Solutions Manual for Glimpses of Algebra and Geometry

Gabor Toth

This is a text file in Latex, the dvi and postscript formats are in solutions.dvi and solutions.ps.

In this manual we solve the hardest problems posed in the Glimpses. We focus on those problems that have been left out from the Hints for Selected Problems, and whose solutions require novel ideas.

For comments/suggestions, please write to Gabor Toth, Department of Mathematics, Rutgers University, Camden, New Jersey, 08102, U.S.A., or send a message through e-mail: gtoth@crab.rutgers.edu

Section 1.

1.(b) Write the number in the form 7k, , , or , take the cubes of these, and consider the remainders modulo 7.

2. By Problem 1(b), we need to show that 7a-1 is not a square. This follows as in 1(b) by squaring the numbers 7k, , , and , and studying the remainders modulo 7.

Section 2.

2. Write and repeat, with appropriate modifications, the proof of irrationality of e as in the remark following Theorem 1.

4. Using the geometric series formula, a Riemann sum can be worked out as

For , we have so that the Riemann sum converges to

8. The pyramidal staircase can be cut into a large square pyramid of volume , and two sets of n triangular prisms of heights , common base area , and total volume . Notice that the two sets of prisms overlap in n small square pyramids with total volume , and this has to be subtracted form the volume. The volume of the pyramidal staircase is thus . Now use Problem 5.

10. Write with . Notice that iff is odd.

Section 3.

3. For a,b,c consecutive, we have a+c=2b. Substituting this into the Pythagorean equation, we obtain 4a=3b. Since a,b are relatively prime a=3 and b=4 follow.

8. Let , , be a rational point on the Bachet's curve given by . Implicit differentiation gives so that the slope m of the tangent line at is . We need to couple the equation of the tangent line with Bachet's equation to obtain the coordinates of the intersection point. Since , we write Bachet's equation as . After factoring, we have . Substituting and canceling the factor , we obtain

Substituting again , we finally arrive at

This is a quadratic equation in x that has as a root since . Factoring, we obtain

Bachet's formula follows.

13. and . In particular, a,b are relatively prime since is not a square. Setting a=2st, , and , we obtain with t>s relatively prime and of different parity. This is impossible.

Section 4.

5. Let P and Q be polynomials of degrees m and n with rational coefficients and leading coefficient one such that . Let P and Q have roots and . Consider the polynomial . The coefficients of R are symmetric in and , and therefore, by the fundamental theorem of symmetric polynomials, they can be expressed as polynomials (with integral coefficients) in the coefficients of P and Q. In particular, the coefficients of R are rational. Since , is algebraic.

6. By the addition formulas

For , these rewrite as

Since and , these recurrence relations imply that and are polynomials. Now is algebraic since it is a root of the polynomial .

12. The existence and uniqueness of follow from the properties of the exponential function. Let m be rational. If is rational then so is , and this contradicts to the corollary to Theorem 1 in Section 2. Transcendentality of for algebraic m follows from Lindemann's theorem.

Section 5.

2. The equation of the circle is

Setting x=1 and solving for y, we obtain the quadratic formula for and . A generalization of this construction to cubic polynomials would mean that the roots of a cubic polynomial with constructible coefficients would be constructible. This is false (cf. Problem 3 on page 19).

Section 6.

4. The two lines are parallel iff is real. Thus, we may assume that the two lines intersect. The condition is invariant under translation (that amounts to add a constant to each variable) so that we may assume that the lines intersect at the origin. Then and are real constant multiples of each other, and the condition reduces to . Now write and in polar form and consider the argument of the ratio .

5. As in Problem 4, the equation is invariant under adding a constant to each variable so that we may assume that the centroid of the triangle is at the origin, . Squaring this, we see that the condition is equivalent to . On the other hand, . Thus, . Similarly, and . In particular, and hence . Now write , , and in polar form.

7. The condition remains invariant when we translate, rotate, and scale each triangle individually. (For example, translation corresponds to the row operation of adding a constant multiple of the first row to the row the triangle is represented by.) We can thus reduce the problem to the case when and . The condition now reduces to .

