Canadian Mathematical Society
Canadian Mathematical Society

Solutions to problems

Determine a value of the parameter θ so that

f(x)cos2 x+cos2 (x+θ)-cosxcos(x+θ)

is a constant function of x.

Solution 1.

f(x) =cos2 x+(cosxcosθ-sinxsinθ)2 -cosx(cosxcosθ-sinxsinθ) =cos2 x(1+cos2 θ-cosθ)+(1-cos2 x)(sin2 θ)-sinxcosxsinθ(2cosθ-1) =sin2 θ+cos2 x(1+cos2 θ-cosθ-1+cos2 θ)- 1 2 sin2xsinθ(2cosθ-1) =sin2 θ+(2cosθ-1)(cos2 xcosθ-sin2xsinθ).

The function f(x) is constant when 2cosθ-1=0, or when θ=π/3, and its constant value in this case is 3/4.
Solution 2.

f(x) = 1+cos2x 2 + 1+cos2(x+θ) 2 - 1 2 (cos(2x+θ)-cosθ) = 1 2 [2-cosθ+cos2x(1+cos2θ-cosθ)+sin2x(sinθ-sin2θ)] = 1 2 [2-cosθ+(2cosθ-1)(cos2xcosθ+sin2xsinθ].

When θ=π/3, cosθ=1/2 and the function is the constant 3/4.
Solution 3. First, note the identity

cos2 A+cos2 B=1+ 1 2 (cos2A+cos2B)=1+cos(A+B)cos(A-B).

Applying this yields

f(x) =1+cos(2x+θ)cosθ- 1 2 (cos(2x+θ)+cosθ) = (1- 1 2 cosθ )+cos(2x+θ) [cosθ- 1 2 ].

Hence, f(x) is the constant 3/4 when θ=π/3.
Solution 4. We have that

f(x)= 3 4 cos2 x+[cos(x+θ)- 1 2 cosx]2 .

The function f(x) can be made to take the constant value 3/4 if we can find a parameter θ for which

[cos(x+θ)- 1 2 cosx]2 = 3 4 sin2 x

for all x. This is equivalent to

cos(x+θ)= 1 2 cosx± 3 2 sinx=cos(x± π 3 )

for all x. Thus, if θ=±π/3, then f(x) is constantly equal to 3 4 .
Comment. Some students started by showing that f(0)=f( π 2 ) implies that 1+cos2 θ-cosθ=cos2 ( π 2 +θ)=1-cos2 θ, or 0=2cos2 θ-cosθ=cosθ(2cosθ-1). This tells us that θ π 2 (mod π) and θ± π 3 (mod 2π) 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.

Prove that there exists exactly one sequence { xn } of positive integers for which

x1 =1, x2 >1, xn+1 3 +1= xn xn+2

for n1.
Solution. Let x2 =u. Then the first four terms of the sequence are

1,u, u3 +1, u8 +3 u5 +3 u2 +(2/u),

so for the whole sequence to consist of positive integers, we must have that u=2. Now for any n3,

xn = xn-1 3 +1 xn-2 = xn-2 9 +3 xn-2 6 +3 xn-2 3 +1+ xn-3 3 xn-2 xn-3 3             (1).

>From the given condition, it can be seen that any consecutive pairs of terms in the sequence, if integers, are coprime. We know that x1 , x2 , x3 are integers. Let n4. Suppose that it has been shown that xm is an integer for 1mn-1. Then ( xn-3 3 +1)/ xn-2 = xn-4 is an integer, as is ( xn-2 3 +1)/ xn-3 = xn-1 and its cube. Since the numerator of (1) is a multiple of each of xn-2 and xn-3 3 separately, and since these two divisors are coprime, xn must be an integer. The result follows by induction.
Solution 2. [M. Mika] As before, we see that x2 must be 2. It can be checked that x3 and x4 are integers. For any integer n4, we have that

xn 3 +1 = ( xn-1 3 +1)3 xn-2 3 +1 = ( xn-1 3 +1)3 xn-1 xn-3 -1 +1 = xn-1 ( xn-1 8 +3 xn-1 5 +3 xn-1 2 + xn-3 ) xn-1 xn-3 -1 .


( xn 3 +1)( xn-1 xn-3 -1)= xn-1 ( xn-1 8 +3 xn-1 5 +3 xn-1 2 + xn-3 ).

Supposing, as an induction hypothesis,that x1 ,, xn are integers, we see that xn-1 and xn-1 xn-3 -1 are coprime, so we must have that xn-1 divides xn 3 +1. Thus, xn+1 is an integer.

