Problem #430

Public 11/01/17 8xp Programming 13.3%

Let fib(n) be nth 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 vM+N be the vector of f(i) values for i=1 to M+N. For example v2+2=[2, 2, 3, 4]

Let TM be the sum of the first M elements of VM+N and TN be the sum of the last N. For example TM=4 and TN=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 VM+N starting from A to B.
|AB|j = vM+N[j]*|AB|/TM 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 VM+N starting from B to C.
|BC|i = vM+N[M+i]*|BC|/TN for i=1 to N

So now we have M*N subareas named aij 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: areafirst,areamedian,arealast 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

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