ACHS Math Team
Solutions to Math League Contest 2 of 20 Nov 2007
Peter S. Simon
Upcoming Contests
◮
Contest 3: December 18, 2007
◮
Contest 4: January 15, 2008
◮
Contest 5: February 5, 2008
◮
AMC 10/12: February 12, 2008
◮
Contest 6: March 18, 2008
Note
I’ll be having surgery on Jan 17, 2008, and will not be able to attend Math
Team meetings for at least several weeks thereafter. Are any other
parents willing to coach the Math Team for several weeks if I prepare
materials in advance?
Problem 2-1
If (x − 10)(x + 10) = 0, what is the value of (x − 1)(x + 1)?
Problem 2-1
If (x − 10)(x + 10) = 0, what is the value of (x − 1)(x + 1)?
(x − 10)(x + 10) = x 2 − 102 = x 2 − 100 = 0 ⇐⇒ x 2 = 100,
Problem 2-1
If (x − 10)(x + 10) = 0, what is the value of (x − 1)(x + 1)?
(x − 10)(x + 10) = x 2 − 102 = x 2 − 100 = 0 ⇐⇒ x 2 = 100,
so
(x − 1)(x + 1) = x 2 − 1 = 100 − 1 = 99
Problem 2-2
If the least common multiple (LCM) of the first 2006 positive integers is
m, and the LCM of the first 2007 positive integers is km, what is the value
of k?
3
Problem 2-2
If the least common multiple (LCM) of the first 2006 positive integers is
m, and the LCM of the first 2007 positive integers is km, what is the value
of k?
Let m = LCM(1, 2, 3, . . . , 2006). Then
km = LCM(1, 2, . . . , 2006, 2007)
3
Problem 2-2
If the least common multiple (LCM) of the first 2006 positive integers is
m, and the LCM of the first 2007 positive integers is km, what is the value
of k?
Let m = LCM(1, 2, 3, . . . , 2006). Then
km = LCM(1, 2, . . . , 2006, 2007) = LCM(m, 2007)
3
Problem 2-2
If the least common multiple (LCM) of the first 2006 positive integers is
m, and the LCM of the first 2007 positive integers is km, what is the value
of k?
Let m = LCM(1, 2, 3, . . . , 2006). Then
km = LCM(1, 2, . . . , 2006, 2007) = LCM(m, 2007) = LCM(m, 32 ×2231 )
3
Problem 2-2
If the least common multiple (LCM) of the first 2006 positive integers is
m, and the LCM of the first 2007 positive integers is km, what is the value
of k?
Let m = LCM(1, 2, 3, . . . , 2006). Then
km = LCM(1, 2, . . . , 2006, 2007) = LCM(m, 2007) = LCM(m, 32 ×2231 ) = m,
the last equality because both 223 and 9 are already factors of m.
3
3
Problem 2-2
If the least common multiple (LCM) of the first 2006 positive integers is
m, and the LCM of the first 2007 positive integers is km, what is the value
of k?
Let m = LCM(1, 2, 3, . . . , 2006). Then
km = LCM(1, 2, . . . , 2006, 2007) = LCM(m, 2007) = LCM(m, 32 ×2231 ) = m,
the last equality because both 223 and 9 are already factors of m. Since
km = m, then k = 1.
Problem 2-3
If 2100 words fill any page that a 23-page local newspaper typesets with
large type, and 2800 words fill any page that it typesets with small type,
how many pages must the paper typeset with small type so that an article
of 56,000 words will exactly fill all 23 pages of some future issue?
Problem 2-3
If 2100 words fill any page that a 23-page local newspaper typesets with
large type, and 2800 words fill any page that it typesets with small type,
how many pages must the paper typeset with small type so that an article
of 56,000 words will exactly fill all 23 pages of some future issue?
Let m be the number of pages set with small type.
Problem 2-3
If 2100 words fill any page that a 23-page local newspaper typesets with
large type, and 2800 words fill any page that it typesets with small type,
how many pages must the paper typeset with small type so that an article
of 56,000 words will exactly fill all 23 pages of some future issue?
Let m be the number of pages set with small type. Then (23 − m) is the
number of pages set with large type.
Problem 2-3
If 2100 words fill any page that a 23-page local newspaper typesets with
large type, and 2800 words fill any page that it typesets with small type,
how many pages must the paper typeset with small type so that an article
of 56,000 words will exactly fill all 23 pages of some future issue?
