Length of the red line

Since a square is symmetrical about its center, we can consider the problem with half of the square. Let us consider the left half and further break down the problem to considering red lines $AE$ and $CE$ first. Let us fixed the distance of point $E$ to $CA$ to say $d$ . Then the locus of point $E$ is a line parallel and distance $d$ from $CA$ (orange dash line). The shortest path from $A$ to $E$ to $C$ is one that is travelled by light where the orange dash line serves as a mirror. This follows the Fermat's principle of least time . We can of course prove this mathematically but I am not going to do it here. The light path is one when $AE=CE$ and $\triangle ACE$ is isosceles. So we can expect the shortest five red lines to occurs when $\triangle ACE$ and $\triangle BDF$ are congruent isosceles triangles. Then the shortest $EF$ is also apparent.

Let the base angle of the isosceles triangles be $\theta$ . The sum of lengths of the five red lines is then given by:

$\begin{aligned} S & = 4\cdot BF + 2\cdot FO \\ & = 4\sec \theta + 2(1-\tan \theta) & \small \blue{\text{To find the minimum}} \\ \frac {dS}{d\theta} & = 4\sec \theta \tan \theta - 2\sec^2 \theta & \small \blue{\text{Equating to 0}} \\ 2 \tan \theta & = \sec \theta & \small \blue{\text{For }\sec \theta \ne 0} \\ \sin \theta & = \frac 12 \\ \implies \theta & = 30^\circ \\ \implies \min (S) & = \frac 8{\sqrt 3} + 2 - \frac 2{\sqrt 3} = 2 + \sqrt{12} \end{aligned}$

Therefore $a+b = 2+12=\boxed{14}$ .

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Discussion: If the problem is not restricted to a square. When $EF = AF$ , then all the five sides have the same length and are sides of a regular hexagon. In fact, the five sides are exactly a unit building block of beehive. Of course, natural always preserves resources and follows the principle of least action , a present day's spinoff of the Fermat's principle. The moral of this problem is we can always look at nature to solve mathematical problems.

While I can't prove it, the minimum solution should be symmetric. Let C be (0,0). Then E and F must lie on the line y = 1, and they must reflect each other about the line x = 1. Thus the four segments from the vertices have the same length and the red line can be expressed as $4AE + EF$ . Let E and F be (t, 1) and (2-t, 1), respectively. The total length can now be expressed in terms of t and differentiated.

$L = 4AE + EF = 4\sqrt{t^2+1} + (2 - 2t)$

$\dfrac{dL}{dt} =\dfrac{4t}{ \sqrt{t^2+1} }-2$

Setting the first derivative = 0:

$\dfrac{dL}{dt} = 0 \implies \dfrac{4t}{\sqrt{t^2+1}} = 2 \implies t = \dfrac{\sqrt{3}}{3}$

So $AE = \dfrac{2\sqrt{3}}{3}, \hspace{3mm} EF = 2 - \dfrac{2\sqrt{3}}{3} ,$ $\hspace{3mm} \text{and } L = \dfrac{4\cdot2\sqrt{3}}{3} + \left(2-\dfrac{2\sqrt{3}}{3}\right) = 2+2\sqrt{3} = 2+\sqrt{12}$

The solution is $2 + 12 = \boxed{14}$

One can derive two types of minimum, (1) based on all lines of equal length which gives total length as (10/3)[√7-1]~5.485 and (2) based on all angles on two junctions as equal to 120° each which gives total length as 2(1+√3)~5.464 a global minimum.

If E and F are not at the same altitude, then replace the one which has more perimeter from the other two edges connected to it (one of them if they are equal) with a point in the same altitude as the other point, but in the same latitude as the original point. Now if they are not at the middle-line altitude they are not minimal, so they are on the line paralel to AB, passing through the square's midpoint. It's not hard to see that E is on the left half of the square and F is on the right. From this, we see that EF passes through the squares mindpoint O, and E and F are entirely independent, minimalising the sum of 5 edges is equivalent to minimalising AE+EO+ED, and similarly to F, so E stands where that sum is minimal and F is symmetric to it. Thus it's minimising [\2\sqrt{x^2+1}+1-x]\ and multiplying 2 times, you can see the derivative in the other comments and the minimal value.