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\begin{center}
{\bf {\Large Solutions Manual}}
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{\large for}
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{\it {\Large Glimpses of Algebra and Geometry}}
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\begin{center}
{\sl Gabor Toth}
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This is a text file in Latex, the dvi and postscript formats are in 
solutions.dvi and solutions.ps.

\noindent 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.

\noindent 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

\noindent{\bf Section 1.}

\noindent 1.(b) Write the number in the form $7k$, $7k\pm 1$,$7k\pm 2$, or 
$7k\pm 3$, take the cubes of these, and consider the remainders modulo $7$.

\noindent 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 $7k$, $7k\pm 1$,$7k\pm 2$, and 
$7k\pm 3$, and studying the remainders modulo $7$.

\noindent{\bf Section 2.}

\noindent 2. Write $\cos 1 = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}$
and repeat, with appropriate modifications, the proof of irrationality of $e$ 
as in the remark following Theorem 1.

\noindent 4. Using the geometric series formula, a Riemann sum can be worked 
out as
\begin{eqnarray*}
& & \sum_{k=1}^m (\alpha^k)^n(\alpha^k-\alpha^{k-1}) = 
\frac{\alpha -1}{\alpha} \sum_{k=1}^m \alpha^{(n+1)k} \\
& & \quad  = (\alpha -1) \alpha^n \frac{\alpha^{m(n+1)}-1}{\alpha^{(n+1)}-1} 
= \frac{a^{n+1}-1}{\alpha^n+\alpha^{n-1}+\ldots + \alpha +1} .
\end{eqnarray*}
For $m\to \infty$, we have $\alpha =\sqrt[m]{a} \to 1$ so that the Riemann 
sum converges to 
\[
\frac{a^{n+1}-1}{n+1}.
\]  

\noindent 8. The pyramidal staircase can be cut into a large square pyramid
of volume $\frac{n^3}{3}$, and two sets of $n$ triangular prisms of heights 
$1,2,\ldots,n$, common base area $\frac{1}{2}$, and total volume 
$\frac{1+2+\ldots +n}{2}$. 
Notice that the two sets of prisms overlap in $n$ small square pyramids with total 
volume $\frac{n}{3}$, and this has to be subtracted form the volume. The volume 
of the pyramidal staircase is thus $\frac{n^3}{3}+(1+2+ \ldots +n) -
\frac{n}{3}$. Now use Problem 5.

\noindent 10. Write $a=\sum_{n=0}^N c_n2^n$ with $c_n\in \{ 0,1 \}$. Notice 
that $c_n=1$ iff $a_n$ is odd.

\noindent{\bf Section 3.}

\noindent 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.

\noindent 8. Let $(x_0,y_0)$, $y_0\neq 0$, be a rational point on the 
Bachet's curve given by $y^2=x^3-a$. Implicit differentiation gives 
$2yy'=3x^2$ so that the slope $m$ of the tangent line at $(x_0,y_0)$ is
$m=\frac{3x_0^2}{2y_0}$. We need to couple the equation of the tangent
line $y-y_0=m(x-x_0)$ with Bachet's equation $y^2=x^3-a$ to obtain the 
coordinates of the intersection point. Since $y_0^2=x_0^3-a$, we write
Bachet's equation as $y^2-y_0^2=x^3-x_0^3$. After factoring, we have
$(y-y_0)(y+y_0)=(x-x_0)(x^2+xx_0+x_0^2)$. Substituting $y-y_0=m(x-x_0)$
and canceling the factor $x-x_0$, we obtain
\[
m(y+y_0) = x^2+xx_0+x_0^2.
\]
Substituting again $y-y_0=m(x-x_0)$, we finally arrive at
\[
m(2y_0+m(x-x_0)) = x^2+xx_0+x_0^2.
\]
This is a quadratic equation in $x$ that has $x_0$ as a root since 
$m=\frac{3x_0^2}{2y_0}$. Factoring, we obtain
\[
(x-x_0)\biggl( x- \frac{x_0^4+8ax_0}{4y_0^2} \biggr) =0.
\]
Bachet's formula follows. 

