3. Suppose $A B C D$ is a rectangle whose diagonals meet at $E$. The perimeter of triangle $A B E$ is $10 \pi$ and the perimeter of triangle $A D E$ is $n$. Compute the number of possible integer values of $n$.
Answer: 47
Solution: For each triangle $\mathcal{T}$, we let $p(\mathcal{T})$ to denote the perimeter of $\mathcal{T}$.
First, we claim that $\frac{1}{2} p(\triangle A B E)<p(\triangle A D E)<2 p(\triangle A B E)$. To see why, observe that
$$
p(\triangle A D E)=E A+E D+A D<2(E A+E D)=2(E A+E B)<2 p(\triangle A B E),
$$

Similarly, one can show that $p(\triangle A B E)<2 p(\triangle A D E)$, proving the desired inequality.
This inequality limits the possibility of $n$ to only those in $(5 \pi, 20 \pi) \subset(15.7,62.9)$, so $n$ could only range from $16,17,18, \ldots, 62$, giving 47 values. These values are all achievable because
- when $A D$ approaches zero, we have $p(\triangle A D E) \rightarrow 2 E A$ and $p(\triangle A B E) \rightarrow 4 E A$, implying that $p(\triangle A D E) \rightarrow \frac{1}{2} p(\triangle A B E)=5 \pi$
- similarly, when $A B$ approaches zero, we have $p(\triangle A D E) \rightarrow 2 p(\triangle A B E)=20 \pi$; and
- by continuously rotating segments $A C$ and $B D$ about $E$, we have that $p(\triangle A D E)$ can reach any value between $(5 \pi, 20 \pi)$.

Hence, the answer is 47 .