In the figure is a rectangle ABCD with sides AB = 6, BC = 8. Suppose a dynamic point P starts to move along AC from point A to C at speed of 2 units/second. In the mean time, another dynamic point Q starts to move along CB from point C to B at speed of 1 unit/second. The motion will end when either of them reach the terminal. Find the time of t when \triangle PQC is an isosceles triangle during the moving of point P and Q.
Drag the slider to observe the motion of point P and Q. During the moving, there are 3 times when \triangle PQC could form an isosceles triangle .
First, let’s calculate the diagonal line of the rectangle AC using Pythagorean Theorem.
AC=\sqrt{AB^2+BC^2}
=\sqrt{6^2+8^2} =10
1. When QP=QC, QF is the height from Q to the base PC.
Let x=QC
FC=QC\cos \angle QCF
=\dfrac{BC}{AC}x=\dfrac{4}{5} x
Noticed
2FC+AP=AC
and
AP = 2QC
Then
2\cdot \dfrac{4}{5}x+2x=10
Solve for x
x=\dfrac{25}{9}
2. When QC=PC, noticed
AP+PC=AC
2x+x=10
x=\dfrac{10}{3}
3. When QP=PC,
PC = \dfrac{EC}{\cos \angle ACB}
=\dfrac{\dfrac{QC}{2} }{\dfrac{4}{5} } =\dfrac{5}{8}x
Since AP+PC=AC
2x+\dfrac{5}{8}x = 10
x=\dfrac{80}{21}
Since the speed of point Q is 1unit/s, the time t when \triangle PQC is formed as an isosceles triangle is the length of QC or x, that is
t = \begin{cases} \dfrac{25}{9}s &\text{if } QP=QC \\ \\ \dfrac{10}{3}s &\text{if } QC=PC \\ \\ \dfrac{80}{21}s &\text{if } QP=PC \\ \end{cases}