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# Geometrical Situation: Strikes Again.

Let, $$\displaystyle{x,y\in { R }^{ + }\\ E=\sqrt { { x }^{ 2 }+144 } +\sqrt { { y }^{ 2 }+25 } +\sqrt { { (5-x) }^{ 2 }+{ (12-y) }^{ 2 } } }$$

Then Find Minimum Value of ' E ' .

This situation is arise after solving this Algebraic question This is open Discussion , If you Got anything related to it then share it with us , we all together complete this. I also done lot of work on this Geometrical Situation and still working on this.

###### Note

Here only Geometrical procedure are welcomed ,But Algebraic Solution are not required.

You Can use any tool of mathematics and physics (Except Algebra and Calculus )

Note by Deepanshu Gupta
1 year, 10 months ago

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Let's see here shall we...

What we need to find is the minimum perimeter of triangle which has vertices $$(x,12);(5,y);(0,0)$$

This is something that is best left to reflection. Some say that a ray travels fastest between two given points. I'm going to say something here, it may sound stupid, but I think it is right. The triangle with least perimeter is obtained if we do the following:

1. Send a ray from $$(0,0)$$ to mirror on line $$x=5$$.

2. The reflected ray will get reflected again by mirror on $$y=12$$.

3. The final ray reflected should pass through origin.

Here, our mirrors are perpendicular. Hence, the angle of deviation is always $$\pi$$. Therefore, the only ray that satisfies the above conditions is the ray from origin to $$(5,12)$$.

Thus, our answer must be $$26$$.... @Deepanshu Gupta @Mvs Saketh , can you verify using some other method? · 1 year, 10 months ago

Verified with $$C++$$ code. The answer is correct.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 #include #include #include #include #include int main() { clrscr(); double x,y,sm=100; for(x=0;x<=100;x+=0.1) { for(y=0;y<=100;y+=0.1) { double k; k=sqrt((x*x)+144)+sqrt((y*y+25)); k+=sqrt((5-x)*(5-x)+(12-y)*(12-y)); if(k
· 1 year, 10 months ago

ohh Looping , Really great , I have too IP in mah school day's :) . But It's long time so I forget almost . I know only Java and don't know about C++ . So This goes over to my head :) · 1 year, 10 months ago

Raghav I've posted the solution in my profile you can see that. · 1 year, 10 months ago

But then, how does RMS AM not work, @Raghav Vaidyanathan , does that mean that in this question there are no real values for RMS AM to hold?

Also, please check my response to your report on my electricity problem · 1 year, 10 months ago

I think we made a mistake while putting in boundary values for x and y in RMS-AM. Actually, somewhere we missed out on a negative sign. And yes, in this problem, RMS AM inequality does hold. But we cant claim equality. We can look at this note of Trishit: here

I'll look at your electric field question. But I don't think I can read the reply to my report unless i attempt the question. · 1 year, 10 months ago

But I have done this problem by Cauchy-Schwarz inequality and my answer came 17\sqrt{2}. And now I'm approaching this problem geometrically taking the answer as mine. So you can check this algebrically if I'm not wrong. · 1 year, 10 months ago

Can you find value of x and y which gives the value 17 root 2? · 1 year, 10 months ago

I'm trying. · 1 year, 10 months ago

I am sorry, but I do not know this inequality. Is it the one that Saketh calls dot product inequality? · 1 year, 10 months ago

I'M POSTING THE SOLUTION IN MY PROFILE YOU CAN SEE THAT. · 1 year, 10 months ago

Please don't capitalise all words , because it's generally means that you are shouting . · 1 year, 10 months ago

you can see the inequality in this site Cauchy-Schwarz · 1 year, 10 months ago

Thank you! Please also reply to my previous comment. · 1 year, 10 months ago

The answer isnt 26 which can be verified using RMS - AM inequality,

rather it is 17(root(2)) · 1 year, 10 months ago

Ok, I thought so too. Can you please give me the values for x and y which give answer as $$17(2)^{0.5}$$? · 1 year, 10 months ago

x=12 and y=5 · 1 year, 10 months ago

Hello???? · 1 year, 10 months ago

Sorry, i was wrong, wait let me check · 1 year, 10 months ago

But if we substitute said values into the given equation, we do not get $$17(2)^{0.5}$$. Please check again. · 1 year, 10 months ago

sum of sides of triangle, points are (x,12) (5,y) and (0,0) , sum is minimum means mininum perimeter,, it can then be evaluated . · 1 year, 10 months ago

I thought the same too. We can now do it by using the concept of reflection. · 1 year, 10 months ago

Yes Good Going , you are on right path , do you get more info about it ? · 1 year, 10 months ago

Sorry brothers , I'am not replying earlier beacuse I did not use brilliant after 8.00 PM to 11.30 PM since I gave old year paper in this time.

