GEOMETRY GEOMETRY GEOMETRY

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3001 posts

#1 h Jul 20, 2021, 8:47 PM • 25 Y

Y by fishy15, centslordm, anantmudgal09, Bubu-Droid, buratinogigle, tree_3, HWenslawski, gadamus003, p_square, RedFlame2112, megarnie, jhu08, HamstPan38825, amar_04, son7, psi241, Ballocoma, StefanSebez, PHSH, Aryan-23, TheHU-1729, Rounak_iitr, ehuseyinyigit, Funcshun840, MS_asdfgzxcvb

Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.

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Mindstormer

102 posts

Mindstormer

#2 h Jul 20, 2021, 8:58 PM • 28 Y

Y by Kagebaka, centslordm, Mathscienceclass, turko.arias, buratinogigle, p_square, richrow12, 554183, RedFlame2112, jhu08, tenebrine, megarnie, OlympusHero, Ballocoma, pikapika007, math31415926535, PHSH, amar_04, Quidditch, ihatemath123, MS_Kekas, LLL2019, TheHU-1729, ehuseyinyigit, Rounak_iitr, bin_sherlo, cosdealfa, Funcshun840

This problem was proposed by my friend Mykhailo Shtandenko from Ukraine (who might be the youngest IMO author in history — he turns 18 on 25th of July)

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mshtand1

82 posts

mshtand1

#3 h Jul 20, 2021, 9:00 PM • 147 Y

Y by Mindstormer, Kagebaka, Vlados021, MS_Kekas, 62861, ABCDE, mijail, samrocksnature, timon92, IndoMathXdZ, mira74, insertionsort, franchester, OlympusHero, juliankuang, 606234, tigerzhang, Dadgarnia, khina, tree_3, centslordm, smartguy888, Aryan-23, sotpidot, anantmudgal09, Mathscienceclass, IMOStarter, AOPqghj, mango5, HamstPan38825, john0512, Aimingformygoal, TheMath_boy, Mathmick51, Hyperbolic_, turko.arias, NJOY, ineqcfe, Math_olympics, buratinogigle, Aritra12, Chesssaga, KST2003, yalex999, megarnie, V-217, tapir1729, p_square, primesarespecial, richrow12, CaptainLevi16, TienLQD, judgefan99, pavel kozlov, mathapple101, Cosmobius, KPBY0507, 554183, Inconsistent, Aniruddha07, SMSGodslayer, barcelona, Geometry285, buffet, VipMath, ninjasrule34, whiteguard, N1RAV, andyloo666, Mathphile01, RedFlame2112, jackson46, TETris5, vsamc, mathtiger6, jhu08, Didier, mathleticguyyy, ZHEKSHEN, L567, kante314, BVKRB-, amar_04, tenebrine, ratatuy, PIartist, jasperE3, Gaussian_cyber, phoenixfire, 799786, MathLuis, pog, proxima1681, fjm30, rcorreaa, hakN, CyclicISLscelesTrapezoid, MatBoy-123, Seicchi28, Quidditch, chystudent1-_-, StefanSebez, rayfish, pikapika007, minusonetwelth, StopSine, happymathEZ, David-Vieta, rg_ryse, coolmath_2018, PHSH, IMUKAT, lrjr24, trying_to_solve_br, BIGFLIPPA, egxa, wenwenma, fuzimiao2013, lethan3, RetroTurtle, TheStrayCat, oVlad, SatisfiedMagma, ihatemath123, Triangle_Center, mathmax12, nguyenducmanh2705, LLL2019, thdnder, OronSH, TheHU-1729, Maths_Girl, trk08, Assassino9931, sabkx, Kingsbane2139, EpicBird08, ehuseyinyigit, Rounak_iitr, ohiorizzler1434, Funcshun840, aidan0626, MS_asdfgzxcvb, pingupignu, Soupboy0, jkim0656, giangtruong13

OHMYGOD, This is my problem. I'm quite surprised to see it on IMO. I even doubted that this problem will be on the Shortlist. The story of how I came up with it is quite lengthy, and I don't remember all the details, but I got chain of problems, each time improving former version by doing inversion, watching some new stuff in construction in Geogebra, until finally, in few months came up with the final version of this problem. I actually sent slightly different proposal, and in that statement the fact of this problem was very unexpected for me, so I decided to send it on IMO. Thanks to Fedir Yudin, who first solved this problem and to Anton Trygub who first solved it with the nice solution which I present you here.

Attachments:

1626811062325_1626811061020_1626811059832_1626811058569_1626811056839_1626811022116_1626811019045_2_5451728741887643403.pdf (220kb)

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MS_Kekas

275 posts

MS_Kekas

#4 h Jul 20, 2021, 9:06 PM • 16 Y

Y by centslordm, trying_to_solve_br, Mindstormer, Fedor Bakharev, richrow12, 554183, RedFlame2112, jhu08, ZHEKSHEN, amar_04, megarnie, pog, PHSH, LLL2019, Rounak_iitr, Funcshun840

I am very happy that a problem by Mykhailo Shtandenko made it to IMO. He really deserved it with all the hard work he has put into problemsetting. I am sure that in next few years you will have many more chances to enjoy his problems on various international competitions, including IMO, RMM, and EGMO.

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mshtand1

82 posts

mshtand1

#5 h Jul 20, 2021, 9:17 PM • 6 Y

Y by centslordm, jhu08, ZHEKSHEN, megarnie, A64298347, Funcshun840

Oh, thanks, MS_Kekas, you are also a great proposer!

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juckter

323 posts

juckter

#6 h Jul 20, 2021, 9:34 PM • 5 Y

Y by centslordm, Kagebaka, jhu08, hakN, Rounak_iitr

Will make a diagram later.

Let $P$ be the point where $AD$ intersects $(BDC)$ for the second time. Then $\angle APC = \angle DBC = \angle FDA$ . As $\angle FAD = \angle CAP$ it follows that $\triangle AFD \sim \triangle ACP$ , thus $AD/AP = AF/AC$ . Similarly $AE/AB = AD/AP$ , so $AB \cdot AF = AC \cdot AE$ and $BCEF$ is cyclic. Now we prove that $(DEF)$ and $(BDC)$ are tangent. Let $AD$ intersect $BC$ at $K$ , then

$\angle FED = \angle AED - \angle AEF = \angle ABP - \angle ABC = \angle PBC = \angle KDC$
Now $\angle FED + \angle DCB = \angle DKB$ and some more easy angle chasing shows $\angle DKB = \angle FDB$ , so these circumcircles are indeed tangent. Let $EF$ and $BC$ intersect at $T$ , then $T$ is the radical center of $(DEF)$ , $(DBC)$ and $(BCEF)$ , so $D$ lies on the circle $\Omega$ with center $T$ that is orthogonal to $(BCEF)$ . To finish it is enough to prove that the second intersection of $(ADC)$ and $(XDE)$ also lies on $\Omega$ . We will actually generalize this, and prove that it holds for any point $D$ on $\Omega$ .

Now we invert at $D$ , with the inverted diagram looking as follows:

IMO 3, inverted wrote:

Let $BCEF$ be an isosceles trapezoid with $BC \parallel EF$ and let $\Omega$ be its axis of symmetry. Let $D$ be any point on the plane, and $A$ be the second intersection of $(BDF)$ and $(CDE)$ . Let $X$ be a point on $(CDE)$ such that $BD/DC = BX/XC$ . Then $AC$ and $EX$ intersect on $\Omega$ .

Now let $P = BF \cap CE$ , then $A, D, P$ are collinear on account of $AD$ being the radical axis of $(BDF)$ and $(CDE)$ , and $P$ lies on $\Omega$ . Rewording the problem wrt the circle $(CDE)$ and relabeling points, we get the following:

IMO 3, inverted and reworded wrote:

Let $ABCDE$ be a cyclic pentagon, let $P = AB \cap CD$ and $Q = BD \cap CE$ , and let $X$ be the reflection of $D$ wrt $PQ$ . Then $AX/AD = EX/ED$ .

To finish, notice that the required statement is equivalent to $AX/XE = AD/DE$ , which is equivalent to $X$ lying on the $D$ -Apollonian circle of $\triangle ADE$ . Let $DD \cap AE = R$ , then by Pascal on $DDBAEC$ we get $P, Q, R$ collinear, implying that $RD = RX$ , which finishes as $R$ is the center of the Apollonian circle.

