • 9 Y

Y by Wizard_32, a_friendwr_a, megarnie, Adventure10, Mango247, Tastymooncake2, ehuseyinyigit, cubres, and 1 other user

Problem with a lot of points on a circumference calls for
Pascal theorem.

Historical note:
he was 13 y.o. when he discovered it, probably solving something similar
to this Problem #2 at French National math olympiad years ago.

Mr. T

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abacadaea

2176 posts

abacadaea

#8 h Jul 8, 2010, 1:27 AM • 4 Y

Y by a_friendwr_a, Adventure10, Mango247, cubres

Click to reveal hidden text

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Sung-yoon Kim

324 posts

Sung-yoon Kim

#9 h Jul 8, 2010, 10:38 AM • 4 Y

Y by Adventure10, Mango247, cubres, and 1 other user

We show that $\triangle{AFI_a}$ and $\triangle{AIE}$ are similar. Then we have $\angle{AEI}=\angle{AI_a F}=\angle{ADG}$ and we're done. To show that, inverse the plane with regard to $A$ with the radius $\sqrt{bc}$ where $b=AC,c=AB$ . Then we have another figure which can be also obtained by reflecting the original figure. Note that $E,F$ are mapped to $F,E$ resp. Hence $AE \cdot AF = bc = AI \cdot AI_a$ , which implies directly that $\triangle{AFI_a}$ and $\triangle{AIE}$ are similar, as desired.

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April

1270 posts

April

#10 h Jul 8, 2010, 12:02 PM • 3 Y

Y by Chokechoke, Adventure10, cubres

orl wrote:

Given a triangle $ABC$ , with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$ . Let $E$ be a point on the arc $BDC$ , and $F$ a point on the segment $BC$ , such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$ . If $G$ is the midpoint of $IF$ , prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$ .

Let $D'$ be a point on $AI$ such that $DI = DD'$ . Notice that $\angle IDG = \angle AD'F$ . So we only have to prove that $\angle AD'F = \angle AEI$ .

We have $\triangle ABF\sim\triangle AEC$ , therefore $AB\cdot AC = AE\cdot AF\quad (1)$ .
$\triangle ABI\sim\triangle AD'C$ $\Longrightarrow AB\cdot AC = AI\cdot AD'\quad (2)$ .
Combine $(1)$ and $(2)$ , we have $AE\cdot AF = AI\cdot AD'$ , i.e. $\dfrac{AF}{AD'}=\dfrac{AI}{AE}$ . On the other hand, $\angle FAD' = \angle IAE$ , so $\triangle FAD'\sim\triangle IAE$ . It follows $\angle AD'F=\angle AEI$ , which completes our solution.

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feliz

290 posts

feliz

#11 h Jul 8, 2010, 8:01 PM • 3 Y

Y by Adventure10, Mango247, cubres

To add a little algebraic point of view...

The points where $EI$ and $DG$ meet $\Gamma$ are functions of $\angle AEI$ and $\angle ADG$ , which we must prove are equal. Let $DG$ meet $AF$ at $T$ . Since $\angle TAD = \angle IAE$ , we must prove that $TAD$ and $IAE$ are congruent. To do this, we need a spiral similarity between $ATI$ and $ADE$ . On the other hand, naming $\theta = \angle BAF$ , we see that $\angle AFC = \angle ADE = \angle ABC + \theta$ . Thus, if $\{P\} = FC \cap AD$ , we have $AFP$ similar to $ADE$ , and it remains to prove $AFP$ is similar to $ATI$ , or $TI \parallel FP$ . If this happens, let ${Q} = BC \cap DG$ . Since $G = \frac{F + I}{2}$ , $TIQF$ must be a parallelogram. Reversely, if $QI \parallel AF$ , we are going to have that parallelogram. In other words, we need to prove $QIP$ is similar to $FAP$ .

We prove this happens in the degenerate cases $F = B$ and $F = C$ . When $F$ varies on (in? at?) $BC$ , $G$ varies linearly, and so does $P$ , because it is a mean of $G$ and $D$ , with weights that depend only on the ratio $IP/PD$ . So our linearity ends the argument. If $F = B$ (draw another figure!), however, $BID$ is isosceles, so that $DG$ is a simmetry axis, which leads to $\angle PIQ = \angle DIQ = \angle DBQ = \frac{1}{2}\angle BAC = \angle PAB$ . And here is the similarity!

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Zhero

2043 posts

Zhero

#12 h Jul 8, 2010, 9:04 PM • 2 Y

Y by Adventure10, cubres

Let $M$ be the midpoint of $AI$ . We want to show that $\angle GDM = \angle IEA$ , that is, $\triangle MGD \sim \triangle AIE$ .