\noindent 13. $78=2\cdot 3\cdot 13 = \frac{ab}{2}$ and $a^2+b^2=c^2$.
In particular, $a,b$ are relatively prime since $6^2+26^2=712$ is not a square.
Setting $a=2st$, $b=t^2-s^2$, and $c=t^2+s^2$, we obtain
$2\cdot 3\cdot 13= st(t-s)(t+s)$ with $t>s$ relatively prime and of different
parity. This is impossible. 

\noindent{\bf Section 4.}

\noindent 5. Let $P$ and $Q$ be polynomials of degrees $m$ and $n$ with
rational coefficients and leading coefficient one such that $P(c_1)=Q(c_2)=0$. 
Let $P$ and $Q$ have roots $c_1=\alpha_1,\ldots,\alpha_m$ and 
$c_2=\beta_1,\ldots,\beta_n$. 
Consider the polynomial $R(x)=\prod_{i=1}^m \prod_{j=1}^n 
(x-(\alpha_i+\beta_j))$.
The coefficients of $R$ are symmetric in $\alpha_i$ and $\beta_j$, 
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 $R(c_1+c_2)=0$, 
$c_1+c_2$ is algebraic.

\noindent 6. By the addition formulas
\begin{eqnarray*}
\cos((n+1)\alpha ) &=& \cos(n\alpha )\cos(\alpha )-\sin(n\alpha )\sin(\alpha ) \\
\sin((n+2)\alpha ) &=& \sin((n+1)\alpha )\cos(\alpha )+\cos((n+1)\alpha )
\sin(\alpha )
\end{eqnarray*}
For $\alpha = \cos^{-1}(x)$, these rewrite as
\begin{eqnarray*}
T_{n+1}(x) &=& xT_n(x) -(1-x^2)U_{n-1}(x) \\
U_{n+1}(x) &=& xU_n(x) +T_{n+1}(x).
\end{eqnarray*}
Since $T_1(x)=x$ and $U_0(x)=1$, these recurrence relations imply that 
$T_n$ and $U_n$ are polynomials. Now $\cos (\pi/n)$ is algebraic
since it is a root of the polynomial $T_n +1$.  

\noindent 12. The existence and uniqueness of $x_0$ follow from the properties
of the exponential function. Let $m$ be rational. If $x_0$ is rational then
so is $e^{x_0}=mx_0+1$, and this contradicts to the corollary to Theorem 1
in Section 2. Transcendentality of $x_0$ for algebraic $m$ follows from
Lindemann's theorem.  

\noindent{\bf Section 5.}

\noindent 2. The equation of the circle is
\[
\biggl( x-\frac{a-c}{2a} \biggr)^2 + \biggl( y+\frac{b}{2a} \biggr)^2 
= \frac{(a-c)^2+b^2}{4a^2}.
\]
Setting $x=1$ and solving for $y$, we obtain the quadratic formula for
$r_1$ and $r_2$. 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).

\noindent{\bf Section 6.}

\noindent 4. The two lines are parallel iff $\frac{z_1-z_2}{z_3-z_4}$ is
real. Thus, we may assume that the two lines intersect. The condition 
$\Re \bigl( \frac{z_1-z_2}{z_3-z_4} \bigr) =0$ 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 $z_1,z_2$ and $z_3,z_4$ are real constant
multiples of each other, and the condition reduces to $\Re(z_1/z_3)=0$.
Now write $z_1$ and $z_3$ in polar form and consider the argument of the 
ratio  $z_1/z_3$.   

\noindent 5. As in Problem 4, the equation is invariant under adding a
constant to each variable $z_1,z_2,z_3$ so that we may assume that the centroid
of the triangle is at the origin, $z_1+z_2+z_3=0$. Squaring this, we see that
the condition is equivalent to $z_1^2+z_2^2+z_3^2= z_1z_2+z_2z_3+z_3z_1 =0$.
On the other hand, $z_1z_2+z_2z_3+z_3z_1 =z_1(z_2+z_3)+z_2z_3=-z_1^2+z_2z_3$.
Thus, $z_1^2=z_2z_3$. Similarly, $z_2^2=z_1z_3$ and $z_3^2=z_1z_2$. In
particular, $z_1^3=z_2^3=z_3^3$ and hence 
$|z_1|=|z_2|=|z_3|$. Now write $z_1$, $z_2$, and $z_3$ in polar
form.  