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Dadgarnia

164 posts

Dadgarnia

#7 h Jul 20, 2021, 9:36 PM • 2 Y

Y by centslordm, jhu08

mshtand1 wrote:

OHMYGOD, This is my problem. I'm quite surprised to see it on IMO. I even doubted that this problem will be on the Shortlist. The story of how I came up with it is quite lengthy, and I don't remember all the details, but I got chain of problems, each time improving former version by doing inversion, watching some new stuff in construction in Geogebra, until finally, in few months came up with the final version of this problem. I actually sent slightly different proposal, and in that statement the fact of this problem was very unexpected for me, so I decided to send it on IMO. Thanks to Fedir Yudin, who first solved this problem and to Anton Trygub who first solved it with the nice solution which I present you here.

Congratulations! I really enjoyed your problem. Here goes my solution:
Let $D'$ be the isogonal conjugate of $D$ with respect to $ABC$ . Since $\angle ADE=\angle DCB=\angle ACD'$ , $ED'DC$ is cyclic and similarly $FD'DB$ is cyclic. From POP we get $AE\cdot AC=AD'\cdot AD=AF\cdot AB$ therefore $BFEC$ is cyclic, too. Let $AD$ intersects $BC$ and $EF$ at $R$ and $S$ . Notice that $\angle EDC=\angle ADC-\angle ADE=\angle ADC-\angle DCB=\angle CRA\ (1)$ . Hence
$\angle EFD+\angle DBC=\angle EFD+\angle ADF=\angle ASF=\angle CRA=\angle EDC$ so $\odot (EDF)$ and $\odot (BDC)$ are tangent. It yields that $EF$ , $BC$ and the common internal tangent of $\odot (EDF)$ and $\odot (BDC)$ are concurrent. Let $K$ be the concurrency point. Similarly we can prove that $KD'$ is tangent to both $\odot (ED'F)$ and $\odot (BD'C)$ , so $KD'^2=KB\cdot KC=KD^2$ or $KD'=KD$ . Suppose that $\odot (EDX)$ and $\odot (ADC)$ intersects again at $P$ . We just need to prove that $K$ is the circumcenter of $\odot (PD'D)$ , since $O_1O_2$ is the perpendicular bisector of $PD$ . It's sufficient to prove that $\angle D'PD=90^\circ-\angle D'DK$ or $\angle APD'=\angle APD+\angle D'DK-90^\circ$ . Notice that
$\angle D'DK=\angle D'DC-\angle KDC=\angle D'DC-\angle DBC=\angle D'DC-\angle ABD'$ $\Longrightarrow \angle APD+\angle D'DK=\angle ACD+\angle D'DC-\angle ABD'=180-\frac12 \angle A-\angle ABD'=\angle AD'B$ hence we need to prove that $\angle APD'=\angle AD'B-90^\circ=\angle AFD-90^\circ$ .
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Now perform an inversion centered at $A$ that fixed $\odot (BFEC)$ . The images of the points are denoted by primes. It's obvious that $X'$ lies on $AC$ and $\angle X'FE=180^\circ-\angle C$ . Also $P'$ lies on $D'E$ and $\odot (X'D'C)$ . We want to prove that $\angle ADP'=\angle AFD-90^\circ$ or $P'D$ passes through the circumcenter of $\odot (AFD)$ . We claim that $\triangle FX'P'\sim \triangle DRB$ . Notice that $\angle FX'E=\angle C-\angle AEF=\angle C-\angle B$ and $\angle P'X'C=\angle ED'C\overset{(1)}{=}\angle CRA$ , hence $\angle FX'P'=\angle CRA+\angle C-\angle B=\angle ARB$ . So to prove our claim we need to show that $\frac{X'F}{X'P'}=\frac{RD}{RB}$ . It's clear that $\triangle X'EP'\sim \triangle D'EC$ and $\triangle D'EC\sim \triangle RDC$ , therefore
$\triangle X'EP'\sim \triangle RDC \Longrightarrow X'P'=X'E\cdot\frac{CR}{RD}.$ So
$\frac{X'F}{X'P'}=\frac{RD}{RB} \Longleftrightarrow \frac{X'F}{X'E}=\frac{RC}{RB}=\frac{AC}{AB}$ The last equality holds by law of sines, hence the claim is proved and it yields $\angle X'P'F=\angle DBC$ . Let $J$ be the second intersection of $AC$ and $\odot (AFD)$ . We have $\angle AJF=\angle ADF=\angle DBC=\angle X'P'F$ so $X'P'JF$ is cyclic. On the other hand, we proved that $\angle P'X'C=\angle CRA=\angle B+\frac12 \angle A$ and $\angle FX'E=\angle C-\angle B$ , therefore $2\angle P'X'C+\angle FX'E=\angle A+\angle B+\angle C=180^\circ$ . It means that $X'P'$ is the external angle bisector of $\angle FX'E$ , hence $P'F=P'J$ . Also it's clear that $DF=DJ$ so $P'D$ is the perpendicular bisector of $FJ$ and passes through the circumcenter of $\odot (AFD)$ , as desired.

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kvanta

113 posts

kvanta

#8 h Jul 20, 2021, 10:02 PM • 3 Y

Y by centslordm, jhu08, Kingsbane2139

Михайло Штанденко -- well-done mate. Your problem is definitely the hardest from this IMO...

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mshtand1

82 posts

mshtand1

#9 h Jul 20, 2021, 10:05 PM • 2 Y

Y by centslordm, jhu08

Thank you!

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A02

21 posts

A02

#10 h Jul 20, 2021, 10:12 PM • 11 Y

Y by centslordm, Kagebaka, archp, mira74, Fedor Bakharev, PSJL, SK_pi3145, jhu08, lahmacun, Dukejukem, BaishuakRayimbek

Here is a sketch of my solution:

Begin by proving $BCEF$ cyclic (e.g. by introducing isogonal conjugate of $D$ as other solutions have).

Let $EF$ meet $BC$ at $Y$ , by angle chasing one can show $YD^2 = YB \cdot YC$ .

Notice $(BFY)$ and $AC$ are inverses wrt the circle $\omega$ centred at $Y$ passing through $D$ .

Let $(BFY)$ meet $(AC)$ at $(P, Q)$ .

Angle chasing gives $XB$ tangent to $(BFY)$ .

DIT applied to $FYBB$ gives that $(P, Q), (X, E), (A, C)$ are pairs of an involution on $AC$ .

Hence $\omega$ , $(ADC)$ and $(DEX)$ are coaxial (since $D$ also lies on radical axis) so we are done.

This post has been edited 4 times. Last edited by A02, Jul 26, 2021, 8:47 PM

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anantmudgal09

1980 posts

anantmudgal09

#11 h Jul 20, 2021, 10:50 PM • 3 Y

Y by SK_pi3145, Aimingformygoal, jhu08

Kagebaka wrote:

Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.

The solution proceeds in three steps.

Step 1. Identifying the point of concurrence.

Let $D^\ast$ be the isogonal conjugate of $D$ in $\triangle ABC$ . Note that $D^\ast$ lies on $\odot(CDE)$ and $\odot(FBD)$ , and $A$ lies on $\overline{DD^\ast}$ hence $BFEC$ is cyclic by the converse of radical axis theorem. Note also that $\angle BDF+\angle CDE=180^{\circ}$ hence the circles $\odot(BDC)$ and $\odot(EDF)$ are tangent at $D$ , so combining them with $(BFEC)$ and applying radical axes, we get $\overline{BC}, \overline{EF},$ and the line through $D$ tangent to $\odot(BDC)$ concur (say at point $T$ ).

Step 2. Reformulating line through centres.

