Let $F$ be the midpoint of $AB$ and let $(I)$ touch $BC$ at $P.$ If $BI$ cuts $EF$ at $L,$ we have $\widehat{FLB}=\widehat{LBC}=\widehat{FBL}$ $\Longrightarrow$ $\triangle FBL$ is F-isosceles $\Longrightarrow$ $FA=FB=FL$ $\Longrightarrow$ $\widehat{ALB}=90^{\circ}$ $\Longrightarrow$ $AQLI$ is cyclic $\Longrightarrow$ $\widehat{CQL}=\widehat{AIL}=90^{\circ}-\tfrac{1}{2}\widehat{ACB}=\widehat{CQP}$ $\Longrightarrow$ $L \in PQ,$ i.e. $L$ is on the polar of $C$ WRT $(I)$ $\Longrightarrow$ $CK \perp IL$ is the polar of $L$ WRT $(I)$ $\Longrightarrow$ $ELF \perp IK$ is the polar of $K$ WRT $(I)$ $\Longrightarrow$ $KQ$ is the polar of $E$ WRT $(I)$ $\Longrightarrow$ $KQ \perp IE,$ as desired.

let $P$ is tangency point of B-excircle to $AC$. Then $ \angle KIQ=\angle C=\angle BCP$ and \[ \frac{BC}{CP}=\frac{KI}{IQ}. \] So $ \triangle CBP \sim \triangle QKI $ and $ \angle IKQ=\angle CBP$ $ \Rightarrow $ $ KQ\perp BP $. We have that $BP\parallel IE$, now $KQ\perp IE$.

This problem can also be solved using barycentric coordinates:using the fact that $I=(a,b,c);I_a=(-a,b,c)$ we see that the circumcenter of $\triangle{BIC}$ has coordinates $\frac{I+I_a}{2}$.Then $K=B+C+I-2O$ where $B=(0,1,0),C=(0,0,1)$ and $O$ is the circumcenter of $\triangle{BIC}$.Also note that $Q=(s-c,0,s-b);E=(1,0,1)$.Thus the coordinates of points $K,Q,I,E$ are known to us,and all that is required is testing the criteria for $KQ \perp IE$. (I have wrtten all the coordinates in normalized form)

Since I solved this as I woke up, Ill make it intuitive. So, we want the polar of $K$ to contain $E$, equivalent is that its inverse $K'$ lies on the midline. If $ID$ intersects the parallel to $BC$ at $A$ to make $P$, then we must show $K'$ is the midpoint of $PD$. $P_{\infty}$ be the point at infinity on $ID \implies (P, D; E, P_{\infty}) = -1$ or $(P', D; E, I) = -1$. Obviously, $P' = ID \cap EF$, which means if the perpendiculars from $B$ to $IC$ and $C$ to $IB$ are $X, Y$, then $XY \in EF$. But every good geometer should have in his repertoire that, if $IC \cap EF = X'$, then $CX'F \sim CIB \implies CX'B \sim CFI \implies X'=X$. Similarly for $Y$, so we are done.

I have another geometrical approach to this. The circumdiameter of $\triangle{BKC}$ is $\frac{a}{cos\frac{A}{2}}$ and consequently $KI=atan\frac{A}{2}$.Thus $\frac{KI}{IQ}=\frac{atan\frac{A}{2}}{r}=\frac{tan\frac{A}{2}}{tan\frac{B}{2}}+\frac{tan\frac{A}{2}}{tan\frac{C}{2}}=\frac{s-b}{s-a}+\frac{s-c}{s-a}=\frac{a}{s-a}$. Now let $AI \cap BC=Y$ and $IE \cap BC=X$.Applying Menelaus' theorem in $\triangle{AYC}$ with transversal $XIE$ we get $\frac{CX}{YX} \cdot \frac{YI}{AI} \cdot \frac{AE}{CE}=1$ $\implies \frac{YX}{CX}=\frac{YI}{AI}=\frac{BY}{AB}=\frac{a}{b+c}$ $\implies \frac{CY}{CX}=\frac{b+c-a}{b+c}$ $\implies CX=\frac{ab}{b+c-a}$ So $\frac{CX}{CE}=\frac{2a}{b+c-a}=\frac{a}{s-a}=\frac{KI}{IQ}$.Also note that $\angle{KIQ}=\angle{BCE}=\angle{C}$.Thus $\triangle{XCE} \sim \triangle{KIQ}$ and the result is immediate.

What do you think will be the rating of this problem on ISL scale Like G1, G2 or what ?

