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Circle Packing

N circles have been packed inside a square of side length R unit and radius of the circle is to be calculated.

Case 1

When n = 1

 

Obviously in this case, diameter of the circle = side of the square.
 
So, the radius of the circle Description: 5764.png R unit
 
Description: P-342-6.tif
 
Description: P-343-1.tif

Case 2

When n = 2

Consider this figure
 
Description: P-343-2.tif

This is a right-angled triangle with sides r, r and r √2. Hence the diagonal of the square = 4r + 2 r √2 = √2 R
 
So, Description: 5732.png

Case 3

When n = 4
 
Description: P-343-3.tif

Obviously, in this case 2 × diameter of a circle = side of square.
 
So, the radius of a circle Description: 5717.png R

Case 4

When n = 5
 
Description: P-343-4.tif

Consider this figure:
 
Description: P-343-5.tif

If the radius of the circle is r, then the distance be­tween the centre of the circle and the vertex of a square = r √2
 
Description: P-343-6.tif
 
Now, consider this figure, the triangle formed here is a right-angled triangle:
 
1/√2(1/2 of the diagonal of the square) = r + r + r √2
 
= r(2 + √2), so, Description: 5685.png

Case 5

When n = 8
 
Description: 23298.png

We will find out the value of r in 3 steps here:
 
Description: P-343-8.tif
 
Step 1 Find 1 - 2 = 5 - 6
 
Step 2 Find 2 - 3 = 4 - 5
 
Step 3 Find 3 - 4

Step 1 1 - 2 = 5 - 6 = r + r √2
 
Step 2 For 2 - 3
 
This is an equilateral triangle with side length = 2r. Height of this triangle r √3
 
Description: P-344-1.tif
 
So, 2 - 3 = 4 - 5 = r√3 – r

Step 3
 
Description: P-344-2.tif
 
This is a square with side length 2 r and diagonal = 2r √2
 
So, 3 - 4 = 2r
 
Hence 1 – 6 = 2(r + r√2) + 2(r√3 – r) + 2r = R√2. Now r can be calculated.

Case 6

When n = 9
 
Description: P-344-3.tif
 
Radius of circle = r
 
So, the side of a square = 3 × diameter of a circle = 6r
 
So, 6r = R unit, r = R/6 unit

Circles Inside Circles

Radius of the outer circle R = 1 unit and n similar circles of ‘r’ radius have been inserted inside this outer circle.
 
N = 1

Description: P-344-4.tif


Obviously R = r

 
N = 2
Description: P-344-5.tif

Obviously, 4 r = 2 R Hence r = R/2

N = 3
Description: P-344-6.tif

The triangle formed inside will be an equilateral triangle of the side length 2 r. The centre of this circle will be incentre/centroid/circumcentre of this equilateral triangle. So, the distance between the centre of any smaller circle to the centre of the bigger circle = 2/3 median = x = 2/3 [√3/2 × a], where a = 2r.
 
Now, r + x = R. Now put the values of x from the above condition to find the value of r.
 
N = 4
Description: P-345-1.tif

The square formed inside will be of the side length 2r. Now consider this figure,

 
N = 5
Description: P-345-2.tif

The figure inside is a right angled triangle, with base and height being equal to r, and hypotenuse = r√2 = r + x. Hence x = r(√2 – 1)
 
Now 2R = 4 r + 2 x. Hence r = 2(R – x)/4[Now put the value of x from the above condition.]
 
N = 6
Description: P-345-3.tif
Here we can put a similar circle inside all the six circles making it a 7 circle figure. Now, the situation is the same as that of a circle packing with N = 7
 
In this case, 6r = R, so, r = 1/6 R
 
N = 7

Description: P-345-4.tif


In this case, 6r = R, so, r = 1/6 R

Ptolemy’s Theorem of Cyclic Quadrilateral

Description: P-345-5.tif
 
For a cyclic quadrilateral, the sum of the products of the two pair of the opposite sides equal the product of the diagonals.
 
In the figure given above, AB × CD + BC × DA = AC × BD
 
In any cyclic quadrilateral of side lengths abc and d, the length of the diagonals p and q will be equal to: Description: 5578.pngDescription: 5571.png
 
Example
Under some special conditions, it is given that a cyclic quadrilateral ABCD is a parallelogram. What kind of figure will ABCD be?
Solution
ABCD is a parallelogram. Thus a + b = 180°. And ∠A + ∠C = 180°
 
Description: P-345-6.tif
 
This means that a parallelogram drawn inside a circle is always a rectangle.
 




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