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, y Î U, x ¹ y } . 

What is the smallest possible number of distinct elements that
V can contain?
Solution. Let U = { x
_{i} : 1
£ i
£ n } and
x
_{1} < x
_{2} <
¼ < x
_{n}. Then
x_{1} + x_{2} < x_{1} + x_{3} < ¼ < x_{1} + x_{n} < x_{2} + x_{n} < ¼ < x_{n1} + x_{n} 

so that the 2n
 3 sums x
_{1} + x
_{j} with 2
£ j
£ n
and x
_{i} + x
_{n} with 2
£ i
£ n
1 have distinct values.
On the other hand, the set { 1, 2, 3, 4,
¼, 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 2n3
distinct sums occur. Then, by an example, demonstrate that
exactly 2n3 sums are possible. A lot of solvers got into hot
water by combining these two steps.

116.

Prove that the equation
x^{4} + 5x^{3} + 6x^{2}  4x  16 = 0 

has exactly two real solutions.
Solution. In what follows, we denote the given polynomial by p(x).
Solution 1.
p(x) = (x^{2} + 3x + 4)(x^{2} + 2x  4) = 
æ ç
è


æ ç
è

x + 
3 2


ö ÷
ø

2

+ 
7 4


ö ÷
ø


æ ç
è

(x + 1)^{2}  5 
ö ÷
ø

. 

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) = x^{4} + 5x^{3} + 6x^{2}  4x  16 = (x  1)(x+2)^{3}  8.
Let q(x) = f((x+2)) = (x3)(x)^{3}  8 = x^{4} + 3x^{3}  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 2 £ x £ 0, 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) 
æ ç
è


1 sinx

+ 
1 cosx

+ 
1 sinx cosx


ö ÷
ø

= 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
p/2, so we exclude this possibility.
Thus, in what follows, we suppose that a
¹ 1 and
that sinx cosx
¹ 0. [
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
u^{2}  1 = 2sinx cosx, so that
0 = u^{2}  (a1)u  a = (u+1)(ua) . 

Since sinx cosx
¹ 0, u + 1
¹ 0. Thus, u = a, and
sinx, cosx are the roots of the quadratic equation
t^{2}  at + 
a^{2}  1 2

= 0 . 

Hence
(sinx, cosx) = 
æ ç
è


1 2

(a ± 
 _____ Ö2  a^{2}

), 
1 2

(a ± 
 _____ Ö2  a^{2}

) 
ö ÷
ø

. 

For this to be viable, we require that
a
 £ Ö2
and a
¹ 1.
Solution 2. The given equation (since sinx cosx ¹ 0)
is equivalent to
(a1)(sinx + cosx + 1) = 2sinx cosx 

Þ (a1)^{2}(2 + 2sinx + 2cosx + 2sinx cosx) = 4sin^{2} x cos^{2} x 

Û 4(a1)(sinx cosx) +2(a1)^{2}(sinx cosx) = 4sin^{2} x cos^{2} x 

Û 2(a1) + (a1)^{2} = sin2x 

For this to be viable, we require that
a
 £ Ö2.
For all values of x, we have that
2(1 + sinx + cosx) + 2sinx cosx = (sinx + cosx + 1)^{2} , 

so that
(sinx + cosx + 1  1)^{2} = 1 + 2sinx cosx = a^{2} , 

whence
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
(a1)(sinx + cosx + 1) = (a1)(a+1) = a^{2}  1 = 2sinx cosx 

as desired. Taking sinx + cosx =
a, we find that
(a1)(sinx + cosx + 1) = (a  1)(1  a) = (a1)^{2} ¹ a^{2}  1 = 2sinx cosx 

so this does not work. Hence the equation is solvable when
a
 £ Ö2, a
¹ 1, and the solution
is given by x =
^{1}/
_{2}q where sin
q = a
^{2}  1 and sinx + cosx = a.
Solution 3. We have that
(a1) 
æ ç
è

Ö2 cos 
æ ç
è

x  
p 4


ö ÷
ø

+ 1 
ö ÷
ø



 
 
 
 
= 2cos^{2} 
æ ç
è

x  
p 4


ö ÷
ø

 1 . 


