Question

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.

Collected in the board: Rectangle

Steven Zheng posted 1 week ago

Answer

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}


Steven Zheng posted 1 week ago

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