Subareas
Public  11/01/17  8xp  Programming  13.3% 
Let fib(n) be n_{th} Fibonacci number. fib(1)=1, fib(2)=1 and fib(n)=fib(n1)+fib(n2) 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 v_{2+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 M1 points on AB side and N1 points on BC side. Draw M1 line segments from C to the M1 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 N1 line segments from A to the N1 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: a_{11}=3/22, a_{12}=45/154; a_{21}=4/11, a_{22}=16/77 See also: Median  Wikipedia, the free encyclopedia
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