Given is a triangle $\triangle{ABC}$,with $AB<AC<BC$,inscribed in circle $c(O,R)$.Let $D,E,Z$ be the midpoints of $BC,CA,AB$ respectively,and $K$ the foot of the altitude from $A$.At the exterior of $\triangle{ABC}$ and with the sides $AB,AC$ as diameters,we construct the semicircles $c_1,c_2$ respectively.Suppose that $P\equiv DZ\cap c_1 \ , \ S\equiv KZ\cap c_1$ and $R\equiv DE\cap c_2 \ , \ T\equiv KE\cap c_2$.Finally,let $M$ be the intersection of the lines $PS,RT$. i. Prove that the lines $PR,ST$ intersect at $A$. ii. Prove that the lines $PR\cap MD$ intersect on $c$. [asy][asy]import graph; size(8cm); 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 = -4.8592569519241255, xmax = 12.331775417316715, ymin = -3.1864435704043403, ymax = 6.540061585876658; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((0.6699432366054657,3.2576036755978928)--(0.,0.)--(5.,0.)--cycle, aqaqaq); /* draw figures */ draw((0.6699432366054657,3.2576036755978928)--(0.,0.), uququq); draw((0.,0.)--(5.,0.), uququq); draw((5.,0.)--(0.6699432366054657,3.2576036755978928), uququq); draw(shift((0.33497161830273287,1.6288018377989464))*xscale(1.662889476749906)*yscale(1.662889476749906)*arc((0,0),1,78.3788505217281,258.3788505217281)); draw(shift((2.834971618302733,1.6288018377989464))*xscale(2.7093067970187343)*yscale(2.7093067970187343)*arc((0,0),1,-36.95500560847834,143.0449943915217)); draw((0.6699432366054657,3.2576036755978928)--(0.6699432366054657,0.)); draw((-0.9938564482532047,2.628510486065423)--(2.5,0.)); draw((0.6699432366054657,0.)--(0.,3.2576036755978923)); draw((0.6699432366054657,0.)--(5.,3.257603675597893)); draw((2.5,0.)--(3.3807330143335355,4.282570444700163)); draw((-0.9938564482532047,2.628510486065423)--(2.5,4.8400585427926455)); draw((2.5,4.8400585427926455)--(5.,3.257603675597893)); draw((-0.9938564482532047,2.628510486065423)--(3.3807330143335355,4.282570444700163), linewidth(1.2) + linetype("2 2")); draw((0.,3.2576036755978923)--(5.,3.257603675597893), linewidth(1.2) + linetype("2 2")); draw(circle((2.5,1.18355242571055), 2.766007292905304), linewidth(0.4) + linetype("2 2")); draw((2.5,4.8400585427926455)--(2.5,0.), linewidth(1.2) + linetype("2 2")); /* dots and labels */ dot((0.6699432366054657,3.2576036755978928),linewidth(3.pt) + dotstyle); label("$A$", (0.7472169504504719,2.65), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.2,-0.4), NE * labelscalefactor); dot((5.,0.),linewidth(3.pt) + dotstyle); label("$C$", (5.028818057451246,-0.34281415594345044), NE * labelscalefactor); dot((2.5,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.4275434226319077,-0.32665717063401356), NE * labelscalefactor); dot((2.834971618302733,1.6288018377989464),linewidth(3.pt) + dotstyle); label("$E$", (3.073822835009383,1.5637101105701008), NE * labelscalefactor); dot((0.33497161830273287,1.6288018377989464),linewidth(3.pt) + dotstyle); label("$Z$", (0.003995626216375389,1.402140257475732), NE * labelscalefactor); dot((0.6699432366054657,0.),linewidth(3.pt) + dotstyle); label("$K$", (0.6179610679749769,-0.3105001853245767), NE * labelscalefactor); dot((-0.9938564482532047,2.628510486065423),linewidth(3.pt) + dotstyle); label("$P$", (-1.0785223895158957,2.7916409940873033), NE * labelscalefactor); dot((0.,3.2576036755978923),linewidth(3.pt) + dotstyle); label("$S$", (-0.14141724156855653,3.454077391774215), NE * labelscalefactor); dot((5.,3.257603675597893),linewidth(3.pt) + dotstyle); label("$T$", (5.061132028070119,3.3571354799175936), NE * labelscalefactor); dot((3.3807330143335355,4.282570444700163),linewidth(3.pt) + dotstyle); label("$R$", (3.445433497126431,4.375025554412117), NE * labelscalefactor); dot((2.5,4.8400585427926455),linewidth(3.pt) + dotstyle); label("$M$", (2.5567993051074027,4.940520040242407), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]
Problem
Source: 2016 Greece Team Selection Test,Problem 2
Tags: geometry
23.07.2016 13:56
i. Let $C_1, C_2$ be circles with $AB, AC$ as diameters, respectively. Note that $K\in C_1, C_2\Rightarrow SK,TK$ are diameters of $C_1, C_2$, respectively. $\Rightarrow AS//KB$ and $AT//KC$, which implies $A\in ST$ $\angle{PAB}=\frac12(180^o-\angle{PZA})=\frac12(180^o-\angle{BAC})$ Similarly, $\angle{RAC}=\frac12(180^o-\angle{BAC})$ $\Rightarrow \angle{PAR}=\angle{PAB}+\angle{BAC}+\angle{RAC}=180^o$, which implies $A\in PR$ $\Rightarrow$ lines $PR,ST$ intersect at $A$ ii.Let $H$ be the midpoint of arc $BAC$ $\angle{AHD}=\frac12\angle{BAC}+\angle{BCA}$ $\angle{PAK}=(90^o-\angle{ABC})+\frac12(180^o-\angle{BAC})=180^o-\angle{ABC}-\frac12\angle{BAC}=\frac12\angle{BAC}+\angle{BCA}$ $\Rightarrow P,A,H$ are colinear $\Rightarrow P,H,Q$ are colinear $\angle{MST}=\angle{PBA}=\frac12\angle{BAC}=\angle{RCA}=\angle{MTS}$ $\Rightarrow$ triangle $MTS$ is isosceles $\Rightarrow$ triangle $MBC$ is isosceles $\Rightarrow MD\perp BC$ $\Rightarrow H\in MD$ Hence $PR\cap MD$ intersect on $c$
14.06.2019 14:18
Beautiful Problem!! gavrilos wrote: Given is a triangle $\triangle{ABC}$,with $AB<AC<BC$,inscribed in circle $c(O,R)$.Let $D,E,Z$ be the midpoints of $BC,CA,AB$ respectively,and $K$ the foot of the altitude from $A$.At the exterior of $\triangle{ABC}$ and with the sides $AB,AC$ as diameters,we construct the semicircles $c_1,c_2$ respectively.Suppose that $P\equiv DZ\cap c_1 \ , \ S\equiv KZ\cap c_1$ and $R\equiv DE\cap c_2 \ , \ T\equiv KE\cap c_2$.Finally,let $M$ be the intersection of the lines $PS,RT$. i. Prove that the lines $PR,ST$ intersect at $A$. ii. Prove that the lines $PR\cap MD$ intersect on $c$. Solution: Obviously, $BSAK$, $KATC$ and $BSTC$ are rectangles. Also from here, $PD,$ $A-$bisector and $\odot (ABK)$ concur, say at $G$ and $AG, DR$ concur on $\odot (AKC)$ $\implies$ $PR$ is the exterior angle bisector of $\angle BAC$ $\implies$ $PR$, $ST$ concur at $A$ $\implies$ $\angle MTS$ $=$ $\angle \frac{1}{2} \angle BAC$ $=$ $\angle GAC$ $=$ $\angle PGA$ $=$ $\angle MST$ $\implies$ $M$ lies on perpendicular bisector of $BC$ $\implies$ hence, $PR$, $MD$ concure on $\odot (ABC)$ at the midpoint of arc $BAC$