olymmath

DE-BUNDESWETTBEWERB 2013

Bundeswettbewerb Mathematik 2013.

Official paper

The official paper in 2 languages

German (round1) · German (round2)

Round 1

Problem 1

  • Round 1

Is it possible to partition the set S={1,2,…,21}S=\{1,2,\ldots,21\} into subsets that in each of these subsets the largest number is equal to the sum of the other numbers?

Answer, solutions and source

Problem 3

  • Round 1

In the interior of the square ABCDABCD , the point PP lies in such a way that ∠DCP=∠CAP=25∘\angle DCP = \angle CAP=25^{\circ} . Find all possible values of ∠PBA\angle PBA .

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Round 2

Problem 6

  • Round 2

A parallelogram of paper with sides 2525 and 1010 is given. The distance between the longer sides is 66 . The paper should be cut into exactly two parts in such a way that one can stick both the pieces together and fold it in a suitable manner to form a cube of suitable edge length without any further cuts and overlaps. Show that it is really possible and describe such a fragmentation.

Answer, solutions and source

Problem 7

  • Round 2

Let ABCDEFABCDEF be a convex hexagon whose vertices lie on a circle. Suppose that AB⋅CD⋅EF=BC⋅DE⋅FAAB\cdot CD\cdot EF = BC\cdot DE\cdot FA . Show that the diagonals AD,BEAD, BE and CFCF are concurrent.

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Problem 8

  • Round 2

Consider the Pascal's triangle in the figure where the binomial coefficients are arranged in the usual manner. Select any binomial coefficient from anywhere except the right edge of the triangle and labet it CC . To the right of CC , in the horizontal line, there are tt numbers, we denote them as a1,a2,⋯ ,ata_1,a_2,\cdots,a_t , where at=1a_t = 1 is the last number of the series. Consider the line parallel to the left edge of the triangle containing CC , there will only be tt numbers diagonally above CC in that line. We successively name them as b1,b2,⋯ ,btb_1,b_2,\cdots,b_t , where bt=1b_t = 1 . Show that bta1−bt−1a2+bt−2a3−⋯+(−1)t−1b1at=1b_ta_1-b_{t-1}a_2+b_{t-2}a_3-\cdots+(-1)^{t-1}b_1a_t = 1. For example, Suppose you choose (41)=4\binom41 = 4 (see figure), then t=3t = 3 , a1=6,a2=4,a3=1a_1 = 6, a_2 = 4, a_3 = 1 and b1=3,b2=2,b3=1b_1 = 3, b_2 = 2, b_3 = 1 . 111b312b2113b131146a14a21a3………………\begin{array}{ccccccccccc} & & & & & 1 & & & & & & & & & 1 & & \underset{b_3}{1} & & & & & & & 1 & & \underset{b_2}{2} & & 1 & & & & & 1 & & \underset{b_1}{3} & & 3 & & 1 & & & 1 & & \boxed{4} & & \underset{a_1}{6} & & \underset{a_2}{4} & & \underset{a_3}{1} & \ldots & & \ldots & & \ldots & & \ldots & & \ldots & & \ldots \end{array}

Answer, solutions and source