Expert: Scott A Wilson - 8/4/2009

Question
QUESTION: Hi Mr Wilson ,

I have 2 different size circles ,

The radius of a circle is 2 cm , others 1 cm .

At least how many circles of 1 cm radius ones cover(mean totally closed) to 2 cm radius  one ?

Could you draw it and send to me?

Thanks ,

ANSWER: I have thought about this for awhile and realized that was a silly approach that I did.
Now for each of the 1 cm radius circles, there can be a square inscribed in them that has width √2.  The size of the square that it would take to cover the big circle is 4cm x 4cm.

Since the square have a width √2 have a width of around 1.414 each,
the total width of all 3 squares is 3*1.414, which is 4.2... bigger than 4.

This means if the squares on the inside of the smaller circles are laid side by side, then the circles will definitely cover and area that is bigger than 4x4 if they are arranged 3 wide and 3 deep.  The answer, then, is 3x4 = 9.

I don't see how to correct it to a smaller number.


---------- FOLLOW-UP ----------

QUESTION: I have download a drawer program and tried to draw my question.

I think this is the optimum solution . We can cover 7 circles (1 radius ) to main circle ( 2 radius) .

Attached file you can see..

Answer
Yes, you're right.  It does barely cover the circle.  I took a few minutes, wrote something down, and in half a page I got there.

Think of drawing the circle with radius 2 around the origin.
Don't actually do this, but read on.  The equation would be x² + y² = 4.

Think of drawing another circle with radius 1 around the origin,
so it is in the center of the bigger circle.  The equation would be x² + y² = 1.

Consider the line at 60°.  This is the line that would have the last points in either circle.
Points on this line would be at (t,(√3)t).

We need to find a value at (0,k) to put the small circle that is at top so that it includes both of these points.

Note that the points we need to worry about are from where t is 1/2 out to where t = 1,
for at t = 1/2, r = 1, and at t = 1, r = 2.

The two points are (1/2, √3/2), which is on the edge of the inside circle,
and (1, √3), which is on the outside circle.

The distance to the second point is √(1² + (√3-y)²).
For this to be at most 1, y has to be √3 to that the second number is 0.
That makes the √ into √(1²+0) = 1.  So k = √3.

Now that we know what the k value is, we can find the other distance.
It is √((1/2)² + (√3/2 - y²).  Since the value we are using for y is √3,
it can be seen the √3/2 - √3 = -√3/2, and when squared, that gives 3/4.
When 1/2 is squared, the answer is 1/4.  If we add these two values together, we get 1,
and the square-root of 1 is 1, so this point is also on the border of the circle.

Using the same approach, it can be shown the for each outer circles that are next to each other, there are two points in question.  As described above, they can be shown to be on the edge of the circles.

Thus, I have just shown it can be done with 7 circles.
Between the two of us, the problem has been resolved.



Now if the points had to all be inside the circles and not on the edges, one more circle would be needed on the outside of the center one.  It can be seen that if two circles are pushed together by just a fractional value, the points would be inside of both circles.  Pushing them all together in a circular fashion would leave one whole that couldn't be covered, so one more circle would be needed to cover it.