Discussions of these problems have already started on AoPS. Let's start a few over here. :)

Let \(R\) and \(S\) be different points on a circle \(\Omega\) such that \(RS\) is not a diameter. Let \(\ell\) be the tangent line to \(\Omega\) at \(R\). Point \(T\) is such that \(S\) is the midpoint of the line segment \(RT\). Point \(J\) is chosen on the shorter arc \(RS\) of \(\Omega\) so that the circumcircle \(\Gamma\) of triangle \(JST\) intersects \(\ell\) at two distinct points. Let \(A\) be the common point of \(\Gamma\) and \(\ell\) that is closer to \(R\). Line \(AJ\) meets \(\Omega\) again at \(K\). Prove that the line \(KT\) is tangent to \(\Gamma\).

An integer \(N > 2\) is given. A collection of \(N(N + 1)\) soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove \(N(N - 1)\) players from this row leaving a new row of \(2N\) players in which the following \(N\) conditions hold:

(\(1\)) no one stands between the two tallest players,

(\(2\)) no one stands between the third and fourth tallest players,

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(\(N\)) no one stands between the two shortest players.

Show that this is always possible.

An ordered pair \((x, y)\) of integers is a primitive point if the greatest common divisor of \(x\) and \(y\) is \(1\). Given a finite set \(S\) of primitive points, prove that there exist a positive integer \(n\) and integers \(a_0, a_1, \ldots , a_n\) such that, for each \((x, y)\) in \(S\), we have:

\[a_0x^n + a_1x^{n-1} y + a_2x^{n-2}y^2 + \cdots + a_{n-1}xy^{n-1} + a_ny^n = 1.\]

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TopNewestMine is a bit different than Sharky and here is my solution to problem \(4\).

Let \(KS \cap TA = L.\) Join \(RN\) and \(AS.\) Now, \(\angle SRK = \angle SJK = \angle ATS.\) Therefore , \(AT || RK\) by alt. int. \(\angle\). Now, since \(RT || AK\) and \(RS = ST\) \(=> KTLR\) is a \(||gm\). \(=>\) \(\angle LRT = \angle KTR\) \(--1\) and \(\angle TNK = \angle RKL\) also, \(\angle ARS = \angle RKL\) - By Alt. Segment Theorem. \(=>\) \(\angle ARS = \angle TNK => ANSR\) is cyclic \(=> \angle LAS = \angle LRS --2.\) Therefore, by \(1\) and \(2\) we have \(\angle TAS = \angle LRS = \angle RTK.\) So done by the Alt. Segment Theorem.

Comments-- As is obvious from the problem, it is a \(5\) minutes angle chasing. I highly doubt that this question is shortlisted from \(1959\) Romania.[Joke] It's very disappointing especially when there was no geometry in first day.Log in to reply

I remembered seeing a comment saying that "Questions 1 and 4 have become easier in order for people to get the HM". Maybe this will explain.

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How I solved P4:

Firstly, wow, this was an easy geometry problem compared to normal geo questions.

Firstly, we prove \(AT \parallel RK\). Since \(\angle SRK = \angle SJK = \angle STA\), this follows by alternate angles.

Now, the midpoint condition felt weird, so I thought about how I could exploit it. This was through the construction of a parallelogram.

Let \(RK\) intersect the circumcircle of \(STK\) at \(B\). We will show \(BTAR\) is a parallelogram. SInce \(\angle ARS = \angle BKR\) by alternate segment theorem, and \(\angle STB = \angle BKR\), we have \(\angle ARS = \angle STB\), so \(AR \parallel BT\), so \(BTAR\) is a parallelogram.

Since \(S\) is the midpoint of \(RT\), we have \(A, S, B\) collinear. Thus, \(\angle TAB = \angle SBK = \angle STK\), it follows \(KT\) is tangent to \(\Gamma\) by alternate segment theorem.

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