Let $P$ be a point inside an acute triangle $ABC$. Let the lines $BP$ and $CP$ intersect the sides $AC$ and $AB$ at $D$ and $E$, respectively. Let the circles with diameters $BD$ and $CE$ intersect at points $S$ and $T$. Prove that if the points $A$, $S$, and $T$ are colinear, then $P$ lies on a median of $\triangle ABC$.
Problem
Source: 2021 Thailand MO P8
Tags: geometry, power of a point, radical axis
Flash_Sloth
29.11.2021 07:41
Let $X$ the feet from $D$ to $AB$, $Y$ be the feet from $E$ to $AC$, then $A,S,T$ collinear implies that $A$ lies on the radical axis of the two circles, hence by p.o.p $X,Y,C,B$ are concyclic. On the other hand, $X,Y,D,E$ are concyclic. Hence $DE // BC$, implying that $P$ lies on the $A$-median
N1RAV
14.03.2022 06:02
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.25; /* changes label-to-point distance */ pen dps = linewidth(0.35) + fontsize(5); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.75087089900906, xmax = 15.540378998277053, ymin = -7.166713104100008, ymax = 4.756833699627123; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); /* draw figures */ draw((0.14292593947863963,3.875677248981198)--(-3.54,-2.32), linewidth(1.4)); draw((-3.54,-2.32)--(5.48,-2.74), linewidth(1.4)); draw((5.48,-2.74)--(0.14292593947863963,3.875677248981198), linewidth(1.4)); draw((xmin, 0.2647688299948041*xmin-1.382718341818393)--(xmax, 0.2647688299948041*xmax-1.382718341818393), linewidth(1)); /* line */ draw((xmin, -0.3345983548698598*xmin-0.9064010153131684)--(xmax, -0.3345983548698598*xmax-0.9064010153131684), linewidth(1)); /* line */ draw(circle((0.036628253515954556,-1.3730203219902208), 3.6998702915659205), linewidth(1) + blue); draw(circle((1.6140868227496774,-1.4464718108223296), 4.076579432592348), linewidth(1) + blue); draw((xmin, -7.744986277218352*xmin + 4.982636688901803)--(xmax, -7.744986277218352*xmax + 4.982636688901803), linewidth(1) + linetype("4 4")); /* line */ draw((xmin, 21.476195980715772*xmin + 0.806172620745251)--(xmax, 21.476195980715772*xmax + 0.806172620745251), linewidth(1) + linetype("2 2") + red); /* line */ draw((-2.2518263545006456,-0.15294362164465894)--(3.613256507031909,-0.4260406439804419), linewidth(1)); draw((xmin, -0.5944347633802667*xmin + 1.7218046328092795)--(xmax, -0.5944347633802667*xmax + 1.7218046328092795), linewidth(1)); /* line */ draw((xmin, 0.8067313231375155*xmin + 1.663675232797575)--(xmax, 0.8067313231375155*xmax + 1.663675232797575), linewidth(1)); /* line */ draw(circle((0.9972829469161013,-1.9440668673516799), 4.552830137463778), linewidth(1) + qqccqq); draw(circle((0.6807150762656331,-0.28949213281255165), 2.935718777243982), linewidth(1) + qqccqq); /* dots and labels */ dot((0.14292593947863963,3.875677248981198),dotstyle); label("$A$", (0.21010453637888382,4.053270378910304), NE * labelscalefactor); dot((-3.54,-2.32),dotstyle); label("$B$", (-3.863156794077152,-2.7786997617345888), NE * labelscalefactor); dot((5.48,-2.74),dotstyle); label("$C$", (5.597918387118459,-3.204540719010558), NE * labelscalefactor); dot((0.7947003748841812,-1.1723064533638763),dotstyle); label("$P$", (0.876638208635326,-0.9827618114837633), NE * labelscalefactor); dot((3.613256507031909,-0.4260406439804419),linewidth(2pt) + dotstyle); label("$D$", (3.7649507884132425,-0.2051391938493852), NE * labelscalefactor); dot((-2.2518263545006456,-0.15294362164465894),linewidth(2pt) + dotstyle); label("$E$", (-2.60414874648165,0.10961281805024403), NE * labelscalefactor); dot((0.07080022648299775,2.326692160173174),linewidth(2pt) + dotstyle); label("$S$", (-0.2527660693547566,1.8685211198422897), NE * labelscalefactor); dot((-0.2731430021820057,-5.05990002487658),linewidth(2pt) + dotstyle); label("$T$", (-0.19722159666671976,-4.907904548114433), NE * labelscalefactor); dot((0.97,-2.53),linewidth(2pt) + dotstyle); label("$M$", (1.0432716266994366,-2.3898884529173996), NE * labelscalefactor); dot((-0.8404393249619989,2.2213909840785364),linewidth(2pt) + dotstyle); label("$Q$", (-1.0303886869872725,2.553569616329718), NE * labelscalefactor); dot((1.1675546644401351,2.605578152076743),linewidth(2pt) + dotstyle); label("$R$", (1.1358457478461648,2.997925397835077), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $Q$ be second intersection of circle with diameter $BD$ with $AB$ and analogously define $R$. Observe $AQ \cdot AB = AS \cdot AT = AR \cdot AC$, therefore $BQRC$ is concyclic. Also since $BD$ is diameter, $\angle{BQD} = 90^{\circ}$, similarly $\angle{ERC} = 90^{\circ}$. Therefore $EQRD$ is also concyclic. By Converse of Reim's, we get that $DE || BC$. And to finish, use converse of Ceva's Theorem (along with BPT) to get that $P$ lies on the $A$-median