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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## Comments

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TopNewestHow 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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Mine 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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