Let ${c\equiv c\left(O, R\right)}$ be a circle with center ${O}$ and radius ${R}$ and ${A, B}$ be two points on it, not belonging to the same diameter. The bisector of angle${\angle{ABO}}$ intersects the circle ${c}$ at point ${C}$, the circumcircle of the triangle $AOB$ , say ${c_1}$ at point ${K}$ and the circumcircle of the triangle $AOC$ , say ${{c}_{2}}$ at point ${L}$. Prove that point ${K}$ is the circumcircle of the triangle $AOC$ and that point ${L}$ is the incenter of the triangle $AOB$. Evangelos Psychas (Greece)
Problem
Source: 2015 JBMO Shortlist G3
Tags: geometry, JBMO
navi_09220114
08.10.2017 13:33
suffices to prove KA=KC=KO. KCA=AOB/2=AKB/2 => KA=KC. KA=KO is obvious since K is midpoint of minor arc OA in (AOB). It is also a well known lemma that L is the incenter.
Jzhang21
03.09.2018 02:01
Note that by construction, $\angle LBO=\angle LBA$ and $LACO$ is cyclic. Since $OB=OC$, $\angle LBO=\angle LCO=\angle LAO.$ Because $OA=OB$, $\angle LAO+\angle LAB=\angle OAB=\angle OBA=\angle LBO+\angle LBA \implies \angle LAB=\angle LAO=\angle LBO=\angle LBA$ so $L$ is the incenter of $\triangle ABO.$ Note that $AKOB$ is cyclic and $\angle AOK=\angle ABK=\angle KBO=\angle OAK \implies AK=KO.$ Also, $AO=OC$ so $\angle OCK+\angle ACK=\angle ACO=\angle CAO=\angle OAK+\angle KAC$. Since $\angle OCK=\angle OBK=\angle KAO$, then $\angle CAK=\angle KCA \implies KA=KC$. Hence, $KA=KO=KC$ so $K$ is the circumcenter of $\triangle ACO$. $\blacksquare$
TheUltimate123
03.09.2018 02:30
First, notice that$$\measuredangle OAL=\measuredangle OCL=\measuredangle OBL\implies\measuredangle LAB=\measuredangle ABL,$$so $L$ is the intersection of the perpendicular bisector of $\overline{AB}$ and the angle bisector of $\angle ABO$, which is the incenter as $\triangle OAB$ is isosceles. Then, by the Incenter-Excenter Lemma, $(AOC)=(AOL)$ has center $K$, the midpoint of $\widehat{AO}$, so we are done. $\square$
mathlogician
29.05.2020 06:30
Notice that $\angle COA = 2\angle CBA = 2\angle KBA = 2\angle KOA,$ so $K$ is on the angle bisector of $\angle COA.$ But since $OA=OC,$ $K$ is also on the perpendicular bisector of $AC.$ But also note that $\angle KAO = \angle KBO = \angle KBA = \angle KOA,$ so $AK=AO$ and $K$ lies on the perpendicular bisector of $AO,$ which implies the desired result for (a). For (b), note that $K$ is the intersection of $BL$ with $(AOB),$ and also that $AK=KL=KO$ by part (a), so $L$ is the incenter.
Com10atorics
01.09.2020 18:13
We not that $K$ is the midpoint of arc $AO$ not containing $B$ so we want to prove $KA=KC$.Now we have, $\angle AKC=180-\angle AKB=180-\angle AOB$ $\angle KCA=\angle BCA=\tfrac{1}{2} \angle AOB$ So $2\angle KCA+\angle AKC=180$ which proves it. So $C$ is the $B-excenter$ of $\triangle AOB$ and $L$ is obviously gonna be the incenter of this triangle.
Math_legendno12
10.07.2023 04:58
This is too easy for a G3