Prove that within a unit cube, one can place two regular unit tetrahedra that have no common point.
Solution 1. Let ABCDEFGH be the cube, with ABCD the top face, EFGH the lower face and AE, BF, CG, DH edges. Let O be the centre of the cube, and let P, Q, R be the midpoints of AB, DH, FG respectively.
The centre O lies on the diagonal CE, which is the axis of a rotation that takes BDG, AHF, PQR, so ΔPQR is equilateral with centre O and CEPQR.
Using Pythagoras' Theorem, we calculate some lengths:




[As a check that O is the centre of ΔPQR, we can compute PO=QO=RO=1/2=[1/3]PQ.]
Since the height of a regular tetrahedron with side s is s2/3, we can construct a regular tetrahedron CUVW with apex C, base UVW homothetic to PQR with centre O, height (3)/2, and side length [(3)/2][3/2]=3/(22)>1. Since 3/(22)=9/8<3/2, triangle UVW lies within triangle PQR, and so the tetrahedron lies within the cube. Shrink the tetrahedron by a homothety with factor 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 (ε,η,ζ) where ε,η,ζ=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 1/2 about the origin. Four of its vertices, (0,0,0), (0,1/2,1/2), (1/2,1/2,0), (1/2,0,1/2), 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 RTAC and MNAC. Observe that RS=CS=3/2, ST=1/(23), TC=1/3 and RT=(2/3). Since ST:SN=RT:MN, SN=1/(22) and MG=NC=((3)/2)-(1/(22)). Hence EX+MG=3-1/(22)<2=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.

Find all pairs (x,y) of positive real numbers for which the least value of the function

f(x,y)= x4 y4 + y4 x4 - x2 y2 - y2 x2 + x y + y x

is attained. Determine that minimum value.
Solution 1. Observe that

f(x,y)-2= ( x2 y2 -1 )2 + ( y2 x2 -1 )2 + ( x y - y x )2 + ( x y -2+ y x ) (x-y)2 xy 0

with equality if and only if x=y. The required minimum is 2.
Solution 2. Let u=( x2 / y2 )+( y2 / x2 ) and v=(x/y)+(y/x) Note that u,v2 with equality if and only if x=y. Then f(x,y)= u2 -u-2+v=(u-2)(u+1)+v2 with equality if and only if x=y. The desired minimum is 2.
Solution 3. Let v=(x/y)+(y/x). Then

f(x,y)=[ v2 -2]2 -2-[ v2 -2]+v= v4 -5 v2 +v+4=(v-2)( v3 +2 v2 -v-1)+2.

Note that v2 and so v3 +2 v2 -v-1=( v3 -v)+(2 v2 -1)>0. The desired result now follows.
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.

Given positive numbers ai with a1 < a2 << an , for which permutation ( b1 , b2 ,, bn ) of these numbers is the product

Πi=1 n ( ai + 1 bi )

Solution 1. By the arithmetic-geometric means inequality, we have that, for each i, 2 ai bi ai 2 + bi 2 , so that

( ai bi +1)2 = ai 2 bi 2 +2 ai bi +1 ai 2 bi 2 + ai 2 + bi 2 +1=( ai 2 +1)( bi 2 +1).


Πi=1 n( ai bi +1) Πi=1 n( ai 2 +1) Πi=1 n( bi 2 +1)= Πi=1 n( ai 2 +1).

Equality occurs if and only if bi = ai for each i.

Πi=1 n ( ai + 1 bi ) = Πi=1 n( ai bi +1) Πi=1 n bi = Πi=1 n( ai bi +1) Πi=1 n ai .

Thus, the given expression is maximized &lrArr; Πi=1 n( ai bi +1) is maximized &lrArr; ai = bi for each i&lrArr;( b1 , b2 ,, bn ) is obtained from ( a1 , a2 ,, an ) by the identity permutation.
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 ( b1 , b2 ,, bn ) is a permutation for which there is a pair i,j of indices for which ai < aj while bi > bj . Then

( ai + 1 bj ) ( aj + 1 bi )- ( ai + 1 bi ) ( aj + 1 bj )=( aj - ai ) ( 1 bj - 1 bi )>0.

with the result that the product can be made larger by interchanging the positions of bi and bj . The result follows.

Determine all real-valued functions f(x) of a real variable x for which

f(xy)= f(x)+f(y) x+y

for all real x and y for which x+y0.
Solution 1. Setting y=1 yields that (x+1)f(x)=f(x)+f(1) so that xf(x)=f(1) for x-1. Set x=0 to obtain f(1)=0, so that, for x0, (x+1)f(x)=f(x). >From this, we deduce that, as long as x0,-1, we have that f(x)=0.
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)0.
Solution 3. Taking y=0 leads to (x-1)f(0)=f(x) for all x0. Taking y=1 leads to xf(x)=f(1) for all x-1. Hence, for x0,1, we have that x(x-1)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 x0.

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