DISINI test_08_solutions
UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-FIRST ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 12, 2008
1) At the beginning of the semester, Mr. Kost had 25 students in his class. At the end of the semester, he had 20 students in
his class. What was the percent decrease in the number of students in Mr. Kost's class over the semester?
20 = 25(x)
x=
20
25
=
4
5
Decrease = 20%
1
2) On a map, a line segment of length 1 4 inches represents 40 miles. What distance is represented by a line segment of
length 2 inches?
5
inch
4
= 40 miles ï
40 miles
5
inch
4
5
ï 40 ·
= 32
miles
inch
ï 2 inches = 64 miles
4
3) Simplify 5 – 2 – 5 – 1 6 – 2 – 6 – 1 . Express your answer as a rational number in lowest terms.
1
52
–
1
5
1
62
−
1
6
=
1
25
–
4) What rational number between
q–
1
2
2
3
=3
−q
ï q–
1
2
1
5
1
2
1
36
and
2
3
–
1
6
4
5
= − 25
· − 36
is three times as far from
= 2 – 3q ï 4q = 2 +
1
2
=
5
2
=
1
45
1
2
as it is from
ïq=
2
?
3
5
8
5) Each small square in the given 10 x 10 square has an
area of one square unit.
Find the area of the shaded region.
A=
1
(1)(10)
2
+
1
(9)(10)
2
+ 10 = 5 + 45 + 10(1) = 60
6) If a, b and c are integers such that a + b + c = 91 and abc = 729, determine the value of a2 + b2 + c2 .
abc = 729 = 36 ï a = 34 b = 32 c = 1
(a + b + c
2
(81 + 9 + 1) = 91
= a2 + b2 + c2 + 2(ab + ac + bc)
912 = a2 + b2 + c2 + 2(ab + ac + bc) = a2 + b2 + c2 + 2(81 · 9 + 81 · 1 + 9 · 1)
912 = a2 + b2 + c2 + 2(819) ï a2 + b2 + c2 = 912 – 2(819) = 6643
7) The ratio of boys to girls in Mrs. Brown's class is 3 to 2. In Mrs. Smith's class, the ratio of boys to girls is 4 to 3. If there
are 30 students in Mrs. Brown's class and 28 students in Mrs. Smith's class, what is the total number of boys in the two
classes?
In Mrs. Brown's class
B
G
=
3
2
3
2
G + G = 30 ï
5
2
G = 30 ï G = 12 ï B =
3
(12)
2
= 18
7
3
G = 28 ï G = 12 ï B =
4
(12)
3
= 16
In Mrs. Smith's class
B
G
=
4
3
4
3
G + G = 28 ï
Boys = 18 + 16 = 34
8) The sixth grade class at Williston Central School is raffling off a turkey as a fundraising project. If the turkey costs $22 and
the tickets are sold at 75 cents each, how many tickets will have to be sold for the class to make a profit of $20 ?
3
4
x = 42
x=
4
3
42 = 4(14) = 56
9) A sweater, originally priced at $100, is successively discounted by 10%, 15% and 20%. What is the price of the sweater
after the third discount?
9 85 80
10 100 100
P = 100
=
9 ⋅ 17 ⋅ 8
20
= 61.2
10) An integer n > 1 is abundant if the sum of its proper divisors (positive integer divisors smaller than n) is greater than n.
Find the smallest abundant integer.
n
Divisors
Sum
2
1
1
3
1
1
4
1,2
3
5
1
1
6
1,2,3
6
7
1
1
8
1,2,4
7
9
1,3
4
10
1,2,5
8
11
1
1
12
1,2,3,4,6
16
n = 12
11) A binary operation
b = a2 b − b2 a. Find 2 (3 4).
is defined by a
2 (3 4) = 2 (9 · 4 – 16 · 3) = 2 (–12) = 4(–12) – 144(2) = – 48 – 288 = – 336
12) Larry's retirement portfolio consists of stocks and bonds. In 2000, 30% of the value of his portfolio was in stocks.
By 2008, the value of his stocks has doubled, while the value of his bonds has decreased by
1
.
3
What fraction of the
value of Larry's 2008 retirement portfolio is in bonds? Express your answer as a rational number in lowest terms.
2
P=
3
60 +
70
2
3
=
70
140
180 + 140
=
14
18 + 14
=
14
32
=
7
16
13) A day of skiing at Misty Valley costs $40. However, if you purchase a Misty Pass for $100, you receive a 30% discount on
each day that you ski. Suppose you purchase a Misty Pass at the beginning of the ski season. What is the fewest number of
days that you must ski so that your total cost is less than it would have been if you had not purchased a Misty Pass?
Let n = number of days.
No pass Cn = 40n
With pass Cw = 100 + .7(40)n = 100 + 28n
40n – (100 +28n) > 0
12n > 100 ï n >
25
3
ï n=9
log3 x3 – 8 .
14) Find all real numbers x such that log3 x – 2 =
Let y = log3 (x)
y–2=
3y–8
y2 – 4y + 4 = 3y – 8
y2 – 7y + 12 = 0 ï (y – 4)(y – 3) = 0 ï y = 3 , 4 ï x = 33 , 34 ï x = 27 , 81
15) If x4 + x2 + 1 = 0, what is the value of x2 +
x2 +
3
1
x2
x2 +
=
3
1
x2
x4 + 1
x2
=
–x2
x2
3
3
1
x2
3
?
x4 + x2 + 1 = 0 ï x4 + 1 = – x2
=–1
16) Let f(x) be a function such that f(x) + 2 f(3 – x) = 4x + 5 for every real number x. What is f(1)?
f 1 + 2 f 2 = 4+5 = 9
f 2 + 2 f 1 = 4 + 5 = 13
f 1 + 2f 2 =9
– 4 f 1 –2 f 2 = 13
– 3f(1) = – 17 ï f(1) =
17
3
17) Let p(x) be a cubic polynomial such that p(n) = n for n = 0, 1, – 1 and p(2) = 100. What is p(3)?
