2015 JBMO Shortlist G3

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Source: 2015 JBMO Shortlist G3

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30630 posts

#1 h Oct 8, 2017, 10:27 AM • 3 Y

Y by Adventure10, Mango247, lian_the_noob12

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)

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#2 h Oct 8, 2017, 10:33 AM • 2 Y

Y by Adventure10, Mango247

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.

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#3 h Sep 2, 2018, 11:01 PM • 2 Y

Y by Adventure10, Mango247

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 $\triangle ACO$ . $\blacksquare$

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#4 h Sep 2, 2018, 11:30 PM • 2 Y

Y by Adventure10, Mango247

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. $\square$

This post has been edited 2 times. Last edited by TheUltimate123, Sep 2, 2018, 11:30 PM

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#5 h May 29, 2020, 3:30 AM

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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.

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#8 h Sep 1, 2020, 3:13 PM

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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.

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Math_legendno12

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Math_legendno12

#9 h Jul 10, 2023, 1:58 AM

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This is too easy for a G3

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