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Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .

Attempt

Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as $$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$ Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.

But from here I am not able to get the value of x and hence $\alpha$. Any suggestions?

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    $\begingroup$ Too bad you haven't really read other solutions. $\endgroup$ – Aqua Dec 2 '18 at 17:06
  • $\begingroup$ @greedoid Why you said that? $\endgroup$ – jayant98 Dec 2 '18 at 18:41
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    $\begingroup$ Because you upvote all solution imediatly after accepting first answer which does not tell you actualy how to find a solution and why to take tis substitution $t=x-1$... and last solution is not really a solution. $\endgroup$ – Aqua Dec 2 '18 at 18:58
  • $\begingroup$ Okay. Understood your point. But, how come you can say that I didn't get the answers from another's method. In case of Robert, he used the concepts that I have been taught previously. In case of Cesareo,Boshu and farruhota, the method is almost same and the same thinking I have used. Also in your case, you have also used geometric approach but by triangle inequality taking another point. But now for selecting answer I would upvote the answer which I think gave another view also. Hope it clears all the "fogs". BTW thanks for suggesting me another method. $\endgroup$ – jayant98 Dec 2 '18 at 19:06
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By letting $x=t+1$ we get an even function $$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}.$$ Now we show that the minimum value is attained at $t=0$: we have to verify $$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}\geq \sqrt{20}$$ or, after squaring, $$4t^2+10+2\sqrt{4t^4-16t^2+25}\geq 20$$ that is $$\sqrt{(5-2t^2)^2+4t^2}\geq 5-2t^2$$ which trivially holds.

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    $\begingroup$ $\alpha$=20....? $\endgroup$ – jayant98 Dec 2 '18 at 16:33
  • $\begingroup$ Yes, now try to prove it. $\endgroup$ – Robert Z Dec 2 '18 at 16:36
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    $\begingroup$ $\alpha=20$ is correct by I'd prefer an algebraic approach. $\endgroup$ – Robert Z Dec 2 '18 at 16:45
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    $\begingroup$ The symmetry in $t$ gives you the minimum candidate. The proof is up to you (algebraic or geometric). $\endgroup$ – Robert Z Dec 2 '18 at 16:51
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    $\begingroup$ @jayant98 See my (algebraic) conclusion. $\endgroup$ – Robert Z Dec 2 '18 at 16:56
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There are a number of ways to do this, including brute force and calculus, but since you already found out the rather nice geometric interpretation as the sum of the distances from the given points, let's do that.

A few things will come in handy here.

  1. The line on which $A$ and $B$ lie is $y=x+3$, which is parallel to your line of interest (i.e. $y=x+2$).
  2. Both lines have slope $1$.

Here's the general idea: Note that the sum of the distances is minimum along the perpendicular bisector of the line segment $AB$; the actual minimum is at the point of intersection of the bisector and $y=x+3$, but since you have an additional constraint, you find the intersection of the bisector with $y=x+2$, call it $C$.

Note that if you drop perpendiculars to $x$ and $y$ axes respectively from $A$ and $B$, they intersect at $A'=(0,2)$ and $B'=(2,4)$. You'll notice that their mid point is $C$. Then $C$ turns out to be $(1,3)$. That gives us that $\alpha=20$.

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  • $\begingroup$ Ah, I solved through this method. See the comments on @Robert . Thanks for the help. $\endgroup$ – jayant98 Dec 2 '18 at 16:48
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Your method is also really good. Consider the two points $A(-1,2)$ and $B(2,5)$. We need to find the point $(x,x+2)$ on the line $y=x+2$ for which the sum $AD+BD$ is minimum (see the graph below).

$\hspace{3cm}$enter image description here

You reflect the point $A$ over the line $y=x+2$ to find the point $C(0,1)$. The line through $C$ and $B$ is $y=2x+1$. The two lines intersect at $D(1,3)$, which is the solution of the given problem. Hence, when $x=1$, the sum will be minimum $\sqrt{20}$.

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  • $\begingroup$ Thanks for another method (2nd para). $\endgroup$ – jayant98 Dec 2 '18 at 16:59
  • $\begingroup$ You are welcome. It is actually the method you started and asked suggestion to continue. Good luck. $\endgroup$ – farruhota Dec 2 '18 at 17:06
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Hint.

Calling $f(x) = \sqrt{x^2+(x+1)^2}$ we seek for

$$ \min_x f(x) + f(x-3) $$

enter image description here

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If we write this as $$\sqrt{x^2+(x+1)^2}+\sqrt{(x-2)^2+(x-3)^2}$$ then we are searching for a point $T$ on line $y=x$ for which $TA+TB$ takes minumim where $A(0,-1)$ and $B(3,2)$.

Now this is well know problem froma ancient greek. Reflect $A$ acros this line and get $A'(-1,0)$. By triangle inequality we can see that $TA+TB\geq A'B$ and that minimum is achieved at intersection of lines $y=x$ and line $A'B$ which is $y = {x +1 \over 2}$.

So $\alpha = A'B^2 = 20$.

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