PRMO 2017 question 13

In a rectangle ABCD, E is the midpoint of AB; F is a point on AC such that BF is perpendicular to AC; and FE perpendicular to BD. Suppose BC = 8sqrt(3). Find AB.

  Solution:
1. Set up a simple grid. Put the rectangle so that \[ A=(0,0),\quad B=(x,0),\quad C=(x,y),\quad D=(0,y), \] where \(x=AB\) (unknown) and \(y=BC=8\sqrt3\).

2. Locate the midpoint \(E\). Since \(E\) is the midpoint of \(AB\), \[ E =\Bigl(\tfrac x2,\,0\Bigr). \]

3. Find the foot \(F\) of the perpendicular from \(B\) onto \(AC\). - The diagonal \(AC\) has slope \[ m_{AC} \;=\;\frac{y-0}{x-0}\;=\;\frac{y}{x}\,, \] so any line through \(B\) perpendicular to \(AC\) must have slope \[ m_{BF} \;=\;-\frac1{m_{AC}}\;=\;-\frac{x}{y}\,. \] - Thus the line \(BF\) has equation \[ y \;=\;-\,\frac{x}{y}\,\bigl(X - x\bigr) \quad\bigl(\text{since it passes through }(x,0)\bigr). \] - Meanwhile \(AC\) itself is \[ y \;=\;\frac{y}{x}\,X. \] - Solving \[ -\frac{x}{y}\,(X - x) \;=\;\frac{y}{x}\,X \] for \(X\) gives \[ X_F \;=\;\frac{x^3}{\,x^2+y^2\,}, \quad Y_F \;=\;\frac{y}{x}\,X_F \;=\;\frac{x^2\,y}{\,x^2+y^2\,}. \] So \(\displaystyle F=\Bigl(\frac{x^3}{x^2+y^2}\,,\,\frac{x^2y}{x^2+y^2}\Bigr).\)

4. Write the slope of \(BD\). The diagonal \(BD\) goes from \(B(x,0)\) to \(D(0,y)\), so \[ m_{BD} \;=\;\frac{y-0}{0-x} \;=\;-\frac{y}{x}\,. \]

5. Impose \(FE\perp BD\). - The slope of the line \(FE\) is \[ m_{FE} \;=\; \frac{Y_F - 0}{\,X_F - \tfrac{x}{2}\,} \;=\; \frac{\tfrac{x^2y}{x^2+y^2}}{\;\tfrac{x^3}{x^2+y^2}-\tfrac{x}{2}\,} \;=\; \frac{2xy}{\,x^2-y^2\,}. \] - Two lines are perpendicular exactly when the product of their slopes is \(-1\). So \[ m_{FE}\;\times\;m_{BD} \;=\; \frac{2xy}{x^2-y^2}\;\times\;\Bigl(-\tfrac{y}{x}\Bigr) \;=\;-1. \] Simplifying, \[ -\,\frac{2y^2}{\,x^2-y^2\,}\;=\;-1 \quad\Longrightarrow\quad 2y^2 = x^2 - y^2 \quad\Longrightarrow\quad x^2 = 3y^2. \]

6. Finish. Since \(y=BC=8\sqrt3\), \[ AB \;=\;x \;=\;\sqrt3\;y \;=\;\sqrt3\,(8\sqrt3) \;=\;8\cdot3 \;=\;24. \] --- \(\boxed{24}\)