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Transcript
1. In the diagram, two straight lines are to be
drawn through O(0, 0) so that the lines divide
the figure OPQRST into 3 pieces of equal area.
Find the sum of the slopes of the lines.
y
P(0, 6)
Q(6, 6)
R(6, 2)
O(0, 0)
S(12, 2)
x
T(12, 0)
2. Determine all linear functions f (x) = ax + b
such that if g (x) = f –1(x) for all values of x,
then f(x) – g(x) = 44 for all values of x.
(Note : f –1 is the inverse function of f )
3. (a) Determine all pairs (a, b) of positive
integers for which a3 + 2ab = 2013.
(b) Determine all real values of x for which
log2(2x–1 + 3x+1) = 2x – log2(3x).
4. Integers can be written in bases other than
the usual base 10. For example, the notation
(235)7 stands for the base 7 representation of
the integer 2 × 72 + 3 × 7 + 5 (which equals
124 in base 10). In general, if x, y and z are
integers between 0 and b – 1, inclusive, then
(xyz)b = xb2 + yb + z.
Find all triples (x, y, z) for which (xyz)10 = 2(xyz)7,
where each of x, y and z comes from the list
1, 2, 3, 4, 5, 6.
5. Vernon starts with a first number n with
0 < n < 1. Vernon enters the first number into
a machine to produce a second number. He
then enters the second number back into the
machine to produce a third number. Vernon
continually enters the result back into the
machine giving a chain of numbers.
When Vernon enters a number x into the
machine,
1
zz if x ≤ , the machine outputs 2x, and
2
1
zz if x > , the machine outputs 2(1 – x).
2
If the machine ever produces the number 1,
Vernon stops the process.
(a) A chain starts with 3 . This gives
11
3
6
10
2
→ → → → .... . What are the
11 11 11 11
next four numbers in this chain?
(b) Vernon enters a number x with 0 < x < 1
into the machine and the machine produces
the number x. Determine the value of x.
(c) The fourth number in a chain is 1.
Determine all of the possible values of the
first number in this chain.
Solutions
1. Extend QR downwards to meet the x-axis at
U(6, 0).
y
P(0, 6)
Q(6, 6)
R(6, 2)
O(0, 0)
U
S(12, 2)
x
T(12, 0)
The area of figure OPQRST equals the sum of
the areas of square OPQU (which has side length
6, so area 36) and rectangle RSTU (which has
height 2 and width 6, so area 12).
Thus, the area of figure OPQRST is 48. If we are
to divide the figure into three pieces of equal
area, then each piece has area 16.
Let V be the first point on the perimeter
Mathematics TODAY | NOVEMBER ’14
73
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(measured clockwise from P) so that the line
through O and V cuts off an area of 16.
Note that the area of DOPQ is half of the area
of square OPQU, or 18, so V is to the left of
Q on PQ.
Thus, V has coordinates (v, 6) for some number v.
y
V
P(0, 6)
Q(6, 6)
R(6, 2)
O(0, 0)
U
S(12, 2)
x
T(12, 0)
Consider DOPV having base OP of length 6
and height PV of length v.
Since the area of DOPV is 16, then
1
16
(6)(v) = 16 or 3v = 16 or v = .
2
3
Thus, the coordinates of W are (8, 2), and so
2 1
the slope of OW is = . 8 4
Thus, the sum of the two required slopes is
9 1 11
+ = .
8 4 8
2. Since f (x) = ax + b, we can determine an
expression for g(x) = f –1(x) by letting y = f (x)
and to obtain y = ax + b. We then interchange x
and y to obtain x = ay + b which we solve for y
x b
to obtain ay = x – b or y = − .
a a
x
b
Therefore, f –1 (x) = − .
a a
LED
I
A
ET ER
D
E
EF OF
6 9 OR
R
Therefore, the slope of OV is
=M
.
N
16 8
FOR3 LUTIO ISSUE DAY
Let W be the second desired point.
ER S TO
SO
B
M
C
1
E
(OT)(T S) AT I
Since the area of DOTS is OV
N2
M
E
1
1
H
theT
total area)
= (12) (2) = 12 (less than ofA
2
3
M
1
and the area of trapezoid ORST is (RS + OT)
2
1
1
of
(ST) = (6 + 12) (2) = 18 (more than
2
3
the total area), then W lies on RS.
y
P(0, 6)
V
Q(6, 6)
R
O(0, 0)
U
W
S(12, 2)
x
T(12, 0)
Suppose that W has coordinates (w, 2) for
some number w.
We want the area of trapezoid WSTO to be
16.
