Let $ABC$ be an acute,non-isosceles triangle with $AB<AC<BC$.Let $D,E,Z$ be the midpoints of $BC,AC,AB$ respectively and segments $BK,CL$ are altitudes.In the extension of $DZ$ we take a point $M$ such that the parallel from $M$ to $KL$ crosses the extensions of $CA,BA,DE$ at $S,T,N$ respectively (we extend $CA$ to $A$-side and $BA$ to $A$-side and $DE$ to $E$-side).If the circumcirle $(c_{1})$ of $\triangle{MBD}$ crosses the line $DN$ at $R$ and the circumcirle $(c_{2})$ of $\triangle{NCD}$ crosses the line $DM$ at $P$ prove that $ST\parallel PR$.
Problem
Source: Greek TST 2014-Pr.3.
Tags: geometry, circumcircle, trapezoid, geometry proposed
thugzmath10
26.06.2014 06:28
Since $D, E, Z$ are the midpoints of $BC, AC, AB$ we have $DE\parallel BA$ and $DZ\parallel CA$. Thus, we have $\angle{EDC}=\angle{ABC}=\angle{AKL}=\angle{AST}$, so $S, N, C, D$ are concyclic. Similarly, we have $\angle{ZDB}=\angle{ACB}=\angle{ALK}=\angle{ATS}$, so $T, M, B, D$ are concyclic. Thus, by POP, we have $PN\cdot DN=NT\cdot NM$ and $MR\cdot DM=MS\cdot MN$, which gives $\frac{PN}{MR}\cdot\frac{DN}{DM}=\frac{NT}{MS}$.
Observe that $DE\parallel BA\implies ZT\parallel DN$ and $DZ\parallel CA\implies ES\parallel DM$, so by Thales' Theorem we have $\frac{MT}{MN}=\frac{MZ}{MD}\implies \frac{NT}{MN}=\frac{ZD}{MD}$ and $\frac{NS}{NM}=\frac{NE}{ND}\implies \frac{MS}{NM}=\frac{DE}{DN}$. Thus we get $\frac{NT}{MS}=\frac{ZD}{DE}\cdot\frac{DN}{DM}$, which follows that $\frac{PN}{MR}=\frac{ZD}{DE}$.
Furthermore, we have $\angle{ABC}=\angle{AST}=\angle{NMD}$ and $\angle{ACB}=\angle{ATS}=\angle{MND}$, so $\triangle{ABC}\sim\triangle{DMN}$ by AA Similarity. Thus we have $\frac{AC}{AB}=\frac{DN}{DM}$. This implies that $\frac{PN}{MR}=\frac{ZD}{DE}=\frac{2ZD}{2DE}=\frac{AC}{AB}=\frac{DN}{DM}$. We then have $\frac{PN}{DN}=\frac{MR}{DM}\implies \frac{DP}{DN}=\frac{DR}{DM}$, so therefore $PR\parallel MN$ and consequently we get $ST\parallel PR$.
Vo Duc Dien
17.08.2014 04:34
Should be simple. $BCKL$ is cyclic and because $ST || LK$, $BCST$ is also cyclic and $\angle ATS = \angle BTM = \angle C = \angle BDM$ which implies that $BDTM$ is also cyclic. Similarly, $CDSN$ is cyclic. Also because $DR || BT$ and $DP || CS$, $TR =BD = CD = SP$ and $\angle BTR = \angle B, \angle MTR = \angle B + \angle C = \angle CDN + \angle C = \angle CSN + \angle CSP = \angle NSP$. Therefore, $STPR$ is an isosceles trapezoid and $ST || PR$.
liberator
22.08.2014 00:41
[asy][asy] /* Free script by liberator */ unitsize(0.7cm); defaultpen(fontsize(11pt)); pen c = rgb(0.9,0,0); pen d = rgb(0.4,0.6,0.8); /* Initialize objects */ pair A = (3,6); pair B = origin; pair C = (10,0); pair D = midpoint(B--C); pair E = midpoint(C--A); pair Z = midpoint(A--B); pair K = foot(B,C,A); pair L = foot(C,A,B); pair M = (-2,6); pair T1 = rotate(180,A)*B; pair T = intersectionpoint(Z--T1, circumcircle(B,D,M)); pair S1 = rotate(180,A)*C; pair S = intersectionpoint(M--T, C--S1); pair N1 = rotate(180,T)*M; pair N2 = rotate(180,N1)*M; pair N = intersectionpoint(T--N2, circumcircle(C,D,S)); pair R = intersectionpoint(D--N, circumcircle(B,D,M)); pair P = intersectionpoint(D--Z, circumcircle(C,D,S)); /* Draw objects */ draw(A--B--C--cycle, d+linewidth(1)); draw(D--E--Z--cycle, d); draw(K--L, d); draw(M--N, d); draw(M--Z, d); draw(N--E, d); draw(A--S, d); draw(A--T, d); draw(P--R, d); draw(T--R, d); draw(S--P, d); draw(circumcircle(B,D,M), c); draw(circumcircle(C,D,S), c); /* Place dots on and label each point */ dot(A); label("$A$", A, dir(100)); dot(B); label("$B$", B, dir(200)); dot(C); label("$C$", C, dir(340)); dot(D); label("$D$", D, dir(-90)); dot(E); label("$E$", E, 1.5*dir(100)); dot(Z); label("$Z$", Z, dir(180)); dot(K); label("$K$", K, dir(40)); dot(L); label("$L$", L, dir(-60)); dot(M); label("$M$", M, dir(180)); dot(N); label("$N$", N, dir(90)); dot(S); label("$S$", S, dir(120)); dot(T); label("$T$", T, dir(90)); dot(P); label("$P$", P, dir(-120)); dot(R); label("$R$", R, dir(0)); [/asy][/asy] $\angle BTM = \angle ALK = \angle BDM$, so $BDTM$ is cyclic. Similarly, $CDSN$ is cyclic. As $DR \parallel BT, DP \parallel CS$, $TR =BD=CD = SP$. Hence $\angle MTR = \angle B + \angle C = \angle CSN + \angle CSP = \angle NSP$. It follows $STRP$ is an isosceles trapezoid, and $ST \parallel PR$.