The condition that we are being asked to prove is equivalent to proving that $EF$ is antiparallel to $BC$ for every $(E,F)$ that is interesting. So basically we need to prove that if $(E,F)$ is interesting then $EFBC$ is cyclic. This is an example of a geometry problem that can be solved methodically by using complex numbers. For every point $X$ in the plane, we denote $x$ to be the associated complex number, and $\bar{x}$ its conjugate. Let's first review some properties: 1- $x = e^{i\varphi}|x|$ where $\varphi$ is the argument of $x$ and $|x|$ is its distance to the origin (absolute value) 2- $x$ and $\bar{x}$ have the same absolute value and oposite argument 3- $x \in \mathbb{R}$ if and only if $x$ = $\bar{x}$ (this follows from 1 and 2) 4- $x\bar{x} = |x|^2$ (this follows from 1 and 2) 5- $\frac{x}{\bar{x}} = e^{2i\varphi}$ where $\varphi$ is the argument of $x$ (this follows from 1 and 2) 6- $x$ belongs to the unit circle if and only if $\bar{x} = \frac{1}{x}$ (this follows from 4) 7- $\varphi = \angle{ACB}$ (from $A$ to $B$ in positive direction) if and only if $\frac{b-c}{|b-c|} = e^{i\varphi}\frac{a-c}{|a-c|}$ (this follows from 1) 8- $\frac{a-b}{\bar{a}-\bar{b}} = e^{2i\varphi}\frac{c-d}{\bar{c}-\bar{d}}$ where $\varphi$ is the angle between $CD$ and $AB$ (from $CD$ to $AB$ in positive direction) (this follows from 5) 9- $AB \parallel CD$ if and only if $\frac{a-b}{\bar{a}-\bar{b}} = \frac{c-d}{\bar{c}-\bar{d}}$ (this follows from 8) 10- $AB \perp CD$ if and only if $\frac{a-b}{\bar{a}-\bar{b}} = -\frac{c-d}{\bar{c}-\bar{d}}$ (this follows from 8) 11- $A,B,C$ are colinear if and only if $\frac{a-b}{\bar{a}-\bar{b}} = \frac{a-c}{\bar{a}-\bar{c}}$ (this follows from 9) 12- $A,B,C,D$ belong to a circle if and only if $\frac{a-c}{b-c}:\frac{a-d}{b-d} \in \mathbb{R}$ (this follows from 7) 13- $A,B,C,D$ belong to a circle if and only if $\frac{a-c}{\bar{a}-\bar{c}}:\frac{b-c}{\bar{b}-\bar{c}} = \frac{a-d}{\bar{a}-\bar{d}}:\frac{b-d}{\bar{b}-\bar{d}}$ (this follows from 3 and 12. It also follows directly from 8) 14- For a chord $AB$ of the unit circle, we have $\frac{a-b}{\bar{a}-\bar{b}} = -ab$ (this follows from 6) 15- If $C$ belongs to the chord $AB$ of the unit circle, then $\bar{c} = \frac{a+b-c}{ab}$ (this follows from 6 and 11) Now, back to the problem. We can assume without loss of generality that the circumcircle of $\triangle{ABC}$ is the the unit circle in the complex plane. The strategy to solve a problem like this using complex numbers is by first computing all the points(complex numbers) that we need in terms of some reduced set of points(complex numbers). The conditions of the problem will be reduced to some equation(s); what we are asked to prove will also be reduced to some equation(s); and all we need to prove is that one equation implies the other. So, here we are going to try to express everything in terms of $a,b,c,e,f$ We can express $\bar{a},\bar{b},\bar{c},\bar{e},\bar{f}$ in terms of $a,b,c,e,f$ using 6 and 15 We can express $m$ in terms of $e,f$ since $m = \frac{e+f}{2}$ and therefore $\bar{m}$ can be expressed in terms of ${a,b,c,e,f}$ too. We can express $\bar{k}$ in terms of $b,c,k$ using 15 $EF \perp MK$, so, using 10 we get an equation involving $k,m,e,f$ and their conjugates. All of those can be expressed in terms of $a,b,c,e,f$ and $k$. So, we solve for $k$, and we will also get $k$ expressed in terms of $a,b,c,e,f$ and therefore $\bar{k}$ Let $R$ be the middle point of $MK$. We can express $r$ in term of $m,k$ and therefore in terms of $a,b,c,e,f$. Therefore $\bar{r}$ too. We can express $\bar{s}$ in terms of $s,a,c$ using 15 $SR \parallel EF$, so, using 9 we get an equation involving $s,r,e,f$ and their conjugates. All of those can be expressed in terms of $a,b,c,e,f$ and $k$. So, we solve for $s$ and we get $s$ expressed in terms of $a,b,c,e,f$ and therefore $\bar{s}$ In the same way we get $t$ and $\bar{t}$ expressed in terms of $a,b,c,e,f$. Actually, we should be able to get the formulas for $t$ and $\bar{t}$ from the ones from $s$ and $\bar{s}$ interchanging $b$ with $c$ and $e$ with $f$. The beautiful thing up to this point, is that all the equations we get are simple linear equations, so they can be easily solved. This is because all the geometric constructions up to this point only involve straight lines. The condition of $KSAT$ being cyclic is translated to an equation (using 13) involving $k,s,a,t$ and their conjugates. All of them we have been able to express in terms of $a,b,c,e,f$. So, this is an equation that relates $a,b,c,e,f$. Let's call this equation $(I)$. We need to prove that $EFBC$ is cyclic. This is also equivalent to an equation (using 13) involving $e,f,b,c$ and their conjugates; all of which we have expressed also in terms of $a,b,c,e,f$. Let's call this second equation $(II)$. All we need to prove is that $(I)$ implies $(II)$ I carried out the computations to get equation $(II)$ and I got: $(II): (b-c)(fbc- ebc - fac + eab - a^2b + a^2c) = 0$ I didn't carry out the computations to get equation $(I)$, I gave up before finishing. But I bet that it is something like: $(I): a^{?}(e-f)^{?}(b-c)^{?}(fbc- ebc - fac + eab - a^2b + a^2c) = 0$ So that $(I)$ would clearly imply $(II)$