8. .

Section 7.

4. We have

and

The stated inequality is thus equivalent to .

Section 8.

4.(a) If P is an odd degree polynomial with real coefficients then , where is the sign of the leading coefficient of P. By continuity, P must have a real zero. (b) The polynomial , , has coefficients that are, up to sign, the elementary symmetric polynomials in the variables , . But an elementary symmetric polynomial in these variables is symmetric in , , and hence it can be written as a real polynomial in the coefficients of P. Thus the coefficients of are real. The degree of is , . Since is odd, the induction hypothesis applies. Thus, for each k, there are indices such that the root is a complex number. Since there are finitely many roots but infinitely many choices of k, for some , and are both complex numbers. The polynomial has complex coefficients, so that, by the complex form of the quadratic formula, and are complex. (c) If P is a complex polynomial then has real values, in particular, its coefficients must be real. By the above, has a complex root. Since , it is also a complex root of P.

Section 9.

1. The regular hexagon is obtained by slicing the cube with a plane through the origin and normal vector (1,1,1). The plane extensions of the three sides of the cube meeting at (1,1,1) cut an equilateral triangle out of this plane, and the plane extensions of the other three sides of the cube meeting at (-1,-1,-1) further truncate this triangle to a regular hexagon.

3.(a) The vertices of are those of plus the midpoints of the circular arcs over the sides of . Thus, half of a side of and a side of meeting at a vertex of are two sides of a right triangle whose third side is . We thus have

This gives

In particular, since , we have

with n-1 nested square roots. (b) Let denote the area of . Since half of the sides of serve as heights of the 2n isosceles triangles that make up , we have . In particular, we have

with n-1 nested square roots.

Section 10.

1. Since G is finite it cannot contain translations. The argument on pp. 75-76 shows that if and are two rotations about different centers and different rotation angles then their compositions and are also rotations with different centers but the same rotation angle. Then is a translation.

2. Type 5.

4. Type 2.

5. We may assume that , . The rotation leaves invariant since it is multiplication by . If is a lattice, then is in the point group . By the crystallographic restriction, n=3,4 or 6.

Section 11.

2. From the geometry of the stereographic projection it follows that, given a circle S on the extended plane, is a great circle on iff contains an antipodal pair of points. The rest follows from the remark on pp. 102-103.

3. Let , be great circles on meeting in an angle at . If p is the South Pole then angle preservance of is clear since is the dihedral angle between the vertical planes that contain and , and the -images of and are the intersections of these planes with . In general, let R be a reflection to a great circle C that carries p into the South Pole. Then and meet in angle at the South Pole, and, by the previous argument, and also meet in angle . Now write these circles as and and use the fact that the reflection to the circle is angle preserving. (This can be seen directly from the remark on pp. 102-103; see also the argument on pp. 112-113.)

Section 12.

7. Setting , the image of a circle |z|=r<1 is an ellipse with semi-major axis r+1/r and semi-minor axis r-1/r. The image of a half-line emanating from the origin is a hyperbola. The ellipses and the hyperbolas are confocal.

8.(a) If the two circles intersect then apply a Möbius transformation that sends one of the intersection points to . It thus remains to consider two disjoint nonconcentric circles and . Applying a suitable isometry, we may assume that the line through their centers is the real axis. Show that there is a (unique) circle C that intersects both circles and , and the real axis at right angles (see Problem 2 of Section 13). Choose a linear fractional tranformation with real coefficients that maps C to a vertical line. Since the real axis is preserved, the images of and intersect the real axis and the vertical line at right angles. Hence they must be concentric.

Section 13.

1. Problem 4 of Section 7 implies that the given linear fractional transformations are self maps of the unit disk . Theorem 9 asserts that the group of linear fractional transformations of onto itself is 3 dimensional (with being the parameters subject to the constraint ad-bc=1). Since and are equivalent through a linear fractional transformation (Problem 5 of Section 12), it follows that the group of linear fractional transformations that leave invariant is also 3 dimensional. On the other hand, the linear fractional transformations given in the problem form a group and they depend on 3 parameters, , and . Thus, these give all the linear fractional transformations preserving .

