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
|
|
n
å
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+ 1
n
|
|
|
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
and
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
|
|
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
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
|
|
aa1
2
|
+ |
aa2
2
|
+ ¼+ |
aan
2
|
= |
a
2
|
|
n
å
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
|
, |
|
from which
|
|
n
å
i=1
|
|
1
ai
|
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 = 2p.
Now
|
|
|
= 2cos |
a+ b
2
|
cos |
a- b
2
|
+ 2cos2 |
g
2
|
- 1 |
| |
|
= -2 cos |
g
2
|
cos |
a- b
2
|
+ 2cos2 |
g
2
|
- 1 |
| |
| = |
1
2
|
|
é
ê
ë
|
2cos |
g
2
|
- cos |
a- b
2
|
|
ù
ú
û
|
2
|
+ |
1
2
|
sin2 |
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
|
|
y2 + z2 - 1
2yz
|
+ |
x2 + z2 - 1
2xz
|
+ |
y2 + x2 - 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 - x2
|
+ 1 - x |
2
|
|
|
| |
|
= |
æ
ú
ú
Ö
|
|
( |
| ____
Ö1 + x
|
+ |
| ____
Ö1 - x
|
)2 |
2
|
|
|
| |
|
|
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
|
) |
|
|
| |
|
= |
|
( |
| ____
Ö1 + x
|
+ |
| ____
Ö1 - x
|
)( |
| ____
Ö1 + x
|
- |
| ____
Ö1 - x
|
)(1 + x + |
| _____
Ö1 - x2
|
+ 1 - x) |
|
|
| |
|
= |
|
(1 + x - 1 + x)(2 + |
| _____
Ö1 - x2
|
) |
|
|
| |
|
|
-
184.
-
Using complex numbers, or otherwise, evaluate
Solution. Let z = cos20° + isin20°,
so that 1/z = cos20° - isin20°.
Then, by De Moivre's Theorem, z9 = -1. Now,
|
sin70° = cos20° = |
1
2
|
(z + |
1
z
|
) = |
z2 + 1
2z
|
, |
|
|
sin50° = cos40° = |
1
2
|
(z2 + |
1
z2
|
) = |
z4 + 1
2z2
|
, |
|
and
|
sin10° = cos80° = |
1
2
|
(z4 + |
1
z4
|
) = |
z8 + 1
2z4
|
. |
|
Hence
|
|
|
= |
z2 + 1
2z
|
· |
z4 + 1
2z2
|
· |
z8 + 1
2z4
|
|
| |
|
= |
1 + z2 + z4 + z6 + z8 + z10 + z12 + z14
8z7
|
|
| |
| |
| |
|
|
