Giza-IQ Test - Solution 10

Each of the two ants has 4 edges to choose. This gives a total of 4² = 16 possible choices. Of those choices, we have to discard the four cases in which the ants choose the same edge, because then they would meet between their starting vertices after walking 1/2 edge each.

We are left with 12 possible choices. This result can also be obtained by noting that if we let one of the ants choose the edge it likes, then the second ant is only left with three possible choices: 4*3 = 12.

After walking their first edge, in 4 of the 12 possible cases, the ants find themselves still on opposite vertices, while in the other 8 cases, they are at two ends of the same edge.

Now, if they are at the two ends of an edge, it is for them impossible to meet after walking exactly a further edge without one of them having to walk back to its vertex of origin. As the question states that the ants never double back, these possibilities have to be discarded.

As a result of the considerations above, the only cases that satisfy all the conditions are those in which the ants move from a pair of opposite edges to a different pair of opposite edges. The probability is 4/16 = 1/4.

In its new position, each ant can choose one of 3 possible edges (not 4, because ants never double back). Therefore, there are 3² = 9 possibilities. Of these, only 2 lead to a meeting satisfying the required conditions: those in which both ants walk to one of the two vertices that were so far unoccupied by either one of them. The probability is therefore 2/9.

The total probability is given by the product of the probabilities calculated for the two edges: P = 1/4 * 2/9 = 1/18 = 0.0555 .

Giza-IQ Test - Solution 9

 

 It is a regular Octahedron.

Giza-IQ Test - Solution 8

If you lie the octahedron on its ADE face, its top view is as follows:

The area of the smallest enclosing rectangle is given by BC, which is 1, multiplied by the projection of DF onto the horizontal plane, which is twice the distance between the centre of a face and one of its vertices = 2 * 2/3 * sqrt(3)/ 2 = 2/3 * sqrt(3)

The height of our box coincides with the distance between the faces ADE and BCF, which is twice the radius of the inscribed sphere.

Using the coordinate axes shown in the figure below, I write the equation of the sphere centred in the origin that has on its surface the centre of one of the faces. If you find a better method for calculating the radius, I’d be happy to hear from you. I just use the first method that comes to mind.

Considering that CDEF is a square of side 1, the distance of each vertex from the centre of the octahedron is sqrt(2)/2. Therefore, the coordinates of the vertices B, C, and F are:

B: (0, 0, sqrt(2)/2)

C: (0, sqrt(2)/2, 0)

F: (sqrt(2)/2, 0, 0)

To calculate the coordinates of the centre of the BCF face, I just average the coordinates of its vertices:

centre of BCF: (sqrt(2)/6, sqrt(2)/6, sqrt(2)/6)

The equation of a sphere centred in the origin is simply x² + y² + z² = r², and r turns out to be sqrt(6)/6.

We can finally calculate the volume of our box as:

V = area of base * height = [2/3 * sqrt(3)] * [sqrt(6) / 3] = 2/3 * sqrt(2)

Giza-IQ Test - Solution 7

The cubic root of 2 is 1.259921049894873 = ~1.26

The following table shows how many 2-cubes (i.e., cubes of volume 2) fit side-by-side into the length occupied by a given number of 1-cubes:

1-cubes 2-cubes

   1      0

   2      1   1.26

   3      2   2.52

   4      3   3.78

   5      3   3.78

   6      4   5.04

   7      5   6.30

   8      6   7.56

   9      7   8.82

  10      7   8.82

  11      8  10.08

For example, six 2-cubes need just a bit less space than eight 1-cubes.

To solve the problem, we only need try possible box dimensions starting from the smallest one that can contain a 2-cube. We can afford to do so, because we expect to find a solution pretty soon.

1-cubes   1-vol   2-cubes   2-vol   ratio

2x2x2       8     1x1x1       2      1/4

2x2x3      12     1x1x2       4      1/3

2x3x3      18     1x2x2       8      4/9

3x3x3      27     2x2x2      16     16/27

2x2x4      16     1x1x3       6      3/8

2x3x4      24     1x2x3      12      1/2

3x3x4      36     2x2x3      24      2/3

3x4x4      48     2x3x3      36      3/4

4x4x4      64     3x3x3      54     27/32

2x2x5      20     1x1x3       6      3/10

2x3x5      30     1x2x3      12      2/5   ***

Giza-IQ Test - Solution 6

If you open the icosahedron, you can easily see that the shortest line between the geometrical centres of two opposite faces is the straight line. This is the geodesic line, which you would obtain in absence of friction if you stretched between the two points an elastic band. If you establish the Cartesian axes as shown in the figure, the coordinates of the two points are (0, 1/3) and (5/2, 2/3).

Therefore, the solution is given by:

d = sqrt((5/2) ² + (1/3)²) = sqrt(226) / 6 = 2.505549396395485.

Giza-IQ Test - Solution 5

The volume of a tetrahedron is baseArea * height / 3. If we extend exactly two edges, they can only be adjacent. If you choose two that are not part of the base, the maximum height is obtained when the rotating edge is perpendicular to the base.

Then, the height coincides with the length of the edge.

The area of the base is sqrt(3)/4. Therefore, the maximum volume of the tetrahedron is:

Vmax = sqrt(3) / 4 * 1/3 = sqrt(3) / 12 = 0.144337567297406.

