Discussions of these problems have already started on AoPS. Let's start a few over here. :)
Let and be different points on a circle such that is not a diameter. Let be the tangent line to at . Point is such that is the midpoint of the line segment . Point is chosen on the shorter arc of so that the circumcircle of triangle intersects at two distinct points. Let be the common point of and that is closer to . Line meets again at . Prove that the line is tangent to .
An integer is given. A collection of soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove players from this row leaving a new row of players in which the following conditions hold:
() no one stands between the two tallest players,
() no one stands between the third and fourth tallest players,
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.
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() no one stands between the two shortest players.
Show that this is always possible.
An ordered pair of integers is a primitive point if the greatest common divisor of and is . Given a finite set of primitive points, prove that there exist a positive integer and integers such that, for each in , we have:
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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∥RK. Since ∠SRK=∠SJK=∠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 ∠ARS=∠BKR by alternate segment theorem, and ∠STB=∠BKR, we have ∠ARS=∠STB, so AR∥BT, so BTAR is a parallelogram.
Since S is the midpoint of RT, we have A,S,B collinear. Thus, ∠TAB=∠SBK=∠STK, it follows KT is tangent to Γ by alternate segment theorem.
Mine is a bit different than Sharky and here is my solution to problem 4.
Let KS∩TA=L. Join RN and AS. Now, ∠SRK=∠SJK=∠ATS. Therefore , AT∣∣RK by alt. int. ∠. Now, since RT∣∣AK and RS=ST =>KTLR is a ∣∣gm. => ∠LRT=∠KTR −−1 and ∠TNK=∠RKL also, ∠ARS=∠RKL - By Alt. Segment Theorem. => ∠ARS=∠TNK=>ANSR is cyclic =>∠LAS=∠LRS−−2. Therefore, by 1 and 2 we have ∠TAS=∠LRS=∠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.
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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.