My solution : Lemma : Let $ E , F $ be the points on $ CA, AB $, respectively . Let $ M $ be the midpoint of $ EF $ and $ K \in BC $ be the point s.t. $ KE=KF $ . Let $ \ell $ be the perpendicular bisector of $ MK $ and $ S \equiv \ell \cap CA, T \equiv \ell \cap AB $ . If $ A, K, S, T $ are concyclic, then $ AK $ is A-symmedian of $ \triangle AEF $ Proof : Let $ X \equiv AM \cap \odot (AKST) $ . From $ EF \parallel ST\Longrightarrow AM $ is A-median of $ \triangle AST $ , so notice $ M, K $ are symmetry WRT $ ST $ we gte $ M $ is the reflection of $ X $ in the midpoint of $ ST $ , hence $ STKX $ is an Isosceles trapezoid $\Longrightarrow XK \parallel ST \parallel EF \Longrightarrow AK $ is A-symmedian of $ \triangle AEF $ . ____________________________________________________________ Back to the main problem : Let the perpendicular bisector of $ E_iF_i $ cuts $ BC $ at $ K_i $ ($i=1,2$) . Let $ H $ be the orthocenter of $ \triangle ABC $ and $ B' \equiv BH \cap CA, C' \equiv CH \cap AB $ . Let $ R $ be the midpoint of $ BC $ and $ S \equiv \odot (AH) \cap AR $ (i.e. $ AB'SC' $ is a harmonic quadrilateral) . From $ AS \cdot AR =AC' \cdot AB=AB' \cdot AC \Longrightarrow S $ is the Miquel point of $ R, B', C' $ WRT $ \triangle ABC $ . $ (\star) $ From $ RB', RC' $ are the tangents of $ \odot (AB'C') $ (well-knwon) $ \Longrightarrow \angle RB'C'=\angle RC'B'=\angle BAC $ . From the lemma $ \Longrightarrow AK_1 $ is A-symmedian of $ \triangle AE_1F_1 $ , so combine $ K_1E_1=K_1F_1 \Longrightarrow K_1E_1, K_1F_1 $ are the tangents of $ \odot (AE_1F_1) $ , hence we get $ \angle K_1E_1F_1=\angle K_1F_1E_1=\angle BAC \Longrightarrow \triangle RB'C' \sim \triangle K_1E_1F_1 $ . Similarly, we can prove $ \triangle RB'C' \sim \triangle K_2E_2F_2 \Longrightarrow \triangle RB'C' \sim \triangle K_1E_1F_1\sim \triangle K_2E_2F_2 $ , so from $ (\star) \Longrightarrow S $ is the center of spiral similarity that maps $ \triangle RB'C'\mapsto\triangle K_1E_1F_1 \mapsto \triangle K_2E_2F_2 $ , hence we get $ E_1E_2:F_1F_2=SE_1:SF_1=SE_2:SF_2=SB':SC'=AB':AC'=AB:AC $ . Q.E.D

From 2015 Taiwan TST Round 3 Mock IMO Day 2 Problem 1, it follows that $K_1 \in BC$ is intersection of the tangents of $\odot(AE_1F_1)$ at $E_1,F_1$ and similarly $K_2 \in BC$ is intersection of the tangents of $\odot(AE_2F_2)$ at $E_2,F_2.$ This is also the lemma in Telv's previous solution. Let $P$ be the 2nd intersection of $\odot(AE_1F_1)$ and $\odot(AE_2F_2);$ center of the spiral similarity carrying $E_1E_2$ into $F_1F_2.$ Since the isosceles $\triangle K_1E_1F_1$ and $\triangle K_2E_2F_2$ are directly similar, then $K_1$ is the image of $K_2$ under the referred spiral similarity $\Longrightarrow$ $P$ is Miquel point of $\{AB,BC,K_2F_2,K_1F_1 \}$ $\Longrightarrow$ $BK_2PF_2$ is cyclic $\Longrightarrow$ $\angle PBK_2=\angle PF_2K_2=\angle PAF_2$ $\Longrightarrow$ $\odot(ABP)$ touches $BC$ and similarly $\odot(ACP)$ touches $BC$ $\Longrightarrow$ $AP$ is radical axis of $\odot(ABP),\odot(ACP)$ bisecting their common tangent $\overline{BC},$ i.e. $AP$ is A-median of $\triangle ABC.$ Therefore $AB:AC=\text{dist}(P,E_1E_2):\text{dist}(P,F_1F_2)=E_2E_2:F_1F_2.$

I also used the same lemma as Telv Cohl did. Let $L, M$ be the intersections of $CF_1$ and $BE_1, CF_2$ and $BE_2.$ Since $B,K_1,C$ are collinear then using the converse of Pascal theorem for 6 points $A,F_1,F_1,L,E_1,E_1$ we get $L$ lies on $(AE_1F_1)$. Similarly $M$ lies on $(AE_2F_2).$ Therefore $\angle F_2F_1C=\angle E_2E_1B, \angle F_1F_2C=\angle E_1E_2B$, which follows that $\triangle CF_1F_2\sim\triangle BE_1E_2$. Then $\frac{E_1E_2}{F_1F_2}=\frac{dis(B,AC)}{dis(C,AB)}=\frac{AB}{AC}.$ Thus $\frac{E_1E_2}{AB}=\frac{F_1F_2}{AC}.$