\noindent 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
$z_1=w_1=0$ and $z_2=w_2=1$. The condition now reduces to $z_3=w_3$. 

\noindent 8. $2\cdot 3 = (1+\sqrt{5} i)(1-\sqrt{5} i)$.

\noindent{\bf Section 7.}
 
\noindent 4. We have 
\[
|z-w|^2=(z-w)({\bar z}-{\bar w}) = |z|^2-z{\bar w}-{\bar z}w+|w|^2
\]
and
\[
|1-z{\bar w}|^2=(1-z{\bar w})(1-{\bar z}w)=1-z{\bar w}-{\bar z}w+|z|^2|w|^2.
\]
The stated inequality is thus equivalent to $0<(1-|z|^2)(1-|w|^2)$.


\noindent{\bf Section 8.}

\noindent 4.(a) If $P$ is an odd degree polynomial with real coefficients 
then $\lim_{x\to \pm \infty}P(x) = \pm \,{\rm sgn}\,(c_n)\, \infty$, where
$\,{\rm sgn}\,(c_n)$ is the sign of the leading coefficient $c_n$ of $P$.
By continuity, $P$ must have a real zero. (b) The polynomial $Q_k$, $k \in 
{\bf Z}$, has coefficients that are, up to sign, the elementary symmetric
polynomials in the variables $\alpha_i+\alpha_j+k\alpha_i\alpha_j$,
$1\leq i<j \leq n$. But an elementary symmetric polynomial in these variables
is symmetric in $\alpha_i$, $i=1,\ldots,n$, and hence it can be written as a 
real polynomial in the coefficients of $P$. Thus the coefficients of $Q_k$ are
real. The degree of $Q_k$ is $\frac{n(n-1)}{2} = 2^{m-1}a(2^ma-1)$, $m\geq 1$. 
Since $a(2^ma-1)$ is odd, the induction hypothesis applies. Thus, for 
each $k$, there are indices $1\leq i<j \leq n$ such that the root
$\alpha_i+\alpha_j+k\alpha_i\alpha_j$ is a complex number. Since there are finitely 
many roots $\alpha_i$ but infinitely many choices of $k$, for some 
$1\leq i<j \leq n$, $\alpha_i+\alpha_j$ and $\alpha_i\alpha_j$ are both
complex numbers. The polynomial $z^2-(\alpha_i+\alpha_j)z+\alpha_i\alpha_j =
(z-\alpha_i)(z-\alpha_j)$ has complex coefficients, so that, by the complex form 
of the quadratic formula, $\alpha_i$ and $\alpha_j$ are complex. (c) If $P$
is a complex polynomial then $P{\bar P}$ has real values, in particular,
its coefficients must be real. By the above, $P{\bar P}$ has a complex root.
Since $|P|^2 = P{\bar P}$, it is also a complex root of $P$. 

\noindent{\bf Section 9.}

\noindent 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.

\noindent 3.(a) The vertices of $P_{2n}$ are those of $P_n$ plus the
midpoints of the circular arcs over the sides of $P_n$. Thus, half of a side of
$P_n$ and a side of $P_{2n}$ meeting at a vertex of $P_n$ are two sides of
a right triangle whose third side is $1-\sqrt{1-(s_n/2)^2}$. We thus have
\[
s_{2n}^2 = \biggl( \frac{s_n}{2} \biggr)^2 + \biggl( 1- \sqrt{1- 
\biggl( \frac{s_n}{2} \biggr)^2}\biggr)^2.
\]
This gives
\[
s_{2n} = \sqrt{2-\sqrt{4-s_n^2}} .
\]
In particular, since $s_4=\sqrt{2}$, we have
\[
s_{2^n} = \sqrt{2-\sqrt{2+\sqrt{2+\ldots +\sqrt{2}}}},
\]
with $n-1$ nested square roots. 
(b) Let $A_n$ denote the area of $P_n$.
Since half of the sides of $P_n$ serve as heights of the $2n$ isosceles
triangles that make up $P_{2n}$, we have $A_{2n} =\frac{ns_n}{2}$. In 
particular, we have
\[
A_{2^n} =2^{n-1} s_{2^n} = 2^{n-1} \sqrt{2-\sqrt{2+\sqrt{2+\ldots +\sqrt{2}}}} ,
\]         
with $n-1$ nested square roots. 