Let t = cos(x
 p/4). Then

= 2t^{2}  Ö2(a  1)t  a 
 


Since x cannot be a multiple of
p/2, t cannot equal
1/
Ö2. Hence cos(x + [(
p)/4]) = t = a/
Ö2,
so that x = [(
p)/4] +
f where cos
f = a/
Ö2.
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
a
 £ Ö2 and a
¹ 1.
Solution 4. Note that sinx + cosx = 1 implies that
2 sinx cosx = 0. Since we are assuming that sinx + cosx ¹ 0, we multiply the equation (a1)(sinx + cosx + 1) = 2sinx cosx by sinx + cosx  1 to obtain the equivalent
equation
(a1)[(sinx + cosx)^{2}  1] = 2sinx cosx(sinx + cosx  1) 

Û (a1)2sinx cosx = 2sinx cosx(sinx + cosx 1) 

Û sin 
æ ç
è

x + 
p 4


ö ÷
ø

= 
1 Öa



where sin
q = a/
Ö2.
We have the same conditions on a as before.
Solution 5. The equation is equivalent to
(a1)(sinx + 1) = cosx (2sinx + 1  a) . 

Squaring, we obtain
(a1)^{2} (sinx + 1)^{2} = (1  sin^{2} x)[4 sin^{2} x + 4(1a)sinx + (1  a)^{2}] . 

Dividing by sinx + 1 yields
2sin^{2} x  2asinx + (a^{2}  1) = 0 Þsinx = 
1 2

(a ± 
 _____ Ö2  a^{2}

) . 

[Note that the equation cosx = a
 sinx in Solution 3 leads
to the equation here.] Thus
sin^{2} x = 
2

and cos^{2} x = 
2

. 

Thus
(sinx, cosx) = 
æ ç
è


1 2

(a ± 
 _____ Ö2  a^{2}

),± 
1 2

(a ± 
 _____ Ö2  a^{2}

) 
ö ÷
ø

. 

Note that
0
£ sin
^{2} x
£ 1 requires 0
£ 1
±a
Ö[(2
 a
^{2})]
£ 2, or equivalently
a
 £ Ö2 and
a
^{2}(2
 a
^{2})
£ 1
Û a
 £ Ö2 and (a
^{2}  1)
^{2} ³ 0
Û
a
 £ Ö2.
We need to check for extraneous roots. If
(sinx , cosx) = ((1/2)(a ± 
 _____ Ö2  a^{2}

),(1/2)(a ± 
 _____ Ö2  a^{2}

) , 

then
(a1)(sinx + cosx + 1) = (a1)(a+1) = a^{2}  1 = (1/2)[a^{2}  (2a^{2})] = 2sinx cosx 

as desired. On the other hand, if
(sinx , cosx) = ((1/2)(a ± 
 _____ Ö2  a^{2}

),(1/2)(a ± 
 _____ Ö2  a^{2}

) , 

then
(a1)(sinx + cosx + 1) = (a1)( 
 ____ Ö2a^{2}

+ 1) 

while
2sinx cosx = (1/2)[a^{2}  (2  a^{2})] = (a^{2}  1) . 

These are not equal when a
¹ 1. Hence
(sinx , cosx) = ((1/2)(a ± 
 _____ Ö2  a^{2}

),(1/2)(a ± 
 _____ Ö2  a^{2}

) . 

Solution 6. Let u = sinx + cosx, so that
u^{2} = 1 + 2sinx cosx. Then the equation is equivalent to
(a1)(u+1) = u^{2}  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 sinx cosx

= 
2 a1

, 

we have that

æ ç
è


1 sinx

+ 
1 cosx


ö ÷
ø

2

= 
æ ç
è


2 a1

 
1 sinx cosx


ö ÷
ø

2



Û 
1 sin^{2} x cos^{2} x

+ 
2 sinx cosx

= 
4 (a1)^{2}

 
4 (a1)sinxcosx

+ 
1 sin^{2} x cos^{2} x



Û 
1 sinx cosx


æ ç
è

2 + 
4 a1


ö ÷
ø

= 
4 (a1)^{2}



Û 2sinx cosx = a^{2}  1 Ûsin2x = a^{2}  1 . 

We check this solution as in Solution 2.
Solution 8. [S. Patel] Let z = cosx + isinx. Note that
z ¹ 0, ±1, ±i. Then
sinx = 
1 2i


æ ç
è

z  
1 z


ö ÷
ø

= 
z^{2}  1 2iz



and
cosx = 
1 2


æ ç
è

z + 
1 z


ö ÷
ø

= 
z^{2} + 1 2z

. 

