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?

  • $\begingroup$ I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter. $\endgroup$ – fleablood Jul 28 '18 at 15:25
  • $\begingroup$ 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)? $\endgroup$ – Malkin Jul 28 '18 at 15:25
  • $\begingroup$ @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. $\endgroup$ – 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 $4.$

enter image description here


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.