Let $ABC$ be an acute triangle. $PQRS$ is a rectangle with $P$ on $AB$, $Q$ and $R$ on $BC$, and $S$ on $AC$ such that $PQRS$ has the largest area among all rectangles $TUVW$ with $T$ on $AB$, $U$ and $V$ on $BC$, and $W$ on $AC$. If $D$ is the point on $BC$ such that $AD\perp BC$, then $PQ$ is the harmonic mean of $\frac{AD}{DB}$ and $\frac{AD}{DC}$. What is $BC$?

Note: The harmonic mean of two numbers $a$ and $b$ is the reciprocal of the arithmetic mean of the reciprocals of $a$ and $b$.

[i]2017 CCA Math Bonanza Lightning Round #4.4[/i]
1. **Maximizing the Area of Rectangle \(PQRS\)**:
   - We need to show that the area of \(PQRS\) is maximized when \(Q\) and \(R\) are the midpoints of \(BD\) and \(CD\) respectively.
   - Let \(D\) be the foot of the altitude from \(A\) to \(BC\). Then, \(AD \perp BC\).
   - Let \(PQ = x \cdot AD\) and \(QR = y \cdot BC\).
   - The area of \(PQRS\) is given by \(PQ \cdot QR = x \cdot AD \cdot y \cdot BC\).

2. **Expressing \(PQ\) and \(QR\) in terms of \(x\)**:
   - Let \(PQ = x \cdot AD\). Then, \(\frac{PQ}{AD} = x\).
   - Since \(Q\) is on \(BD\), we have \(BQ = BD \cdot x\) and \(QD = BD \cdot (1 - x)\).
   - Similarly, \(DR = CD \cdot (1 - x)\) and \(QR = BC \cdot (1 - x)\).

3. **Maximizing the Area**:
   - The area of \(PQRS\) is given by:
     \[
     [PQRS] = PQ \cdot QR = (x \cdot AD) \cdot (BC \cdot (1 - x)) = AD \cdot BC \cdot x \cdot (1 - x)
     \]
   - The area of \(\triangle ABC\) is fixed, and we need to maximize \(x(1 - x)\).
   - The function \(f(x) = x(1 - x)\) is a quadratic function, and its maximum value occurs at \(x = \frac{1}{2}\).

4. **Harmonic Mean Condition**:
   - Given that \(PQ\) is the harmonic mean of \(\frac{AD}{DB}\) and \(\frac{AD}{DC}\), we need to find \(BC\).
   - The harmonic mean \(H\) of two numbers \(a\) and \(b\) is given by:
     \[
     H = \frac{2ab}{a + b}
     \]
   - Let \(a = \frac{AD}{DB}\) and \(b = \frac{AD}{DC}\). Then:
     \[
     PQ = \frac{2 \cdot \frac{AD}{DB} \cdot \frac{AD}{DC}}{\frac{AD}{DB} + \frac{AD}{DC}}
     \]
   - Since \(PQ = \frac{1}{2} AD\), we have:
     \[
     \frac{1}{2} AD = \frac{2 \cdot \frac{AD}{DB} \cdot \frac{AD}{DC}}{\frac{AD}{DB} + \frac{AD}{DC}}
     \]

5. **Solving for \(BC\)**:
   - Let \(DB = x\) and \(DC = y\). Then:
     \[
     \frac{1}{2} AD = \frac{2 \cdot \frac{AD}{x} \cdot \frac{AD}{y}}{\frac{AD}{x} + \frac{AD}{y}}
     \]
   - Simplifying, we get:
     \[
     \frac{1}{2} AD = \frac{2 \cdot \frac{AD^2}{xy}}{\frac{AD(x + y)}{xy}} = \frac{2 \cdot AD^2}{AD(x + y)} = \frac{2 \cdot AD}{x + y}
     \]
   - Therefore:
     \[
     \frac{1}{2} = \frac{2}{x + y} \implies x + y = 4
     \]
   - Since \(BC = DB + DC = x + y\), we have:
     \[
     BC = 4
     \]

The final answer is \(\boxed{4}\)