sorry, i don't understand!

But almost all of the participants understood!

I am 'sympathizing' with Mathuz... Yes, it's difficult to understand partly because of the word 'lie' used in this situation. When reading the phrase "the points $ B_1 $ and $ B_2 $ lie in the figure bounded by the plane $ A_1A_2A_3 $" I immediately thought that those points lie on the interior of triangle $ A_1A_2A_3 $. But in this case, the way I understand correctly is that those points are $ floating $ in space bounded by the plane that contains triangle $ A_1A_2A_3 $ and the spheres of radius $ 1 $ and centers $ A_1 $, $ A_2 $ and $ A_3 $. If you replace the word 'lie' with the word 'are', it will help clarifying the problem a bit. However, the space bounded by the spheres of radius $ 1 $ and centers $ A_1 $, $ A_2 $ and $ A_3 $ are on either side of triangle $ A_1A_2A_3 $, and both points $ B_1 $ and $ B_2 $ could be on either side of $ A_1A_2A_3 $. Anyway, this looks similar to an old USAMO problem.

Well,I don't think that this problem consists of something more than saying that the sphere $k$ with radius $max(B_1A_1,B_1A_2,B_1A_3,B_1A_4)$ and centre $B_1$ includes the whole figure except some of the vertices,just because the radius of $k$ is the longest segment with ends $B_1$ and another lieing incide the figure from the problem.

the points lie inside the figure bounded by the spheres in the half-plane determined by the given plane and $ A_{4} $...they said us during the contest.

Let's say $H_1, H_2$ are the foot of altitudes of $B_1, B_2$ to the plane $A_1 A_2 A_3$, and $G$ is the center of $A_1 A_2 A_3$. When $H_1 B_1 = h_1$, $H_2 B_2 = h_2$, $H_1 G_1 = d_1$, $H_2 G_2 = d_2$, the inequality $\bar{B_1 B_2 }^2 \le (d_1 + d_2 )^2 + (h_1 - h_2 )^2$ holds. It is easily shown that $d_i = \sqrt{\frac{3}{4} - h_i^2} - \frac{\sqrt{3}}{2}$. $\bar{B_1 B_2} ^2 \le ( (\sqrt{\frac{3}{4} - h_1^2 } - \sqrt{3}) - (-\sqrt{\frac{3}{4} - h_2^2}))^2 + (h_1 - h_2)^2$. This can be interpreted as the distance between two points $P(\sqrt{\frac{3}{4}-h_1^2 } - \sqrt{3} , h_1)$ and $Q(\sqrt{\frac{3}{4}-h_2^2 } , h_2)$, which move along the perimeter of the quarter-circles $(x-\sqrt{3})^2 +y^2 = 3/4$ and $x^2 + y^2 = 3/4$ respectively. That distance is max when $\angle POQ$ is max $\implies$ $Q=(-\sqrt{\frac{3}{4}}, 0)$ or $Q = (0, \sqrt{\frac{3}{4}})$ $\implies$ $B_2 = A_1 , A_2 , A_3$ or $B_2 = A_4$.