But really Raghav +1 for your approach bro ! But can you please tell in more detail what do you mean by " the angle of deviation is always $$\pi$$ " .

And I also did it by using optics , I done in this way : Taking Mirror images of respective sides in given line mirrors as show in figure.

So required sum is , $$E=OA+OB+AB\equiv FA+AB+BD$$

Now For to Minimise this 3 different segments of side , They should follow straight line path , Hence F, A,B,D should be collinear. Hence : $${ E }_{ min }=FD=\sqrt { { 10 }^{ 2 }+{ 24 }^{ 2 } } \\ \boxed { { E }_{ min }=26 }$$ .

Verifying Values of x and y. by equating slopes : $$m=\cfrac { -24 }{ 10 } =\cfrac { 12-24 }{ x } =\cfrac { 0-y }{ 10-5 } \\ \boxed { x=5\quad \& \quad y=12 }$$ .

Raghav Vaidyanathan Mvs Saketh what do you think ?

EDIT : Okay I also Get Proof for my claim for straight line . which is by using polygon inequality( here in this case Quadrilateral inequality ) .

Proof : $$\displaystyle{E=OA+OB+AB\equiv FA+AB+BD\ge FD}$$ since sum of three sides is always greater than 4th side , and equality case holds when quadriletral is degenerate means when it turns into straight Line. Hence Proved. · 1 year, 10 months ago

What I meant by saying that the angle of deviation is $$\pi$$ is this.

Consider any ray of light falling on two perpendicular mirrors. The direction of this ray will be completely reversed after two reflections. This means that the angle of deviation is $$\pi$$.

Also, this is a great approach. @Deepanshu Gupta . What I did was actually the same thing.

An analogy to this will be the method to find triangle of least perimeter whose vertices lie on the sides of another triangle. We can do it using your reasoning as above. But also, as I have noticed in many cases, sending a light ray from a point such that it comes back to that point and retraces it's path also works. In both cases, we can see that the triangle of least perimeter is formed by joining the feet of the altitudes, i.e the pedal triangle. · 1 year, 10 months ago

thanks ! I think You are talking of this situation which involves pedal triangle. You Can try This problem Triangle in Triangle which i had posted a while ago . · 1 year, 10 months ago

@Deepanshu Gupta - but bro can you explain why RMS fails here? · 1 year, 10 months ago

I Think I got it , here after applying RMS>AM inequality, and while deciding minimum conditions are :

" Equality of variables must be attained " .

$$\displaystyle{\bullet { x }^{ 2 }=144\Rightarrow x=12\quad .\quad .\quad .\quad .(1)\quad (\because x>0)\\ \bullet { y }^{ 2 }=25\Rightarrow y=5.\quad .\quad .\quad .\quad .(2)\quad (\because y>0)\\ \bullet { (5-x) }^{ 2 }={ (12-y) }^{ 2 }\Rightarrow 5-x=\pm (12-y)\\ case(a):\quad (+)sign\\ y-x=7\quad .\quad .\quad \quad .\quad .(3)\\ 5-12=7\\ -7\neq 7\quad ,\quad Hence\quad \times \\ case(b):\quad (-)sign\\ x+y=17\\ 5+12=17\quad }$$

But in $$case(b)$$ if we are dealing with -ve sign , then It means , question is phrased wrongly. So It should be rephrased as :

Question:

Let, $$\displaystyle{x,y\in { R }^{ + }\\ E=\sqrt { { x }^{ 2 }+144 } +\sqrt { { y }^{ 2 }+25 } +\sqrt { { (5-x) }^{ 2 }+{ (y-12) }^{ 2 } } }$$

Then Find Minimum Value of ' E ' .

Solution:

Using RMS>AM inequality : $$\displaystyle{E\ge \cfrac { [(x+12)]+[(y+5)]+[(5-x)+(y-12)] }{ \sqrt { 2 } } \\ E\ge \cfrac { 10+2y }{ \sqrt { 2 } } }$$ which contradicts. Means we can't proceed further , Atleast I don't .

@Mvs Saketh what do you think ? · 1 year, 10 months ago

Okay Let me check . I'am also amazed , this is one reason I hate inequalities. · 1 year, 10 months ago

Exactly, so RMS simply specifies one of the lower bounds and not the correct answer, but it is holding in that question · 1 year, 10 months ago

Yes while using inequalities we should careful about equality conditions. So checking equality conditions are must in inequalities. After solving that question I already checked the equality condition. And also i Posted equality conditions in that question's comment also. So be careful Next time while using inequalities, atleast check it once . :) · 1 year, 10 months ago

got it thanks :) · 1 year, 10 months ago

We can Try RMS >= AM inequality. · 1 year, 10 months ago