The points $P$ and $Q$ on the side $AC$ of the non-isosceles $\Delta ABC$ are such that $\angle ABP=\angle QBC<\frac{1}{2}\angle ABC$. The angle bisectors of $\angle A$ and $\angle C$ intersect the segment $BP$ in points $K$ and $L$ and the segment $BQ$ in points $M$ and $N$, respectively. Prove that $AC$,$KN$, and $LM$ are concurrent.
Problem
Source: IV International Festival of Young Mathematicians Sozopol 2013, Theme for 10-12 grade
Tags: geometry, concurrency, moving points
nguyenvanthien63
25.01.2020 12:39
[asy][asy] import graph; size(11.457692805668595cm); 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 = -4.348279642518577, xmax = 7.109413163150018, ymin = -1.0150355590086655, ymax = 2.710775338949761; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((0.16326251200405917,0.6780267914880583)--(-0.1773497434766288,-0.38857177970591983)--(0.23746425867650936,-0.002911357734459174)--cycle, linewidth(1.) + zzttqq); draw((1.3754765903737423,1.0551202269589035)--(6.219175160982959,-0.24019877934474354)--(2.6722394785434744,0.29760406400775763)--cycle, linewidth(1.) + zzttqq); /* draw figures */ draw((0.036966812600625426,1.8370232257117245)--(-0.1773497434766288,-0.38857177970591983), linewidth(1.)); draw((-0.1773497434766288,-0.38857177970591983)--(6.219175160982959,-0.24019877934474354), linewidth(1.)); draw((6.219175160982959,-0.24019877934474354)--(0.036966812600625426,1.8370232257117245), linewidth(1.)); draw((0.036966812600625426,1.8370232257117245)--(0.2783378081379181,-0.3780017076323865), linewidth(1.)); draw((0.036966812600625426,1.8370232257117245)--(3.6931858705311558,-0.29879131443254325), linewidth(1.)); draw((xmin, 0.12342635134862133*xmin-0.03222070475860592)--(xmax, 0.12342635134862133*xmax-0.03222070475860592), linewidth(1.)); /* line */ draw((xmin, 0.311078251110564*xmin + 0.6272393747819182)--(xmax, 0.311078251110564*xmax + 0.6272393747819182), linewidth(1.)); /* line */ draw((xmin, 0.023195876288659836*xmin-0.3844579969964103)--(xmax, 0.023195876288659836*xmax-0.3844579969964103), linewidth(1.)); /* line */ draw((0.16326251200405917,0.6780267914880583)--(-0.1773497434766288,-0.38857177970591983), linewidth(1.)); draw((1.3754765903737423,1.0551202269589035)--(6.219175160982959,-0.24019877934474354), linewidth(1.)); draw((0.8567746864765389,0.5728732582065204)--(-0.1773497434766288,-0.38857177970591983), linewidth(1.)); draw((0.8567746864765389,0.5728732582065204)--(6.219175160982959,-0.24019877934474354), linewidth(1.)); draw((0.036966812600625426,1.8370232257117245)--(0.8567746864765389,0.5728732582065204), linewidth(1.)); draw((0.16326251200405917,0.6780267914880583)--(-0.1773497434766288,-0.38857177970591983), linewidth(1.) + zzttqq); draw((-0.1773497434766288,-0.38857177970591983)--(0.23746425867650936,-0.002911357734459174), linewidth(1.) + zzttqq); draw((0.23746425867650936,-0.002911357734459174)--(0.16326251200405917,0.6780267914880583), linewidth(1.) + zzttqq); draw((1.3754765903737423,1.0551202269589035)--(6.219175160982959,-0.24019877934474354), linewidth(1.) + zzttqq); draw((6.219175160982959,-0.24019877934474354)--(2.6722394785434744,0.29760406400775763), linewidth(1.) + zzttqq); draw((2.6722394785434744,0.29760406400775763)--(1.3754765903737423,1.0551202269589035), linewidth(2.) + zzttqq); /* dots