$MG$ is parallel to $AF$ , so $\angle GMD = \angle FAD = \angle IAE$ . Hence, $\triangle MGD \sim \triangle AIE$ if and only if $\frac{MG}{MD} = \frac{AI}{AE}$ . $MG = \frac{AF}{2}$ , so this is equivalent to $2 AI \cdot MD = AE \cdot AF$ .

We claim that $AE \cdot AF$ is fixed. It is then sufficient to show that the result is true for some choice of $E$ and $F$ (namely, when $E=C$ and $F=B$ ), as it would imply that $2 AI \cdot MD = AE \cdot AF$ for some choices of $E$ and $F$ , and thus for all choices of $E$ and $F$ .

Showing that $AE \cdot AF$ is not difficult. $\angle ABF = \angle AEC$ and $\angle BAF = \angle CAE$ , so $\triangle ABF \sim \triangle AEC$ , so $AE \cdot AF = AB \cdot AC$ .

That the result is true when $F=B$ and $E=C$ is not difficult to show either. In this case, $G$ is the midpoint of the base of isosceles $\triangle IBD$ , so $DG$ is the bisector of $\angle BDA$ , so it meets $\Gamma$ on the midpoint of minor arc $AB$ . On the other hand, $CI$ trivially meets $\Gamma$ on the midpoint of minor arc $AB$ as well, so our proof is complete.

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Mithril

28 posts

Mithril

#13 h Jul 9, 2010, 2:57 PM • 4 Y

Y by thunderz28, Adventure10, Mango247, cubres

Let $EI$ intersect $\Gamma$ again at $P$ . We'll prove that $DP$ meets $FI$ at its midpoint.

Let $DP$ cut $AF$ and $BC$ at $X$ and $Y$ , resp. Notice that $APXI$ is cyclic, because $\angle XAI = \angle DAE = \angle XPI$ . We'll see that $XFYI$ is a parallelogram.

Lemma 1: $XI$ is parallel to $FY$ .
Proof: As $APXI$ is cyclic, we have $\angle PAI = \angle DXI$ . But $\angle PAI$ is equal to the angle between $DP$ and the tangent to $\Gamma$ through $D$ . Then, it follows that $XI$ and that tangent are parallel. But as $D$ is the midpoint of arc $BC$ , $XI$ must be parallel to $BC$ too, and the lemma follows.

Lemma 2: $D$ is the circumcenter of $BIC$ .
Proof: As $D$ is the midpoint of arc $BC$ , we have $DB = DC$ . Now, let the circle with center $D$ which passes through $B$ and $C$ meet $AD$ at $I'$ .
We know that $\angle I'CB = \angle BDI'/2 = \angle BCA/2$ . Thus $I'$ is also on the bisector of $\angle C$ , so $I'$ is the incenter, and the lemma follows.

Lemma 3: $FX$ is parallel to $YI$ .
Proof: Consider the inversion with center $D$ and radius $DB$ . It maps $BC$ into $\Gamma$ , and, by lemma 2, fixes $I$ .
Now, as $Y$ is mapped to $P$ and $I$ is fixed, we have $\angle YID = \angle DPI$ . But we know that $XPAI$ is cyclic, thus $\angle YID = \angle XAI$ , and the result follows.

Now, by lemmas 1 and 3, we get that $XFYI$ is a parallelogram. Thus $DP$ meets $FI$ at its midpoint, which is what we wanted to prove.

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wenxin

24 posts

wenxin

#14 h Jul 10, 2010, 1:11 PM • 2 Y

Y by Adventure10, cubres

orl wrote:

Given a triangle $ABC$ , with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$ . Let $E$ be a point on the arc $BDC$ , and $F$ a point on the segment $BC$ , such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$ . If $G$ is the midpoint of $IF$ , prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$ .

2010 IMO Problem 2 was proposed by Tai Wai Ming (2008 Hong Kong IMO team member) and Wang Chongli

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lchserious

80 posts

lchserious

#15 h Jul 10, 2010, 3:02 PM • 4 Y

Y by RudraRockstar, Adventure10, Mango247, cubres

wenxin wrote:

orl wrote:

Given a triangle $ABC$ , with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$ . Let $E$ be a point on the arc $BDC$ , and $F$ a point on the segment $BC$ , such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$ . If $G$ is the midpoint of $IF$ , prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$ .

2010 IMO Problem 2 was proposed by Tai Wai Ming (2008 Hong Kong IMO team member) and Wang Chongli

Oh really !?

Tai Wai Ming was my teammate but we have not kept in contact since IMO 2008

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HamstPan38825

8864 posts

HamstPan38825

#77 h Jul 29, 2023, 1:46 PM • 1 Y

Y by cubres

Very enjoyable and nice configuration in general.

