My solution : Since $ \measuredangle FEI=\measuredangle ICB=\measuredangle FIK $ , so $ KI $ is tangent to $ \odot (EFI) $ at $ I \Longrightarrow KI^2 =KE \cdot KF $ , hence $ K $ lie on the radical axis $ \tau $ of $ \odot (ABC) $ and degenerate circle $ \odot I $ . Similarly, we can prove $ L \in \tau $ and $ M \in \tau \Longrightarrow K, L, M $ are collinear at $ \tau $ . Q.E.D ____________________________________________________________ Remark : From this proof we get $ \overline{KLM} \perp OI $ where $ O $ is the center of $ \odot (ABC) $

Dear Mathlinkers, 1. (XYZ) the perspectrix of ABC and DEF 2. by Desargues theorem (two times), we can prove that K, L, M are collinear and (KLM) parallel to (XYZ). Sincerely Jean-Louis

really easy proof: we know by Desargues, BC\capEF=K' , CA\cap FD=L', AB\cap DE=M' concur, then from DM/ME=(DM'/M'E)(DI/IA)(BE/IE) (due to parrallel), multiply analogously, inverse menelaus done.

Generalization : Let $ P $ be a point and $ \triangle DEF $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC $ . Let $ K \in EF $ be a point such that $ PK \parallel BC $ and define $ L \in FD, M\in DE $ similarly . Then $ K, L, M $ are collinear . _____________________________________________________________ The proof of the generalization is as same as my proof for the original problem in my previous post

@Telv, i believe my method also works for your generalization right?

can someone please add a diagram i am having a hard time to make one diagram

Nice problem. Desargue's theorem killed this one.

It's some old problem !

I have a trig bash to contribute (well sine rule so that's kind of ok) Some angle chasing shows that $\angle KIE = \angle DFE$ and similarly elsewhere (should use directed angles). Sine rule on triangle $KIE$ gives $\frac{EK}{\sin{\angle DFE}}=\frac{EK}{\sin{\angle KIE}}=\frac{IE}{\sin{\angle EKI}}$ Multiplying this for $KIE, LIF, MID$ and dividing for $KIF, LID, MIE$ gives. $\frac{EK\times FL \times DM \times \sin{\angle FED} \times \sin{\angle DFE} \times \sin{\angle EDF}}{FK \times DL \times ME \times \sin{ \angle DFE} \times \sin{\angle EDF} \times \sin{\angle FED}}=\frac{IE \times IF \times ID \times \sin{\angle EKI} \times \sin{\angle FLI} \times \sin{\angle DMI}}{IE \times IF \times ID \times \sin{\angle FKI} \times \sin{\angle DLI} \times \sin{\angle EMI}}=1$ Stuff cancels and by Menelaos we are done.

Ridiculously easy. Assume $K$ is on $E$'s side of $EF$.By trivial angle-chasing $\triangle{IEK} \sim \triangle{FIK}$ from which $\frac{KE}{KF}=(\frac{IE}{IF})^2$;this is symmetrical in $E,F$,so it holds no matter which side $K$ is on.Multiply the 3 obtained relations($K,L,M$) and apply Menelaus' ,Q.E.D.

I found a solution using barycentric.

Solution: Lemma: Let $P $ be a point inside $\Delta ABC$. Suppose the tangent to $\odot (BPC) $ at $P $ intersect $BC $ at $A'$. Define $B'$ and $C'$ respectively. Then $A'$, $B'$, and $C'$ are collinear. Proof of the lemma: Let $A_1$, $B_1$, and $C_1$ be on $BC $, $CA $, and $AB $ respectively such that $PA_1$, $PB_1$, and $PC_1$ are symmedians of $\Delta PBC$, $\Delta PCA $, and $\Delta PAB $ respectively. Then, we have, $\frac {BA_1}{A_1C}=\frac{BP^2}{CP^2} $ and so on. Thus, by Ceva's Theorem, $AA_1$, $BB_1$, and $CC_1$ are concurrent, say at $Q$. So, $A'$, $B'$, and $C'$ are collinear on the trilinear polar $\mathcal {T}_p$ of $Q$ w.r.t $\Delta ABC$. Main problem: Observe that $I$ is the orthocenter of $\Delta DEF$. Then, simply apply the above lemma taking $P\equiv I $ and $\Delta DEF$ as the reference triangle and the result follows immediately.

