Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.

For a positive number \omega , determine the range of \omega for which the curve y = \sin(\omega x) has no inscribed right triangle.
Could someone help me figure out how to approach this?

Find all such that
Proposed by Ngo Van Trang, Vietnam

Find positiv integer numbers s.t. and be positiv integer numbers.

In acute triangle , points and are the feet of the angle bisector and altitude from , respectively. Suppose that and . Compute .

it is given that a and b are coprime to each other and a, b belong to N*

Let be a positive integer.Prove that there is a polynomial with integer coefficients so that ,then.

Find the number of six-digit palindromic numbers that are divisible by .

Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.
NOTE: The probability of each point getting chosen don’t need to be uniform (as long as it is symmetric about the origin), you just need a way to choose points in the infinite plane (such as a normal distribution)

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)

Let be a trapezoid. By we denote the angle bisector of angle . Let and . Prove that is cyclic.

Let be a triangle such that and . Let be the foot of the interior bissector of (its intersection with ).
If calculate the sidelength .