Subfield: Geometry


Question: Given noncollinear points $A, B, C$, segment $\overline{A B}$ is trisected by points $D$ and $E$, and $F$ is the midpoint of segment $\overline{A C} . \overline{D F}$ and $\overline{B F}$ intersect $\overline{C E}$ at $G$ and $H$, respectively. If $[D E G]=18$, compute $[F G H]$.

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Solution:
Compute the desired area as $[E G F B]-[E H B]$. To compute the area of concave quadrilateral $E G F B$, draw segment $\overline{B G}$, which divides the quadrilateral into three triangles, $\triangle D E G, \triangle B D G$, and $\triangle B G F$. Then $[B D G]=[D E G]=18$ because the triangles have equal bases and heights. Because $D, G$, and $F$ are collinear, to compute $[B G F]$ it suffices to find the ratio $D G / G F$. Use Menelaus's Theorem on $\triangle A D F$ with Menelaus Line $\overline{E C}$ to obtain

$$
\frac{A E}{E D} \cdot \frac{D G}{G F} \cdot \frac{F C}{C A}=1
$$



Because $E$ and $F$ are the midpoints of $\overline{A D}$ and $\overline{C A}$ respectively, $A E / E D=1$ and $F C / C A=$ $1 / 2$. Therefore $D G / G F=2 / 1$, and $[B G F]=\frac{1}{2}[B D G]=9$. Thus $[E G F B]=18+18+9=45$.

To compute $[E H B]$, consider that its base $\overline{E B}$ is twice the base of $\triangle D E G$. The ratio of their heights equals the ratio $E H / E G$ because the altitudes from $H$ and $G$ to $\overleftrightarrow{B E}$ are parallel to each other. Use Menelaus's Theorem twice more on $\triangle A E C$ to find these values:

$$
\begin{gathered}
\frac{A D}{D E} \cdot \frac{E G}{G C} \cdot \frac{C F}{F A}=1 \Rightarrow \frac{E G}{G C}=\frac{1}{2} \Rightarrow E G=\frac{1}{3} E C, \text { and } \\
\frac{A B}{B E} \cdot \frac{E H}{H C} \cdot \frac{C F}{F A}=1 \Rightarrow \frac{E H}{H C}=\frac{2}{3} \Rightarrow E H=\frac{2}{5} E C .
\end{gathered}
$$

Therefore $\frac{E H}{E G}=\frac{2 / 5}{1 / 3}=\frac{6}{5}$. Thus $[E H B]=\frac{6}{5} \cdot 2 \cdot[D E G]=\frac{216}{5}$. Thus $[F G H]=45-\frac{216}{5}=\frac{9}{5}$.
The method of mass points leads to the same results as Menelaus's Theorem, but corresponds to the physical intuition that masses on opposite sides of a fulcrum balance if and only if the products of the masses and their distances from the fulcrum are equal (in physics-speak, the net torque is zero). If a mass of weight 1 is placed at vertex $B$ and masses of weight 2 are placed at vertices $A$ and $C$, then $\triangle A B C$ balances on the line $\overleftrightarrow{B F}$ and also on the line $\overleftrightarrow{C E}$. Thus it balances on the point $H$ where these two lines intersect. Replacing the masses at $A$ and $C$ with a single mass of weight 4 at their center of mass $F$, the triangle still balances at $H$. Thus $B H / H F=4$.

Next, consider $\triangle B E F$. Placing masses of weight 1 at the vertices $B$ and $E$ and a mass of weight 4 at $F$, the triangle balances at $G$. A similar argument shows that $D G / G F=2$ and that $E G / G H=5$. Because $\triangle D E G$ and $\triangle F H G$ have congruent (vertical) angles at $G$, it follows that $[D E G] /[F H G]=(D G / F G) \cdot(E G / H G)=2 \cdot 5=10$. Thus $[F G H]=[D E G] / 10=$ $\frac{18}{10}=\frac{9}{5}$.

Final answer: $\frac{9}{5}$