Question

In the country of Four, there are four cities, A, B, C and D. \(B\) is to the East of \(A\), \(C\) is to the South of \(B\), \(D\) is to the West of \(C\), and \(A\) is to the North of \(D\). The Government plans to connect these four cities by road so that one can go from any city to any other and the total road length is minimum. The distances \(AB,\,BC,\,CD,\,DA\) are all \(10\) km. What should be the total length of the road?

• A. 26.64 km
• B. 27.32 km
• C. 28.30 km
• D. 30 km
• E. 36 km

Hint:

For symmetric point sets (e.g., a square), try connecting with symmetry-respecting segments (like diagonals). They often attain natural lower bounds from required opposite-corner paths.

Answer 1

The Correct Option is C

Step 1: The description places the four cities at the vertices of a square of side \(10\) km. The straight-line distance between opposite cities is each diagonal: \[ AC=BD=10\sqrt{2}\ \text{km}. \] Step 2: Laying two roads along the diagonals \(AC\) and \(BD\) connects all four cities (they meet at the center). Total length \[ L=AC+BD=2\times 10\sqrt{2}=20\sqrt{2}\approx 28.28\ \text{km}\ (\text{rounded } 28.30\ \text{km}). \] Any connected network must contain a path from \(A\) to \(C\) (length at least \(AC\)) and from \(B\) to \(D\) (length at least \(BD\)); the two-diagonal construction attains this lower bound, hence is minimal.
Therefore the minimum total road length is \(\boxed{28.30\ \text{km}}\).