Let p(x) = ax3 + bx2 + cx + d
p(0) = 0 ï d = 0
p 1 = 1
p −1 = −1
a+b+c = 1
ï 2b = 0 ï b = 0
−a + b − c = −1
p 2 = 100
p 1 = 1
100 = 8 a + 2 c
ï
1 = a + c
1=a+c ï c=1–
p(x) =
49
3
x3 –
46
3
49
3
ï c=–
46
3
49
3
3
ï p(3) =
3
–
46
3
100 = 8 a + 2 c
−2 = − 2 a − 2 c
ï 98 = 6a ï a =
49
3
(3) = 9(49) – 46 = 395
18) Ship A leaves port at 6:00 a.m. and travels due north at a steady speed of 40 mph. Sometime later, Ship B leaves the same
port and travels due east at a steady speed of 36 mph. At 10:00 a.m., the two ships are 200 miles apart.
How many minutes after Ship A left port did Ship B leave port?
Let t be the time n hours ship B has been traveling at 10:00.
4 40
2
+ 36 t
2
= 2002
362 t2 = 4 100
t2 =
10 –
4 100 36
362
10
3
=
20
3
2
=
2
3
=6
– 16(16)(100) = 4(100)(100 – 64)
400
36
ït=
20
6
=
10
3
ï t = 6:40 or ship B left
2
3
hour after ship A or 40 minutes
19) What is the largest positive integer n such that 2008! is an integer multiple of 15n ?
If 15n | 2008! then 3n | 2008! and 5n | 2008! Need to find the number times 5 divides the integers § 2008.
2008
5
= 401
2008
25
= 80
2008
125
= 16
2008
625
=3
401 + 80 + 16 + 3 = 500
20) Find all ordered pairs of real numbers (x,y) satisfying both of the equations 3x 9y = 81 and
2x
1
=
.
8y
128
3x 9y = 81 ï 3x 32 y = 34 ï x + 2y = 4
2x
1
=
ï 2x 2−3 y = 2−7 ï x – 3y = – 7
8y
128
x+2 y = 4
ï 5y = 11 ï y =
x − 3 y = −7
11
5
ï x=4–
22
5
21) A sequence an is defined by a0 = 3, a1 = 6, and an =
=
–2
5
(x , y) = (–
2
5
,
11
)
5
an−1
for n ≥ 2. Find a2008 .
an−2
a0 = 3
a1 = 6
a2 =
a3 =
6
3
2
6
=2
=
1
3
=
1
6
=
1
2
1
a4 =
3
2
1
a5 =
6
1
3
1
a6 =
2
1
=3
6
a2008 = a2008 mod 6 = a4 =
1
6
22) A cube is inscribed in a sphere of radius 1. Find the surface area of the cube.
If the center of the sphere is at the origin, the equation of the sphere is x2 + y2 + z2 = 1. A vertex of the cube ia at (x,x,x) ï
3 x2 = 1 ï x2 =
1
3
The length of a side of the cube is 2x =
a =1
. The surface area of the cube is 6
3
23) Find all ordered pairs of real numbers (a , b) such that
a+b –
2
square
a+b −
1
2
b2
2
3
a =1
−4a = 1
.
2
=6
4
3
=8
a a+b
a+b–2
2a + b – 1 = 2
+a=1
a a+b
square
4 a2 + b2 + 1 +4ab – 4a – 4b = 4 a2 + ab
b2 + 1 – 4a – 2b = 0
b2 + 1 + 1 –
4a =
1
2
1
2
b2 – 1 ï 4a =
24) Express tan arctan
1
2
1
(4)
2
1
2
+ arctan
+ arctan
+ arctan
1
4
+ arctan
1
3
+ arctan
1
4
+ arctan
1
5
=
1
2
+ arctan
1
3
=
1
5
1
2
1–
1
4
+ arctan
+ arctan
1
3
+ arctan
1
5
1
4
=
1–
1 1
2 3
2008
n
2007
n
2007
n
2008!
n! 2008−n !
=
2008!
n! 2007 − n !
n! 2008−n !
2007!
1 1
4 5
=
+ arctan
1
2
1
+ tan arctan
3
+arctan
1
3
tan arctan
1
4
1
4
+ arctan
+arctan
1
5
1
5
6
5
=1
9
=
5
20
19
=
9
19
20
1
5
=
2
6
1
+ arctan
1
5
=
3
+
4
tan arctan
=
1+
1−
9
19
9
28
=
19
19
10
=
28
10
=
14
5
19
2008
2007
is divisible by
where
n
n
25) Find the number of positive integers n < 2008 such that
=
2
as a rational number in lowest terms.
1 – tan arctan
1
+
2
1
tan arctan
2008
n
1
,
4
(a , b) =
1
3
1
tan arctan
tan arctan
1
4
–1=1 ï a=
=0 ï b=2
tan α +tan β
1 – tan α tan β
tan(a+b) =
tan arctan
2
b2 – 2b = 0 ï b2 – 4b + 4 = 0 ï (b – 2
k
k!
=
.
j! k – j !
j
2007!
n! 2007−n !
2008
2008 − n
2008 = 23 · 251 Number of divisors = (3 + 1)(1 + 1) = 8
Since n >0 must exclude 2008 as a divisor, so 7 values of n.
26) In the next football season, the Flounders will play 16 games and the outcomes of the games will be recorded as a 16-letter
sequence of W's and L's (a W for each win and an L for each loss; there will be no ties). How many such sequences
correspond to winning exactly six games with no consecutive wins?
Arrange the 6 wins and 5 of the losses as follows:
WL
WL
WL
WL
WL
W
Can arrange he remainig 5 L's in the 7 spaces in
5+7
5
1
=
11!
5! × 6!
= 462 ways
27) If f(x) = a x4 + b x2 + 3 x + 7 and f(– 4) = 2008, determine the value of f(4) .
4
f(–4) = a –4
4
f(4) = a 4
+ b –4
+b 4
2
2
– 12 + 7 = 2008 ï 44 a + 42 b = 2008 +12 – 7 = 2013
+ 12 + 7 = 2013 + 19 = 2032
28) Find the area of the region in the xy plane consisting of all points (x,y) whose coordinates satisfy the inequalities
| y | – | x | § 2 and | x | § 3.
