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

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

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

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

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

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30629 posts

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

Y by

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 $AK$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

386 posts

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

Y by

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.567472173179354, xmax = 14.83309735695828, ymin = -8.148721280388122, ymax = 9.5969940600116; /* image dimensions */ /* draw figures */ draw((0.2,6.46)--(-5.46,-0.86), linewidth(0.8) + blue); draw((-5.46,-0.86)--(3.22,-4.92), linewidth(0.8) + blue); draw((0.2,6.46)--(3.22,-4.92), linewidth(0.8) + blue); draw(circle((0.4334836078837503,0.4312408168549142), 6.033278700964063), linewidth(0.8) + red); draw((0.2,6.46)--(-1.6403202157335612,-2.646624415221399), linewidth(0.8) + blue); draw(circle((-1.7416676072407848,2.1131200351069457), 4.7608233034000325), linewidth(0.8) + red); draw((-5.46,-0.86)--(0.5303744975125011,-2.07057010643378), linewidth(0.8) + blue); draw((0.2,6.46)--(0.5303744975125011,-2.07057010643378), linewidth(0.8) + blue); /* dots and labels */ dot((0.2,6.46),dotstyle); label("$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 $ABC$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a