Let $Y=\odot(ADC) \cap \odot(EXD)$ with $Y \ne D$ . It suffices to show that the perpendicular bisector of $\overline{DY}$ passes through $T$ . Thus, we wish to understand point $Y$ . Let $M$ be the midpoint of arc $\widehat{BAC}$ . Notice that $\angle AYX=\angle DYX-\angle DYA=\angle DEX-\angle DCA=\angle CDE=\angle ADC-\angle ADE=\angle(AD, BC)$ hence $AMXY$ cyclic. Also, $\angle CYD=\tfrac{1}{2}\angle A$ ; which characterises $Y$ completely. Note also that $\overline{CD}$ meets the perpendicular bisector $\ell$ of $\overline{BC}$ at $S$ , then $DSMY$ is cyclic since $\angle DYM=\angle DYA+\angle AYM=\angle DCA+(90^{\circ}-\angle C)=90^{\circ}-\angle SCB$ , as claimed. In order to show $TD=TY$ , since $TD^2=TB \cdot TC$ , it suffices to show that $Y$ lies on the $D$ -Appolonian circle $\omega$ in $\triangle BDC$ .

Step 3. Changing reference to $\triangle BDC$ and moving points.

We employ the method of moving points (or the rather cheekily named "MMP").

Fix $\triangle BDC$ and move point $M$ along $\ell$ ; let $Y_1=\odot(DSM) \cap \omega$ with $Y_1 \ne D$ and $Y_2$ be the point on $\omega$ with $\angle CY_2D=\angle (CM, \ell)$ . So $M \mapsto Y_1$ and $M \mapsto Y_2$ are projective mappings, hence it suffices to check $Y_1=Y_2$ for three choices of $M$ . When $M$ is the point at infinity on $\ell$ , we get $Y_1, Y_2$ are both the second intersection of ray $\overrightarrow{DC}$ with $\omega$ . When $M=\ell \cap \overline{DB}$ , we see that $Y_1=Y_2=A$ . Finally, when $M$ coincides with the reflection of $S$ in $\overline{BC}$ , we get $Y_1, Y_2$ both as the reflection of $D$ in $\overline{BC}$ .

Thus, we have solved the problem.

Remark. The strategy for the final moving points steps after finding the right "base-change" is inspired from this solution to IMO 2018 P6

This post has been edited 1 time. Last edited by anantmudgal09, Jul 20, 2021, 10:52 PM
Reason: "Three steps"

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ltf0501

191 posts

ltf0501

#12 h Jul 21, 2021, 12:04 AM • 2 Y

Y by Cindy.tw, jhu08

Solution finished by Li4 and me!! I think it is a medium 3,6 level problem. (~MOHS 40)

First we prove that $B$ , $C$ , $E$ , $F$ are concyclic. Let $D^*$ be the isogonal conjugate of $D$ w.r.t $\triangle ABC$ , which lies on $AD$ . Since $\measuredangle BDC + \measuredangle EDF = 180^{\circ}$ , $D$ also has an isogonal conjugate w.r.t the complete quadrilateral $\mathcal{Q}\{BC, CA, AB, EF\}$ , which is exactly $D^*$ . Therefore, $\measuredangle ADF = \measuredangle CBD = \measuredangle D^*BF$ , which implies that $B$ , $D^*$ , $D$ , $F$ are concyclic. Similarly, $C$ , $D^*$ , $D$ , $E$ are concyclic. Hence

$AF \cdot AB = AD \cdot AD^* = AE \cdot AC,$ implying that $B$ , $C$ , $E$ , $F$ are concyclic.

Let $Y = AD \cap EF$ , and $Z = AD \cap BC$ . Then from $\triangle ABC \stackrel{-}{\sim} \triangle AEF$ we have

$\triangle EYD \stackrel{+}{\sim} \triangle DZC \quad \text{and} \quad\triangle FYD \stackrel{+}{\sim} \triangle DZB.$ This means
$\measuredangle CDE = \measuredangle CZD = \measuredangle CBD + \measuredangle BDZ = \measuredangle CBD + \measuredangle DFE,$ implying that $\odot(BDC)$ is tangent to $\odot(EDF)$ . By radical axis theorem we have $SD$ is the common tangent line of $\odot(BDC)$ and $\odot(EDF)$ .

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Let $S = BC \cap EF$ , and let $M$ be the Miquel point of complete quadrilateral $\mathcal{Q}\{BC, CA, AB, EF\}$ . Then $A$ , $S$ , $M$ are collinear as $B$ , $C$ , $E$ , $F$ are concyclic. Let $T \neq D$ be another intersection of $\odot(ADC)$ and $\odot(EDX)$ . Consider the inversion transformation $\varphi_S$ with center $S$ and power $SC \cdot SB$ , then $D$ is fixed under this transformation. Therefore to prove that $S$ , $O_1$ , $O_2$ are collinear, it suffices to show that $T$ is also fixed under the transformation $\varphi_S$ .

Notice that $\varphi_S(\odot(ADC)) = \odot(MDB)$ . It suffices to show that $\odot(EDX)$ , $\odot(ADC)$ , and $\odot(MDB)$ are concurrent at a point $T$ . Let $K = BM \cap AC$ . Then

$\operatorname{Power}(K, \odot(BDM)) = KB \cdot KM = KA \cdot KC = \operatorname{Power}(K, \odot(ADC)).$ On the other hand,

$\measuredangle BME = \measuredangle AME - \measuredangle AMB = \measuredangle AFE - \measuredangle SFB = 2\measuredangle AFE = 2 \measuredangle BCA = \measuredangle BXE,$ so $B$ , $E$ , $M$ , $X$ are concyclic, implying that

$\operatorname{Power}(K, \odot(BDM)) = KB \cdot KM = KX \cdot KE = \operatorname{Power}(K, \odot(EDX)).$ Hence, $K$ lies on the radical axis $DT$ of $\odot(ADC)$ and $\odot(EDX)$ , and also $B$ , $D$ , $M$ , $T$ are concyclic, as desired.

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Remark

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hukilau17

283 posts

hukilau17

#13 h Jul 21, 2021, 12:16 AM • 17 Y

Y by 606234, Kagebaka, rocketsri, Kamran011, Bubu-Droid, tapir1729, jhu08, HamstPan38825, 554183, Assassino9931, rayfish, ZIXILI, Quidditch, Helixglich, kamatadu, NO_SQUARES, EpicBird08

complex bash

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USJL

535 posts

USJL

#14 h Jul 21, 2021, 12:28 AM • 2 Y

Y by jhu08, Kingsbane2139

This is probably the only problem I like in Day 1, and I think this geo is pretty neat!

Lemma 1. $BCEF$ is concyclic.
Proof

Now we invert about $D$ and label the image of a point $T$ as $T'$ .

Lemma 2. $B'C'E'F'$ is a trapezoid with $B'F'=C'E'$ and $B'C'\parallel E'F'$ .
Proof

Lemma 3. Let $M$ be the Mique point of $(AB, BC, CA, EF)$ . Then $BEMX$ are concyclic.
Proof

Now we can put everything together. Assume that $BC$ intersects $EF$ at $P$ . Then $P'$ is the intersection of $DB'C'$ and $DE'F'$ , showing that $P'$ is the reflection of $D$ w.r.t. the common perpendicular bisector $\ell$ of $EF$ and $BC$ . Also $M'$ is the intersection of $A'B'C'$ and $A'E'F'$ , showing that $M'$ is the reflection of $A'$ w.r.t. $\ell$ .

The statement is equivalent to showing that $DP'O_1'O_2'$ are concyclic. It is well-known that $O_1'$ is the reflection of $D$ w.r.t. $AC$ and $O_2'$ is the reflection of $D$ w.r.t. $EX$ . Thus it is equivalent to showing that $\ell, AC, EX$ are concurrent. However, since $B'M'E'X'$ , $B'M'A'C'$ and $A'C'E'X'$ are concyclic, their radical axes are concurrent. They are exactly $\ell, A'C'$ and $E'X'$ , as desired.