1
Therefore, (WS + OT) (ST) = 16
2
1
⇒ (12 – w + 12) (2)= 16
2
⇒ 24 – w = 16 ⇒ w = 8
74
Note that a ≠ 0. (This makes sense since
the function f(x) = b has a graph which is a
horizontal line, and so cannot be invertible.)
Therefore, the equation f(x) – g(x) = 44
 x b
becomes (ax + b) –  −  = 44 or
a a
1
b


= 0x + 44, and this
 a − a  x +  b + a  = 44
equation is true for all x.
Since the equation
1
b


 a − a  x +  b + a  = 0x + 44
is true for all x, then the coefficients of the
linear expression on the left side must match
the coefficients of the linear expression on the
right side.
1
b
Therefore, a − = 0 and b + = 44.
a
a
From the first of these equations, we obtain
1
or a2 = 1, which gives a = 1 or a = – 1.
a
b
If a = 1, the equation b + = 44 becomes
a
b + b = 44, which gives b = 22.
a=
b
= 44 becomes
a
b – b = 44, which is not possible.
Therefore, we must have a = 1 and b = 22, and
so f(x) = x + 22.
If a = – 1, the equation b +
Mathematics TODAY | NOVEMBER ’14
Page 74
3. (a) First, we factor the left side of the given
equation to obtain a(a2 + 2b) = 2013.
Next, we factor the integer 2013 as
2013 = 3 × 671 = 3 × 11 × 61. Note
that each of 3, 11 and 61 is prime, so we
can factor 2013 no further (We can find
the factors of 3 and 11 using tests for
divisibility by 3 and 11, or by systematic
trial and error.)
Since 2013 = 3 × 11 × 61, then the positive
divisors of 2013 are
1, 3, 11, 33, 61, 183, 671, 2013
Since a and b are positive integers, then a
and a2 + 2b are both positive integers.
Since a and b are positive integers, then
a2 ≥ a and 2b > 0, so a2 + 2b > a.
Since a(a 2 + 2b) = 2013, then a and
a2 + 2b must be a divisor pair of 2013
(that is, a pair of positive integers whose
product is 2013) with a < a2 + 2b.
We make a table of the possibilities:
a
a2 + 2b
2b
b
1
2013
2012
1006
3
671
662
331
11
183
62
31
33
61
–1028
N/A
Note that the last case is not possible, since
b must be positive.
Therefore, the three pairs of positive integers
that satisfy the equation are (1, 1006),
(3, 331), (11, 31).
(We can verify by substitution that each
is a solution of the original equation.)
(b) We successively manipulate the given
equation to produce equivalent equations :
log2(2x–1 + 3x+1) = 2x – log2(3x)
log2(2x–1 + 3x+1) + log2(3x) = 2x
log2((2x–1 + 3x+1)3x) = 2x
(using log2 A + log2 B = log2 AB)
(2x–1 + 3x+1)3x = 22x
(exponentiating both sides)
2–12x3x + 313x3x = 22x
1 x x
⋅ 2 3 + 3 ⋅ 32x = 22x
2
2x3x + 6 ⋅ 32x = 2 ⋅ 22x (multiplying by 2)
2x3x2–2x + 6 ⋅ 32x2–2x = 2
(dividing both sides by 22x ≠ 0)
2–x3x + 6 ⋅ 32x ⋅ 2–2x = 2
x
 3
 3
 2 + 6  2
2x
=2
x
 3
Next, we make the substitution t =   ,
2
2
2x 
x
 3
 3
noting that   =     = t 2. Thus,
 2 
2
we obtain the equivalent equations
t + 6t2 = 2
6t2 + t – 2 = 0
(3t + 2) (2t – 1) = 0
1
2
Therefore, t = − or t = .
2
3
x
 3
Since t =   > 0, then we must have
x 2
1
 3
t=  = .
2
2
Thus,
log(1 / 2) log 1 − log 2
=
log(3 / 2) log 3 − log 2
log 2
− log 2
=
=
log 3 − log 2 log 2 − log 3
x = log 3/2(1 / 2) =
4. Since we are told that (xyz)b = xb2 + yb + z, then
(xyz)10 = 102x + 10y + z = 100x + 10y + z and
(xyz)7 = 72x+ 7y + z = 49x + 7y + z.
From the given information
(xyz)10 = 2(xyz)7
100x + 10y + z = 2(49x + 7y + z)
100x + 10y + z = 98x + 14y + 2z
2x = 4y + z
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Since the left side of this equation is an even
integer (2x), then the right side must also be
an even integer.
Since 4y is an even integer, then for 4y + z to
be an even integer, it must be the case that z
is an even integer.
This gives us three possibilities: z = 2, z = 4
and z = 6.
Case 1 : z = 2
Here, 2x = 4y + 2 or x = 2y + 1.
