Please answer ASAP Randomly pick 6 points from a square of side = 1. Show that you can always find 2 points from these 6 that their distance is less or equal to

Randomly pick 5 points from a sphere. Show that you can always find a closed semi-sphere ( half a sphere and boundary) that contains 4 points.

Problem 1.

My thinking is that the furthest you can get is have two points at each opposite corner, so the distance between them is sqrt(2). If we have two other points with this property, then all four corners are filled up. It is possible to pick two points where the distance is 1 unit.

Then a fifth point can be placed at the center such that the distance from it to any of the corners is sqrt(2)/2. We placed the fifth point at the center to try to get as far away as possible from the other four points.

Basically we're trying to find the worst case scenario (leading to the largest distance possible) and seeing how we can fill up the square. This establishes the upper bound. Any other kind of scenario will have a distance less than the upper bound.

Problem 2.

For this one, I'm not sure what to make of it. The terminology is a bit strange so I'm not going to be fairly helpful here. Sorry about that.

If I had to guess, I'd assume it has something to do with the fact that a plane is uniquely defined by 3 points. That fourth point is not coplanar with the other three, which helps define the semi-spherical portion. The fifth point is just extra. The points can't be all collinear or else a plane won't form. Though to be honest, I'm still not sure about problem 2. I'd get a second opinion.

answer:   rotate 270° clockwise about (7, 1) and reflect across the line x = 7

step-by-step explanation:

because i took test

care to share were the trapezoid shown is ?