IMO 2018 Problem 1

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322 posts

#1 h Jul 9, 2018, 11:20 AM • 40 Y

Y by ssk9208, l1090107005, Kayak, falantrng, fattypiggy123, Hypernova, YRNG-BCC168, sa2001, Wizard_32, Supercali, Kunihiko_Chikaya, Carpemath, Muradjl, opptoinfinity, Amir Hossein, MexicOMM, Durjoy1729, Medjl, e_plus_pi, Davrbek, NahTan123xyz, nguyendangkhoa17112003, AlastorMoody, Richangles, magicarrow, Aryan-23, Lcz, centslordm, HamstPan38825, megarnie, ImSh95, rayfish, This_deserves_a_like, Adventure10, deplasmanyollari, Rounak_iitr, aidan0626, ItsBesi, cubres, Sedro

Let $\Gamma$ be the circumcircle of acute triangle $ABC$ . Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$ . The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece

This post has been edited 2 times. Last edited by djmathman, Jun 16, 2020, 4:02 AM
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UlanKZ

38 posts

UlanKZ

#2 h Jul 9, 2018, 11:22 AM • 9 Y

Y by Carpemath, bucketboy, Amir Hossein, centslordm, megarnie, ImSh95, Adventure10, Mango247, cubres

The coincides WLOG if $AB<AC$ and $B=D$

This post has been edited 1 time. Last edited by UlanKZ, Jul 9, 2018, 11:33 AM
Reason: Good luck

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orthocentre

72 posts

orthocentre

#3 h Jul 9, 2018, 11:23 AM • 9 Y

Y by NotKris, Carpemath, Amir Hossein, centslordm, megarnie, ImSh95, Adventure10, Mango247, cubres

"Prove that lines $DE$ and $FG$ are either parallel or they coincide."

Prove that they are either parallel or not parallel...?

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SHARKYKESA

436 posts

SHARKYKESA

#4 h Jul 9, 2018, 11:24 AM • 6 Y

Y by centslordm, ImSh95, Adventure10, Mango247, sabkx, cubres

jdevine wrote:

"Prove that lines $DE$ and $FG$ are either parallel or they coincide."

Prove that they are either parallel or not parallel...?

No, coincide = they are the same line.

This post has been edited 1 time. Last edited by SHARKYKESA, Jul 9, 2018, 11:24 AM

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orthocentre

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orthocentre

#5 h Jul 9, 2018, 11:24 AM • 11 Y

Y by Snakes, s333neb, samoha, socr4tes, danepale, centslordm, megarnie, ImSh95, Adventure10, Mango247, cubres

Oh... xD thanks.

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Amin_Hashemi-1379

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Amin_Hashemi-1379

#7 h Jul 9, 2018, 11:28 AM • 9 Y

Y by MK4J, APWLGeometry, MarkBcc168, Luchitha2, centslordm, ImSh95, Quidditch, Adventure10, cubres

novus677 wrote:

Answer appears to be hi from geogebra observations...

This app has some bugs. Not working good always

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SHARKYKESA

436 posts

SHARKYKESA

#8 h Jul 9, 2018, 11:28 AM • 7 Y

Y by Makorn, centslordm, IAmTheHazard, ImSh95, TomMae10, Adventure10, cubres

novus677 wrote:

Answer appears to be hi from geogebra observations...

You can also observe from the question...

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TheDarkPrince

3042 posts

TheDarkPrince

#10 h Jul 9, 2018, 11:35 AM • 8 Y

Y by thewizardofmath, FadingMoonlight, centslordm, ImSh95, Adventure10, Mango247, cubres, pokpokben

Complex bash!!

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orthocentre

72 posts

orthocentre

#11 h Jul 9, 2018, 11:35 AM • 5 Y

Y by centslordm, ImSh95, Adventure10, Mango247, cubres

Just what I'm doing right now.

