Assignment 7 ANSWERS

Practice:  Learn how to read off the barycentric coordinates of a point P given lattice information or ratio information as in the practice problems in Lab 7 and on the web.  There will be some trapezoid practice also.  There will be more information on the web for study suggestions for the quiz next week.

(7.1 – 10 points) Finding barycentric coordinates from cevian ratios

Given a triangle ABC and points A' on BC and B' on CA, let P be the intersection of AA' and BB'.  Suppose the ratio BA'/A'C = 2/3 and ratio CB'/B'A = 1/5.

As usual, let (x,y,z) be the barycentric cooridnate with respect to ABC. We are given two ratios: z/y = 2/3 and x/z = 1/5.

• What are the barycentric coordinates of P?

Substitute x = (1/5)z, y = (3/2)z into x+y+z = 1 to get ((1/5)+(3/2) + 1)z = 1. So z = 10/27 and x = (1/5)(10/27)=(2/27) and y=(3/2)(10/27) = 15/27. So (x,y,z) = (1/27)(2, 15, 10).

• If C' is the intersection of CP with AB, what is the ratio AC'/C'B?

Ratio = y/x = 15/2 from the relation between barycentric coordinates and centers of mass.

• If Q is the intersection of line A'B' with line AB, what is the ratio AQ/QB?

Ratio = -15/2. This follow from Menelaus. If ratio = r, then (2/3)(1/5)r = -1, so r = -15/2.

(7.2 – 15 points) Plotting points from barycentric coordinates

Construct an equilateral triangle ABC and construct the points P, Q, R if

• the barycentric coordinates of P are (1/7, 4/7, 2/7)
• the barycentric coordinates of Q are (2/(3+sqrt (2)), 1/(3+sqrt (2)), sqrt(2)/(3+sqrt (2)))
• the barycentric coordinates of R are (-1/3, 6/3, -2/3)

There are a number of ways to do this with more or less efficiency, but the straightforward approach has two parts. Let the coordinates in general be (x, y, z) (1) Construct the lengths and lay them side by side on a line. So for exampl, in the first case, construct any segment of length s, then on the same line lay out a segment of length 4s and then 2s. If there are square roots, as in the second case, the roots can be constructed using Pythagoras or the geometric mean (mean proportional) construction. If there are negative lengths, then reverse direction when you lay out the lengths. (2) Use this segment, of whatever length, divided into the correct ratios x, y, z, with transversals as in B&B to construct segments with the same proportions on a segment of length AB. (3) Transfer these lengths using the compass to construct the appropriate points on the side of triangle ABC. You only need two lines to construct a point such as P, so you will not need all of your information. Note: This problem is simpler than the general case because all side lengths of ABC are the same. In general, you would need to repeat (2) for any side (actually two sides would do) to construct the points on that particular side.

(7.3 – 10 points) Coordinate and cubes

In (x,y,z) space, the 6 points e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) and –e1, -e2, -e3 form the vertices of a regular octahedron.  The faces are equilateral triangles.  The centers of these triangles are the vertices of a cube.

• What are the coordinates of the vertices of this cube?

One vertex is (1/3,1/3,1/3). The others are the (x,y,z) for all possibilities where x, y, and z is either +1/3 or -1/3.

One method for finding from basic principles is to let M be the midpoint of e1e2. The Me3 is a median of triangle e1e2e3 and the centroid C of this triangle is one of the cube vertices. We know MC/Me3 = 1/3. If O = (0,0,0), then the triangle MOe3 is a right triangle, and the line through C and through (0,0,1/3) is parallel to MO by transversals (or Thales). But this parallel says that the z-coordinage of C is 1/3. Using the same method with the other coordinates gives (1/3,1/3,1/3).

Applying the same method to each of the 8 faces of the octahedron gives all the vertices as described.

A more advanced but quicker method is the one involving centers of mass. The centroid of a triangle in coordinate 3-space = (1/3)(A+B+C). This gives the result immediately.

• What are the coordinates of the vertices of the tetrahedron formed from 4 vertices of the cube (two choices of tetrahedron)?

One such tetrahedron has vertices (1/3,1/3,1/3), (1/3,-1/3,-1/3), (-1/3,-1/3,1/3), (-1/3,1/3,-1/3). There is another tetrahedron with the other four vertices not in this list.

These are all the same distance apart (use the distance formula).

• What are the coordinates of the vertices of the octahedron whose vertices are the midpoints of the edges of the tetrahedron?

These are the midpoints of the edges of the tetrahedron above. All of these edges are diagonals of the square faces of the cube.

Use the formula for the midpoint of AB, which is (1/2)(A+B). So, because two of the signs are changed, each midpoint will have one non-zero coordinate. For example the midpoint of (1/3,1/3,1/3)(1/3,-1/3,-1/3) is (1/3,0,0) and the midpoint of (-1/3,-1/3,1/3)(-1/3,1/3,-1/3) is (-1/3,0,0). The six vertices are the multiples of e1, e2, e3 by +1/3 or -1/3.

Notice you can find the volumes directly from formulas for the octahedra and the cube and also know the ratios of volumes of the tetrahedron in the cube, the octahedron in the tetrahedron, etc. from earlier work.