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teddy8732

31 posts

teddy8732

#15 h Jul 21, 2021, 12:28 AM • 14 Y

Y by daniel05, space10, chrono223, ilesejin, aqua4689, fet2019, geo0420, imn8128, Paradoxes, jhu08, KPBY0507, F_Xavier1203, I-love-K2I, Kingsbane2139

Here is K2I 's Solution :

Part 1 : $B,C,E,F$ cyclic
pf) Let the mid point of arc $(BC)$ of $(ABC)$ is $M$
noticing that radical axis of $(DFB), (DEC)$ is $DM$ , so $AB*AF=AC*AE$

Part 2 : $PD^2=PC*PB$
Let $EF$ meets $BC$ at $P$ .
pf) By simple angle chase, $DE, DB$ is isogonal w.r.t $\angle FDC$
So, by Isogonality lemma on $(DF,DC)$ and $(DE,DB)$
$DA$ and $DP$ is isogonal w.r.t $\angle FDC$
so $\angle CDP=\angle FDA=\angle DBC$

Part 3 : Finish
Let Miquel Point of $BCEF$ be $K$ (so, $K$ lies on circumcircle of $(AEF),(ABC),(BFP),(CEP)$ )
Since $BCEF$ is cyclic, well known that $K$ lies on $PA$
$BK$ meets $AC$ , $(AEF)$ at $S,T$ respectively.
$\angle KTE= \angle KAE = \angle KBC$ . So $TE//BC$ .
also note that $\angle ATE= \angle AFE = \angle ACB= \angle XBC$ .
So $\triangle ATE$ and $\triangle XBC$ are homothetic, and the homothetic center is $S$ .
$SA:SX=SE:SC$ . So $SA*SC=SE*SX...(1)$ .
Let the circle centered at $P$ and radius $PD$ be $w$ .
Let $(ABC)$ meets $w$ at $U,V$ .
Easy to see that $U,V,S$ colinear (육점공선 또는 Pole & Polar), which gives $US*VS=AS*SC ... (2)$
In conclusion, powers of $S$ in three circle $(DXE), (DAC), w$ are all same by (1) and (2)
and note that three circle all passes $D$ .
So three circles are coaxial. So centers of three circle, which is $O_1,O_2,P$ are colliner.
$Q.E.D$

This post has been edited 3 times. Last edited by teddy8732, May 6, 2023, 4:41 AM
Reason: .

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aleksijt

44 posts

aleksijt

#67 h Mar 18, 2023, 11:48 PM • 1 Y

Y by TheHU-1729

Here's an isogonality and inversion only solution. Probably similar to something above, but haven't checked.

Claim 1: $BCEF$ is cyclic.
Proof: Consider the isogonal conjugate $D^*$ of $D$ w.r.t. $\triangle ABC$ . By the angle conditions, we get that $BFD^*D$ and $CED^*D$ are cyclic. PoP at $A$ finishes.

Let $R=EF\cap BC$ , and $K$ be the second intersection of $(EDX)$ and $(CAD)$ . Note that proving $RD=RK$ suffices. Also, let $RA\cap (ABC)=\left\{S,A\right\}$ .

Claim 2: $RD$ is tangent to $(BDC)$ and $(EFD)$ , or the inversion $\mathcal{J}$ at $R$ with radius $RD$ fixes $(BFEC)$ .
Proof: Note that $DF$ and $DC$ are isogonal w.r.t. $\angle EDB$ , as are $DB$ and $DE$ themselves. By the isogonality lemma, so are $DR$ and $DA$ . The claim follows.

To prove $RD=RK$ , we'll show that $\mathcal{J}$ fixes $K$ by proving that $(SBD)$ , $(ACD)$ and $(XED)$ are coaxial, because $(SBC)\stackrel{\mathcal{J}}{\longleftrightarrow}(ACD)$ .

Claim 3: $(SBD)$ , $(ACD)$ and $(XED)$ are coaxial.
Proof: We invert about $A$ , fixing $(BCEF)$ . Because of angle chasing, or noting that $S$ is the spiral center $FE\rightarrow BC$ , $RBFS$ is cyclic. Hence, this inversion swaps $S\longleftrightarrow R$ .
After seeing where everything goes, it boils down to showing $RCX^*F$ is cyclic, where $X^*$ is the image of $X$ . This follows by direct angle chasing.

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awesomeming327.

1699 posts

awesomeming327.

#68 h Jun 17, 2023, 4:43 AM • 2 Y

Y by Inconsistent, TheHU-1729

Let $D'$ be the isogonal conjugate of $D$ with respect to $\triangle ABC$ , let $P$ be the intersection of $EF$ and $BC$ , let $Y$ and $Z$ be the intersection of $AD$ with $BC$ and $EF$ , respectively. Let $S$ be the Miquel point of complete quadrilateral $BCEFAP$ , and $G$ be the intersection of $BS$ and $AC$ .

$~$
We have $\angle ABD'=\angle CBD=\angle ADF$ , so $BDD'F$ is cyclic. Similarly, $CDD'E$ is cyclic. We have $AF\cdot AB=AD'\cdot AD=AE\cdot AC$ so $BCEF$ is cyclic. Thus, $AEZF\sim ABYC$ . In particular, $\angle PYZ=\angle PZY$ . We have
$\angle CDE=180^\circ-\angle CDY-\angle ADE=180^\circ-\angle CDY-\angle DCY=\angle DYC$ so let $P'$ be on the tangent to $(BDC)$ at $D$ on the same side as $E$ and $C$ as $AD$ , then $\angle P'DC=\angle DBY$ and so
$\angle P'DE=\angle CDE-\angle P'DC = \angle DYC-\angle DBY = \angle BDY$ On the other hand,
$\angle DFZ=\angle DZE-\angle FDZ=\angle DYC-\angle DBY=\angle BDY$ so $P'D$ is tangent to $(DEF)$ . Thus, by radical center theorem, $DP'$ , $EF$ , and $BC$ concur, at $P$ . Since $BCEF$ is cyclic, $S$ lies on $AP$ and the following are concyclic:

$ABCS$
$SAFE$
$SECP$
$SFBP$

We have $\begin{align*} \angle SEX &= \angle SFA \\ &= \angle AFE - \angle SFE \\ &= \angle ACB - \angle SBC \\ &= \angle XBC - \angle SBC \\ &= SBX \end{align*}$ so $SBEX$ is cyclic. We have $PD^2 = PE\cdot PF = PC\cdot PB=PA\cdot PS$ so the inversion centered at $P$ fixing $D$ switches $A$ and $S$ , $B$ and $C$ , $E$ and $F$ . Thus, $(ACD)$ is taken to $BDS$ . We have $GA\cdot GC = GS\cdot GB=GE\cdot GX$ so $G$ has equal power with respect to $(ACD)$ , $(DEX)$ , and $(BDS)$ . Thus, they are coaxial, and so $P$ is equidistant from the two common intersection points. Thus, the centers of $(ACD)$ and $(DEX)$ are collinear with $P$ , as desired.

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signifance

140 posts

signifance

#70 h Oct 8, 2023, 6:26 PM

Y by

Alright, another 45 MOHS... great. Why do I feel like I received the short end of 45 MOHS again and not actually solving the hard ones... But nonetheless one of my favorite problems, and wth SO MANY CIRCLESSSSS. What a journey it has been (2.5 hrs is not really considered a journey but wtv)

Let $W=AD\cap BC,Y=AD\cap(BDC),Z=BC\cap EF$ ; note that $AYC=DBC=ADF,AYB=DCB=ADE\implies\frac{AF}{AC}=\frac{AD}{AP}=\frac{AE}{AB},$ so BCEF is cyclic. Furthermore, let T be a point on the tangent to (DEF). Then $DEF=AED-AEF=ABY-ABC=CBY=CDW\text{ and }FBD+FDB=AFD=180-FAD-ADF=180-BAW-DBW=ABD+DWB$ $\implies DEF+DCB=CDW+DCB=DWB=FDB=FDT+TDB\Rightarrow TDB=DCB,$ so TD is also tangent to (CDB); in particular, the radcenter Z of (BDC),(DEF),(BCEF) means that the common tangent through D to (DEF) and (BDC) also passes through that point. Define M as the Miquel point of BCEF (so it lies on AZ, and ABCM is cyclic (!)).

Consider the inversion centered at Z with radius DZ, meaning $ZA\cdot ZM=ZE\cdot ZF=ZD^2=ZB\cdot ZC\implies(ADC)\leftrightarrow(MDB)$ . Let $U=(EDX)\cap(ADC),V=BM\cap AC$ ; observe that $\bullet\quad\operatorname{pow}_{(BDM)}V=BV\cdot VM\stackrel{(!)}{=}AV\cdot VC=\operatorname{pow}_{(ADC)}$ , so V lies on the radax UD; furthermore, $UV\cdot VD=AV\cdot VC=BV\cdot VM$ , so BDMU is cyclic. This means U is fixed under this inversion; in particular, letting $O_3$ be the center of (BDMX), this means $ZO_1O_3$ is a line (since the circles invert to each other), but by coaxiality (DEUX is coaxial) $O_2$ lies on this line as well!

