IMO 1973 · Problem 4

A soldier needs to check on the presence of mines in a region having the shape of an equilateral triangle. The radius of action of his detector is equal to half the altitude of the triangle. The soldier leaves from one vertex of the triangle. What path should he follow in order to travel the least possible distance and still accomplish his mission?

Official text of the problem. Official problem paper (PDF)

Answer

Let the triangle be \(ABC\), with side length \(a\), and let \(A\) be the starting vertex. Put \(r=\sqrt3a/4\).

Let \(D\) be the midpoint of \(AC\), let \(X\) be the midpoint of \(BD\), and choose \(Y\) on \(XC\) so that \(CY=r\). Follow the broken line \[ \boxed{A\longrightarrow X\longrightarrow Y}. \] Its reflection interchanging \(B\) and \(C\) is equally good. The minimum distance is \[ \boxed{\frac a4\left(2\sqrt7-\sqrt3\right)}. \]

Solution

Write \(h=\sqrt3a/2\) for the altitude, so that the detector radius is \(r=h/2\).

We first show that the proposed route checks the entire triangle. Since \(BD=h\) and \(X\) is its midpoint, \[ XB=XD=r. \] Also, the right triangles \(ADX\) and \(CDX\) give \[ AX=CX=\sqrt{\left(\frac a2\right)^2+\left(\frac h2\right)^2} =\frac{\sqrt7\,a}{4}>r. \] Thus \(Y\) exists strictly between \(X\) and \(C\), and the route has length \[ AX+XY=AX+XC-r =\frac a4(2\sqrt7-\sqrt3). \]

The region checked while traversing any straight segment is convex. Indeed, if points \(Z_1,Z_2\) are detected from positions \(W_1,W_2\) on that segment, then the corresponding intermediate position also lies on the segment, and \[ \begin{aligned} &\left|(1-t)Z_1+tZ_2-\bigl((1-t)W_1+tW_2\bigr)\right|\\ &\qquad\le (1-t)|Z_1-W_1|+t|Z_2-W_2| \le r \end{aligned} \] for every \(0\le t\le1\).

While traversing \(AX\), the soldier detects \(A\), and from \(X\) he detects both \(B\) and \(D\). By this convexity, he checks all of triangle \(ABD\). While traversing \(XY\), he detects \(B,D\) from \(X\) and \(C\) from \(Y\), so he checks all of triangle \(BCD\). These two triangles form \(ABC\), proving that the route succeeds.

It remains to prove optimality. Consider any successful route of finite length \(L\). It must enter the closed disks of radius \(r\) centred at \(B\) and \(C\). These disks are disjoint because \[ BC=a>2r=\frac{\sqrt3\,a}{2}. \] After interchanging \(B,C\) if necessary, choose positions \(P,Q\), visited in that order, such that \[ PB\le r,\qquad QC\le r. \] The lengths of the corresponding portions of the route are at least their endpoint distances. Consequently, \[ L\ge AP+PQ\ge AP+PC-QC\ge AP+PC-r. \]

Define \(D,X\) as above for this labelling. Let \(\ell\) be the line through \(X\) parallel to \(AC\), and let \(C'\) be the reflection of \(C\) across \(\ell\).

The disk centred at \(B\) with radius \(r=BX\) is tangent to \(\ell\) at \(X\), and lies entirely on the same side of \(\ell\) as \(C'\). Since \(\ell\) is the perpendicular bisector of \(CC'\), every point \(P\) of that disk satisfies \[ PC\ge PC'. \] Moreover, the midpoint theorem in triangle \(ACC'\) shows that \(X\) is the midpoint of \(AC'\): \(D\) is the midpoint of \(AC\), while \(DX\parallel CC'\) and \(DX=CC'/2\). Therefore \[ AC'=2AX. \] Applying the triangle inequality now gives \[ L\ge AP+PC-r \ge AP+PC'-r \ge AC'-r =2AX-r =\frac a4(2\sqrt7-\sqrt3). \]

The proposed route attains this bound, so it is shortest possible.

The solution was checked against John Scholes' IMO solution archive (Kalva).