$13$ birds are landing somewhere. Destiny wants that each of the $5$ birds that you choose $4$ of them must stand on a circle. Prove that there at least $6$ birds on the same circle.
I don't know how use the property of the circle in the proof. But I tried something else which is to use the pigeon hole directly but that didn't help.
Can somebody give an explanation on how to tackle this problem?
Let us consider, among all birds, $4$ that are not on the same circle. Let us call these birds $A, B, C, D\,\,$. Within this group, we can identify $4$ possible subsets of $3$ birds ($ABC\,\,$, $ABD\,\,$, $ACD\,\,$, $BCD\,\,$). Each of these subsets identifies a unique circle. Let us trace these $4$ circles.
Now let us consider the groups of $5$ birds that can be obtained by taking $A,B,C,D\,\,$ plus a fifth bird given by any of the remaining $13-4=9\,\ $ ones. In each of these $9$ groups, the fifth bird must necessarily stand on one of the above mentioned $4$ circles.
So we have to place $9$ birds on $4$ circles that already contain $3$ birds each. Even if we homogeneously distribute the first $8$ of these birds among the $4$ circles (i.e. $2$ birds for each circle), we are left with $4$ circles containing $5$ birds each, and one last bird to be placed on one of these. Thus, there must necessarily be a circle that contains at least $6$ birds.