\noindent{\bf Section 10.}

\noindent 1. Since $G$ is finite it cannot contain translations. The
argument on pp. 75-76 shows that if $R_1$ and $R_2$ are two rotations about 
different centers and different rotation angles then their compositions
$R_1 \circ R_2$ and $R_2 \circ R_1$ are also rotations with different centers
but the same rotation angle. Then 
$R_2^{-1} \circ R_1^{-1} \circ R_2 \circ R_1$ 
is a translation.   

\noindent 2. Type 5. 

\noindent 4. Type 2.

\noindent 5. We may assume that $\omega = z(2\pi/n)$, $n\geq 3$. The rotation
$R_{2\pi/n}$ leaves ${\bf Z}[\omega]$ invariant since it is multiplication
by $\omega$. If $L={\bf Z}[\omega]$ is a lattice, then $R_{2\pi/n}$ is in 
the point group ${\bar G}$. By the crystallographic restriction, $n=3,4$ or $6$.

\noindent{\bf Section 11.}

\noindent 2. From the geometry of the stereographic projection it follows that,
given a circle $S$ on the extended plane, $h_N^{-1}(S)$ is a great circle
on $S^2$ iff $S \cap S^1$ contains an antipodal pair of points. The rest
follows from the remark on pp. 102-103.

\noindent 3. Let $C_1$, $C_2$ be great circles on $S^2$ meeting in an angle
$\alpha$ at $p\in S^2$. If $p$ is the South Pole then angle preservance of
$h_N$ is clear since $\alpha$ is the dihedral angle between the 
vertical planes that 
contain  $C_1$ and $C_2$, and the $h_N$-images of $C_1$ and $C_2$ are the 
intersections of these planes with ${\hat \co}$. In general, let $R$ be a
reflection to a great circle $C$ that carries $p$ into the South Pole.
Then $R(C_1)$ and  $R(C_2)$ meet in angle $\alpha$ at the South Pole, and, 
by the previous argument, $h_N(R(C_1))$ and  $h_N(R(C_2))$ also meet in angle
$\alpha$. Now write these circles as $(h_N \circ R \circ h_N^{-1})(h_N(C_1))$
and $(h_N \circ R \circ h_N^{-1})(h_N(C_2))$ and use the fact that the
reflection  $h_N \circ R \circ h_N^{-1}$ to the circle $h_N(C)$ is
angle preserving. (This can be seen directly from the remark on pp. 102-103;
see also the argument on pp. 112-113.)


\noindent{\bf Section 12.}

\noindent 7. Setting $z=|z|z(\theta )$, 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.

\noindent 8.(a) If the two circles intersect then apply a M\"obius 
transformation that sends one of the intersection points to $\infty$.  
It thus remains to consider two disjoint nonconcentric circles $C_1$ and $C_2$. 
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 $C_1$ and $C_2$, 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 $C_1$ and $C_2$ intersect the real axis
and the vertical line at right angles. Hence they must be concentric. 


\noindent{\bf Section 13.}

\noindent 1. Problem 4 of Section 7 implies that the given linear
fractional transformations are self maps of the unit disk $D^2$. Theorem 9 
asserts that the group of linear fractional transformations 
of $H^2$ onto itself is 3 dimensional
(with $a,b,c,d\in \br$ being the parameters subject to the constraint
$ad-bc=1$). Since $D^2$ and $H^2$ are equivalent through a linear fractional 
transformation (Problem 5 of Section 12), 
it follows that the group of linear fractional 
transformations that leave $D^2$ 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, $\Re (w)$, $\Im (w)$ and $\theta$. Thus, these
give all the linear fractional transformations preserving $D^2$. 

