How Many Cricket Balls Are In The Pyramid?

Cool problem!

First we need to find how much height each layer adds. We have four externally tangent spheres whose centers form a square of sidelength 7. In the middle is situated another sphere of the same radius. If we observe from a top-down perspective, the center of the middle sphere is in the middle of the square. The goal of this is to create a right triangle: the hypotenuse is the distance from the center of one of the spheres to the center of the middle sphere, with the legs being the x and y components of this hypotenuse.

In this 2-D sense, the distance from this center to the center of any other sphere is equal to $\frac{7 \sqrt{2}}{2}$ . In addition, the hypotenuse is of length 7, leaving the y-component. We have a right triangle, so it's natural to use pythagorean theorem. $\frac{49}{2} + y^2 = 49$ , so $y = \frac{7 \sqrt{2}}{2}$ . Thus, for each layer we add, we add this much to the height. In addition, we will also have another $7$ centimeters from the bottom layer sphere's radius and the top layer's radius. Thus, we can find the number of layers by $7 + n (\frac{7 \sqrt{2}}{2}) = 180$ , where there are $n+1$ layers. Solving, we get $\lceil n \rceil = 36$ . Thus, we require $6 * 37 * 73 = 16206$ balls, and $\lceil \frac{16206}{100} \rceil = 163$ boxes.

We determine first the least number of layers the pyramid will have.

We consider a small pyramid having 2 layers of balls with radius $r$ . This pyramid will have 4 balls in its base and one ball placed on top of these four balls. To find the height of such pyramid, we connect the centers of these balls, forming a square pyramid having base length of $2r$ , and a slant height of $2r$ , along its face's edge. Let $d$ be the diagonal length of the square base and $h$ be the height of such pyramid. We then use pythagorean theorem to find the height $h$ .

For the right triangle formed when the pyramid base is cut along its diagonals,

$(\frac{d}{2})^2 + (\frac{d}{2})^2 = (2r)^2$

and for the right triangle formed by connecting the vertex, the corner and the base's center,

$h^2 + (\frac{d}{2})^2 = (2r)^2$

we deduce that,

$h = \frac{d}{2}$ .

But,

$d = 2\sqrt{2} r$

so,

$h = \sqrt{2} r$ .

For the height $H_2$ of a 2-layered pyramid, we have:

$H_2 = r + h + r$ ,

where the first r is the radius of the top sphere and the last r is the radius of the bottom sphere.

Simplifying, we have

$H_2 = 2r + \sqrt{2} r$ .

If we have a 3-layered pyramid, the height becomes

$H_3 = 2r + 2\sqrt{2} r$

For an $n$ -layered pyramid, we will have,

$H_n = 2r + (n-1)\sqrt{2} r$ .

Applying the formula for the $n$ -layered pyramid in our problem:

$180 cm \leq H_n = 2(3.5cm) + (n-1)\sqrt{2} (3.5cm)$ .

We arrive at the inequality: $n \geq 35.95...$

Taking $n = 36$ to be the least number of layers for the pyramid, the number of balls $b$ to be used is $b = 1^2 + 2^2 + 3^2 + .. + 36^2 = 16206$ . Dividing this number by $100$ and rounding up, we obtain the answer : $\boxed{163}$ .

To get the number of layers we should notice first layer centers of balls are at $3.5 sm$ while each next layer is $+3.5 \sqrt{2} sm$ and the last layer of 1 ball adds $+3.5 sm$ height (one radius more of the last ball) to the pyramid and we have:

$3.5 + (n - 1) \times 3.5\sqrt{2} + 3.5 >= 180$ which gives minimum $n = \boxed{36}$ .

The number of balls is

$N = \frac{n \times (n+1) \times (2n + 1)}{6} = 16206$ thus $\boxed{163}$ boxes are necessary.

A ball on the next layer is placed in the space between four balls in the layer below. You need to find the increase in height of the arrangement when a ball is placed on a layer.

To find this, try to visualize 4 balls as they are depicted in this problem's description. Now imagine placing a ball between them. Connect the centers of the 4 balls to form a square of side equal to diameter of any ball here. Next join each of these vertices of this square to the center of the ball placed above it. Just consider these 5 vertices; they form a square pyramid.

It isn't difficult to derive the height of this pyramid, so I ll just provide the result: The height of this is $x/\sqrt{2}$ ; where $x$ is the dimension of the pyramid, and the diameter of each ball

This is the height of the center of the top ball from the centers of the balls below. The total height of both layers would thus be $x/2$ (ground to center of first layer) + $x/\sqrt{2}$ (height between two layer centers)+ x/2(2nd layer center to top) which is $x + x/\sqrt{2}$

The first layer by itself has height x, thus each layer increases height by $x/\sqrt{2}$ Thus if k layers are needed in total and substituting x as 7

$7 + (k-1)7/\sqrt{2} = 180$

Solve this to get k = 36 layers

Each layer n has n^2 balls by problem description.

So we need to find

$1^2 + 2^2 + ... + 36^2$ Which has the sum $k(k+1)(2k+1)/6$ where k=36

This is evaluated to 16206 balls. Each box has 100 balls, thus we need 163 boxes in total.