Let m be the number of pages set with small type. Then (23 − m) is the
number of pages set with large type. The total number of words typeset
is then
2800m + 2100(23 − m) = 56000
Problem 2-3
If 2100 words fill any page that a 23-page local newspaper typesets with
large type, and 2800 words fill any page that it typesets with small type,
how many pages must the paper typeset with small type so that an article
of 56,000 words will exactly fill all 23 pages of some future issue?
Let m be the number of pages set with small type. Then (23 − m) is the
number of pages set with large type. The total number of words typeset
is then
2800m + 2100(23 − m) = 56000
700m = 56000 − 48300 = 7700
Problem 2-3
If 2100 words fill any page that a 23-page local newspaper typesets with
large type, and 2800 words fill any page that it typesets with small type,
how many pages must the paper typeset with small type so that an article
of 56,000 words will exactly fill all 23 pages of some future issue?
Let m be the number of pages set with small type. Then (23 − m) is the
number of pages set with large type. The total number of words typeset
is then
2800m + 2100(23 − m) = 56000
700m = 56000 − 48300 = 7700
m = 11
Problem 2-4
The diagonals of a convex quadrilateral Q are perpendicular. If three
consecutive sides of Q have respective lengths 3, 9, and 19, then how
long is the fourth side?
5
Problem 2-4
The diagonals of a convex quadrilateral Q are perpendicular. If three
consecutive sides of Q have respective lengths 3, 9, and 19, then how
long is the fourth side?
3
9
x
19
5
Problem 2-4
The diagonals of a convex quadrilateral Q are perpendicular. If three
consecutive sides of Q have respective lengths 3, 9, and 19, then how
long is the fourth side?
Use the Pythagorean Thm in each of the
four right triangles seen at right:
9
3
b
2
2
2
2
2
2
3 + 19 = a + b + c + d
a
c
92 + x 2 = b 2 + c 2 + a2 + d 2
x
d
19
5
Problem 2-4
The diagonals of a convex quadrilateral Q are perpendicular. If three
consecutive sides of Q have respective lengths 3, 9, and 19, then how
long is the fourth side?
Use the Pythagorean Thm in each of the
four right triangles seen at right:
9
3
b
2
2
2
2
2
2
3 + 19 = a + b + c + d
a
c
92 + x 2 = b 2 + c 2 + a2 + d 2
so
2
x
2
2
2
2
2
2
2
x +9 = 3 +19 =⇒ x = 3 +19 −9 = 289.
5
d
19
Problem 2-4
The diagonals of a convex quadrilateral Q are perpendicular. If three
consecutive sides of Q have respective lengths 3, 9, and 19, then how
long is the fourth side?
Use the Pythagorean Thm in each of the
four right triangles seen at right:
9
3
b
2
2
2
2
2
2
3 + 19 = a + b + c + d
a
c
92 + x 2 = b 2 + c 2 + a2 + d 2
so
2
x
2
2
2
2
2
2
2
x +9 = 3 +19 =⇒ x = 3 +19 −9 = 289.
√
We conclude that x = 289 = 17.
5
d
19
Problem 2-5
I wrote a sequence of n integers. In this sequence, the sum of any 3
consecutive terms is positive, while the sum of any 4 consecutive terms
is negative. What is the largest possible value of n?
6
Problem 2-5
I wrote a sequence of n integers. In this sequence, the sum of any 3
consecutive terms is positive, while the sum of any 4 consecutive terms
is negative. What is the largest possible value of n?
Here are some 5-term sequences that satisfy the given requirements:
−2,
−2,
−7,
6
−3,
−2,
−4,
6,
5,
12,
−2,
−2,
−6,
−3
−2
5
Problem 2-5
I wrote a sequence of n integers. In this sequence, the sum of any 3
consecutive terms is positive, while the sum of any 4 consecutive terms
is negative. What is the largest possible value of n?
Here are some 5-term sequences that satisfy the given requirements:
−2,
−2,
−7,
−3,
−2,
−4,
6,
−2,
5,
12,
−2,
−6,
−3
−2
5
Are there any such 6-term sequences? Suppose c1 , c2 , c3 , c4 , c5 , c6 is a
six-term sequence such that
c1 + c2 + c3 > 0,
6
c2 + c3 + c4 > 0,
c3 + c4 + c5 > 0,
c4 + c5 + c6 > 0
Problem 2-5
I wrote a sequence of n integers. In this sequence, the sum of any 3
consecutive terms is positive, while the sum of any 4 consecutive terms
is negative. What is the largest possible value of n?