\noindent 2. Applying a suitable isometry if necessary, $l_1$ can be assumed
to be vertical with ideal point $c\in \br$. Then $l_2$ is a semi-circle.
Draw a Euclidean line from $c$ tangent to $l_2$. The hyperbolic line
perpendicular to both $l_1$ and $l_2$ is represented by the semi-circle
with center at $c$ that contains the point of tangency.  

\noindent 4. We may assume that the triangle has vertices $i$, $ti$, $t>1$, and
$z(\theta )$, $0< \theta < \frac{\pi}{2}$. Let $\alpha$ and $\beta$ be the angles
at $z(\theta )$ and at $ti$. (a) From the hyperbolic distance formula, we obtain
\[
\cosh a = \frac{t^2+1}{2t}, \,\, \cosh b = \frac{1}{\sin \theta },\,\,
\cosh c = \frac{t^2+1}{2t\sin \theta }.
\]
The Pythagorean theorem follows. (b) Let $r>0$ be the radius and $c\in \br$ the
center of the semi-circle that represents the hyperbolic line through $ti$
and $z(\theta )$. Let $\delta$ be the angle at $c$ between by the radial 
segment connecting $c$ and $z(\theta )$, and the real axis. We have
\[
\sin \beta = \frac{t}{r},\,\, r\cos \delta = r\cos \beta +\cos \theta,\,\,
r\sin \delta = \sin \theta .
\]
The identity $\cos^2 \delta +\sin^2 \delta =1$ gives
\[
r^2 = r^2\cos^2 \beta + 2r\cos \beta \cos \theta +1
\]
so that we have
\[
t^2 =  r^2 \sin^2 \beta = 2r \cos \beta \cos \theta +1.
\]
Using this and $\alpha = \theta - \delta$, we compute
\begin{eqnarray*}
\cos \alpha &=& \cos (\theta -\delta) = \cos \theta \cos \beta + \frac{1}{r} \\
&=& \frac{t}{r} \biggl( \frac{t^2-1}{2t} + \frac{1}{t} \biggr) \\
&=& \sin \beta \cosh a.
\end{eqnarray*}
Swithching the roles of $a,b$ and $\alpha,\beta$, we also have
\[
\cos \beta = \sin \alpha \cosh b.
\]
Now the identity
\[
\sinh a \tan \beta = \tanh b
\]
follows by eliminating $\alpha$. 

\noindent{\bf Section 14.}

\noindent 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
$H^2$. Conjugating $g$ with an isometry that carries this endpoint to $\infty$,
the two hyperbolic lines become vertical Euclidean lines, and the 
conjugated $g$ has
the form $z\mapsto z+a$, $a\in \br$. Now conjugate this with $z\mapsto
\frac{1}{a} z$ to obtain $z\mapsto z+1$. (b) This is given on pp. 127-128.
(c) Two nonintersecting hyperbolic lines with no common endpoints on the
boundary of $H^2$ can be mapped to concentric Euclidean semi-circles by
an isometry of $H^2$ (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 $z\mapsto kz$, $k\neq 0$.   

\noindent 2. Parabolic isometries have a unique fixed point on $\br \cup
\{ \infty \}$. Elliptic isometries have a unique fixed point in $H^2$.
Hyperbolic isometries have two fixed points on $\br \cup \{ \infty \}$.

\noindent 3. Since a linear fractional transformation determines the 
corresponding matrix in $SL(2,\co )$ up to sign, ${\rm trace}^2$ is well
defined.  Given an isometry $g$, the condition 
\[
g(z) = \frac{az+b}{cz+d} =z, \quad a,b,c,d \in \br ,
\]
for $z$ to be a fixed point of $g$ amounts to solve a quadratic equation 
with discriminant $(a-d)^2-4bc = (a+d)^2-4= \,{\rm trace}^2(g)-4$.
If ${\rm trace}^2(g)=4$ then $g$ has a unique fixed point on 
$\br \cup \{ \infty \}$. Conjugating $g$ by a suitable isometry, we may assume 
that this fixed point is $\infty$. This means that $c=0$ and $b\neq 0$ (by
unicity of the fixed point). As in Problem 1, we obtain that $g$ is parabolic.
If $0\leq \,{\rm trace}^2(g)<4$ then there are two fixed points of $g$ and 
they are conjugate complex numbers. Thus there is a unique fixed point in
$H^2$. The argument on pp. 127-128 shows that $g$ is elliptic. Finally,
if ${\rm trace}^2(g)>4$ then $g$ has two fixed points on $\br \cup \{ \infty \}$
that may be assumed to be $0$ and $\infty$. We thus have $b=c=0$, and $g$ is
hyperbolic. 