Let $I_A$ be the $A$ -excenter, and $H = \overline{EI} \cap (ABC)$ . It suffices to show that $\overline{HD} \parallel \overline{FI_A}$ by homothety reasons (then it passes through $G$ .)

On the other hand, notice that $AIE$ and $AFI_A$ are spirally similar as $AE \cdot AF = AB \cdot AC = AI \cdot AI_A$ . This implies $L = \overline{FI_A} \cap \overline{IE}$ lies on $(AIF)$ . So $\measuredangle IHD = \measuredangle EAD = \measuredangle DAF = \measuredangle ILF,$ implying the result.

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huashiliao2020

1292 posts

huashiliao2020

#78 h Aug 26, 2023, 2:09 AM • 1 Y

Y by cubres

Wasted a couple hours trying to find a synthetic without constructing anything, i seriously need to get better at this; also i am NOT going to type in latex properly

Construct the A-excenter I_a, and let $H=DG\cdot IE$ ; we want to prove $AI_aF=ADG=AEI$ , and since we know $FAI_a=EAI_a$ , we're motivated to try to prove that $AFI_a\similar AIE$. Indeed, we have

$$BAI=I_aAC,ABI=IBC=II_aC\implies ABI\similar AI_aC\implies AB\cdot AC=AI\cdot AI_a\implies ABF\similar AEC\implies AE\cdot AF=AB\cdot AC=AI\cdot AI_a,$$

upon which dividing yields SAS similarity and we're done. $\blacksquare$

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Zhaom

5124 posts

Zhaom

#79 h Oct 8, 2023, 5:33 PM • 1 Y

Y by cubres

Assume $AB\le{}AC$ . The case where $AB>AC$ is similar. Let $K'$ be the intersection of $\overline{EI}$ and $\Gamma$ , let $X$ be the intersection of $\overline{AF}$ and $\overline{K'D}$ , and let $Y$ be the intersection of $\overline{K'D}$ and $\overline{BC}$ . We have that $\angle{}DK'E=\angle{}DAE=\angle{}FAD$ , so $K'AIX$ is cyclic. This means that $\angle{}AXI=\angle{}AK'E=\angle{}ABE=\angle{}AFC$ since $\triangle{}ABE\sim\triangle{}AFC$ , so $\overline{XI}\parallel\overline{BC}$ . It is therefore sufficient to prove that $\overline{AF}\parallel\overline{IY}$ .

Let the circumcircle of $\triangle{}AK'I$ intersect $\overline{AB}$ and $\overline{AC}$ at points $M$ and $N$ and let the intersection of $\overline{MN}$ and $\overline{AI}$ be $W$ . We have that $\angle{}K'MN=180^\circ-\angle{}K'AC=\angle{}K'BC$ and $\angle{}K'NM=\angle{}K'AB=\angle{}K'CB$ , so $\triangle{}K'MN\sim\triangle{}K'BC$ . The spiral similarity mapping $\triangle{}K'MN$ to $\triangle{}K'BC$ maps $I$ to $D$ and maps $A$ to a point $Z$ . We claim that $\overline{DZ}\parallel\overline{AF}$ . We will do this by showing that $\angle{}AWN=\angle{}AFC$ . Notice that
$\begin{align*} \angle{}AWN&=\angle{}AMN+\angle{}BAD\\ &=\angle{}AMI-\angle{}IMN+\angle{}BAD\\ &=\angle{}AMI-\angle{}IAN+\angle{}BAD\\ &=\angle{}AMI-\frac{\angle{}CAB}{2}+\frac{\angle{}CAB}{2}\\ &=\angle{}AMI\\ &=\angle{}AXI\\ &=\angle{}AFC \end{align*}$ since $K'AIX$ is cyclic and $\overline{XI}\parallel\overline{BC}$ , proving the claim.

It is then sufficient that $\angle{}YID=\angle{}IDZ$ . However, since $\triangle{}K'AI\sim\triangle{}K'ZD$ , we have that
$\begin{align*} \angle{}IDZ&=\angle{}K'DZ-\angle{}K'DA\\ &=\angle{}K'IA-\angle{}K'DA\\ &=\angle{}EK'D, \end{align*}$ so it is sufficient that $\triangle{}DYI\sim\triangle{}DIK'$ , or $DY\cdot{}DK'=DI^2$ . However, by the Incenter-Excenter Lemma, we have that $DI^2=DB^2$ , so it is sufficient to prove that $DY\cdot{}DK'=DB^2$ , or $\triangle{}DYB\sim\triangle{}DBK'$ . Therefore, it is sufficient that $\angle{}YBD=\angle{}BK'D$ , but this is true since $\angle{}BK'D=\angle{}BAD=\angle{}CAD=\angle{}CBD=\angle{}YBD$ .

