Shortest Pentaway

A Steiner point is a road junction where three roads meet, each making an angle of $120^\circ$ with the others. The shortest road network is one containing $3 = 5-2$ Steiner points. One symmetrical network is below.

Using the Sine Rule four times, we have $\begin{array}{rcl} a & = & \displaystyle \frac{2 \sin 42^\circ}{\sin 120^\circ} \\ b & = & \displaystyle \frac{2 \sin 18^\circ}{\sin 120^\circ} \\ b + c & = & \displaystyle \frac{2\sin 54^\circ}{\sin 60^\circ} \\ c + d & = & \displaystyle \frac{2\sin 66^\circ}{\sin 60^\circ} \end{array}$ and so the shortest road network has length $2a + 2b + 2c + d = \boxed{7.78231}$ km.

It is known that the steiner tree for 5 points has 3 steiner points. So we know that the steiner tree would look something like this:

To find the steiner tree, we make use of transformations:

Our primary objective is to find the position of $O$

First, rotate line $AB$ 60 degrees about point $A$ . Since we want to minimise $OP + PA + PB$ for any randomly chosen $O$ , $OP+PA+PB=OJ$ (a straight line)

Do the same for side $CD$ .

Now the problem is reduced to finding the steiner tree of a triangle. This would enable us to find $O$ :

To do so, we can employ similar techniques. Rotate line $KJ$ about point $K$ by 60 degrees

It can be seen that the length of steiner tree would be the length of $ML$ , which is the height of equilateral triangle $KJM$ plus the height of triangle $LKJ$ . Through symmetry and some trigonometry, we can find that the length of the steiner tree is:

$\sqrt{3}\left(1+2\cos 12\right)+\sqrt{5+2\sqrt{5}}-2\sin 12=7.78231364665$

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In response to Mark's comment below:

Point $O$ , such that $L$ , $O$ , $O'$ and $M$ are collinear can be found through the intersection of line $LM$ and the excircle of triangle $KJM$ . This is because $OKMJ$ is a cyclic quadrilateral, concluded by noticing that $\angle KMO=\angle OJK$ through triangle congruity. The same technique can be applied to find the positions of the rest of the steiner points