From what I have seen recently, probably G4 or G3

Dear Mathlinkers, a proof involving only two harmonic pencils with three couples of perpendicular rays leeds to the result. Sincerely Jean-Louis

mathuz wrote: \[ \frac{BC}{CP}=\frac{KI}{IQ}. \] Can you explain more on this step? I do not get it.

Finally bashed it... Take in-circle to be the unit circle and then note that $K \longrightarrow k=\frac{2d^2(q+f)}{(q+d)(f+d)}$ Now its simple since we know $E \longrightarrow e= \frac{qf}{q+f} +\frac{qd}{q+d}$ and so $z= \frac{e-k}{q-i}$ and we get $z=\frac{(q+f)(q-d)}{q(d+f)}$(after some smart manipulations) which is clearly imaginary after conjugating. (Note:- $I,F \longrightarrow i, f$ where $F$ is the in-touch point on $AB$ respectively.)

Just a note: Using the well-known Lemma 8 here, one can show that the polar of $K$ w.r.t. the incircle is the $A$-midline $l$ of $\triangle ABC.$ Consequently, $E \equiv l \cap AC$ is the intersection of the polars of $K$ and $Q$, which is precisely the pole of $KQ.$ The desired result follows.

Let $QI$ intersect the incircle at $X$.Let $BX$ further intersect the circle at $Y$.It is well known that $BY\parallel IE$.Let incircle touch $BC$ at $P$ .As $QY\perp XY$ it is sufficient to prove that $PI\cap YQ=K'$ is indeed $K$.But this is very easy,as we get $K'YPB$ to be cyclic,so $\angle{BKY}=\angle{YPC}=\angle{PQY}$ which further implies $K'B\parallel PQ$.Now as $CI\perp PQ,\Rightarrow CI\perp K'B$ and indeed $K'$ is the orthocentre of $BIC$,which is $K$.QED

Coordinate Let $P,Q,R$ be the $A,B,C$ intouch points. Let $Q'$ be the reflection of $Q$ in $I$. It is well known that $IE || BQ'$ So showing Let the incentre I be $(0,0)$ and $P$ be $(0,-1)$ Let $Q=(\cos(x),\sin(x))$ and $R=(\cos(y),\sin(y))$ Then it is obvious that $C=(\frac{1+\sin x}{\cos x},-1)$ and $B=(\frac{1+\sin y}{\cos y},-1)$ Since $IP \perp BC$, $K \in IP$ That is the $x-$coordinate of $K$ is zero. Let $K$ be $(0,m)$ Since $K$ is the orthocentre $KB \perp IC$, $$(\frac{m+1}{-\frac{1+\sin y}{\cos y}}) \cdot (\frac{-1}{\frac{1+\sin x}{\cos x}})=-1$$This yields, $$m=-(\frac{(1+\sin x)(1+\sin y)}{\cos x \cos y}+1)$$ To show now that $KQ \perp BQ'$ is just easy computation.

Luis González wrote: Let $F$ be the midpoint of $AB$ and let $(I)$ touch $BC$ at $P.$ If $BI$ cuts $EF$ at $L,$ we have $\widehat{FLB}=\widehat{LBC}=\widehat{FBL}$ $\Longrightarrow$ $\triangle FBL$ is F-isosceles $\Longrightarrow$ $FA=FB=FL$ $\Longrightarrow$ $\widehat{ALB}=90^{\circ}$ $\Longrightarrow$ $AQLI$ is cyclic $\Longrightarrow$ $\widehat{CQL}=\widehat{AIL}=90^{\circ}-\tfrac{1}{2}\widehat{ACB}=\widehat{CQP}$ $\Longrightarrow$ $L \in PQ,$ i.e. $L$ is on the polar of $C$ WRT $(I)$ $\Longrightarrow$ $CK \perp IL$ is the polar of $L$ WRT $(I)$ $\Longrightarrow$ $ELF \perp IK$ is the polar of $K$ WRT $(I)$ $\Longrightarrow$ $KQ$ is the polar of $E$ WRT $(I)$ $\Longrightarrow$ $KQ \perp IE,$ as desired. Could someone specify how the last step is reached when we already have QLP collinear and two perpendicular relationship?

\bmp.....