I have Another solution: I use the following lemma that has been proved in TelvCohl's solution in the post #3#. Lemma: the tangents from $E_1,F_1$ to $\odot (\triangle AE_1F_1)$ intersect each other at $K_1$. Back to the main problem: From the lemma $K_1E_1,K_1F_1$ are tangents to $\odot (\triangle AE_1F_1)$. So $E_1F_1$ is polar of $K_1$ WRT $\odot (\triangle AE_1F_1)$ let $BE_1\cap \odot (\triangle AE_1F_1)=X$ and $AX\cap E_1F_1=Y$. since $B,K_1,C$ are collinear polar of $B,K_1,C$ are concurrent. Notice that $Y$ lies on polar of $K_1,B$ so $Y$ also belongs to the polar of $C$. hence $C,X,F_1$ are collinear. Let $H$ be the orthocenter of $\triangle ABC$ and $AH\cap BC=D$ and let $R$ be the projection of $H$ on the line $AN$ where $N$ is midpoint of $BC$. since $\odot (BC)$ is orthogonal to cyclic quadrilateral $AE_1XF_1$ (well known) we get that $\odot (AE_1XF_1)$ passes throw $R$ ($\because$ the inverse of $\odot (AE_1XF_1)$ passes throw $N$ under the inversion $\Psi$ with center $A$ and power $AH.AD$). Similarly $\odot (\triangle AE_2F_2)$ passes throw $R$. hence $R$ is miquel point of $E_1E_2F_1F_2\Longrightarrow \triangle RE_1E_2\sim \triangle RF_1F_2\Longrightarrow \frac{E_1E_2}{F_1F_2}=\frac{\text{dis}(R,AC)}{\text{dis}(R,AB)}=\frac{AB}{AC}$. DONE

Remark: In My solution in the previous post the fact that $\odot (\triangle AE_1XF_1)$ passes throw $R$ is old. For more solutions see Turkey's TST 2010 Q5

Let $M'$ be the reflection of $M$ in the midpoint of $ST$. Then as noted before, $\angle SM'T=\angle SMT=\angle SKT$, so $M'$ lies on $(ASTK)$, and since $KM' \parallel ST$, $\angle SAK=\angle TAM$, which means that $K$ is the point of intersection of the tangents at $E,F$ of $w$, the circumcircle of $\triangle AEF$. Let $X = BE \cap w$. Then applying Pascal to hexagon $AFFXEE$, we get that $C,F,X$ are collinear. Let $Y= BC \cap (BFX)$. Then $\angle BYX =\angle AFX =\angle XEC$, so $C,E,X,Y$ are concyclic. Then if $AE=x,AF=y$ (signed lengths), we have that $a *BY =BX*BE= c(c-y)$, $a*CY=CX*CF=b(b-x)$, which leads to $bx+cy=b^2+c^2-a^2$. The desired result follows easily.

I use the same lemma as Telv. Let $(E_1,F_1)$ and $(E_2, F_2)$ be to interesting pairs and $K_1, K_2$ the corresponding points $K$. Let furthermore $X=F_1K_1 \cap F_2K_2$, $Y=E_1K_1 \cap E_2K_2$ and $Z=E_1F_1 \cap E_2F_2$. Since $\triangle E_1F_1K_1 \sim \triangle E_2F_2K_2$, it is easy to see that these lines cannot be parallel. Obiously $F_1F_2XZ$ and $E_1E_2YZ$ are cyclic quadrilaterals, so sine law yields $$\frac{F_1F_2}{XZ} = \frac{\sin \angle F_1ZF_2}{\sin \alpha}, \quad \frac{E_1E_2}{YZ} = \frac{\sin \angle E_1ZE_2}{\sin \alpha} \Longrightarrow \frac{F_1F_2}{XZ}=\frac{E_1E_2}{YZ}.$$ It remains to prove that $\frac{XZ}{YZ} = \frac{AC}{AB}$. But by simple angle chasing ($XYK_1K_2$ is also cyclic) we get $\angle ZYX = \beta$ and $\angle YXZ = \gamma$. This implies that the triangles $XYZ$ and $ABC$ are similar, and thus the desired result.

