Subareas

Let fib(n) be n_th Fibonacci number. fib(1)=1, fib(2)=1 and fib(n)=fib(n-1)+fib(n-2) for n>2. Let f(n)=1+(fib(n)%5) Let v_M+N be the vector of f(i) values for i=1 to M+N. For example v₂₊₂=[2, 2, 3, 4] Let T_M be the sum of the first M elements of V_M+N and T_N be the sum of the last N. For example T_M=4 and T_N=7 for M=2 and N=2. Consider an ABC triangle with M-1 points on AB side and N-1 points on BC side. Draw M-1 line segments from C to the M-1 points on the AB side thereby creating M line segments on the AB side proportional to the first M values of V_M+N starting from A to B. |AB|_j = v_M+N[j]*|AB|/T_M for j=1 to M Similarly draw N-1 line segments from A to the N-1 points on the BC side thereby creating N line segments on the BC side proportional to the last N values of V_M+N starting from B to C. |BC|_i = v_M+N[M+i]*|BC|/T_N for i=1 to N So now we have M*N subareas named a_ij where 1 ≤ i ≤ N and 1 ≤ j ≤ M Let's assume that the area of ABC triangle is equal to 1. Find the values of these M*N subareas for M=19 and N=23, sort them in ascending order and submit the answer as per the format given below. Answer format: area_first,area_median,area_last where areas are in the form of p/q (reduced fraction). You are given that the answer = 3/22,1/4,4/11 for M=2 and N=2. Areas: a₁₁=3/22, a₁₂=45/154; a₂₁=4/11, a₂₂=16/77 See also: Median - Wikipedia, the free encyclopedia