This post has been edited 3 times. Last edited by Zhaom, Feb 23, 2024, 4:59 AM

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AngeloChu

471 posts

AngeloChu

#80 h Jan 5, 2024, 11:50 PM • 1 Y

Y by cubres

first off, let $AF$ intersect $\Gamma$ at $Q$ , let $EI$ intersect $\Gamma$ at $P$ , let $PD$ intersect $AQ$ at $L$ , let $BC$ intersect $PD$ at $M$ , and let $PD$ intersect $IF$ at $J$
it is obvious that $QD=ED$ and $BQ=EC$ , so $QE$ is parallel to $BC$
our problem is equivalent to proving $DP$ bisects $IF$ , or $J$ is the midpoint of $IF$
we have $QPD=EPD=QAD=DAE$ , $PQA=PDA=PEA$ , $DAB=DCB=DAC$ , $PQA=PDA=PEA$ , so $PQL$ , $PDI$ , $ADL$ and $AEI$ are all similar
from this, we can get that $APLI$ are concyclic, and $LI$ is parallel to $QE$ and $BC$
we can then get that $FJM$ and $LJI$
(to be finished later)

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Aiden-1089

285 posts

Aiden-1089

#81 h Mar 9, 2024, 1:49 PM • 1 Y

Y by cubres

Let $X$ and $P$ be the second intersections between $\Gamma$ and $EI$ , $AF$ respectively. Let $Y=DX \cap AP$ , $M=DX \cap IF$ . We need to show that $M$ is the midpoint of $IF$ .
Note that $BC$ , $EP$ , and the tangent to $\Gamma$ and $D$ are parallel.
By Pascal's theorem on $ADDXEP$ , $IY$ is parallel to $BC$ .

We proceed by length chasing. Let $Q=IY \cap DF$ , $H=AD \cap BC$ . By Ceva's theorem on the cevians $DM$ , $FA$ , $IQ$ of $\Delta DFI$ , intersecting at $Y$ , $IM=MF \iff \frac{QF}{QD}=\frac{AI}{AD} \iff \frac{DH}{HI}=\frac{DI}{AI}$ .
For convenience let $\angle BAC = 2\alpha$ , $\angle ABC = 2\beta$ , $\angle ACB = 2\gamma$ .
$\frac{DH}{HI}=\frac{\frac{BH}{\sin 2\gamma} \cdot \sin \alpha}{\frac{BH}{\sin (90^{\circ}-\gamma)} \cdot \sin \beta}=\frac{\sin \alpha}{2\sin\beta \sin\gamma}.$ $\frac{DI}{AI}=\frac{BD}{\frac{AB}{\sin(90^{\circ}-\gamma)}\cdot \sin \beta}=\frac{BD}{AB} \cdot \frac{\cos \gamma}{\sin \beta}=\frac{\sin \alpha}{\sin 2\gamma} \cdot \frac{\cos \gamma}{\sin \beta}=\frac{\sin \alpha}{2\sin\beta \sin\gamma}.$ This concludes the proof. $\square$

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bryanguo

1032 posts

bryanguo

#82 h Mar 31, 2024, 8:24 PM • 1 Y

Y by cubres

Delete the $<\tfrac{1}{2} \angle BAC$ condition. We prove a stronger version of the problem statement with MMP.

Define $K_1=\Gamma \cap \overline{EI}$ and $K_2=\Gamma \cap \overline{DG}.$ Fix $\triangle ABC$ and animate $F$ linearly along $\overline{BC}.$ Then $I,D$ are fixed and $E,K_1,K_2,G$ vary as a function of $F.$

Note $F \mapsto E$ is a projective map since rotations preserve cross ratios. Furthermore, $E \mapsto K_1$ is a projective map, so $F \mapsto K_1$ is projective. On the other hand, by homothety about $I$ with ratio $1/2,$ we know $G$ moves linearly. Taking perspectivity at $D,$ the map $G \mapsto K_2$ is projective. It suffices to show these maps coincide for $3$ values of $F.$ This is not hard to do. Check $F=B, F=C,$ and $F=\overline{AI} \cap \overline{BC}.$

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joshualiu315

2534 posts

joshualiu315

#83 h Jun 4, 2024, 12:24 PM • 1 Y

Y by cubres

Let $X = \overline{EI} \cap \overline{DG}$ and denote the $A$ -excenter of $\triangle ABC$ as $I_A$ It suffices to show $\angle ADX = \angle AEX$ .

