Let $A,B,C,D$ be an arbitrary quadrilateral (all points are distinct) with
midpoint $M_1$ of side $AB$, $M_2$ of side $BC$, $M_3$ of side $CD$,
$M_4$ of side $DA$.

(a) Prove that $M_1M_2M_3M_4$ is a parallelogram.

(b) Discuss the extreme cases $A = B$ and $A = B = C$ (is the statement of (a) still meaningful
in these cases?).

Sample LaTeX code

\textbf{Problem 1.7.} Let \$A,B,C,D\$ be an arbitrary quadrilateral (all points are distinct) with
midpoint \$M_1\$ of side \$AB\$, \$M_2\$ of side \$BC\$, \$M_3\$ of side \$CD\$,
\$M_4\$ of side \$DA\$.

\vspace{10pt}

(a) Prove that \$M_1M_2M_3M_4\$ is a parallelogram.

\vspace{10pt}

\textbf{Proof.} Answer here.

\vspace{10pt}

(b) Discuss the extreme cases \$A = B\$ and \$A = B = C\$ (is the statement of (a) still meaningful in these cases?).

\vspace{10pt}

\textbf{Answer.}

\vspace{10pt}

(a) Prove that \$M_1M_2M_3M_4\$ is a parallelogram.

\vspace{10pt}

\textbf{Proof.} Answer here.

\vspace{10pt}

(b) Discuss the extreme cases \$A = B\$ and \$A = B = C\$ (is the statement of (a) still meaningful in these cases?).

\vspace{10pt}

\textbf{Answer.}

(a) Consider triangle $\triangle ABC$ and midpoints $A', B', C'$
of the sides $BC, CA$, and $AB$ respectively. Show that
$(A'B')$ || $(AB)$, $(B'C')$ || $(BC)$, $(C'A')$ || $(CA)$,
and that $A'-B'=C'-A=\frac{1}{2}(B-A)$.

(b) Let $A',B',C'$ be three non-collinear points.
Find all triangles $\triangle ABC$ for which $\triangle A'B'C'$ is
the triangle on the midpoints of the sides of triangle
$\triangle ABC$.

Sample LaTeX code

\noindent \textbf{Problem 1.8.} (a) Consider triangle \$\triangle ABC\$ and midpoints \$A',B',C'\$
of the sides \$BC,CA\$, and \$AB\$ respectively. Show that
\$(A'B') \| (AB), (B'C') \| (BC), (C'A') \| (CA)\$,
and that

\begin{displaymath}

A'-B'=C'-A=\tfrac{1}{2}(B-A).

\end{displaymath}

\vspace{10pt}

\textbf{Answer.}

\begin{displaymath}

A'-B'=C'-A=\tfrac{1}{2}(B-A).

\end{displaymath}

\vspace{10pt}

\textbf{Answer.}

Consider the same construction carried out also for the
quadrilateral $ACBD$ (watch the order). This results in
a second parallelogram $N_1N_2N_3N_4.$ Note that $N_2=M_2$
and $N_4=M_4$, so that the two resulting parallelograms
have one diagonal in common.

Use KSEG to make a construction of
of this situation and wiggle the given points. What theorem
can you deduce from these experiments?

Sample LaTeX code

\noindent \textbf{Problem 1.9.} Consider the same construction carried out also for the
quadrilateral \$ACBD\$ (watch the order). This results in
a second parallelogram \$N_1N_2N_3N_4.\$ Note that \$N_2=M_2\$
and \$N_4=M_4\$, so that the two resulting parallelograms
have one diagonal in common.

\vspace{10pt}

Use KSEG to make a construction of of this situation and wiggle the given points. What theorem can you deduce from these experiments?

\vspace{10pt}

\textbf{Answer.}

\vspace{10pt}

Use KSEG to make a construction of of this situation and wiggle the given points. What theorem can you deduce from these experiments?

\vspace{10pt}

\textbf{Answer.}

Prove that the centroid of a triangle is also the centroid of
the triangle of the midpoints of its sides.

Sample LaTeX code

\noindent \textbf{Problem 1.11.} Prove that the centroid of a triangle is also the centroid of
the triangle of the midpoints of its sides.