So I finally tried this again and found a nice synthetic way. India IMO TST 2014 wrote: In a triangle $ABC$, let $I$ be its incenter; $Q$ the point at which the incircle touches the line $AC$; $E$ the midpoint of $AC$ and $K$ the orthocenter of triangle $BIC$. Prove that the line $KQ$ is perpendicular to the line $IE$. Let $L$ be on the incircle $\omega$ such that $\overline{EL}$ touches $\omega$. Let $N$ be on line $\overline{IK}$ such that $\overline{EN} \parallel \overline{BC}$. Note that $\overline{QL} \perp \overline{IE}$ so it suffices to show that $K$ lies on $\overline{QL}$. To see this, note that $N$ is the inverse of $K$ in $\omega$ so $N \in \odot(IE) \Longrightarrow N \in \odot(IQL) \Longrightarrow K \in \overline{QL}$ as desired.

The polar of $K$ is $A$-midline and that of $Q$ is $AC$ $\implies$ the pole of $KQ$ is $E$ and hence $IE\perp KQ$.

Ah, I see no one has posted a barycentric solution yet! India TST 2014/1 wrote: In a triangle $ABC$, let $I$ be its incenter; $Q$ the point at which the incircle touches the line $AC$; $E$ the midpoint of $AC$ and $K$ the orthocenter of triangle $BIC$. Prove that the line $KQ$ is perpendicular to the line $IE$. Let $\triangle ABC$ be reference; then $I=(a:b:c), E=(\tfrac{1}{2}, 0, \tfrac{1}{2}), Q=(s-c:0:s-a)$. Let $J=(-a:b:c)$ be the $A$-excenter and $M=(0,\tfrac{1}{2}, \tfrac{1}{2})$ be the midpoint of $\overline{BC}$. Observe $BKCJ$ is a parallelogram. Reflect $Q$ about $M$ to get $Q'=\left(\frac{-(s-c)}{b}, 1, \frac{s-c}{b}\right)$; so $\overline{KQ} \parallel \overline{JQ'}$. Now we see \begin{align*}\overrightarrow{Q'J} &= \left(\frac{a}{2(s-a)}-\frac{(s-c)}{b}, \frac{c-a}{2(s-a)}, \frac{-c}{2(s-a)}+\frac{(s-c)}{b} \right)\\ &=\left(ab-2(s-a)(s-c):b(c-a):2(s-a)(s-c)-ac\right)\\ &=\left(2ab-b^2+(c-a)^2: 2b(c-a): b^2-2bc-(c-a)^2\right) \\ &\overset{\text{def}}{=} (x_1:y_1:z_1)\end{align*}and that \begin{align*}\overrightarrow{IE} &=\left(\frac{1}{2}-\frac{a}{2s}, \frac{-b}{2s}, \frac{1}{2}-\frac{c}{2s}\right)\\ &=(b+c-a:-2b:b-c+a)\\ &\overset{\text{def}}{:=} (x_2,y_2,z_2). \end{align*}It is now time to apply EFFT, so behold: \begin{align*}y_1z_2+y_2z_1&=2b(c-a)(b-(c-a))+(b^2-2bc-(c-a)^2)(-2b)\\ &= 2b^2\left[3c-a-b\right] \end{align*}\begin{align*} x_1z_2+x_2z_1&=(2ab-b^2+(c-a)^2)(b-c+a)+(b^2-2bc-(c-a)^2)(b+c-a) \\ &= 2(c-a)\left[(c-a)^2-b(a+c)\right]\end{align*}\begin{align*} x_1y_2+x_2y_1&= (2ab-b^2+(c-a)^2)(-2b)+2b(c-a)(b+c-a) \\ &= 2b^2\left[b+c-3a\right] \end{align*}hence \begin{align*}\sum_{\text{cyc}} a^2(y_1z_2+y_2z_1) &= 2a^2b^2\left[3c-a-b\right]+2b^2(c-a)\left[(c-a)^2-b(a+c)\right]+2b^2c^2\left[b+c-3a\right] \\&= 2b^2\left(a^2(3c-a-b)+(c-a)^2-b(c^2-a^2)+c^2(b+c-3a)\right) \\ &= 2b^2\left(a^2(3c-a)-c^2(3a-c)+(c-a)^3\right) \\ &= 0 \end{align*}hence we're done!

