Three snooker balls (of diameter 1, and weight 1N each) are placed inside hollow glass prism?

Question

The base of the prism is vertical equilateral triangle of side 1 + √3, and its depth is 1.
[The three balls can just tightly fit inside such prism , touching each other and the faces of the prism]

All surfaces are frictionless.
What is magnitude of force applied by the balls to the bottom face of the prism?

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Answer 1

3.333 N. The weight of the bottom 2 balls exerts only a vertical force of 2 N on the bottom. On each bottom ball, the weight of the top ball exerts a force vector of 1/sqrt(3) N at 30 deg from the vertical. This vector is the intermediate-length side of a 30-60-90 triangle representing the three forces applied to each bottom ball, the other two being perpendicular from bottom and side to center of ball. The vertical force on each bottom ball is the hypotenuse of this triangle or 2/3 N. Total = 3.333 N.

Answer 2

4/3 N