


Solution 2.

Solution 3. First, note the identity


Solution 4. We have that



Comment. Some students started by showing that f(0) = f([(p)/2]) implies that 1 + cos^{2} q cosq = cos^{2}([(p)/2] + q) = 1  cos^{2} q, or 0 = 2cos^{2} q cosq = cosq(2cosq 1). This tells us that q º [(p)/2] (mod p) and q º ±[(p)/3] (mod 2p) are the only possibilities. However, the first of these turns out to be extraneous. It yields f(x) = 1 ±^{1}/_{2}sin2x, which is not constant.


 (1). 
Solution 2. [M. Mika] As before, we see that x_{2} must be 2. It can be checked that x_{3} and x_{4} are integers. For any integer n ³ 4, we have that


The centre O lies on the diagonal CE, which is the axis of a rotation that takes B ® D ® G, A ® H ® F, P ® Q ® R, so DPQR is equilateral with centre O and CE ^PQR.
Using Pythagoras' Theorem, we calculate some lengths:



Since the height of a regular tetrahedron with side s is sÖ[(2/3)], we can construct a regular tetrahedron CUVW with apex C, base UVW homothetic to PQR with centre O, height (Ö3)/2, and side length [(Ö3)/2][Ö[(3/2)]] = 3/(2Ö2) > 1. Since 3/(2Ö2) = Ö[(9/8)] < Ö[(3/2)], triangle UVW lies within triangle PQR, and so the tetrahedron lies within the cube. Shrink the tetrahedron by a homothety with factor Ö[(8/9)] about its centre to get one of the desired tetrahedra.
The second tetrahedra can be found in a similar way from EUVW (which is congruent to CUVW). The two tetrahedra are strictly separated by the plane of PQR.
Solution 2. Let the cube have vertices at the eight
points (e, h, z) where e, h, z = 0, 1. The plane of equation x + y + z = 3/2 passes through
(0, 1, ^{1}/_{2}), (^{1}/_{2}, 1, 0),
(1, ^{1}/_{2}, 0), (1, 0, ^{1}/_{2}),
(^{1}/_{2}, 0, 1) and (0, ^{1}/_{2}, 1) at the
middle of various edges of the cube, and bisects the cube into
two congruent halves. Consider the cube reduced by a homothety
of factor 1 / Ö2 about the origin. Four of its
vertices, (0, 0, 0), (0, 1/Ö2, 1/Ö2),
(1/Ö2, 1/Ö2, 0), (1/Ö2, 0,1/Ö2), constitute the four vertices of a regular
unit tetrahedron contained in the original cube.
Since the sum of the coordinates of all of
these points is less than 3/2, they all lie on the same side of the
plane bisecting the cube, as does the whole tetrahedron.
Its image reflected in the centre of the cube is a second tetrahedron
contained in the upper portion of the cube.
Solution 3. [D. Tseng] Consider unit tetrahedra CPQR and EUVW, each sharing a vertex and a face with the cube and directed inwards with the face diagonal AC intersecting PQ in its midpoint S and face diagonal EG intersecting UV in its midpoint X. (Each tetrahedron is carried into the other by a reflection in the centre of the cube.) Planes PQR and UVW are parallel. These tetrahedra intersect the internal plane ACGE in two triangles EXW and CSR. Let SR produced meet EG in M, and let T and N be be points on AC for which RT ^AC and MN ^AC. Observe that RS  = CS  = Ö3/2, ST  = 1/(2Ö3), TC  = 1/Ö3 and RT  = Ö[((2/3))]. Since ST:SN = RT:MN, SN  = 1/(2Ö2) and MG  = NC  = ((Ö3)/2)  (1/(2Ö2)). Hence EX + MG  = Ö3  1/(2Ö2) < Ö2 = EG . This means that the parallel lines WX and RS have a region of ACGE between them that do not intersect triangles EXW and CSR, and so the two tetrahedra are separated by the slab between the parallel planes PQR and UVW that passes through the centre of the cube.


Solution 2. Let u = (x^{2}/y^{2}) + (y^{2}/x^{2}) and
v = (x/y) + (y/x) Note that u, v ³ 2 with equality
if and only if x = y. Then
f(x, y) = u^{2}  u  2 + v = (u  2)(u + 1) + v ³ 2
with equality if and only if x = y. The desired minimum is 2.
Solution 3. Let v = (x/y) + (y/x). Then

Comment. Several solvers did this problem by calculus. The most important thing you need to know about calculus is when not to use it. Calculus provides very general algorithms for doing optimization problems, and such algorithms often have two undesirable characteristics: (1) they may not provide the quickest and most convenient approach in particular cases; (2) they tend to operate as ``black boxes'', preventing the solver from appreciating the essence of the problem or the significance of the answer. When you have a problem of this type, you should check to see whether it can be handled without calculus, and use calculus only as a last resort, or at least when it is clear that every other approach is messier.



Now

Solution 2. There are finitely many permutations of the numbers, so that there must be a permutation which maximizes the value of the given expression. We show that it is the identity permutation, by showing that, for any other permutation, we can find a permutation that yields a larger value.
Suppose that (b_{1}, b_{2}, ¼, b_{n}) is a permutation for which there is a pair i, j of indices for which a_{i} < a_{j} while b_{i} > b_{j}. Then


For each nonzero value of x, xf(0) = f(x) + f(0), so that (x  1)f(0) = f(x). Taking x = 2 gives f(0) = f(2) = 0. Finally, 2f(1) = 2f(1) = 0, so f(1) = 0. Hence, f(x) must be indentically equal to 0.
Solution 2. [S.E. Lu] For all nonzero x, we have that f(x) = (x  1)f(0). The equality f(x) + f(y) = (x + y)f(xy) leads to either f(0) = 0 or x + y  2 = (xy  1)(x + y). The latter simplifies to (x + y)(2  xy) = 2 for all nonzero x, y, which is patently false. Hence f(0) = 0, so f(x) º 0.
Solution 3. Taking y = 0 leads to (x1)f(0) = f(x) for all x ¹ 0. Taking y = 1 leads to xf(x) = f(1) for all x ¹ 1. Hence, for x ¹ 0, 1, we have that x(x1)f(0) = f(1). This holds for infinitely many x if and only if f(0) = f(1) = 0. It follows that f(x) = 0 for all real x.
Comment. Suppose that the given condition is weakened to hold only when both x and y are nonzero. Then we get xf(x) = f(1) so that f(x) = f(1)/x for all nonzero x. it can be checked that, for any constant c, f(x) = c/x for is a solution for x ¹ 0.