Thanks to all who have provided solutions or commented.

Here is my comment.

First, the problem was, in my opinion, relatively simple and straight forward. For perspective I asked this problem to to my son (a relatively good math student, post-doc in Physics) over phone and got the answer while I was on the phone. I also asked this problem to some bright middle school students (I volunteer to coach math teams) and the students who had no background in spherical trigonometry were able to solve this using ordinary trigonometry and algebra.

(in about 15 minutes using scientific calculators)

Here is one of those methods.

Work in rectangular coordinates..

Assume the radius is 1, we can easily

find (x,y,z) coordinates of A (Arunadhati) and V (Vasistha).Now all you have to do is to find the

equation of the plane which passes through center of earth (0,0,0) and the points V & AFind the line where it

intersects the plane of arctic circle (z=sin (66.5)).You are done!****

This is even simpler ( really a one-two liner) using vectors... or matrices/determinants..For those who may like to see here are some details... (Actually doing it takes much less time than writing this post

First: X (any point V or A) = cos (lat) cos (long)

Y = cos (lat) sin (long)

Z= sin (lat)

For arctic circle, we have Z= sin (66.5 deg) ... and if we find the long(let us call it x) we are done..

as all we have to do is to convert it to in years knowing 360 degrees is 26000 years etc)

Equation of the plane going through (0,0,0)(x1,y1,z1)(x2,y2,z2) is

det |x, y, z |

|x1,y1,z1| = 0

|x2,y2,z2|

Where this meets the arctic circle is now trivial, we put x=cos(66.5)cos (x), y=cos(66.5) sin (x) and

z = sin (66.5)

So we know every value in above except "x", solve for it !!!

(Actually it boils down to - simplifying by dividing each row by cos (lat) etc.) we get

det | cos(x) ; sin(x) ; tan(66.5) |

| cos (long-V); sin (long-V); tan (lat-V)| = 0

| cos (long-A); sin (long -A) ; tan (lat -A)|

equate this to zero, and you

can get the value of x..

A simple trick, favorite of mine when a trigonometric equation has to be solved, is to assume:

tan(x/2)= t , we can get

cos(x) = (1-t^2)/(1+t^2) and

sin(x) = 2t/(1+t^2)

The above equation is quadratic in t, giving two values of t...

Hope this helps.