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{\Large \bf MATH 476 -- College Geometry}
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\noindent
{\large \bf Homework Assignment 1}\hfill {\large \bf Solutions}\\
{\large \bf and Proofs 1}

\vspace{.25in}
{\bf Homework:}
\be

\item {\bf Section 2.1, page 59:} 5, 11, 13

\be
\item[(5)] ``The medians of a triangle are concurrent lines'' translates to, ``If three lines are the
      three medians of a triangle, then they are concurrent.'' The converse is, ``If three (distinct) lines are
      concurrent, then they are the three medians of a triangle.''  This is true, although the proof is not easy.
\item[(11)] I will just fill in the missing steps: (a) by (1) and Theorem 1; (b) Theorem 2; (c) steps
     (2) through (5) show that  $r \implies s,$ so by the definition, $a$ is $b.$
\item[(13)] Prove that $p \implies q.$

\begin{tabular}{lll}
&{\sc Conclusions} &{\sc Justifications}\\
(1) & $p$ & Given\\
(2) & $r$ or $s$ & These are all of the possibilities\\
(3) & $s$ & The first possibility\\
(4) & $y$ & Theorem 3\\
(5) & $\sim p$ & Theorem 1\\
   & Contradiction! & We assumed $p$\\
(6) & $r$ & The only remaining possibility\\
(7) & $x$ & Theorem 2\\
(8) & $q$ & Theorem 4\\
\end{tabular}

Therefore, $p\implies q.$
\ee

\item {\bf Section 2.2, page 67:} 6, 9

\be \item[(6)] This is only one pair of possibly many pairs of solutions.  Yours may be different.
\begin{center}
\mbox{\pict{1.5}{1.5}{hw1snf1}\hspace{2in}\pict{2.25}{1.875}{hw1snf2}}
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In the first figure, there are exactly three points, and exactly two points per line.  In the
second figure, there are 5 points, and one of the lines has three points, while the other two
have two points each.  Verify for yourself that both pictures satisfy the axioms.

    \item[(9)] \bob
\ee

\item {\bf Section 2.3, page 74:} 1, 2, 4, 6

\be
\item[(1)] Axiom I-1: We must check that every pair of numbers in $S$ corresponds to a unique line.
     The pair $\{1,2\}$ belongs to exactly one line: the line $\{1,2,3\}.$  The pair $\{1,3\}$
     belongs to the same line (and only that line).  The pair $\{2,3\}$ also gives that line, and
     only that line.  All of the other pairs appear explicitly in the list of lines (and only
     appear once).

     Axiom I-2: The possible sets of three points are: $P_1=\{1,2,3\}, P_2=\{1,2,4\}, P_3=\{1,2,5\},
     P_4=\{1,3,4\}, P_5=\{1,3,5\}, P_6=\{1,4,5\}, P_7=\{2,3,4\}, P_8=\{2,3,5\}, P_9=\{2,4,5\},$
     and $P_{10}=\{3,4,5\}.$  Now it is just a
     matter of making sure that every set of three (noncollinear) points occurs in
     exactly one of the 5 specified
     planes, which I will call $P_1, \ldots P_5,$ in the same order they are printed in the book.
     Then $P_1, P_2, P_4, P_7 \subset P_1; P_1, P_3, P_5, P_8 \subset P_2; P_6\subset P_3;
     P_9\subset P_4; P_{10}\subset P_5.$  Thus, every set of three points does give rise to a plane.
     However, one set of three points seems to give rise to two different planes: The points
     $\{1,2,3\}$ belong to both $P_1$ and $P_2.$  However, since these points are collinear, the
     axiom allows them to belong to more than one plane.

     Axiom I-3: In our case, this is easy to verify, since all lines except the line $\{1,2,3\}$
     consist of exactly 2 points, so if two points lie in a plane, then automatically the line
     determined by them does as well.  The only line we must check is the line $\{1,2,3\}.$
     Since the planes $P_1$ and $P_2$ both contain all three points, and no other plane contains
     any two of them, this axiom is also satisfied.

     Axiom I-4: We must check all possible intersections of planes.  $P_1\cap P_2=\{1,2,3\}$ is a
     line.  $P_1\cap P_3=\{1,4\}$ is a line.  $P_1\cap P_4 =\{2,4\}$ is a line.  $P_1\cap P_5=
     \{3,4\}$ is a line. [Bored yet?] $P_2\cap P_3=\{1,5\}$ is a line.  $P_2\cap P_4=\{2,5\}$ is a
     line.  $P_2\cap P_5=\{3,5\}$ is a line.  $P_3\cap P_4=\{4,5\}=P_3\cap P_5=P_4\cap P_5$ are
     all lines.  Whew!

     Axiom I-5: The points $\{1,2,3,4,5\}$ are noncoplanar (and there are more than 4 of them).
     Just by looking at the way the lines and planes are defined, we can see that Axiom I-5 is
     satisfied.

     What a hoot!  Here's my picture; yours may differ.

     \cpict{2}{1.6}{hw1snf3}


    \item[(2)] I-2, I-3, and I-4 all fail: Points 2, 3, and 4 are non-collinear, but they do not determine a unique plane.  For I-3,
    points 2 and 3 lie in the plane $\{2,3,4,5\},$ but the line $\{1,2,3\}$ does not.  Finally, the intersection of the first and
    third planes is $\{2,3,4\}.$  This contains only part of the line $\{1,2,3\},$ and contains the whole line $\{2,4\}.$  This
    contradicts axiom I-4.
    \item[(4)] By  process similar to what we did in problem 1, we see that axiom I-1 is satisfied (just
    check each possible pair of points).  Each set of three noncollinear points does determine a
      plane: the possibilities are $\{1,2,3\}, \{1,2,4\}, \{1,3,4\},$ and $\{2,3,4\}.$  All of them
      except $\{1,2,4\}$ appear in the list of planes, but since 1,2, and 4 are collinear, the
      points satisfy axiom I-2.  However, axiom I-3 fails:  the points 1 and 2 lie in the first
      plane, but the line they determine, $\{1,2,4\},$ does not.  Axiom I-4 is satisfied since
      the intersections are all 2-point lines. Axiom I-5 is also satisfied: we have four points,
      not all coplanar; every line contains at least two distinct points, and every plane contains
      three noncollinear points.
    \item[(6)] All axioms are satisfied except I-5 since the 7 points are all coplanar.
\ee

\item {\bf Section 2.4, page 87:} 3, 4, 8, 10, 11
\be
 \item[(3)]
 \be \item[(a)] \bob
     \item[(b)] $AC+CD=5<6=AD.$
 \ee
 \item[(4)] (a) 2, (b) 5, (c) 1, (d) 3, (e) 4, (f) 3.
 \item[(8)]
 \be \item[(b)] $x\le 3.$
     \item[(c)] $x\in \R.$
 \ee
 \item[(10)]
 \be \item[(a)] $x\ge -3.$
     \item[(b)] $x\ge -1.$
     \item[(c)] $5\le x\le 6.$
     \item[(d)] $x\ge -3$ or $x\le 6.$  This is equivalent to $x\in \R.$
 \ee
 \item[(11)] Since $WZ=23$ and $XZ=21,$ $WX=2.$  With $WY=17,$ this gives $XY=15.$
 \ee

\ee
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