finding the minimal distance of 2 cars without trigonometry

Question

Given $\angle ABC =120^{\circ}$, and the distance $AB=700\,\text{km}$:

A car is driving from point $A$ to point $B$ at the speed of $80\,\mathrm{km}\cdot\mathrm{h}^{-1}$.

A second car is driving (started same time when the first car) from point $B$ to point $C$ at the speed of $100\,\mathrm{km}\cdot\mathrm{h}^{-1}$.

For how long since the two cars started driving, their distance will be minimal?

Remark: no trigonometry is allowed.

I tried to build right triangles but it didn't work for me.

Thanks.

Answer 1

For $\angle ABC=90^o$: $$\min_{0 \leq t \leq \frac{70}{8}}{(700-80t)^2+(100t)^2}$$ this can be rewritten as $$700^2+\min_{0 \leq t \leq \frac{70}{8}}{(80^2+100^2)t^2-2\cdot80\cdot700t}$$ As a quadratic with roots $t=0$ and $t=\frac{2\cdot80\cdot700}{80^2+100^2}$ it is minimized at the midpoint of the roots, that is in $$t^*=\frac{80\cdot700}{80^2+100^2}=3.41$$

Answer 2

Hints

Take a look at the following figure. I think that it will help you to go through.

$1.$ Write the coordinates of the cars at time $t$.

$2.$ Compute their distance $d=d(t)$ using their coordinates.

$3.$ Minimize the function $d^2=d^2(t)$ where $0 \le t \le \frac{700}{80}$.

Answer 3

Say we assume the x-axis to be along $BC$. The origin coincides with $B$.

Thus equation of $AB$ is $y=\tan 120^\circ \cdot x$ i.e. $y=-\sqrt{3}x$

So by the problem, $x^2+3x^2=700^2 \Rightarrow x=\pm 350$

As evident from this problem, here $x=-350$

Hence the co-ordinates of $A=(-350,350\sqrt{3})$

So the instantaneous co-ordinates of 1st car are $(80 \cos 60^\circ t - 350, 350\sqrt{3} - 80 \sin 60^\circ t)$ and that of 2nd car is $(100t,0)$.

So you have to minimise the function $f(x)=\sqrt{(80 \cos 60^\circ t - 350 - 100t)^2+(350\sqrt{3} - 80 \sin 60^\circ t)^2}$

or in other words,

you have to minimise the function $$g(x)=f^2(x)=(80 \cos 60^\circ t - 350 - 100t)^2+(350\sqrt{3} - 80 \sin 60^\circ t)^2$$ $$=(40t - 350 - 100t)^2+(350\sqrt{3} - 40\sqrt{3} t)^2$$ $$=( 350 + 60t)^2+(350\sqrt{3} - 40\sqrt{3} t)^2$$

Hope this helps.