2. Applying a suitable isometry if necessary, can be assumed to be vertical with ideal point . Then is a semi-circle. Draw a Euclidean line from c tangent to . The hyperbolic line perpendicular to both and is represented by the semi-circle with center at c that contains the point of tangency.

4. We may assume that the triangle has vertices i, ti, t>1, and , . Let and be the angles at and at ti. (a) From the hyperbolic distance formula, we obtain

The Pythagorean theorem follows. (b) Let r>0 be the radius and the center of the semi-circle that represents the hyperbolic line through ti and . Let be the angle at c between by the radial segment connecting c and , and the real axis. We have

The identity gives

so that we have

Using this and , we compute

Swithching the roles of a,b and , we also have

Now the identity

follows by eliminating .

Section 14.

1. (a) A parabolic isometry g is the composition of two reflections in hyperbolic lines that meet at a common endpoint on the boundary of . Conjugating g with an isometry that carries this endpoint to , the two hyperbolic lines become vertical Euclidean lines, and the conjugated g has the form , . Now conjugate this with to obtain . (b) This is given on pp. 127-128. (c) Two nonintersecting hyperbolic lines with no common endpoints on the boundary of can be mapped to concentric Euclidean semi-circles by an isometry of (see Problem 2 of Section 13). Now a horizontal translation brings this configuration to concentric circles centered at the origin. The corresponding composition of reflections is of the form , .

2. Parabolic isometries have a unique fixed point on . Elliptic isometries have a unique fixed point in . Hyperbolic isometries have two fixed points on .

3. Since a linear fractional transformation determines the corresponding matrix in up to sign, is well defined. Given an isometry g, the condition

for z to be a fixed point of g amounts to solve a quadratic equation with discriminant . If then g has a unique fixed point on . Conjugating g by a suitable isometry, we may assume that this fixed point is . This means that c=0 and (by unicity of the fixed point). As in Problem 1, we obtain that g is parabolic. If then there are two fixed points of g and they are conjugate complex numbers. Thus there is a unique fixed point in . The argument on pp. 127-128 shows that g is elliptic. Finally, if then g has two fixed points on that may be assumed to be 0 and . We thus have b=c=0, and g is hyperbolic.

4. This follows form Problems 2-3.

6. If is the unique fixed point of the parabolic then, by the stated commutativity, is left fixed by and we must have . Thus is either parabolic or hyperbolic. If were another fixed point of then would also be left fixed by . This is a contradiction since .

8. This follows from Example 8 on pp. 142-143 by minor modifications.

Section 15.

1. Using the Euler formula for complex exponents, we have

The complex extension of cosine is therefore defined by

Since the exponential function is periodic with period , the cosine function is periodic with period . We now claim that it is one-to-one on any vertical strip , , and each strip is mapped onto the whole complex plane with cuts and along the real axis. Indeed, using the trigonometric identity

we see that iff , for some . If and are in a vertical strip then follows. For surjectivity, notice that the equation

is quadratic in so that

(The arguments are reciprocals of each other.)

Section 17.

8. By the construction of the dodecahedron on pp. 186-187, every vertex of the dodecahedron is the vertex of at most two cubes. The 5 inscribed cubes have the total of 40 vertices. Since the dodecahedron has 20 vertices, it follows that every vertex of the dodecahedron is the vertex of exactly two cubes. These two cubes have a common diagonal. There are exactly 20/2=10 diagonals. 10 is also the number of different pairs of cubes. Thus the statement is true.

10. (a) The extensions of the sides of a spherical triangle give 3 great circles that divide the sphere into 8 regions; the original triangle whose area we denote by A and its opposite triangle, 3 spherical triangles with a common side to the original triangle whose areas we denote by X, Y and Z, and their opposites. Let , , and denote the spherical angles of the original triangle opposite to the sides that are common with the other 3 triangles with areas X, Y, and Z. Notice that A+X is the area of a spherical wedge with spherical angle . We thus have . Similarly, and . Adding, we obtain . On the other hand, , the area of the entire sphere. Subtracting, the statement follows.