(7.4 – 20 points) Roof of a dodecahedron

This problem is about the "roof" on a square base that is used to construct a dodecahedron.

Setup: Constructing the roof

• Take two regular pentagons with side s and diagonal length d.  We have shown earlier that d/s = golden ratio.
• Cut each pentagon along one diagonal so that the 2 pieces are an isosceles triangle and an isosceles trapezoid.
• Take a square of side d.  Attach to two opposite sides of the square the trapezoids (along the side of length d).  Attach to the other two opposite sides the triangles.  These attached polygons fit together to form a pentahedron (5-sided polyhedron) with a square base.  It resembles a tent or a roof.  We will call it the roof.

Problem 1.  What is the height of the two vertices of the roof above the square base, i.e., what is the distance of each of these points from the square base?

If the square is ABCD, then in the (x,y,0) plane, we can take A = (d/2, d/2, 0), B = (-d/2,d/2,0), etc. The points below the edge PQ where the trapzoids join in an edge of length s are P' = (s/2,0,0) and Q' = (s/2,0,0). This means that if the height of the tent is h, then P = (s/2,0,h).

But we know s = AP, since AP is a side of the pentagon. So we can write s^2 = (1/4)((s-d)^2 + d^2) + h^2. At this point we can divide by s^2 to get an equation involving d/s = g = golden ratio and h/s, which we will call H.

Then the equation is 1 = (1/4)((g-1)^2 + g^2) + H^2.

It is important to remember that g^2 = g+1 to simplify to get H = 1/2 and h = s/2.

NOTE: Instead of using the distance formula directly, it is possible to use Pythagoras twice. Compute h from triangle PP'A. For this we know PA = s, but it is necessary to computer P'A from a right triangle in the base, such as AMP' (see definition of M below) or the distance formula in the plane. It turns out that this distance = s* sqrt (3/4), which is the altitude of an equilateral triangle with side s.

Problem 2: Find the exact dihedral angle (and a decimal degree approximation) of 2 dihedral angles in the roof. 

• the angle that the triangle makes with the square base and
• the angle that the trapezoid makes with the square base. 

In the base square, drop perpendiculars to construct points M on AB and N on DA so that AMP'N is a rectangle.

The angle N in triangle NPP' is the dihedral angle that the triangle makes with the square base.

This is a right triangle with legs NP' = (d-s)/2 and PP' = s/2. We can compute the angle in the similar triangle where s = 1 and d = g. Angle N = angle PNP' = arctan (PP'/NP') = 1/(g-1).

The angle M in triangle MPP' is the dihedral angle that the trapezoid makes with the square base.

This is a right triangle with legs MP' = d/2 and PP' = s/2. We can compute the angle in the similar triangle where s = 1 and d = g. Angle M = angle PMP' = arctan (PP'/MP') = 1/g.

The decimal approximations of angles N and M are 58.28 and 31.72.

Problem 3:  Use your answer to Problem 2 to show why 6 of these roof pieces can be attached to the faces of a cube to form a regular dodecahedron.

The decimal approximations of angles N and M are 58.28 and 31.72. Thus they appear to be complementary, with sum = 90 degrees. But this is just approximate.

For the angles to be complementary, this must mean that angle P' in triangle MPP' = angle N in triangle NPP'. In other words, the right triangles NPP' and PMP' must be similar.

This would be true if tan angle N = PP'/NP' = MP'/PP" = cot angle M. In other words, is it true that 1/(g-1) equals 1/(1/g) = g? As it turns out, this is true. To check, just take the product g(g-1) = g^2 - g = g+1-g = 1.

So the angles are exactly complementary! This means that if two tents are built on adjacent faces of a cube so that one triangular face and one trapezoidal face share an edge, then the dihedral angles add up to 90+(0 = 180 degrees, so there the a straight angle between the triangle and the trapezoid. The both lie in the same plane and thus define a pentagonal face.

Hints and suggestions:  You can use a coordinate-free approach, but it is also OK to use coordinates, with the base in the (x, y, 0) plane with sides parallel to the axes, so that the heights are z-coordinates.  You also have the distance formula available.

(7.5 – 10 to 20 points) Seeking Archimedean Polyhedra

Use a search engine such as Google or the Math Archives or the Math Forum or other means, and find a site that gives a good explanation, with examples, of what is meant by an Archimedean Polyhedron (some may spell it Archimedian) and with other information about polyhedra.

• Send an email to the Math 444 list with the web address of the page.
• In the email, write a one-paragraph review telling what you found valuable (or not so valuable) in the site.
• Name and describe your favorite Archimedean polyhedron, giving info about faces, edges, vertices so that one could recognize it from your description
• Optional extra credit: make a model of your polyhedron.  You can ask for some cardboard shapes by email.  Be sure to say how many and what kind.

Note:  It may be difficult to find 31 different sites, but more credit will given for originality in finding the site, interesting and insightful reviews, and clear descriptions.  In other words, original comments about the same site are OK, but finding good, different sites is storngly encouraged. All this information will be collected on a web page for the class.

See the web page link for the responses.