IMO 1971 · Problem 4

All the faces of tetrahedron \(ABCD\) are acute-angled triangles. We consider all closed polygonal paths of the form \(XYZTX\) defined as follows: \(X\) is a point on edge \(AB\) distinct from \(A\) and \(B\); similarly, \(Y,Z,T\) are interior points of edges \(BCCD,DA\), respectively. Prove:

(a) If \(\angle DAB+\angle BCD\ne\angle CDA+\angle ABC\), then among the polygonal paths, there is none of minimal length.

(b) If \(\angle DAB+\angle BCD=\angle CDA+\angle ABC\), then there are infinitely many shortest polygonal paths, their common length being \(2AC\sin(\alpha/2)\), where \[ \alpha=\angle BAC+\angle CAD+\angle DAB. \]

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

Solution

The successive edges containing \(X,Y,Z,T\) are \(AB,BC,CD,DA\). Write \[ a=\angle DAB,\qquad b=\angle ABC,\qquad c=\angle BCD,\qquad d=\angle CDA. \] All angles of every face lie strictly between \(0\) and \(\pi/2\).

(a) Suppose that a shortest admissible path exists. Each of its vertices can be moved slightly in either direction along its edge.

Keeping \(Y,Z,T\) fixed and differentiating the length with respect to \(AX\), we obtain \[ 0=\frac{d}{d(AX)}(XY+XT) =-\cos\angle BXY+\cos\angle AXT. \] Thus \(\angle BXY=\angle AXT\). Applying the same argument at the other three vertices gives \[ \begin{aligned} \angle BXY&=\angle AXT=u,\\ \angle BYX&=\angle CYZ=v,\\ \angle CZY&=\angle DZT=w,\\ \angle DTZ&=\angle ATX=s. \end{aligned} \] The angle sums in the four triangles \(BXY,CYZ,DZT,ATX\) now give \[ u+v+b=\pi,\quad v+w+c=\pi,\quad w+s+d=\pi,\quad s+u+a=\pi. \] Adding the first and third equations and comparing with the sum of the other two yields \[ b+d=a+c. \] Consequently, if the stated angle sums are unequal, no admissible path can have minimum length.

(b) Assume henceforth that \[ a+c=b+d. \]

Unfold the faces successively into a plane as \[ ABC,\qquad BCD,\qquad CDA',\qquad DA'B', \] placing consecutive triangles on opposite sides of their common edge. Here \(A'\) and \(B'\) are copies of \(A\) and \(B\). Append one further triangle \(A'B'C'\), a copy of \(ABC\), across \(A'B'\).

Choose the orientation so that the first triangle \(ABC\) is clockwise. The successive signed turns between the directed vectors \[ \overrightarrow{AB},\quad \overrightarrow{CB},\quad \overrightarrow{CD},\quad \overrightarrow{A'D},\quad \overrightarrow{A'B'} \] are \(b,-c,d,-a\). Their sum is zero. Therefore \(AB\) and \(A'B'\) have the same direction, and their lengths are equal, so \[ \overrightarrow{A'B'}=\overrightarrow{AB}. \] Both copies of \(ABC\) have the same orientation. Hence the second is a translate of the first: for some vector \(\mathbf v\), \[ A'=A+\mathbf v,\qquad B'=B+\mathbf v,\qquad C'=C+\mathbf v. \]

Put \[ \omega=\angle ACB+\angle BCD+\angle DCA. \] In the unfolding, the rays \(CA,CB,CD,CA'\) turn successively clockwise through a total angle \(\omega\). Moreover, \[ \begin{aligned} \alpha+\omega &=(\angle BAC+\angle ACB) +(\angle CAD+\angle DCA)+a+c\\ &=(\pi-b)+(\pi-d)+a+c\\ &=2\pi. \end{aligned} \] Since \(CA=CA'=AC\) and \(0<\omega<3\pi/2\), the chord formula gives \[ |\mathbf v|=AA' =2AC\sin(\omega/2) =2AC\sin(\alpha/2)>0. \]

Any admissible path unfolds into a polygonal line \(XYZTX'\), where \(X'\) is the corresponding point on \(A'B'\). In particular, \(X'=X+\mathbf v\). Unfolding preserves lengths, so the triangle inequality gives the universal lower bound \[ XY+YZ+ZT+TX \ge XX' =|\mathbf v| =2AC\sin(\alpha/2). \]

It remains to prove that infinitely many admissible paths attain this bound.

We use an elementary planar observation. Suppose the rays \[ OP,\ OQ,\ OR,\ OS \] occur successively clockwise, with the three intervening angles \(p,q,r\) all acute, and suppose \(OP=OS\). Orient the plane so that \(\overrightarrow{PS}\) points horizontally to the right. Then \(Q\) and \(R\) lie strictly above \(O\).

Indeed, because \(OP=OS\), the directions of \(OQ\) and \(OR\), measured counterclockwise from the positive horizontal direction, are \[ \frac{\pi-p+q+r}{2}, \qquad \frac{\pi-p-q+r}{2}. \] Both belong to \((0,\pi)\), since \(p,q,r\) are acute. This proves the observation, including the case \(p+q+r=\pi\).

Now orient our unfolding so that \(\mathbf v\) points horizontally to the right. Apply the observation:

  • At \(C\), to the rays \(CA,CB,CD,CA'\). It shows that \(B\) and \(D\) lie strictly above \(C\).
  • At \(A'\), to the rays \(A'C,A'D,A'B',A'C'\). It shows that \(D\) and \(B'\) lie strictly above \(A'\).

The required equalities of lengths are \(CA=CA'=AC\) and \(A'C=A'C'=AC\); all intervening angles are face angles and hence acute.

Translation by \(\mathbf v\) preserves height. Thus \(A,A'\) have the same height, as do \(B,B'\). Consequently, every point in \[ \{A,C,A'\} \] lies strictly below every point in \[ \{B,D,B'\}. \] There is therefore a nonempty open horizontal strip between these two sets of vertices.

Every horizontal line in this strip intersects \[ AB,\quad BC,\quad CD,\quad DA',\quad A'B' \] at interior points \(X,Y,Z,T,X'\), respectively. Its intersection with each of the four developed triangles is the segment joining the corresponding two points. Since consecutive triangles lie on opposite sides of their common edge, these four segments concatenate without reversal. They form the straight segment \(XX'\).

Folding the triangles back gives an admissible closed path \(XYZTX\), whose length is \[ XX'=|\mathbf v|=2AC\sin(\alpha/2). \] It therefore attains the lower bound and is shortest.

Finally, the strip contains infinitely many horizontal lines, and distinct lines give distinct points \(X\) on \(AB\), hence distinct paths. This proves (b).

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