$\textbf{Remark.}$ There were two main steps here. The first was defining Y to prove BCEF cyclic, then using a good diagram to conjecture the tangent circles; a flurry of information was suddenly figured out. The second step was observing all the cyclic quads, and btw idt the last few concyclicities was required: for example, u can also prove BEMX cyclic!

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starchan

1605 posts

starchan

#71 h Nov 17, 2023, 9:12 PM • 3 Y

Y by Sanjana42, Siddharth03, mxlcv

this is an insane problem, congratulations to the author
solution
diag?

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popop614

271 posts

popop614

#72 h Nov 24, 2023, 3:02 AM

Y by

Pure unfiltered heat. Awesome.

Let $K$ be the intersection of $AD$ and $BC$ .

$\textbf{Claim.}$ $BCEF$ is cyclic.
$\textit{Proof.}$ Let the circle tangent to $CD$ at $D$ passing through $E$ intersect $AD$ at $P$ . Let $EP$ intersect $AB$ at $F'$ . Note now that it is sufficient to prove that $(BDK)$ is tangent to $DF$ .

Note that $\measuredangle APE = \measuredangle DPE = \measuredangle CDE = \measuredangle CKD = \measuredangle CKA = \measuredangle BKA,$ whereby we conclude that $\measuredangle AEF' = \measuredangle CBA$ . Therefore $BCEF'$ is cyclic. Also note that from a relatively easy angle chase $\triangle DPE \sim \triangle CKD$ .

Now,
$\frac{PD}{PF'} = \frac{PD}{\frac{AF' \cdot PE}{AE}} = \frac{PD}{PE} \cdot \frac{AB}{AC} = \frac{CK}{DK} \cdot \frac{BK}{CK} = \frac{BK}{DK},$ and since $\angle BKA = \angle F'PD$ we conclude that $\triangle F'PD \sim \triangle DKB$ . This establishes that $F' = F$ . $\square$

Define $T = \overline{EF} \cap \overline{BC}$ and $M = \overline{AT} \cap (ABC)$ . Note that $M$ is the miquel point of complete quadrilateral $BCEFAT$ .

$\textbf{Claim.}$ $MBEX$ is cyclic.
$\textit{Proof.}$ $\measuredangle BME = \measuredangle BMT + \measuredangle AME = \measuredangle BFT + \measuredangle AFE = 2\measuredangle BCA = \measuredangle BXE. \square$

Let $W = \overline{BM} \cap \overline{AC}$ . By radical axes on $(ABC)$ , $(BMEX)$ , and $(EXD)$ , and also $(ABC)$ , $(EXD)$ , $(ADC)$ , we conclude that the radical axis of $(ADC)$ and $(EXD)$ passes through $W$ . Let $K$ be the second intersection of $(EXD)$ and $(ADC)$ , and hene $MKBD$ is cyclic.

$\textbf{Claim.}$ $TD^2 = TB \cdot TC$ .
$\textit{Proof.}$ Suffices to prove that $\measuredangle FED + \measuredangle DCB = \measuredangle FDB$ , as then the two circles $(DEF)$ and $(BCD)$ are tangent.
This is equivalent to showing
$\measuredangle ADC + \measuredangle DCB = \measuredangle AKB = \measuredangle FDB$ but this is very easy since we know that $\measuredangle DBC = \measuredangle FDA$ . $\square$

Now,
$\textbf{Invert about $T$ with radius $\sqrt{TB \cdot TC}$}.$ (incredible!!!) $D$ stays fixed, $(B,C)$ , $(E,F)$ $(M,A)$ swap. Then $(MBD)$ and $(ADC)$ swap, which implies that $K$ is fixed. Thus $TK = TD$ , and so the perpendicular bisector of $KD$ , which is $O_1O_2$ , passes through $T$ . Whew.

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thdnder

194 posts

thdnder

#74 h Dec 28, 2023, 9:32 AM • 1 Y

Y by TheHU-1729

What an adventure!

Let $D'$ be the isogonal conjugate of $D$ wrt $\triangle ABC$ . Consider the following claim:

Claim: Quadrilateral $BCEF$ is cyclic.

Proof. $\angle D'DE = \angle ADE = \angle BCD = \angle ACD' = \angle ECD'$ , thus $CDD'E$ is cyclic. Similarly $BDD'F$ is cyclic. Hence we get $AB \cdot AF = AD \cdot AD' = AE \cdot AC$ , so we're done. $\blacksquare$

Claim: Circles $(BCD), (EFD)$ are tangent to each other.

Proof. It suffices to prove that $\angle EFD + \angle CBD = \angle CDE$ . Note that $\angle EDF + \angle BDC = 180^{\circ}$ , so $\angle CDE = 180^{\circ} - \angle BDF = 180^{\circ} - \angle BD'F = \angle BFD' + \angle D'BF = \angle BFD' + \angle CBD$ . Therefore we only have to prove $\angle EFD = \angle BFD'$ , or equivalently $\angle EFD' = \angle BFD$ . Now compute $\angle BFD' + \angle BFD = 180^{\circ} - \angle ADB + \angle BD'D = \angle BAD + \angle DBA + \angle BAD' + \angle ABD' = \angle BAC + \angle ABC = 180^{\circ} - \angle BCE = \angle BFE$ . Therefore $\angle EFD' = \angle BFD$ , as desired. $\blacksquare$

Let $P = EF \cap BC$ . Then we have $PE \cdot PF = PC \cdot PB$ , so $P$ lies on the radical axis of circles $(EFD), (BCD)$ . Thus $PD$ tangents $(EFD), (BCD)$ . Hence we get $PE \cdot PF = PC \cdot PB = PD^2$ . Now perform an inversion centered $P$ with radius $PD$ . Let $A', X'$ be the images of $A, X$ under the inversion, respectively. Then since $PA \cdot PA' = PD^2 = PB \cdot PC$ , so $A'$ is Miquel point of $BCEF$ . The inversion swaps $(ADC)$ to $(A'DB)$ , so $(ADC) \cap (A'DB)$ is fixed under the inversion. Let $Y$ be the intersection of $(ADC)$ and $(A'DB)$ other than $D$ . Then $PD = PY$ .

Claim: $Y$ lies on $(EXD)$ .

Proof. We only need to prove that $Y$ lies on $(FX'D)$ . Note that $(A'EB)$ passes through the center of $(BCEF)$ , so $X$ lies on $(A'EB)$ . Therefore $X'$ lies on $(AFC)$ . Since $X$ lies on $AC$ , therefore $X'$ lies on $(PA'B)$ . Hence $X' = (AFC) \cap (PA'B)$ . Now note that the radical axes of circles $(ACD), (A'DB), (ABC)$ are concurrent, so $AC, A'B, DY$ are concurrent. Let $R = AC \cap A'B \cap DY$ . The radical axes of circles $(AFCX'), (PX'A'FB), (ABC)$ are concurrent, so $FX', AC, AB$ are concurrent, which means $FX'$ passes through $X$ . Therefore $YR \cdot RD = AR \cdot RC = FR \cdot X'R$ . Hence $FDX'Y$ is cyclic, as needed. $\blacksquare$

Since $Y$ lies on $(EXD), (ADC)$ and $PD = PY$ means $P$ lies on $O_1O_2$ . This completes proof. $\blacksquare$

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math_comb01

662 posts

math_comb01

#75 h Jan 13, 2024, 4:10 PM

Y by

Here's a sketch:
This felt staightforward
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CLaim 1: $BCEF$ is cyclic
PRoof
Claim 2 $(EDF),(DBC)$ are tangent to $D$ ; $EF \cap BC \cap DD = T$
PROof
Let $(AEF) \cap (ABC) = I$ , clearly $T-I-A$ , let $H = (DIB) \cap (DEC)$ , $U=BI \cap AC \cap DH$
Claim 3: $IEXB$ is cyclic
PROOf
Claim 4: $ADHC$ is cyclic
PROOF
By this we conclude $TD=TH$ and we're done.

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TestX01

339 posts

TestX01

#76 h Jul 9, 2024, 2:19 AM

Y by

lol

Let $D'$ be the isogonal conjugate of $D$ in $\triangle ABC$ . From our angle condition, clearly $BD'DF$ and $CD'DE$ are cyclic. Radical axis theorem, as $D'$ still lies on the angle bisector, implies $BFEC$ is cyclic.