This is a nice problem! We shall proceed with the same notation as in the problem statement. Lemma 1. Lines $KE,KF$ are tangents to $(AEF)$ if and only if $K,S,A,T$ are on a circle. Proof of Lemma 1. Notice that $ST \parallel EF$. Now, $M$ is the mid-point of $EF$ implies that $AM$ is a median in triangle $AST$ as well. Now, the condition is equivalent to saying that reflection of the median $AM$ in side $ST$ of triangle $AST$ meets its circumcircle again at point $K$. It is well-known that the only such point satisfies that $AK$ is a symmedian in triangle $AST$ and so, it is a symmedian in $AEF$. Now, the intersection of the symmedian with the perpendicular bisector of the opposite side is the intersection of tangents at the end-points of the sides, our lemma holds. Lemma 2. Lines $BE$ and $CF$ meet on $(AEF)$. Proof of Lemma 2. Let line $BE$ meet $(AEF)$ again at point $P$. We prove that $C,F,P$ are on a line. For this, we apply Pascal's Theorem on the degenerate cyclic hexagon $(AEEPFF)$ we get that $K=EE \cap FF$, $B=EP \cap FA$, $C_0=AE \cap PF$ are collinear. Therefore, we get $C=C_0$ which proves the claim. Lemma 3. $(AEF)$ passes through a fixed point $X \not= A$. Proof of Lemma 3. We claim that the fixed point $X$ is in fact the intersection of the $A$ appolonian circle wrt $ABC$ with the median from vertex $A$. This follows by applying inversion about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the bisector of angle $BAC$, the result to be proven is just Pascal's Theorem applied as done in the previous paragraph. For more reference, see ELMO 2013 G3 proposed by Allen Liu. We now proceed with the final proof. Indeed, notice that $X$ is the center of a spiral similarity sending $F_1F_2$ to $E_1E_2$. Therefore, $\frac{E_1E_2}{F_1F_2}$ is constant and so, we only want to prove that $\frac{MB'}{MC'}=\frac{AB}{AC}$ where points $B',C'$ are the feet of altitudes from $B,C$ onto sides $AC,AB$. Noting that $AB'MC'$ is a harmonic quadrilateral and so \begin{align*} \frac{AB}{AC}=\frac{AB'}{AC'}=\frac{MB'}{MC'} \end{align*} This concludes our proof.

Bad solution, but it works I guess:

I use Telv's Lemma. From here we obtain $\angle KEF=\angle KFE=\angle BAC$. Let $N$ be the second intersection of $BC$ and $(EFK)$. Then $\angle ENK=180^{\circ}-\angle EFK=180^{\circ}-\angle BAC$, so $AENB$ is cyclic. Similarly $AFNC$ is cyclic. Hence $BF\cdot BA+CE\cdot CA=BN\cdot BC+CN\cdot CB=BC^2$ Hence $BF_1\cdot BA+CE_1\cdot CA=BF_2\cdot BA+CE_2\cdot CA$ So $AB\cdot F_1F_2=AC\cdot E_1E_2$ as desired.

Here is a similar solution to the above. Note by Taiwan TST 2015 that $KE, KF$ are tangent to $(AEF)$; let $(KEF)$ meet $BC$ at $X$, then $\measuredangle FXB = \measuredangle FXK = \measuredangle FAB$, so $(ABFX)$ is cyclic. Similarly $(ACEX)$ is cyclic, so by ELMO SL 2013 $(AEF)$ passes through the $A$-HM point $P$ of $ABC$. Then, since there is a spiral similarity centered at $R$ taking $E_1F_1\to E_2F_2$ and $E_2F_2\to EF$, where $E,F$ are the feet of the $B$, $C$ altitudes, \[ \frac{E_1E_2}{F_1F_2} = \frac{RE_2}{RF_2}, \]but $\frac{RE_2}{RF_2} = \frac{RE}{RF} = \frac{AE}{AF} = \frac{AB}{AC}$, since $R$ is on the $A$-symmedian of $AEF$. So, $\frac{E_1E_2}{F_1F_2} = \frac{AB}{AC}$ as desired. $\blacksquare$

Even quicker finish. First, we prove that $\angle KEF = \angle KFE = \angle A$. Note that $M$ lie on $A$-median of $\triangle AST$. Thus the reflection $M'$ of $M$ across midpoint of $ST$ lie on $\odot(ASKT)$. Moreover, $SKM'T$ is isosceles trapezoid thus lines $AM, AK$ are isogonal w.r.t. $\triangle ABC$. This means $AK$ is $A$-symmedian of $\triangle AEF$. Combining with the symmedian lemma gives the conclusion. Now, let $X$ be the second intersection of $\odot(KEF)$ and $BC$. Then $\angle BXF = \angle KEF = \angle BAC$ or $\triangle BXF\sim\triangle BCA$. Similarly, $\triangle CXE\sim\triangle CBA$. Hence if $(E_1,F_1)$ and $(E_2,F_2)$ are interesting, then $E_1X_1\parallel E_2X_2$ and $F_1X_1\parallel F_2X_2$ so $$E_1E_2 : X_1X_2 : F_1F_2 = \frac{1}{AC} : \frac{1}{BC} : \frac{1}{AB}$$hence the result follows.

Solution with Jeffrey Kwan based heavily on the HM point: Let's give a self-contained proof of the main lemma. Claim: Lines $KE$ and $KF$ are tangents to $(AEF)$, hence $\angle KEF = \angle KFE = \angle A$. Proof. Notice that $AM$ is the $A$-median of $\triangle AST$ and moreover lies on the reflection of $(AST)$ over $\overline{ST}$. As $M$ lies inside $\triangle AST$ it follows (from known lemmas) that $M$ is the $A$-HM point of $\triangle AST$. Then $ATKS$ is harmonic, as $K$ is the reflection of $M$ across $\overline{ST}$. So $\overline{AK}$ is a symmedian of $\triangle AST$ and hence of $\triangle AFE$, proving the claim. $\blacksquare$ So $\triangle KEF$ is similar to a triangle of fixed shape $\Delta$, namely the triangle with angles $(A, A, 180^{\circ}-2A)$, as $E$ and $F$ vary over interesting pairs. Claim: Let $ABC$ be a triangle and let $\triangle D_1E_1F_1 \sim \triangle D_2E_2F_2$ both be inscribed in $ABC$. Then the Miquel points of $D_1 E_1 F_1$ and $D_2 E_2 F_2$ coincide at the center of spiral similarity sending $D_1E_1F_1$ to $D_2E_2F_2$. Proof. Let $S$ be the center of spiral similarity taking $D_1 E_1 F_1$ to $D_2 E_2 F_2$, say. Then $S$ is the Miquel point of $E_1 F_1 F_2 E_2$, so it lies on $(AE_1F_1)$ and $(AE_2F_2)$. Similarly it lies on all four of $(BF_1D_1)$, $(BF_2D_2)$, $(CD_1E_1)$, $(CD_2E_2)$. So $S$ is the desired point. $\blacksquare$ Now let $Y$ and $Z$ are the feet of the $B$ and $C$ altitudes and $W$ is the midpoint of $BC$. Let $S$ denote the $A$-HM point of $ABC$. We use the following known lemma. Lemma: In $\triangle WYZ$ we also have $\angle Y = \angle Z = A$, $\angle W = 180^{\circ} - 2A$. Moreover the Miquel point of $\triangle WYZ$ is the $A$-HM point of $ABC$. Now if $\triangle K_1 E_1 F_1$ and $\triangle K_2 E_2 F_2$ are given as in the problem, then $S$ is their spiral center and \[ \frac{E_1E_2}{F_1F_2} = \frac{\operatorname{dist}(S,\overline{AC})}{\operatorname{dist}(S,\overline{AB})} = \frac{AB}{AC} \]with the last equality since $S$ lies on the $A$-median.