\noindent 4. This follows form Problems 2-3.

\noindent 6. If $z_0\in \br \cup \{ \infty \}$ is the unique fixed point of the 
parabolic $g_1$ then, by the stated commutativity, $g_2(z_0)$ is left fixed by 
$g_1$ and we must have $g_2(z_0)=z_0$. Thus $g_2$ is either parabolic or 
hyperbolic.
If $w_0 \in \br \cup \{ \infty \}$ were another fixed point of $g_2$
then $g_1(w_0)$ would also be left fixed by $g_2$. This is a contradiction
since $g_1(w_0) \neq w_0,z_0$.

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


\noindent{\bf Section 15.}

\noindent 1. Using the Euler formula for complex exponents, we have 
\[
\cos \theta = \frac{e^{i\theta}+ e^{-i\theta}}{2} .
\]
The complex extension of cosine is therefore defined by
\[
\cos z = \frac{e^{iz}+ e^{-iz}}{2}.
\]
Since the exponential function is periodic with period $2\pi i$, the cosine
function is periodic with period $2\pi$. We now claim that it is one-to-one 
on any vertical strip $k\pi < \Re (z) <(k+1)\pi$, $k\in {\bf Z}$, and each strip
is mapped onto the whole complex plane with cuts
$(-\infty ,-1)$ and $(1,\infty )$ along the real axis. Indeed, using the 
trigonometric identity 
\[
\cos z_1 - \cos z_2 = -2 \sin \biggl( \frac{z_1-z_2}{2} \biggr) 
\sin \biggl( \frac{z_1+z_2}{2} \biggr) ,
\]
we see that $\cos z_1 = \cos z_2$ iff $z_1\pm z_2=2m\pi$, for
some $m \in {\bf Z}$.
If $z_1$ and $z_2$ are in a vertical strip then $z_1=z_2$ follows. For
surjectivity, notice that the equation
\[
\cos z = \frac{e^{iz}+ e^{-iz}}{2} =w
\]
is quadratic in $e^{iz}$ so that
\[
z= -i \log (w\pm \sqrt{w^2 -1}) = \pm i \log (w+ \sqrt{w^2 -1}).
\]
(The arguments $w\pm \sqrt{w^2 -1}$ are reciprocals of each other.) 

\noindent{\bf Section 17.}

\noindent 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. 

\noindent 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 $\alpha$, $\beta$, and $\gamma$ 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
$\alpha$. We thus have $A+X=2\alpha$. Similarly, $A+Y=2\beta$ and $A+Z=2\gamma$.
Adding, we obtain $3A+X+Y+Z=2(\alpha + \beta + \gamma )$.
On the other hand, $2(A+X+Y+Z)=4\pi$, the area of the entire sphere.
Subtracting, the statement follows.

\noindent 17.(a) In Figure 17.28 the triangles $\triangle p_0p_3p_4$ and 
$\triangle p_1p_3p_4$ are similar (by the definition of the golden section). 
Thus all three angles at $p_3$ are congruent
and thereby equal to $\frac{\pi}{5}$. It follows that the segment connecting
$p_0$ and $p_4$ is the side of a decagon inscribed in a unit circle.

\noindent 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 $\frac{1}{2}$.

\noindent 19. The 20 vertices of the 5 tetrahedra are vertices of a
regular polyhedron with symmetry group ${\cal A}_5$. 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.     

