Solutions: Math League Contest #2,
November 18, 2008
ACHS Math Competition Team
18 Nov 2008
Problem 2-1
What is the only negative integer x for which |x + 2| = |x + 4|?
Problem 2-1
What is the only negative integer x for which |x + 2| = |x + 4|?
|x + 2| = |x − (−2)| is the distance of x from −2.
Problem 2-1
What is the only negative integer x for which |x + 2| = |x + 4|?
|x + 2| = |x − (−2)| is the distance of x from −2.
|x + 4| = |x − (−4)| is the distance of x from −4.
Problem 2-1
What is the only negative integer x for which |x + 2| = |x + 4|?
|x + 2| = |x − (−2)| is the distance of x from −2.
|x + 4| = |x − (−4)| is the distance of x from −4.
So we’re looking for a number x which is equally distant from −2
and −4, which is, of course, their average, x = −3.
Problem 4-2
Into how many unit squares can we
partition the 36-sided equilateral polygon
shown if its perimeter is 36 and if each of
its sides is perpendicular to both sides
adjacent to it?
Problem 4-2
Into how many unit squares can we
partition the 36-sided equilateral polygon
shown if its perimeter is 36 and if each of
its sides is perpendicular to both sides
adjacent to it?
The total number of squares is
N = (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7) + 9
Problem 4-2
Into how many unit squares can we
partition the 36-sided equilateral polygon
shown if its perimeter is 36 and if each of
its sides is perpendicular to both sides
adjacent to it?
The total number of squares is
N = (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7) + 9
= (1 + 3 + 5 + 7) + (7 + 5 + 3 + 1) + 9
Problem 4-2
Into how many unit squares can we
partition the 36-sided equilateral polygon
shown if its perimeter is 36 and if each of
its sides is perpendicular to both sides
adjacent to it?
The total number of squares is
N = (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7) + 9
= (1 + 3 + 5 + 7) + (7 + 5 + 3 + 1) + 9
= [(1 + 7) + (3 + 5) + (5 + 3) + (7 + 1)] + 9
Problem 4-2
Into how many unit squares can we
partition the 36-sided equilateral polygon
shown if its perimeter is 36 and if each of
its sides is perpendicular to both sides
adjacent to it?
The total number of squares is
N = (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7) + 9
= (1 + 3 + 5 + 7) + (7 + 5 + 3 + 1) + 9
= [(1 + 7) + (3 + 5) + (5 + 3) + (7 + 1)] + 9
=4×8+9
Problem 4-2
Into how many unit squares can we
partition the 36-sided equilateral polygon
shown if its perimeter is 36 and if each of
its sides is perpendicular to both sides
adjacent to it?
The total number of squares is
N = (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7) + 9
= (1 + 3 + 5 + 7) + (7 + 5 + 3 + 1) + 9
= [(1 + 7) + (3 + 5) + (5 + 3) + (7 + 1)] + 9
=4×8+9
= 41
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45.
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
d
q
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
d
0
q
0
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
7
d
0
1
q
0
0
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
7
5
d
0
1
2
q
0
0
0
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
7
5
3
d
0
1
2
3
q
0
0
0
0
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
7
5
3
1
d
0
1
2
3
4
q
0
0
0
0
0
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
7
5
3
1
4
d
0
1
2
3
4
0
q
0
0
0
0
0
1
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
7
5
3
1
4
2
d
0
1
2
3
4
0
1
q
0
0
0
0
0
1
1
Problem 2-3
I have 10 nickels, 10 dimes, and 10 quarters. In how many different
ways can I pay for a 45 cent item? (Note: Two ways are different if
and only if the number of nickels, dimes, or quarters used is
different.)
Let n, d, and q be the number of nickels, dimes, and quarters
used, respectively. Then 5n + 10d + 25q = 45. Let’s make a table,
starting with use of the lowest denomination coins:
n
9
7
5
3
1
4
2
0
So there are 8 different ways.
d
0
1
2
3
4
0
1
2
q
0
0
0
0
0
1
1
1
Problem 2-4
I cut a roll of red tape into 251 pieces, and then I cut each of the
251 pieces into 8 smaller pieces. At most how many cuts did I
need to make to produce the resulting 2008 pieces?
