MathDB

14

Part of 2011 AIME Problems

Problems(2)

Octagon midpoints

Source:

3/18/2011
Let A1A2A3A4A5A6A7A8A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 be a regular octagon. Let M1M_1, M3M_3, M5M_5, and M7M_7 be the midpoints of sides A1A2‾\overline{A_1 A_2}, A3A4‾\overline{A_3 A_4}, A5A6‾\overline{A_5 A_6}, and A7A8‾\overline{A_7 A_8}, respectively. For i=1,3,5,7i = 1, 3, 5, 7, ray RiR_i is constructed from MiM_i towards the interior of the octagon such that R1⊥R3R_1 \perp R_3, R3⊥R5R_3 \perp R_5, R5⊥R7R_5 \perp R_7, and R7⊥R1R_7 \perp R_1. Pairs of rays R1R_1 and R3R_3, R3R_3 and R5R_5, R5R_5 and R7R_7, and R7R_7 and R1R_1 meet at B1B_1, B3B_3, B5B_5, B7B_7 respectively. If B1B3=A1A2B_1 B_3 = A_1 A_2, then cos⁡2∠A3M3B1\cos 2 \angle A_3 M_3 B_1 can be written in the form m−nm - \sqrt{n}, where mm and nn are positive integers. Find m+nm + n.
trigonometrysymmetryanalytic geometrygraphing linesslopePythagorean Theoremgeometry
Permutations [2011.II.14]

Source:

3/31/2011
There are NN permutations (a1,a2,…,a30)(a_1,a_2,\dots,a_{30}) of 1,2,…,301,2,\dots,30 such that for m∈{2,3,5}m\in\{2,3,5\}, mm divides an+m−ana_{n+m}-a_n for all integers nn with 1≤n<n+m≤301\leq n <n+m\leq 30. Find the remainder when NN is divided by 1000.
AIME2011 AIME II