MHB Can 101 Discs of Radius 1/2 Cover a Rectangle That 25 Discs of Radius 1 Can?

[caffeinemachine]

Hello MHB.
I am having trouble with the following quesion.

Let $R=\{(x,y)\in\mathbb R^2:A\leq x\leq B, C\leq y\leq D\}$ be a rectangle in $\mathbb R^2$ which can be covered (overlapping allowed) with $25$ discs of radius $1$ each.
Then $R$ can be covered with $101$ rectangles of radius $1/2$ each.

I think the question is wrong. I think that a rectangle having one side $1.01$ units and the other side just long enough so that $25$ discs of radius $1$ units can cover it will not be coverable by 101 discs of radius $1/2$.
I don't know how to prove this.

Can anybody help?

 

[caffeinemachine]

Hello MHB.
I am having trouble with the following quesion.

Let $R=\{(x,y)\in\mathbb R^2:A\leq x\leq B, C\leq y\leq D\}$ be a rectangle in $\mathbb R^2$ which can be covered (overlapping allowed) with $25$ discs of radius $1$ each.
Then $R$ can be covered with $101$ rectangles of radius $1/2$ each.

I think the question is wrong. I think that a rectangle having one side $1.01$ units and the other side just long enough so that $25$ discs of radius $1$ units can cover it will not be coverable by 101 discs of radius $1/2$.
I don't know how to prove this.

Can anybody help?

I discussed this with one of my friends and he solved it.

We note that any $a\times b$ rectangle is a union of 4 rectangles of dimension $(a/2)\times (b/2)$.

$R$ is a $A\times B$ rectabgle.
Say $R'$ denotes a copy of $(A/2)\times (B/2)$ rectangle.
Since $R$ can be covered by $25$ discs of unit radius, $R'$ can be covered by $25$ discs of radius $1/2$. Using $4$ different $R'$s we can cover $R$. Thus $R$ can be covered by a $100$ discs fo radius $1/2$. The $101$th disc is just for fun.