One way to pack a 100 by 100 square with 10000 circles, each of diameter 1, is to put them in 100 rows with 100 circles in each row and 100 in each vertical column. If the circles are repacked so that the centres of any three tangent circles form an equilateral triangle, how many additional circles can be packed
Answers
ᴀɴꜱᴡᴇʀ:-
• When the circles are packed so that the centres of any three tangent circles form and equlateral triangle, than alternate rows of circles will contain 100 or 99 circles. We need to determine how many rows of 100 and how many rows of 99 we can fit into the 100 cm width if we pack them in this way.
For that what we need to do is to remove one circle from every second and shift it to form given configuration in the attachment.
‣ Since each circle has a diameter1, △PQR and △PXY are equlateral triangle with side of length1.
M △PQR altitude PS bisects side QR .
‣ Use pythagoras theorem to find PS:-
H²=P²+B²
PS²=(1)²-(1/2)²=(1-1)/4=3/4
Therefore, PS=√3/2
‣ Similarly, XZ=√3/2
‣ Since, all radii have length=1/2,
PU=ZT=XZ-XT
= √3/2-1/2
‣ and, PV= PS+SV
=PS+RW
=√3/2+1/2
‣ UV=PU+PV
=√3/2-1/2+√3/2+1/2=√3
‣ This tells us that two rows of circles require a height of √3 before a third row begins.
‣ Therefore, 100/√3=57.7, we can pack 57 double rows, each containing 100+99=199 circles.
‣ The given square(100×100) our configuration of 57 double rows requires a height of 57√3 before the next row begins. So, 100-57√3>1 , and since the circles each have diameter1. . . . . . we can put a final row of 100 circles.
‣ Thus we have 57 double rows containing 199 circles each and one containing 100 circles.
The number of cirvles used in this new packing is = (57×199)+100 =11443
‣ So, the maximum numver of additional circles that can be packed in a square is 11443-10000 = 1443
Answer:
1443 is the answer
Step-by-step explanation:
I hope the answer is useful for you