Let $BD$ be the external bisector of a triangle $ABC$ with $AB > BC$; $K$ and $K_1$ be the touching points of side $AC$ with the incircle and the excircle centered at $I$ and $I_1$ respectively. The lines $BK$ and $DI_1$ meet at point $X$, and the lines $BK_1$ and $DI$ meet at point $Y$. Prove that $XY \perp AC$.
Problem
Source: Sharygin 2018
Tags: geometry
Vrangr
04.04.2018 22:32
I'll post my complete solutions later, but essentially. Prove these $3$ claims by simple angle chasing: $I, K, D, B$ are concyclic. $I_1, K_1, B, D$ are concyclic. $X, Y, B, D$ are concylic.
Drunken_Master
05.04.2018 08:09
Let $L=BK_1 \cap IK$. Its well known that $I$ is midpoint of $KL$. Now let $XI \cap BK_1=E$ and $YI \cap BK=F$. Use projections to establish that $(B,K;F,X)=(B,L;Y,E)=-1$. Due to equality in cross ratios, $EF, KL,XY$ concur, say at $P$. Now, if $M=BI \cap XY$ then by ceva+menelaus, $(X,Y;M,P)=-1$. Taking a projection, $(K,L;I,P)=-1$. Thus $P$ is point at infinity along $KL$. So, $IK \parallel XY$. The result follows. $\square$
Vrangr
05.04.2018 08:28
Expanding my previous sketch.
$\measuredangle$ refers to directed angles, $\angle$ refers to non-directed angle.
First note that $B, I, I_1$ are collinear.
\[\measuredangle DBI = \measuredangle DBA + \measuredangle ABI = 90^{\circ} - \measuredangle ABI + \measuredangle ABI = 90^{\circ} = \measuredangle DKI\]\[\measuredangle DBI_1 = \measuredangle DBI = 90^{\circ} = \measuredangle DK_1I_1\]Therefore, $I, B, D, K$ and $I_1, B, D, K_1$ are concyclic.
\begin{align*}
\measuredangle BYD &= \measuredangle K_1YD = \measuredangle YK_1D + \measuredangle K_1DY = \measuredangle BK_1D + \measuredangle KDI \\
&= \measuredangle BI_1D + \measuredangle KBI = \measuredangle BI_1X + \measuredangle XBI_1 = \measuredangle BXI_1\\
&= \measuredangle BXD
\end{align*}Therefore, $X, Y, B, D$ are concylic.
\[\measuredangle DYX = \measuredangle DBX = \measuredangle DBK = \measuredangle DIK\]Therefore, $XY \parallel IK$.
Since, $IK \perp BC$ and $XY \parallel IK$, $XY \perp BC$.
TheDarkPrince
05.04.2018 09:45
[asy][asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10cm);
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 = -18.297065589106747, xmax = 17.18860204391376, ymin = -12.52516414389673, ymax = 7.137386118334379; /* image dimensions */
pair B = (0.,5.), C = (-1.,0.), A = (6.,0.), D = (-14.164923100695951,0.), I = (1.144384918843065,1.7579751253983609), I_1 = (3.8556150811569347,-5.922898226094311), K_1 = (3.855615081156934,0.), K = (1.144384918843065,0.), X = (2.389726929001948,-5.441097613458075), Y = (2.389726929001947,1.9009783410681198);
draw(B--C--A--cycle, linewidth(1.));
/* draw figures */
draw(B--C, linewidth(1.));
draw(A--B, linewidth(1.));
draw(D--A, linewidth(1.));
draw(B--D, linewidth(1.));
draw(B--I_1, linewidth(1.));
draw(circle(I_1, 5.922898226094311), linewidth(1.));
draw(circle(I, 1.7579751253983609), linewidth(1.));
draw(B--K_1, linewidth(1.));
draw(I_1--D, linewidth(1.));
draw(Y--D, linewidth(1.));
draw(B--X, linewidth(1.));
/* dots and labels */
dot(B,dotstyle);
label("$B$", B,dir(90));
dot(C,dotstyle);
label("$C$", C, dir(-90));
dot(A,dotstyle);
label("$A$", A,dir(0));
dot(D,linewidth(4.pt) + dotstyle);
label("$D$", D, dir(180));
dot(I,linewidth(4.pt) + dotstyle);
label("$I$", I, dir(45));
dot(I_1,linewidth(4.pt) + dotstyle);
label("$I_1$", I_1, dir(0));
dot(K_1,linewidth(4.pt) + dotstyle);
label("$K_1$", K_1, dir(-90));
dot(K,linewidth(4.pt) + dotstyle);
label("$K$", K, dir(-135));
dot(X,linewidth(4.pt) + dotstyle);
label("$X$", X, dir(-90));
dot(Y,linewidth(4.pt) + dotstyle);
label("$Y$", Y, dir(-120));
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy][/asy]
We know that $\angle DBI = 90^{\circ}=\angle IKD$ and $\angle DBI_1 = 90^{\circ}=\angle DK_1I_1$. So, $DBIK$ and $DBK_1I_1$ is cyclic.
So we have,
\begin{align*}
\angle XBY &= \angle XBI + \angle I_1BY\\
&= \angle KBI + \angle I_1BK_1\\
&= \angle KDI + \angle I_1DK_1\\
&= \angle XDY.
\end{align*}
So, $XDBY$ is cyclic.
$\Rightarrow \angle YXB = \angle YDB = \angle IDB = \angle IKB$. So, $YX||IK$. We also know that $IK \perp AC$, so $XY \perp AC$.
Proved.
math_pi_rate
27.05.2018 08:27
This is just Pappus' theorem on $B, I, I_1$ and $D, K_1, K$ in that order; and using the fact that $IK$ & $I_1K_1$ meet at the point at infinity on these lines.
Mahdi_Mashayekhi
12.12.2021 20:53
∠DBI = 90 = ∠IKD ---> DBIK is cyclic ∠I1K1D = 90 = ∠I1BD ---> I1K1BD is cyclic ∠BYD = ∠BK1D + ∠KDI = ∠BI1D + ∠KBI = ∠BXD ---> XYBD is cyclic ∠BYX = 180 - ∠BDX = ∠BK1I1 ---> XY || K1I1 ---> XY is perpendicular to AC