$DE$ meets $BC,AC$ at $X,Y$. $DF$ meets $BC,AB$ at $R,Q$. $EF$ meets $AC,AB$ at $Z,P$. Simple angle chasing gives $Q,I,Y$ collinear and $QY \parallel BC$, similar for the others. It is well known that the hexagon $XYZPQR$ lies on a conic. Thus Pascal's Theorem on $Q,Y,X,P,Z,R$ gives the result$\blacksquare$

Tumon2001 wrote: Thus, by Ceva's Theorem, $AA_1$, $BB_1$, and $CC_1$ are concurrent. So, $A'$, $B'$, and $C'$ are collinear on the trilinear polar $\mathcal {T}_p$ of $P$ w.r.t $\Delta ABC$. This is wrong because $A',B',C'$ lie on the trilinear polar of $AA_1\cap BB_1\cap CC_1$ and not on the trilinear polar of $P$.Your argument is true when $P$ is one of the foci of steiner inellipse.

Wictro wrote: $DE$ meets $BC,AC$ at $X,Y$. $DF$ meets $BC,AB$ at $R,Q$. $EF$ meets $AC,AB$ at $Z,P$. Simple angle chasing gives $Q,I,Y$ collinear and $QY \parallel BC$, similar for the others. It is well known that the hexagon $XYZPQR$ lies on a conic. Thus Pascal's Theorem on $Q,Y,X,P,Z,R$ gives the result$\blacksquare$ Can somebody give me a resource on where to learn conics? I basically know nothing about them... any books, handouts (preferably olympiad oriented) would be appreciated

da-rong_wae wrote: Wictro wrote: $DE$ meets $BC,AC$ at $X,Y$. $DF$ meets $BC,AB$ at $R,Q$. $EF$ meets $AC,AB$ at $Z,P$. Simple angle chasing gives $Q,I,Y$ collinear and $QY \parallel BC$, similar for the others. It is well known that the hexagon $XYZPQR$ lies on a conic. Thus Pascal's Theorem on $Q,Y,X,P,Z,R$ gives the result$\blacksquare$ Can somebody give me a resource on where to learn conics? I basically know nothing about them... any books, handouts (preferably olympiad oriented) would be appreciated Bump bump ... i also have the very same question

Fumiko wrote: Tumon2001 wrote: Thus, by Ceva's Theorem, $AA_1$, $BB_1$, and $CC_1$ are concurrent. So, $A'$, $B'$, and $C'$ are collinear on the trilinear polar $\mathcal {T}_p$ of $P$ w.r.t $\Delta ABC$. This is wrong because $A',B',C'$ lie on the trilinear polar of $AA_1\cap BB_1\cap CC_1$ and not on the trilinear polar of $P$.Your argument is true when $P$ is one of the foci of steiner inellipse. Typo! Thanks for pointing it out. I have fixed it. @2above.....try A.V. Akopyan's "Geometry of Conics".

