Canadian Mathematical Society
Canadian Mathematical Society
Solutions for July-August, 2002 problems

Prove that if the quadratic equation x2 +ax+b+1=0 has nonzero integer solutions, then a2 + b2 is a composite integer.
Solution. Suppose that the integer roots are u and v. Then u+v=-a and uv=b+1, whence

a2 + b2 =(u+v)2 +(uv-1)2 = u2 + v2 + u2 v2 +1 =( u2 +1)( v2 +1).

Since each factor exceeds 1, a2 + b2 is composite.
Comment. You should make sure that you are familiar with the relationship between the coeffients and roots of polynomials; a poor way to solve this problem is to use the quadratic formula. Note that, if, say, u=0, then a2 + b2 can be prime; for example, you get the value 5 when (u,v)=(0,2).

Let f(x) be a polynomial with real coefficients for which the equation f(x)=x has no real solution. Prove that the equation f(f(x))=x has no real solution either.
Solution 1. Let g(x)=f(x)-x. Then, g(x) is a polynomial that never vanishes. We argue that it must always have the same sign. Suppose if possible that g(a)<0<g(b) for some reals a and b. Since g(x), being a polynomial, is continuous, the Intermediate Value Theorem applies and there must be a number c between a and b for which g(c)=0, yielding a contradition.
Thus, either g(x)>0 for all x or else g(x)<0 for all x. Then

f(f(x))-x =f(f(x))-f(x)+f(x)-x =g(f(x))+g(x)

for all real x. Since g never changes sign, both g(x) and g(f(x)) have the same sign (either positive or negative) and so their sum cannot vanish. Hence f(f(x))x for any real x.
Solution 2. Suppose, if possible, that f(f(a))=a. Let b=f(a). Then f(b)=a. By hypothesis, ba. Wolog, suppose that a<b. Then f(a)-a>0 and f(b)-b<0. Since f(x)-x is a polynomial, it is continuous and so the Intermediate Value Theorem applies on the closed interval [a,b]. As it has opposite signs at the endpoints of [a,b], it must vanish somewhere in the interior of the interval. But then this contradicts the hypothesis. Hence f(f(x))=x can have no real solutions.
Comment. It is important to highlight that f(x) is a continuous function, as this is key to the result. (Can you construct a counterexample where it is not assumed that f is a polynomial or continuous?) Several of you tried to give a geometric argument for this, and apart from mangling the terminology (e.g., not distinguishing between a function and its graph, points and values), operated at too intuitive a level. Notice that the statement `` f(x)>0 or f(x)<0 for all x'' lacks a certain precision, as it could be interpreted to mean that at any individual point x, one of the two alternatives occurs (the function does not vanish), but that different alternative might occur at different points.

Let 0a4. Prove that the area of the bounded region enclosed by the curves with equations




cannot exceed 1 3 .
Solution. In the situation that 0a1, the two curves intersect in the points (a/3,a/3) and (a,a), and the bounded region is the triangle with these two vertices and the vertex (a/2,0). This triangle is contained in the triangle with vertices (0,0), (1/2,0) and (1,1) with area 1/4. Hence, when 0a1, the area of the bounded region cannot exceed 1/4.
Let 1a3. In this case, the bounded region is a quadrilateral with the four vertices (a/3,a/3), (a/2,0), ((a+2)/3,(4-a)/3) and (1,1). Noting that this quadrilateral is the result of removing two smaller triangles from a larger one (draw a diagram!), we find that its area is

