CMS/SMC
Canadian Mathematical Society
www.cms.math.ca
Canadian Mathematical Society
  location: 
       
Solutions
115.
Let U be a set of n distinct real numbers and let V be the set of all sums of distinct pairs of them, i.e.,

V={x+y:x,yU,xy}.

What is the smallest possible number of distinct elements that V can contain?
Solution. Let U={ xi :1in} and x1 < x2 << xn . Then

x1 + x2 < x1 + x3 << x1 + xn < x2 + xn << xn-1 + xn

so that the 2n-3 sums x1 + xj with 2jn and xi + xn with 2in-1 have distinct values. On the other hand, the set {1,2,3,4,,n} has the smallest pairwise sum 3=1+2 and the largest pairwise sum 2n-1=(n-1)+n, so there are at most 2n-3 pairwise sums. Hence V can have as few as, but no fewer than, 2n-3 elements.
Comment. This problem was not well done. A set is assumed to be given, and so you must deal with it. Many of you tried to vary the elements in the set, and say that if we fiddle with them to get an arithmetic progression we minimize the number of sums. This is vague and intuitive, does not deal with the given set and needs to be sharpened. To avoid this, the best strategy is to take the given set and try to determine pairwise sums that are sure to be distinct, regardless of what the set is; this suggests that you should look at extreme elements - the largest and the smallest. To make the exposition straightforward, assume with no loss of generality that the elements are in increasing or decreasing order. You want to avoid a proliferation of possibilities.
Secondly, in setting out the proof, note that it should divide cleanly into two parts. First show that, whatever the set, at least 2n-3 distinct sums occur. Then, by an example, demonstrate that exactly 2n-3 sums are possible. A lot of solvers got into hot water by combining these two steps.


116.
Prove that the equation

x4 +5 x3 +6 x2 -4x-16=0

has exactly two real solutions.
Solution. In what follows, we denote the given polynomial by p(x).
Solution 1.

p(x)=( x2 +3x+4)( x2 +2x-4)= ( (x+ 3 2 )2 + 7 4 ) ((x+1)2 -5 ).

The first quadratic factor has nonreal roots, and the second two real roots, and the result follows.
Solution 2. Since p(1)=p(-2)=-8, the polynomial p(x)+8 is divisible by x+2 and x-1. We find that

p(x)=(x+2)3 (x-1)-8.

When x>1, the linear factors are strictly increasing, so p(x) strictly increases from -8 unboundedly, and so vanishes exactly once in the interval (1,). When x<-2, the two linear factors are both negative and increasing, so that p(x) strictly decreases from positive values to -8. Thus, it vanishes exactly once in the interval (-,-2). When -2<x<1, the two linear factors have opposite signs, so that (x+2)3 (x-1)<0 and p(x)<-8<0. The result follows.
Solution 3. [R. Mong] We have that p(x)= x4 +5 x3 +6 x2 -4x-16=(x-1)(x+2)3 -8. Let q(x)=f(-(x+2))=(-x-3)(-x)3 -8= x4 +3 x3 -8. By Descartes' Rule of Signs, p(x) and q(x) each have exactly one positive root. (The rule says that the number of positive roots of a real polynomial has the same parity as and no more than the number of sign changes in the coefficients as read in descending order.) It follows that p(x) has exactly one root in each interval (-,-2) and (0,). Since p(x)-8 for -2x0, the desired result follows.
Solution 4. Since the derivative p'(x)=(x+2)2 (4x-1), we deduce that p'(x)<0 for x< 1 4 and p'(x)>0 for x> 1 4 . It follows that p(x) is strictly decreasing on (, 1 4 ) and strictly increasing on ( 1 4 ,). Since the leading coefficient is positive and p( 1 4 )<0, p(x) has exactly one root in each of the two intervals.


117.
Let a be a real number. Solve the equation

(a-1) ( 1 sinx + 1 cosx + 1 sinxcosx )=2.

Solution. First step. When a=1, the equation is always false and there is no solution. Also, the left side is undefined when x is a multiple of π/2, so we exclude this possibility. Thus, in what follows, we suppose that a1 and that sinxcosx0. [Comment. This initial clearing away the underbrush avoids nuisance situations later and makes the exposition of the core of the solution go easier.]
Solution 1. Let u=sinx+cosx. Then u2 -1=2sinxcosx, so that

(a-1)(u+1)= u2 -1&lrArr;


0= u2 -(a-1)u-a=(u+1)(u-a).

Since sinxcosx0, u+10. Thus, u=a, and sinx, cosx are the roots of the quadratic equation

t2 -at+ a2 -1 2 =0.

Hence

(sinx,cosx)= ( 1 2 (a±2- a2 ), 1 2 (a2- a2 ) ).

