Let $C_1(O_1)$ and $ C_2(O_2)$ be two circles such that $C_1$ passes through $O_2$. Point $M$ lies on $C_1$ such that $M \notin O_1O_2$. The tangents from $M$ at $O_2$ meet again $C_1$ at $A$ and $B$. Prove that the tangents from $A$ and $B$ at $C_2$ - others than $MA$ and $MB$ - meet at a point located on $C_1$.
Problem
Source: 2002 Romania JBMO TST4 p3
Tags: geometry, circles, Tangents, concurrent, concurrency
parmenides51
14.08.2024 15:51
If you cannot solve a specific problem., solve an equivalent problem equivalent problem Quote: Let $C_1(O_1)$ and $ C_2(O_2)$ be two intersecting circles. Point $M$ lies on $C_1$ such that $M \notin O_1O_2$. The tangents from $M$ at $O_2$ meet again $C_1$ at $A$ and $B$. Let the tangents from $A$ and $B$ at $C_2$ - others than $MA$ and $MB$ meet at a point $N$ on $C_2$ . Proce that point $C_2$ passes through $O_1$.
$NA, NB$ are tangents to $C_1$ so $O_1$ lies on angle bisector of $\angle ANB$
$MA, MB$ are tangents to $C_1$ so $O_1$ lies on angle bisector of $\angle AMB$
So $O_1$ lies on intersection of angle bisectors of $\angle ANB$ and $\angle AMB$ (1)
But angle bisector of $\angle ANB$ cuts $C_2$ at arc midpoint of arc $AB$ (that doesn't contain $N$)
Similarly angle bisector of $\angle AMB$ cuts $C_2$ at arc midpoint of arc $AB$ (that doesn't contain $N$)
So intersection of these two angles bisectors is arc midpoint of $AB$ of $ C_2$ (that doesn't contain $N$) and according to (1)
that arc midpoint is $O_1$, so $O_1$ as an arc midpoint of $ C_2$, lies on $C_2$