and labels */ dot((0.036966812600625426,1.8370232257117245),linewidth(4.pt) + dotstyle); label("$B$", (0.10291036831670461,1.9689103371438799), NE * labelscalefactor); dot((-0.1773497434766288,-0.38857177970591983),linewidth(4.pt) + dotstyle); label("$A$", (-0.5070675220570191,-0.7347754472153326), NE * labelscalefactor); dot((6.219175160982959,-0.24019877934474354),linewidth(4.pt) + dotstyle); label("$C$", (6.285118716699039,-0.10831166791258826), NE * labelscalefactor); dot((3.6931858705311558,-0.29879131443254325),linewidth(4.pt) + dotstyle); label("$Q$", (3.762777710559047,-0.742552236286666), NE * labelscalefactor); dot((0.2783378081379181,-0.3780017076323865),linewidth(4.pt) + dotstyle); label("$P$", (0.2347974797488611,-0.8336907807894501), NE * labelscalefactor); dot((0.23746425867650936,-0.002911357734459174),linewidth(4.pt) + dotstyle); label("$K$", (0.25128336867788065,0.18843433280976432), NE * labelscalefactor); dot((1.3754765903737423,1.0551202269589035),linewidth(4.pt) + dotstyle); label("$L$", (1.438267371567289,1.1940735574799592), NE * labelscalefactor); dot((0.16326251200405917,0.6780267914880583),linewidth(4.pt) + dotstyle); label("$M$", (0.1523680351037633,0.8478698899705478), NE * labelscalefactor); dot((2.6722394785434744,0.29760406400775763),linewidth(4.pt) + dotstyle); label("$N$", (2.5098501519535605,-0.058854001125529486), NE * labelscalefactor); dot((0.8567746864765389,0.5728732582065204),linewidth(4.pt) + dotstyle); label("$I$", (0.9272048147676826,0.6994968896093715), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let I be the incenter of triangle $ABC.$ We have $\angle MBA = \angle LBC, \angle MBC = \angle LBA.$ In triangle $ABC$ with $AM, BM, CM$ are concurrent, we have $\frac{\sin \angle MAC}{\sin \angle MAB}. \frac{ \sin C/2}{\sin C/2}. \frac{\sin \angle MBA}{ \sin \angle MBA} = 1$ Implying $\dfrac{\sin \angle MAC}{\sin \angle MAB} = \dfrac{\sin \angle MBC}{\sin \angle MBA}$ similarly $\dfrac{\sin \angle LCA}{\sin \angle LCB} = \dfrac{\sin \angle LBA}{\sin \angle LBC}$ In triangle $ABC$ with $AM, BI, CL$ , we have $\frac{\sin \angle MAB}{\sin \angle MAC}. \frac{ \sin B/2}{\sin B/2}. \frac{\sin \angle LCA}{ \sin \angle LCB} = \frac{\sin \angle MBA}{\sin \angle MBC}. \frac{\sin \angle LBA}{ \sin \angle LBC} = 1$ so $AM, CL, BI$ are concurrent, then $AM \cap CL = X \in BI$ Consider two triangles $AMK, CLN$, we have $AM \cap CL = X, AK \cap CN = I, KM \cap LN = B$ and $X, I, B$ are colinear then $AC, ML, KN$ are concurrent.
math_pi_rate
25.01.2020 14:10
Animate $P$ linearly on $AC$ keeping $\triangle ABC$ fixed. Then $P \mapsto Q$ is a projective map by isogonality in $\angle ABC$. Also, perspectivity from $B$ gives the maps $P \mapsto K,L$ and $Q \mapsto M,N$. But, when $P$ is the foot of the $A$-internal angle bisector, the points $K,L,M,N$ all coincide with the incenter. This means that degree of lines $KN$ and $LM$ are atmost $1$ (since degree of each of the points $K,L,M,N$ is $1$). Thus, it suffices to prove $AC,KN,LM$ concurrent for $(0+1+1)+1=3$ positions of $P$. Taking $P=A,C$ and $P$ as the foot of the $A$-external angle bisector trivially work.