\vspace{10pt}

\textbf{Proof.}

\vspace{10pt}

\textbf{Proof.}

Consider Figure 1.10 (a) State and (b) prove the theorem suggested by the figure. (See link for picture.)

Figure 1.10 consists of a triangle $\triangle ABC$ with medians crossing
at the centroid $G$. The quadrilateral connecting the midpoint
of $AB$ to the midpoint of $BG$ to the midpoint of $CG$ to the midpoint
of $CA$ is drawn into the figure. If you make this construction
in KSEG and wiggle the figure the solution of this problem is easy.

Sample LaTeX code

\noindent \textbf{Problem 1.12.} Consider Figure 1.10 (a) State and (b) prove the theorem suggested by the figure.

\vspace{10pt}

(a) \textbf{Theorem.}

\vspace{10pt}

(b) \emph{Proof.}

\vspace{10pt}

(a) \textbf{Theorem.}

\vspace{10pt}

(b) \emph{Proof.}

Let $ABCD$ be a parallelogram. Prove that the lines joining $A$ to the
midpoints of the opposite sides trisect the diagonal not through $A$.

Sample LaTeX code

\noindent \textbf{Problem 1.13.} Let \$ABCD\$ be a parallelogram. Prove that the lines joining \$A\$ to the
midpoints of the opposite sides trisect the diagonal not through \$A\$.

\vspace{10pt}

\textbf{Proof.}

\vspace{10pt}

\textbf{Proof.}

Let $ABCD$ be a quadrilateral. Consider the points

$E = \frac{1}{3}(A+2B),\ \ F=\frac{1}{3}(2B+C),\ \ G=\frac{1}{3}(C+2D),\ \ H=\frac{1}{3}(2D+A)$

and prove that $EFGH$ is a parallelogram.
Sample LaTeX code

\textbf{Problem 1.16a.} Let \$ABCD\$ be a quadrilateral. Consider the points

\begin{displaymath}

E = \frac{1}{3}(A+2B),\ \ F=\frac{1}{3}(2B+C),\ \ G=\frac{1}{3}(C+2D),\ \ H=\frac{1}{3}(2D+A)

\end{displaymath}

and prove that \$EFGH\$ is a parallelogram.

\vspace{10pt}

\textbf{Proof.}

\begin{displaymath}

E = \frac{1}{3}(A+2B),\ \ F=\frac{1}{3}(2B+C),\ \ G=\frac{1}{3}(C+2D),\ \ H=\frac{1}{3}(2D+A)

\end{displaymath}

and prove that \$EFGH\$ is a parallelogram.

\vspace{10pt}

\textbf{Proof.}

Prove that the points with barycentric coordinates $a$ = constant are
the points on the line $\ell$ parallel to $\ell_{BC}$.

Sample LaTeX code

\noindent \textbf{Problem 1.17.} Prove that the points with barycentric coordinates \$a\$ = constant are
the points on the line \$\ell\$ parallel to \$\ell_{BC}\$.

\vspace{10pt}

\textbf{Proof.}

\vspace{10pt}

\textbf{Proof.}

Let $A$ be the color red, $B$ the color blue, $C$ the color yellow. If one
mixes $a$ quarts of red paint with $b$ quarts of blue paint and $c$ quarts
of yellow paint, one gets $a+b+c$ quarts of paint of color defined by
the formula

$\frac{1}{a+b+c}(aA+bB+cC)$.

E.g. for $a=b=\frac{1}{2}$ but $c=0$ the result is one quart of purple paint.
For $a=0$ but $b=c=\frac{1}{2}$ the result is one quart of green paint.

Mix one quart of paint of color $\frac{1}{6}(A+2B+3C)$ with an unknown quantity
of paint composed of a mixture of only two colors. This results in
an unknown quantity of mixed paint of color $\frac{1}{5}(2A+2B+C)$. Determine all
the unknown quantities and color.

Sample LaTeX code

\noindent \textbf{Problem 1.18.} Let \$A\$ be the color red, \$B\$ the color blue, \$C\$ the color yellow. If one
mixes \$a\$ quarts of red paint with \$b\$ quarts of blue paint and \$c\$ quarts
of yellow paint, one gets \$a+b+c\$ quarts of paint of color defined by
the formula

\begin{displaymath}

\frac{1}{a+b+c}(aA+bB+cC).

