Altitudes $AA_1$ and $BB_1$ of triangle ABC meet in point $H$. Line $CH$ meets the semicircle with diameter $AB$, passing through $A_1, B_1$, in point $D$. Segments $AD$ and $BB_1$ meet in point $M$, segments $BD$ and $AA_1$ meet in point $N$. Prove that the circumcircles of triangles $B_1DM$ and $A_1DN$ touch.
Problem
Source: Sharygin 2011 Final 9.1
Tags: geometry, circumcircle, tangent circles, altitudes, semicircle
GeoMetrix
27.12.2019 20:36
Trivial problem . Personally felt that this was easy for a #1. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -15.24, xmax = 15.24, ymin = -10.01, ymax = 10.01; /* image dimensions */ /* draw figures */ draw((-5.14,7.75)--(-9.58,-2.21), linewidth(1.2)); draw((-9.58,-2.21)--(5.76,-2.35), linewidth(1.2)); draw((5.76,-2.35)--(-5.14,7.75), linewidth(1.2)); draw(circle((-7.36,2.77), 5.452412310161438), linewidth(1.2) + linetype("4 4")); draw((-5.14,7.75)--(-2.1462580857534674,1.174476496058773), linewidth(1.2)); draw((-9.58,-2.21)--(-2.1462580857534674,1.174476496058773), linewidth(1.2)); draw(circle((-2.8903303677570005,-1.2569517056117647), 2.5427321252394575), linewidth(1.2) + linetype("4 4")); draw(circle((-1.4787528626467332,3.355704131435718), 2.281078082882492), linewidth(1.2) + linetype("4 4")); draw((-3.5621919363176793,4.284466128132404)--(0.604686211024213,2.4269421347390323), linewidth(1.2)); draw((-5.212753042963228,-0.22165485039934896)--(-0.5679076925507723,-2.2922485608241785), linewidth(1.2)); draw((-2.1462580857534674,1.174476496058773)--(0.604686211024213,2.4269421347390323), linewidth(1.2)); draw((-2.1462580857534674,1.174476496058773)--(-0.5679076925507723,-2.2922485608241785), linewidth(1.2)); draw((-9.58,-2.21)--(-2.563331219998188,5.3624445249524495), linewidth(1.2)); draw((5.76,-2.35)--(-7.089019132961977,3.37787599903124), linewidth(1.2)); draw((-5.14,7.75)--(-5.231261825685237,-2.2496886143679315), linewidth(1.2)); /* dots and labels */ dot((-5.14,7.75),dotstyle); label("$A$", (-5.06,7.95), NE * labelscalefactor); dot((-9.58,-2.21),dotstyle); label("$B$", (-9.5,-2.01), NE * labelscalefactor); dot((5.76,-2.35),dotstyle); label("$C$", (5.84,-2.15), NE * labelscalefactor); dot((-5.231261825685237,-2.2496886143679315),linewidth(4pt) + dotstyle); label("$A_1$", (-5.16,-2.09), NE * labelscalefactor); dot((-2.563331219998188,5.3624445249524495),linewidth(4pt) + dotstyle); label("$B_1$", (-2.48,5.53), NE * labelscalefactor); dot((-5.187637672090114,2.530272215269085),linewidth(4pt) + dotstyle); label("$H$", (-5.1,2.69), NE * labelscalefactor); dot((-2.1462580857534674,1.174476496058773),linewidth(4pt) + dotstyle); label("$D$", (-2.06,1.33), NE * labelscalefactor); dot((-3.5621919363176793,4.284466128132404),linewidth(4pt) + dotstyle); label("$M$", (-3.48,4.45), NE * labelscalefactor); dot((-5.212753042963228,-0.22165485039934896),linewidth(4pt) + dotstyle); label("$N$", (-5.14,-0.07), NE * labelscalefactor); dot((0.604686211024213,2.4269421347390323),linewidth(4pt) + dotstyle); label("$E$", (0.68,2.59), NE * labelscalefactor); dot((-0.5679076925507723,-2.2922485608241785),linewidth(4pt) + dotstyle); label("$F$", (-0.48,-2.13), NE * labelscalefactor); dot((-7.089019132961977,3.37787599903124),linewidth(4pt) + dotstyle); label("$G$", (-7,3.53), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $AD \cap BC=F$ and let $BD \cap AC=E$. Note that $\angle FDN+\angle FA_1N=90^\circ+90^\circ=180^\circ \implies F \in \odot(DNA_1)$. Similiarly $E \in BMD_1$. Now note that $M$ is the orthocentre of $\Delta EAB \implies EM \perp AB$. Also $N$ is the orthocentre of $\Delta FAB \implies FN \perp AB \implies EM \parallel FN$. So by converse of reims theorem we are done $\blacksquare$.
mathsworm
27.12.2019 21:15
The problem itself is quite easy, but I like the configuration Here's my solution $$\angle DEM=\angle DB_{1}M=\angle DAB=\angle DA_{1}F=\angle DNF$$Hence the triangles $EDM$ and $NDF$ are cyclic. $$\frac{DE}{DN}=\frac{DM}{DF}$$The result follows from a homothety at D.