\noindent 22. Inscribe a Euclidean dodecahedron in $D^3$ with vertices
$v_1,\ldots,v_{20}\in S^2$. By the hyperbolic geometry of $D^3$, for $0<t<1$,
there is a unique hyperbolic dodecahedron in $D^3$ with vertices $tv_1,\ldots,
tv_{20}$. Let $\delta (t)$ be the common dihedral angle of this hyperbolic
dodecahedron. $\delta$ is a continuous decreasing function on $(0,1)$. We have
$\lim_{t\to 1} \delta (t) < \pi /2$, and the limit $\lim_{t\to 0} \delta (t)$ is the
dihedral angle $>\frac{\pi}{2}$ of the Euclidean dodecahedron. Actually, this
dihedral angle is $\delta = \pi - 2\psi$, where
\[
\tan \psi = \frac{\tau}{\sqrt{\tau^2+1}}
\]
(see Problem 20).
By continuity, there is a value $t_0\in (0,1)$ for which $\delta (t_0) =
\frac{\pi}{2}$. The fact that hyperbolic dodecahedra with this dihedral angle
tesselate $D^3$ follows since at each vertex the three faces
are mutually prependicular.      


\noindent{\bf Section 19.}

\noindent{\sl The solutions in this section were written by Joseph Gerver.} 

\noindent 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.

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

\noindent 14. Six. 

\noindent 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. 

\noindent 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. 

\noindent 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.)

\noindent 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. 

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

\noindent{\bf Section 21.}

\noindent 5. This follows by adjusting the argument on pp. 262-263 to the
quaternionic case. The action of $S^3 \subset {\bf H}$ on ${\bf H}^2$
is given by $g:(p,q)\mapsto (p \cdot g^{-1},q \cdot g^{-1})$, $g \in S^3$.

\noindent{\bf Section 22.}

\noindent 2. The binary octahedral group ${\bf O}^*$ contains ${\bf T}^*$
as an index 2 subgroup. Multiplication by $\frac{1+i}{\sqrt{2}}$
has the effect of adding $\frac{\pi}{4}$ to the coordinates of the points on 
the middle Clifford torus corresponding to the elements of ${\bf T}^*$ (see
Figure 22.3).

\noindent 3. A rotation generating the orbit of the 5 tetrahedra circumscribed 
around the colored icosahedron has angle $\frac{2\pi}{5}$ and 
its axis lies in the coordinate plane orthogonal to the first axis and the 
slope of the axis is the golden section $\tau$ (see Figure 17.29). This is
because at the common vertex $(0,1,\tau )$ of the icosahedron and a
golden rectangle all 5 colors are represented. This axis goes through 
\[
\biggl( 0, \frac{1}{\sqrt{\tau^2 +1}},\frac{\tau}{\sqrt{\tau^2 +1}} \biggr) .
\] 
Using the notations in Problem 6 of Section 21, we have 
$\theta = \frac{2\pi}{5}$ and
\begin{eqnarray*}
a &=& \cos \left( \frac{\pi}{5} \right) = \frac{\tau}{2} , \\ 
b &=& \frac{\tau}{\sqrt{\tau^2 +1}} \sin \left( \frac{\pi}{5} \right) =
\frac{1}{2} , \\ 
c &=& \frac{1}{\sqrt{\tau^2 +1}} \sin \left( \frac{\pi}{5} \right) =
\frac{1}{2\tau} , \\
d &=& 0. 
\end{eqnarray*}
Hence the pair of quaternions that corresponds to this rotation is 
\[
\pm \frac{1}{2} \biggr( \tau + i+ \frac{j}{\tau} \biggr).
\]