My solution: The problem is equivalent to the new problem below: Problem: Given $\Delta ABC$, altitudes $AD,BE,CF$ concur at $H$. Let $M_A,M_B,M_C$ be the midpoint of $AH,BH,CH$. The parallel line from $H$ to $EF,FD,DE$ intersect $M_BM_C,M_CM_A,M_AM_B$ at $X,Y,Z$. Prove $\overline{X,Y,Z}$ Lemma: see here Prove: Let the line pass $H$ and parallel to $EF$ intersect $BC$ at $X'$. We define similarly with $Y',Z'$. Since there is a homothety at $H$ with factor $k=2$, send $X \mapsto X'$, so it suffices to prove $\overline{X',Y',Z'}$. Let $M,N,P$ be the midpoint of $EF,FD,DE$ and $M'=HM \cap BC,N'=HN \cap AC,P'=HP \cap AB$. By using the lemma, we have $AM',BN',CP'$ concur. (1) Note that $(X'M',BC)=-1 \Rightarrow \frac{X'B}{X'C}=-\frac{M'B}{M'C}$ and then we do similarly to point $Y',Z'$ (2). Now, just use Ceva for (1) and using (2), and Menelaus for $\Delta ABC$ with $X',Y',Z'$ we have $\overline{X',Y',Z'}$ (Q.E.D) P/S: Can't believe this is a geometry topic but there is no figure

Can we solve it with using pole and polar?