Problem 2-4
I cut a roll of red tape into 251 pieces, and then I cut each of the
251 pieces into 8 smaller pieces. At most how many cuts did I
need to make to produce the resulting 2008 pieces?
Method I
Each cut through a single thickness of tape increases the number
of pieces by 1. We started with 1 piece, we ended with 2008
pieces, so the number of cuts was at most 2007.
Problem 2-4
I cut a roll of red tape into 251 pieces, and then I cut each of the
251 pieces into 8 smaller pieces. At most how many cuts did I
need to make to produce the resulting 2008 pieces?
Method I
Each cut through a single thickness of tape increases the number
of pieces by 1. We started with 1 piece, we ended with 2008
pieces, so the number of cuts was at most 2007.
Method II
To get 251 pieces requires 250 cuts, at most. To cut each of these
pieces into 8 smaller pieces requires 7 cuts, at most. So the total
number of cuts needed is at most 250 + 251 × 7 = 2007.
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b.
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b. Then
f (f (x )) =
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b. Then
f (f (x )) = a(ax + b) + b
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b. Then
f (f (x )) = a(ax + b) + b = a2 x + ab + b
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b. Then
f (f (x )) = a(ax + b) + b = a2 x + ab + b = 4x + 3,
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b. Then
f (f (x )) = a(ax + b) + b = a2 x + ab + b = 4x + 3,
so that a2 = 4 and (a + 1)b = 3.
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b. Then
f (f (x )) = a(ax + b) + b = a2 x + ab + b = 4x + 3,
so that a2 = 4 and (a + 1)b = 3. The two solutions are
(a1 , b1 ) = (2, 1) and (a2 , b2 ) = (−2, −3).
Problem 4-5
If exactly two different linear functions f and g satisfy
f (f (x )) = g (g (x )) = 4x + 3, what is the product of f (1) and g (1)?
If f is a linear function, it can be written as f (x ) = ax + b. Then
f (f (x )) = a(ax + b) + b = a2 x + ab + b = 4x + 3,
so that a2 = 4 and (a + 1)b = 3. The two solutions are
(a1 , b1 ) = (2, 1) and (a2 , b2 ) = (−2, −3). Then
f (1) g (1) = (a1 + b1 )(a2 + b2 ) = (2 + 1)(−2 + (−3)) = −15
Problem 4-6
Two triangles which are not congruent can actually have five pairs of
congruent parts! For example, triangles with side-lengths 8, 12 18, and
12, 18, 27 have five pairs of congruent parts (two sides and three
angles). If two non-congruent right triangles have five pairs of congruent
parts, what is the ratio of the length of the hypotenuse of either triangle to
the length of that triangle’s shorter leg?
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
c
b
a
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
c
b
a
b
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
c
b
c
b
=
a
b
a
d
c
b
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
c
b
c
b
=
=⇒ ac = b2
a
b
a
d
c
b
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
c
b
c
b
=
=⇒ ac = b2 = c 2 −a2
a
b
a
d
c
b
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
2
c
c
c
b
=
=⇒ ac = b2 = c 2 −a2 =⇒
= 2 −1
a
a
b
a
c
b
a
d
c
b
Problem 4-6
If two non-congruent right triangles have five pairs of congruent parts,
what is the ratio of the length of the hypotenuse of either triangle to the
length of that triangle’s shorter leg?
All 3 angles must be congruent, since otherwise
all 3 sides would be and the triangles would be
congruent. So the triangles must be similar.
2
c
c
c
b
=
=⇒ ac = b2 = c 2 −a2 =⇒
= 2 −1
a
a
b
a
2
c
b
a
d
c
b
Let x = c/a. Then x − x − 1 = 0. From the quadratic formula, we find
that
√
c
1+ 5
x= =
a
2
(the positive root is required since c/a > 0)
This value is known as the Golden Ratio.