17.(a) In Figure 17.28 the triangles and are similar (by the definition of the golden section). Thus all three angles at are congruent and thereby equal to . It follows that the segment connecting and is the side of a decagon inscribed in a unit circle.

18. The vertices of the octahedron are the midpoints of the faces of the cube circumscribed around the reciprocal pair of tetrahedra. The vertex figure at a vertex v of the octahedron is parallel to the face of the cube whose midpoint is v. Homothety of the two cubes follows. Since the edges of the reciprocal cube bisect those of the octahedron perpendicularly, the ratio of magnification is .

19. The 20 vertices of the 5 tetrahedra are vertices of a regular polyhedron with symmetry group . Thus, it must be a dodecahedron. At a fixed vertex v of the colored icosahedron all 5 colors are represented. Consider the 5 triangular faces, one for each tetrahedron, that are extensions of the 5 icosahedral faces meeting at v. The 5 vertices closest to v in each of these triangular faces form a regular pentagon since the order 5 rotation with axis through v permutes these vertices cyclically. The pentagon is perpendicular to this axis and hence parallel to the vertex figure of the icosahedron at v. Homothety of the two dodecahedra follows.

22. Inscribe a Euclidean dodecahedron in with vertices . By the hyperbolic geometry of , for 0<t<1, there is a unique hyperbolic dodecahedron in with vertices . Let be the common dihedral angle of this hyperbolic dodecahedron. is a continuous decreasing function on (0,1). We have , and the limit is the dihedral angle of the Euclidean dodecahedron. Actually, this dihedral angle is , where

(see Problem 20). By continuity, there is a value for which . The fact that hyperbolic dodecahedra with this dihedral angle tesselate follows since at each vertex the three faces are mutually prependicular.

Section 19.

The solutions in this section were written by Joseph Gerver.

12. If we assume that no more than three countries touch at one point (so that the corresponding graph is a triangulation) then a necessary and sufficient condition is that each country have an even number of neighbors.

13. Four colors are needed for a checker board, but no finite number will suffice for every map.

14. Six.

15. Take a complete graph on 27 vertices, and color each edge red or green. Then we can find a subgraph which is either (i) a triangle with all edges colored green, or (ii) a complete graph on eight vertices with all edges colored red.

16. Use a stereographic projection to go between maps on the sphere and maps on the plane. One point (the point at infinity) will be missing from the sphere, but that does not affect the number of colors, because countries that only touch at one point are not required to have different colors.

17. Establish customs stations at every frontier between neighboring countries, and build roads from each customs station to the capitals of both countries. If two roads should cross, rebuild them by changing the connections so that they do not cross. The capitals are the vertices and the roads are the edges of a planar graph. (The roads should not cross any frontiers except at the customs stations. This can be done provided every country is pathwise connected.)

20. There cannot be a vertex of degree 3; else we could remove that vertex, color the remaining graph with four colors, replace the vertex, and color it differently from its three neighbors. If there is a vertex of degree 4, and we use four colors to color every other vertex, then the four neighbors of the vertex of degree 4 must require four different colors, say red, green, yellow, and blue; otherwise the missing color could be used to color the vertex of degree 4. We can now use the same argument as in the proof of the five color theorem, looking at red-green paths or yellow-blue paths, and swapping either red and green or yellow and blue for part of the graph.

21. The red-blue path might cross the red-green path in Figure 19.6.

Section 21.

5. This follows by adjusting the argument on pp. 262-263 to the quaternionic case. The action of on is given by , .

Section 22.

2. The binary octahedral group contains as an index 2 subgroup. Multiplication by has the effect of adding to the coordinates of the points on the middle Clifford torus corresponding to the elements of (see Figure 22.3).

