Concurrency(little hard)

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Source: Rioplatense Olympiad 2000

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#1 h Feb 28, 2018, 10:20 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle with $AB < AC$ , let $L$ be midpoint of arc $BC$ (the point $A$ is not in this arc) of the circumcircle $w$ ( $ABC$ ). Let $E$ be a point in $AC$ where $AE = \frac{AB + AC}{2}$ , the line $EL$ intersects $w$ in $P$ .
If $M$ and $N$ are the midpoints of $AB$ and $BC$ , respectively, prove that $AL, BP$ and $MN$ are concurrents

This post has been edited 1 time. Last edited by mathisreal, Feb 28, 2018, 10:21 PM

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#2 h Mar 1, 2018, 7:10 AM • 2 Y

Y by Adventure10, Mango247

mathisreal wrote:

Let $ABC$ be a triangle with $AB < AC$ , let $L$ be midpoint of arc $BC$ (the point $A$ is not in this arc) of the circumcircle $w$ ( $ABC$ ). Let $E$ be a point in $AC$ where $AE = \frac{AB + AC}{2}$ , the line $EL$ intersects $w$ in $P$ .
If $M$ and $N$ are the midpoints of $AB$ and $BC$ , respectively, prove that $AL, BP$ and $MN$ are concurrents

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.1278216169548168, xmax = 24.89223801199639, ymin = -9.684722053341803, ymax = 2.9366375930857926; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((9.64,0.62)--(7.28,-4.62)--(15.683526658917778,-4.859318254818102)--cycle, linewidth(2.) + zzttqq); draw(arc((11.48176332945889,-4.739659127409051),0.41838761258876006,137.80321076350407,178.36875380244265)--(11.48176332945889,-4.739659127409051)--cycle, linewidth(2.) + qqwuqq); draw(arc((15.683526658917778,-4.859318254818102),0.41838761258876006,137.80321076350407,178.36875380244265)--(15.683526658917778,-4.859318254818102)--cycle, linewidth(2.) + qqwuqq); draw(arc((11.394747461520922,-7.795172659050729),0.41838761258876006,101.77864104115174,142.34418408009034)--(11.394747461520922,-7.795172659050729)--cycle, linewidth(2.) + qqwuqq); draw((11.186038505653338,-4.731237383585054)--(11.177616761829341,-5.026962207390604)--(11.473341585634893,-5.035383951214602)--(11.48176332945889,-4.739659127409051)--cycle, linewidth(2.) + qqwuqq); draw((10.528392979698486,-3.6404309036991123)--(10.23877770011074,-3.7008220207046185)--(10.299168817116247,-3.9904373002923648)--(10.588784096703993,-3.9300461832868585)--cycle, linewidth(2.) + qqwuqq); /* draw figures */ draw((9.64,0.62)--(7.28,-4.62), linewidth(2.) + zzttqq); draw((7.28,-4.62)--(15.683526658917778,-4.859318254818102), linewidth(2.) + zzttqq); draw((15.683526658917778,-4.859318254818102)--(9.64,0.62), linewidth(2.) + zzttqq); draw(circle((11.520529508887835,-3.378406420033452), 4.418556906226584), linewidth(2.)); draw((8.46,-2.)--(11.48176332945889,-4.739659127409051), linewidth(2.)); draw((11.394747461520922,-7.795172659050729)--(15.903879749662734,-2.8217325475898916), linewidth(2.)); draw((9.64,0.62)--(11.394747461520922,-7.795172659050729), linewidth(2.)); draw((7.28,-4.62)--(15.903879749662734,-2.8217325475898916), linewidth(2.)); draw((11.394747461520922,-7.795172659050729)--(11.48176332945889,-4.739659127409051), linewidth(2.)); draw((7.28,-4.62)--(11.394747461520922,-7.795172659050729), linewidth(2.)); draw((11.394747461520922,-7.795172659050729)--(15.683526658917778,-4.859318254818102), linewidth(2.)); /* dots