Note that the angle condition is equivalent to, if we let $P$ be the foot of the bisector from $A$ , $ED$ tangent to $PDC$ and $FD$ tangent to $PDB$ at $D$ . Hence, $\measuredangle CDE=\measuredangle CPD=\measuredangle BPD=\measuredangle BDF$ thus $D$ lies on the isoptic cubic of $BCEF$ thus it has an isogonal conjugate.

Let $D'$ is already the intersection of two isogonal lines, hence $D'$ is precisely the isogonal conjugate. This also implies that $D'$ is the Clawson-Schmidt conjugate of $D$ in $BCEF$ , and from its definition, we conclude that $(FDE)$ is tangent to $(BDC)$ . Radical axis theorem on $(BDC), (FDE)$ and $(BFEC)$ gives $EF,BC,$ and the tangent to $(BDC)$ at $D$ concur, say at $Q$

Now invert at $Q$ fixing $(BCEF)$ . $D$ is fixed by Power of a Point. Now the line $AO_1$ must contain $O_1'$ thus we want to show $O_1',O_2,Q$ collinear. Let $\omega$ be the circle we are inverting about. We want to show that $\omega, (ADC), (XED)$ are coaxial. as then their centers are collinear. First of all, all of these circles pass through $D$ , hence we only need to characterize some other point that has equal power to these three circles.

$XB$ is tangent to $(BFQ)$ as $\measuredangle XBC=\measuredangle BCE=\measuredangle BFE$ , now use Desargues Involution Theorem on degenerate quadrilateral $BBFQ$ and line $AC$ . $BF$ intersects $AC$ at $A$ , $BQ$ intersects $AC$ at $C$ . Hence $\{A,C\}$ is one of the pairs of involution. Now, consider the intersection of $(BFQ)$ with $\omega$ , these intersections lie on $AC$ as $AC$ is the inverse of $(BFQ)$ upon inversion. Hence, letting these be $M,N$ in some order, $\{M,N\}$ is another pair of involution. Finally, $BB\cap AC=X$ , and $FQ\cap AC=E$ , hence $\{E,X\}$ are also a pair of involution.

Such pairs are an involution on a line hence an inversion about some fixed point. That point has equal power with respect to the three circles, so we are done.

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awesomehuman

496 posts

awesomehuman

#77 h Aug 1, 2024, 6:28 PM • 2 Y

Y by crazyeyemoody907, john0512

Let $\Omega$ be the circumcircle of $\triangle ABC$ .
Let $M$ be the arc midpoint of $BC$ .
Let $N$ be the arc midpoint of $BAC$ .
Let $K$ be the intersection of $MC$ with $(ADC)$ .
Let $Z$ be the intersection of $KA$ with $\Omega$ .
Let $\ell_C$ be the tangent to $\Omega$ at $C$ .
Let $U$ be the intersection of $ZN$ with $\ell_C$ .
Then, we have
$X = AC \cap NM$ $U = \ell_C \cap ZN$ $K = MC \cap AZ.$ By Pascal's theorem on $NMCCAZ$ , $X$ , $U$ , and $K$ are colinear.

Let $T$ be on $MN$ such that $\angle DCT = \angle MAC$ .
Then,
$\angle ZKC = \angle AKC = \angle ADC = \angle MDC$ $\angle CZK = \angle CZA = \angle CMA = \angle CMD$ Therefore, $\triangle ZKC \sim \triangle MDC$ . We have
$\angle TCM = \angle DCM + \angle TCD = \angle KCZ + \angle CAM =\angle KCZ + \angle UCK = \angle UCZ$ $\angle CMT = \angle CMN = \angle CZN = \angle CZU.$ Therefore, $\triangle MTC \sim \triangle ZUC$ .
Therefore, $MTDC\sim ZUKC$ .
Let $R$ be the intersection of $TD$ with $XK$ . Then,
$\angle RKC = \angle UKC = \angle TDC = \angle RDC.$ So, $R$ is on $(KCD) = (ADC)$ .
We have $\angle DCT = \angle MAC = \angle DAC$ .
So, $TC$ is tangent to $(ADC)$ . Therefore,
$TB^2 = TC^2 = TD \cdot TR.$ So, inversion at $T$ with radius $TC$ swaps $D$ and $R$ while keeping $B$ and $C$ fixed.
So, it swaps $(BRC)$ with $(BDC)$ .
Let $P$ be the point such that $PD$ is tangent to $(BDC)$ and $PR$ is tangent to $(BRC)$ .
Because inversion preserves angles, $\angle TDP = \angle PRT$ . So, $\triangle RPD$ is isosceles.
Let $\omega$ be the circle tangent to $PD$ at $D$ and $PR$ at $R$ .
By the radical center theorem on $(BDC)$ , $(BRC)$ , and $\omega$ , $P$ lies on $BC$ .
We have
$\angle XRD = \angle KRD = \angle KCD = \angle MCD = \angle MAB + \angle BCD$ $= \angle CAM + \angle ADE = \angle EAD+\angle ADE = \angle AED = \angle XED.$ Therefore, $R$ is on $(EXD)$ . So, $RD$ is the radical axis of $(EXD)$ and $(ADC)$ .
Therefore, $O_1O_2$ is the perpendicular bisector of $RD$ .
Because $PR = PD$ and $P$ is on $BC$ , $P=O_1O_2\cap BC$ .

Let $D'$ be the point on $AD$ such that $DBD'C$ is cyclic. Then,
$\angle ADE = \angle BCD =\angle BD'D = \angle BD'A.$ Because $\angle EAD = \angle D'AB$ , $\triangle BAD'\sim \triangle EAD$ .
Similarly, $\triangle CAD'\sim \triangle FAD$ , so $AEDF\sim ABD'C$ .
So, $\triangle AEF\sim \triangle ABC$ .
Therefore, $EFBC$ is cyclic.
We have
$\angle PDE = \angle PDD' + \angle ADE = \angle DBD' + \angle BCD = \angle DBD' + \angle BD'D = \angle BDD' = \angle BCD' = \angle DFE.$ So, $PD$ is tangent to $(DEF)$ .
By the radical center theorem on $(FECB)$ , $(BCD)$ , and $(EFD)$ , $P$ lies on $EF$ . Therefore, $BC$ , $EF$ , and $O_1O_2$ concur at $P$ .

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cursed_tangent1434

597 posts

cursed_tangent1434

#78 h Oct 6, 2024, 6:30 AM • 1 Y

Y by Narwhal234

Another IMO3 YAY! My first actual solution involving some non-trivial isogonal conjugate argument. We denote by $R$ the intersection of lines $\overline{EF}$ and $\overline{BC}$ . We start off by making the following observation.

Claim : Points $B$ , $C$ , $E$ and $F$ are concyclic.

Proof : Note that since
$\measuredangle CDB + \measuredangle FDE = \measuredangle CBD + \measuredangle DCB + \measuredangle FDA + \measuredangle ADE = \pi$ it follows that $D$ has an isogonal conjugate $D'$ with respect to quadrilateral $BFEC$ . Further note that since $CD$ and $CD'$ are isogonal with respect to $\angle C$ , and $BD$ and $BD'$ are isogonal with respect to $\angle B$ , $D'$ is also the isogonal conjugate of $D$ with respect to $\triangle ABC$ . Since $D$ lies on the internal $\angle A-$ bisector this implies that $D'$ also lies on the internal $\angle A-$ bisector. Now, note that if we let $D'' = \overline{AD} \cap (CDE)$ and $K = \overline{BC} \cap (CED)$ , we must have $AD \parallel EK$ since $\measuredangle KED = \measuredangle KCD = \measuredangle BCD = \measuredangle ADE$ so
$\measuredangle BCD'' = \measuredangle KCD'' = \measuredangle KED'' = \measuredangle DD''E = \measuredangle DCE$ so lines $CD$ and $CD''$ are isogonal in $\angle BCA$ . Combined with the fact that $D''$ lies on the internal $\angle A-$ bisector this implies that $D''$ is the isogonal conjugate of $D$ with respect to $\triangle ABC$ , so $D''\equiv D'$ and thus $D'$ lies on $(DEC)$ . An entirely similar argument shows that $D'$ also lies on $(BDF)$ which implies that $(BDF) \cap (CED)$ lies on $\overline{AD}$ . Thus, $\overline{AD}$ must be the radical axis of $(BDF)$ and $(CDE)$ implying,
$AF \cdot AB = AD \cdot AD' = AE \cdot AC$ which concludes that $BFEC$ is a cyclic quadrilateral as desired.