Probably a different solution. MathKnight16 wrote: Let $\triangle{ABC}$ be a scalene triangle with incentre $I$ and circumcircle $\omega$. Lines $AI, BI, CI$ intersect $\omega$ for the second time at points $D, E, F$, respectively. The parallel lines from $I$ to the sides $BC, AC, AB$ intersect $EF, DF, DE$ at points $K, L, M$, respectively. Prove that the points $K, L, M$ are collinear. (Cyprus) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.15, xmax = 3.0233333333333356, ymin = -1.3033333333333228, ymax = 7.563333333333344; /* image dimensions */ pen ttffqq = rgb(0.2,1,0); pen ffdxqq = rgb(1,0.8431372549019608,0); pen qqttcc = rgb(0,0.2,0.8); /* draw figures */ draw(circle((-2.280860402066602,2.2233127014169525), 2.5951607419999174), linewidth(0.8) + red); draw((-3.47,4.53)--(-4.71,1.31), linewidth(1.2) + ttffqq); draw((-3.47,4.53)--(0.17,1.37), linewidth(1.2) + ffdxqq); draw((0.17,1.37)--(-4.71,1.31), linewidth(1.2) + qqttcc); draw((-3.47,4.53)--(-2.3522245885399875,-0.37086663504121775), linewidth(1.2)); draw((-4.71,1.31)--(-0.6123592093501464,4.211015658782898), linewidth(1.2)); draw((0.17,1.37)--(-4.716969010258008,3.1178707632068448), linewidth(1.2)); draw((-0.6123592093501464,4.211015658782898)--(-2.3522245885399875,-0.37086663504121775), linewidth(1.2)); draw((-4.716969010258008,3.1178707632068448)--(-0.6123592093501464,4.211015658782898), linewidth(1.2)); draw((-4.716969010258008,3.1178707632068448)--(-2.3522245885399875,-0.37086663504121775), linewidth(1.2)); draw((-1.2297409472574308,2.5851597234432644)--(-9.896666666666667,2.2433333333333443), linewidth(1.2) + qqttcc); draw((-4.716969010258008,3.1178707632068448)--(-9.896666666666667,2.2433333333333443), linewidth(1.2)); draw((-1.6999003063954472,1.3470094224623508)--(-7.363333333333333,6.47), linewidth(1.2) + ffdxqq); draw((-7.363333333333333,6.47)--(-4.716969010258008,3.1178707632068448), linewidth(1.2)); draw((-3.501624138803824,1.3248570802606083)--(xmax, 2.4181783593857986*xmax + 9.79240879541895), linewidth(1.2) + ttffqq); /* ray */ draw((-2.3522245885399875,-0.3708666350412171)--(xmax, 2.633469433110764*xmax + 5.823644918690384), linewidth(1.2)); /* ray */ draw((-9.896666666666667,2.243333333333341)--(xmax, 1.6684210526315748*xmax + 18.75514035087716), linewidth(0.8)); /* ray */ /* dots and labels */ dot((-3.47,4.53),dotstyle); label("$A$", (-3.4166666666666656,4.656666666666678), NE * labelscalefactor); dot((-4.71,1.31),dotstyle); label("$B$", (-4.976666666666666,1.23), NE * labelscalefactor); dot((0.17,1.37),dotstyle); label("$C$", (0.3033333333333349,1.3233333333333441), NE * labelscalefactor); dot((-3.01,2.51),dotstyle); label("$I$", (-2.87,2.1766666666666774), NE * labelscalefactor); dot((-2.3522245885399875,-0.37086663504121775),dotstyle); label("$D$", (-2.39,-0.6233333333333227), NE * labelscalefactor); dot((-0.6123592093501464,4.211015658782898),dotstyle); label("$E$", (-0.4566666666666652,4.19), NE * labelscalefactor); dot((-4.716969010258008,3.1178707632068448),dotstyle); label("$F$", (-5.056666666666666,3.1766666666666774), NE * labelscalefactor); dot((-3.933444189459237,3.326540088662306),dotstyle); label("$Y$", (-4.003333333333333,3.523333333333344), NE * labelscalefactor); dot((-2.5179687263082795,3.7035113118500456),dotstyle); label("$X$", (-2.63,3.8966666666666776), NE * labelscalefactor); dot((-4.268558214520283,2.456324636487652),dotstyle); label("$Z$", (-4.563333333333333,2.2033333333333442), NE * labelscalefactor); dot((-3.501624138803824,1.3248570802606086),dotstyle); label("$P$", (-3.6033333333333326,1.03), NE * labelscalefactor); dot((-1.6999003063954472,1.3470094224623508),dotstyle); label("$Q$", (-1.63,1.07), NE * labelscalefactor); dot((-1.2297409472574308,2.5851597234432644),dotstyle); label("$R$", (-1.11,2.59), NE * labelscalefactor); dot((-9.896666666666667,2.2433333333333443),dotstyle); label("$K$", (-9.99,2.3766666666666776), NE * labelscalefactor); dot((-7.363333333333333,6.47),dotstyle); label("$L$", (-7.6033333333333335,6.523333333333344), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Define $\{AB\cap FD=Z\},\{AB\cap FE=Y\},\{AC\cap EF=X\},\{AC\cap ED=R\},\{BC\cap DF=P\},\{BC\cap DE=Q\}$. Now define the lines parallel to $\{AB,BC,CA\}$ be $\{\ell_1,\ell_2,\ell_3\}$ respectively. It's well known that $I$ is the orthocenter of $\triangle DEF$. Now by Pascal's Theorem on $ABFDEC$, we get that $\{\overline{Z-I-R}\},\{\overline{Y-I-Q}\},\{\overline{X-I-P}\}$.So, $\angle AZR=\frac{B}{2}+\frac{B}{2}=\angle B\implies ZR\|BC$. Similarly we get that $YQ\|AC$ and $XP\|AB$. So, $\ell_1=XP,\ell_2=ZR,\ell_3=YQ$. Now consider the Inversion $\Psi$ centered at $D$ with radius $DB$. $\Psi:P\leftrightarrow F$ and $\Psi:Q\leftrightarrow E$. So, $DP\cdot DF=DQ\cdot DE\implies FPQE$ is a cyclic quadrilateral and as $AR\|BC$. By Reim's Theorem we get that $FZRE$ is a cyclic quadrilateral. Similarly we get that $FYQD,XPDE$ are cyclic quadrilaterals. Now we can restate the problem in terms of $\triangle DEF$. Restated Problem wrote: $DEF$ is a triangle with orthocenter $I$. A line passing through $I$ is antiparallel to $FD$ and intersects $EF,ED$ at $Y,Q$ respectively. A line antiparallel to $ED$ passing through $I$ intersects $EF,FD$ at $X,P$ respectively and another line passing through $I$ antiparallel to $EF$ intersects $FD,ED$ at $Z,R$ respectively. Then $\{ZR\cap EF\},\{DF\cap YQ\},\{XP\cap ED\}$ are collinear. Introduce that Orthic triangle $A_1B_1C_1$ of $\triangle DEF$. As $\{A_1B_1C_1,DEF\}$ are pairs of perspective triangles. Hence, by Desargues Theorem we get that $\{B_1C_1\cap EF\},\{A_1B_1\cap ED\},\{A_1C_1\cap DF\}$ are collinear. Define $\{K,L,M\}$ as defined in the problem. Now notice that $KL\|\text{ Orthic Axis of }\triangle ABC$ also $LM\|\text{ Orthic Axis of }\triangle ABC$. So, $\overline{K-L-M}$. $\blacksquare$