It is very clear that $\overline{DG} \parallel \overline{I_AF}$ . Then, observe that

$\begin{align*} \angle ADX = \angle AEX & \iff \angle AI_AF = \angle AEI \\ & \iff \triangle AEI \sim \triangle AI_AF \\ & \iff \frac{AE}{AI} = \frac{AI_A}{AF} \\ & \iff AE \cdot AF = AI \cdot AI_A. \end{align*}$
However, $\sqrt{bc}$ inversion swaps $I, I_A$ and $E, F$ so the conclusion follows immediately. $\square$

Remarks: The inversion is not necessary to prove $AE \cdot AF = AB \cdot AC$ ; in fact, I realized that similar triangles are way easier to prove in retrospect. A more rigorous way to show that $E$ and $F$ are swapped is to realize that they lie on isogonal conjugates in $\angle BAC$ and that $\overline{BC}$ and $(ABC)$ are swapped under the inversion.

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P2nisic

406 posts

P2nisic

#84 h Jun 23, 2024, 9:28 AM • 1 Y

Y by cubres

orl wrote:

Given a triangle $ABC$ , with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$ . Let $E$ be a point on the arc $BDC$ , and $F$ a point on the segment $BC$ , such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$ . If $G$ is the midpoint of $IF$ , prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$ .

Proposed by Tai Wai Ming and Wang Chongli, Hong Kong

Let $I_A$ be the $A$ excenter then:
By symetric invertion with center $A$ we get that $AF\cdot AE=AI\cdot AI_A\Rightarrow \frac{AF}{AI}=\frac{AI_A}{AE}$ and with $\angle FAI_A=\angle IAE$ we get that $AFI_A\approx AIE$

Now since $D$ is the midpoint of $II_AS$ and $G$ the midpoint of $IF$ we get that:
$\angle IDG=\angle II_AF=\angle AEI\Rightarrow EI\cap DG\varepsilon (ABC)$