Now, let $M_Q$ denote the Miquel Point of cyclic quadrilateral $BFEC$ . We then know that $\overline{AM_Q}$ also passes through $R$ . Now, we can notice the following further cyclic quadrilaterals.

Claim : Quadrilaterals $BEM_QX$ and $BDM_QS$ are cyclic.

Proof : The first one is immediate. Note that,
$\measuredangle EM_QB = \measuredangle EM_QA+ \measuredangle AM_QB = \measuredangle EFA + \measuredangle ACB = 2\measuredangle ACB = \measuredangle CXB = \measuredangle EXB$ which implies that $BEM_QX$ is indeed cyclic. Further, let $L = \overline{AC} \cap \overline{BM_Q}$ . Note that,
$LX \cdot LE = LB \cdot LM_Q = LA \cdot LC$ which implies that $L$ lies on the radical axis of circles $(XED)$ and $(ADC)$ . Letting $S = (XED) \cap (ADC) \ne S$ , this then implies $LB \cdot LM_Q = LD \cdot LS$ so quadrilateral $BDM_QS$ must be cyclic as well, as desired.

We now prove a claim which will come in very important later on.

Claim : Line $\overline{DL}$ is a common tangent to circles $(DBC)$ , $(DEF)$ and $(DAM_Q)$ .

Proof : We start off by showing that circles $(DBC)$ and $(DEF)$ are tangent to each other. Note that it suffices to check that $\measuredangle EDC = \measuredangle DBC + \measuredangle FDE$ which we shall do as follows. Let $P = \overline{AD} \cap \overline{BC}$ .
$\begin{align*} \measuredangle EFD + \measuredangle DBC & = \measuredangle D'FB + \measuredangle DBC\\ & = \measuredangle D'DB + \measuredangle DBP \\ &= \measuredangle DPB \\ & = \measuredangle EKB \\ &= \measuredangle EKC \\ &= \measuredangle EDC \end{align*}$ so indeed circles $(BDC)$ and $(EDF)$ are tangent to each other at $D$ . Further, by Radical Center theorem on circles $(DEF)$ , $(BCD)$ and $(BCEF)$ their common tangent must pass through $R$ . Again,
$AR^2 = RB \cdot RC = RA \cdot RM_Q$ which further implies that $\overline{RD}$ is tangent to $(ADM_Q)$ finishing the proof of the claim.

To finish off we need to show the following key claim.

Claim : Circles $(SBC)$ and $(SAM_Q)$ are tangent at $S$ , with common tangent $\overline{SR}$ .

Proof : As before, we first show that circles $(SBC)$ and $(SAM_Q)$ are tangent at $S$ , for which it suffices to check $\measuredangle SAM_Q + \measuredangle CBS = \measuredangle CSM_Q$ . This is a really long angle chase. I will only include a sketch here but it is detailed enough for anyone interested to be able to fill in the details. Note that since
$\measuredangle CSM_Q = \measuredangle CSD + \measuredangle DSM_Q = \measuredangle CAD + \measuredangle DBM_Q = \measuredangle CAD + \measuredangle CBM_Q + \measuredangle DBC$ it suffices to show that,
$\measuredangle SAC + \measuredangle CBS = \measuredangle CAD + \measuredangle CBS$ which we shall do as follows.

Note that since due to our previous claim that $\overline{DR}$ is tangent to $(AM_QR)$ and $(DBC)$ , we have $\measuredangle M_QDR = \measuredangle M_QAD$ and $\measuredangle RDC = \measuredangle DBC$ . Thus,
$\measuredangle DBC = \measuredangle M_QDR + \measuredangle RDC + \measuredangle DAM_Q = \measuredangle M_QDC + \measuredangle DAM_Q$ Further,
$\measuredangle M_QDC + \measuredangle CBM_Q = \measuredangle M_QAD + \measuredangle CAM_Q + \measuredangle DBC$ which implies $\measuredangle CAD + \measuredangle DBC = \measuredangle M_QDC + \measuredangle CBM_Q$ . Now we are almost done as $(BDM_QS)$ cyclic implies,
$\measuredangle CAD + \measuredangle DBC = \measuredangle SDM_Q + \measuredangle M_QDC + \measuredangle M_QBS + \measuredangle CBM_Q = \measuredangle SAC + \measuredangle CBS$ which is a result which we have already confirmed. Thus, circles $(SBC)$ and $(SAM_Q)$ are indeed tangent at $S$ . Further, by Radical Center Theorem on circles $(SBC)$ , $(SAM_Q)$ and $(ABCM_Q)$ it follows that the common tangent to $(SBC)$ and $(SAM_Q)$ at $S$ passes through $R$ .

Now we simply need to connect all the pieces. Note that due to the last two claims,
$RS^2 = RA \cdot RM_Q = RD^2$ so $R$ must lie on the perpendicular bisector of segment $SD$ , which is well known to be the line joining the centers of the circles $(XED)$ and $(ADC)$ which is simply line $\overline{O_1O_2}$ implying that this line indeed passes through $R$ as we set out to show.

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Narwhal234

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Narwhal234

#79 h Oct 7, 2024, 6:20 AM

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The above admitted orz while solving

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alinazarboland

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alinazarboland

#80 h Oct 16, 2024, 12:57 PM

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We'll use Animation. But first let's make a proper setup.
One can easily observe that if $D'$ is the isogonal conjugate of $D$ wrt $ABC$ , then $C,D',D,F$ and $B,D',D,E$ and so $B,C,E,F$ are concyclic and if $T=EF \cap BC$ then $TD$ is tangent to $(BDC)$ , all by radical axis theorem. Let $S$ be the midpoint of arc $(BAC)$ in $(ABC)$ . Let $J = (ADC) \cap (ASX)$ . Now let's start animating:
Move $D$ on $BC$ with degree 1. Then by inverting through $A$ , $J$ has degree 2. And by inverting through $B,C$ , both $E,F$ have degree 2. We'll now prove that $D,E,X,J$ are concyclic. By inverting through $X$ , $E*,D*$ have degree 2 and $J*$ has degree 1 so we need 6 special cases to check the claimed concyclicity which are the items 1,2,3,4,7 below+ the case for $D$ where $E$ goes to infinity. Now , It's enough to show that $T$ lies on the perpendicular bisector of $DJ$ . But before that , note that there's two cases where $E=F$ so the line $EF$ has degree at most $2 = deg(E)+deg(F)-k$ where $k = 2$ is the number of coincidences of $E,F$ . Let $M$ be the midpoint of $DJ$ which has degree at most $deg(D)+deg(J)=3$ and note that since there exists one case(no.8 below) where $D=J$ , $DJ$ has degree at most $2 =deg(D)+deg(J)-k$ .So perpendicular bisector of $DJ$ , which is the perpendicular through $M$ to $DJ$ has degree at most $3+2=5$ and $O_1O_2 \cap BC$ would have degree at most 5. So to prove that the recent intersection , is $T$ , a degree 2 point we need $8=5+2 +1$ special cases which are:
1,2,3,4 . The infinity point of the direction $l$ , $l \cap (ABC)$ , $l \cap BC$ , $A$
5,6 . The two cases , where $TJ$ turns to be tangent to $(DEX)$ which exists by IVT
7. The one case where the perpendicular bisector of $DJ$ coincides with $BC$ which exists by IVT
8.The one case where $D=J$ which exists by IVT.
Where $l$ is the angle-bisector of $A$ ...And we're done.