For this to be viable, we require that a2 and a1.
Solution 2. The given equation (since sinxcosx0) is equivalent to

(a-1)(sinx+cosx+1)=2sinxcosx


(a-1)2 (2+2sinx+2cosx+2sinxcosx)=4sin2 xcos2 x


&lrArr;4(a-1)(sinxcosx)+2(a-1)2 (sinxcosx)=4sin2 xcos2 x


&lrArr;2(a-1)+(a-1)2 =sin2x


&lrArr;sin2x= a2 -1.

For this to be viable, we require that a2.
For all values of x, we have that

2(1+sinx+cosx)+2sinxcosx=(sinx+cosx+1)2 ,

so that

(sinx+cosx+1-1)2 =1+2sinxcosx= a2 ,

whence

sinx+cosx=±a.

Since we squared the given equation, we may have introduced extraneous roots, so we need to check the solution. Taking sinx+cosx=a, we find that

(a-1)(sinx+cosx+1)=(a-1)(a+1)= a2 -1=2sinxcosx

as desired. Taking sinx+cosx=-a, we find that

(a-1)(sinx+cosx+1)=(a-1)(1-a)=-(a-1)2 a2 -1=2sinxcosx

so this does not work. Hence the equation is solvable when a2, a1, and the solution is given by x= 1 2 θ where sinθ= a2 -1 and sinx+cosx=a.
Solution 3. We have that

(a-1) (2cos (x- π 4 )+1 ) =(a-1)(sinx+cosx+1) =2sinxcosx=sin2x =cos ( π 2 -2x ) =cos2 (x- π 4 ) =2cos2 (x- π 4 )-1.

Let t=cos(x-π/4). Then

0 =2 t2 -2(a-1)t-a =(2t-a)(2t+1).

Since x cannot be a multiple of π/2, t cannot equal 1/2. Hence cos(x+ π 4 )=t=a/2, so that x= π 4 +φ where cosφ=a/2. Since the equation at the beginning of this solution is equivalent to the given equation and the quadratic in t, this solution is valid, subject to a2 and a1.
Solution 4. Note that sinx+cosx=1 implies that 2sinxcosx=0. Since we are assuming that sinx+cosx0, we multiply the equation (a-1)(sinx+cosx+1)=2sinxcosx by sinx+cosx-1 to obtain the equivalent equation

(a-1)[(sinx+cosx)2 -1]=2sinxcosx(sinx+cosx-1)


&lrArr;(a-1)2sinxcosx=2sinxcosx(sinx+cosx-1)


&lrArr;sinx+cosx=a


&lrArr;sin (x+ π 4 )= 1 a


&lrArr;x=θ- π 4

where sinθ=a/2. We have the same conditions on a as before.
Solution 5. The equation is equivalent to

(a-1)(sinx+1)=cosx(2sinx+1-a).

Squaring, we obtain

(a-1)2 (sinx+1)2 =(1-sin2 x)[4sin2 x+4(1-a)sinx+(1-a)2 ].

Dividing by sinx+1 yields

2sin2 x-2asinx+( a2 -1)=0sinx= 1 2 (a±2- a2 ).

[Note that the equation cosx=a-sinx in Solution 3 leads to the equation here.] Thus

sin2 x= 1±a2- a2 2 andcos2 x= 1a2- a2 2 .

Thus

(sinx,cosx)= ( 1 2 (a±2- a2 ),± 1 2 (a2- a2 ) ).

Note that 0sin2 x1 requires 01±a2- a2 2, or equivalently a2 and a2 (2- a2 )1 &lrArr; a2 and ( a2 -1)2 0 &lrArr; a2.
We need to check for extraneous roots. If

(sinx,cosx)=((1/2)(a±2- a2 ),(1/2)(a2- a2 ),

then

(a-1)(sinx+cosx+1)=(a-1)(a+1)= a2 -1=(1/2)[ a2 -(2- a2 )]=2sinxcosx

as desired. On the other hand, if

(sinx,cosx)=((1/2)(a±2- a2 ),-(1/2)(a2- a2 ),

then

(a-1)(sinx+cosx+1)=(a-1)(2- a2 +1)

while

2sinxcosx=-(1/2)[ a2 -(2- a2 )]=-( a2 -1).

These are not equal when a1. Hence

(sinx,cosx)=((1/2)(a±2- a2 ),(1/2)(a2- a2 ).

Solution 6. Let u=sinx+cosx, so that u2 =1+2sinxcosx. Then the equation is equivalent to (a-1)(u+1)= u2 -1, whence u=1 or u=a. We reject u=1, so that u=a and we can finish as in Solution 3.
Solution 7. [O. Bormashenko] Since

1 sinx + 1 cosx + 1 sinxcosx = 2 a-1 ,

we have that

( 1 sinx + 1 cosx )2 = ( 2 a-1 - 1 sinxcosx )2


&lrArr; 1 sin2 xcos2 x + 2 sinxcosx = 4 (a-1)2 - 4 (a-1)sinxcosx + 1 sin2 xcos2 x


&lrArr; 1 sinxcosx (2+ 4 a-1 )= 4 (a-1)2


&lrArr;2sinxcosx= a2 -1&lrArr;sin2x= a2 -1.

