Solutions.

178.

Suppose that n is a positive integer and that
x_{1}, x_{2}, ¼, x_{n} are positive real
numbers such that x_{1} + x_{2} + ¼+ x_{n} = n. Prove that

n å
i=1

Ö[n ]ax_{i} + b £ a + b + n  1 

for every pair a, b of real numbers with each ax_{i} + b nonnegative.
Describe the situation when equality occurs.
Solution. Regarding ax
_{i} + b as
a product with n
1 ones, we use the arithmeticgeometric
means inequality to obtain that
Ö[n ]ax_{i} + b £ 
(ax_{i} + b) + 1 + ¼+ 1 n



for 1
£ i
£ n, with equality if and only if x
_{i} = (1
b)/a.
Adding these n inequalities yields the desired result.

179.

Determine the units digit of the numbers a^{2},
b^{2} and ab (in base 10 numeration), where
a = 2^{2002} + 3^{2002} + 4^{2002} + 5^{2002} 

and
b = 3^{1} + 3^{2} + 3^{3} + ¼+ 3^{2002} . 

Solution. Observe that, for positive integer
k, 2
^{4k} º 6 and
3
^{4k} º 1, modulo 10, so that
2
^{2002} º 6 ·4
º 4, 3
^{2002} º 9
and 4
^{2002} º 6, modulo 10. Hence
a
º 4 + 9 + 6 + 5
º 4 and a
^{2} º 6, modulo 10.
Note that b = (1/2)(3
^{2003}  3), and that
3
^{2003}  3
º 7
 3 = 4, modulo 10. Since b is the sum
of evenly many factors, it is even, and so b
º 2 and
b
^{2} º 4, modulo 10. Finally, ab
º 4 ·2 = 8,
modulo 10. Hence the units digits of a
^{2}, b
^{2} and ab are
respectively 6, 4 and 8.

180.

Consider the function f that takes the set of
complex numbers into itself defined by f(z) = 3z + z .
Prove that f is a bijection and find its inverse.
Solution. Injection (oneone). Suppose that
z = x + yi and w = u + vi, and that f(z) = f(w). Then
3x + 3yi + 
 ______ Öx^{2} + y^{2}

= 3u + 3vi + 
 ______ Öu^{2} + v^{2}

. 

Equating imaginary parts yields that y = v, so that
3(x  u) = 
 ______ Öu^{2} + y^{2}

 
 ______ Öx^{2} + y^{2}

= (u^{2}  x^{2})/( 
 ______ Öu^{2} + y^{2}

+ 
 ______ Öx^{2} + y^{2}

) . 

Suppose, if possible, that u
¹ x. Then
3( 
 ______ Öx^{2} + y^{2}

+ x) =  [3( 
 ______ Öu^{2} + y^{2}

+ u] . 

Since
Ö[(x
^{2} + y
^{2})]
³ x
, and
Ö[(u
^{2} + y
^{2})]
³ u
, we see that,
unless x = y = u = v = 0, this equation
is impossible as the left side is positive and the right is negative
Thus, x = u.
Surjection (onto). Let a + bi be an arbitrary complex
number, and suppose that f(x + yi) = a + bi. It is straightforward to
see that f(z) = 0 implies that z = 0, so we may assume that
a^{2} + b^{2} > 0. We must have that
3x + 
 ______ Öx^{2} + y^{2}

= a 

and
Substituting y = b/3 into the first equation yields

 _______ Ö9x^{2} + b^{2}

= 3a  9x . 

For this equation to be solvable, it is necessary that 3x
£ a.
Squaring both sides of the equation leads to
72x^{2}  54ax + 9a^{2}  b^{2} = 0 . 

When x = a/3 is substituted into the left side of the equation,
we obtain 8a
^{2}  18a
^{2} + 9a
^{2}  b
^{2} =
(a
^{2} + b
^{2}) < 0.
This means that the two roots of the equation straddle a/3,
so that exactly one of the roots satisfies the necessary condition
3x
£ a. Hence, we must have
(x, y) = 
æ ç
è


9a  
 ________ Ö9a^{2} + 8b^{2}

24

, 
b 3


ö ÷
ø

. 

Thus, the function is injective and surjective, and so it is a
bijection.

181.

Consider a regular polygon with n sides,
each of length a, and an
interior point located at distances a_{1}, a_{2}, ¼,
a_{n} from the sides. Prove that
Solution. By constructing triangles from bases along the
sides of the polygons to the point A, we see that the area
of the polygon is equal to

aa_{1} 2

+ 
aa_{2} 2

+ ¼+ 
aa_{n} 2

= 
a 2


n å
i=1

a_{i} . 