1- 1 2 · a 3 · a 2 - 1 2 · (4-a) 3 · (2- a 2 ) =1- a2 12 - 1 12 (4-a)2 =- a2 -4a+2 6 = 1 3 - (a-2)2 6 ,

whence we find that the area does not exceed 1/3 and is equal to 1/3 exactly when a=2.
The case 3a4 is the symmetric image of the case 0a1 and we find that the area of the bounded region cannot exceed 1/4.
Comment. This is not a difficult problem, but it does require a lot of care in keeping the situation straight and the computations sound. Always keep in mind completion of the square when it is a matter of optimizing a quadratic. Use of calculus is more cumbersome in such situations, and begs the question as to whether the optimum is a maximum or a minimum. (You can lose points for not explicitly resolving this issue.) R. Barrington Leigh solved the problem by looking at the area between the graph of y=1-x-1-2x-a and the x-axis. The equation of this curve with the absolute value signs is equal to y=3x-a for xa/2, to y=a-x when a/2x1 and a+2-3x when x1, with suitable modification when a2. Because at each value of x, the difference in the ordinates remains the same as in the original problem, the area here is the same as the area between the two graphs of the problem. You might want to try the problem in this way.

Let I be the incentre of the triangle ABC and D be the point of contact of the inscribed circle with the side AB. Suppose that ID is produced outside of the triangle ABC to H so that the length DH is equal to the semi-perimeter of ΔABC. Prove that the quadrilateral AHBI is concyclic if and only if angle C is equal to 90ˆ .
Note. In the solutions that follow, we use the standard notation that a, b, c are the respective lengths of the sides BC, CA, AB, r is the inradius of the triangle and s is its semi-perimeter (so that 2s=a+b+c). We note that the area of the triangle is rs, that the distances from the vertices A, B, C to the tangent points of the incircle are respectively s-a, s-b, s-c, and that r=(s-c)tan(C/2). If you are not familiar with these relationships, then regard them as exercises.
Solution 1. The quadrilateral AHBI is concyclic if and only if DI:DH=DB:DA, or equivalently,

rs=(s-a)(s-b)= s2 -s(a+b)+ab= s2 -s(2s-c)+ab=s(c-s)+ab.

The area of the triangle is rs= 1 2 absinC=abt(1+ t2 )-1 and r=(s-c)t, where t=tanC/2. Then the concylic condition is equivalent to

rs=- rs t + rs(1+ t2 ) t =rst,

or t=1. This is equivalent to C/2= 45ˆ , and the result follows.
Solution 2. The concyclic condition is equivalent to

rs=(s-a)(s-b)&lrArr;4rs=(c+b-a)(c+a-b)= c2 - b2 - a2 +2ab.

We have, for any triangle, that 2absinC=4rs (four times the area) and c2 = a2 + b2 -2abcosC. Hence the concyclic condition is equivalent to


If sinC+cosC=1, then, by squaring, 1+2sinCcosC=1sin2C=0 so that C= 90ˆ . On the other hand, if C= 90ˆ , then sinC+cosC=1. The result follows.
Comment. We can end the solution, keeping the equivalences to the end, by noting that

sinC+cosC=1&lrArr;sin(C+ 45ˆ )=2&lrArr;C= 90ˆ .

Solution 3. [A. Feizmohammadi]

AIB= 180ˆ -(BAI+ABI)= 180ˆ - A+B 2 = 90ˆ + C 2 .


tanAHB= (s-a)/s+(s-b)/s 1-(s-a)(s-b)/ s2 = (2s-a-b)s (a+b)s-ab = cs (2s-c)s-ab = cs 2 s2 -cs-ab .

Suppose that C is right. Then AIB= 135ˆ and

c2 = a2 + b2 (a+b)2 - c2 =2ab(a+b-c)(a+b+c)=2abab=2s(s-c),

so that

tanAHB= cs 2 s2 -cs-2 s2 +2cs =1AHB= 45ˆ .

Since angles AHB and AIB are supplementary, AHBI is concyclic.
On the other hand, suppose that AHBI is concyclic. Then

s(s-a)(s-b)(s-c) =(s-a)(s-b)=rs(s-a)(s-b)=s(s-c) ab=(a+b-c)s2ab=(a+b)2 - c2 a2 + b2 = c2 ,

so that ACB is right.

Let a,b,c be positive real numbers for which a+b+c=1. Prove that

a3 a2 + b2 + b3 b2 + c2 + c3 c2 + a2 1 2 .