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Ywgh1

139 posts

Ywgh1

#85 h Jun 26, 2024, 12:38 PM • 1 Y

Y by cubres

Solved with SuperHmm7 and Ammh4

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.7819946725137cm); 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 = -7.110549584948537, xmax = 22.671445087565164, ymin = -7.116149568571489, ymax = 6.9983233667149; /* image dimensions */ pen qqttzz = rgb(0.,0.2,0.6); pen qqffff = rgb(0.,1.,1.); pen wwzzff = rgb(0.4,0.6,1.); pen wwccff = rgb(0.4,0.8,1.); pen wwccqq = rgb(0.4,0.8,0.); pen qqzzqq = rgb(0.,0.6,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen wwffqq = rgb(0.4,1.,0.); /* draw figures */ draw(circle((6.42,0.24), 5.874316981573262), linewidth(1.) + qqttzz); draw((1.52,-3.)--(11.478869149674523,-2.745873896612408), linewidth(1.)); draw((11.478869149674523,-2.745873896612408)--(4.101180235347286,5.637284029866867), linewidth(1.)); draw((4.101180235347286,5.637284029866867)--(1.52,-3.), linewidth(1.)); draw((4.101180235347286,5.637284029866867)--(1.52,-3.), linewidth(1.) + qqffff); draw((1.52,-3.)--(11.478869149674523,-2.745873896612408), linewidth(1.) + qqffff); draw((11.478869149674523,-2.745873896612408)--(4.101180235347286,5.637284029866867), linewidth(1.) + qqffff); draw((4.101180235347286,5.637284029866867)--(10.490979088141643,-3.9949414711319724), linewidth(1.) + wwzzff); draw((2.5703092250260915,-4.197057689176556)--(4.101180235347286,5.637284029866867), linewidth(1.) + wwzzff); draw((6.569849493777312,-5.632405395509976)--(4.101180235347286,5.637284029866867), linewidth(1.) + wwccff); draw((2.7615820377563653,-2.9683178480860586)--(5.3512791992080295,-0.06952623869681132), linewidth(1.) + wwccqq); draw(circle((4.2071782883836155,2.6701786632459257), 2.9689981212318606), linewidth(1.) + blue); draw((3.2042965798356873,-0.12431200994271625)--(5.3512791992080295,-0.06952623869681132), linewidth(1.)); draw((3.2042965798356873,-0.12431200994271625)--(2.7615820377563653,-2.9683178480860586), linewidth(1.) + qqzzqq); draw((2.7615820377563653,-2.9683178480860586)--(4.908564657128707,-2.9135320768401525), linewidth(1.) + qqzzqq); draw((4.908564657128707,-2.9135320768401525)--(5.3512791992080295,-0.06952623869681132), linewidth(1.) + qqzzqq); draw((5.3512791992080295,-0.06952623869681132)--(3.2042965798356873,-0.12431200994271625), linewidth(1.) + qqzzqq); draw(circle((5.327719697263722,1.4395707887620663), 4.373236273766241), linewidth(1.) + qqwuqq); draw((1.2629522122759738,3.0529091907011026)--(6.569849493777312,-5.632405395509976), linewidth(1.) + wwccqq); draw((2.5703092250260915,-4.197057689176556)--(10.490979088141643,-3.9949414711319724), linewidth(1.) + wwzzff); draw((1.2629522122759738,3.0529091907011026)--(5.3512791992080295,-0.06952623869681132), linewidth(1.) + wwffqq); /* dots and labels */ dot((1.52,-3.),dotstyle); label("$B$", (1.613930449484157,-2.776748504324864), NE * labelscalefactor); dot((11.478869149674523,-2.745873896612408),dotstyle); label("$C$", (11.571713944281683,-2.5255200216579543), NE * labelscalefactor); dot((4.101180235347286,5.637284029866867),dotstyle); label("$A$", (4.194732135062415,5.856375718228946), NE * labelscalefactor); dot((6.569849493777312,-5.632405395509976),linewidth(4.pt) + dotstyle); label("$D$", (6.6613390557920775,-5.448906001781996), NE * labelscalefactor); dot((10.490979088141643,-3.9949414711319724),dotstyle); label("$E$", (10.589638966583761,-3.7588234820227844), NE * labelscalefactor); dot((2.5703092250260915,-4.197057689176556),linewidth(4.pt) + dotstyle); label("$F_{1}$", (2.6645222860912354,-4.010051964689694), NE * labelscalefactor); dot((2.7615820377563653,-2.9683178480860586),linewidth(4.pt) + dotstyle); label("$F$", (2.8472339098489883,-2.776748504324864), NE * labelscalefactor); dot((5.3512791992080295,-0.06952623869681132),linewidth(4.pt) + dotstyle); label("$I$", (5.450874548396965,0.12379852282945857), NE * labelscalefactor); dot((4.056430618482198,-1.518922043391435),linewidth(4.pt) + dotstyle); label("$G$", (4.149054229122977,-1.3378944672325623), NE * labelscalefactor); dot((1.2629522122759738,3.0529091907011026),linewidth(4.pt) + dotstyle); label("$X$", (0.9972787193017416,3.0471845029535003), NE * labelscalefactor); dot((3.2042965798356873,-0.12431200994271625),linewidth(4.pt) + dotstyle); label("$S$", (3.30401296924337,0.05528166392030134), NE * labelscalefactor); dot((4.908564657128707,-2.9135320768401525),linewidth(4.pt) + dotstyle); label("$Q$", (4.994095489002583,-2.731070598385426), NE * labelscalefactor); dot((5.968345846524453,-2.8864890401201238),linewidth(4.pt) + dotstyle); label("$P$", (6.067526278579381,-2.7082316454157067), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

A sketch of the solution:
Label the points as above.
Let $X$ be the intersection of $EI$ with $(ABC)$ , we will show that $X-G-D$ are collinear.

1) Show that $XAIS$ cyclic.

2) Show that $IQSF$ is a parallelogram, this implies that $XD$ bisects segment $IF$ which means that $G$ Lies on $XD$ .

proof: By shooting lemma and some ratios.

This post has been edited 8 times. Last edited by Ywgh1, Jul 28, 2024, 8:08 PM

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Anancibedih

18 posts

Anancibedih

#86 h Aug 1, 2024, 10:06 PM • 1 Y

Y by cubres

Let $EI\cap \Gamma=T, AF\cap \Gamma=K$ . If we show that $DT$ bisects $IF$ , we were done. Let $DT\cap AL=L$ . Since $\angle{LTI}=\angle{IAE}=\angle{IAL}$ , $LIAT$ is a cyclic quadrilateral Since $\angle {ETA}=\angle{ALI}=\angle{AKI}$ it is $LI||KE$ . Also, since $KE||BC$ , $LI||BC$ . Let $TL\cap BC=S$ . It is clear that " $DT$ bisects $[IF]$ $\Longleftrightarrow$ $SILF$ is a parallelogram". We have to prove that $\frac{|DI|}{|DA|}=\frac{|DS|}{|DL|}$ . Let $AD\cap BC=R$ . We know that $|DI|=|BD|$ (the rule that $D$ is the center of $(BIC)$ .) In last equation, the expression on the right is equal to $\frac{|DR|}{|DI|}$ from $CR||IL$ . If we substitute, the expression we have to prove is $\frac{|DR|}{|DI|}=\frac{|DI|}{|DA|}$ . Since $\angle{CBD}=\angle{DAC}=\angle{DAB}$ , $\triangle {DRB}\sim \triangle {DBA}$ . $\frac{|BD|}{|DA|}=\frac{|DR|}{|BD|}$ . With the equation $|BD|=|DI|$ we just said, the expression goes into $\frac{|DR|}{|DI|}=\frac{|DI|}{|DA|}$ . We're done.

