Taking a break from Project Euler I have come across this problem recently

It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 units without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?

Looking at figure 1, we can safely assume that the first circle should have two sides of the square as tangents. We can minimise the area of the bounding square by positioning the other two circles on either side of the diagonal from the top left to the bottom right of the square. This diagonal will divide the square into two symmetrical halves.

Initially I tried this problem using co-ordinate geometry. In figure 2, set the origin to be the top-left corner of the square. The points A, B and C are the centres of the respective circles, which we shall call a, b and c. The point X is the point on the common tangents of circles b and c.

- The line through A and X will pass through the top-left and bottom-right of the square, therefore we can establish the equation for it. This will be y = -x
- The point A is (0.5, -0.5)
- The line AC is 1 unit long
- The line CX is 0.5 units
- The angle CXA is 90 degrees
- Using Pythagoras, we can calculate length AX
- We know the equation of the line through A and X and the coordinates of the point A. Now from the length of AX we can calculate the coordinates of the point X
- The line through B and X is perpendicular to the line through A and X, and we know a point on it (X). We can therefore calculate the equation for this line
- Knowing the equation of this line and the point X on it we can calculate the coordinates of the point C because it is a known distance from X
- Adding 0.5 to the absolute value of the y co-ordinate of C will give us the height of the square, which will give us the area

I then tried to find a solution without using co-ordinate geometry. Starting with figure 3 with the equilateral triangle of side 1 unit.

I then constructed another two right-angled triangles (ADB and ACE) with hypotenuses (hypotenusii?) of AB and AC and angle EAD = 90 degrees. From here it was plain sailing.

From figure 4:

- Angle CAB is 60 degrees (it is an angle in an equilateral triangle)
- Since angle CAB is 60, the sum of angle EAC and angle BAD is 30
- Since these two angles must be equal, angle EAC is 15 degrees
- Cos EAC = AE/AC
- AE = Cos EAC * AC
- AC =1 therefore AE = Cos EAC
- EAC is 15 degrees therefore AE = Cos 15
- The length of the line from the radius of circle A to the top of the square is 0.5
- The length of the line from the radius of circle c to the bottom of the square is 0.5
- The length of the square is therefore 0.5 + Cos 15 + 0.5 and the area is this expression squared

Tags: Mathematics

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