The given equation is equivalent to

= (a  1) 
é ê
ë


iz z^{2}  1

+ 
z z^{2} + 1

+ 
2iz^{2} z^{4}  1


ù ú
û


 
= (a  1) 
é ê
ë


iz z^{2}  1

+ 
z z^{2} + 1

+ 
i z^{2}  1

+ 
i z^{2} + 1


ù ú
û


 
= (a  1) 
é ê
ë


i(z+1) z^{2}  1

+ 
z + i z^{2} + 1


ù ú
û


 
= (a  1) 
é ê
ë


i z  1

+ 
1 z  i


ù ú
û


 
= (a  1) 
é ê
ë


(i+1)z z^{2}  (1+i)z + i


ù ú
û

. 


This is equivalent to
z^{2}  (1 + i)z + i = (a  1)[(1+i)z]Û z^{2}  (1 + i)az + i = 0 . 

Hence

= 
(1+i)a ± 
 ________ Ö2ia^{2}  4i

2


 
= 
(1+i)a ± 
 __ Ö2i


 _____ Öa^{2}  2

2


 
= 
æ ç
è


1+i 2


ö ÷
ø

(a ± 
 _____ Öa^{2}  2

) . 


Suppose that a^{2} > 2. Then z ^{2} = ^{1}/_{2}[a ±Ö[(a^{2}  2)]]^{2} = (a^{2}  1) ±aÖ[(a^{2}  2)]. Since
z  = 1, we must have a^{2}  2 = ±aÖ[(a^{2}  2)],
whence a^{4}  4a^{2} + 4 = a^{4}  2a^{2} or a^{2} = 2, which we do not have.
Hence, we must have a^{2} £ 2, so that
z = 
æ ç
è


1+i 2


ö ÷
ø

(a ± 
 _____ Öa^{2}  2

) . 

Therefore
and
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 D. Then

1 r

= 
s D

= 
1 + sinx + cosx sinx cosx

= 
2 a1



so that 1
£ a. 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 = 90
^{°}, we
have that
r = (s1)tan45^{°} = s  1 

whence

a1 2

= 
1 2

(sinx + cosx  1)Û sinx + cosx = a . 


118.

Let a, b, c be nonnegative real numbers.
Prove that
a^{2}(b + c  a) + b^{2}(c + a  b) + c^{2}(a + b  c) £ 3abc . 

When does equality hold?
Solution 1. Observe that
3abc  [a^{2}(b+ca) + b^{2}(c+ab) + c^{2}(a+bc)] = abc  (b+ca)(c+ab)(a+bc) . 
 (*) 
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
arithmeticgeometric means inequality yields that
2a ³ 2 
 ____________ Ö(c+ab)(a+bc)



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 a £ b £ c.
Then

 [a^{2}(b+ca) + b^{2}(c+ab) + c^{2}(a+bc)] 
 
= a(bc  ab  ac + a^{2}) + b(ac  bc  ab + b^{2}) + c(ab  ac  bc + c^{2}) 
 
= a(ba)(ca)  b(cb)(ba)  c(ca)(cb) 
 
= a(ba)(ca) + (cb)[ab  b^{2} + c^{2}  ca] 
 
= a(ba)(ca) + (cb)^{2}(c+ba) ³ 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:
[a^{2}  (bc)^{2}][b^{2}  (ac)^{2}][c^{2}  (ab)^{2}] £ a^{2}b^{2}c^{2} . 

Since b
£ a+c and c
£ a+b, we find that
b
c
 £ a, whence (b
c)
^{2} £ a
^{2}.
Thus, a
^{2}  (b
c)
^{2} £ a
^{2}, 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 åf(a, b, c) denote the
cyclic sum f(a, b, c) + f(b, c, a) + f(c, a, b). Suppose that
u = a^{3} + b^{3} + c^{3} = åa^{3} and v = å(bc)a^{2}.
Then
u+v = 
å
 a^{3} + (bc)a^{2} = 
å
 a^{2}(a + b  c) 

and
uv = 
å
 a^{3}  (bc)a^{2} = 
å
 a^{2}(a  b + c) = 
å
 a^{2}(c + a  b) = 
å
 b^{2}(a+bc) . 

Then

å
 a^{2}(a+bc) 
å
 b^{2}(a+bc) = u^{2}  v^{2} £ u^{2} . 

By the CauchySchwarz Inequality,

å
 ab(a+bc) £ 
 ______ Öu^{2}  v^{2}

£ a^{3} + b^{3} + c^{3} . 

(Note that the left side turns out to be positive; however, the
result would hold anyway even if it were negative, since a
^{3} + b
^{3}+ c
^{3} is nonnegative.)
Then
a^{2}b + ab^{2}  abc + b^{2}c + bc^{2}  abc + a^{2}c + ac^{2}  abc £ a^{3} + b^{3} + c^{3} 

Þ a^{2}(b+ca) + b^{2}(c+ab) + c^{2}(a+bc) £ 3abc 

as desired.
Equality holds if and only if
a^{2}(a+bc) : b^{2}(b+ca) : c^{2}(c+ab) = b^{2}(a+bc) : c^{2}(b+ca) : a^{2}(c+ab) . 