3. A rotation generating the orbit of the 5 tetrahedra circumscribed around the colored icosahedron has angle and its axis lies in the coordinate plane orthogonal to the first axis and the slope of the axis is the golden section (see Figure 17.29). This is because at the common vertex of the icosahedron and a golden rectangle all 5 colors are represented. This axis goes through

Using the notations in Problem 6 of Section 21, we have and

Hence the pair of quaternions that corresponds to this rotation is

4. Throughout this problem we use the notations and the results in Problem 5 of Section 11. In particular, we write the elements of the tetrahedral, octahedral and icosahedral groups as linear fractional transformations. We choose the regular tetrahedron such that its vertices are alternate vertices of the cube, and the cube is inscribed in such that its faces are orthogonal to the coordinate axes. We also agree that the positive octant contains a vertex of the tetrahedron. This vertex must be

The other 3 vertices are

The 3 half-turns around the coordinate axes plus the identity give the dihedral group , a subgroup of the tetrahedral group:

Working out the 8 linear fractional transformations corresponding to the order 3 rotations, we obtain

Putting these together, we arrive at the 12 elements of the tetrahedral group:

We choose the octahedron such that its vertices are the six intersections of the coordinate axes with . This octahedron is homothetic to the intersection of the tetrahedron above and its reciprocal. The octahedral group is generated by the tetrahedral group and a symmetry of the octahedron that interchanges the tetrahedron with its reciprocal. An example to the latter is the qurter-turn around the first axis. This quarter-turn corresponds to the linear fractional transformation . Thus the 24 elements of the octahedral group are as follows:

Finally, we work out the icosahedral group. We inscribe the icosahedron in such that the North and South Poles become vertices. The icosahedron can now be considered as being made up of a northern and southern pentagonal pyramid separated by a pentagonal antiprism. The order 5 rotations with axis through the poles are symmetries of the icosahedron and they correspond to the linear fractional transformations

where is the fifth root of unity. We still have the freedom to rotate the icosahedron around this vertical axis for a convenient position. We fine tune the position of the icosahedron by agreeing that the second coordinate axis must go through the midpoint of one of the cross edges of the pentagonal antiprism. The half-turn U around this axis thus becomes a symmetry of the icosahedron and it corresponds to the linear fractional transformation

The rotations listed above do not generate the entire symmetry group of the icosahedron because they both leave the equator invariant. We choose for another generator the half-turn V whose axis is orthogonal to the axis of U and goes through the midpoint of an edge in the base of the upper pentagonal pyramid. Since U and V are (commuting) half-turns with orthogonal rotation axes, their composition W = UV is also a half-turn whose rotation axis is orthogonal to those of U and V. With the identity, U, V, and W form a dihedral subgroup of the icosahedral group. Using the explicit form of the vertices of the icosahedron projected to in p. 208, the axis of the rotation V goes through the point

With this, the linear fractional transformation corresponding to the half-turn V finally writes as

Composing this with U, we obtain the linear fractional transformation corresponding to W=UV:

Making all possible combinations of these linear fractional transformations, we finally arrive at the 60 elements of the icosahedral group:

Section 23.

1. A vertex figure of the 4-dimensional cube is a regular 3-dimensional polyhedron with four vertices since exactly four edges meet at a vertex. It is thus a regular tetrahedron. The 3-dimensional hyperplane through the origin with normal vector (1,1,1,1) gives an octahedral slice of the cube; the four 3-dimensional cubic faces that meet at (1,1,1,1) intersect this hyperplane in a regular tetrahedron while the other four cubic faces meeting at (-1,-1,-1,-1) further truncate this tetrahedron to obtain the octahedron. For an analogy, consider Problem 1 in Section 9.

2.(a) Take a good look at Figure 23.2. The Schläfli symbol means that the 3-dimensional faces of the 4-dimensional cube are ordinary 3-dimensional cubes with Schläfli symbol , and each edge is surrounded with 3 of these. (b) Let , , and denote the number of vertices, edges, and faces of the n-dimensional cube. We have , , , and , . These recurrences can be solved easily and they give and .

3.(a) A 2-dimensional projection of the 4-dimensional tetrahedron is the pentagram star (top of Figure 19.4). (b) Each edge is surrounded by 3 ordinary tetrahedral faces; thus the Schläfli symbol is

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