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\centerline{MATH 392 -- Geometry Through History}
\centerline{Selected Solutions for Problem Set 3}
\centerline{March 1, 2016}
\vskip 10pt
\noindent
I.  Show under the assumption of HAA that given any angle $\angle ABC < 180^\circ$, there
is a line $\ell$ that is simultaneously parallel to $\overrightarrow{BA}$ in the 
direction of $A$ and to $\overrightarrow{BC}$ in the direction of $C$.  Hint: use point (3)
on page 47 of McCleary. 
\vskip 10pt
\noindent
{\it Solution:}  First construct an angle bisector $\overrightarrow{BD}$ for the angle $\angle ABC$.
By the assumption, both of the ``halves,'' the angles $\angle ABD = \angle CBD = \theta$ are \emph{acute}.
(Recall that this construction uses only Postulates I - IV in Book I of Euclid, so it is 
valid in the hyperbolic case as well.)  Now point (3) on page 47 of McCleary says that every
acute angle is the angle of parallelism $\Pi(x)$ for some $x > 0$.  Let $D'$ be on the 
ray $\overrightarrow{BD}$ such that $BD' = x$ where $\Pi(x) = \theta$ for the $\theta$ above (one half the original 
angle).  Then if we erect a perpendicular
$\overrightarrow{D'E}$ in the direction of $A$ to $\overrightarrow{BD}$, we will have
that $\overrightarrow{D'E}$ and $\overrightarrow{BA}$ are parallel in the direction of $A$.
Similarly, if we erect a perpendicular
$\overrightarrow{D'F}$ in the direction of $C$ to $\overrightarrow{BD}$, we will have
that $\overrightarrow{D'F}$ and $\overrightarrow{BC}$ are parallel in the direction of $C$.
Finally the points $E,D',F$ must all be along a single line since $\angle ED'B = \angle FD'B$
are both right angles.  The line $\ell$ is $\overleftrightarrow{EF}$.  
\vskip 10pt
\noindent 
II.  Note:  All of the following facts about 3-dimensional geometry
are proved in McCleary.  Suggestion:  try to 
work out proofs for yourselves using just properties following
from Euclid's Postulates I-IV (not V), then use the text to check your work.  
\begin{enumerate}
\item[(C)]  If $T$ and $T'$ are perpendicular planes, and $\ell$ is a line on $T$, then 
$\ell \perp T'$ is equivalent to $\ell \perp m$ for $m = T\cap T'$.
\item[] {\it Solution:}  First note that the way this is stated 
means that we must prove two implications (``is equivalent to'' means the same
thing as ``if and only if'').   So to begin, assume that $T$ and $T'$ are perpendicular
planes and $\ell$ is a line on $T$ with $\ell \perp T'$.  Suppose $\ell \cap T' = P$.
The point $P$ must be a point on the line $T \cap T' = m$.   By definition (Definition 4.14
in McCleary), 
$\ell$ meets every line through $P$ in $T'$ at a right angle.  Since $m$ is one of 
those lines, $\ell \perp m$.  Conversely, suppose $\ell$ is a line on $T$ that
meets $m = T\cap T'$ at a right angle.   Since the planes $T$ and $T'$ are assumed
perpendicular, there is a line $\ell'$ in $T'$ that meets $T$ in a right angle (see figure on 
right on page 59 in McClearly).  Take any point $R$ on $\ell'$ not lying in $T$
and join it to $P = \ell \cap m$ with a line.  That line $\overleftrightarrow{RP}$
also meets $\ell$ at a right angle 
by part (B).  But then $\ell$ meets two different lines through $P$ 
in $T'$ at a right angle, so $\ell$ is perpendicular to $T'$ by part (A).  
\item[(D)]  If $\ell_1 \perp T$ and $\ell_2 \perp T$, then $\ell_1,\ell_2$
are coplanar and non-intersecting (but not necessarily parallel -- this is a
difference in the hyperbolic case!)
\item[] {\it Solution:}  We want to assume that $\ell_1$ and $\ell_2$ are \emph{distinct} 
lines, of course, so they must meet $T$ at two distinct points.
Let $\ell_1 \cap T = P$ and $\ell_2\cap T = Q$.  Let $T'$
be the plane containing $Q$ and $\ell_1$.  Then $T$ and $T'$ are perpendicular planes
by definition.  We want to show that $T'$ must contain
$\ell_2$ as well.   Note that $T' \cap T$ contains both 
points $P$ and $Q$, so $T'\cap T = \overleftrightarrow{PQ}$.  Let $\ell'$ 
be the line through $Q$ perpendicular to $T'\cap T$ in $T'$.   Then $\ell'$ is 
also perpendicular to $T$ by part (C).  But there is only one line through $Q$ that is
perpendicular to the plane $T$.  Hence $\ell' = \ell_2$, and $\ell_2$ must also 
lie in $T'$.     
\item[(E)]
Given a line $\ell$ and a plane $T$ not containing $\ell$, there exists a unique plane
$T'$ containing $\ell$ that is perpendicular to $T$.  
\item[] {\it Solution:}  Note that we really need to assume \emph{in addition} that 
$\ell$ is not itself perpendicular to $T$ (see the class notes where we discuss
the analog of the Playfair Postulate for 3-dimensional hyperbolic geometry).  
If that is the case, since $T$ does not contain $\ell$, there is a point $P$
on $\ell$ that is not in $T$.  Let $m$ be the line through $P$ perpendicular to $T$.
The plane containing $\ell$ and $m$ contains $\ell$ and is perpendicular to $T$
by definition.  This shows existence.  Any second such plane would have
to contain $\ell$ and some line perpendicular to $T$ passing through a point $Q$
in $\ell$.  But then the two planes must coincide because of part (D) -- any two
lines distinct lines perpendicular to the same plane are coplanar.  
\end{enumerate}
\vskip 10pt
\noindent
IV.  Why is the formula
$$\cosh(c/k) = \cosh(a/k)\cosh(b/k)$$
called the hyperbolic form of the Pythagorean theorem?  Hint: Use III, part (C) and
think about what happens as $k \to \infty$.  
\vskip 10pt 
\noindent
{\it Solution:}  Part (C) of III should show that the Taylor series of $\cosh(t)$
at $t = 0$ looks like this:
$$\cosh(t) = \sum_{m=0}^\infty \frac{t^{2m}}{(2m)!} = 1 + \frac{t^2}{2!} + \frac{t^4}{4!} + \cdots .$$
This series converges to $\cosh(t)$ for all real $t$ (for the same reason that the Taylor
series for $e^t$ converges for all $t$).  If we use this on the given equality we get
$$1 + \frac{(c/k)^2}{2!} + \frac{(c/k)^4}{4!} + \cdots = 
(1 + \frac{(a/k)^2}{2!} + \frac{(a/k)^4}{4!} + \cdots)(1 + \frac{(b/k)^2}{2!} + \frac{(b/k)^4}{4!} + \cdots).$$
Subtracting 1 from both sides, multiplying out the right side and rearranging a bit yields 
$$\frac{c^2}{2k^2} = \frac{a^2}{2k^2} + \frac{b^2}{2k^2} + \frac{1}{k^4}({\rm other\ terms}).$$
Hence if we multiply both sides by $2k^2$ and let $k \to \infty$ we get
$$c^2 = a^2 + b^2.$$
Recall that $a,b,c$ were the lengths of the sides in a (hyperbolic) right triangle.  Hence
if $k \to \infty$, or equivalently if $a,b,c\to 0$, in the limit, we have the Euclidean 
Pythagorean relation.  
\end{document}