4
2
x y
+
+
−
−
+
−
+
−
x
+
y
=2
y−x = 2
−y − x = 2
y+x = 2
−y + x = 2
−3
−2
−1
1
2
3
−2
−4
A=2
4+10
2
(3) = 42
Area of two trapezoids.
d
d
d
d
d
d
6
= 2 + 3 + 4 + 5 + 6 + 7 , where each dj is an integer such that 0 § dj < j .
7
2!
3!
4!
5!
6!
7!
Find d2 + d3 + d4 + d5 + d6 + d7 .
29) Suppose that
d
d
d
d
d
d
6
= 2 + 3 + 4 + 5 + 6 + 7
7
2!
3!
4!
5!
6!
7!
Multiply by 7! and simplify:
4320 = 2520 d2 + 840 d3 + 210 d4 + 42 d5 + 7 d6 + d7
At each step pick the largest possible value for each di
d2 = 1
1800 = 840 d3 + 210 d4 + 42 d5 + 7 d6 + d7
d3 = 2
120 = 210 d4 + 42 d5 + 7 d6 + d7
d4 = 0
120 = 42 d5 + 7 d6 + d7
d5 = 2
36 = 7 d6 + d7
d6 = 5
1 = d7
d2 + d3 + d4 + d5 + d6 + d7 = 1 + 2 + 0 + 2 + 5 + 1 = 11
30) Let q be an acute angle such that 8cos(2q) + 8sec(2q) = 65. Find the value of cos(q).
8cos(2q) +
8
y
8y +
8
cos 2 θ
= 65
Let y = cos(2q)
= 65 ï 8 y2 – 65y + 8 = 0 ï (8y – 1)(y – 8) = 0 ï y =
cos(2q) =
1
8
ï 2 cos2 θ
cos2 θ =
9
16
−1 =
y = 8 (not possible for cosine)
1
8
3
4
ï cos( ) =
31) Let Cn be a circle of radius
1
8
2
3
n
centered at the
origin, for n = 0, 1, 2, ... and let An be the area of
the region that is inside circle C2 n and outside
circle C2 n+1 , for n = 0, 1, 2, ... The first three
regions are shown (shaded) in the figure.
∞
An .
Find
n= 0
A0 = p (1 –
2
2
3
A1 = p
2
3
4
A2 = p
2
3
8
2
3
4n
An =
∞
n= 0
5
p
9
An = 59 p
5
p
9
)=
−
2
3
6
−
2
3
10
1
1−
2
3
4
=p
=
2
3
4
8
2
3
=p
1–
5
1
p
16
9
1−
1–
=
2
3
2
3
2
=
2
5
81
p 81 − 16
9
5
p
9
2
3
=
5
p
9
2
3
=
5
81
p 65
9
4
8
=
9π
13
81
32) In the plane, point A has coordinates 0, 2 and point B has coordinates 0, 2 . The region S is the set of points inside
the square with vertices at 5, ≤5 and 5, ≤5 , and the region T consists of all points X in the plane such that the
triangle A B X is obtuse. Find the area of the region common to S and T.
−
4
4
2
2
−
2
4
−
−
2
−
−
−
−
4
−4
4
4
2
2
−2
A=p 2
2
2
4
−4
−2
2
−2
−2
−4
−4
4
+ 2(3)(10) = 4 + 60
33) A standard die with sides labeled 1 through 6 is rolled three times. Find the probability that the product of the three numbers
rolled is divisible by 8. Express your answer as a rational number in lowest terms.
Three 2' s 2 × 2 × 2
1
3
Two 2' s
2×2×4
Two 2' s
2×2×6
3
One 2
2 × 4 odd 3 × 3! = 18
3
One 2
2×4×4
One 2
2×4×6
6
One 2
2×6×6
3
No 2
4 × 4 odd 3 3 = 9
No 2
4×4×4
1
No 2
4×4×6
3
No 2
4 × 6 odd 3 × 3! = 18
No 2
4×6×6
3
No 2
6×6×6
1
Total = 1 + 3 + 3 + 18 + 3 + 6 + 3 + 9 + 1 + 3 + 18 + 3 + 1 = 72
p=
72
63
=
1
3
34) Suppose that a, b, c are real numbers such that
a + b + c = 10
. Find the maximum value of a.
+ b2 + c2 = 50
a2
c = 10 – a – b
b2 + c2 = 50 – a2
b2 + (10 – a – b
2
= 50 – a2
b2 + 100 + a2 + b2 – 20a – 20b + 2ab = 50 – a2
2 b2 + b(2a – 20) + 50 – 20a + 2 a2 = 0 If b is to be real then
(2a – 20
2
– 4(2)(50 – 20a + 2 a2 ) ¥ 0
4 a2 – 80a + 400 – 400 + 160a – 16 a2 ¥ 0
– 12 a2 + 80a ¥ 0 ï 4a(20 – 3a) ¥ 0
20
3
Largest a =
35) Let r1 , r2 , r3 , r4 , r5 be the five roots of the equation x5 – 6 x4 + 6 x 3 + 7x – 5 = 0 .
5
5
Determine the value of
ri – r j
2
.
i= 1 j= 1
5
5
5
ri = – (–6) = 6
ri rj = 6
i= 1
i = 1 j = i+1
2
ri – rj
= ri 2 – 2 ri rj + rj 2
5
5
There are 20 non-zero terms in
2
ri – r j
. These include 40 ri 2 terms, so each of the ri 2 terms occurs 8 times.
i= 1 j= 1
There are 20 terms ri rj , so each of the 5 pairs occurs 4 times.
2
r1 + r2 + r 3 + r4 + r5
62 =
5
5
ri 2 + 2(6) ï
i= 1
5
= r1 2 + r2 2 + r3 2 + r4 2 + r5 2 + 2 r1 r2 + r1 r3 + · · · r4 r5 )
ri 2 = 36 – 12 = 24
i= 1
5
ri – r j
2
5
ri 2 – 4
=8
i= 1 j= 1
i= 1
5
5
ri rj = 8(24) – 4(6) = 168
i = 1 j = i+1
36) Jack and Doug run two races on a 200 yard track. For the first race, Doug lets Jack start 8 yards ahead and also gives him
a two second head start. Doug wins this race by 2 seconds. For the second race, Doug lets Jack start 16 yards ahead and
also gives him a 5 second head start. Doug loses this race by 20 yards. How many seconds does it take Doug to run the
200 yards?
