# finding the minimal distance of 2 cars without trigonometry

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.

• The angle is $120$ not $90$ that's why i don't understand how to solve without using trigonometry.
– roy
Dec 23 '15 at 17:47
• What do you mean by trigonometry here? All answers are going to implicitly build triangles at some point - even the distance formula is just the Pythagorean Theorem. Dec 24 '15 at 8:24
• i mean, i can use Pythagorean Theorem which is not involving the need of trigonometry. distance formula as well. it's a question that given before learning trigonometry at all. only plain simple distance formula /Pythagorean Theorem or any basic geometric use.
– roy
Dec 25 '15 at 20:59

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 $$t^*=\frac{80\cdot700}{80^2+100^2}=3.41$$

• i didn't understand your hint, and it should be, according the answer : $t=2.5$ and in your function it's $t=3.41$
– roy
Dec 23 '15 at 17:42
• oh i made a mistake.. see my edit qustion.
– roy
Dec 23 '15 at 17:46
• @roy, good luck Dec 23 '15 at 17:48
– roy
Dec 23 '15 at 17:49

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.

• thank you but i wanted to know if it solvable without using trigonometry.
– roy
Dec 23 '15 at 18:03
• @roy I used co-ordinate geometry and calculus. Where did you find trigo? Dec 23 '15 at 18:04
• hmm $sin,cos$..
– roy
Dec 23 '15 at 18:05
• @roy That's elementary usage.... Anyways.. Dec 24 '15 at 6:01

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 $0 \le t \le \frac{700}{80}$. • i feel bad for the effort and time, but the angle is $120$ not $90$, but thank you!
– roy
Dec 23 '15 at 17:58
• @roy: Take a look at the new pic. :) Hope this help. Take care when you write questions. :) Dec 23 '15 at 18:10
• thank you:) i will next time! i didn't get how did you now those values and also how you continue from there.
– roy
Dec 23 '15 at 18:13
• @roy: Try to write down the coordinates of the cars at time $t$. It is a hint and you should discover the details by yourself! :) Dec 23 '15 at 18:16
• ok, ill try just it seems like you used parametric line by vector and i don't know that material.
– roy
Dec 23 '15 at 18:19