Three circles, 5 points of intersection, prove that two circles are tangent There are 3 circles and 5 points of intersection (point of intersection is a point where at least 2 circles meet). Prove that two circles are tangent (it means that they intersect in a single point)
 A: Let $C_1,C_2,C_3$ be the circles. Suppose, towards contradiction, that there is no two of the circles are tangent to each other, that is no two circles intersecting in exactly one point. Since in total we have $5$ intersection points, $C_1,C_2$ have to intersect in two points, say $p_1,p_2$. Furthermore, to accomplish the total $5$ intersection points, $C_3$ must intersect with $C_1,C_2$ in $3$ points other than $p_1,p_2$. Without loss of generality, these three points, say $p_3,p_4,p_5$, are distributed as $p_3,p_4$ are in $C_1$ and $p_5$ is in $C_2$. Now, $C_3$ must contain $p_1$ or $p_2$, otherwise, it is tangent to $C_2$. However, $C_3$ will then intersect with $C_1$ in $3$ points, which is impossible. Hence, the result follows.
A: Three circle make 6 points of intersection when no two circles have a common tangent . If two circles make internal tangential contact even then there are 6 points, however these are 4 single points and one double point at the point of tangency as shown.
If it was required to prove that there are three points of intersection, it should make external contact with one tangent double point among the four points.

