MathDB

4

Part of 2012 Indonesia MO

Problems(2)

Maximum Number of Shadow

Source: Indonesia National Science Olympiad 2012 D1 P4

9/4/2012
Given 20122012 distinct points A1,A2,…,A2012A_1,A_2,\dots,A_{2012} on the Cartesian plane. For any permutation B1,B2,…,B2012B_1,B_2,\dots,B_{2012} of A1,A2,…,A2012A_1,A_2,\dots,A_{2012} define the shadow of a point PP as follows: Point PP is rotated by 180∘180^{\circ} around B1B_1 resulting P1P_1, point P1P_1 is rotated by 180∘180^{\circ} around B2B_2 resulting P2P_2, ..., point P2011P_{2011} is rotated by 180∘180^{\circ} around B2012B_{2012} resulting P2012P_{2012}. Then, P2012P_{2012} is called the shadow of PP with respect to the permutation B1,B2,…,B2012B_1,B_2,\dots,B_{2012}. Let NN be the number of different shadows of PP up to all permutations of A1,A2,…,A2012A_1,A_2,\dots,A_{2012}. Determine the maximum value of NN.
Proposer: Hendrata Dharmawan
rotationcomplex numberscombinatorics proposedcombinatorics
B,P,Q,R are collinear

Source: Indonesia National Science Olympiad D2 P4

9/5/2012
Given a triangle ABCABC, let the bisector of ∠BAC\angle BAC meets the side BCBC and circumcircle of triangle ABCABC at DD and EE, respectively. Let MM and NN be the midpoints of BDBD and CECE, respectively. Circumcircle of triangle ABDABD meets ANAN at QQ. Circle passing through AA that is tangent to BCBC at DD meets line AMAM and side ACAC respectively at PP and RR. Show that the four points B,P,Q,RB,P,Q,R lie on the same line.
Proposer: Fajar Yuliawan
geometrycircumcirclegeometry proposed