whoa the second claim does not need to be that complicated???

Solved with naman12. 2015 Taiwan TST Round 3 Mock IMO Day 2 Problem 1 wrote: Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$. Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$, respectively. If the quadrilateral $KSAT$ is cycle, prove that $\angle{KEF}=\angle{KFE}=\angle{A}$. From now on we will only be using the fact that $K$ is the polar of $EF$ wrt $(AEF)$. Define $M = (AEF) \cap (BKF) \cap (CKE)$ and let $A_0$ be the point on $(AEF)$ such that $A_0,K,M$ are collinear. Angle chase gives $A_0A||BC$. Projecting onto line $BC$ from $A$ gives $AM$ bisecting $BC$. Moreover, in directed angles, $\angle MCT = \angle MEK = \angle TAC$ so $M$ is independent of $EF$. Note that for $K = M$ we can take $E,F$ to be the feet of altitudes from $B,C$, while for $K = B$ we can take $F = A$ and $E$ the point on $AC$ other than $A$ with $BA = BE$, and there is a similar choice of $(E,F)$ for $K = C$. For any other good pair $(E_1,F_1)$ the problem statement now follows from spiral similarity. $\blacksquare$

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Lemma. If $(E,F)$ is interesting then $A,E,F,D$ are concyclic. Proof. Notice that $K$ is the point on the circumcircle such that its reflection in $TS$ lies on the $A$-median, therefore, $K$ is the second intersection of $(ATS)$ and the $A-$ symmedian in $\triangle ATS$. Suppose the tangent to $(ATS)$ at $A$ intersect $BC$ at $K'$, then $$(K',K;B,C)\overset{A}{=}(A,K;T,S)=-1$$Moreover, $M$ is the $A-$humpty point of $\triangle ATS$ Now suppose $AD$ intersect $(ATS)$ again at $G$. We first show the following. CLAIM. $\frac{TF}{TA}=\frac{GD}{GA}$ Proof. By spiral sim. lemma we have that $K$ is the center of spiral similarity sending $\overline{DG}$ to $\overline{AK'}$, hence $$\frac{DG}{AG}=\frac{AK'}{DK'}\cdot\frac{KD}{KA}$$On the other hand, \begin{align*} \frac{TF}{TA}&=\frac{IM}{IA}\\ &=\frac{IT^2}{IA^2}\\ &=\frac{KS^2}{AS^2}\\ &=\left(\frac{AK'}{AK}\cdot\frac{KC}{K'C}\right)^2 \end{align*}For each point $X$ define $f(X)=\pm\frac{KX}{K'X}$, where $f$ is negative if and only if $X$ lies inside segment $KK'$ or it lies below $KK'$, then it suffices to show $$f(A)f(D)=f(C)^2$$by ratio lemma, $f(A)f(D)=f(N)$ where $N$ is the midpoint of $BC$. Moreover, $$K'B\times K'C=K'K\times K'N$$$$KB\times KC=KN\times KK'$$from the harmonic conditions, hence $$f(N)=f(B)f(C)=f(C)^2$$as desired. $\blacksquare$ Now, a homothety at $A$ which sends $\triangle ATS$ to $\triangle AFE$ will send $G$ to $I$, this completes the proof of the lemma. $\blacksquare$ [asy][asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.760433772373943, xmax = 19.375392351983418, ymin = -27.005110383071003, ymax = 11.432112570887039; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); draw((-2.5744297462234242,-5.259500087187417)--(-7.088141678673,-5.076104623147327)--(-6.12886969510659,-2.768890110948807)--cycle, linewidth(0.8) + zzttqq); draw((-2.5744297462234242,-5.259500087187417)--(0.12615653932945348,-3.642990092573564)--(-1.1176249307501425,-2.325277556090636)--cycle, linewidth(0.8) + zzttqq); /* draw figures */draw(circle((-2.8060639557129097,-1.8084335226740398), 3.4588314268995517), linewidth(0.8) + blue); draw(circle((-4.717476877953287,-2.366750011988238), 3.600090833073584), linewidth(0.8) + blue); draw((-6.12886969510659,-2.768890110948807)--(0.12615653932945348,-3.642990092573564), linewidth(0.8)); draw((-7.088141678673,-5.076104623147327)--(-1.1176249307501425,-2.325277556090636), linewidth(0.8)); draw((-4.468445123604657,1.2247172636118744)--(-9.093671122213715,-9.899748715801291), linewidth(0.8)); draw((-9.093671122213715,-9.899748715801291)--(6.935129321241813,-10.856692025858337), linewidth(0.8)); draw((6.935129321241813,-10.856692025858337)--(-4.468445123604657,1.2247172636118744), linewidth(0.8)); draw((-2.5744297462234242,-5.259500087187417)--(-7.088141678673,-5.076104623147327), linewidth(0.8) + zzttqq); draw((-7.088141678673,-5.076104623147327)--(-6.12886969510659,-2.768890110948807), linewidth(0.8) + zzttqq); draw((-6.12886969510659,-2.768890110948807)--(-2.5744297462234242,-5.259500087187417), linewidth(0.8) + zzttqq); draw((-2.5744297462234242,-5.259500087187417)--(0.12615653932945348,-3.642990092573564), linewidth(0.8) + zzttqq); draw((0.12615653932945348,-3.642990092573564)--(-1.1176249307501425,-2.325277556090636), linewidth(0.8) + zzttqq); draw((-1.1176249307501425,-2.325277556090636)--(-2.5744297462234242,-5.259500087187417), linewidth(0.8) + zzttqq); /* dots and labels */dot((-4.468445123604657,1.2247172636118744),dotstyle); label("$A$", (-4.8671715027950935,1.6601322286974445), NE * labelscalefactor); dot((-9.093671122213715,-9.899748715801291),dotstyle); label("$B$", (-9.25316167388989,-11.255418405048774), NE * labelscalefactor); dot((6.935129321241813,-10.856692025858337),dotstyle); label("$C$", (7.232582079622686,-10.979493991639378), NE * labelscalefactor); dot((-2.5744297462234242,-5.259500087187417),linewidth(4pt) + dotstyle); label("$D$", (-2.315322675976303,-6.0700650443587595), NE * labelscalefactor); dot((-6.12886969510659,-2.768890110948807),dotstyle); label("$F_{1}$", (-7.658256157128147,-2.5233107007782247), NE * labelscalefactor); dot((-7.088141678673,-5.076104623147327),dotstyle); label("$F_{2}$", (-8.57532682926615,-5.23465007927319), NE * labelscalefactor); dot((0.12615653932945348,-3.642990092573564),linewidth(4pt) + dotstyle); label("$E_1$", (1.1137241850614472,-3.79923511418762), NE * labelscalefactor); dot((-1.1176249307501425,-2.325277556090636),linewidth(4pt) + dotstyle); label("$E_2$", (-0.88118133170029703,-1.9252211319925705), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Now we return to the original problem, then by spiral similarity lemma, $D$ is the center of spiral similarlity sending $\overline{F_1F_2}$ to $\overline{E_1E_2}$, hence $$\frac{F_1F_2}{E_1E_2}=\frac{DF_1}{DE_1}=\frac{\sin\angle BAD}{\sin\angle CAD}=\frac{AC}{AB}$$as desired.