This post has been edited 1 time. Last edited by alinazarboland, Oct 16, 2024, 12:58 PM

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kmh1

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kmh1

#81 h Nov 21, 2024, 7:59 PM

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Let $M$ be the intersecting point of lines $EF$ and $BC$ , and $K$ be the intersecting point of lines $AD$ and $BC$ . Since $\angle CDE=\angle ADC-\angle ADE=\angle ADC-\angle DCB=\angle DKC$ , we have $\frac {CE}{EA}=\frac {DCsinCDE}{DAsinADE}=\frac {DCsinDKC}{DAsinDCB}$ . Similarly we have $\frac{BF}{FA}=\frac{DBsinDKB}{DAsinDBC}$ . Using Menelaus theorem we get $\frac{CM}{MB}=\frac{CE}{EA}\cdot\frac{FA}{BF}=\frac{DCsinDKC}{DAsinDCB}\cdot\frac{DAsinDBC}{DBsinDKB}=\frac{DCsinDBC}{DBsinDCB}=\frac{DC^2}{DB^2}$ From this, it is easy to show that line $MD$ is tangent to the circumcircle of $DBC$ .

We need to show that $M$ is on line $O_1O_2$ . Define $\omega$ as the circle centered at $M$ with radius $MD$ . It suffices to show that $\omega$ and the circumcircles of triangles $ADC$ and $EXD$ are coaxial. We can show this by finding a point $T(\neq D)$ such that the power of $T$ to the three circles are equal.

Define $T$ as the point on segment $AE$ such that $XT\cdot TE=AT\cdot TC$ . We will prove that $MT^2-MD^2=-AT\cdot TC$ . Note that this solves the problem, because $T$ has same power on $\omega$ and the circumcircles of triangles $ADC$ and $EXD$ . Clearly $T\neq D$ , so the three circles would be coaxial.

We have $\frac{TA}{TE}=\frac{TX}{TC}=\frac{XA}{CE}$ , so $TA=\frac{XA\cdot AE}{XA+CE}$ and $TC=AC-TA=\frac{XA\cdot AC+CE\cdot AC-XA\cdot AE}{XA+CE}=\frac{CE\cdot XC}{XA+CE}$ . We have $MT^2-MD^2=TC^2+MC^2+2TC\cdot MCcosC-MB\cdot MC=TC^2+2TC\cdot MCcosC-MC\cdot BC$ , so we have
$MT^2-MD^2=-AT\cdot TC \iff TC^2+2TC\cdot MCcosC-MC\cdot BC=-AT\cdot TC \iff TC\cdot AC+2TC\cdot MCcosC=MC\cdot BC$ Using $TC=\frac{CE\cdot XC}{XA+CE}$ as we noted above, and $2XCcosC=BC$ which is obvious, we get
$TC\cdot AC+2TC\cdot MCcosC=MC\cdot BC \iff \frac{CE\cdot XC\cdot AC+BC\cdot CE\cdot MC}{XA+CE}=MC\cdot BC$ $\iff CE\cdot XC\cdot AC=\cdot XA\cdot MC\cdot BC\iff \frac{CE}{MC}=\frac{XA\cdot BC}{XC\cdot AC}$
So proving $\frac{CE}{MC}=\frac{XA\cdot BC}{XC\cdot AC}$ will solve the problem.

RHS is easy to calculate. $XC=\frac{a}{2cosC}=\frac{a^2b}{a^2+b^2-c^2}$ and $XA=XC-b=\frac{(c^2-b^2)b}{a^2+b^2-c^2}$ , so $\frac{XA\cdot BC}{XC\cdot AC}=\frac{c^2-b^2}{ab}$ .

For LHS, we will show that $\angle CEM=\angle B$ . (The motivation for this is that LHS has to be fixed as $D$ moves on the bisector of $\angle A$ , which means $\angle CEM$ has to be fixed. Considering the case $D=A$ , we can see that this fixed angle has to be $\angle CEM=\angle B$ .)

Many people in this forum have proved that $BCEF$ is cyclic, e.g. using the isogonal conjugate of $D$ , so it follows that $\angle CEM=\angle AEF=\angle B$ . However I failed to see this, so below is what I did. It is not elegant, but it is quite straightforward.
Click to reveal hidden text

We can easily finish off. We have $\frac{CE}{MC}=\frac{sinECM}{sinCEM}=\frac{sin(C-B)}{sinB}=cosB\cdot\frac{c}{b}-cosC=\frac{c^2-b^2}{ab}$ , therefore $\frac{CE}{MC}=\frac{XA\cdot BC}{XC\cdot AC}$ as desired. So $T$ is indeed a point which the power to the three circles are equal. The problem is solved.

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shanelin-sigma

155 posts

shanelin-sigma

#82 h Feb 17, 2025, 5:18 AM

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Very short solution. Hope this isn’t fake solve……

Kagebaka wrote:

Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.

1. Let $EF\cup BC=T$ construst a circle $\omega$ centered at $T$ with ratio $\overline{DT}$
Our goal is to show that $O_1-O_2-T$ and this is equivalent to show $\odot (ADC),\odot(DEX),\omega$ are coaxial
Since we’ve known that $D$ is on these circles so we only has to find another point $P$ such that it has same power to them.
———————————————————————————————————————
2. Claim: $XB$ is tangent to $\odot(BET)$
Proof: Notice that $\odot(BCEF)$ cyclic (many of above have shown that and I’m too lazy to type again)
then $\measuredangle XBT= \measuredangle XBC= \measuredangle BCX= \measuredangle BCE= \measuredangle BFE= \measuredangle BFT$
———————————————————————————————————————
3. Consider line $AC$ and $BBET$ applying DIT to show that $(X;T) (E;X) (U;V)$ are some pairs of an involution (where $U,V=\odot (BET)\cup AC$ )
Since an involution on a line must be an inversion so $\exists P\in AC$ such that $PX\cdot PT=PE\cdot PX=PU\cdot PV$
Namely $pow_{\odot(ADC)}(P)=pow_{\odot(DEX)}(P)=pow_{\omega}(P)$ so we are done

This post has been edited 3 times. Last edited by shanelin-sigma, Feb 18, 2025, 3:44 AM

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EeEeRUT

64 posts

EeEeRUT

#84 h Apr 1, 2025, 2:52 AM

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Wow, this is a wonderful problem.

Claim: $E, F, B, C$ are concyclic.
Proof: Let $Z$ be isogonal conjuagate of $D$ , hence $\angle ADE \angle DCB = \angle ZCA$ Thus, $EDZC$ is cyclic. Similarly, $FDZB$ is cyclic.
By radical axis, $EFBC$ is cyclic.

Let $EF$ intersects $BC$ at $O$ .
Claim: $(BCD)$ and $(DEF)$ are tangent. Moreover, $O$ lies on their common tangent line.
Proof: By the claim above, $O$ lies on radical axis of $(BCD)$ and $(DEF)$ .
Hence, we need to show that $(BCD)$ and $(DEF)$ are tangent.
Consider
$\begin{align*} \angle FDB &= 180^{\circ} - \angle DFB - \angle DBF\\ &= \angle AFD - \angle FBC + \angle DBC\\ &=\angle AFD + \angle FDA - \angle FBC\\ &=180^{\circ} - \angle FAD - \angle FBC\\ &=\angle ADE + \angle FED\\ &= \angle DCB + \angle FED \end{align*}$ Hence, $(BCD)$ and $(DEF)$ are tangent.

Let the second intersection of $(DEX)$ and $(ADC)$ be $P$ .
We will instead show that $OP = OD$ .
Let $S$ be miquel point of $\square FECB$ .

Claim: $S, E, X, B$ are concyclic.
Proof: Let $G$ be circumcenter of $(BCEF)$ .
Consider $\begin{align*} \angle BSE &= \angle FOB + \angle FAE\\ &= 360^{\circ} - \angle OBF - \angle AEF - 2\angle OFB\\ &= 360^{\circ} - 180^{\circ} - 2\angle ECB\\ &= 180^{\circ} - \angle EGB \end{align*}$ That is $SEBG$ is cyclic. And since, $\angle EXB = 2\angle ECB = \angle EGB$ .
We have $SEXBG$ cyclic.

Let $SB$ intersect $AC$ at $K$ . Note that $K$ on the radical axis of $(SEXB)$ , $(SACB)$ .
Hence, $KA \cdot KC = KE \cdot KX$ . Thus, $K$ is on radical axis of $(PEXD)$ and $(APDC)$ .
Hence, $SPDB$ is cyclic.
Note that inversion at $O$ with radius $OD$ swap $(SPDB)$ and $(APDC)$ . This mapping has $2$ fix point $D$ and $P$ , hence we have $OP= OD \blacksquare$ .

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