My attempt at Question 15 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 15 from this test.

The question is about packing spheres in a rectangular box.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 15.

In my previous blog post, I promised that this blog post would be about this question. This is because both the question I examined in the previous blog post and the question I examine in this blog post are asked and answer in the same book and the same chapter. And their answers are given on the same page! Imagine that! The authors of this test appear to be completely deranged!

The book I am talking about is Martin Garner’s “New Mathematical Diversions”. The chapter I am talking about is chapter 7, “Packing Spheres”. The page this question is asked is page 86. The page this question is answered is 90. Below I provide the whole chapter, for your convenience:

The whole chapter is fascinating, but before I discuss the question further, I would like to draw attention to page 86, right after the question is posed, where Martin Gardner answers a very interesting question: how come honeycombs are hexagonal.

Ok, back to the question at hand. The answer is that the maximum number of spheres is 594 and Martin Gardner provides this answer and the explanation on page 90. The explanation Martin Gardner provides is extremely brief, to the point of it being difficult to be understood. In the rest of this blog post, I would like to make this answer a little more understandable.

Here is the answer provided by Martin Gardner:

Turn the box on its side and form the first layer by making a row of five, then a row of four, then of five, and so on.

It is possible to make eleven rows (six rows of five each, five rows of four each), accommodating 50 balls and leaving a space of more than .3 inch to spare.

The second layer also will take eleven rows, alternating four and five balls to a row, but this time the layer begins and ends with four-ball rows so that the number of balls in the layer is only 49.

(The last row of four balls will project .28+ inch beyond the edge of the first layer, but because this is less than .3 inch, there is space for it.)

Twelve layers (with a total height of 9.98+ inches) can be placed in the box, alternating layers of 50 balls with layers of 49, to make a grand total of 594 balls.

The above answer is provided by *the* man Martin Gardner, and I use bold italic font for this. I took the liberty to divide the answer in paragraphs. The first two paragraphs talk about the first layer of spheres, the second two paragraphs talk about the second layer of spheres and the fifth and last paragraph talk about all layers of spheres. Below, I will divide the answer into these three steps: studying the first layer, then the second layer, then all layers.

So, let me begin. All lengths below are in inches. I will use the Pythagorean theorem and the fact that when two spheres touch, the line segment that connects their centers and is the distance between their centers is equal to the sum of their radii. And since we are dealing with spheres of the same radius, when two spheres touch, their distance is twice their radius.

I will first study the first layer. We have the box sitting flat on a horizontal table and the box’s dimensions are length=10, width = 5 and height = 10. So the height is one of the two large edges. In the following image, I put some spheres, not all, only the first five rows. The image depicts the first layer, as we look down from above.

In the following image, which is the same as the image above, I calculate the distance between two rows of spheres. The distance is the red vertical line segment. Each dotted line passes through the centers of the spheres in a row, so the red vertical line segment is indeed the distance between two consecutive rows.

Let me denote with x the vertical line segment which is the distance between two rows of spheres. From the right triangle we have:

x^2 + r^2 = (2r)^2 =>
x^2 = (2r)^2 – r^2 =>
x^2 = (2r + r) (2r – r) =>
x^2 = 3r^2 =>
x = r * sqrt(3) =>
x = 0.5 * 1.7321 =>
x = 0,866

So how many spheres can be packed in the first layer? I made the following table, adding distances, remembering to account for the initial and final 0.5 inches of the first and last row, and being careful not to exceed the 10 inches which is the length of the box. The table is the following:

As far as the previous table is concerned, in the first column I denote lengths  and in the second column I denote number of spheres in a row. Red color denotes a sum.

This table explains what Martin Gardner meant when he wrote:

Turn the box on its side and form the first layer by making a row of five, then a row of four, then of five, and so on.

It is possible to make eleven rows (six rows of five each, five rows of four each), accommodating 50 balls and leaving a space of more than .3 inch to spare.

Martin Gardner was describing the arrangement of the first layer as we look down on the box from above and I just explained these two paragraphs of Martin Gardner’s text.

Moving on, I will now study the second layer as we look down from above. The following image presents the layout of the second layer of spheres as we look down from above.

I will now superimpose the second layer on top of the first layer, thus producing the following image.

In the image above, the first layer is depicted containing blue spheres and the second layer is depicted containing white translucent spheres. So, the white translucent spheres are above the blue spheres and we are looking from above the white translucent spheres.

