3. Let time t BE=4-T BP=3-T PM, starting from point P and perpendicular to m be: (radical number 3 (3-t)/2) 2+((5-t)/2) 2 = t 2.

The solution is T= 13/7, so when point p moves to 13/7cm, APEQ is a diamond.

4. impossible. If and only if T= 13/7, PE = EQ If PE is perpendicular to EQ and T=2, then PE can't be perpendicular and equal to EQ at the same time, so it can't be a square.