circumcenter lies on AK, intersection of altitude with bisector and circumcircle

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circumcenter lies on AK, intersection of altitude with bisector and circumcircle

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Source: 2015 SPbU finals , grades 6-9 p3 v2 - Saint Petersburg State University School Olympiad

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#1 h Jan 3, 2021, 9:23 AM

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The angle bisector $AL$ is drawn in triangle $ABC$ . From point $B$ the altitude was drawn on $AL$ , that cuts the side $AL$ at point $H$ , and the circumcribed circle of triangle $ABL$ at point $K$ . Prove that the center of the circumscribed circle of triangle $ABC$ lies on line $AK$ .

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#2 h Jan 4, 2021, 11:56 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.567472173179354, xmax = 14.83309735695828, ymin = -8.148721280388122, ymax = 9.5969940600116; /* image dimensions */ /* draw figures */ draw((0.2,6.46)--(-5.46,-0.86), linewidth(0.8) + blue); draw((-5.46,-0.86)--(3.22,-4.92), linewidth(0.8) + blue); draw((0.2,6.46)--(3.22,-4.92), linewidth(0.8) + blue); draw(circle((0.4334836078837503,0.4312408168549142), 6.033278700964063), linewidth(0.8) + red); draw((0.2,6.46)--(-1.6403202157335612,-2.646624415221399), linewidth(0.8) + blue); draw(circle((-1.7416676072407848,2.1131200351069457), 4.7608233034000325), linewidth(0.8) + red); draw((-5.46,-0.86)--(0.5303744975125011,-2.07057010643378), linewidth(0.8) + blue); draw((0.2,6.46)--(0.5303744975125011,-2.07057010643378), linewidth(0.8) + blue); /* dots and labels */ dot((0.2,6.46),dotstyle); label("$A$", (0.28012578073089145,6.645562324526709), NE * labelscalefactor); dot((-5.46,-0.86),dotstyle); label("$B$", (-5.3814256615514475,-0.6680483910773606), NE * labelscalefactor); dot((3.22,-4.92),dotstyle); label("$C$", (3.287244907451347,-4.733227951273531), NE * labelscalefactor); dot((-1.6403202157335612,-2.646624415221399),linewidth(4pt) + dotstyle); label("$L$", (-1.5575581300427201,-2.505732301850972), NE * labelscalefactor); dot((-1.4433052518697453,-1.6717172959345352),linewidth(4pt) + dotstyle); label("$H$", (-1.3719334925908402,-1.5219217233560083), NE * labelscalefactor); dot((0.5303744975125011,-2.07057010643378),linewidth(4pt) + dotstyle); label("$K$", (0.5956876643990874,-1.930295925750144), NE * labelscalefactor); dot((0.4334836078837503,0.4312408168549142),linewidth(4pt) + dotstyle); label("$O$", (0.5028753456731474,0.5756366798502349), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Quick one

$\color{black}\rule{25cm}{1pt}$
Let $\alpha = \angle BAC$ and let $\beta = \angle CBA$ .
We have that $\angle HBA = 90-\frac{1}{2}\alpha$ . This implies that $\angle LBK = \beta + \frac{1}{2}\alpha-90$
Since $\angle LBK = \angle LAK$ , we have that:
$\angle CAK = \angle CAL - \angle KAL = 90-\beta = \angle CAO$ thus we have that $O$ is on $AK$ , where $O$ is the circumcenter of $ABC$ .

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