Covering Table with Coins

Question

Yesterday Professor Pheno Menon took me and one of my friends to his house to demonstrate one of his most recent mathematical discoveries. When we got there, he showed us a rectangular table. He gave us a packet containing quite a few coins and said:

"Friends, see that table over there? Last night I tried almost all possible configurations and came to my phenomenal conclusion: this table cannot be completely covered with $400$ coins like these."

I was well aware of the phenomenal mistakes he often makes; besides, the tables seemed too small for such a bold statement. Seeing disbelief in my eyes, he went on:

"Don't think it's true? OK, go ahead, here are your $400$ coins. These are all yours if you can cover this table with these."

Now I closely noticed the coins. They were all identical and perfectly circular disks, with a very small thickness. So I immediately got to work. After half an hour of effort (while my friend and Professor were lazing on the couch); I came up with a configuration.

"Look Professor, I knew you were wrong," I said, "I didn't even need all the coins. I have arranged $100$ coins on the table such that none of them overlap, and it's impossible to put another coin without causing overlap."

"No no, my boy, you've got me wrong." he replied. "You need to cover the table entirely; there must not be gaps. Every point on the table must lie below some coin. Your coins may, however, overlap."

My friend finally spoke out, "Then, Professor, you're certainly wrong."

How did my friend know?

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Notes:

1. For a coin to lie on the table, its center must lie on the table.
2. This is not lateral thinking; you must supply a valid mathematical proof that $400$ coins are enough to cover the table using the given information.
3. This was recently posted (with a different wording) on another online forum by a user who claimed to have heard this from his friend. The exact source is unknown to me.

Answer 1

Let the radius of a coin be $r$.

"it's impossible to put another coin without causing overlap" implies that the table does not contain any point $P$, where $P$ is at least $2r$ from every coin's centre. (If such a $P$ existed, we could put a coin with its centre at $P$.)

So if the 100 coins were replaced by big coins of radius $2r$, they'd cover the table -- every point is within $2r$ of at least one coin. Very well then, shrink this arrangement of big coins linearly by a factor of 2. This shows how 100 of the original 400 coins may cover, say the bottom left-hand quadrant of the table. Do likewise for the table's other 3 quadrants.

Answer 2

Let's say that the coin has the radius r.
And 100 coins are placed in such a way that no other coins can be placed. And not of them are overlapping.
This means that the centers of any 2 coins are at least 2r apart.
We put (in our imagination) a 2r radius coin over each 1r coin with matching centers.
If they don't cover the whole table it means that there are 2 big coins close to each other where the distance between the centers is more than 4r.
This means that we can fit a 1r coin (with diameter of 2r) between the 2 original 1r coins. And this should not possible.
So this means that 100 2r coins cover the table completely.

Let's say that the table has the dimensions $x\times y$.
Scaling by a factor of $\frac{1}{2}$ it means that 100 1r coins can cover $\frac{x}{2} \times \frac{y}{2}$ table. Which is $\frac{1}{4}$ of the table.
This means that 400 1r coins can cover the full table.

Answer 3

I have arranged 100 coins on the table such that none of them overlap, and it's impossible to put another coin without causing overlap:

so he got something like that:

we can see that coins can't overlap:

we will first replace coins in a more suitable way:

I see two possibilities for him to do so:

We will chose the first possibility (unless he already did the second from the start)
For each of these possibilities we have filled empty space with coins, but the space filled is smaller than a coin (or we would have been able to fill 1 more coin on the table).
Plus, to get such a pattern we need 4 x coin/4 = 1coin:

Which means for each coin we add less than a full coin: We got less than 200 coins on the table.

We now need to fill the entire table.

to do so we need to add exactly 1 coin for each space:

As we can see such a coin will fill the entire empty space. And each space is made from 4 x coin/4 = 1 coin. So for each coin on the table we will add 1 more coin:
We had less than 200 coins, so we have now less than 2*200=400 coins on the table.