We check this solution as in Solution 2.
Solution 8. [S. Patel] Let z=cosx+isinx. Note that z0,±1,±i. Then

sinx= 1 2i (z- 1 z )= z2 -1 2iz

and

cosx= 1 2 (z+ 1 z )= z2 +1 2z .

The given equation is equivalent to

1 =(a-1) [ iz z2 -1 + z z2 +1 + 2 iz2 z4 -1 ] =(a-1) [ iz z2 -1 + z z2 +1 + i z2 -1 + i z2 +1 ] =(a-1) [ i(z+1) z2 -1 + z+i z2 +1 ] =(a-1) [ i z-1 + 1 z-i ] =(a-1) [ (i+1)z z2 -(1+i)z+i ].

This is equivalent to

z2 -(1+i)z+i=(a-1)[(1+i)z]&lrArr; z2 -(1+i)az+i=0.

Hence

z = (1+i)a±2 ia2 -4i 2 = (1+i)a±2i a2 -2 2 = ( 1+i 2 )(a± a2 -2).

Suppose that a2 >2. Then z2 = 1 2 [a± a2 -2]2 =( a2 -1)±a a2 -2. Since z=1, we must have a2 -2=±a a2 -2, whence a4 -4 a2 +4= a4 -2 a2 or a2 =2, which we do not have. Hence, we must have a2 2, so that

z= ( 1+i 2 )(a± a2 -2).

Therefore

cosx=\frakRez= a2- a2 2

and

sinx=\frakImz= a±2- a2 2 .

Comment. R. Barrington Leigh had an interesting approach for solutions with positive values of sinx and cosx. Consider a right triangle with legs sinx and cosx, inradius r, semiperimeter s and area Δ. Then

1 r = s Δ = 1+sinx+cosx sinxcosx = 2 a-1

so that 1a. We need to determine right triangles whose inradius is 1 2 (a-1). Using the formula r=(s-c)tan(C/2) with c=1 and C= 90ˆ , we have that

r=(s-1)tan 45ˆ =s-1

whence

a-1 2 = 1 2 (sinx+cosx-1)&lrArr;sinx+cosx=a.



118.
Let a,b,c be nonnegative real numbers. Prove that

a2 (b+c-a)+ b2 (c+a-b)+ c2 (a+b-c)3abc.

When does equality hold?
Solution 1. Observe that

3abc-[ a2 (b+c-a)+ b2 (c+a-b)+ c2 (a+b-c)]=abc-(b+c-a)(c+a-b)(a+b-c).             (*)

Now

2a=(c+a-b)+(a+b-c)

with similar equations for b and c. These equations assure us that at most one of b+c-a, c+a-b and a+b-c can be negative. If exactly one of these three quantities is negative, than (*) is clearly nonnegative, and is equal to zero if and only if at least one of a, b and c vanishes, and the other two are equal. If all the three quantities are nonnegative, then an application of the arithmetic-geometric means inequality yields that

2a2(c+a-b)(a+b-c)

with similar inequalities for b and c. It follows from this that (*) is nonnegative and vanishes if and only if a=b=c, or one of a, b, c vanishes and the other two are equal.
Solution 2. Wolog, suppose that abc. Then

3abc -[ a2 (b+c-a)+ b2 (c+a-b)+ c2 (a+b-c)] =a(bc-ab-ac+ a2 )+b(ac-bc-ab+ b2 )+c(ab-ac-bc+ c2 ) =a(b-a)(c-a)-b(c-b)(b-a)-c(c-a)(c-b) =a(b-a)(c-a)+(c-b)[ab- b2 + c2 -ca] =a(b-a)(c-a)+(c-b)2 (c+b-a)0.

Equality occurs if and only if a=b=c or if a=0 and b=c.
Solution 3. [S.-E. Lu] The inequality is equivalent to

(a+b-c)(b+c-a)(c+a-b)abc.

At most one of the three factors on the left side can be negative. If one of them is negative, then the inequality is satisfied, and equality occurs if and only if both sides vanish (i.e., one of the three variables vanishes and the others are equal).
Otherwise, we can square both sides to get the equivalent inequality:

[ a2 -(b-c)2 ][ b2 -(a-c)2 ][ c2 -(a-b)2 ] a2 b2 c2 .

