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

"Prove that lines $DE$ and $FG$ are either parallel or they coincide." Prove that they are either parallel or not parallel...?

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.

Oh... xD thanks.

novus677 wrote:

Answer appears to be hi

from geogebra observations... This app has some bugs. Not working good always

novus677 wrote:

Answer appears to be hi

from geogebra observations... You can also observe from the question...

Complex bash!!

Just what I'm doing right now.

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

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 $DE // FG$

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

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.

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.

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.)

Here’s a pretty rotation proof. Let $M$, $N$, $M'$ and $N'$ be the midpoints of minor arc $AB$, minor arc $AC$, arc $ACB$, and arc $ABC$. and $O$ be the center of $\Gamma$. Let $K \neq F$ be on $\Gamma$ such that $FK$ and $MM'$ are parallel, and $L \neq G$ on $\Gamma$ such that $GL$ and $NN'$ are parallel. Observe that the distance of $FK$ to $MM'$ is half the length of $AD$, which is half the length of $AE$, which is the distance of $GL$ to $NN'$. Thus we can rotate isosceles trapezoid $MM'KF$ about $O$ to get isosceles trapezoid $N'NGL$, and so $MF = M'K = NG$. So $MNGF$ is an isosceles trapezoid. But it can be easily shown that both $MN$ and $DE$ are perpendicular to the bisector of $\angle BAC$.

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Can’t wait for someone to say that this was the easiest IMO problem ever and that it shouldn’t have been on the test at all.

Wictro wrote: Can’t wait for someone to say that this was the easiest IMO problem ever and that it shouldn’t have been on the test at all. It's universally understood.

Wictro wrote: Can’t wait for someone to say that this was the easiest IMO problem ever and that it shouldn’t have been on the test at all. It is not true

Let $ AB<AC,$ Wlog: $B\not= D.$ We have $\angle FBD+\angle AFG-\angle AFD=\angle FBD+\angle DFG$ $$=\angle FDB+\angle DFG=^{?} \angle ADE=\angle AED=^{?} \angle GEC+\angle EGF$$$=\angle GCE+\angle EGF=\angle GCE+\angle AGF-\angle AGE.$ Then to prove that $\angle AFD=\angle AGE.$ This is easy apply sine rule in $\triangle AFD,\triangle AEG,\triangle AFG.$ If $B\equiv D ,$ then $F\equiv B.$ Let $G'=BE\cap \Gamma,$ then $\angle GEC=\angle AEB=\angle ABF=\angle ACG'\to G'E=G'C,$ then $G\equiv G'.$

Let $A'$ be the intersection of the perpendicular bisectors. I claim the locus of $A$ is the line through the circumcenter $O$ parallel to the angle bisector of $\angle BAC$. Clearly $O$ is on the locus because of the case $D=E=A$. Also note that the locus of $A'$ is the set of points such that the lengths of the projections of $AB'$ and $AC'$ have constant difference where $B'$ is the projection of $A'$ onto $AB$ and $C'$ is the projection of $A'$ onto $AC$. Since this is a linear function, and the locus has dimension $1$, and adding a vector parallel to the bisector leaves the difference unchanged, the locus is a line parallel to $\angle BAC$. Now let $X$ be the intersection of this line and the circumcircle. Let $Y$ be the point on the circumcircle such that $XY || AB$. Similaraly define $Z$ so $XZ || AC$. Now observe that $F$ is the point on minor arc $XY$ so $A'F \perp XY$. Similarly $A'G \perp XZ$ for $Z$ on minor arc $XZ$. This means by symmetry that $A'F=A'G$. Then we have $FG \perp A'O$. Done.

Let $D'$ be the reflection of $D$ over $AF$, notice by angle chase that $BFD'A$ is cyclic. Define $E'$ similarly, so $D',E'\in (ABC)$ and now notice that $D'E'\perp AO$ and now we're done by some more angle chase. hehaw

Let $H=GE\cap \Gamma$, $I=FD\cap \Gamma$. Claim: $HDEF$ is cyclic with a center of $A$ Proof: We can say: \[\angle HAE=\angle HAC=\angle HGC=\angle EFC,\]\[\angle HEA=\angle GEC,\]so $\triangle AEF\cong \triangle CGE$ by $AA$ similarity. Therefore: \[AH=AE=AD.\]We use a similar approach for $AI$, implying the desired result $\square$ We can now say: \[\angle EFG=\angle IHG=\angle IDE,\]so $DE\parallel FG$ $\blacksquare$

Let $\infty_A$ denote the point at infinity along the tangent to the circumcircle at $A$. Let $\infty_B, \infty_C$ be the points at infinity along the perpendicular bisectors of $\overline{AC}, \overline{BC}$. Define $M, N$ to be the midpoints of $\overline{BD}, \overline{CE}$. Move $D$ projectively along line $AB$. Note that the map $$\overline{AB} \to \overline{AB} \to (ABC) \to (ABC) \quad\text{by}\quad D \mapsto M \mapsto F \mapsto G$$is projective where the first map follows by homothety and the last two maps follow by projections through $\infty_C,\infty_A$ respectively. Also note that the map $$\overline{AB} \to \overline{AC} \to \overline{AC} \to (ABC) \quad\text{by}\quad D \mapsto E \mapsto N \mapsto G$$is projective where the first map follows by Fixed Similarity Lemma, the second map follows by homothety and the last map follow by a projection through $\infty_B$. Therefore we only have to verify these coincide for three values of $D$ but this is easy by picking $D = A, D = B$ and $D = X$ where $AX = AC$.

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.

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.

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

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.

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.

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 $DE//FG$

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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. $\blacksquare$

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 $AL$

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 $\huge \blacksquare$.

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.

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.