olymmath

OMCC 2012

Centroamerican and Caribbean Mathematical Olympiad 2012.

Day 1

Problem 2

  • Day 1

Let γ\gamma be the circumcircle of the acute triangle ABCABC . Let PP be the midpoint of the minor arc BCBC . The parallel to ABAB through PP cuts BC,ACBC, AC and γ\gamma at points R,SR,S and TT , respectively. Let K≡AP∩BTK \equiv AP \cap BT and L≡BS∩ARL \equiv BS \cap AR . Show that KLKL passes through the midpoint of ABAB if and only if CS=PRCS = PR .

Answer, solutions and source

Problem 3

  • Day 1

Let a,b,ca,b,c be real numbers that satisfy 1a+b+1b+c+1a+c=1\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} =1 and ab+bc+ac>0ab+bc+ac >0 .

Show that a+b+c−abcab+bc+ac≥4a+b+c - \frac{abc}{ab+bc+ac} \ge 4

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Trilandia is a very unusual city. The city has the shape of an equilateral triangle of side lenght 2012. The streets divide the city into several blocks that are shaped like equilateral triangles of side lenght 1. There are streets at the border of Trilandia too. There are 6036 streets in total. The mayor wants to put sentinel sites at some intersections of the city to monitor the streets. A sentinel site can monitor every street on which it is located. What is the smallest number of sentinel sites that are required to monitor every street of Trilandia?

Attachments:

trilandia.png

Answer, solutions and source

Problem 5

  • Day 2

Alexander and Louise are a pair of burglars. Every morning, Louise steals one third of Alexander's money, but feels remorse later in the afternoon and gives him half of all the money she has. If Louise has no money at the beginning and starts stealing on the first day, what is the least positive integer amount of money Alexander must have so that at the end of the 2012th day they both have an integer amount of money?

Answer, solutions and source

Problem 6

  • Day 2

Let ABCABC be a triangle with AB<BCAB < BC , and let EE and FF be points in ACAC and ABAB such that BF=BC=CEBF = BC = CE , both on the same halfplane as AA with respect to BCBC .

Let GG be the intersection of BEBE and CFCF . Let HH be a point in the parallel through GG to ACAC such that HG=AFHG = AF (with HH and CC in opposite halfplanes with respect to BGBG ). Show that ∠EHG=∠BAC2\angle EHG = \frac{\angle BAC}{2} .

Answer, solutions and source