We try the possible values for y :
zz y = 1 gives x = 2(1) + 1 = 3; this gives the
triple (x, y, z) = (3, 1, 2)
zz y = 2 gives x = 2(2) + 1 = 5; this gives the
triple (x, y, z) = (5, 2, 2)
zz If y is at least 3, then x = 2y + 1 is at least 7,
which is impossible.
Therefore, there are two triples that work when
z = 2.
Case 2: z = 4
Here, 2x = 4y + 4 or x = 2y + 2.
We try the possible values for y :
zz y = 1 gives x = 2(1) + 2 = 4; this gives the
triple (x, y, z) = (4, 1, 4)
zz y = 2 gives x = 2(2) + 2 = 6; this gives the
triple (x, y, z) = (6, 2, 4)
zz If y is at least 3, then x = 2y + 2 is at least 8,
which is impossible.
Therefore, there are two triples that work when
z = 4.
Case 3: z = 6
Here, 2x = 4y + 6 or x = 2y + 3.
We try the possible values for y :
zz y = 1 gives x = 2(1) + 3 = 5; this gives the
triple (x, y, z) = (5, 1, 6)
zz If y is at least 2, then x = 2y + 3 is at least 7,
which is impossible.
Therefore, there is one triple that works when
z = 6.
Finally, the triples that satisfy the equation
are (x, y, z) = (3, 1, 2), (5, 2, 2), (4, 1, 4),
(6, 2, 4), (5, 1, 6).
5. (a) Since the fourth number in the chain is
2
1
2
and
is less than , then the fifth
11
2
11
76
 2 4
number is 2   = .
11 11
4
Since the fifth number in the chain is
11
4
1
is less than , then the sixth
and
11
2
 4 8
number is 2   = .
11 11
8
Since the sixth number in the chain is
11
8
1
is larger than , then the seventh
and
11
2
8

 3 6
number is 2  1 −  = 2   = .
11
11 11
6
Since the seventh number in the chain is
11
1
6
is larger than , then the eighth
and
2
11
6

 5  10
number is 2  1 −  = 2   = .
11
11
11
Therefore, the next four numbers in the
4 8 6 10
chain are
, ,
, .
11 11 11 11
(b) Therefore are two possibilities :
1
1
x≤
and x > .
2
2
1
If x ≤ and x is entered into the machine,
2
then the machine outputs 2x.
Since the input is to be the same as the
output, then x = 2x or x = 0.
This is impossible since x > 0.
1
If x > and x is entered into the machine,
2
then the machine outputs 2(1 – x).
Since the input is to be the same as the
output, then x = 2(1 – x) or x = 2 – 2x.
2
Thus, 3x = 2 and so x = , which satisfies
3
the restrictions.
Therefore, if x is entered into the machine
2
and x is produced, then x = .
3
(c) Suppose that the chain is a, b, c, 1.
In this part, we have to produce the chain
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77
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backwards (that is, we have to determine
c, b and a).
We make the following general observation:
Suppose that x is entered into the machine
and y is produced.
1
If x ≤ , then y = 2x. Solving for x in
2
1
terms of y, we obtain x = y. (This gives
2
us the input in terms of the output)
1
If x > , then y = 2(1 – x). Solving for
2
x in terms of y, we obtain y = 2 – 2x or
1
2x = 2 – y or x = (2 − y).
2
Since the third number in the chain is
c and the fourth is 1, then from above,
1
1
1
1
c = (1) = or c = 2 (2 − 1) = 2 .
2
2
1
In either case, the chain is a, b, , 1.
2
Since the second number in the chain is
1
b and the third is , then from above,
2
1  1 1
1
1 1  3 3
b =   = or b =  2 −  =   = .
2 2
4
2
2
2 2
4
78
Therefore, the chain is either a,
1 1
, ,1
4 2
1 1
3 1
, , 1. If the chain is a, , , 1,
4 2
4 2
then the first number is a and the second
1
number is .
4
1  1 1
Thus, either a =   = or
2 4
8
1
1 1  7 7
a = 2 −  =   = .
2
4
2 4
8
or a,
3 1
, , 1, then the first
4 2
number is a and the second number is
3
.
4
1  3 3
Thus, either a =   = or
2 4
8
1
3 1  5 5
a = 2 −  =   = .
2
4
2 4
8
If the chain is a,
Therefore, the possible first numbers in
1 3 5 7
the chain are , , , .
8 8 8 8
We can double check that each of these
gives a fourth number of 1:
1 1 1
3 3 1
→ → →1
→ → →1
8
4 2
8
4 2
5 3 1
7
1 1
→ → →1
→ → →1
8
4 2
8
4 2
nn
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