Since ba+c and ca+b, we find that b-ca, whence (b-c)2 a2 . Thus, a2 -(b-c)2 a2 , with similar inequalties for the other two factors on the left. It follows that the inequality holds with equality when a=b=c or one variable vanishes and the other two are equal.
Solution 4. [R. Mong] Let f(a,b,c) denote the cyclic sum f(a,b,c)+f(b,c,a)+f(c,a,b). Suppose that u= a3 + b3 + c3 = a3 and v=(b-c) a2 . Then

u+v= a3 +(b-c) a2 = a2 (a+b-c)

and

u-v= a3 -(b-c) a2 = a2 (a-b+c)= a2 (c+a-b)= b2 (a+b-c).

Then

a2 (a+b-c) b2 (a+b-c)= u2 - v2 u2 .

By the Cauchy-Schwarz Inequality,

ab(a+b-c) u2 - v2 a3 + b3 + c3 .

(Note that the left side turns out to be positive; however, the result would hold anyway even if it were negative, since a3 + b3 + c3 is nonnegative.)
Then

a2 b+ ab2 -abc+ b2 c+ bc2 -abc+ a2 c+ ac2 -abc a3 + b3 + c3


a2 (b+c-a)+ b2 (c+a-b)+ c2 (a+b-c)3abc

as desired.
Equality holds if and only if

a2 (a+b-c): b2 (b+c-a): c2 (c+a-b)= b2 (a+b-c): c2 (b+c-a): a2 (c+a-b).

Suppose, if possible that a+b=c, say. Then b+c-a=2b, c+a-b=2a, so that b3 : c2 a= c2 b: a3 and c2 =ab. This is possible if and only if a=b=0. Otherwise, all terms in brackets are nonzero, and we find that a2 : b2 : c2 = b2 : c2 : a2 so that a=b=c.
Solution 5. [A. Chan] Wolog, let abc, so that b=a+x and c=a+x+y, where x,y0. The left side of the inequality is equal to

3 a3 +3(2x+y) a2 +(2 x2 +2xy- y2 )a-( y3 +2 xy2 )

and the right side is equal to

3 a3 +3(2x+y) a2 +(3 x2 +3xy)a.

The right side minus the left side is equal to

( x2 +xy+ y2 )a+( y3 +2 xy2 ).

Since each of a,x,y is nonnegative, this expression is nonnegative, and it vanishes if and only if each term vanishes. Hence, the desired inequality holds, with equality, if and only if y=0 and either a=0 or x=0, if and only if either ( a=0 and b=c) or ( a=b=c).


119.
The medians of a triangle ABC intersect in G. Prove that

AB2 +BC2 +CA2 =3(GA2 +GB2 +GC2 ).

Solution 1. Let the respective lengths of BC, CA, AB, AG, BG and CG be a,b,c,u,v,w. If M is the midpoint of BC, then A,G,M are collinear with AM=(3/2)AG. Let θ=AMB. By the law of cosines, we have that

c2 = 9 4 u2 + 1 4 a2 - 3 2 aucosθ


b2 = 9 4 u2 + 1 4 a2 + 3 2 aucosθ

whence

b2 + c2 = 9 2 u2 + 1 2 a2 .

Combining this with two similar equations for the other vertices and opposite sides, we find that

2( a2 + b2 + c2 )= 9 2 ( u2 + v2 + w2 )+ 1 2 ( a2 + b2 + c2 )

which simplifies to a2 + b2 + c2 =3( u2 + v2 + w2 ), as desired.
Solution 2. We have that

AG= 1 3 (AB+AC).

Hence

AG2 = 1 9 [AB2 +AC2 +2(AB·AC)].

Similarly

BG2 = 1 9 [BA2 +BC2 +2(BA·BC)],

and

CG2 = 1 9 [CA2 +CB2 +2(CA·CB)].

Therefore

9 [AG2 +BG2 +CG2 ] =2[AB2 +BC2 +CA2 ]+AB·(AC+CB)+AC·(AB+BC)+BC·(BA+AC) =3[AB2 +BC2 +CA2 ,

as desired.


120.
Determine all pairs of nonnull vectors x, y for which the following sequence { an :n=1,2,} is (a) increasing, (b) decreasing, where

an =x-ny.

Solution 1. By the triangle inequality, we obtain that

x-ny+x-(n-2)y2x-(n-1)y,

whence

x-ny-x-(n-1)yx-(n-1)y-x-(n-2)y,

for n3. This establishes that the sequence is never decreasing, and will increase if and only if

x-2yx-y0.

This condition is equivalent to

x·x-4x·y+4y·yx·x-2x·y+y·y

or 3y2 2x·y.
Solution 2.

an 2 =x2 -2n(x·y)+ n2 y2 =y2 [ (n- x·y y2 )2 ]+ [x2 - (x·y)2 y4 ].

This is a quadratic whose nonconstant part involves the form (n-c)2 . This is an increasing function of n, for positive integers n, if and only if c3/2. Hence, the sequence increases if and only if

x·y y2 3 2 .


© Canadian Mathematical Society, 2014 : https://cms.math.ca/