# Intersecting Circles.

Let $O_1$ , $O_2$ be the centers of circles $C_1$ , $C_2$ in a plane respectively, and the circles meet at two distinct points $A$ , $B$ . Line $O_1$$A meets the circle C_1 at point P_1 , and line O_2$$A$ meets the circle $C_2$ at point $P_2$. Determine the maximum number of points lying in a circle among these 6 points $A$, $B$, $O_1$ , $O_2$ , $P_1$ and $P_2$.

I drew the two circles, but both the points $P_1$ and $P_2$ coincide with the point $A$. Is this method of solving the question correct?

• I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter. – fleablood Jul 28 '18 at 15:25
• Does the question suggest that line $O_1A$ should be extended to meet the circle $C_1$ again at $P_1$ (so that $AP_1$ is a diameter of the circle)? – Malkin Jul 28 '18 at 15:25
• @EMalkin I don't see how it can mean anything else. $O1$ is the center of the $C_1$ and $A$ is also on $C_1$. Any line through a center, which $O_1A$ is intersects the circle exactly twice (forming a diameter). The implication is $A$ is a different point than $P_1$ so it must be the other point. Even if $A$ could be the same as $P_1$ the question asks for the maximum number of $6$ possible points so it seems we must consider the points being different to have more of them. – fleablood Jul 28 '18 at 15:29

# Analysis & Solution

Notice that $$\angle ABP_1+\angle ABP_2=90^o+90^o=180^o$$. Hence, $$P_1, B, P_2$$ are collinear. Thus, we have three groups of collinear points, which are $$(A, O_1, P_1), (A, O_2, P_2)$$ and $$(P_1, B,P_2).$$

On one hand，we may claim that, no matter what $$5$$ points we pick among the given $$6$$ points, or we pick all of them, there necessarily exists at least one group of collinear points, which is not concylic. For this reason, we obtain the fact,there exists no more than $$4$$ concyclic points among the given $$6$$ points.

One the other hand, we may pick $$4$$ points such that they are concylic, for example, we want $$O_1, O_2, P_1, P_2$$ are concylic. For this purpose, we only need $$c_1$$ and $$c_2$$ are equal circles. (Can you prove this?)

Now, we may conclude, the maximum number of points lying in a circle among the six is $$4.$$

If the correction as described in the comments is taken to be true, then you can prove that $P_1, B$ and $P_2$ are collinear. (Sketch it out and first prove that $ABP_1$ is a right angle.)

After proving this, you can reason that a maximum of four of the six points can lie on a circle.

• angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ... – Math Tise Jul 29 '18 at 2:00
• That's correct! Easy peasy. – Malkin Jul 29 '18 at 17:54