\end{displaymath}

\vspace{10pt}

\noindent Mix one quart of paint of color \$\frac{1}{6}(A+2B+3C)\$ with an unknown quantity of paint composed of a mixture of only two colors. This results in an unknown quantity of mixed paint of color \$\frac{1}{5}(2A+2B+C)\$. Determine all the unknown quantities and color.

\vspace{12pt}

\textbf{Answer.}

\begin{displaymath}

\frac{1}{a+b+c}(aA+bB+cC).

\end{displaymath}

\vspace{10pt}

\noindent Mix one quart of paint of color \$\frac{1}{6}(A+2B+3C)\$ with an unknown quantity of paint composed of a mixture of only two colors. This results in an unknown quantity of mixed paint of color \$\frac{1}{5}(2A+2B+C)\$. Determine all the unknown quantities and color.

\vspace{12pt}

\textbf{Answer.}

Let $(A A'),(B B'),(C C')$ be lines concurring in $G$, as in Ceva's theorem. Prove the following identity

$\frac{G-A}{A'-A}\ +\ \frac{G-B}{B'-B}\ +\ \frac{G-C}{C'-C}\ =\ 2$

Sample LaTeX code

\noindent \textbf{Problem 1.20.} Let \$(A A'),(B B'),(C C')\$ be lines concurring in \$G\$, as in Ceva's theorem.
Prove the following identity

\begin{displaymath}

\frac{G-A}{A'-A}+\frac{G-B}{B'-B}+\frac{G-C}{C'-C}=2.

\end{displaymath}

\vspace{10pt}

\textbf{Proof.}

\begin{displaymath}

\frac{G-A}{A'-A}+\frac{G-B}{B'-B}+\frac{G-C}{C'-C}=2.

\end{displaymath}

\vspace{10pt}

\textbf{Proof.}

Let $(A A'),(B B'),(C C')$ be lines concurring in $G$, as in Ceva's theorem. For which $G$ inside $\triangle ABC$ is the number

$\frac{G-A}{G-A'}\ +\ \frac{G-B}{G-B'}\ +\ \frac{G-C}{G-C'}\ =\ x$

a maximum? (Note that each of these numbers is negative.)

Sample LaTeX code

\noindent \textbf{Problem 1.21.} Let \$(A A'),(B B'),(C C')\$ be lines concurring in \$G\$, as in Ceva's theorem.
For which \$G\$ inside \$\triangle ABC\$ is the number

\begin{displaymath}

\frac{G-A}{G-A'}+\frac{G-B}{G-B'}+\frac{G-C}{G-C'}=x

\end{displaymath}

a maximum? (Note that each of these numbers is negative.)

\vspace{10pt}

\textbf{Proof.}

\begin{displaymath}

\frac{G-A}{G-A'}+\frac{G-B}{G-B'}+\frac{G-C}{G-C'}=x

\end{displaymath}

a maximum? (Note that each of these numbers is negative.)

\vspace{10pt}

\textbf{Proof.}

Consider the indicated ratios in Figure 1.14 (described below). Determine
$X$ as a point of $(CD)$ and $(BE)$, i.e. $X=(CD)(BE)$, by applying Menelaus'
theorem.

Figure: On $\triangle ABC$, the point $D$ lies on side $AB$ with the numbers
3 and 2 printed along $AD$ and $DB$ respectively. These indicate a ratio, not
absolute distances. (Review the difference between proportions of
collinear line segments and measuring distances in the class notes.)
The numbers 4 and 1 are printed along $AE$ and $EC$ respectively.

Sample LaTeX code

\noindent \textbf{Problem 1.22.} Consider the indicated ratios in Figure 1.14. Determine
\$X\$ as a point of \$(CD)\$ and \$(BE)\$, i.e. \$X=(CD)(BE)\$, by applying Menelaus'
theorem.

\vspace{10pt}

\textbf{Proof.}

\vspace{10pt}

\textbf{Proof.}