Here are some 5-term sequences that satisfy the given requirements:
−2,
−2,
−7,
−3,
−2,
−4,
6,
−2,
5,
12,
−2,
−6,
−3
−2
5
Are there any such 6-term sequences? Suppose c1 , c2 , c3 , c4 , c5 , c6 is a
six-term sequence such that
c1 + c2 + c3 > 0,
c2 + c3 + c4 > 0,
c3 + c4 + c5 > 0,
c4 + c5 + c6 > 0
If we add all these inequalities together we get
(c1 + c2 + c3 + c4 ) + (c2 + c3 + c4 + c5 ) + (c3 + c4 + c5 + c6 ) > 0
which can not be true if each of the sums in parentheses is negative.
6
Problem 2-5
I wrote a sequence of n integers. In this sequence, the sum of any 3
consecutive terms is positive, while the sum of any 4 consecutive terms
is negative. What is the largest possible value of n?
Here are some 5-term sequences that satisfy the given requirements:
−2,
−2,
−7,
−3,
−2,
−4,
6,
−2,
5,
12,
−2,
−6,
−3
−2
5
Are there any such 6-term sequences? Suppose c1 , c2 , c3 , c4 , c5 , c6 is a
six-term sequence such that
c1 + c2 + c3 > 0,
c2 + c3 + c4 > 0,
c3 + c4 + c5 > 0,
c4 + c5 + c6 > 0
If we add all these inequalities together we get
(c1 + c2 + c3 + c4 ) + (c2 + c3 + c4 + c5 ) + (c3 + c4 + c5 + c6 ) > 0
which can not be true if each of the sums in parentheses is negative.
Therefore n = 5 is the largest possible value.
6
Problem 2-6
The degree-measure of each
base angle of an isosceles
triangle is 40. The bisector of
one of the base angles is
extended through point P on the
leg opposite that base angle so
that PA = PB, as shown. What
is m∠A?
B
100
C
40
40
D
Problem 2-6
The degree-measure of each
base angle of an isosceles
triangle is 40. The bisector of
one of the base angles is
extended through point P on the
leg opposite that base angle so
that PA = PB, as shown. What
is m∠A?
B
A
100
P
C
40
20
20
D
Problem 2-6
The degree-measure of each
base angle of an isosceles
triangle is 40. The bisector of
one of the base angles is
extended through point P on the
leg opposite that base angle so
that PA = PB, as shown. What
is m∠A?
′
B
A
100
P
C
20
20
40
Find image B of point B through line DA.
B
′
D
Problem 2-6
The degree-measure of each
base angle of an isosceles
triangle is 40. The bisector of
one of the base angles is
extended through point P on the
leg opposite that base angle so
that PA = PB, as shown. What
is m∠A?
′
B
A
100
P
C
100
40
Find image B of point B through line DA.
B
′
20
20
D
Problem 2-6
The degree-measure of each
base angle of an isosceles
triangle is 40. The bisector of
one of the base angles is
extended through point P on the
leg opposite that base angle so
that PA = PB, as shown. What
is m∠A?
′
B
A
100
P
C
40
Find image B of point B through line DA.
80
B
100
′
20
20
D
Problem 2-6
The degree-measure of each
base angle of an isosceles
triangle is 40. The bisector of
one of the base angles is
extended through point P on the
leg opposite that base angle so
that PA = PB, as shown. What
is m∠A?
′
B
A
100
P
60 60
60 60
C
40
Find image B of point B through line DA.
80
B
100
′
20
20
D
Problem 2-6
The degree-measure of each
base angle of an isosceles
triangle is 40. The bisector of
one of the base angles is
extended through point P on the
leg opposite that base angle so
that PA = PB, as shown. What
is m∠A?
′
B
A
80
C
40
40
100
P
60 60
60 60
80
B
100
20
20
′
Find image B of point B through line DA.
∼ △CPB ′ by the side-angle-side (SAS) theorem, so that
Note that △CPA =
◦
m∠A = 80
D