Let x = time for Doug to run 200 yds and y = time for Jack to run 200 yds.
Race 1 Jack runs 200 – 8 –
200
y
192 –
× 4 yds in
800
y
seconds
200
y
Doug runs 200 yds
in
200
seconds
200
x
192 –
800
y
200
=
y
200
200
ï 192y – 800 = 200x
x
Race 2 Jack runs 200 – 16 –
200
y
× 5 yds in
184 –
1000
200
y
seconds
y
Doug runs 200 – 20 yds
in
180
200
seconds
x
184 –
1000
200
y
y
=
180
200
ï 184y – 1000 = 180x
x
Solving gives y = 25 and x = 20
A
37) ABC is an equilateral triangle of side length 1.
Points D, E and F are chosen so that
AD = BE = CF =
1
.
4
Lines CD, AE and BF
D
G
are drawn. These lines intersect at points G, H
and I. Find the ratio of the area of DGHI to the
area of DABC. Express your answer as a rational
number in lowest terms.
F
H
I
B
C
E
Area GHI
Area ABC
=
GH2
12
= GH2
GH = CD – (DG + CH) = CD – (DG + AG)
2
1
4
By the Law of Cosines CD2 =
+ 12 – 2
1
4
π
3
1 cos
=
13
16
ï CD =
13
4
—ACH = —DAG and —DAC = —DGA ï triangles ADG, CDA and CFH are all similar
Let —ACH = —DAG = a , —DAC = —DGA = b and —ADG = —CFH = g
sin α
DG
By similar triangles
sin β
DA
=
=
sin γ
AG
sin α
DA
and
1
From the above
sin β
sin α
=
DA
DG
=
CD
DA
ï DG =
DA2
CD
=
sin β
CD
=
sin γ
AC
2
4
=
b
1
Also from the above
sin β
sin γ
GH = CD – (DG + AG) = b –
Thus
Area GHI
Area ABC
= GH2 =
=
DA
AG
=
1
16 b
CD
AC
+
ï AG =
1
4b
DA ⋅ AC
CD
=
13
4
and b =
4
b
=
1
4b
ï GH =
2
13
4
13
B
38) ABC is an isosceles right triangle with CB = CA = 1.
Point N is chosen inside the triangle so that
—NBC = —NCA = —NAB. Find the area of DNBC.
γ
N
C
From
BAN, g +
π
4
− γ + —BNA = p ï —BNA =
From NBC, CN = sin(g)
3π
.
4
γ
Similarly for —CNA. Then —BNC =
γ
A
π
.
2
By the Law of Sines
sin γ
AN
=
By the Law of Sines
sin γ
BN
=
Area of
1
2
=
1
·
2
ABC = Area of
AN sin(g) ·
ï AN =
4
1
sin
ï BN = 2 sin(g)
4
2
1
2
2 +
ABC =
1
2
ANC + Area of
· CN sin(g) · 1 +
1
2
CNB
· BN sin(g) · 1
2 + sin(g) sin(g) · 1 + 2 sin(g) sin(g) · 1
1 = 2 sin2 γ + sin2 γ + 2 sin2 γ
Area of
2 sin(g)
3π
ANB + Area of
2 sin(g) sin(g) ·
1=
3π
sin
· CN · BN =
1
sin
2
ï sin2 γ =
γ · 2 sin(γ = sin2 γ =
39) ABCD is a rectangle with AB = 4 and AD = 3. Let P
1
5
x
A
be any point on side AB. Let E be the point on
diagonal DB such that PE
1
5
P
4 – x
B
α
DB. Let F be the point
on diagonal AC such that PF
Find PE + PF.
AC.
E
F
D
From D ABC sin(a) =
3
5
From D APF sin(a) =
PF
x
From D ABC sin(a) =
3
=
3
5
PE
4–x
=
PE + PF = 5 (4 – x + x) =
ï PF =
3
5
3
5
x
3
ï PE = 5 (4 – x)
12
5
40) If a + b + c = 3 , ab + ac + bc = 4 and abc = 5 , find
1
a2
+
1
b2
+
1
1
1
+ 2 + 2.
a2
b
c
b 2 c2 + a 2 c2 + a 2 b 2
1
=
c2
abc 2
(ab + ac + bc 2 = 42 = a2 b2 + a2 c2 + b2 c2 + 2 a2 bc + ab2 c + abc2
16 = a2 b2 + a2 c2 + b2 c2 + 2 abc a + b + c
C
16 = a2 b2 + a2 c2 + b2 c2 + 2 5 3
a2 b2 + a2 c2 + b2 c2 = 16 – 2 5 3 = – 14
1
a2
+
1
b2
+
–14
1
=
=–
2
c
52
14
25
41) Lines l1 and l2 intersect in the plane as shown. Circles
A and B, each of radius 1 cm, are drawn tangent to l1
and l2 as shown. Circle C is then constructed tangent
to l1 , l2 , and A (externally), and circle D is constructed
tangent to l1 , l2 , and B (externally), as shown. If the
radius of circle C is 4 cm, find the radius of circle D,
in cm.
l2
D
C
A
B
l1
C
H
A
B
Y
G
E
Let GX = y, YX = t, ZX = w, EX = x,
4
y
By similar triangles CGX and AEX,
Z
X
—CXG = —YXH = a
=
1
x
ï y = 4x
H
A
B
Y
Z
α β
α
E
β
X
F
F
and
—HXZ = —ZXF = b.
Given CG = 4.
H
A
B
Y
Z
α β
α
β
E
X
By similar triangles CGX and AEX,
2a + 2b = p ï a + b =
sin(b) =
and cos(b) =
t+1
x
4+2+t
y
=
ï
t+1
x
=
6+t
4x
ï 4t + 4 = 6 + t ï t =
2
3
p
2
From triangle AEX, sin(a) =
4
5
F
1
1+
ï sin(a) =
2
3
5
ï cos(a) = 5 . Since a and b are complimentary,
4
=
4
5
3
3
.
5
From triangle XBF, sin(b) =
1
1+w
=
4
5
1
4
ï w=
So if R is the radius of circle D, sin(b) =
R
R+2+
1
4
ï
4
5
=
R
R+
9
4
ï 4R + 9 = 5R ï R = 9
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-FIRST ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 12, 2008
1) At the beginning of the semester, Mr. Kost had 25 students in his class. At the end of the semester, he had 20 students in
his class. What was the percent decrease in the number of students in Mr. Kost's class over the semester?
