MathDB

2012 May Olympiad

Part of May Olympiad

Subcontests

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6 collinear point, triangle ineq. May Olympiad (Olimpiada de Mayo) 2012 L2 P4

Six points are given so that there are not three on the same line and that the lengths of the segments determined by these points are all different. We consider all the triangles that they have their vertices at these points. Show that there is a segment that is both the shortest side of one of those triangles and the longest side of another.

111 blue + 88 white chips, game May Olympiad (Olimpiada de Mayo) 2012 L2 P4

Pedro has 111111 blue chips and 8888 white chips. There is a machine that for every 1414 blue chips , it gives 1111 white pieces and for every 77 white chips, it gives 1313 blue pieces. Decide if Pedro can achieve, through successive operations with the machine, increase the total number of chips by 3333, so that the number of blue chips equals 53\frac53 of the amount of white chips. If possible, indicate how to do it. If not, indicate why.
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May Olympiad 2012, Level 2, Problem 3

Let ABCABC be a triangle such that ∠ABC=2∠BCA\angle{ABC} = 2\angle{BCA} and ∠CAB>90∘\angle{CAB}>90^\circ. Let MM be the midpoint of BCBC. The line perpendicular to ACAC that passes through CC cuts the line ABAB at point DD. Show that ∠AMB=∠DMC\angle{AMB} = \angle{DMC}.

paper folding for half area of ABCD (May Olympiad 2012 L1)

From a paper quadrilateral like the one in the figure, you have to cut out a new quadrilateral whose area is equal to half the area of the original quadrilateral.You can only bend one or more times and cut by some of the lines of the folds. Describe the folds and cuts and justify that the area is half. https://2.bp.blogspot.com/-btvafZuTvlk/XNY8nba0BmI/AAAAAAAAKLo/nm4c21A1hAIK3PKleEwt6F9cd6zv4XffwCK4BGAYYCw/s400/may%2B2012%2Bl1.png
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May Olympiad 2012, Level 2, Problem 2

The vertices of two regular octagons are numbered from 11 to 88, in some order, which may vary between both octagons (each octagon must have all numbers from 11 to 88). After this, one octagon is placed on top of the other so that every vertex from one octagon touches a vertex from the other. Then, the numbers of the vertices which are in contact are multiplied (i.e., if vertex AA has a number xx and is on top of vertex A′A' that has a number yy, then xx and yy are multiplied), and the 88 products are then added. Prove that, for any order in which the vertices may have been numbered, it is always possible to place one octagon on top of the other so that the final sum is at least 162162.
Note: the octagons can be rotated.

sum of digits alternate's sum May Olympiad (Olimpiada de Mayo) 2012 L1 P2

We call S (n)(n) the sum of the digits of the integer nn. For example, S(327)=3+2+7=12S (327)=3+2+7=12. Find the value of A=S(1)−S(2)+S(3)−S(4)+...+S(2011)−S(2012).A=S(1)-S(2)+S(3)-S(4)+...+S(2011)-S(2012). (AA has 20122012 terms).
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27 boxes, each has >=12 marbles May Olympiad (Olimpiada de Mayo) 2012 L1 P4

There are 2727 boxes located in a row; each contains at least 1212 marbles. The allowed operation is transfer a ball from a box to its neighbor on the right, as long as said neighbor contains more pellets than the box from which the transfer will be made. We will say that a distribution initial of the balls is happy if it is possible to achieve, by means of a succession of permitted operations, that all the balls are in the same box. Determine what is the smallest total number of marbles with the that you can have a happy initial layout.

Number of friends

There are 12 people such that for every person A and person B there exists a person C that is a friend to both of them. Determine the minimum number of pairs of friends and construct a graph where the edges represent friendships.