Putnam 1999, Problem B-1

Right triangle \(ABC\) has right angle at \(C\) and \(\angle BAC = \theta\); the point \(D\) is chosen on \(AB\) so that \(|AC| = |AD| = 1\); the point \(E\) is chosen on \(BC\) so that \(\angle CDE = \theta\). The perpendicular to \(BC\) at \(E\) meets \(AB\) at \(F\). Evaluate \(\lim_{\theta \to 0} |EF|\).

Solution:
The answer is 1/3. Let $G$ be the point obtained by reflecting $C$ about the line $AB$. Since $\angle ADC = \frac{\pi-\theta}{2}$, we find that $\angle BDE = \pi - \theta - \angle ADC = \frac{\pi-\theta}{2} = \angle ADC = \pi - \angle BDC = \pi - \angle BDG$, so that $E,D,G$ are collinear. Hence
$$|EF| = \frac{|BE|}{|BC|} = \frac{|BE|}{|BG|} = \frac{\sin(\theta/2)}{\sin(3\theta/2)},$$
where we have used the law of sines in $\triangle BDG$. But by l'Hôpital's Rule,
$$\lim_{\theta \to 0} \frac{\sin(\theta/2)}{\sin(3\theta/2)} = \lim_{\theta \to 0} \frac{\cos(\theta/2)}{3\cos(3\theta/2)} = 1/3.$$