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6

Part of 1992 Hungary-Israel Binational

Problems(1)

rectangle has coordinates of vertices are Fibonacci numbers

Source: 3-rd Hungary-Israel Binational Mathematical Competition 1992

5/24/2007
We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof.
The coordinates of all vertices of a given rectangle are Fibonacci numbers. Suppose that the rectangle is not such that one of its vertices is on the xx-axis and another on the yy-axis. Prove that either the sides of the rectangle are parallel to the axes, or make an angle of 45∘45^{\circ} with the axes.
geometryrectangleanalytic geometryalgebra proposedalgebra