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africanboy

7 posts

africanboy

#87 h Oct 30, 2024, 11:27 PM • 1 Y

Y by bo18

And one solution with 0 bash and 0 smilarities.
$K = EI \cap \Gamma$ ,
$L = AI \cap BC$ ,
$F' = AF \cap \Gamma$ ,
$X = KD \cap AF'$ ,
Let $I_A$ be the $A$ -excenter.
We just have to show that $D, G, X$ are colinear.
Now by pascal's theorem on $AF'EKDD$ we have $IP \parallel BC$
By menelaus on $AIF$ and $D-G-X$ we have to show that
$\frac{AD}{ID}\frac{IG}{FG}\frac{FX}{AX}=1\Longleftrightarrow\frac{AD}{ID}=\frac{AX}{FX}$
But from $IP \parallel BC$ we have $\frac{FX}{AX}=\frac{LI}{IA}$ .
Let $a=AI, b=IL, c=LD$ .
We have to show that $AI.ID=AD.IL$ or $a(b+c)=(a+b+c)b$ or $ac=bc+b^2$ .
By power of Point in $ABDC$ and $CIBI_A$ we have $AL.LD=CL.LB=IL.LI_A$ or $(a+b)c=b(2c+b)$ or $ac=bc+b^2$ and we are happy

.

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SimplisticFormulas

116 posts

SimplisticFormulas

#88 h Feb 6, 2025, 5:41 PM

Y by

finally broke my geo-failure streak

soln

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lian_the_noob12

173 posts

lian_the_noob12

#90 h Mar 6, 2025, 8:27 PM

Y by

Let $EI \cap \Gamma \equiv P, DP \cap AF \equiv Q, AF \cap \Gamma \equiv F_1$ Pascal on $DPEF_1AD \implies IQ \parallel DD \parallel BC$
$-1=(A,AI \cap BC; I,I_A) \overset{F}{=} (Q,\infty;I,FI_A \cap IQ) \implies IG=GF \blacksquare$
$[asy] import olympiad; import cse5; size(8cm); defaultpen(fontsize(9pt)); pair A = dir(110); dot("$A$", A, dir(A)); pair B = dir(210); dot("$B$", B, dir(B)); pair C = dir(330); dot("$C$", C, dir(C)); pair I = incenter(A, B, C); dot("$I$", I, dir(45)); pair O = origin; dot("$O$", O, dir(45)); pair D = extension(A, I, O, B+C); dot("$D$", D, dir(225)); pair E = dir(310); dot("$E$", E, dir(E)); pair F_1 = B*C/E; dot("$F_1$", F_1, dir(F_1)); pair F = extension(B, C, A, F_1); dot("$F$", F, dir(F)); pair G = midpoint(I--F); dot("$G$", G, dir(-15)); pair P = extension(E, I, D, G); dot("$P$", P, dir(P)); pair I_A = 2*D-I; dot("$I_A$", I_A, dir(I_A)); pair Q = extension(D, P, A, F); dot("$Q$", Q, dir(230)); pair Z = extension(I, Q, I_A, F); dot(Z); draw(I--Z--I_A); dot(extension(A, I, B, C)); draw(A--B--C--cycle, blue); draw(unitcircle, red); draw(A--F_1--E--P--D--cycle, magenta); draw(A--E, dotted+green); draw(I--F); draw(D--I_A, dotted); /* Source generated by TSQ: A = dir 110 B = dir 210 C = dir 330 I = incenter A B C 45 O = origin 45 D = extension A I O B+C 225 E = dir 310 F_1 = B*C/E F = extension B C A F_1 G = midpoint I--F -15 P = extension E I D G I_A = 2*D-I Q = extension D P A F 230 Z := extension I Q I_A F := Z I--Z--I_A := extension A I B C A--B--C--cycle blue unitcircle red A--F_1--E--P--D--cycle magenta A--E dotted green I--F D--I_A dotted */ [/asy]$

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Nari_Tom

117 posts

Nari_Tom

#91 h Mar 29, 2025, 9:44 AM

Y by

Let $I_A$ be the A-excircle. Since $DI=DI_A$ , we just need to prove $GD \parallel I_AF$ , were $G=EI \cap (ABC)$ . Now let's use $\sqrt{bc}$ inversion, then we just need to prove that $(IAF), (AHG)$ are tangent to each other. Since $\triangle AIE \sim \triangle AI_AF$ , we get that $\angle AEI=\angle AI_AF$ . Then easy angle chase gives the conclusion.