Solution. Observe that, by the arithmetic-geometric means inequality ab 1 2 ( a2 + b2 ), so that

a3 a2 + b2 =a-b ab a2 + b2 a- b 2

with a similar inequality for the other two terms on the left side. Adding these inequalities and using a+b+c=1 yields the desired result.
Comment. R. Furmaniak noted that

2 a3 a2 + b2 -2a+b= b(b-a)2 a2 + b2 0,

which again yields the result by concluding as in the solution.

Let A and B be fixed points in the plane. Find all positive integers k for which the following assertion holds:
among all triangles ABC with AC=kBC, the one with the largest area is isosceles.
Solution 1. In the case that k=1, all triangles satisfying the condition are isosceles with AC=BC, the locus of C is the right bisector of AB and there is no triangle with largest area (the area can be arbitrarily large). Hence k2. Since, by the triangle inequality,


we must have that AB>BC. Therefore, the only way a triangle satisfying the condition can be isosceles is for AB=AC=kBC. Since BC<AC, CAB cannot exceed 60ˆ , so that when k2, there is no isosceles triangle in the set.
Solution 2. As in Solution 1, we need consider only the case that k1, and we can dispose of the case k=1. Let D be the foot of the perpendicular from C to AB (possibly produced; note that D and B lie on the same side of A) and let the respective lengths of AB, AD and CD be c, x and v. We wish to maximize v. The given condition implies that

x2 + v2 =k((x-c)2 + v2 )

so that

v2 =- x2 + 2 k2 c k2 -1 x- k2 c2 k2 -1 .

The maximum value of v is equal to

kc k2 -1

assumed when x= k2 c k2 -1 >c. If this maximum value of v is assumed when AC=AB, then we have x2 + v2 = c2 which leads to k=1/3. If the maximum is assumed when BC=AB, then (x-c)2 + v2 = c2 , which leads to k=3. As both solutions are extraneous, there is no positive integer value of k for which the area is maximized when the triangle is isosceles.
Solution 3. As before, we can deal with the k=1 case. Let k1. Let the sides of the triangle be (a,ka,c) where c is the length of AB; let C be the angle opposite c. Then

c2 =( k2 +1) a2 -2 ka2 cosC

whence a2 = c2 ( k2 +1-2kcosC)-1 . The area of the triangle is equal to

1 2 ka2 sinC= 1 2 kc2 ( k2 +1-2kcosC)-1 sinC.

The derivative of sinC( k2 +1-2kcosC)-1 with respect to C is equal to [(cosC)( k2 +1)-2k][ k2 +1-2kcosC]-2 , from which we can read off that area is maximized when cosC=2k( k2 +1)-1 . At this maximum value, ( k2 +1) c2 =( k2 -1)2 a2 . The triangle maximizing the area can be isosceles in two ways. The case c=a occurs when k=3, and the case c=ka occurs when k=1/3. Neither of these is an integer.
Alternatively, the case c=a corresponds to cosC=k/2, and equating k/2 and 2k/( k2 +1) leads to k2 =3. The case c=ka corresponds to cosC=1/2k, and this leads to k2 =1/3. We conclude as before.
Solution 4. Dispose of the k=1 case as before. Let k1. Let c=AB be fixed, and let the other two sides have length a and la. Sixteen times the square of the area of the triangle is, by Heron's rule,

[(1+k)a+c][(1+k)a-c] [c+(1-k)a][c-(1-k)a] =[(1+k)2 a2 - c2 ][ c2 -(1-k)2 a2 ]=2 a2 c2 (1+ k2 )-( k2 -1)2 a4 - c4 = [ ( k2 +1)2 ( k2 -1)2 c4 - c4 ]- [( k2 -1) a2 - ( k2 +1) ( k2 -1) c2 ]2 = 4 k2 c4 ( k2 -1)2 -( k2 -1) [ a2 - k2 +1 ( k2 -1)2 c2 ].

