AoPS - Problem

Source: Baltic Way 2008, Problem 5

Tags: geometry, 3D geometry, tetrahedron, symmetry, combinatorics unsolved, combinatorics

April

23.11.2008 18:52

Suppose that Romeo and Juliet each have a regular tetrahedron to the vertices of which some positive real numbers are assigned. They associate each edge of their tetrahedra with the product of the two numbers assigned to its end points. Then they write on each face of their tetrahedra the sum of the three numbers associated to its three edges. The four numbers written on the faces of Romeo's tetrahedron turn out to coincide with the four numbers written on Juliet's tetrahedron. Does it follow that the four numbers assigned to the vertices of Romeo's tetrahedron are identical to the four numbers assigned to the vertices of Juliet's tetrahedron?

A pretty hard question this one! Only Germany managed to solve it ...! It's equivalent to the following $ a_1,a_2,a_3,a_4,b_1,b_2,b_3,b_4$ are positive reals such that $ a_1a_2+a_2a_3+a_3a_1 = b_1b_2+b_2b_3+b_3b_1 \ \ \ (1)$, $ a_1a_2+a_2a_4+a_4a_1 = b_1b_2+b_2b_4+b_4b_1 \ \ \ (2)$, $ a_1a_3+a_3a_4+a_4a_1 = b_1b_3+b_3b_4+b_4b_1 \ \ \ (3)$, $ a_2a_3+a_3a_4+a_4a_2 = b_2b_3+b_3b_4+b_4b_2 \ \ \ (4)$. Is it nescessary that $ \{a_1,a_2,a_3,a_4\} = \{b_1,b_2,b_3,b_4\}$ ? Assume $ a_1 \ge a_2 \ge a_3 \ge a_4$. Then we easily get $ b_1 \ge b_2 \ge b_3 \ge b_4$. First: Obviously $ a_1 > b_1, a_2 > b_2, a_3> b_3$ gives $ a_1a_2+a_2a_3+a_3a_1 > b_1b_2+b_2b_3+b_3b_1$, so it is false. Second: Assume $ a_1 > b_1, a_2 > b_2$ gives $ b_3 > a_3$ from $ (1)$ and $ b_4 > a_4$ from $ (2)$. Adding $ (3)$ and four gives $ (a_1+a_2)(a_3+a_4) + 2a_3a_4 = (b_1+b_2)(b_3+b_4) + 2b_3b_4$. So $ (a_1+a_2)(a_3+a_4) > (b_1+b_2)(b_3+b_4)$. Adding $ (1)$ and $ (2)$ gives $ 2a_1a_2 + (a_1+a_2)(a_3+a_4) = 2b_1b_2 + (b_1+b_2)(b_3+b_4)$, yielding $ (a_1+a_2)(a_3+a_4) < (b_1+b_2)(b_3+b_4)$. So it cannot be true either. Third: Then, by symmetry, at most two of the pairs have numbers that differ. If at most one pair, it is obvious that they all are the same. Now assume $ a_1 > b_1$ and $ a_4 < b_4$. Then by $ (1)$ we get $ a_1 = b_1$. So they are the same. So the numbers have to be the same!