\noindent 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 $S^2$ 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 
\[
\biggl( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\biggr) .
\] 
The other 3 vertices are
\[
\biggl( \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\biggr), 
\biggl( -\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\biggr),
\biggl( -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\biggr).
\]
The 3 half-turns around the coordinate axes plus the identity give
the dihedral group $D_2$, a subgroup of the tetrahedral group:
\[
z \mapsto \pm z, \, \pm \frac{1}{z} .
\]
Working out the 8 linear fractional transformations corresponding to
the order 3 rotations, we obtain
\[
z \mapsto \pm i \frac{z +1}{z -1},
\pm i \frac{z -1}{z +1}, \pm \frac{z +i}{z -i},
\pm \frac{z -i}{z +i}.
\]
Putting these together, we arrive at the 12 elements of the tetrahedral group:
\[
z \mapsto \pm z, \pm \frac{1}{z}, \pm i \frac{z +1}{z -1},
\pm i \frac{z -1}{z +1}, \pm \frac{z +i}{z -i},
\pm \frac{z -i}{z +i}.
\]
We choose the octahedron such that its vertices are the 
six intersections of the coordinate axes with $S^2$. 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 
$z \mapsto i z $. 
Thus the 24 elements of the octahedral group are as follows:
\begin{eqnarray*}
& & z \mapsto i^k z,  \frac{i^k}{z}, i^k \frac{z +1}{z -1},
i^k \frac{z -1}{z +1}, i^k \frac{z +i}{z -i},
i^k \frac{z -i}{z +i}, \\
& & \qquad k=0,1,2,3.
\end{eqnarray*} 
Finally, we work out the icosahedral group.
We inscribe the icosahedron in $S^2$ 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
\begin{equation}
z \mapsto \omega^j z, \quad j=0,\ldots,4,
\end{equation}
where $\omega = e^{i \frac{2\pi}{5} }$ 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 
\[
U: \,\,\, z \mapsto -\frac{1}{z} .
\] 
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 $D_2$ of 
the icosahedral group. Using the explicit form of the vertices of the icosahedron
projected to ${\hat \co}$ in p. 208, the axis of the rotation $V$ goes through
the point 
\[
\biggl( \frac{\tau}{\sqrt{\tau^2 +1}} , 0, \frac{1}{\sqrt{\tau^2 +1}} .
\biggr)
\]
With this, the linear fractional transformation corresponding to the half-turn 
$V$ finally writes as
\[
V: \,\,\, z \mapsto \frac{ (\omega^2-\omega^3) z + (\omega - 
\omega^4)}{(\omega -\omega^4)z - (\omega^2 -\omega^3)}.
\]
Composing this with $U$, we obtain the linear 
fractional transformation corresponding to $W=UV$:
\[
W: \,\,\, z \mapsto \frac{-(\omega -\omega^4)z 
+ (\omega^2 -\omega^3)}{ (\omega^2-\omega^3) z + (\omega - \omega^4)} . 
\]
Making all possible combinations of these linear fractional transformations,
we finally arrive at the 60 elements of the icosahedral group:
\begin{eqnarray*}
z & \mapsto & \omega^j z, \,\, -\frac{1}{\omega^j z} ,  \\
& &  \omega^j
\frac{ -(\omega-\omega^4)\omega^k z + (\omega^2 - \omega^3)}{(\omega^2
-\omega^3)\omega^k z + (\omega -\omega^4)} , \\
& & \omega^j
\frac{ (\omega^2-\omega^3) \omega^k z + (\omega - \omega^4)}{(\omega
-\omega^4)\omega^k z - (\omega^2 -\omega^3)} ,  \\
& &  \qquad j,k = 0,\ldots,4. 
\end{eqnarray*}

\noindent{\bf Section 23.}

\noindent 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.

\noindent 2.(a) Take a good look at Figure 23.2. The Schl\"afli symbol
$\{4,3,3\}$ means that the 3-dimensional faces of the 4-dimensional
cube are ordinary 3-dimensional cubes with Schl\"afli symbol
$\{4,3\}$, and each edge is surrounded with 3 of these.
(b) Let $V_n$, $E_n$, and $F_n$ denote the number of vertices, edges, and
faces of the $n$-dimensional cube. We have $V_n=2^n$, $E_{n+1}=2E_n+V_n$,
$E_1=1$, and $F_{n+1}=2F_n+E_n$, $F_1=0$. These recurrences can be solved easily
and they give $E_n=n2^{n-1}$ and $F_n=\frac{n(n-1)}{2} 2^{n-2}$.  

\noindent 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 Schl\"afli symbol is 
$\{3,3,3\}$



\end{document}
