MathLabs

Problem 4

A soldier must investigate mines in an equilateral triangular region. His detector has radius equal to one-half the triangle’s altitude, and he starts at one vertex. Determine the shortest path that checks the whole region.
Step 4 of 4: Length and coverage
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In plain words

The detector disks along this broken line fill the triangular region.

L=72−34=27−34L=\frac{\sqrt7}{2}-\frac{\sqrt3}{4}=\frac{2\sqrt7-\sqrt3}{4}
Detailed analysis

Let YY be the point on segment XCXC with CY=rCY=r. Then XY=XC−rXY=XC-r, so the path A→X→YA\to X\to Y has length AX+XC−r=LAX+XC-r=L. The two-segment tube of radius rr covers the triangle: the distances from the omitted vertices to the relevant segment are at most rr (with XB=CY=rXB=CY=r), and the remaining boundary is contained between these segments. Thus this path checks the whole region; reversing the roles of B,CB,C gives the symmetric path.