MathDB

2012 BAMO

Part of BAMO Contests

Subcontests

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2012 BAMO D reconstruction of parallelogram with a magical ruler

Laura won the local math olympiad and was awarded a "magical" ruler. With it, she can draw (as usual) lines in the plane, and she can also measure segments and replicate them anywhere in the plane; but she can also divide a segment into as many equal parts as she wishes; for instance, she can divide any segment into 1717 equal parts. Laura drew a parallelogram ABCDABCD and decided to try out her magical ruler; with it, she found the midpoint MM of side CDCD, and she extended CBCB beyond BB to point NN so that segments CBCB and BNBN were equal in length. Unfortunately, her mischievous little brother came along and erased everything on Laura's picture except for points A,MA, M, and NN. Using Laura's magical ruler, help her reconstruct the original parallelogram ABCDABCD: write down the steps that she needs to follow and prove why this will lead to reconstructing the original parallelogram ABCDABCD.

2012 BAMO12 4 intersecting circumcircles, 2 equilateral triangles

Given a segment ABAB in the plane, choose on it a point MM different from AA and BB. Two equilateral triangles △AMC\triangle AMC and △BMD\triangle BMD in the plane are constructed on the same side of segment ABAB. The circumcircles of the two triangles intersect in point MM and another point NN. (The circumcircle of a triangle is the circle that passes through all three of its vertices.) (a) Prove that lines ADAD and BCBC pass through point NN. (b) Prove that no matter where one chooses the point MM along segment ABAB, all lines MNMN will pass through some fixed point KK in the plane.
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2012 BAMO C/2 Two infinite rows of evenly-spaced dots, at most 2012 dots in

Two infinite rows of evenly-spaced dots are aligned as in the figure below. Arrows point from every dot in the top row to some dot in the lower row in such a way that: [*]No two arrows point at the same dot. [*]Now arrow can extend right or left by more than 1006 positions. https://cdn.artofproblemsolving.com/attachments/7/6/47abf37771176fce21bce057edf0429d0181fb.png Show that at most 2012 dots in the lower row could have no arrow pointing to them.

2012 BAMO12 3 permutations of a sequence, scramble, two-two

Let x1,x2,...,xkx_1,x_2,...,x_k be a sequence of integers. A rearrangement of this sequence (the numbers in the sequence listed in some other order) is called a scramble if no number in the new sequence is equal to the number originally in its location. For example, if the original sequence is 1,3,3,51,3,3,5 then 3,5,1,33,5,1,3 is a scramble, but 3,3,1,53,3,1,5 is not.
A rearrangement is called a two-two if exactly two of the numbers in the new sequence are each exactly two more than the numbers that originally occupied those locations. For example, 3,5,1,33,5,1,3 is a two-two of the sequence 1,3,3,51,3,3,5 (the first two values 33 and 55 of the new sequence are exactly two more than their original values 11 and 33).
Let n≥2n\geq 2. Prove that the number of scrambles of 1,1,2,3,...,n−1,n1,1,2,3,...,n-1,n is equal to the number of two-twos of 1,2,3,...,n,n+11,2,3,...,n,n+1.
(Notice that both sequences have n+1n+1 numbers, but the first one contains two 1s.)
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