Giza-IQ Test - Solution 4

The volume of a tetrahedron is baseArea * height / 3. If we only extend one of the edges that don’t form the base, as shown in the figure, the maximum height is obtained when the rotating face is perpendicular to the base.

Then, the height coincides with that of an equilateral triangle of side 1: sqrt(3)/2.

The area of the base is sqrt(3)/4. Therefore, the maximum volume of the tetrahedron is:

   Vmax = sqrt(3)/4 * sqrt(3)/2 / 3 = 1/8 = 0.1250.

Just for fun, we can compare this volume with that of a regular tetrahedron. Its height can be calculated as follows:

   h = sqrt( 1² - (2/3)²) = sqrt(5)/3

Therefore, the volume of the regular tetrahedron of unitary edge is

   sqrt(3)/4 * sqrt(5)/3 / 3 = 1/12 * sqrt(5/3) = 0.1076...

Giza-IQ Test - Solution 3

Cut the cone along the line formed by the vertical segment of the string, and open it onto a flat surface.  You obtain a triangle with a rounded base.  The string slips off the cone when the angle at the top of the opened surface of the cone reaches 180 degrees.  That is, when the opened cone looks like a semicircle.

If we take the radius R of the semicircle to be 1, the circumference of the cone base is given by:
  C = 2*pi*R / 2 = pi

The radius of the cone base is then given by
  r = C/(2*pi) = 1/2

The aperture α of the cone is easily calsulated as follows:
  α = 2*arcsin(r/R) = 2*arcsin(1/2) = 60 degrees.

Giza-IQ Test - Questions 9 and 10

Three more geometrical puzzles.  Don’t you like polyhedra?  These are the last two geometrical questions.  Then, I will have for you numerical ones.

1.09    Four regular tetrahedrons are placed inside a regular tetrahedron that has edges of double length.  This is done in such a way that each vertex of the large tetrahedron coincides with one of the vertices of the small tetrahedrons, so that an empty space is left in the middle of the large tetrahedron.  Please describe in words the shape of the empty space.

1.10    Two ants are on the opposite vertices of a regular octahedron.  They choose at random one of the edges and walk on it at the same uniform speed.  Every time each of the ants encounters a vertex, it immediately chooses at random a new edge (that is, it never doubles back onto the edge it has just come from) and walks on it.  What is the probability that the two ants meet after each ant has walked on exactly two edges?

Giza-IQ Test - Solution 2

This is the solution to the second of the two problems I posted on July 11.

Giza-IQ Test - Solution 1

Today, instead of proposing more puzzles, I have decided to begin giving you solutions.

This is the solution to the first of the two problems I posted on July 11.

Giza-IQ Test - Questions 7 and 8

Today’s puzzles are about fitting regular polihedra into boxes.

1.07    A rectangular box can be filled completely with cubes of dimensions 1 x 1 x 1.  When the unit cubes are removed and larger cubes, each having a volume of 2, are placed in the box with their edges parallel to the edges of the box, it turns out that they can only fill 40% of the box. What is a possible set of dimensions for such a box?

1.08    What are the dimensions of the smallest rectangular box that can contain a regular octahedron with unitary edge length?

Giza-IQ Test - Questions 5 and 6

To all French visitors: Happy Bastille Day!  For those who don’t know, on July 14th of 222 years ago, the French Revolution exploded.  That’s why today is the French equivalent of U.S.A’s  July 4th.  I wonder what fraction of such momentous events occurred during summer.  The Russian national day is (was?) October 25th, and the Chinese one is October 1st, but Italy’s is June 2nd, and the Australian one January 26th (which is during the southern summer)...

Anyhow, today I am going to propose two more of my mathematical puzzles.  These two are about Platonic solids: a Tetrahedron and an Icosahedron.  Wikipedia says that many viruses, including the herpes virus, have icosahedral shells. Fascinating  (as an old friend with pointy ears would say)!

1.05    In a regular tetrahedron with unitary edge length, two (and exactly two) edges are increased in length, thereby increasing the volume of the solid.  What is the maximum volume that can be obtained in this way?

1.06    An ant walks on a regular icosahedron with unitary edge length.  It goes from the geometric centre of one face to the geometric centre of the opposite face (the geometric centre of a triangle is the centre of the circumscribed circle).  What is the minimum distance that the ant needs to cover?

Java - Formatting a Sudoku for the Web

I just wrote a small method in Java that somebody might find useful. It converts a Sudoku string into HTML.

Giza-IQ Test - Questions 3 and 4

Here are two more questions to test your intelligence with mathematical and geometrical problems.  The first problem I propose to you today was one of the problems I encountered when in 1967 I attended in Rome the selections for the Italian National Math championship.  I qualified for the national championship and there I reached the fifth place, thanks to which I became part of the Italian team sent to the IX International Mathematics Olympiad.

Giza-IQ Test - Questions 1 and 2

As you might know, in 2010 I joined several High-IQ societies. half a year later, I started working on an Intelligence Test. After developing twenty-three questions, as it often happens to me, I lost interest and left it unfinished. As I don’t think I will ever complete it, I will publish the questions in this blog. To see the correct answers, you will have to send me an email or comment online.