The maximum area occurs when ( k2 -1)2 a2 =( k2 +1) c2 . This occurs when a=c exactly when ( k2 -1)2 = k2 +1, so k2 =3, and occurs when ka=c exactly when ( k2 -1)2 = k4 + k2 , so 3 k2 =1. We conclude as before.
Comments. Several students committed the fallacy of setting, for example, AC=AB, noting that the locus of C in this case was a circle with diameter containing AB, claiming that the area was maximized when C= 90ˆ and so BC was equal to 2AC. They deduced that 2 and 1/2 were the only values of k that allowed the area to be maximized by an isosceles triangle. However, this puts things the wrong way around. The value k is fixed, and then we look at all appropriate triangles. All the solvers did was to maximize the area over all isosceles triangles for which AC=AB; as C traces out the set of points satisfying these conditions, of course, the ratio of the lengths of AC and BC vary. Those solvers who made this mistake are invited to investigate the case k=2 in more detail, and this will help them understand where they went wrong.
If we coordinatize the situation with A~(0,0) and B~(1,0), we see that the locus of C is given by

( k2 -1)( x2 + y2 )-2 k2 x+ k2 =0.

This is the equation of a circle (of Apollonius) with centre at ( k2 /( k2 -1),0) and with radius k( k2 -1)2 . Thus, the area is maximized when C is located at the point ( k2 ( k2 -1)-1 ,k( k2 -1)-1 ). Checking when this yields an isosceles triangle leads to the two values of k that we have already seen: 3 and 1/3

Let Ri and ri re the respective circumradius and inradius of triangle Ai Bi Ci ( i=1,2). Prove that, if C1 = C2 and R1 r2 = r1 R2 , then the two triangles are similar.
Solution. Let Γ be the circumcircle of Δ A1 B1 C1 . Scale Δ A2 B2 C2 so that A2 = A1 =A, B2 = B1 =A and C2 and C1 lie on the same side of AB. Then Γ is the circumcircle of ABC2 , so that R1 = R2 and r1 = r2 . Using the conventional notation, we have that

( s1 - c1 )cot( C1 /2)= r1 = r2 =( s2 - c2 )cot( C2 /2)

whence, as c1 = c2 and C1 = C2 , s1 = s2 . Therefore a1 + b1 = a2 + b2 . Since r1 s1 = r2 s2 , the two triangles have the same area, so that a1 b1 sin C1 = a2 b2 sin C2 and thus a1 b1 = a2 b2 . Since the pairs ( a1 , b1 ) and ( a2 , b2 ) have the same sum and product, they are roots of the same quadratic equation, and so we must have that ( a1 , b1 )=( a2 , b2 ) or ( a1 , b1 )=( b2 , a2 ). In either case, the triangles are congruent, so that triangle A1 B1 C1 is a scaled version of the original triangle A2 B2 C2 . The result follows.
Comment. There were many variations using various trigonometric identities.