and labels */ dot((9.64,0.62),dotstyle); label("$A$", (9.328218823694515,0.7889145151301575), NE * labelscalefactor); dot((7.28,-4.62),dotstyle); label("$B$", (7.054979461962252,-4.803533239806269), NE * labelscalefactor); dot((15.683526658917778,-4.859318254818102),linewidth(3.pt) + dotstyle); label("$C$", (15.799280565067338,-5.138243329877277), NE * labelscalefactor); dot((11.394747461520922,-7.795172659050729),linewidth(4.pt) + dotstyle); label("$L$", (11.085446796567307,-8.080902871751556), NE * labelscalefactor); dot((8.46,-2.),linewidth(4.pt) + dotstyle); label("$M$", (8.22646477721078,-1.930604966696783), NE * labelscalefactor); dot((11.48176332945889,-4.739659127409051),linewidth(4.pt) + dotstyle); label("$N$", (11.489888155403108,-5.0266732998536074), NE * labelscalefactor); dot((10.588784096703993,-3.9300461832868585),linewidth(4.pt) + dotstyle); label("$X$", (10.63916667647263,-3.8133492233462034), NE * labelscalefactor); dot((15.903879749662734,-2.8217325475898916),linewidth(4.pt) + dotstyle); label("$P$", (15.966635610102841,-2.7673801918743033), NE * labelscalefactor); dot((14.79054742616288,-4.049705310695909),linewidth(4.pt) + dotstyle); label("$E$", (14.962505339889818,-4.036489283393542), NE * labelscalefactor); label("$l$", (11.657243200438613,-6.267889883866929), NE * labelscalefactor); label("$m$", (9.188756286164928,-6.3794599138905985), NE * labelscalefactor); dot((10.753280105501249,-4.718913154807559),linewidth(4.pt) + dotstyle); label("$D$", (10.527596646448961,-4.956942031088815), NE * labelscalefactor); label("$n$", (13.651557487111702,-6.504976197667227), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
First, assume that $\overline{AL}$ intersects $\overline{BC}$ and $\overline{MN}$ at $D$ and $X$ , respectively.
Let $\overline{BX}$ intersects $\omega$ at $P$ , $\overline{LP}$ intersects $\overline{AC}$ at $E$ .
We'll prove that $AE=\frac{AB+AC}{2}$ .
Notice that $\angle XNB=\angle ACB=\angle XLB$ , so quadrilateral $XNLB$ is concyclic, implying that $\angle BXL=\angle BNL=90$ .
But $\triangle ABD\sim \triangle ALC$ , also $\angle ABX=\angle ALE$ ; then $\frac{AE}{EC}=\frac{AX}{XD}$ .
By Menelaus's theorem wrt $\triangle ABD$ , $\overline{MXN}$ , $\frac{AM}{MB}\cdot \frac{BN}{ND}\cdot \frac{DX}{XA}=1$ .
If $BC=a,CA=b,AB=c$ , then $BD=\frac{ac}{b+c}$ , $BN=\frac{a}{2}$ .
So $\frac{AE}{EC}=\frac{AX}{XD}=\frac{b+c}{b-c}$ : $AE=\frac{b+c}{2}$ as we wanted.
Done

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#3 h Mar 1, 2018, 2:25 PM • 1 Y

Y by Adventure10

Different solution:
Assume that $MN,BP$ and $AL$ are concurrents and prove that $AE = \frac{AB + AC}{2}$
Let $AL,BP$ and $MN$ intersects at $K$ .In circle $w$ note that $\angle BPL=\angle BAL=\angle LAC$ $\implies$ quadrilateral $AKEP$ is cyclic and $\angle KEA=\angle KPA=\angle BCA$ $\implies$ $KE$ is parallel to $BC$ and by Thales $GE=EC$ (where $G$ is intersection point of $BP$ and $AC$ .Now let $AB=a$ and $AC=c$ in isosceles triangle $BAG$ have that $AG=AB=a$ and also have that $GE=\frac{GC}{2}=\frac{c-a}{2}$ and $AE=a+\frac{c-a}{2}=\frac{a+c}{2}=\frac{AB+AC}{2}$
Done.

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