The Fibonacci sequence is given by equalities $$F_1=F_2=1, F_{k+2}=F_k+F_{k+1}, k\in N$$. a) Prove that for every $m \ge 0$, the area of the triangle $A_1A_2A_3$ with vertices $A_1(F_{m+1},F_{m+2})$, $A_2 (F_{m+3},F_{m+4})$, $A_3 (F_{m+5},F_{m+6})$ is equal to $0.5$. b) Prove that for every $m \ge 0$ the quadrangle $A_1A_2A_4$ with vertices $A_1(F_{m+1},F_{m+2})$, $A_2 (F_{m+3},F_{m+4})$, $A_3 (F_{m+5},F_{m+6})$, $A_4 (F_{m+7},F_{m+8})$ is a trapezoid, whose area is equal to $2.5$. c) Prove that the area of the polygon $A_1A_2...A_n$ , $n \ge3$ with vertices does not depend on the choice of numbers $m \ge 0$, and find this area.
Problem
Source: 2017 XX All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p5
Tags: geometry, lattice points, area of a triangle, areas, Ukrainian TYM
MelonGirl
24.05.2022 11:57
Recall that Cassini identity says $F_{n}^2 - F_{n+r} F_{n-r} = (-1)^{n-r}.$ (a) Area by shoelace is $\frac{1}{2} | F_{m+1} F_{m+4} + F_{m+3} F_{m+6} + F_{m+5} F_{m+2} - F_{m+2} F_{m+3} - F_{m+4} F_{m+5} - F_{m+6} F_{m+1}|$ Note that: $F_{m+1} F_{m+4} - F_{m+2} F_{m+3} = (F_{m+3} - F_{m+2})(F_{m+3} + F_{m+2}) - F_{m+2} F_{m+3} = F_{m+3}^2 - F_{m+2}^2 - F_{m+2} F_{m+3}$ $= F_{m+3} F_{m+1} - F_{m+2}^2 = (-1)^{m+2},$ by Cassini. Similarly, $F_{m+3} F_{m+6} - F_{m+4} F_{m+5} = (-1)^{m+4}.$ Finally, $F_{m+5} F_{m+2} - F_{m+6} F_{m+1} = (F_{m+4} + F_{m+3})(F_{m+4} - F_{m+3}) - (F_{m+4} + F_{m+5})(F_{m+3} - F_{m+2})$ $ = (F_{m+4}^2 - F_{m+3}^2 - F_{m+4} F_{m+3}) - (-F_{m+4} F_{m+2} + F_{m+5} F_{m+3} - F_{m+5} F_{m+2})$ $ = (-1)^{m+3} - (- (F_{m+3}^2 + (-1)^{m+3}) + F_{m+4}^2 + (-1)^{m+3} - F_{m+5} F_{m+2})$ but $F_{m+4}^2 - F_{m+3}^2 - F_{m+5} F_{m+2} = F_{m+4}^2 - F_{m+3}^2 - (F_{m+3} + F_{m+4})(F_{m+4} - F_{m+3}) = 0.$ So this equals $(-1)^{m+3}.$ Finally, $[ \triangle A_1 A_2 A_3 ] = \frac{1}{2} | (-1)^{m+2} + (-1)^{m+4} + (-1)^{m+3}| = \boxed{1}.$ (b) We get similar calculations: $F_{m+1} F_{m+4} - F_{m+2} F_{m+3} = (-1)^{m+2}.$ $F_{m+3} F_{m+6} - F_{m+4} F_{m+5} = (-1)^{m+4}.$ $F_{m+5} F_{m+8} - F_{m+6} F_{m+7} = (-1)^{m+6}.$ And we have to calculate $F_{m+7} F_{m+2} - F_{m+8} F_{m+1}$ But note that $(F_{m+7} F_{m+2} - F_{m+8} F_{m+1}) + (F_{m+6} F_{m+1} - F_{m+7} F_{m}) = F_{m+7} F_{m+1} - F_{m+7} F_{m+1} = 0.$ So by induction we get that it equals $(-1)^m (F_2 F_7 - F_8 F_1) = (-1)^m \cdot (F_7 - F_8) = (-1)^{m+1} \cdot 8.$ Thus, $[A_1 A_2 A_3 A_4] = \frac{1}{2} |(-1)^{m+2} + (-1)^{m+4} + (-1)^{m+6} + (-1)^{m+1} \cdot 8| = \boxed{\frac{5}{2}}.$ (c) In the general case we have $[A_1 A_2 \ldots A_n] = \frac{1}{2} |(-1)^{m+2} + \cdots + (-1)^{m+2(n-1)} + F_{m+(2n-1)} F_{m+2} - F_{m+2n} F_{m+1} |,$ but by the same reasoning as before, $F_{m+2n-1} F_{m+2} - F_{m+2n} F_{m+1} = (-1)^m (F_{2n-1} F_{2} - F_{2n} F_{1}) = (-1)^{m+1} F_{2n-2}.$ The area is then $\boxed{\frac{F_{2n-2} - n+1}{2}}.$