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MS_Kekas

275 posts

MS_Kekas

#12 h Jul 9, 2018, 11:38 AM • 6 Y

Y by Amir Hossein, centslordm, ImSh95, Adventure10, Mango247, cubres

Obviously, we have to show that $GF$ is perpendicular to internal bisector of $BAC$

Let $F_1,G_1$ be points on smaller arc $BC$ chosen so that $BF_1 = CG_1 = AD = AE$ . Than we easily see that proection of $F$ onto the line $AB$ divides line $F_1-B-A$ onto two equal parts. That means that $F$ is the midpoint of arc $ABF_1$ (Archimed Lemma). Simialrly $G$ is the midpoint of arc $ACG_1$ . The left is angle chasing

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TacH

64 posts

TacH

#13 h Jul 9, 2018, 11:38 AM • 10 Y

Y by Maths01, Amir Hossein, nguyendangkhoa17112003, Smkh, pipitchaya.s4869, myh2910, centslordm, Adventure10, Mango247, cubres

Just some sketch then

Extends $FD$ to cut the circumcircle at $X$ , also $GE$ to cut the circumcircle at $Y$
Then we can angle chasing to get that $X,Y,D,E$ are cyclic (of course with center $A$ )
Then, $DE$ is anti-parallel to $XY$
and of course $XY$ is anti parallel to $GF$ (since $X,Y,F,G$ are also cyclics)
hence $DE // FG$

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RC.

439 posts

RC.

#14 h Jul 9, 2018, 11:39 AM • 8 Y

Y by MATH1945, rocketscience, Amir Hossein, square_root_of_3, Assassino9931, centslordm, Adventure10, cubres

Note that it suffices to prove that $\angle AFD = \angle AGE$ Given $AD = AE$ . Which follows from sine rule in $\Delta AFB, \Delta AGC$

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MarkBcc168

1594 posts

MarkBcc168

#16 h Jul 9, 2018, 11:41 AM • 10 Y

Y by IMO2019, Amir Hossein, dchenmathcounts, e_plus_pi, bashee_wang, Richangles, Aryan-23, centslordm, Adventure10, cubres

Let $G'$ be the point on $\Gamma$ such that $FG'\parallel DE$ and let $E'$ be the point such that $G'E' = G'C$ . It suffices to prove that $E'=E$ .

Let $FG'$ intersects $AB, AC$ at $P, Q$ . Notice that $AP=AQ$ so $\angle FDP = \angle FBA = \angle AG'Q$ and $\angle AQG' = \angle DPF$ . Thus $\triangle DPF\sim\triangle G'QA$ . Similarly, $\triangle E'QG'\sim\triangle FPA$ . Thus,
$DP\cdot AQ = PF\cdot G'Q = E'Q\cdot AQ$ so $DP=E'Q\implies AD=AE'\implies E=E'$ as desired.

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WizardMath

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WizardMath

#17 h Jul 9, 2018, 11:42 AM • 8 Y

Y by rocketscience, Amir Hossein, soryn, SHREYAS333, pipitchaya.s4869, centslordm, Adventure10, cubres

Choose points $Y, Z$ on arc $BC$ such that $AD = BY = CZ$ . Then we have $FBY$ and $FDA$ congruent, so $F$ is the midpoint of arc $AY$ . Since $BC$ and $YZ$ are parallel, $FG$ is parallel to the external bisector of $A$ , so done.

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Orestis_Lignos

555 posts

Orestis_Lignos

#18 h Jul 9, 2018, 11:44 AM • 4 Y

Y by Amir Hossein, centslordm, Adventure10, cubres

Let the two bisectors be cut at $K$ and $L \equiv FK \cap BD, M \equiv EC \cap GK$ . Let $I$ be the incenter of the triangle $ABC$ . Then, we obviously have $AI \perp DE$ . Let $T \equiv AI \cap DE$ . So, $ATLF$ is cyclic, so $\angle LAT=\angle LFT=\angle LFG \Rightarrow \angle KFG=\dfrac{\angle A}{2}$ , similarly $\angle FGK=\dfrac{\angle A}{2}$ . So, if $P \equiv EG \cap DB$ , we have $\angle APG=90-\dfrac{\angle A}{2}=\angle ADE \Rightarrow DE \parallel FG$ , done.

(Oviously, when $B \equiv D$ , the lines coincide.)

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Assassino9931

1238 posts

Assassino9931

#170 h Feb 16, 2024, 11:23 PM • 1 Y

Y by cubres

If you prefer to not add new points in the diagram, then trig bash for the win!