We will prove that the points $K,L,M$ lie on the radical axis of $\omega$ and the degenerate circle centered at $I$. Note that $\angle FIK = \angle FCB = \angle FEB = \angle KEI$, so $\triangle IFK \sim \triangle EIK$, and so $KF \cdot KE = KI^2$, as desired. Similar relations hold for $L$ and $M$, and we are done.

It's a cute and easy one here is my solution. Note that $DE$ is the perpendicular bisector of $CI$ so $MC=MI$. similary, $LB=LI , KA=KI$ using the fact that $MI\parallel AB$ we can deduce that $\angle{MCI}=\angle{MIC}=\angle{B}+\frac12 \angle{C}=\angle{FBC}$ so $MC$ is tangent to $(ABC)$ by the law of sins on $\Delta EMC:$ $\frac{EM}{MC}=\frac{sin\frac{\angle{B}}{2}}{sin\frac{\angle{A}}{2}}$ so $\frac{sin^2\frac{\angle{B}}{2}}{sin^2\frac{\angle{A}}{2}}=\frac{EM^2}{MC^2}=\frac{EM^2}{EM.MD}=\frac{EM}{MD}$ similary , $ \frac{sin^2\frac{\angle{A}}{2}}{sin^2\frac{\angle{C}}{2}}=\frac{DL}{LF} , \frac{sin^2\frac{\angle{C}}{2}}{sin^2\frac{\angle{B}}{2}}=\frac{FK}{KE}$ So $\frac{EM}{MD}.\frac{DL}{LF}.\frac{FK}{KE}=-1$ and by applying Menelaus's Theorem , we're clear

Mop2018 wrote: Can we solve it with using pole and polar? We can. Let $X_A$ be the $A$-mixtilinear point. Let $P$, $Q$ be the points of intersection of the $A$-mixtilinear incircle with $AB$, $AC$ resp. It is well known that $X_A$, $P$, $F$ and $X_A$, $Q$, $E$ are collinear. Let $N$ be the nagel point of $ABC$. $KI$=$KA$ because $EF$ is perpendicular bisector of $AI$, and angle chasing gives us that $KA$ is tangent to $\omega$. Generalized bisector theorem on $X_APQ$ and angle chasing gives us that $AFX_AE$ is a harmonic quadrilateral, therefore $KX_A$ is also tangent to $\omega$, and $AX_A$ is the polar of $K$. It is well known that $AX_A$ is the isogonal conjugate of $AN$ wrt. $ABC$, therefore $K$, $L$, $M$ all lie on the polar of $N'$, the isogonal conjugate of $N$.

This is really amazing configuration - Although mine is trig bash - Main question lies in Angle chasing , so from a simple angle chase and using the fact that $I$ is the orthocenter of $\triangle DEF$ we get $$\angle FIK = \frac{C}{2} , \angle IFE = \frac{B}{2} ,\angle KEI = \frac{C}{2}$$, Now from Menelaus theorem the required result is equivalently to prove that $$ \frac{KE}{KF}. \frac{LF}{LD}. \frac{MD}{ME} = 1$$, but from sine rule in $\triangle KFI$ and $\triangle KIE$ , we get $$\dfrac{KE}{KF}=\left ( \dfrac{\sin{\frac{B}{2}}}{\sin{\frac{C}{2}}}\right )^2$$, so similarly yielding another ratio we get all term cancelled out and we remains with a $1$ which is desired. $\blacksquare$

