IMO 1971 SL 7

Given a tetrahedron ABCD whose all faces are acute-

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IMO 1971 SL 7

Origin: NET

Problem

Given a tetrahedron ABCD whose all faces are acute- angled triangles, set \sigma = ∡DAB + ∡BCD −∡ABC −∡CDA. Consider all closed broken lines XY ZTX whose vertices X, Y, Z, T lie in the interior of segments AB, BC, CD, DA respectively. Prove that: (a) if \sigma ̸= 0, then there is no broken line XY ZT of minimal length; (b) if \sigma = 0, then there are infinitely many such broken lines of minimal length. That length equals 2AC sin(\alpha/2), where \alpha = ∡BAC + ∡CAD + ∡DAB.

Solution

(a) Suppose that X, Y, Z are fixed on segments AB, BC, CD. It is proven in a standard way that if \angleATX ̸= \angleZTD, then ZT + TX can be re- duced. It follows that if there exists a broken line XY ZTX of minimal length, then the following conditions hold: \angleDAB = \pi −\angleATX −\angleAXT, \angleABC = \pi −\angleBXY −\angleBY X = \pi −\angleAXT −\angleCY Z, \angleBCD = \pi −\angleCY Z −\angleCZY, \angleCDA = \pi −\angleDTZ −\angleDZT = \pi −\angleATX −\angleCZY. Thus \sigma = 0. (b) Now let \sigma = 0. Let us cut the surface of the tetrahedron along the edges AC, CD, and DB and set it down into a plane. Con- sider the plane figure S = ACD′BD′′C′ thus obtained made up of triangles BCD′, ABC, ABD′′, and AC′D′′, with Z′, T ′, Z′′ respec- tively on CD′, AD′′, C′D′′ (here C′ corresponds to C, etc.). Since

\angleC′D′′A + \angleD′′AB + \angleABC + \angleBCD′ = 0 as an oriented angle (because \sigma = 0), the lines CD′ and C′D′′ are parallel and equally oriented; i.e., CD′D′′C′ is a parallelogram. The broken line XY ZTX has minimal length if and only if Z′′, T ′, X, Y, Z′ are collinear (where Z′Z′′ \parallel CC′), and then this length equals Z′Z′′ = CC′ = 2AC sin(\alpha/2). There is an infinity of such lines, one for every line Z′Z′′ parallel to CC′ that meets the interiors of all the segments CB, BA, AD′′. Such \alpha D′ C C′ D′′ A B Z′′ Z′ Y X T ′ Z′Z′′ exist. Indeed, the triangles CAB and D′′AB are acute-angled, and thus the segment AB has a common interior point with the par- allelogram CD′D′′C′. Therefore the desired result follows.