By the Sine Law in triangles $AFD$ , $AEG$ and the circumcircle of $ABC$ with radius $R$ we have
$\frac{AD}{\sin \angle AFD} = \frac{AF}{\sin \angle ADF} = \frac{AF}{\sin \angle BDF} = \frac{AF}{\sin \angle ABF} = 2R$ $\frac{AE}{\sin \angle AGE} = \frac{AG}{\sin \angle AFG} = \frac{AG}{\sin \angle GFC} = \frac{AG}{\sin \angle ACG} = 2R$ and since $AD = AE$ from the problem condition, we obtain $\sin \angle AFD = \sin \angle AGE$ . As $\angle AFD + \angle AGE < \angle AFG + \angle AGF < 180^{\circ}$ , it follows that $\angle AFD = \angle AGE$ . Denote their common value by $z$ .

Denote $P = AB \cap FG$ and $Q = AC \cap FG$ . Since $GC = GF$ from the perpendicular bisector, we compute
$\angle APQ = \angle AFG + \angle BAF = \angle ACG + \angle BAF = \angle AFG + \angle BAF$ $= \angle GAF + \angle AGF + \angle BAF = x + y + z.$ Analogously $\angle AQP = x + y + z$ . Therefore $AP = AQ$ , so $\angle ADE = 90^{\circ} - \frac{1}{2}\angle ABC = \angle APQ$ and the result follows.

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peppapig_

279 posts

peppapig_

#171 h Mar 9, 2024, 10:28 PM • 1 Y

Y by cubres

We use a phantom point type argument - we start with assuming that $FG$ is such that if $FG\cap AB=X$ and $FG\cap AC=Y$ , then $AX=AY$ . This ensures that no matter what $D$ , $E$ we choose such that $AD=DE$ , $FG$ is parallel to $DE$ . Now, define $D'$ to be the point on segment $AB$ (meaning on line $AB$ between $A$ and $B$ ) such that $BF=FD'$ and define $E'$ similarly to be on segment $AC$ with $CG=GE'$ . I now claim that $AD'=AE' \rightarrow D'E'\parallel FG$ .

WLOG, assume that the radius of the circumcircle of $\triangle ABC$ is $\frac{1}{2}$ . This means that for any chord of the circle that bisects a minor arc with angle $\theta$ , the length of the chord is $\sin \theta$ . Now, let $\angle BAC=a$ , $\angle ABC=b$ , $\angle ACB=c$ , and $\angle GCA=x$ . Through angle chasing, we get that
$\angle FBA=\angle FBC-b=(180-\angle FGC)-b=(\angle GYC+\angle GCA)-b=(90-\frac{a}{2})+x-b=\frac{-b+c}{2}+x,$ which then gives that
$\angle FAB= \frac{b+c}{2}-x,$ and
$\angle CAG=b-x.$
Now, for the trig. By our assumption about the circumcircle's radius, we have that
$CG=\sin \angle GAC=\sin (b-x),$ and since $GC=CE'$ , using the isosceles triangle, we get that
$CE'=2\sin (b-x)\cos x,$ which is equal to $\sin b+\sin (b-2x)$ by Product-to-Sum formulas. This means that
$AE'=AC-(\sin b+\sin (b-2x))=\sin b-(\sin b+\sin (b-2x))=-\sin(b-2x).$ Similarly, we get that
$BD'=2\sin \left(\frac{b+c}{2}-x\right)\cos x,$ which is equal to $\sin c+\sin (b-2x)$ by Product-to-Sum formulas. This means that
$AD'=AB-(\sin c+\sin (b-2x))=\sin c-(\sin c+\sin (b-2x))=-\sin(b-2x),$ which is indeed equal to $AE'$ , as desired.

Varying the line $FG$ along the circle, we see that all possible lines $DE$ are covered using this "backwards" approach (i.e. for any two points $D$ , $E$ , chosen on $AB$ and $AC$ such that $AD=AE$ , we can find line $FG$ such that $D'=D$ and $E=E'$ ). Therefore, $DE\parallel FG$ or $DE$ and $FG$ coincide, as desired.

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Reason: Bad latex

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idkk

118 posts

idkk

#172 h Mar 12, 2024, 11:01 AM • 1 Y

Y by cubres

Supercali wrote:

PavelMath wrote:

Here is my solution.
Let points $D$ and $E$ move linearly along the lines $AB$ and $AC$ respectively. Then the lines $(DE)$ and $(FG)$ also move linearlly. Hence its enough to prove that $(DE)||(FG)$ for three positions of the points $D$ and $E$ . We can take $D=E=A$ , $D=B$ and $E=C$ . For all these cases our statment $(DE)||(FG)$ is obviously true. Hence it is true for arbitrary point $D$ .