In the solution below, $d(\text{point},\text{line})$ is the distance from the point to the line, and $d(\text{line},\text{line})$ is the distance between two parallel lines. First, we characterize interesting pairs as follows: Claim. If $(E,F)$ is interesting, then $KE,KF$ are tangents to $(AEF)$. Proof. This is actually not so easy. We first prove its converse— Lemma wrote: Let two circles be internally tangent to each other at point $P$. Let $X$ be a point on the outer circle, and suppose $A,B$ are on the inner circle such that $XA,XB$ are tangent to the inner circle. Let $PA,PB$ cut the outer circle at $C,D$. Then $CD$ bisects $XA,XB$. To prove the lemma, we extend $XA,XB$ to intersect the outer circle again at $Y,Z$. Let $M$ be the midpoint of $AB$. Then we can view the inner circle as the $X$-mixtilinear incircle of $\triangle XYZ$, so it follows that $M$ is the incenter of $\triangle XYZ$. By shooting lemma, $C,D$ are the midpoints of arcs $XY,XZ$, so $C,M,Z$ are collinear, which implies $\angle MCD = \angle XCD$. Similarly $\angle MDC = \angle XDC$, so $\triangle MCD \cong \triangle XCD$, which gives the desired conclusion (notice that $MX\perp AB$). Now let us return to the claim. Observe that $EF\parallel ST$ implies that $(AEF)$ is tangent to $(ASKT)$ at $A$, so if we draw tangents from $K$ to $(AEF)$ (say the tangent points are $E',F'$) and let $AE',AF'$ cut $(ASKT)$ at $S',T'$, then the Lemma says that $d(K,S'T') = 2d(E'F',S'T')$. Also by our conditions, $d(K,ST) = 2d(EF,ST)$, so $d(EF,E'F') = 2d(ST,S'T')$. But by homothety $d(EF,E'F') \le d(ST,S'T')$, so we must have $EF=E'F'$. This proves the claim. $\square$ We are now halfway through. Notice that the Claim implies all triangles $EFK$ are similar: they are isosceles with base angle equal to $\angle BAC$. Thus, given any $K$ on $BC$, we can uniquely construct $\triangle KEF$ by rotating line $BA$ clockwise $\theta:=(\pi-2\angle BAC)$ units and intersect it with line $BC$. Now, suppose we constructed interesting pairs $\triangle KEF$ and $\triangle K'E'F'$. Rotate line $AB$ clockwise $\theta$ units around $K$, and let the image intersect $AB$ at $X$. Similarly, define $Y$ with respect to $K'$. Let circle $(E'F'YK)$ intersect $BC$ again at $P$ (the circle exists by definition). Finally, suppose we rotate $F'$ around $K$ clockwise for $\theta$ units, and call the image $L$. Picture is attached. Claim. Circle $(F'XK)$ also passes through $P$, and $L,E',P$ are collinear. Proof. This is a simple angle chase: $\angle F'PK = \angle F'E'K' = \angle BAC = \angle FEK = \angle BXK$, so $F',X,K,P$ are concyclic. Then by the converse of Reim's theorem ($\overline{F'XY}$ and $\overline{PE'L}$ cutting the circles $YF'E'K'P$ and $XF'LPK$), points $P,E',L$ are collinear. $\square$ This implies that $\angle ELE' = \angle XLP = \angle XF'P$. But since $\angle BPF' = \angle K'E'F' = \angle BAC$, we have $A,F',P,C$ concyclic, so $\angle XF'P = \angle ACB$. Combined with $\angle LEE' = \angle BAC$, this implies that \[\triangle ELE' \sim \triangle ACB\]so \[\frac{EE'}{AB} = \frac{EL}{AC} = \frac{FF'}{AC}.\]

