equation solving - Is there a better way to get the eight points of intersection?

I want to get the eight points of intersection from the equations 2 Abs[x] + Abs[y] == 1 and Abs[x] + 2 Abs[y] == 1. To solve these equations, I tried

Solve[{2 Abs[x] + Abs[y] == 1, Abs[x] + 2 Abs[y] == 1}, {x, y}]

but could only get the four points. I then tried

y /. Quiet@Solve[#, y] /. 
   Abs[x_] -> {x, -x} & /@ {2 Abs[x] + Abs[y] == 1,  Abs[x] + 2 Abs[y] == 1}

p = ({x, y} /. Solve[y == #] & /@ {{1 - 2 x, (1 - x)/2}, {(1 - x)/2, (

      1 + x)/2}, {(1 + x)/2, 1 + 2 x}, {1 + 2 x, -1 - 2 x}, {-1 - 2 x,
       1/2 (-1 - x)}, {1/2 (-1 - x), 
      1/2 (-1 + x)}, {1/2 (-1 + x), -1 + 2 x}, {-1 + 2 x, 1 - 2 x}})~
  Flatten~1

{{1/3, 1/3}, {0, 1/2}, {-(1/3), 1/3}, {-(1/2), 0}, {-(1/3), -(1/3)}, {0, -(1/2)}, {1/3, -(1/3)}, {1/2, 0}}

It works, but I don't like it. Could you recommend a better method?

Answer

Actually, the way I interpret your ContourPlot, there are only 4 points of intersection (red/blue curves). So I do interpret your question such that you want to include intersection with the coord axes as well. I hope this helps:

eq1 = 2 Abs[x] + Abs[y] == 1;
eq2 = Abs[x] + 2 Abs[y] == 1;

p = {x, y} /.
(Solve[#, {x, y}] & /@
 {{eq1, eq2},
  {x == 0, eq2},
  {eq1, y == 0}})~Flatten~1