Let n be a positive integer and X a set with n distinct elements. Suppose that there are k distinct subsets of X for which the union of any four contains no more that n-2 elements. Prove that k 2n-2 .
Solution 1. [R. Furmaniak] We may assume that n2. We give a proof by contradiction. Suppose the S is a family of k subsets of X, where k> 2n-2 . Let X={ x1 , x2 ,, xn } where the first element is selected so that x1 belongs to some member of S but not to all. (Why can you do this?) For each i, 1in-1, let Ai be the class of all subsets of { x1 , x2 ,, xi } which are contained in some subset of S, and Bi be the class of all subsets of { xi+1 ,, xn } which are contained in some subset of S.
Note that 2n-2 <k Ai Bi , since each set of S is the union of a set in Ai and a set in Bi . ( · refers to the number of elements.) Therefore, either Ai > 2i-1 or Bi > 2n-i-1 . In the former case, Ai must contain a complementary pair of subsets (use the pigeonhole principle on complementary pairs of sets) of { x1 ,, xi }; the the latter, Bi must contain a complementary pair of sets of { xi+1 ,, xn }.
By our choice of x1 , we see that A1 ={, x1 }, so that A1 =2. Therefore there are values of the index i for which Ai > 2i-1 . If it turns out that An-1 > 2n-2 , then An-1 contains two sets whose union is { x1 ,, xn-1 } and so S contains two sets whose union has at least n-1 elements. In this case, S fails to satisfy the hypotheses of the problem.
Suppose that An-1 2n-2 . Then select the smallest index k for which Ak 2k-1 . Then k2 and Ak-1 > 2k-2 while Bk > 2n-k-1 . Therefore, Ak-1 contains two sets whose union is { x1 , x2 ,, xk-1 } and Bk contains two sets whose union is { xk+1 ,, xn }. We conclude that S must contain four sets whose union contains the n-1 elements of X apart from xk , and so S again fails to satisfy the hypothesis of the problem.
We conclude that any family of more than 2n-2 sets fails so satisfy the hypothesis of the problem, and so that every family that does satisfy the hypothesis must have at more 2n-2 elements.
Solution 2. Let P be a class of k subsets for which the union of no four contains more than n-2 elements. Suppose that A1 and A2 are two members for which the cardinality m= A1 A2 of the union of a pair is maximum. Since m<n, we can construct a set Y= A1 A2 {y}, where y\not A1 A2 ; thus Y=m+1.
Let Q={YA:AP}. No two subsets in Q are complementary (i.e. have union equal to Y), so that Q 2m . Let Z=XY and R={ZA:AP}. Then Z=n-(m+1)=n-m-1. We prove that R has no two sets that are complementary, i.e., whose union is Z. For, otherwise, suppose that ( A3 Z)( A4 Z)=Z, for two sets A3 and A4 in P. Then Z=( A3 A4 )Z, so that

Z A3 A4 A1 A2 A3 A4 (YZ){y}

, whence we see that A1 A2 A3 A4 =n-1, contrary to hypothesis. Hence R 2n-m-2 .
Since each set in P is uniquely determined by its intersections with Y and Z, we have that

k=PQR 2m 2n-m-2 = 2n -2,

as desired.
Comment. Several solvers pointed out that if you took all the subsets of n-2 of the elements of X, then we got an ``extreme'' case of a family of sets that satisfied the hypothesis and the conclusion, and essentially said that if we tried to stuff in any more sets, we would contradict the hypothesis. However, this begs the question as to whether you could drop some of these sets and replace than by a larger number of sets while still keeping the hypothesis.

Let n be a positive integer. Determine all n-tples { a1 , a2 ,, an } of positive integers for which a1 + a2 ++ an =2n and there is no subset of them whose sum is equal to n.
Solution 1. Let sk = a1 + a2 ++ ak for 1kn, By hypothesis, all of these are incongruent modulo n. Now let t1 = a2 , tk = sk for 2kn. Again, the hypothesis forces all to be incongruent. Hence the sets { sk } and { tk } both consitute a complete set of residues, modulo n, so it must happen that s1 t1 a1 (mod n). A similar argument can be marshalled when the pair ( a1 , a2 ) is replaced by any other pair of elements. Hence, each of the terms in the set { ai } is congruent to some number m (mod n), where 0mn-1. Now

mn a1 + a2 ++ an =2n

from which either m=2, n is odd, and the n-tple is {2,2,2,,2} or else m=1 and the n-tple is {1,1,1,,1,n+1}.
Solution 2. [J.Y. Jin] Wolog, suppose that a1 a2 an . Note that a1 cannot exceed 2. If a1 =2, then each ai is equal to 2 and n is odd. Suppose that a1 =1. Then an 3. Let sk = a1 ++ ak for 1kn. Then sn-1 =2n- an 2n-3. By the hypothesis, we see that the set of the sk with 1kn-1 contains at most one member from each of the n-1 sets:


so it must contain exactly one from each set. In particular, the values n-2 and n-1 are assumed, so that for some i, si =n-2 and si+1 =n-1. Thus ai+1 =1, and a1 = a2 == ai+1 =1. But this means that i=n-2, so that an-1 =1 and an =n+1.
Comment. Some of the solvers tried an induction argument. However, the process was faulty, since they generally started with an n-tple and then tried to build from that an (n+1)-tple. However, for an induction argument to work, you need to start with a general (n+1)-tple that satisfies the condition, and then indicate how to reduce it to the result for a smaller set.