Define $M_1,K_1,S_1,T_1$ and $M_2,K_2,S_2,T_2$ respectively. We will first show the following claim which has been used in other solutions as well. Claim 1: Suppose that $(E,F)$ is interesting. Then $KE$ and $KF$ are tangent to $(AFE)$. Proof. We will consider two cases. (a) If $\triangle AEF$ is not isosceles, let $N$ be the midpoint of $ST$. Then since $EF\parallel ST$, it follows that $A,M,N$ are collinear. Therefore, $AM$ and $KM$ are symmetric over $ST$, and since $ASKT$ is a convex quadrilateral, it follows that $AK$ is the $A$-symmedian of $\triangle AST$, and hence also the $A$-symmedian of $\triangle AEF$. If $\triangle AEF$ is not isosceles, as $K$ lies on the perpendicular bisector of $EF$, it follows that $KE$ and $KF$ are tangent to $(AEF)$ as desired. (b) Now if $\triangle AEF$ is isoceles, it follows that $ASKT$ is a kite. Therefore, $MSKT$ is a rhombus, and angle chasing gives us \[\measuredangle EFK=\measuredangle MFK=\measuredangle MSK=\measuredangle TKS=\measuredangle EAF. \blacksquare\]This lets us completely remove $S_1,S_2$ and $T_1,T_2$ from the diagram. Claim 2: We have $\frac{AF_1}{F_1F_2}=\frac{BK_1}{K_1K_2}$. Proof. Consider a homothety $\mathcal{H}$ centered at $A$ which sends $F_2$ to $F_1$, and let the image of point $X$ be $X'$. Notice that $\triangle F_1E_1K$ and $\triangle F_1E_2'K_2'$ are directly similar. Therefore, by spiral similarity, we have $\triangle F_1K_1K_2'\sim\triangle F_1E_1E_2'$ and hence, \[\measuredangle (F_1K_1,K_1K_2')=\measuredangle F_1E_1E_2'=\measuredangle (K_1F_1,F_1A)\]and so $AB\parallel K_1K_2'$. Finally, \[\frac{AF_1}{F_1F_2}=\frac{AK_2'}{K_2K_2'}=\frac{BK_1}{K_1K_2}\]as desired. $\blacksquare$ Now let $d(X,l)$ denote the distance from point $X$ to line $l$. By claim 2, we see that the desired ratio is \[\frac{F_1F_2}{E_1E_2}=\frac{AF_1\cdot K_1K_2}{BK_1}\div\frac{AE_1\cdot K_1K_2}{CK_1}=\frac{AF_1}{AE_1}\cdot \frac{CK_1}{BK_1}.\]Since $K_1$ lies on the $A$-symmedian, we have \[\frac{AF_1}{AE_1}\cdot \frac{CK_1}{BK_1} = \frac{d(K_1,AB)}{d(K_1,AC)} \cdot \frac{CK_1}{BK_1}=\frac{[ABK_1]}{[ACK_1]}\cdot \frac{AC}{AB}\cdot \frac{CK_1}{BK_1}=\frac{AC}{AB}\]and we're done.

Claim. For every interesting pair $(E,F)$ line $AK$ is a symmedian of $AEF.$ Proof. Construct parallelogram $MSNT.$ Obviously $KN\parallel ST,N\in \odot (ASKT)$ and since $ST\parallel EF$ we deduce $A\in MN,$ which completes $\Box$ Now let $BE_1\cap CF_1=X_1,BE_2\cap CF_2=X_2.$ Converse of Pascal on $AE_1E_1X_1F_1F_1,AE_2E_2X_2F_2F_2$ gives $$X_1\in \odot (AE_1F_1),X_2\in \odot (AE_2F_2)\implies \measuredangle BE_1E_2=\measuredangle CF_1F_2,\measuredangle BE_2E_1=\measuredangle CF_2F_1\implies BE_1E_2\stackrel{+}{\sim} CF_1F_2\implies$$$$\implies \frac{|E_1E_2|}{|F_1F_2|}=\frac{d(B,AC)}{d(C,AB)}=\frac{|AB|}{|AC|}\text{ } \blacksquare$$