20 = 25(x)
x=
20
25
=
4
5
Decrease = 20%
1
2) On a map, a line segment of length 1 4 inches represents 40 miles. What distance is represented by a line segment of
length 2 inches?
5
inch
4
= 40 miles ï
40 miles
5
inch
4
5
ï 40 ·
= 32
miles
inch
ï 2 inches = 64 miles
4
3) Simplify 5 – 2 – 5 – 1 6 – 2 – 6 – 1 . Express your answer as a rational number in lowest terms.
1
52
–
1
5
1
62
−
1
6
=
1
25
–
4) What rational number between
q–
1
2
2
3
=3
−q
ï q–
1
2
1
5
1
2
1
36
and
2
3
–
1
6
4
5
= − 25
· − 36
is three times as far from
= 2 – 3q ï 4q = 2 +
1
2
=
5
2
=
1
45
1
2
as it is from
ïq=
2
?
3
5
8
5) Each small square in the given 10 x 10 square has an
area of one square unit.
Find the area of the shaded region.
A=
1
(1)(10)
2
+
1
(9)(10)
2
+ 10 = 5 + 45 + 10(1) = 60
6) If a, b and c are integers such that a + b + c = 91 and abc = 729, determine the value of a2 + b2 + c2 .
abc = 729 = 36 ï a = 34 b = 32 c = 1
(a + b + c
2
(81 + 9 + 1) = 91
= a2 + b2 + c2 + 2(ab + ac + bc)
912 = a2 + b2 + c2 + 2(ab + ac + bc) = a2 + b2 + c2 + 2(81 · 9 + 81 · 1 + 9 · 1)
912 = a2 + b2 + c2 + 2(819) ï a2 + b2 + c2 = 912 – 2(819) = 6643
7) The ratio of boys to girls in Mrs. Brown's class is 3 to 2. In Mrs. Smith's class, the ratio of boys to girls is 4 to 3. If there
are 30 students in Mrs. Brown's class and 28 students in Mrs. Smith's class, what is the total number of boys in the two
classes?
In Mrs. Brown's class
B
G
=
3
2
3
2
G + G = 30 ï
5
2
G = 30 ï G = 12 ï B =
3
(12)
2
= 18
7
3
G = 28 ï G = 12 ï B =
4
(12)
3
= 16
In Mrs. Smith's class
B
G
=
4
3
4
3
G + G = 28 ï
Boys = 18 + 16 = 34
8) The sixth grade class at Williston Central School is raffling off a turkey as a fundraising project. If the turkey costs $22 and
the tickets are sold at 75 cents each, how many tickets will have to be sold for the class to make a profit of $20 ?
3
4
x = 42
x=
4
3
42 = 4(14) = 56
9) A sweater, originally priced at $100, is successively discounted by 10%, 15% and 20%. What is the price of the sweater
after the third discount?
9 85 80
10 100 100
P = 100
=
9 ⋅ 17 ⋅ 8
20
= 61.2
10) An integer n > 1 is abundant if the sum of its proper divisors (positive integer divisors smaller than n) is greater than n.
Find the smallest abundant integer.
n
Divisors
Sum
2
1
1
3
1
1
4
1,2
3
5
1
1
6
1,2,3
6
7
1
1
8
1,2,4
7
9
1,3
4
10
1,2,5
8
11
1
1
12
1,2,3,4,6
16
n = 12
11) A binary operation
b = a2 b − b2 a. Find 2 (3 4).
is defined by a
2 (3 4) = 2 (9 · 4 – 16 · 3) = 2 (–12) = 4(–12) – 144(2) = – 48 – 288 = – 336
12) Larry's retirement portfolio consists of stocks and bonds. In 2000, 30% of the value of his portfolio was in stocks.
By 2008, the value of his stocks has doubled, while the value of his bonds has decreased by
1
.
3
What fraction of the
value of Larry's 2008 retirement portfolio is in bonds? Express your answer as a rational number in lowest terms.
2
P=
3
60 +
70
2
3
=
70
140
180 + 140
=
14
18 + 14
=
14
32
=
7
16
13) A day of skiing at Misty Valley costs $40. However, if you purchase a Misty Pass for $100, you receive a 30% discount on
each day that you ski. Suppose you purchase a Misty Pass at the beginning of the ski season. What is the fewest number of
days that you must ski so that your total cost is less than it would have been if you had not purchased a Misty Pass?
Let n = number of days.
No pass Cn = 40n
With pass Cw = 100 + .7(40)n = 100 + 28n
40n – (100 +28n) > 0
12n > 100 ï n >
25
3
ï n=9
log3 x3 – 8 .
14) Find all real numbers x such that log3 x – 2 =
Let y = log3 (x)
y–2=
3y–8
y2 – 4y + 4 = 3y – 8
y2 – 7y + 12 = 0 ï (y – 4)(y – 3) = 0 ï y = 3 , 4 ï x = 33 , 34 ï x = 27 , 81
15) If x4 + x2 + 1 = 0, what is the value of x2 +
x2 +
3
1
x2
x2 +
=
3
1
x2
x4 + 1
x2
=
–x2
x2
3
3
1
x2
3
?
x4 + x2 + 1 = 0 ï x4 + 1 = – x2
=–1
16) Let f(x) be a function such that f(x) + 2 f(3 – x) = 4x + 5 for every real number x. What is f(1)?
f 1 + 2 f 2 = 4+5 = 9
f 2 + 2 f 1 = 4 + 5 = 13
f 1 + 2f 2 =9
– 4 f 1 –2 f 2 = 13
– 3f(1) = – 17 ï f(1) =
17
3
17) Let p(x) be a cubic polynomial such that p(n) = n for n = 0, 1, – 1 and p(2) = 100. What is p(3)?
