Question

¿Can someone help me with this?
http://upload.wikimedia.org/wikipedia/commons/thumb/6/69/Circle_packing_(hexagonal).svg/220px-Circle_packing_(hexagonal).svg.png

http://imgur.com/LRe7RWd

http://imgur.com/3qxK8dg

So, I have this hexagonal packing. It's six circles around another circle. I have to find the height of the largest equilateral triangle that can fit in one of the interstices. I already know how to calculate the distance between the center of the interstices. I WOULD REALLY APPRECIATE IT. The answers are in the third link. I have to demonstrate it. The D in the equations is the DIAMETER of the circles. dip is the ditch between the interstices and "a" is the height (i suppose from the second image).

Answer 1

Magnified image of the first link shown above.
AB is the top side of the hexagon = D
C is the center of the middle circle.
DE, DF and EF are the respective tangents to the circles.
AH is the radius of the circle. AH = CG = D/2
DEF is the interstitial equilateral triangle.
The altitude of an equilateral triangle = sqrt(3)/2 * side.
Altitude CJ = sqrt(3)/2 * AB = sqrt(3)/2 * D
The radius of an inscribed circle of an equilateral triangle = sqrt(3)/6 * side.
OJ = sqrt(3)/6 * D
CO = CJ - OJ = sqrt(3)/2 * D - sqrt(3)/6 * D
OG = CO - CG = sqrt(3)/2 * D - sqrt(3)/6 * D - D/2

Answer 2

Here is what I found, though ranga seems to have the answer.
We form an equilateral triangle by connecting the centers of 3 adjacent circles that create an interstice.
The small equilateral triangle inside the interstice is formed by tangent lines. With some work we can show that
PQR is a 30-60-90 triangle, with side PR= D/2. Using the fact that the side opposite the 60º angle has length Hypotenuse * sqr(3)/2, we find
PQ sqr(3)/2 = PR = D/2
and PQ is D/sqr(3).

Let the distance from the side of the small triangle to point Q be called “x”. x = PQ - radius = D/sqr(3) - D/2
We can show that the height of the small triangle will be 3x, or
3*D*(1/sqr(3) - ½). This can be re-written into the form given in the link.

Answer 3

OG = sqrt(3)/2 * D - sqrt(3)/6 * D - D/2
OG = D/2 * (sqrt(3) - sqrt(3)/3 - 1)
OG = D/2 * (sqrt(3) - 1/sqrt(3) - 1)

OG is the the radius of the inscribed circle of the inner equilateral triangle.

If the side of inner equilateral triangle is x, the radius of inscribed circle will be:
sqrt(3) / 6 * x.
Equate this to OG and solve for x
sqrt(3)/6 * x = D/2 * (sqrt(3) - 1/sqrt(3) - 1)
x = 6/sqrt(3) * D/2 * (sqrt(3) - 1/sqrt(3) - 1)
x = 3D/sqrt(3) * (sqrt(3) - 1/sqrt(3) - 1)

The altitude of an equilateral triangle = sqrt(3)/2 * side
Therefore the altitude DG = sqrt(3)/2 * 3D/sqrt(3) * (sqrt(3) - 1/sqrt(3) - 1)
DG = 3D/2 * (sqrt(3) - 1/sqrt(3) - 1)

Answer 4

Thanks for the answer! You saved my thesis.

Answer 5

Glad to be able to help.