Problem of the Week
Problem E and Solution
Zigzagged

Problem

A fence is to be constructed in a zigzag pattern inside a rectangular field, as shown.

The fence will be constructed so that $\mathrm{\angle }AHD=\mathrm{\angle }EHG$, $\mathrm{\angle }AEH=\mathrm{\angle }BEG$, $\mathrm{\angle }EGH=\mathrm{\angle }FGC$, and $CF=12$ m. If $AB=36$ m and $AD=20$ m, determine the total length of fencing required. That is, determine the value of $AH+EH+EG+FG.$

Solution

Since $ABCD$ is a rectangle, then $AB\parallel CD.$ Then $\mathrm{\angle }EAH=\mathrm{\angle }AHD$, $\mathrm{\angle }AEH=\mathrm{\angle }EHG$, and $\mathrm{\angle }BEG=\mathrm{\angle }EGH$. Since $\mathrm{\angle }AHD=\mathrm{\angle }EHG$, $\mathrm{\angle }AEH=\mathrm{\angle }BEG$, and $\mathrm{\angle }EGH=\mathrm{\angle }FGC$, it follows that $\mathrm{\angle }EAH=\mathrm{\angle }AHD=\mathrm{\angle }EHG=\mathrm{\angle }AEH=\mathrm{\angle }BEG=\mathrm{\angle }EGH=\mathrm{\angle }FGC=\theta .$

Let $R$ and $S$ be on $AB$ such that $RH$ and $SG$ are perpendicular to $AB.$ Let $T$ be on $CD$ such that $ET$ is perpendicular to $CD.$ Then $\mathrm{△}ADH$, $\mathrm{△}ARH$, $\mathrm{△}ERH$, $\mathrm{△}HTE$, $\mathrm{△}GTE$, and $\mathrm{△}ESG$ all have equal angles and a height of $20$ m, so they are all congruent. Let $AR=RE=ES=DH=HT=TG=x.$

Since $\mathrm{△}ADH$ and $\mathrm{△}FCG$ have equal angles, it follows that they are similar. Then $$\begin{array}{rl}\frac{GC}{FC}& =\frac{DH}{AD}\\ \frac{GC}{12}& =\frac{x}{20}\\ GC& =\frac{x}{20}\times 12=\frac{3x}{5}\end{array}$$ Since $DH+HT+TH+GC=36$, then $x+x+x+\frac{3x}{5}=36$. Then $\frac{18x}{5}=36$, so $x=10$. Then $GC=\frac{3(10)}{5}=6$. By the Pythagorean Theorem in $\mathrm{△}FCG$, $$\begin{array}{rl}F{G}^2& =F{C}^2+G{C}^2\\ & ={12}^2+6^2\\ & =180\end{array}$$ Then $FG=\sqrt{180}=6\sqrt{5}$, since $FG>0.$

By the Pythagorean Theorem in $\mathrm{△}ADH$, $$\begin{array}{rl}A{H}^2& =A{D}^2+D{H}^2\\ & ={20}^2+{10}^2\\ & =500\end{array}$$ Then $AH=\sqrt{500}=10\sqrt{5}$, since $AH>0.$

Since $\mathrm{△}ADH$, $\mathrm{△}HTE$, and $\mathrm{△}ETG$ are congruent, it follows that $AH=EH=EG=10\sqrt{5}.$ The total length of fencing required is equal to $AH+EH+EG+FG$, which is $10\sqrt{5}+10\sqrt{5}+10\sqrt{5}+6\sqrt{5}=36\sqrt{5}$ m.