Let p(x) = ax3 + bx2 + cx + d
p(0) = 0 ï d = 0
p 1 = 1
p −1 = −1
a+b+c = 1
ï 2b = 0 ï b = 0
−a + b − c = −1
p 2 = 100
p 1 = 1
100 = 8 a + 2 c
ï
1 = a + c
1=a+c ï c=1–
p(x) =
49
3
x3 –
46
3
49
3
ï c=–
46
3
49
3
3
ï p(3) =
3
–
46
3
100 = 8 a + 2 c
−2 = − 2 a − 2 c
ï 98 = 6a ï a =
49
3
(3) = 9(49) – 46 = 395
18) Ship A leaves port at 6:00 a.m. and travels due north at a steady speed of 40 mph. Sometime later, Ship B leaves the same
port and travels due east at a steady speed of 36 mph. At 10:00 a.m., the two ships are 200 miles apart.
How many minutes after Ship A left port did Ship B leave port?
Let t be the time n hours ship B has been traveling at 10:00.
4 40
2
+ 36 t
2
= 2002
362 t2 = 4 100
t2 =
10 –
4 100 36
362
10
3
=
20
3
2
=
2
3
=6
– 16(16)(100) = 4(100)(100 – 64)
400
36
ït=
20
6
=
10
3
ï t = 6:40 or ship B left
2
3
hour after ship A or 40 minutes
19) What is the largest positive integer n such that 2008! is an integer multiple of 15n ?
If 15n | 2008! then 3n | 2008! and 5n | 2008! Need to find the number times 5 divides the integers § 2008.
2008
5
= 401
2008
25
= 80
2008
125
= 16
2008
625
=3
401 + 80 + 16 + 3 = 500
20) Find all ordered pairs of real numbers (x,y) satisfying both of the equations 3x 9y = 81 and
2x
1
=
.
8y
128
3x 9y = 81 ï 3x 32 y = 34 ï x + 2y = 4
2x
1
=
ï 2x 2−3 y = 2−7 ï x – 3y = – 7
8y
128
x+2 y = 4
ï 5y = 11 ï y =
x − 3 y = −7
11
5
ï x=4–
22
5
21) A sequence an is defined by a0 = 3, a1 = 6, and an =
=
–2
5
(x , y) = (–
2
5
,
11
)
5
an−1
for n ≥ 2. Find a2008 .
an−2
a0 = 3
a1 = 6
a2 =
a3 =
6
3
2
6
=2
=
1
3
=
1
6
=
1
2
1
a4 =
3
2
1
a5 =
6
1
3
1
a6 =
2
1
=3
6
a2008 = a2008 mod 6 = a4 =
1
6
22) A cube is inscribed in a sphere of radius 1. Find the surface area of the cube.
If the center of the sphere is at the origin, the equation of the sphere is x2 + y2 + z2 = 1. A vertex of the cube ia at (x,x,x) ï
3 x2 = 1 ï x2 =
1
3
The length of a side of the cube is 2x =
a =1
. The surface area of the cube is 6
3
23) Find all ordered pairs of real numbers (a , b) such that
a+b –
2
square
a+b −
1
2
b2
2
3
a =1
−4a = 1
.
2
=6
4
3
=8
a a+b
a+b–2
2a + b – 1 = 2
+a=1
a a+b
square
4 a2 + b2 + 1 +4ab – 4a – 4b = 4 a2 + ab
b2 + 1 – 4a – 2b = 0
b2 + 1 + 1 –
4a =
1
2
1
2
b2 – 1 ï 4a =
24) Express tan arctan
1
2
1
(4)
2
1
2
+ arctan
+ arctan
+ arctan
1
4
+ arctan
1
3
+ arctan
1
4
+ arctan
1
5
=
1
2
+ arctan
1
3
=
1
5
1
2
1–
1
4
+ arctan
+ arctan
1
3
+ arctan
1
5
1
4
=
1–
1 1
2 3
2008
n
2007
n
2007
n
2008!
n! 2008−n !
=
2008!
n! 2007 − n !
n! 2008−n !
2007!
1 1
4 5
=
+ arctan
1
2
1
+ tan arctan
3
+arctan
1
3
tan arctan
1
4
1
4
+ arctan
+arctan
1
5
1
5
6
5
=1
9
=
5
20
19
=
9
19
20
1
5
=
2
6
1
+ arctan
1
5
=
3
+
4
tan arctan
=
1+
1−
9
19
9
28
=
19
19
10
=
28
10
=
14
5
19
2008
2007
is divisible by
where
n
n
25) Find the number of positive integers n < 2008 such that
=
2
as a rational number in lowest terms.
1 – tan arctan
1
+
2
1
tan arctan
2008
n
1
,
4
(a , b) =
1
3
1
tan arctan
tan arctan
1
4
–1=1 ï a=
=0 ï b=2
tan α +tan β
1 – tan α tan β
tan(a+b) =
tan arctan
2
b2 – 2b = 0 ï b2 – 4b + 4 = 0 ï (b – 2
k
k!
=
.
j! k – j !
j
2007!
n! 2007−n !
2008
2008 − n
2008 = 23 · 251 Number of divisors = (3 + 1)(1 + 1) = 8
Since n >0 must exclude 2008 as a divisor, so 7 values of n.
26) In the next football season, the Flounders will play 16 games and the outcomes of the games will be recorded as a 16-letter
sequence of W's and L's (a W for each win and an L for each loss; there will be no ties). How many such sequences
correspond to winning exactly six games with no consecutive wins?
Arrange the 6 wins and 5 of the losses as follows:
WL
WL
WL
WL
WL
W
Can arrange he remainig 5 L's in the 7 spaces in
5+7
5
1
=
11!
5! × 6!
= 462 ways
27) If f(x) = a x4 + b x2 + 3 x + 7 and f(– 4) = 2008, determine the value of f(4) .
4
f(–4) = a –4
4
f(4) = a 4
+ b –4
+b 4
2
2
– 12 + 7 = 2008 ï 44 a + 42 b = 2008 +12 – 7 = 2013
+ 12 + 7 = 2013 + 19 = 2032
28) Find the area of the region in the xy plane consisting of all points (x,y) whose coordinates satisfy the inequalities
| y | – | x | § 2 and | x | § 3.
