129 days..
just keep it simple..
wat is the orbital velocity of earth now... GM/R...a1)

then the orbital velocity will be..GM/(R/2).......     b1)

diatance has also changed(assuming. circular)..

 

earlier 2piR   a2)..

now 2pi(R/2)..b2)

 

now a1)/a2) = 365..

then b1)/b2) = x..

 

solve it..now..

 

 

 

 

1..
u need to know the rel. bw.lengths of AF.and AC.
u know the slope of normal.tan(.-X)
where X is angle ACO..

so u now need slope of focal chord..( tan Y)

u have lenghts in those triangles given (at^2..etc.)

use trigo to find AC-AF relation..

all r in t parameter so u get the ratio..

 

another by coordinate..
obtain relation b/w distance frm centre=d & chord length=l...

 

r =  {d^2 + (l/2)^2 }

 

now we want 1/2 l.{r+d} to be max..
since height is (r+d) and base length of chord..

 

y = 1/2 l. ( {d^2 + l/2^2} + d )

its a function..finding its maximum..we find that d should be 1/3  of that of chord length...

 

 

the altitude lines passes through thse pts..i.e

(2,3) & (5,5)
now obtain the equation of this line..

line is y= {2x+ 5 } / 3  ..

this can be written as.. y = {2(x+1) + 3}  / 3

so we want x+1 to be divisible by 3..

 

now it can be 2,5,8,11,14 ...(A.P = 3n-1)  where n  N.

now putting values of each in this equation..u get (y=a) type of line...

 

they also follow a pattern i.e ( 2n+1) where n  N..

 

so in given domain only..3,5,7,9 ..r included so probalility..is 2/5..

 

 

i think its fine..but better if a finer method is presented..

my steps same till sum{ k^-2}....uske baad i culdn't get further..
can u explain wat u did kon!