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## Solutions.

178.
Suppose that n is a positive integer and that x1, x2, ¼, xn are positive real numbers such that x1 + x2 + ¼+ xn = n. Prove that

 nå i=1 Ö[n ]axi + b £ a + b + n - 1
for every pair a, b of real numbers with each axi + b nonnegative. Describe the situation when equality occurs.
Solution. Regarding axi + b as a product with n-1 ones, we use the arithmetic-geometric means inequality to obtain that

 Ö[n ]axi + b £ (axi + b) + 1 + ¼+ 1n
for 1 £ i £ n, with equality if and only if xi = (1-b)/a. Adding these n inequalities yields the desired result.

179.
Determine the units digit of the numbers a2, b2 and ab (in base 10 numeration), where

 a = 22002 + 32002 + 42002 + 52002
and

 b = 31 + 32 + 33 + ¼+ 32002 .
Solution. Observe that, for positive integer k, 24k º 6 and 34k º 1, modulo 10, so that 22002 º 6 ·4 º 4, 32002 º 9 and 42002 º 6, modulo 10. Hence a º 4 + 9 + 6 + 5 º 4 and a2 º 6, modulo 10. Note that b = (1/2)(32003 - 3), and that 32003 - 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 b2 º 4, modulo 10. Finally, ab º 4 ·2 = 8, modulo 10. Hence the units digits of a2, b2 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 (one-one). Suppose that z = x + yi and w = u + vi, and that f(z) = f(w). Then

 3x + 3yi + ______Öx2 + y2 = 3u + 3vi + ______Öu2 + v2 .
Equating imaginary parts yields that y = v, so that

 3(x - u) = ______Öu2 + y2 - ______Öx2 + y2 = (u2 - x2)/( ______Öu2 + y2 + ______Öx2 + y2 ) .
Suppose, if possible, that u ¹ x. Then

 3( ______Öx2 + y2 + x) = - [3( ______Öu2 + y2 + u] .
Since Ö[(x2 + y2)] ³ |x |, and Ö[(u2 + y2)] 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 a2 + b2 > 0. We must have that

 3x + ______Öx2 + y2 = a
and

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

 _______Ö9x2 + b2 = 3a - 9x .
For this equation to be solvable, it is necessary that 3x £ a. Squaring both sides of the equation leads to

 72x2 - 54ax + 9a2 - b2 = 0 .
When x = a/3 is substituted into the left side of the equation, we obtain 8a2 - 18a2 + 9a2 - b2 = -(a2 + b2) < 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 - ________Ö9a2 + 8b2

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 a1, a2, ¼, an from the sides. Prove that

 a nå i=1 1ai > 2p .
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

 aa12 + aa22 + ¼+ aan2 = a2 nå i=1 ai .
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 1/4na2 cot(p/n). Making use of the arithmetic-harmonic means inequality, we find that

a
2
cot p
n
= 1
n
n
å
i=1
ai ³ n
 nå i=1 1/ai
,
from which

 nå i=1 1ai 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

 nå i=1 1ai > 2pa .

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 = 2p. Now

 cosa+ cosb+ cosg
 = 2cos a+ b2 cos a- b2 + 2cos2 g2 - 1
 = -2 cos g2 cos a- b2 + 2cos2 g2 - 1
 = 12 éê ë 2cos g2 - cos a- b2 ùú û 2 + 12 sin2 a- b2 - 32 ³ - 32 .
>From the Law of Cosines applied to the triangles MBC, MCA and MAB, we convert this equation to

 y2 + z2 - 12yz + x2 + z2 - 12xz + y2 + x2 - 12xy ³ - 32 .
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 + _____Ö1 - x2

( (1 + x)   ____
Ö1 + x

- (1 - x)   ____
Ö1 - x

)

 x (2 + _____Ö1 - x2 )
,
where 0 < |x | < 1.
Solution. Observe that

æ
Ö

 1 + _____Ö1 - x2

=   æ
ú
ú
Ö

 1 + x + 2 _____Ö1 - x2 + 1 - x

2

=   æ
ú
ú
Ö

 ( ____Ö1 + x + ____Ö1 - x )2

2

=
 ____Ö1 + x + ____Ö1 - x

Ö2
.
Then, using the formula a3 - b3 = (a - b)(a2 + ab + b2), 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 - x2 )
=
 ( ____Ö1 + x + ____Ö1 - x )( ____Ö1 + x - ____Ö1 - x )(1 + x + _____Ö1 - x2 + 1 - x)

 x Ö2(2 + _____Ö1 - x2 )
=
 (1 + x - 1 + x)(2 + _____Ö1 - x2 )

 x Ö2(2 + _____Ö1 - x2 )
 = 2xxÖ2 = Ö2 .

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, z9 = -1. Now,

 sin70° = cos20° = 12 (z + 1z ) = z2 + 12z ,

 sin50° = cos40° = 12 (z2 + 1z2 ) = z4 + 12z2 ,
and

 sin10° = cos80° = 12 (z4 + 1z4 ) = z8 + 12z4 .
Hence

 sin10° sin50° sin70°
 = z2 + 12z · z4 + 12z2 · z8 + 12z4
 = 1 + z2 + z4 + z6 + z8 + z10 + z12 + z148z7
 = 1 - z168z7(1 - z2)
 = 1 - z7z98(z7 - z9)
 = 1 + z78(z7 + 1) = 18 .
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