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LHE96

2 posts

LHE96

#92 h Apr 27, 2025, 2:48 AM

Y by

$[asy] import geometry; size(8cm); pair A = (2,6); pair B = (8,0); pair C = (0,0); pair F = (6,0); triangle t = triangle(A,B,C); pair I_ = incenter(t); pair D = intersectionpoints(line(A,I_),circumcircle(t))[0]; pair P = intersectionpoints(line(A,F),circumcircle(t))[0]; pair E_ = intersectionpoints(parallel(P,line(B,C)),circumcircle(t))[0]; pair G = (I_+F)/2; pair K = intersectionpoint(line(G,D),line(E_,I_)); pair R = intersectionpoint(line(D,G),line(A,F)); pair J = intersectionpoint(line(B,C),line(A,I_)); pair S_ = intersectionpoint(line(B,C),line(D,G)); pair T = intersectionpoint(line(I_,S_),line(D,F)); draw(circumcircle(A,K,I_),mediumgray); draw(t); draw(circumcircle(t),mediumgray); draw(A--E_); draw(A--P); draw(A--D); draw(I_--F); draw(D--F); draw(R--I_); draw(A--K); draw(I_--T); draw(E_--K,dashed); draw(D--K,dashed); dot("$A$",A,(-0.25,1)); dot("$B$",B,(0.5,-0.75)); dot("$C$",C,(-0.75,-0.75)); dot("$D$",D,(-0.5,-1)); dot("$E$",E_,(-0.75,-0.75)); dot("$F$",F,(0.5,0.75)); dot("$G$",G,(-0.75,-0.75)); dot("$I$",I_,(-1,0)); dot("$J$",J,(-0.75,-0.75)); dot("$P$",P,(0.5,-0.75)); dot("$K$",K,(0.25,1)); dot("$R$",R,(1,0.25)); dot("$S$",S_,(-0.75,-0.75)); dot("$T$",T,(0.5,-0.75)); [/asy]$

Let $K = EI\ \cap\ \Gamma$ (different from $E$ ), $P = AF\ \cap\ \Gamma$ (different from $A$ ), $R = DK\ \cap\ AF$ and $J = BC\ \cap\ AI$ .

:arrow:

Claim 1: $IR \parallel BC$
Proof. Note that $\angle PEB=\angle PAB = \angle CAE = \angle CBE$ $\implies$ $PE\parallel BC$ .
Clearly, $\angle EKD = \angle EAD = \angle DAP$ $\implies$ $AKRI$ is cyclic. Since $AKRI$ is cyclic, we have $\angle ARI =\angle AKI=\angle APE$ $\implies$ $IR \parallel PE$ , which implies $IR \parallel BC$ .

:arrow:

Claim 2: $DG$ passes through $R$
Proof. Let $S = DG\ \cap\ JF$ , $T = IS\ \cap\ DF$ and let $R' = DG\ \cap\ AF$ . Ceva on triangle $\triangle IFD$ gives us $\dfrac{IG}{GF}\cdot\dfrac{FT}{TD}\cdot\dfrac{DJ}{JI}=\dfrac{FT}{TD}\cdot\dfrac{DJ}{JI} = 1$ $\implies$ $\dfrac{FT}{TD}=\dfrac{JI}{DJ}$ $\implies$ $\dfrac{DF}{TD}=\dfrac{DI}{DJ}$ , which implies $JT\parallel IF$ .
Note that $\angle DCB=\angle DAB=\angle CAD =\dfrac{1}{2}\angle A$ $\implies$ $CD$ is tangent to $(AJC)$ at point $C$ . By power of a point, we get $Pow_{(AJC)}D = CD^2 = DJ\cdot DA$ $\implies$ $\dfrac{DA}{CD} = \dfrac{CD}{DJ}$ . It's well known that $CD=DI$ , so $\dfrac{DA}{DI} = \dfrac{DI}{DJ} = \dfrac{DF}{TD}$ $\implies$ $IT\parallel AF$ $\implies$ $\angle TIF = \angle AFI$ , which implies that triangles $\triangle IGS$ and $\triangle FGR'$ are similar ( $IG=GF$ , $\angle SIG = \angle R'FG$ and $\angle IGS = \angle FGR'$ ). This similarity gives us that $GS = GR'$ $\implies$ $IR'FS$ is a parallelogram $\implies$ $IR'\parallel BC$ , and it follows that $R'=R$ $\implies$ $DG$ passes through $R$ .

Since $D$ , $R$ , $K$ and $G$ all lie on the same line, it follows that $EI\ \cap\ DG = EI\ \cap\ DR = K$ , and $K$ lies on $\Gamma$ , as desired.

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