Let AHA,BHB,CHC be the altitudes of triangle ABC. Prove that if HBCAC=HCAAB, then the line symmetric to BC with respect to line HBHC is tangent to the circumscribed circle of triangle HBHCA. (Proposed by Mykhailo Bondarenko)
Problem
Source: Kyiv City MO 2022 Round 2, Problem 10.3
Tags: geometry, ratio
01.02.2022 01:58
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -18.145825013519765, xmax = 38.53620650714312, ymin = -10.429917850596546, ymax = 23.460905173507374; /* image dimensions */ /* draw figures */ draw((5.0130937395214925,13.662649486922337)--(-0.5,-1.58), linewidth(0.4) + blue); draw((-0.5,-1.58)--(18.4099,-1.72114), linewidth(0.4) + blue); draw((18.4099,-1.72114)--(5.0130937395214925,13.662649486922337), linewidth(0.4) + blue); draw((-0.5,-1.58)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((1.6424475820849858,4.3434576955805815)--(18.4099,-1.72114), linewidth(0.4) + blue); draw((1.6424475820849858,4.3434576955805815)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((-3.9735958511073424,7.196086109835047)--(9.717344654058754,20.240716675458682), linewidth(0.4) + blue); draw((-3.9735958511073424,7.196086109835047)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((9.717344654058754,20.240716675458682)--(1.6424475820849858,4.3434576955805815), linewidth(0.4) + blue); draw(circle((4.973871599608791,8.407677519123352), 5.255118339163495), linewidth(0.4) + linetype("4 4") + red); draw(circle((6.852952801866909,3.6601538735235093), 5.255118339163495), linewidth(0.4) + linetype("4 4") + red); draw((1.6424475820849858,4.3434576955805815)--(6.81373066195421,-1.5948180942754766), linewidth(0.4) + blue); draw((6.81373066195421,-1.5948180942754766)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((5.0130937395214925,13.662649486922337)--(4.934649459696092,3.152705551324365), linewidth(0.4) + blue); draw((1.3488303274367524,12.212325273339477)--(1.6424475820849858,4.3434576955805815), linewidth(0.4) + blue); draw((1.3488303274367524,12.212325273339477)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((1.3488303274367524,12.212325273339477)--(6.81373066195421,-1.5948180942754766), linewidth(0.4) + blue); /* dots and labels */ dot((-0.5,-1.58),dotstyle); label("B", (-0.3494430543560244,-1.217237792734236), NE * labelscalefactor); dot((5.0130937395214925,13.662649486922337),dotstyle); label("A", (5.163365494324635,14.026232825094166), NE * labelscalefactor); dot((18.4099,-1.72114),dotstyle); label("C", (18.55690035756221,-1.365232653101502), NE * labelscalefactor); dot((4.934649459696092,3.152705551324365),linewidth(4pt) + dotstyle); label("H", (5.089368064141002,3.444600308834644), NE * labelscalefactor); dot((10.184376819390716,7.72437369706628),linewidth(4pt) + dotstyle); label("HB", (10.343185607178945,8.03244098021989), NE * labelscalefactor); dot((1.6424475820849858,4.3434576955805815),linewidth(4pt) + dotstyle); label("HC", (1.796482420969333,4.628559191772772), NE * labelscalefactor); dot((-3.9735958511073424,7.196086109835047),dotstyle); label("B′", (-3.827322272986776,7.551457684026277), NE * labelscalefactor); dot((9.717344654058754,20.240716675458682),dotstyle); label("C′", (9.862202310985332,20.612004111437503), NE * labelscalefactor); dot((3.227911529694873,7.4648016277396385),linewidth(4pt) + dotstyle); label("H′", (3.387427169917443,7.773449974577176), NE * labelscalefactor); dot((6.81373066195421,-1.5948180942754766),linewidth(4pt) + dotstyle); label("D", (6.976302533823644,-1.291235222917869), NE * labelscalefactor); dot((1.3488303274367524,12.212325273339477),linewidth(4pt) + dotstyle); label("D′", (1.500492700234801,12.509285506329688), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice geo, Assume that the circle (AHBHC) touches B′C′ the reflection of BC across HBHC and let it touch B′C′ in point D′. When we reflect the circle across HBHC we know that the new circle must touch BC, let it touch in point D. Now notice the following: ∠HCDHB=∠HCD′HB=∠HCAHB=α∠HCHHB=180−α=∠HCH′HBwhere H′ is the reflection of H across HBHC. The two conditions layed out give us that HCH′HBD is cyclic, impliy that H is the orthocenter of △HCHBD and that H′ is the orthocenter of △HCD′HB. This implies that D′,H′,H,D all belong to a line. From here we get that ∠HCHBD=γ and that ∠HBHCD=β, these two imply that HBD∥AB and that HCD∥AC, this implies that AHB=DHC and that DHB=AHC. From the angle conditions we have that HCDB∼HBCD∼ACB, which gives us the following relation: ACAB=HBCHBD=HBCAHCwhich implies the following condtion HBCAC=AHCAB. Thus if the last condition holds, we have the tangency condition.
01.02.2022 06:19
Let D lie on BC s.t. AC//DHC. By the condition, we have DHB//AB. This implies (AHBHC) is symmetric to (DHBHC) wrt HBHC. Since AH⊥BC, we have the tangent at D wrt (DHBHC) is parallel to BC, which implies BC is tangent to (DHBHC). So we're done.