Let $ABC$ be a triangle. Let $D$ be the intersection point of the angle bisector at $A$ with $BC$. Let $T$ be the intersection point of the tangent line to the circumcircle of triangle $ABC$ at point $A$ with the line through $B$ and $C$. Let $I$ be the intersection point of the orthogonal line to $AT$ through point $D$ with the altitude $h_a$ of the triangle at point $A$. Let $P$ be the midpoint of $AB$, and let $O$ be the circumcenter of triangle $ABC$. Let $M$ be the intersection point of $AB$ and $TI$, and let $F$ be the intersection point of $PT$ and $AD$. Prove: $MF$ and $AO$ are orthogonal to each other.
Problem
Source: Mediterranean Mathematics Olympiad 2016 P1
Tags: geometry, circumcircle
gavrilos
04.06.2016 16:54
Hello. [asy][asy]import graph; size(12cm); 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 = -10.092441991084792, xmax = 13.488304518991592, ymin = -4.913675218716661, ymax = 8.428062938300267; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((1.190974619658015,4.33314019723913)--(0.,0.)--(6.,0.)--cycle, aqaqaq); /* draw figures */ draw((1.190974619658015,4.33314019723913)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.190974619658015,4.33314019723913), uququq); draw(circle((3.,1.5056836614662303), 3.356647626487826)); draw((-5.581603423878299,0.)--(0.,0.)); draw((-5.581603423878299,0.)--(1.190974619658015,4.33314019723913)); draw((1.190974619658015,4.33314019723913)--(2.4585406870639765,0.)); draw((1.7103684113678874,2.5576066176283265)--(-5.581603423878299,0.)); draw((1.8247576533609955,2.166570098619567)--(-5.581603423878299,0.)); draw((1.8247576533609955,2.166570098619567)--(-2.195314402110148,2.1665700986195615)); draw((2.4585406870639765,0.)--(1.1909746196580149,1.981170635170087)); draw((1.190974619658015,4.33314019723913)--(1.1909746196580149,1.981170635170087)); draw((1.7103684113678874,2.5576066176283265)--(0.4880099266149628,1.7755335796108056), linewidth(1.6) + qqwuqq); draw((1.190974619658015,4.33314019723913)--(3.,1.5056836614662303), linewidth(1.6) + qqwuqq); /* dots and labels */ dot((1.190974619658015,4.33314019723913),linewidth(3.pt) + dotstyle); label("$A$", (0.9222488129640456,4.461001260383274), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.385330175846581,-0.43687935872094685), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.219051835434211,-0.10444402258265134), NE * labelscalefactor); dot((2.4585406870639765,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.318477224744884,-0.4147170029783938), NE * labelscalefactor); dot((-5.581603423878299,0.),linewidth(3.pt) + dotstyle); label("$T$", (-5.615646131089087,0.2501536692981972), NE * labelscalefactor); dot((1.1909746196580149,1.981170635170087),linewidth(3.pt) + dotstyle); label("$I$", (1.1217100146470225,1.3804338121684019), NE * labelscalefactor); dot((0.5954873098290075,2.166570098619565),linewidth(3.pt) + dotstyle); label("$P$", (0.34602756365766774,2.3777398205832885), NE * labelscalefactor); dot((3.,1.5056836614662303),linewidth(3.pt) + dotstyle); label("$O$", (3.0941596757342387,1.6463820810790384), NE * labelscalefactor); dot((1.7103684113678874,2.5576066176283265),linewidth(3.pt) + dotstyle); label("$F$", (1.7,2.688012800979031), NE * labelscalefactor); dot((0.4880099266149628,1.7755335796108056),linewidth(3.pt) + dotstyle); label("$M$", (-0.008570128223180153,1.336109100683296), NE * labelscalefactor); dot((1.8247576533609955,2.166570098619567),linewidth(3.pt) + dotstyle); label("$N$", (2.0082042443491424,2.089629195930099), NE * labelscalefactor); dot((-2.195314402110148,2.1665700986195615),linewidth(3.pt) + dotstyle); label("$S$", (-2.37994219267635,2.4220645320683944), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Consider $N\equiv TI\cap AD$ and $S\equiv PN\cap AT$. First observe that $\angle{TAD}=\angle{TAB}+\frac{\angle{A}}{2}=\angle{C}+\frac{\angle{A}}{2}=\angle{TDA}\Rightarrow \triangle{TAD}$ is isosceles. Since $AI\perp TD$ and $DI\perp AT$,we conclude that $I$ is the orthocenter of $\triangle{TAD}$.Thus $TI\perp AD$,and,since $TA=TD$, $N$ is the midpoint of $AD$.Hence $PN\parallel TC$,whence it follows that $S$ is the midpoint of $AT$. From Ceva's theorem in $\triangle{ATN}$ with respect to $P$,we get $\frac{AS}{ST}\cdot \frac{TM}{NM}\cdot \frac{NF}{AF}=1\Rightarrow \frac{TM}{NM}=\frac{AF}{NF}\Rightarrow MF\parallel AT$, which gives us the desired result.