Tfw you coordinate bash on an accidentally degenerate case and conclude the obvious spiral sim solution fails (it works) Anyways, notice $AK$ is the $A$-symmedian of $\triangle AST$ and $M$ is the respective $A$-HM point. Now let $a, b, c$ be $ST, AS, AT$ respectively. Let $R$ be the midpoint of $ST$. Then $AK \cdot AR = AB \cdot AC$ by sqrtbc so $AM = \frac{2bc}{\sqrt{2b^2+2c^2-a^2}}$ and $AK = AR - RK$ so by radius SR inversion at R, $AK = \frac{\sqrt{2b^2+2c^2-a^2}}{2} - \frac{a^2}{2\sqrt{2b^2+2c^2-a^2}} = \frac{b^2+c^2-a^2}{\sqrt{2b^2+2c^2-a^2}}$ so $\frac{AK}{AM} = \cos A$ which is fixed so it follows that $M$ varies on a line antiparallel to $BC$ as $K$ varies along $BC$ and by affine transforming $AB, AC$ to the x and y axis the desired result follows.

wat oops., Notice that $AK$ and $AM$ are symmedian and median of $\triangle AST$. Now define points $E'\in AB$ and $F'\in AC$ such that $E'F'$ has midpoint $K$; since $E'$ and $F'$ have equal distances to $BC$ they move linearly together. It's clear that $F'F$ is the $F'$-altitude of $\triangle AE'F'$ and similarly for $E'E$, meaning that $E$ and $F$ move linearly as well. Now we just check two cases ($K=B$ and $K=C$ work perfectly).

Here is a solution that uses linearity. Claim: If $KSAT$ is cyclic, then $KE$ and $KF$ are tangents to $(AEF)$. Proof. Since $\overline{EF} \parallel \overline{ST}$, $\overline{AM}$ is a median of $\triangle AST$. Since $M$ is the reflection of $K$ over $ST$, $\overline{AK}$ is the $A$-symmedian of $\triangle AST$, so it is also the $A$-symmedian of $\triangle AEF$. Since $K$ lies on both the $A$-symmedian and perpendicular bisector of $EF$ in $\triangle AEF$, $KE$ and $KF$ are tangent to $(AEF)$, as claimed. Claim: $(AEB)$ and $(AFC)$ intersect at $A$ and some point on $\overline{BC}$. Proof. By a quick angle chase, it suffices to show that the intersection of $\overline{BE}$ and $\overline{CF}$ lies on $(AEF)$. The proof of this is as follows. Let $X'$ denote $\overline{BE} \cap (AEF)$. Then by Pascal's theorem on $AEEX'FF$, we find that $\overline{AE} \cap \overline{X'F}$ lies on $\overline{KB}$. Since $\overline{AE} \cap \overline{X'F}$ lies on both $\overline{KB}$ and $\overline{AE}$, it must be $C$. Thus, $C$, $X'$, and $F$ are collinear, as desired. Now we quote the following lemma: Lemma: For any $\bullet$, we have \[ \text{Pow}(\bullet, (AEF))-\text{Pow}(\bullet, (AEB))=\text{Pow}(\bullet, (AFC))-\text{Pow}(\bullet, (ABC)). \]Proof. Let $f(\bullet)$ denote LHS and $g(\bullet)$ denote the RHS. Note that $f$ and $g$ are linear functions. It suffices to show that $f(\bullet)=g(\bullet)$ holds for three distinct non-collinear positions of $\bullet$, since every point in the plane can be described a linear combination of those points. This task is easy: choose $\bullet = A, E, F$. It is easy to verify that each of these points all work, by the means of radical axis. Consider any two interesting pairs $(E_1, F_1)$ and $(E_2, F_2)$. Let $\theta(\bullet) = \text{Pow}(\bullet, (AE_1F_1)) - \text{Pow}(\bullet, (AE_2F_2))$ for each $\bullet$ in $\mathbb{R}^2$. Realize that \[ \frac{E_1E_2}{AB} = \frac{F_1F_2}{AC} \]holds if and only if $\theta(B)+\theta(C)=0$. Let $X_1$ and $X_2$ denote the intersection points in the second claim for $(AE_1F_1)$ and $(AE_2F_2)$, respectively. By the lemma, we have that \[ \theta(B) + \theta(C) = (BX_1 \cdot BC - BX_2 \cdot BC) + (CX_1 \cdot BC - CX_2 \cdot BC) = BC^2-BC^2 = 0, \]and we are done.

cute! solved with Om245 and rjp08

First note that $AM$ is the median of $\triangle AST$ since $ST \parallel EF$. Then since $ST$ is perpendicular to $MK$ and $K$ lies on the circumcircle of $\triangle AST$ by EGMO $4.26 (g)$, it follows that $AK$ is a symmedian wrt $\triangle AST$. By homothety we also have $AK$ being a symmedian to $\triangle AEF$ and since $KE = KF$ we have $KE$ and $KF$ being tangents to $(AEF)$. Now let $G_1 = BE_1 \cap (AE_1F_1) \neq E_1$ and define $G_2$ similarly. By Pascal's on $AE_1E_1G_1F_1F_1$ we get that $G_1 \in CF_1$ as well. So then $G_2$ is the intersection of $BE_2$ and $CF_2$. So then $\measuredangle BE_1E_2 = \measuredangle BE_1A = \measuredangle G_1F_1A = CF_1F_2$. And similarly $\angle BE_2E_1 = \angle CF_2F_1$ so $\triangle BE_1E_2 \sim \triangle CF_1F_2$. Also note that by LoS we have $\frac{AB}{BE_1} = \frac{\sin AE_1B}{\sin A} = \frac{\sin AF_1C}{\sin A} = \frac{AC}{CF_1}$. Combining the similarity we get that $\frac{E_1E_2}{E_1B} = \frac{F_1F_2}{CF_1}$ which implies the desired.