IMO 2017 Day 2

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

Day 1

Problem 4

Let RR and SS be different points on a circle Ω\Omega such that RSRS is not a diameter. Let \ell be the tangent line to Ω\Omega at RR. Point TT is such that SS is the midpoint of the line segment RTRT. Point JJ is chosen on the shorter arc RSRS of Ω\Omega so that the circumcircle Γ\Gamma of triangle JSTJST intersects \ell at two distinct points. Let AA be the common point of Γ\Gamma and \ell that is closer to RR. Line AJAJ meets Ω\Omega again at KK. Prove that the line KTKT is tangent to Γ\Gamma.

Problem 5

An integer N>2N > 2 is given. A collection of N(N+1)N(N + 1) soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove N(N1)N(N - 1) players from this row leaving a new row of 2N2N players in which the following NN conditions hold:

(11) no one stands between the two tallest players,

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




(NN) no one stands between the two shortest players.

Show that this is always possible.

Problem 6

An ordered pair (x,y)(x, y) of integers is a primitive point if the greatest common divisor of xx and yy is 11. Given a finite set SS of primitive points, prove that there exist a positive integer nn and integers a0,a1,,ana_0, a_1, \ldots , a_n such that, for each (x,y)(x, y) in SS, we have:

a0xn+a1xn1y+a2xn2y2++an1xyn1+anyn=1.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.

Note by Sharky Kesa
4 years ago

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How I solved P4:

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

Firstly, we prove ATRKAT \parallel RK. Since SRK=SJK=STA\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 RKRK intersect the circumcircle of STKSTK at BB. We will show BTARBTAR is a parallelogram. SInce ARS=BKR\angle ARS = \angle BKR by alternate segment theorem, and STB=BKR\angle STB = \angle BKR, we have ARS=STB\angle ARS = \angle STB, so ARBTAR \parallel BT, so BTARBTAR is a parallelogram.

Since SS is the midpoint of RTRT, we have A,S,BA, S, B collinear. Thus, TAB=SBK=STK\angle TAB = \angle SBK = \angle STK, it follows KTKT is tangent to Γ\Gamma by alternate segment theorem.

Sharky Kesa - 4 years ago

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Mine is a bit different than Sharky and here is my solution to problem 44.

Let KSTA=L.KS \cap TA = L. Join RNRN and AS.AS. Now, SRK=SJK=ATS.\angle SRK = \angle SJK = \angle ATS. Therefore , ATRKAT || RK by alt. int. \angle. Now, since RTAKRT || AK and RS=STRS = ST =>KTLR=> KTLR is a gm||gm. =>=> LRT=KTR\angle LRT = \angle KTR 1--1 and TNK=RKL\angle TNK = \angle RKL also, ARS=RKL\angle ARS = \angle RKL - By Alt. Segment Theorem. =>=> ARS=TNK=>ANSR\angle ARS = \angle TNK => ANSR is cyclic =>LAS=LRS2.=> \angle LAS = \angle LRS --2. Therefore, by 11 and 22 we have TAS=LRS=RTK.\angle TAS = \angle LRS = \angle RTK. So done by the Alt. Segment Theorem.

Comments -- As is obvious from the problem, it is a 55 minutes angle chasing. I highly doubt that this question is shortlisted from 19591959 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.

Steven Jim - 4 years ago

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