Divide the square into 4 1X1 squares.

By Pigeon hole principle a square must has 2 points.

Maximum distance is achieved when they are at opposite corner points of the 1X1 squares which is √2

3 votes

Consider a square of side length $2$. We throw five points into the square. Consider the following statements:

- There will always be three points that lie on a straight line.
- There will always be a line connecting a pair of points such that two points lie on one side of the line and one point on the other.
- There will always be a pair of points which are at distance at most $\sqrt{2}$ from each other.

Which of the above is true:

- $(i)$ only.
- $(ii)$ only.
- $(iii)$ only.
- $(ii)$ and $(iii)$.
- None of the above.

2 votes

Best answer

- Not necessarily true. There can be a line connecting a pair of points such that two other points lie on one side of the line and the remaining point on the other.
- This is also not necessarily true as even all the $5$ points can be collinear (lie on same line).
- This is TRUE.

We are given a square of side $2$ and so it takes $2\times 2$ square units of area. This area can be divided into $4$ equal squares of side $1$ taking an area of $1$ square units each. The maximum separation of two points in each square is the diagonal of the small square which is $\sqrt 2.$ This means we cannot put more than $2$ points in each of these squares and since we have $5$ points and only $4$ such squares, at least one square must have more than $1$ points (Pigeonhole Principle) which means there are at most $\sqrt 2$ units apart.

2 votes

(i)Not necessarily true.There will be a line connecting a pair of points such that two points lie on one side of the line and one point on the other.

(ii)There will be a line connecting a pair of points such that two points lie on one side of the line and one point on the other.

But It is not a always case.because There can be three points that lie on a straight line

(iii) There will always be a pair of points which are at distance at most √2 from each other.

√2 is the case when the point in middle .But as there are 1 extra point , the

max among minimum distance will must be less than √2

for 5th point min distance among a pair of vertices must be less than or equal to 1

So, only (iii) is true

Ans will be (C)