I have a similar proof using Perspectivity. This argument can more rigorous.

how u do that? A similar case where moving points was a India TST P1

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ABCDE

1963 posts

ABCDE

#173 h Mar 19, 2024, 6:22 AM • 10 Y

Y by InCtrl, pieater314159, trumpeter, OronSH, anantmudgal09, math_comb01, ihatemath123, kingu, qwerty123456asdfgzxcvb, cubres

We proceed using differential moving points, with thanks to Linus Tang for bringing this problem to my attention. The idea is to compute the derivatives of $F$ and $G$ as a function of $D$ and $E$ and show that their angular velocities are equal so the line through them remains parallel to a fixed line, reducing the problem to checking one special case. Formalizing this idea takes a bit more work, as these derivatives themselves depend on the positions of $F$ and $G$ .

Let $M$ and $N$ be the midpoints of $BD$ and $CE$ , $\ell_F$ and $\ell_G$ be the tangents to $(ABC)$ at $F$ and $G$ , and $\theta_F$ and $\theta_G$ be the angles that $\ell_F$ and $\ell_G$ make with $AB$ and $AC$ . We will parameterize $AB$ and $AC$ with $\mathbb R$ and $(ABC)$ with $\mathbb R/2\pi\mathbb Z$ , though we will be loose with sign conventions to simplify the presentation. Throughout the solution, $\frac{dY}{dX}$ will denote the derivative of the point $Y$ as a function of the point $X$ which varies along a curve parameterized as above.

We have that $\frac{dF}{dM}$ has magnitude $\frac1{\cos\theta_F}$ along $\ell_F$ by linearizing $(ABC)$ at $F$ and $\frac{d\theta_F}{dF}=1$ as the direction of $\ell_F$ is perpendicular to the argument of $F$ on $(ABC)$ . Clearly, $\frac{dM}{dD}=\frac12$ , so $\frac{d\theta_F}{dD}=\frac1{2\cos\theta_F}$ by the chain rule. Repeating the same argument with $N$ , $\ell_G$ , and $\theta_G$ , we find that $\frac{d\theta_G}{dE}=\frac1{2\cos\theta_G}$ . Now, defining $E$ as the point on $AC$ such that $AD=AE$ , we have that $\frac{dE}{dD}=1$ so $\frac{d\theta_G}{dD}=\frac1{2\cos\theta_G}$ as well.

The key claim now is that $\theta_F=\theta_G$ as functions of $D$ . Indeed, $\theta_F(B)=\theta_G(B)$ as when $D=B$ we have that $D,E,G$ are collinear and both angles can be chased to be equal to $\angle C$ . Now, both $\theta_F$ and $\theta_G$ satisfy the differential equation $\frac{d\theta}{dD}=\frac1{2\cos\theta}$ with the initial condition $\theta(B)=\angle C$ . By standard ODE theory, modulo some technical details such as generalizing the problem to allow for $D$ and $E$ to lie slightly outside $AB$ and $AC$ and checking that $\theta\mapsto\frac1{2\cos\theta}$ is "nice" enough, the solution to this differential equation is unique, so indeed we have that $\theta_F=\theta_G$ . (Technically, we might only be able to conclude local uniqueness, but this would amusingly allow us to finish easily with vanilla moving points.)

Finally, $\theta_F=\theta_G$ implies that the angle bisectors of $\angle(\ell_F,\ell_G)$ and $\angle BAC$ are parallel, so $FG$ is perpendicular to the angle bisector of $\angle BAC$ along with $DE$ , as desired.

I would be very interested in any other problems that can be done using this kind of approach. For instance, it should be possible to show the generalized Poncelet's porism with these techniques. If anyone knows of other examples, please let me know.

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mofumofu

179 posts

mofumofu

#174 h Jun 16, 2024, 10:19 AM • 3 Y

Y by G81928128, busy-beaver, cubres

I was at the contest! I thought that the constructions of new points in this problem is not entirely trivial.

I wrote a short guide on Reim and my solving process for this problem (hopefully with some motivations and takeaways) on the blog here.

