2018 JBMO Problems/Problem 4

Problem

Let $\triangle ABC$ and $A'$ , $B'$ , $C'$ the symmetrics of vertex over opposite sides.The intersection of the circumcircles of $\triangle ABB'$ and $\triangle ACC'$ is $A_1$ . $B_1$ and $C_1$ are defined similarly.Prove that lines $AA_1$ , $BB_1$ and $CC_1$ are concurent.

Solution

$Lemma:$ $AA_1$ passes through circumcenter $O$ of $\triangle ABC$ .

$Proof of Lemma:$ Since length of $AB'$ is equal to that of $AB$ , it follows that circumcenter of $\triangle ABB', O1$ lies on side $AC$ of $\triangle ABC$ .

Similarly, circumcenter of $\triangle ACC', O2$ lies on side $AB$ of $\triangle ABC$ .

So the radical axis ( $AC$ ) of circles $ABC$ and $ACC'$ passes through $O1$ . Also note that Mid-point $E$ of $AC$ and center of circle ACC' ( $O2$ ) passes through $O$ and is perpendicular to $AO1$ .

Similarly, radical axis ( $AB$ ) of circles $ABC$ and $ABB'$ passes through $O2$ . Also, Mid-point $D$ of $AB$ and center of circle ABB' ( $O1$ ) passes through $O$ and is perpendicular to $AO2$ .

Thus, circumcenter, $O$ of $\triangle ABC$ is the orthocenter of $\triangle AO1O2$ .

Thus, $AO$ is perpendicular to $O1O2$ . Thus, $AO$ passes through $A1$ , implying $AA1$ passes through circumcenter of $\triangle ABC$ .

Now using $Lemma$ above, it can be proven similarly that $BB1$ and $CC1$ also pass through circumcenter of $\triangle ABC$ .

Thus, lines $AA_1$ , $BB_1$ and $CC_1$ are concurrent at circumcenter of $\triangle ABC$ .

$Kris17$

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2018 JBMO Problems/Problem 4

Problem

Solution