Suppose that f is a real-valued function defined on the reals for which


for all real x and y. Prove that f(x)0 for all real x.
Solution. Suppose x1; let y=x(x-1)-1 . Then xy=y+x, so that

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

For each real z, the equation z=( x2 -2x)(1-x)-1 is equivalent to x2 +(z-2)x-z=0. This quadratic equation (in x) has a discriminant z2 +4 which is positive and therefore it is solvable for all real z. (Note that in no case is x=1 a solution.) Hence f(z)0 for all real z.
Comment. O. Bormashenko noted that when 2x= z2 +4+2-z and 2y= z2 +4+2+z, then xy=x+y and y-x=z.

Let u=(5-2)1/3 -(5+2)1/3 and v=(189-8)1/3 -(189+8)1/3 . Prove that, for each positive integer n, un + vn+1 =0.
Solution. Observe that, if c=a-b, then c3 = a3 - b3 -3abc, so that c3 +3abc-( a3 - b3 )=0. Applying this to u and v in place of c, we obtain that

0= u3 +3u+4=(u-1)( u2 -u+4)=(u+1) [ (u- 1 2 )2 - 15 4 ]


0= v3 +15v+16=(v+1)( v2 -v+16)=(v+1) [ (v- 1 2 )2 - 63 4 ].

Since the expresseions in square brackets are negative, we must have that u=v=-1, from which the result follows.
Comment. Some students observed that

( 5±1 2 )3 =5±2


( 21±1 2 )3 =189±8,

and this immediately leads to the result.

Determine the value of

cos 5ˆ +cos 77ˆ +cos 149ˆ +cos 221ˆ +cos 293ˆ .

Preliminary work. Before getting into the solution, we will discuss how to obtain the trigonometric ratios of certain angles related to 36ˆ . It is useful for you to know some of these techniques, as these angles tend to come up in problems, and to be on the safe side in a contest, you should try to include a justification for assertions that you make about these angles. Here is one way to evaluate t=cos 36ˆ . Observe that

t =-cos 144ˆ =1-2cos2 72ˆ =1-2(2 t2 -1)2 =-8 t4 +8 t2 -1

from which we see that

0=8 t4 -8 t2 +t+1=(2t-1)(t+1)(4 t2 -2t-1).

Since t is equal to neither 1 nor 1 2 , we must have that 4 t2 =2t+1. Solving this equation will give you an actual numerical value (can you justify your choice of root?).
A very useful relation is 4cos 36ˆ cos 72ˆ =1. This can be checked geometrically. Let PQS be a triangle for which P=S= 36ˆ and PQS= 108ˆ . Let R be a point on the side PS for which PQR= 72ˆ and SQR= 36ˆ . Then PQ=PR, PQ=QS and QR=RS; let r be the common length of PQ, PR, QS and let s be the common length of QR and RS. Then cos 72ˆ =s/2r and cos 36ˆ =r/2s and the desired result follows. An algebraic derivation of this result can also be given.

4cos 36ˆ cos 72ˆ = 4sin 36ˆ cos 36ˆ cos 72ˆ sin 36ˆ = 2sin 72ˆ cos 72ˆ sin 36ˆ = sin 144ˆ sin 36ˆ =1.

We can make use of complex numbers. Let ζ=cos 72ˆ +isin 72ˆ so that ζ is a nonreal root of

0= x5 -1=(x-1)( x4 + x3 + x2 +x+1).

Hence 1+ζ+ ζ2 + ζ3 + ζ4 =0; using de Moivre's theorem, and taking the real part of this equation, we find that

1+cos 72ˆ +cos 144ˆ +cos 216ˆ +cos 288ˆ =1+2cos 72ˆ -2cos 36ˆ =0.