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P2nisic

406 posts

P2nisic

#175 h Jul 2, 2024, 1:02 PM • 1 Y

Y by cubres

juckter wrote:

Let $\Gamma$ be the circumcircle of acute triangle $ABC$ . Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$ . The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece

Let $X,Y$ are points on $(ABC)$ such that $AX=AY=AD$
Let $F$ be the center of $(DXB)$ then we get that:
$\angle XFB=360-2\angle XDB=360-2(180-\angle ADA)=2\angle XDA=180-\angle XAB$ so $F$ belongs to $(ABC)$ .
$\angle XFY=2\angle XBA=2\angle XBD=\angle XFD$ so $F,D,Y$ are collinear.
$\angle XGF=\angle XAF=\frac{1}{2}\angle XAD=\angle XED$ so $DE//FG$

Attachments:

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MagicalToaster53

159 posts

MagicalToaster53

#176 h Jul 26, 2024, 4:25 AM • 1 Y

Y by cubres

Let $GE$ and $FD$ meet $\Gamma$ at $X$ and $Y$ respectively. Then it suffices to show by Reim's theorem that $XYED$ is cyclic. We are motivated that $A$ is the circumcenter of the circumcircle of $XYED$ as $\triangle ADE$ is isosceles. We make the following claim:

Claim: $\triangle AXE$ and $\triangle ADY$ are both isosceles.
Proof: We prove for $\triangle AXE$ as the case for $\triangle ADY$ follows by symmetry. We find that $\angle GXA = \angle GCE = \angle GEC = \angle AEX. \phantom{c} \square$
Therefore we have that $AX = AE = AD = AY$ , so that $XYED$ is cyclic, as desired. $\blacksquare$

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Phat_23000245

18 posts

Phat_23000245

#177 h Jul 26, 2024, 5:05 AM • 1 Y

Y by cubres

Cho B', C' là các điểm trên đường tròn ngoại tiếp ABC sao cho BF = B'F và CG = C'G. Vì $\góc B'AF = \góc BAF$, $AB'FD$ là một hình diều. Tương tự, $AC'GE$ là một hình diều, và do đó $AB' = AD = AE = AC'$ . Khi đó $F, A, G$ là các trung điểm của các cung $BB', B'C', C'C$ tương ứng. Lưu ý rằng tổng của ba cung này là $360^\circ - 2\góc A$, do đó một nửa tổng đó là $180^\circ - \góc A$. Nếu $L$ là trung điểm của cung nhỏ $BC$ , phép đuổi góc đơn giản để tìm góc giữa $AL$ và $FG$ là $90^\circ$ . Khi đó $DE$ rõ ràng là vuông góc với $AL$ :fool:

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kinnikuma

9 posts

kinnikuma

#179 h Aug 19, 2024, 10:17 AM • 1 Y

Y by cubres

Construct $H$ so that $ADFH$ is a parallelogram, and construct $I$ in the same manner. Notice, by some angle chasing, that $H$ and $I$ both lie on $\Omega(ABC)$ . Let $J$ be the intersection of $(HF)$ and $(IG)$ . By definition of the parallelogram, we have $(JF) \parallel (AD)$ and $(JG) \parallel (AE)$ . Furthermore, notice that $HF = AD = AE = IG$ , so that $FHIG$ becomes a trapezoid. Therefore $\angle JFG = \angle JGF$ . Consequently, with the parallelisms stated before, $JFG$ and $ADE$ are similar. Hence with the parallelism stated they are necessarily homothetic : the parallelism that was unknown to us is especially $(FG) \parallel (DE)$ , as wanted $\huge \blacksquare$ .

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Jndd

1417 posts

Jndd

#180 h Sep 11, 2024, 1:53 AM • 1 Y

Y by cubres

Construct points $X$ and $Y$ on $\Gamma$ on minor arcs $AB$ and $AC$ respectively, such that $FXAB$ and $GYAC$ are isosceles trapezoids. Then, since $FD$ is the reflection of $FB$ over the line through $F$ perpendicular of $AB$ , and similarly for $GE$ , we have that $XADF$ and $YAEG$ are both parallelograms. Thus, $XF=AD=AE=YG$ , meaning $XYGF$ is an isosceles trapezoid. Additionally, letting $M$ and $N$ being midpoints of minor arcs $AB$ and $AC$ respectively, notice that $M$ and $N$ are also the midpoints of minor arcs $XF$ and $YG$ . Thus, minor arcs $FM$ , $MX$ , $GN$ , and $NX$ are all equal, which now gives that $MNGF$ is an isosceles trapezoid. Thus, $MN\parallel FG$ , so we want to show $MN\parallel DE$ . We have $\angle(DE,BC) = B - \angle{ADE} = B - \left(90 - \frac{A}{2}\right) = \frac{B-C}{2}.$ We will show that this value is equal to $\angle(MN,BC)$ . We see that $\angle(MN,BC) = \angle{NBC} - \angle{MNB} = \frac{B}{2}-\frac{C}{2},$ as desired.