4
2
x y
+
+
−
−
+
−
+
−
x
+
y
=2
y−x = 2
−y − x = 2
y+x = 2
−y + x = 2
−3
−2
−1
1
2
3
−2
−4
A=2
4+10
2
(3) = 42
Area of two trapezoids.
d
d
d
d
d
d
6
= 2 + 3 + 4 + 5 + 6 + 7 , where each dj is an integer such that 0 § dj < j .
7
2!
3!
4!
5!
6!
7!
Find d2 + d3 + d4 + d5 + d6 + d7 .
29) Suppose that
d
d
d
d
d
d
6
= 2 + 3 + 4 + 5 + 6 + 7
7
2!
3!
4!
5!
6!
7!
Multiply by 7! and simplify:
4320 = 2520 d2 + 840 d3 + 210 d4 + 42 d5 + 7 d6 + d7
At each step pick the largest possible value for each di
d2 = 1
1800 = 840 d3 + 210 d4 + 42 d5 + 7 d6 + d7
d3 = 2
120 = 210 d4 + 42 d5 + 7 d6 + d7
d4 = 0
120 = 42 d5 + 7 d6 + d7
d5 = 2
36 = 7 d6 + d7
d6 = 5
1 = d7
d2 + d3 + d4 + d5 + d6 + d7 = 1 + 2 + 0 + 2 + 5 + 1 = 11
30) Let q be an acute angle such that 8cos(2q) + 8sec(2q) = 65. Find the value of cos(q).
8cos(2q) +
8
y
8y +
8
cos 2 θ
= 65
Let y = cos(2q)
= 65 ï 8 y2 – 65y + 8 = 0 ï (8y – 1)(y – 8) = 0 ï y =
cos(2q) =
1
8
ï 2 cos2 θ
cos2 θ =
9
16
−1 =
y = 8 (not possible for cosine)
1
8
3
4
ï cos( ) =
31) Let Cn be a circle of radius
1
8
2
3
n
centered at the
origin, for n = 0, 1, 2, ... and let An be the area of
the region that is inside circle C2 n and outside
circle C2 n+1 , for n = 0, 1, 2, ... The first three
regions are shown (shaded) in the figure.
∞
An .
Find
n= 0
A0 = p (1 –
2
2
3
A1 = p
2
3
4
A2 = p
2
3
8
2
3
4n
An =
∞
n= 0
5
p
9
An = 59 p
5
p
9
)=
−
2
3
6
−
2
3
10
1
1−
2
3
4
=p
=
2
3
4
8
2
3
=p
1–
5
1
p
16
9
1−
1–
=
2
3
2
3
2
=
2
5
81
p 81 − 16
9
5
p
9
2
3
=
5
p
9
2
3
=
5
81
p 65
9
4
8
=
9π
13
81
32) In the plane, point A has coordinates 0, 2 and point B has coordinates 0, 2 . The region S is the set of points inside
the square with vertices at 5, ≤5 and 5, ≤5 , and the region T consists of all points X in the plane such that the
triangle A B X is obtuse. Find the area of the region common to S and T.
−
4
4
2
2
−
2
4
−
−
2
−
−
−
−
4
−4
4
4
2
2
−2
A=p 2
2
2
4
−4
−2
2
−2
−2
−4
−4
4
+ 2(3)(10) = 4 + 60
33) A standard die with sides labeled 1 through 6 is rolled three times. Find the probability that the product of the three numbers
rolled is divisible by 8. Express your answer as a rational number in lowest terms.
Three 2' s 2 × 2 × 2
1
3
Two 2' s
2×2×4
Two 2' s
2×2×6
3
One 2
2 × 4 odd 3 × 3! = 18
3
One 2
2×4×4
One 2
2×4×6
6
One 2
2×6×6
3
No 2
4 × 4 odd 3 3 = 9
No 2
4×4×4
1
No 2
4×4×6
3
No 2
4 × 6 odd 3 × 3! = 18
No 2
4×6×6
3
No 2
6×6×6
1
Total = 1 + 3 + 3 + 18 + 3 + 6 + 3 + 9 + 1 + 3 + 18 + 3 + 1 = 72
p=
72
63
=
1
3
34) Suppose that a, b, c are real numbers such that
a + b + c = 10
. Find the maximum value of a.
+ b2 + c2 = 50
a2
c = 10 – a – b
b2 + c2 = 50 – a2
b2 + (10 – a – b
2
= 50 – a2
b2 + 100 + a2 + b2 – 20a – 20b + 2ab = 50 – a2
2 b2 + b(2a – 20) + 50 – 20a + 2 a2 = 0 If b is to be real then
(2a – 20
2
– 4(2)(50 – 20a + 2 a2 ) ¥ 0
4 a2 – 80a + 400 – 400 + 160a – 16 a2 ¥ 0
– 12 a2 + 80a ¥ 0 ï 4a(20 – 3a) ¥ 0
20
3
Largest a =
35) Let r1 , r2 , r3 , r4 , r5 be the five roots of the equation x5 – 6 x4 + 6 x 3 + 7x – 5 = 0 .
5
5
Determine the value of
ri – r j
2
.
i= 1 j= 1
5
5
5
ri = – (–6) = 6
ri rj = 6
i= 1
i = 1 j = i+1
2
ri – rj
= ri 2 – 2 ri rj + rj 2
5
5
There are 20 non-zero terms in
2
ri – r j
. These include 40 ri 2 terms, so each of the ri 2 terms occurs 8 times.
i= 1 j= 1
There are 20 terms ri rj , so each of the 5 pairs occurs 4 times.
2
r1 + r2 + r 3 + r4 + r5
62 =
5
5
ri 2 + 2(6) ï
i= 1
5
= r1 2 + r2 2 + r3 2 + r4 2 + r5 2 + 2 r1 r2 + r1 r3 + · · · r4 r5 )
ri 2 = 36 – 12 = 24
i= 1
5
ri – r j
2
5
ri 2 – 4
=8
i= 1 j= 1
i= 1
5
5
ri rj = 8(24) – 4(6) = 168
i = 1 j = i+1
36) Jack and Doug run two races on a 200 yard track. For the first race, Doug lets Jack start 8 yards ahead and also gives him
a two second head start. Doug wins this race by 2 seconds. For the second race, Doug lets Jack start 16 yards ahead and
also gives him a 5 second head start. Doug loses this race by 20 yards. How many seconds does it take Doug to run the
200 yards?
