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Difference between revisions of "2018 JBMO Problems/Problem 4"

(Created page with "== Problem == Let <math>\triangle ABC</math> and <math>A'</math>,<math>B'</math>,<math>C'</math> the symmetrics of vertex over opposite sides.The intersection of the circumci...")
 
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Latest revision as of 19:40, 2 February 2019

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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