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ezpotd

1252 posts

ezpotd

#181 h Sep 23, 2024, 3:05 PM • 1 Y

Y by cubres

maturity is realizing that no one cares if the synthetic solution is garbage

We use complex numbers. $D$ is the reflection of $B$ over the foot from $F$ to $AB$ , so we have $d = 2 \frac 12 (f + a + b -\frac{ab}{f}) - b = a+ f- \frac{ab}{f}$ , symmetrically $e = a + g - \frac{ac}{g}$ . The condition that $AD = AE$ can be written as $\mid a + f - \frac{ab}{f} - a \mid = \mid a + g - \frac{ac}{g} - a \mid$ , which simplifies to $\mid f^2 - ab \mid = \mid g^2 - ac \mid$ , using $|x|^2 = x \overline{x}$ gives the condition as $(f^2 - ab)(\frac{1}{f^2} - \frac{1}{ab}) = (g^2 - ac)(\frac{1}{g^2} - \frac{1}{ac})$ , upon expansion and cancelling $2$ on both sides this becomes $\frac{f^2}{ab} + \frac{ab}{f^2} = \frac{g^2}{ac} + \frac{ac}{g^2}$ , shifting terms and common denominator fives $\frac{f^2c - g^2b}{abc} = \frac{acf^2 - abg^2}{f^2g^2}$ , cancelling $f^2c-g^2b$ gives $f^2g^2 = a^2bc$ as the equivalent condition.

Now we desire to prove $\frac{d - e}{f - g}$ is self conjugating, which is equivalent to $\frac{f - g + \frac{ac}{g} - \frac{ab}{f}}{f - g} = \frac{\frac 1f - \frac 1g + \frac{g}{ac} - \frac{f}{ab}}{\frac 1f - \frac 1g}$ , multiplying the numerator and denominator of the right hand side gives $\frac{f -g + \frac{f^2g}{ab} - \frac{g^2f}{ac}}{f - g}$ , so it suffices to prove $\frac{ac}{g} - \frac{ab}{f} = \frac{f^2g}{ab} - \frac{g^2f}{ac}$ , since $f^2g^2 = a^2bc$ , we can write $\frac{f^2g}{ab} = \frac{f^2g^2}{gab} = \frac{a^2bc}{gab} = \frac{ac}{g}$ , so we are done.

This post has been edited 1 time. Last edited by ezpotd, Sep 25, 2024, 6:06 AM

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Osim_09

36 posts

Osim_09

#182 h Feb 6, 2025, 5:24 PM • 1 Y

Y by cubres

Let GE intersect (ABC) at X and FD intersect (ABC) at Y.Then <FBD=<DYA and <FBD=<FDB=<ADY hence AD=AY.
<GCE=<AXE and <GCE=<GEC=<AEX hence AX=AE (since AD=AE) we have XDEY cyclic.So <YXE=<YDE and <YXE=<YFG hence DE//FG.

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Phat_23000245

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Phat_23000245

#183 h Feb 15, 2025, 4:41 AM

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skibidi toilet :yoda:

Quote:

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bjump

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bjump

#185 h Mar 25, 2025, 2:58 PM

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Let $H$ be where the perpendicular bisectors of $DB$ , and $EC$ meet. Let $O$ be the circumcenter of $\triangle ABC$ , let $P$ and $Q$ be the midpoints of $AB$ , and $AC$ respectively, and $K$ , and $L$ be the arcmidpoints of $AB$ , and $AC$ respectively.
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Claim: $ONHM$ is a rhombus.