Let x = time for Doug to run 200 yds and y = time for Jack to run 200 yds.
Race 1 Jack runs 200 – 8 –
200
y
192 –
× 4 yds in
800
y
seconds
200
y
Doug runs 200 yds
in
200
seconds
200
x
192 –
800
y
200
=
y
200
200
ï 192y – 800 = 200x
x
Race 2 Jack runs 200 – 16 –
200
y
× 5 yds in
184 –
1000
200
y
seconds
y
Doug runs 200 – 20 yds
in
180
200
seconds
x
184 –
1000
200
y
y
=
180
200
ï 184y – 1000 = 180x
x
Solving gives y = 25 and x = 20
A
37) ABC is an equilateral triangle of side length 1.
Points D, E and F are chosen so that
AD = BE = CF =
1
.
4
Lines CD, AE and BF
D
G
are drawn. These lines intersect at points G, H
and I. Find the ratio of the area of DGHI to the
area of DABC. Express your answer as a rational
number in lowest terms.
F
H
I
B
C
E
Area GHI
Area ABC
=
GH2
12
= GH2
GH = CD – (DG + CH) = CD – (DG + AG)
2
1
4
By the Law of Cosines CD2 =
+ 12 – 2
1
4
π
3
1 cos
=
13
16
ï CD =
13
4
—ACH = —DAG and —DAC = —DGA ï triangles ADG, CDA and CFH are all similar
Let —ACH = —DAG = a , —DAC = —DGA = b and —ADG = —CFH = g
sin α
DG
By similar triangles
sin β
DA
=
=
sin γ
AG
sin α
DA
and
1
From the above
sin β
sin α
=
DA
DG
=
CD
DA
ï DG =
DA2
CD
=
sin β
CD
=
sin γ
AC
2
4
=
b
1
Also from the above
sin β
sin γ
GH = CD – (DG + AG) = b –
Thus
Area GHI
Area ABC
= GH2 =
=
DA
AG
=
1
16 b
CD
AC
+
ï AG =
1
4b
DA ⋅ AC
CD
=
13
4
and b =
4
b
=
1
4b
ï GH =
2
13
4
13
B
38) ABC is an isosceles right triangle with CB = CA = 1.
Point N is chosen inside the triangle so that
—NBC = —NCA = —NAB. Find the area of DNBC.
γ
N
C
From
BAN, g +
π
4
− γ + —BNA = p ï —BNA =
From NBC, CN = sin(g)
3π
.
4
γ
Similarly for —CNA. Then —BNC =
γ
A
π
.
2
By the Law of Sines
sin γ
AN
=
By the Law of Sines
sin γ
BN
=
Area of
1
2
=
1
·
2
ABC = Area of
AN sin(g) ·
ï AN =
4
1
sin
ï BN = 2 sin(g)
4
2
1
2
2 +
ABC =
1
2
ANC + Area of
· CN sin(g) · 1 +
1
2
CNB
· BN sin(g) · 1
2 + sin(g) sin(g) · 1 + 2 sin(g) sin(g) · 1
1 = 2 sin2 γ + sin2 γ + 2 sin2 γ
Area of
2 sin(g)
3π
ANB + Area of
2 sin(g) sin(g) ·
1=
3π
sin
· CN · BN =
1
sin
2
ï sin2 γ =
γ · 2 sin(γ = sin2 γ =
39) ABCD is a rectangle with AB = 4 and AD = 3. Let P
1
5
x
A
be any point on side AB. Let E be the point on
diagonal DB such that PE
1
5
P
4 – x
B
α
DB. Let F be the point
on diagonal AC such that PF
Find PE + PF.
AC.
E
F
D
From D ABC sin(a) =
3
5
From D APF sin(a) =
PF
x
From D ABC sin(a) =
3
=
3
5
PE
4–x
=
PE + PF = 5 (4 – x + x) =
ï PF =
3
5
3
5
x
3
ï PE = 5 (4 – x)
12
5
40) If a + b + c = 3 , ab + ac + bc = 4 and abc = 5 , find
1
a2
+
1
b2
+
1
1
1
+ 2 + 2.
a2
b
c
b 2 c2 + a 2 c2 + a 2 b 2
1
=
c2
abc 2
(ab + ac + bc 2 = 42 = a2 b2 + a2 c2 + b2 c2 + 2 a2 bc + ab2 c + abc2
16 = a2 b2 + a2 c2 + b2 c2 + 2 abc a + b + c
C
16 = a2 b2 + a2 c2 + b2 c2 + 2 5 3
a2 b2 + a2 c2 + b2 c2 = 16 – 2 5 3 = – 14
1
a2
+
1
b2
+
–14
1
=
=–
2
c
52
14
25
41) Lines l1 and l2 intersect in the plane as shown. Circles
A and B, each of radius 1 cm, are drawn tangent to l1
and l2 as shown. Circle C is then constructed tangent
to l1 , l2 , and A (externally), and circle D is constructed
tangent to l1 , l2 , and B (externally), as shown. If the
radius of circle C is 4 cm, find the radius of circle D,
in cm.
l2
D
C
A
B
l1
C
H
A
B
Y
G
E
Let GX = y, YX = t, ZX = w, EX = x,
4
y
By similar triangles CGX and AEX,
Z
X
—CXG = —YXH = a
=
1
x
ï y = 4x
H
A
B
Y
Z
α β
α
E
β
X
F
F
and
—HXZ = —ZXF = b.
Given CG = 4.
H
A
B
Y
Z
α β
α
β
E
X
By similar triangles CGX and AEX,
2a + 2b = p ï a + b =
sin(b) =
and cos(b) =
t+1
x
4+2+t
y
=
ï
t+1
x
=
6+t
4x
ï 4t + 4 = 6 + t ï t =
2
3
p
2
From triangle AEX, sin(a) =
4
5
F
1
1+
ï sin(a) =
2
3
5
ï cos(a) = 5 . Since a and b are complimentary,
4
=
4
5
3
3
.
5
From triangle XBF, sin(b) =
1
1+w
=
4
5
1
4
ï w=
So if R is the radius of circle D, sin(b) =
R
R+2+
1
4
ï
4
5
=
R
R+
9
4
ï 4R + 9 = 5R ï R = 9