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

1. There will always be three points that lie on a straight line.
2. 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.
3. 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:

1. $(i)$ only.
2. $(ii)$ only.
3. $(iii)$ only.
4. $(ii)$ and $(iii)$.
5. None of the above.
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+1 vote
1. 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.
2. This is also not necessarily true as even all the $5$ points can be collinear (lie on same line).
3. 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.
by Veteran (431k points)

(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)

by Veteran (119k points)
+1

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