Let the circumcircle of a triangle be . The tangents to at meet at point . For a point on line which is not on the segment , let the midpoint of be . Lines meet again at points respectively. Let be the midpoint of . Prove that the points lie on a circle.

Proposed by Mehmet Can Baştemir

Let be a cyclic quadrilateral with . Let be the midpoint of the arc not containing . Suppose there is a point inside such that and .

Prove that lines , and are concurrent.

Passing memes between social platforms (and posting in October)...

Let be the set of rooted binary trees, where we include the empty tree with 0 nodes. There is an obvious bijection between nonempty trees in and pairs of trees in , by taking the left and right subtree of . So there is an obvious bijection between and {the empty tree, plus pairs of trees in }.

Let's write this as , so . In other words, is a sixth root of unity.

So . Unfortunately this is absurd because there is more than one 6-tuple of trees.

But, let's try again and write . And in fact:

• There is a "very explicit" bijection between the set of 7-tuples of trees, and the set of all trees. Here "very explicit" means that to map 7 trees to 1, you just need to inspect the 7 trees down to a fixed finite depth, independent of what the trees actually are, to construct some part of the 1 tree. Then you can paste the subtrees beneath that depth to subtrees of what you constructed.
• This is tight: there exists a very explicit bijection between the set of -tuples of trees and trees iff .

Now the question is, of course: how did treating , an infinite set of combinatorial objects, as a complex number produce a reasonable theorem?

It turns out there's some ring theory you can pull to see why this is the case. Some arXiv papers on the subject: https://arxiv.org/pdf/math/9405205v1.pdf https://arxiv.org/abs/math/0212377

Consider where you adjoin infinitely many free variables. This has infinite Krull dimension because ideals are all prime. It's also local; the ideal generated by all the variables is the unique maximal ideal.

Consider ; think of like an infinity. It is local because is the unique maximal ideal. The ideal generated by all is prime; that maximal ideal divides it infinitely many times.

Consider , the ring of polynomials with integer constant coefficient. Each integer prime or prime integer or something generates a maximal ideal . The intersection of those infinitely many maximal ideals is the prime ideal .

Let be a scalene triangle with orthocenter and circumcenter . Denote by , the midpoints of , . Suppose the circle with diameter meets the circumcircle of at , and meets line at a point . The tangent to at meets line at . Show that the circumcircles of and intersect at a point on .

Proposed by Evan Chen

Suppose that is the midpoint of the arc , containing , in the circumcircle of , and let be the -excircle's center. Assume that the external angle bisector of intersects at . Prove that is perpendicular to , where is the incenter of .

Let be a diameter of a circle , and let be the tangent at . Furthermore, let be a fixed, positive real, and consider all pairs of points and lying on , on opposite sides of , such that . The lines and intersect at points and , respectively. Show that all the lines pass through a common point.