However, by constructing triangles whose bases are the sides of
the polygons and whose apexes are at the centre of the polygon,
we see that the area of the polygon is equal to
^{1}/
_{4}na
^{2} cot(
p/n). Making use of the arithmeticharmonic
means inequality, we find that

a 2

cot 
p n

= 
1 n


n å
i=1

a_{i} ³ 
n

, 

from which

n å
i=1


1 a_{i}

³ 
2n ·tan(p/n) a

. 

Since tanx > x for 0 < x <
p/2, we have that
tan(
p/n) > (
p/n), we obtain that

182.

Let M be an interior point of the equilateral
triangle ABC with each side
of unit length. Prove that
MA.MB + MB.MC + MC.MA ³ 1 . 

Solution. Let the respective lengths of MA, MB and
MC be x, y and z, and let the respective angles
BMC, CMA and AMB be
a,
b and
g.
Then
a+
b+
g = 2
p.
Now

= 2cos 
a+ b 2

cos 
a b 2

+ 2cos^{2} 
g 2

 1 
 
= 2 cos 
g 2

cos 
a b 2

+ 2cos^{2} 
g 2

 1 
 
= 
1 2


é ê
ë

2cos 
g 2

 cos 
a b 2


ù ú
û

2

+ 
1 2

sin^{2} 
a b 2

 
3 2

³  
3 2

. 


>From the Law of Cosines applied to the triangles MBC,
MCA and MAB, we convert this equation to

y^{2} + z^{2}  1 2yz

+ 
x^{2} + z^{2}  1 2xz

+ 
y^{2} + x^{2}  1 2xy

³  
3 2

. 

This simplifies to (x + y + z)(xy + xz + yz)
 (x + y + z)
³ 0.
Since x + y + z
¹ 0, the result follows.

183.

Simplify the expression


æ Ö


( (1 + x) 
 ____ Ö1 + x

 (1  x) 
 ____ Ö1  x

) 

, 

where 0 < x  < 1.
Solution. Observe that

= 
æ ú
ú
Ö

1 + x + 2 
 _____ Ö1  x^{2}

+ 1  x 
2



 
= 
æ ú
ú
Ö

( 
 ____ Ö1 + x

+ 
 ____ Ö1  x

)^{2} 
2



 


Then, using the formula a
^{3}  b
^{3} = (a
 b)(a
^{2} + ab + b
^{2}), we find
that the expression given in the problem is equal to


( 
 ____ Ö1 + x

+ 
 ____ Ö1  x

)( 
 ____ Ö1 + x

3

 
 ____ Ö1  x

3

) 
x Ö2(2 + 
 _____ Ö1  x^{2}

) 


 
= 
( 
 ____ Ö1 + x

+ 
 ____ Ö1  x

)( 
 ____ Ö1 + x

 
 ____ Ö1  x

)(1 + x + 
 _____ Ö1  x^{2}

+ 1  x) 
x Ö2(2 + 
 _____ Ö1  x^{2}

) 


 
= 
(1 + x  1 + x)(2 + 
 _____ Ö1  x^{2}

) 
x Ö2(2 + 
 _____ Ö1  x^{2}

) 


 



184.

Using complex numbers, or otherwise, evaluate
sin10^{°} sin50^{°} sin70^{°} . 

Solution. Let z = cos20
^{°} + isin20
^{°},
so that 1/z = cos20
^{°}  isin20
^{°}.
Then, by De Moivre's Theorem, z
^{9} =
1. Now,
sin70^{°} = cos20^{°} = 
1 2

(z + 
1 z

) = 
z^{2} + 1 2z

, 

sin50^{°} = cos40^{°} = 
1 2

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

) = 
z^{4} + 1 2z^{2}

, 

and
sin10^{°} = cos80^{°} = 
1 2

(z^{4} + 
1 z^{4}

) = 
z^{8} + 1 2z^{4}

. 

Hence
sin10^{°} sin50^{°} sin70^{°} 

= 
z^{2} + 1 2z

· 
z^{4} + 1 2z^{2}

· 
z^{8} + 1 2z^{4}


 
= 
1 + z^{2} + z^{4} + z^{6} + z^{8} + z^{10} + z^{12} + z^{14} 8z^{7}


 
= 
1  z^{16} 8z^{7}(1  z^{2})


 
= 
1  z^{7}z^{9} 8(z^{7}  z^{9})


 
= 
1 + z^{7} 8(z^{7} + 1)

= 
1 8

. 