To calculate distances, I made the following image. In this image, we have three blue spheres from the first layer and a red sphere from the second layer. I changed the color of the second layer sphere from white translucent to red, so it is easier to see.

Here, I want to calculate the horizontal displacement that the center of the second layer sphere has in relation to the center of the first row of spheres from the first layer. This can be found from the green equilateral triangle formed by the centers of the three spheres on the first layer. The barycenter of this equilateral triangle is exactly below the center of the red sphere from the second layer. Thus, the displacement I am looking for is the distance of this barycenter from a side of this equilateral triangle.

If I denote h as one of the heights of the equilateral triangle (all three height being equal in an an equilateral triangle), then the distance of the barycenter (the point where all three heights meet) from one of the sides is (1/3)h.

The displacement I am trying to calculate is shown with the black line segment in the following image:

From the right triangle that is formed from one of the edges of the green equilateral triangle and one of its heights, we can calculate this displacement.

h^2 + r^2 = (2r)^2 =>
h^2 = 4r^2 – r^2 =>
h^2 = 3r^2 =>
h = r * sqrt(3) =>
(1/3)h = r * sqrt(3) / 3 =>
(1/3)h = 0.5 * sqrt(3) / 3 =>
(1/3/)h = 0,2887, almost equal to 0,289.

Adding this horizontal displacement for the second layer, my table for the second layer is as follows:

And this explains the following two paragraphs of what Martin Gardner wrote:

The second layer also will take eleven rows, alternating four and five balls to a row, but this time the layer begins and ends with four-ball rows so that the number of balls in the layer is only 49.

(The last row of four balls will project .28+ inch beyond the edge of the first layer, but because this is less than .3 inch, there is space for it.)

So I have explained the first two layers, and I can now talk about all layers. To do this, I need to calculate the vertical displacement that the second layer has in relation to the first layer, i.e. how much higher are the centers of the spheres of the second layer in relation to the centers of the first layer.

Up until now, we were looking the structure from above and we studied horizontal geometrical shapes and horizontal projections. Now we will need to think vertically. We will need to look the box from the side.

So, I want you to consider the same arrangement of the first and second layer as before:

But this time, imagine that we do not want to look this arrangement from above, but we want to look at it from the side, in order to find how much is the center of the red sphere on the second layer is elevated in relation to the centers of the blue spheres of the first layer.

This elevation, this vertical distance, let’s call x. We can find x from the following right triangle.

This right triangle is formed as follows: The side with length 2r is formed by the center of one of the spheres of the first layer and one of the spheres of the second layer. The side with length x is completely perpendicular. And the side with length (2/3)h is completely horizontal. Imagine that this is a triangle that we can see when we look at the arrangement in the image before that, from the side.

From the right triangle in the last image we have:

x^2 + ((2/3)h)^2 = (2r)^2 =>
x^2 + (4/9)h^2 = 4 r^2

But from the discussion about the second layer, we know that h^2 = 3r^2.

So we have:

x^2 + (4/9) (3r^2) = 4r^2 =>
x^2 + (4/3)r^2 = 4r^2 =>
x^2 = 4r^2 – (4/3)r^2 =>
x^2 = (12/3)r@ – (4/3)r^2 =>
x^2 = (8/3)r^2 =>
x = r * sqrt(8/3) =>
x = 0.5 * sqrt(8/3) =>
x= 0.8165.

And so the table for all the layers that can fit on the box is the following:

And this is what the following last paragraph of Martin Gardner’s text is about:

Twelve layers (with a total height of 9.98+ inches) can be placed in the box, alternating layers of 50 balls with layers of 49, to make a grand total of 594 balls.

Now, if you see at the image provided with the question, the spheres are arranged in a cubic manner, and this in no way will provide the maximum packing. This manner will provide a 5*10*10=500 spheres as the number of spheres that can fit in the box. The authors of the test are setting us up to fail. This is the behavior of a very bad person.

So I am asking: Would you want to be friends and belong to a group of people whose mentality is to judge people on their intelligence instead of their character, and that keep knowledge hidden, and that set others for failure?

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About Dimitrios Kalemis

I am a systems engineer specializing in Microsoft products and technologies. I am also an author. Please visit my blog to see the blog posts I have written, the books I have written and the applications I have created. I definitely recommend my blog posts under the category "Management", all my books and all my applications. I believe that you will find them interesting and useful. I am in the process of writing more blog posts and books, so please visit my blog from time to time to see what I come up with next. I am also active on other sites; links to those you can find in the "About me" page of my blog.
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