Proof: Note that by perpendicularity to $AB$ , and $AC$ we have $ON \parallel HM$ , and $OM \parallel HN$ . Note that by contruction we have $PI=QJ$ therefore by sliding principal we have $ON \sin( 180^\circ-\angle NOM) = OM \sin(180^\circ - \angle NOM) \implies ON=OM$ . Therefore $ON=NH=MH=MO$ . $\square$

This means that $M$ , and $N$ are symmetric about $OH$ , which means that $LG$ and $KF$ are symmetric about $OH$ . Therefore $KLGF$ is a cyclic isosceles trapezoid with $KL \parallel FG$ It is well known that the arcmidpoints of $AB$ , and $AC$ are perpendicular to the angle bisector of $\angle BAC$ , and that $DE$ is also perpendicular to the angle bisector. Therefore we are finished.

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Trasher_Cheeser12321

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Trasher_Cheeser12321

#187 h Apr 4, 2025, 6:06 AM

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Trying to improve my trig bash (guided by OTIS walkthrough).
$[asy] size(7cm); pair A = dir(110); pair B = dir(205); pair C = dir(335); pair D = A+0.4dir(degrees(B-A)); pair E = A+0.4dir(degrees(C-A)); pair X = 0.5(B+D); pair F = intersectionpoints(X+10*dir(degrees(A-B)-90)--X-10*dir(degrees(A-B)-90), circumcircle(A, B, C))[1]; pair Y = 0.5(C+E); pair G = intersectionpoints(Y+10*dir(degrees(A-C)-90)--Y-10*dir(degrees(A-C)-90), circumcircle(A, B, C))[0]; draw(extension(A, dir(-90), F, G)+dir(degrees(G-extension(A, dir(-90), F, G)))*0.06 -- extension(A, dir(-90), F, G)+dir(degrees(G-extension(A, dir(-90), F, G)))*0.06+dir(degrees(A-dir(-90)))*0.06 -- extension(A, dir(-90), F, G)+dir(degrees(A-dir(-90)))*0.06, red+linewidth(0.5)); draw(A--B--C--cycle); draw(D--E^^C--G--F--B); draw(circumcircle(A, B, C)); draw(A--dir(-90), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(160)); dot("$E'$", E, dir(40)); dot("$F$", F, dir(F)); dot("$G'$", G, dir(G)); dot("$P$", extension(A, B, F, G), dir(125)); dot("$Q$", extension(A, C, F, G), dir(60)); [/asy]$
Define $G'$ on $(ABC)$ such that $\overline{FG'} \perp \ell$ where $\ell$ is the angle bisector of $A$ . Also, define the point $E'$ on $\overline{AC}$ such that $GE' = GC$ . The goal is to show that $E = E'$ and $G = G'$ .

Claim: $\angle FBA + \frac{\angle B}{2} = \angle G'CA + \frac{\angle{C}}{2}$ .
Proof: This can be shown by proving $\angle FBA + \frac{\angle B}{2} + \frac{\angle C}{2} = \angle G'CA + \angle C$ :
$\begin{align*} \angle FBA + \frac{\angle B}{2} + \frac{\angle C}{2} &= \angle FBA+\left(90^\circ - \frac{\angle A}{2}\right)\\ &= \angle FBA + \angle APQ\\ &= 180^\circ - \angle BFG' \\ &= \angle G'CB \end{align*}$ which is equal to $\angle G'CA + \angle C$ . $\blacksquare$

The rest can be done with trig bash. Let $x = \angle FBA - \frac{\angle C}{2} = \angle GCA - \frac{\angle B}{2}$ .
$\begin{align*} \frac{BF}{\sin \angle FAB} = 2R \quad \Longrightarrow \quad BF &= 2R \sin \angle FAB\\ &= 2R\sin(\angle C - \angle FBA)\\ &= 2R\sin\left(\frac{\angle C}{2}-x\right).\\\\ BD=BF\cos\angle FBA \quad \Longrightarrow \quad BD &= 4R\sin\left(\frac{\angle C}{2}\right)\cos\left(\frac{\angle C}{2}+x\right)\\ &= 2R\left(\sin(-2x)+\sin\angle C\right) \qquad\qquad\qquad\quad \text{(product to sum)}\\ &= 2R\left(\sin\angle C-\sin(2x)\right).\\\\ AD = AB-BD \quad \Longrightarrow \quad AD &= (2R\sin\angle C) - 2R(\sin\angle C-\sin(2x))\\ &= 2R\sin(2x) \end{align*}$ Doing the same thing on the other side, $AE'$ can also be calculated to be equal to $2R\sin(2x)$ . Therefore, $AD=AE'$ , meaning $E = E'$ . It then also follows that $G=G'$ . $\blacksquare$ .

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