The circle $\omega$ with center at point $I$ inscribed in an triangle $ABC$ ($AB\neq AC$) touches the sides $BC$, $CA$ and $AB$ at points $D$, $E$ and $F$, respectively. The circumcircles of triangles $ABC$ and $AEF$ intersect secondary at point $K.$ The lines $EF$ and $AK$ intersect at point $X$ and intersects the line $BC$ at points $Y$ and $Z$, respectively. The tangent lines to $\omega$, other than $BC$, passing through points $Y$ and $Z$ touch $\omega$ at points $P$ and $Q$, respectively. Let the lines $AP$ and $KQ$ intersect at the point $R$. Prove that if $M$ is a midpoint of segment $YZ,$ then $IR\perp XM$.
Problem
Source: Kazakhstan National Olympiad 2024 (10-11 grade), P6
Tags: geometry
Davsch
21.03.2024 13:29
$X,P,Q$ are collinear, $AKPQ$ concyclic, $DQ\perp EF$, $IDKZQ$ concyclic.
First, note that $(AKEFI)$ has diameter $AI$, so $IK\perp XZ$. In addition, we have $XY=EF\perp IA$, $BC\perp ID$. Since $-1=(Y,Z;M,\infty_{BC})$, it suffices to show $(IA,IK;ID,IR)=-1$. Note that inversion with respect to $\omega$ maps $(ABC)$ to the nine-point circle of $DEF$, so the image $T=IK\cap EF$ of $K$ under this inversion is the foot of $D$ onto $EF$. Since $A,K,Z$ are collinear, their polars concur, so $T$ lies on $DQ$.
Note as well that since $DD,PP,EF$ concur at $Y$, $DPEF$ is harmonic, so $P$ lies on $AD$. Hence, projecting onto line $AP$, we only need to show $-1=(A,IK\cap AP;R,D)$.
Projecting through $K$ onto $QD$, this is equivalent to $-1=(KA\cap QD,T;Q,D)$, i.e. that $KA\cap QD$ lies on the polar of $T$ with respect to $\omega$. But this is obvious since $T$ lies on the polar $EF$ of $A$ and the polar $DQ$ of $Z$, so $AK$ is the polar of $T$.
hukilau17
22.03.2024 02:22
We complex bash with $\omega$ as the unit circle. Rename $I$ to $J$, and let $O, T$ be the circumcenters of $\triangle ABC,\triangle AEF$ respectively. Then $$|d|=|e|=|f|=1$$$$a=\frac{2ef}{e+f}$$$$b=\frac{2df}{d+f}$$$$c=\frac{2de}{d+e}$$$$j=0$$$$o=\frac{2def(d+e+f)}{(d+e)(d+f)(e+f)}$$$$t = \frac{ef}{e+f}$$$$y = \frac{d(de+df-2ef)}{d^2-ef}$$Now $X$ is the reflection of $A$ over line $OT$. Now we have the vectors $$a' = a-t = \frac{ef}{e+f}$$$$o' = o-t = \frac{ef(d^2+de+df-ef)}{(d+e)(d+f)(e+f)}$$Then if $k' = k-t$, we have $$k' = \frac{\overline{a'}o'}{\overline{o'}} = \frac{ef(d^2+de+df-ef)}{(e+f)(de+df+ef-d^2)}$$and $$k = k'+t = \frac{2def}{de+df+ef-d^2}$$Now we have the vector $$k-a = \frac{2ef(d^2-ef)}{(e+f)(de+df+ef-d^2)}$$Thus since $X$ lies on line $AK$, we have $$\frac{x-a}{k-a} \in \mathbb{R}$$$$\frac{(de+df+ef-d^2)(ex+fx-2ef)}{2ef(d^2-ef)} = -\frac{(d^2+de+df-ef)(e\overline{x}+f\overline{x}-2)}{2(d^2-ef)}$$$$(de+df+ef-d^2)(ex+fx-2ef) = -ef(d^2+de+df-ef)(e\overline{x}+f\overline{x}-2)$$$$\overline{x} = \frac{2ef(d^2+de+df-ef) + 2ef(de+df+ef-d^2) - x(e+f)(de+df+ef-d^2)}{ef(e+f)(d^2+de+df-ef)}$$which we simplify to $$\overline{x} = \frac{4def - x(de+df+ef-d^2)}{ef(d^2+de+df-ef)}$$Intersecting this with line $EF$ gives $$\frac{4def - x(de+df+ef-d^2)}{ef(d^2+de+df-ef)} = \frac{e+f-x}{ef}$$$$4def-x(de+df+ef-d^2) = (e+f)(d^2+de+df-ef) - x(d^2+de+df-ef)$$$$x = \frac{d^2e+d^2f+de^2+df^2-2def-e^2f-ef^2}{2(d^2-ef)}$$Intersecting with line $BC$, to find the coordinate of $Z$, gives $$\frac{4def - z(de+df+ef-d^2)}{ef(d^2+de+df-ef)} = \frac{2d-z}{d^2}$$$$4d^3ef - d^2z(de+df+ef-d^2) = 2def(d^2+de+df-ef) - efz(d^2+de+df-ef)$$$$z = \frac{2def(-d^2+de+df-ef)}{d^4-d^3e-d^3f+de^2f+def^2-e^2f^2} = -\frac{2def}{d^2-ef}$$Now we have $$y = \frac{2dp}{d+p}$$so $$p = \frac{dy}{2d-y} = \frac{de+df-2ef}{2d-e-f}$$Similarly, $$q = \frac{dz}{2d-z} = -\frac{ef}{d}$$Also, $$m = \frac{y+z}2 = \frac{d(de+df-4ef)}{2(d^2-ef)}$$Now we just have to find the coordinate of $R$. Now let $AP, KQ$ meet the unit circle again at $A', K'$ respectively. Then $$a' = \frac{p-a}{p\overline{a}-1} = \frac{d(de+df-2ef)}{2d-e-f} = d$$(That is, $A, D, P$ are collinear.) And $$k' = \frac{q-k}{q\overline{k}-1} = \frac{ef(d^2+de+df-ef)}{d(de+df+ef-d^2)}$$Then \begin{align*} r &= \frac{a'p(k'+q) - k'q(a'+p)}{a'p - k'q} \\ &= \frac{\frac{d(de+df-2ef)}{2d-e-f}\left[\frac{ef(d^2+de+df-ef)}{d(de+df+ef-d^2)}-\frac{ef}d\right] + \frac{e^2f^2(d^2+de+df-ef)}{d^2(de+df+ef-d^2)}\left[d+\frac{de+df-2ef}{2d-e-f}\right]}{\frac{d(de+df-2ef)}{2d-e-f} + \frac{e^2f^2(d^2+de+df-ef)}{d^2(de+df+ef-d^2)}} \\ &= \frac{2d^2ef(de+df-2ef)(d^2-ef) + 2e^2f^2(d^2+de+df-ef)(d^2-ef)}{d^3(de+df-2ef)(de+df+ef-d^2) + e^2f^2(2d-e-f)(d^2+de+df-ef)} \\ &= \frac{2ef(d^2-ef)(d^3e+d^3f-d^2ef+de^2f+def^2-e^2f^2)}{-d^6e-d^6f+d^5e^2+4d^5ef+d^5f^2-d^4e^2f-d^4ef^2+d^2e^3f^2+d^2e^2f^3-de^4f^2-4de^3f^3-de^2f^4+e^4f^3+e^3f^4} \\ &= \frac{2ef(d^2-ef)(d^2+ef)(de+df-ef)}{(d^2-ef)(d^2+ef)(-d^2e-d^2f+de^2+4def+df^2-e^2f-ef^2)} \\ &= \frac{2ef(de+df-ef)}{de^2+4def+df^2-d^2e-d^2f-e^2f-ef^2} \end{align*}So we have the vector $$x-m = \frac{de^2+df^2+2def-e^2f-ef^2}{2(d^2-ef)} = \frac{(e+f)(de+df-ef)}{2(d^2-ef)}$$Thus $$\frac{x-m}{r-j} = \frac{(e+f)(de^2+4def+df^2-d^2e-d^2f-e^2f-ef^2)}{4ef(d^2-ef)}$$which is indeed pure imaginary. $\blacksquare$
sami1618
02.06.2024 22:31
It is well known that $A$, $P$, and $D$ are collinear and that $DQ\perp EF$. Let $S$ be the foot of $D$ onto $EF$, which is known to be the inverse of $K$. Notice we have $ID\perp BC$, $IK\perp XZ$, $IA\perp XY$. So since $(Y,Z;M,P_{\infty})$ is a harmonic bundle it is sufficient to show that $(IA,IK;IR,ID)$ is a harmonic bundle. By projecting onto line $AD$ it is sufficient to show that $(KA,KI;KQ,KD)$ is a harmonic bundle. As $\angle AKI=90^{\circ}$ it is sufficient to show that $$\angle QKI=\angle IQS=\angle IZD=\angle IKD$$
axolotlx7
04.06.2024 22:42
Nice Since $DEPF$ is harmonic by construction, $P$ lies on $AD$. Let $ID$ and $IR$ hit $(AI)$ again at $G$, $H$ respectively. We want to show that $AH \parallel XM$. It suffices to show that \[ -1 = (XY,XZ;XM,\ell_{X, \parallel BC})=(AA,AK;AH,AG)_{(AI)} = (A,K;H,G). \]Note that $ZKQID$ is cyclic. Hence let $L = AD \cap IK$, we have \[ (A,K;H,G) \stackrel I= (A,L;R,D) \stackrel K= (Z, I; Q, D) = -1 \]hence proved.