olymmath

PAMO 2015

Pan-African Mathematics Olympiad 2015.

Day 1

Problem 1

  • Day 1

Prove that x−1+2x+9+19−3x<9\sqrt{x-1}+\sqrt{2x+9}+\sqrt{19-3x}<9 for all real xx for which the left-hand side is well defined.

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Problem 2

  • Day 1

A convex hexagon ABCDEFABCDEF is such that AB=BCCD=DEEF=FAAB=BC CD=DE EF=FA and ∠ABC=2∠AEC∠CDE=2∠CAE∠EFA=2∠ACE\angle ABC=2\angle AEC \angle CDE=2\angle CAE \angle EFA=2\angle ACE Show that ADAD , CFCF and EBEB are concurrent.

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Problem 3

  • Day 1

Let a1,a2,...,a11a_1,a_2,...,a_{11} be integers. Prove that there are numbers b1,b2,...,b11b_1,b_2,...,b_{11} , each bib_i equal −1,0-1,0 or 11 , but not all being 00 , such that the number N=a1b1+a2b2+...+a11b11N=a_1b_1+a_2b_2+...+a_{11}b_{11} is divisible by 20152015 .

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Day 2

Problem 4

  • Day 2

For a positive integer nn denote d(n)d(n) its greatest odd divisor. Find the value of the sum d(1008)+d(1009)+...+d(2015)d(1008)+d(1009)+...+d(2015)

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Problem 5

  • Day 2

There are seven cards in a hat, and on the card kk there is a number 2k−12^{k-1} , k=1,2,...,7k=1,2,...,7 . Solarin picks the cards up at random from the hat, one card at a time, until the sum of the numbers on cards in his hand exceeds 124124 . What is the most probable sum he can get?

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Problem 6

  • Day 2

Let ABCDABCD be a quadrilateral (with non-perpendicular diagonals).

The perpendicular from AA to BCBC meets CDCD at KK . The perpendicular from AA to CDCD meets BCBC at LL . The perpendicular from CC to ABAB meets ADAD at MM . The perpendicular from CC to ADAD meets ABAB at NN .

  1. Prove that KLKL is parallel to MNMN .
  2. Prove that KLMNKLMN is a parallelogram if ABCDABCD is cyclic.

Answer, solutions and source