Acute triangle ABC with AB>BC is inscribed in circle Ω. Points D and E, that lie on (BC) and (AB) are the feet of altitudes from A and C in triangle ABC, and M is the midpoint of the segment DE. Half-line (AM intersects the circle Ω for the second time in N. Show that the circumcenter of triangle MDN lies on the line BC.
Problem
Source: Moldova TST 2021
Tags: geometry, circumcircle
19.09.2021 19:05
[asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A = dir(100), B = dir(220), C = dir(320), H = orthocenter(A,B,C), D = foot(A,B,C), E = foot(C,A,B), M = (E+D)/2, N = intersectionpoint(M--M+dir(A--M)*100,unitcircle), Ha = 2D-H, Hb = 2*foot(B,C,A)-H, K = extension(A,A+dir(D--E),Hb,foot(Hb,B,C)); draw(A--B--C--A); draw(unitcircle); draw(A--Ha^^B--Hb^^C--E, gray); draw(D--E^^A--N); draw(N--Hb, dashed); draw(A--K--Hb); dot("A", A, dir(100)); dot("B", B, dir(220)); dot("C", C, dir(320)); dot("HA", Ha, dir(240)); dot("N", N, dir(230)); dot("D", D, dir(315)); dot("E", E, dir(150)); dot("M", M, dir(180)); dot("HB", Hb, dir(30)); dot("K", K, dir(330)); dot("H", H, dir(60)); [/asy][/asy] Let H be the orthocenter, HA=¯AH∩(ABC), and HB=¯BH∩(ABC). Claim: Let K∈(ABC) be such that ¯AK∥¯DE. Then ¯HBK⊥¯BC. Proof. We have ∡KAHA=∡EDA=∡ABH=∡ABHB=∡AHAHBwhich means AHAKHB is an isosceles trapezoid, implying ¯HBK⊥¯BC. ◻ Now, from −1=(DE;M∞DE)A=(HAB;NK)HB=(HA,H;¯HBN∩¯AH,∞AH)we deduce N, D, HB are collinear. It follows that ∡DNM=∡HBNA=∡HBA=90∘−∡BAC=90∘−∡EDB=90∘−∡MDB,which implies the conclusion.
23.02.2022 07:36
Let's hope I didn't make any mistakes We only need to show that ∡DNM=90−∡C, i.e. N,D,B′ collinear where B′ is the second intersection of BH and (ABC). Let BM∩(ABC)={B,B1}. We know that BB1 is symmedian in ABC. Then, reflection of B1 wrt AC is B−humpty point. So, X,Y,B′,B1 cyclic, where X is the midpoint of AC and Y is the foot of altitude from B to AC, because ∠XYB′=90∘=∠HMBX=∠XB1B′(MB is B−humpty point). Let B′B1∩AC=T. Then TX⋅TY=TB′⋅TB1=TA⋅TCwhich means that (A,C;Y,T)=−1, so DE passes through T. Finally, Pascal Thorem on CANB′B1B shows that N,B′,D collinear.