Suppose, if possible that a + b = c, say. Then
b+c
a = 2b, c+a
b = 2a, so that b
^{3}:c
^{2}a = c
^{2}b:a
^{3} and
c
^{2}=ab. This is possible if and only if a = b = 0.
Otherwise, all terms in brackets are nonzero, and we find that
a
^{2}:b
^{2}:c
^{2} = b
^{2}:c
^{2}:a
^{2} so that a = b = c.
Solution 5. [A. Chan] Wolog, let a £ b £ c, so that
b = a+x and c = a + x + y, where x, y ³ 0. The left side
of the inequality is equal to
3a^{3} + 3(2x + y)a^{2} + (2x^{2} + 2xy  y^{2})a  (y^{3} + 2xy^{2}) 

and the right side is equal to
3a^{3} + 3(2x + y)a^{2} + (3x^{2} + 3xy)a . 

The right side minus the left side is equal to
(x^{2} + xy + y^{2})a + (y^{3} + 2xy^{2}) . 

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
AB ^{2} + BC ^{2} + CA ^{2} = 3 (GA ^{2} + GB ^{2} + GC ^{2}) . 

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
q =
ÐAMB. By the law of cosines, we have that
c^{2} = 
9 4

u^{2} + 
1 4

a^{2}  
3 2

aucosq 

b^{2} = 
9 4

u^{2} + 
1 4

a^{2} + 
3 2

aucosq 

whence
b^{2} + c^{2} = 
9 2

u^{2} + 
1 2

a^{2} . 

Combining this with two similar equations for the other vertices
and opposite sides, we find that
2(a^{2} + b^{2} + c^{2}) = 
9 2

(u^{2} + v^{2} + w^{2}) + 
1 2

(a^{2} + b^{2} + c^{2}) 

which simplifies to a
^{2} + b
^{2} + c
^{2} = 3(u
^{2} + v
^{2} + w
^{2}), as
desired.
Solution 2. We have that

®
AG

= 
1 3

( 
®
AB

+ 
®
AC

) . 

Hence
 
®
AG

^{2} = 
1 9

[ 
®
AB

^{2} +  
®
AC

^{2} + 2( 
®
AB

· 
®
AC

)] . 

Similarly
 
®
BG

^{2} = 
1 9

[ 
®
BA

^{2} +  
®
BC

^{2} + 2( 
®
BA

· 
®
BC

)] , 

and
 
®
CG

^{2} = 
1 9

[ 
®
CA

^{2} +  
®
CB

^{2} + 2( 
®
CA

· 
®
CB

)] . 

Therefore

[  
®
A

G ^{2} + 
®
BG

^{2} +  
®
CG

^{2} ] 
 
= 2[  
®
AB

^{2} +  
®
BC

^{2} +  
®
CA

^{2}] + 
®
AB

·( 
®
AC

+ 
®
CB

) + 
®
AC

·( 
®
AB

+ 
®
BC

) + 
®
BC

·( 
®
BA

+ 
®
AC

) 
 
= 3[  
®
AB

^{2} +  
®
BC

^{2} +  
®
CA

^{2} , 


as desired.

120.

Determine all pairs of nonnull vectors
x, y for which the following sequence
{ a_{n} : n = 1, 2, ¼} is (a) increasing,
(b) decreasing, where
Solution 1. By the triangle inequality, we obtain that
x  ny + x (n  2)y  ³ 2 x  (n  1)y  , 

whence
x  ny  x  (n1)y  ³ x  (n1)y x  (n2)y  , 

for n
³ 3. This establishes that the sequence is never
decreasing, and will increase if and only if
x  2y  ³ x  y  ³ 0 . 

This condition is equivalent to
x·x  4x·y + 4y·y ³ x·x  2x·y + y·y 

or 3
y ^{2} ³ 2
x·
y.
Solution 2.

= x ^{2} 2n(x·y) + n^{2} y ^{2} 
 
= y ^{2} 
é ê
ë


æ ç
è

n  
x·y y ^{2}


ö ÷
ø

2


ù ú
û

+ 
é ê
ë

x ^{2}  
(x·y)^{2} y ^{4}


ù ú
û

. 


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 c
£ 3/2. Hence, the sequence
increases if and only if