TelvCohl wrote: My solution : Since $ \measuredangle FEI=\measuredangle ICB=\measuredangle FIK $ , so $ KI $ is tangent to $ \odot (EFI) $ at $ I \Longrightarrow KI^2 =KE \cdot KF $ , hence $ K $ lie on the radical axis $ \tau $ of $ \odot (ABC) $ and degenerate circle $ \odot I $ . Similarly, we can prove $ L \in \tau $ and $ M \in \tau \Longrightarrow K, L, M $ are collinear at $ \tau $ . Q.E.D ____________________________________________________________ Remark : From this proof we get $ \overline{KLM} \perp OI $ where $ O $ is the center of $ \odot (ABC) $ So amazing solution!!! So cool...

who know for which triangles should use dezargue

I actually just started learning about Projective Transformations ( from EGMO), so I would really appreciate it if anyone points out mistakes in this solution and clarifies my concept. SOLUTION: As the problem is purely projective, we can conveniently assume that ABC is an equilateral triangle then, the 6 points JKLMGH in the figure given are all cyclic, thus these points are always conic which means by Pascal's theorem, we can easily prove the required result. NOTE: In the figure, we have already proved that HL II BC GK II AB JM II AC THE POINTS ARE DEFINED QUITE SIMPLY THUS I AM NOT WRITING IT HERE.

We use complex numbers with $(ABC)$ set as the unit circle, the notation is as usual. It's fairly straightforward to get $$k = \frac{a^2(ab+ac+2bc)}{bc-a^2}$$and $l, m$ are defined similarly. Perform a transformation $\tau : x\to x + \sum{ab}$ in the complex plane. Now $$k = \frac{bc(a+b)(a+c)}{bc-a^2};\hspace{0.1 cm} l = \frac{ca(b+a)(b+c)}{ca-b^2};\hspace{0.1 cm} m = \frac{ab(c+a)(c+b)}{ab-c^2}$$And hence $$\frac{k-l}{l-m} = \frac{c(a-b)(a+b)^2(c^2-ab)}{a(b-c)(b+c)^2(a^2-bc)}\in \mathbb{R}$$

Notice $$\angle FIK=\angle FCB=\angle FEB$$so $\triangle FIK\sim\triangle IEK$ Hence, $IK^2=KF\cdot KE$ and $K$ lies on the radical axis of $\omega$ and the circle with radius $0$ at $I.$ Similarly, $L,M$ also lie on the radical axis. $\square$

We have KIF=BCI=BEF, so by sim triangles KIF and KEI we get KI/KF=KE/KI, so K has equal power wrt circumcircle and circle radius 0 at I, analogously L and M do, so theyre collinear on that radical axis. $\blacksquare$

let's donate $(ABC)=\omega$ $\angle KIF=\angle BCI=\angle BCF=\angle BEF=\angle IEK$ from similarity $KI^2=KF.KE$ $KI^2=pow(K,I)$ $KF.KE=pow(K,\omega)$ thus $K$ lies on radical axis line of $I,\omega$ similarly $L,M$ and we are done

storage We denote $\odot(ABC):=\Gamma$. Since $KI \parallel BC \implies \angle{KIF}=\angle{BCI}=\angle{BEF}$ this gives $\triangle{KIF} \sim \triangle{KEI} \implies KI^{2}=KE \cdot KF$ Now we treat $I$ as the degenerate circle , so $\mathrm{pow}(K,I)=KI^2$ and $\mathrm{pow}(K,\Gamma)=KE \cdot KF \implies \mathrm{pow}(K,I)=\mathrm{pow}(K,\Gamma)$ gives $K \in \text{radical axis of }$ $\Gamma$ $\text{and}$ $I$ , similarly we can get $L,M \in \text{radical axis of }$ $\Gamma$ $\text{and}$ $I$. so $\overline{K-L-M}$. $\square$