1971 IMO Problems/Problem 4

Problem

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 $BC, CD, DA$ , respectively. Prove:

(a) If $\angle DAB + \angle BCD \neq \angle CDA + \angle ABC$ , 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 $\alpha = \angle BAC + \angle CAD + \angle DAB$ .

Solution

Rotate the triangle $BCD$ around the edge $BC$ until $ABCD$ are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting $X$ and $Z$ . Therefore, $XYB=ZYC$ . Summing the four equations like this, we get exactly $\angle ABC+\angle ADC=\angle BCD+\angle BAD$ .

Now, draw all four faces in the plane, so that $BCD$ is constructed on the exterior of the edge $BC$ of $ABC$ and so on with edges $CD$ and $AD$ .

The final new edge $AB$ (or rather $A'B'$ ) is parallel to the original one (because of the angle equation). Call the direction on $AB$ towards $B$ "right" and towards $A$ "left". If we choose a vertex $X$ on $AB$ and connect it to the corresponding vertex $X'$ on A'B'. This works for a whole interval of vertices $X$ if $C$ lies to the left of $B$ and $D$ and $D$ lies to the right of $A$ . It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.

Finally, regard the sine in half the isosceles triangle $ACA'$ which gives the result with the angles around $C$ instead of $A$ , but the role of the vertices is symmetric.

1971 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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1971 IMO Problems/Problem 4

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