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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.

Rectangle ABCD with top left vertex A, length AB, and width AD. E is on AB, H and G are on CD, with H to the left of G, and F is on BC. The fence starts at vertex A and goes from A to H to E to G to F.

The fence will be constructed so that \(\angle AHD=\angle EHG\), \(\angle AEH=\angle BEG\), \(\angle EGH=\angle FGC\), 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 \(\angle EAH=\angle AHD\), \(\angle AEH=\angle EHG\), and \(\angle BEG=\angle EGH\). Since \(\angle AHD=\angle EHG\), \(\angle AEH=\angle BEG\), and \(\angle EGH=\angle FGC\), it follows that \(\angle EAH=\angle AHD=\angle EHG=\angle AEH=\angle BEG=\angle EGH=\angle FGC=\theta.\)

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 \(\triangle ADH\), \(\triangle ARH\), \(\triangle ERH\), \(\triangle HTE\), \(\triangle GTE\), and \(\triangle ESG\) all have equal angles and a height of \(20\) m, so they are all congruent. Let \(AR=RE=ES=DH=HT=TG=x.\)

Since \(\triangle ADH\) and \(\triangle FCG\) have equal angles, it follows that they are similar. Then \[\begin{aligned} \frac{GC}{FC} &= \frac{DH}{AD}\\ \frac{GC}{12} &= \frac{x}{20}\\ GC &= \frac{x}{20} \times 12=\frac{3x}{5} \end{aligned}\] 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 \(\triangle FCG\), \[\begin{aligned} FG^2 &= FC^2+GC^2\\ &= 12^2 + 6^2\\ &=180 \end{aligned}\] Then \(FG=\sqrt{180}=6\sqrt{5}\), since \(FG>0.\)

By the Pythagorean Theorem in \(\triangle ADH\), \[\begin{aligned} AH^2 &= AD^2+DH^2\\ &= 20^2 + 10^2\\ &=500 \end{aligned}\] Then \(AH=\sqrt{500}=10\sqrt{5}\), since \(AH>0.\)

Since \(\triangle ADH\), \(\triangle HTE\), and \(\triangle ETG\) 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.