MathDB

4

Part of 2018 Stanford Mathematics Tournament

Problems(2)

SMT 2018 Geometry #4 regular dodecagon

Source:

8/8/2023
Let a1,a2,...,a12a_1, a_2, ..., a_{12} be the vertices of a regular dodecagon D1D_1 (1212-gon). The four vertices a1a_1, a4a_4, a7a_7, a10a_{10} form a square, as do the four vertices a2a_2, a5a_5, a8a_8, a11a_{11} and a3a_3, a6a_6, a9a_9, a12a_{12}. Let D2D_2 be the polygon formed by the intersection of these three squares. If we let[A] [A] denotes the area of polygon AA, compute [D2][D1]\frac{[D_2]}{[D_1]}.
geometry
F_{n (k+1)} = b_n F_{n k} + c_n F_{n (k-1)} , Fibonacci numbers

Source: 2018 SMT - Stanford Math Tournament , Team Round, Proof Question 4

1/26/2022
Let FkF_k denote the series of Fibonacci numbers shifted back by one index, so that F0=1F_0 = 1, F1=1,F_1 = 1, and Fk+1=Fk+Fk−1F_{k+1} = F_k +F_{k-1}. It is known that for any fixed n≥1n \ge 1 there exist real constants bnb_n, cnc_n such that the following recurrence holds for all k≥1k \ge 1: Fn⋅(k+1)=bn⋅Fn⋅k+cn⋅Fn⋅(k−1).F_{n\cdot (k+1)} = b_n \cdot F_{n \cdot k} + c_n \cdot F_{n\cdot (k-1)}. Prove that ∣cn∣=1|c_n| = 1 for all n≥1n \ge 1.
FibonacciFibonacci sequenceFibonacci Numbersnumber theory