Given an acute-angled triangle ABC with D is the foot of the altitude from A. The perpendicular lines to BC through B and C intersect the altitudes from C and B at points M and N, respectively. Show that AD = BC if and only if A,M,N and D lie on the same circle.
Problem
Source: Colombia 2020
Tags: Columbia, geometry
25.12.2023 00:07
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.114cm); 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 = -12.82, xmax = 17.56, ymin = -6.25, ymax = 8.15; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); draw((-1.42,6.35)--(-7.54,-3.35)--(2.14,-3.33)--cycle, linewidth(0.7) + rvwvcq); draw((-7.115736836841633,-3.349123423216615)--(-7.116613413625019,-2.9248602600582485)--(-7.540876576783385,-2.925736836841634)--(-7.54,-3.35)--cycle, linewidth(0.7)); draw((2.139123423216615,-2.905736836841634)--(1.7148602600582488,-2.906613413625019)--(1.715736836841634,-3.330876576783385)--(2.14,-3.33)--cycle, linewidth(0.7)); draw((-1.400861465175194,-2.91305085520608)--(-1.8251246283335603,-2.913927431989465)--(-1.824248051550175,-3.338190595147831)--(-1.399984888391809,-3.337314018364446)--cycle, linewidth(0.7)); /* draw figures */ draw((-1.42,6.35)--(-7.54,-3.35), linewidth(0.7) + rvwvcq); draw((-7.54,-3.35)--(2.14,-3.33), linewidth(0.7) + rvwvcq); draw((2.14,-3.33)--(-1.42,6.35), linewidth(0.7) + rvwvcq); draw(circle((-2.7072050445591582,1.506817568342667), 5.0113184785097005), linewidth(0.7) + dtsfsf); draw((-7.552676403593336,2.785379339174353)--(-7.54,-3.35), linewidth(0.7) + dbwrru); draw((2.132691503792112,0.2073121646177605)--(2.14,-3.33), linewidth(0.7) + dbwrru); draw((-7.54,-3.35)--(2.132691503792112,0.2073121646177605), linewidth(0.7) + wvvxds); draw((-7.552676403593336,2.785379339174353)--(2.14,-3.33), linewidth(0.7) + wvvxds); draw((-1.42,6.35)--(-1.399984888391809,-3.337314018364446), linewidth(0.7) + wvvxds); /* dots and labels */ dot((-1.42,6.35),dotstyle); label("A", (-1.34,6.55), NE * labelscalefactor); dot((-7.54,-3.35),dotstyle); label("B", (-8.04,-3.29), NE * labelscalefactor); dot((2.14,-3.33),dotstyle); label("C", (2.22,-3.13), NE * labelscalefactor); dot((-1.399984888391809,-3.337314018364446),linewidth(4.pt) + dotstyle); label("D", (-1.18,-3.97), NE * labelscalefactor); dot((2.132691503792112,0.2073121646177605),linewidth(4.pt) + dotstyle); label("N", (2.46,0.53), NE * labelscalefactor); dot((-7.552676403593336,2.785379339174353),linewidth(4.pt) + dotstyle); label("M", (-8.12,2.61), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Toss the figure on the coordinate plane. Let D=(0,0), A=(0,a), B=(b,0), C=(c,0). Since △ABC is acute, we may assume a>0,b<0,c>0. Now, M has x−coordinate equal to b, and is on the line y=ab(x−c), thus M=(b,b(b−c)a). Symmetry in b and c implies that N=(c,c(c−b)a). The circumcentre of a triangle with vertices (0,0), (0,a), (p,q) has coordinates (p2+q2−aq2p,a2). Now, A,M,D,N are concyclic iff △AMD and △AND have the same circumcentre. Therefore, the problem asks us to show that a=c−b if and only if (a2b+b(b−c)2−a2(b−c)2a2,a2)=(a2c+c(c−b)2−a2(c−b)2a2,a2). But this is clear, since the difference of the x−coordinates is 1a2(c−b)((c−b)2−a2) so this is 0 iff c−b=a. (recall that a>0,b<0,c>0)
26.12.2023 07:57
Observes △ADB≅△CBM.Therefore we get ∠BMD=∠MDB=45 which give us ∠MDA=45 Similarly we get ∠NDA=45 which give us ∠MDN=90. Now drop perpendicular from M to AD at P. we will get MP=BD and AP=DC, similary NQ⊥AD which give us AQ=BD,NQ=DC Now as △APM≅△NQA we get ∠MAN=90 hence M,A,N,D cyclic. Second part we get △ADB∼△CBM and △ADC∼△BCN but as ADBC is always constant. We get △BMD∼△CND so ∠BMD=∠CND=∠MDA=∠NDA which give us AD is angle bisector of ∠MDN Now from AM=AN we get AD=BC.