Restated problem wrote: In a triangle $ABC$, let $I$ be its incenter; $\omega$ be its incircle; $\triangle DEF$ be its intouch triangle; $M$ be the midpoint of $AC$ and $K$ be the orthocenter of triangle $BIC$. Prove that the line $KE$ is perpendicular to the line $IM$. Note that, as $E$ lies on the polar of $M$, it suffices to show that $K$ lies on the polar of $M$. Let $CI \cap DF=T$, and let $\infty_{BC}$ denote the point at infinity on $BC$. From here, we get that $T$ lies on the circle with $AC$ as diameter. So we have $\angle CTM=\angle MCT=\frac{C}{2}=\angle BCT$, and so $MT$ and $BC$ are parallel. Also, as $BK \perp IT$, and as $B$ lies on the polar of $T$ wrt $\omega$ (cause $D,F,T$ are collinear), we get that $K$ also lies on the polar of $T$ wrt $\omega$. And as $T,\infty_{BC},Q$ are collinear, their polars wrt $\omega$ must be concurrent, which gives that $BK,ID$ and the polar of $M$ concur. As $K$ lies on $ID$ also, we get that $K$ lies on the polar of $M$. $\blacksquare$

hajimbrak wrote: In a triangle $ABC$, let $I$ be its incenter; $Q$ the point at which the incircle touches the line $AC$; $E$ the midpoint of $AC$ and $K$ the orthocenter of triangle $BIC$. Prove that the line $KQ$ is perpendicular to the line $IE$. Lemma:- The Polar of the Orthocenter of $BIC$ where $I$ is the incenter of $\triangle ABC$ WRT $\odot(I)$ is the $A-\text{Midline}$ of $\triangle ABC$ (Well known) So by Lemma $E\in\text{ Polar of }K$ WRT $\odot(I)$. So, by La Hire's Theorem we get that $K\in\text{ Polar of }E$, So, $KQ$ is the Polar of $E$ WRT $\odot(I)$. Hence, $KQ\perp IE$.

hajimbrak wrote: In a triangle $ABC$, let $I$ be its incenter; $Q$ the point at which the incircle touches the line $AC$; $E$ the midpoint of $AC$ and $K$ the orthocenter of triangle $BIC$. Prove that the line $KQ$ is perpendicular to the line $IE$. We use Lemma from Lemmas in Euclidean Geometry by Yufei Zhao to get that the polar of $K$ with respect to incircle is $A$-midline of $\triangle ABC$. Now, we see that by La Hire's Theorem, $K$ lies on Polar of $E$ and $EQ$ is tangent to incircle too so $KQ$ is Polar of $E$ with respect to incircle, so now $IE \perp KQ$ as desired [asy][asy] import graph; size(7cm); 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 = -2.626666666666665, xmax = 10.413333333333343, ymin = 1.15333333333333, ymax = 9.20666666666666; /* image dimensions */ pen qqzzcc = rgb(0,0.6,0.8); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((7.68,9.46)--(-1.94,5.52)--(1.72,1.4)--cycle, linewidth(2) + qqzzcc); /* draw figures */ draw((7.68,9.46)--(-1.94,5.52), linewidth(2) + qqzzcc); draw((-1.94,5.52)--(1.72,1.4), linewidth(2) + qqzzcc); draw((1.72,1.4)--(7.68,9.46), linewidth(2) + qqzzcc); draw((1.571769153993443,4.705643666991477)--(4.7,5.43), linewidth(2) + fuqqzz); draw((2.723898686205071,5.729137377645496)--(3.247884617813999,3.466233224761884), linewidth(2) + fuqqzz); draw((-1.94,5.52)--(1.571769153993443,4.705643666991477), linewidth(2) + fuqqzz); draw((1.571769153993443,4.705643666991477)--(1.72,1.4), linewidth(2) + fuqqzz); draw(circle((1.571769153993443,4.705643666991477), 2.0845865998720594), linewidth(2)); /* dots and labels */ dot((7.68,9.46),linewidth(4pt) + dotstyle); label("$A$", (7.373333333333341,8.993333333333327), NE * labelscalefactor); dot((-1.94,5.52),linewidth(4pt) + dotstyle); label("$B$", (-1.8933333333333313,5.633333333333328), NE * labelscalefactor); dot((1.72,1.4),linewidth(4pt) + dotstyle); label("$C$", (1.7733333333333374,1.513333333333331), NE * labelscalefactor); dot((1.571769153993443,4.705643666991477),linewidth(4pt) + dotstyle); label("$I$", (1.6266666666666707,4.806666666666662), NE * labelscalefactor); dot((3.247884617813999,3.466233224761884),linewidth(4pt) + dotstyle); label("$Q$", (3.3066666666666715,3.566666666666663), NE * labelscalefactor); dot((4.7,5.43),linewidth(4pt) + dotstyle); label("$E$", (4.746666666666672,5.54), NE * labelscalefactor); dot((2.723898686205071,5.729137377645496),linewidth(4pt) + dotstyle); label("$K$", (2.773333333333338,5.833333333333329), NE * labelscalefactor); dot((2.87,7.49),linewidth(4pt) + dotstyle); label("$D$", (2.92,7.5933333333333275), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]