(Note that taking the imaginary part yields a triviality.) Another way to look at this result is to note that the vectors represented by the five roots of unity sum bound a closed regular pentagon and so sum to zero.
Solution 1. Let cos 36ˆ =t. Then

cos 5ˆ +cos 77ˆ +cos 149ˆ +cos 221ˆ +cos 293ˆ =[cos 5ˆ +cos 293ˆ ]+[cos 77ˆ +cos 221ˆ ]+cos 149ˆ =cos 149ˆ [2cos 144ˆ +2cos 72ˆ +1] =cos 149ˆ [-2cos 36ˆ +2cos 72ˆ +1]=0.

Alternatively, this is seen to be equal to

cos 149ˆ [-2t+2(2 t2 -1)+1]=cos 149ˆ [-2t+4 t2 -1]=0.

Solution 2. [C. Huang]

cos 5ˆ +cos 77ˆ +cos 149ˆ +cos 221ˆ +cos 293ˆ =cos 5ˆ +2cos 185ˆ cos 108ˆ +2cos 185ˆ cos 36ˆ =cos 5ˆ [1+2(cos 72ˆ -cos 36ˆ )]=cos 5ˆ [1-4sin 18ˆ sin 54ˆ ] =cos 5ˆ [1-4cos 72ˆ cos 36ˆ ]=0.

Comment. There were other solutions with similar manipulations. A quick solution can also be realized using complex numbers. The given expression is equal to the real part of

(cosθ+isinθ)(1+ζ+ ζ2 + ζ3 + ζ4 )=0,

where ζ is an imaginary fifth root of unity and θ is the angle whose degree measure is 5ˆ .

Prove that, for each positive integer n exceeding 1,

1 2n + 1 21/n <1.

Solution 1. For positive integer n>1 and -1<x, x0, we have that (1+x)n >1+nx (use induction). Hence, for n2,

(1- 1 2n )n >1- n 2n 1 2 .

(The latter inequality is left as an an easy induction exercise.) Hence

1- 1 2n > 1 21/n

and the result follows.
Solution 2. [R. Furmaniak] By the arithmetic-geometric means inequality, we have that

1- 1 2n = 1 2 ·1+ 1 4 ·1++ 1 2n-1 ·1+ 1 2n-1 · 1 2 > ( 1 2 )1/( 2n-1 ) .

Now, 2n-1 =1+1+2++ 2n-2 1+1+1++1=n, so that 1/( 2n-1 )<1/n and (1/2)1/( 2n-1 ) >(1/2)1/n =1/( 21/n ). The result follows.
Solution 3. [S. Wong] By the arithmetic-geometric means inequality

1 21/n = ( 1 2 ·1·11 )1/n < n-1+ 1 2 n =1- 1 2n

1 2n + 1 21/n <1+ 1 2n - 1 2n .

When n=2, 2n= 2n , while if n3,

2n =(1+1)n =1+n++( n n-1 )+1>2n

The desired inequality follows.

Solve, for real x,

x· 21/x + 1 x · 2x =4.

Solution 1. If x<0, the left side is negative and so there is no negative value of x which satisfies the equation. If x=0, then the left side is undefined. Suppose that x>0. Then by the arithmetic-geometric means inequality, we have that

4=x· 21/x + 1 x · 2x 2 2(1/x)+x 2 22 =4,

since (1/x)+x2. Since the outside members of this inequality are equal, we must have equality everywhere, and this occurs if and only if x=1. Hence, x=1 is the sole solution of the equation.
Solution 2. As before, we see that x must be positive. The equation is equivalent to

0 = x2 21/x -4x+ 2x =(x· 21/(2x) -2· 2-1/(2x) )2 -(4· 2-(1/x) - 2x ).

Since the first member of the right side is nonnegative, we must have that 4· 2-1/x 2x , which is equivalent to 4 2x+(1/x) or x+(1/x)2. But the last can occur if and only if x=1. Indeed, x=1 satisfies the equation.

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