DISINI test_12_sol
UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-FIFTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 14, 2012
1) What is the smallest positive integer n such that 2n + 1 and 3n + 1 are both primes?
n
2n+1
1
3
2
5
3n+1
4
7
n=2
2) Every student in Ms. Math’s class took the AP Calculus Exam and received a score of either a 4 or a 5. If 32 of the students earned
a 5 and that was 80% of the class, how many students earned a 4 on the AP Exam?
Let n = number of students.
0.8 n = 32 ï n =
32
0.8
=
320
8
= 40
Number scoring 4 = n – 32
1
3) Express
2+
40 – 32 = 8
as a rational number in lowest terms.
3
5
4+ 6
1
=
3
2+
5
4+ 6
1
3
2 + 29
=
1
2+
18
29
=
1
76
=
29
76
29
6
4 M , A, B and N are collinear with MA = AB = BN. Arcs MA , AN , NB and MB
are all semicircles. If MN = 9, what is the perimeter of the shaded region ?
P =2p
3
2
+ 2p(3) = 3p + 6p = 9
5) If p and q are the roots of the equation x2 – 6 x + 2 = 0, find the value of
1
1
+
.
p
q
(x – p)(x – q) = x2 – p + q x + pq = x2 – 6 x + 2 ï pq = 2 and p + q = 6
1
1
+
=
p
q
p+q
pq
=
6
2
=3
6) Among the 41 students in a class,
12 have a cat,
5 have a dog,
8 have a robot,
2 have a cat and a dog,
6 have a cat and a robot,
3 have a dog and a robot, and
1 has a cat, a dog and a robot.
How many students have neither a cat nor a dog nor a robot?
Dog
Cat
1
1
5
1
2
5
0
Robot
None = 41 – (1 + 1 + 1 + 2 + 5 + 5) = 41 – 15 = 26
7) The total cost of tickets to the school play for one adult and 3 children is $27. If the cost of a ticket for an adult is $1 more than
the cost of a ticket for a child, what is the cost of an adult ticket?
A = cost of adult ticket and C = cost of child ticket.
A + 3C = 27 ï A + 3(A – 1) = 27 ï 4A = 30 ï A =
30
4
= 7.50
8) A line passing through the points (3 , – 6) and (9 , 3) intersects the x-axis at the point (a , 0). What is the value of a ?
9,3
a,0
3,−6
3
9–a
=
6
a–3
9) Let f (n) =
ï 3a – 9 = 54 – 6a ï 9a = 63 ï a = 7
f 4
3 n!
. Find
.
3n !
f 5
3 ÿ 4!
f 4
3 ÿ 4!
15!
15 ÿ 14 ÿ 13
12!
= 3 ÿ 5! =
·
=
= 3 ÿ 14 ÿ 13 = 546
f 5
12!
3 ÿ 5!
5
15!
10) For non-zero real numbers a and b, define a * b = a b –
1 * 2 = 1(2) –
1
2
1
* 3 = 2 (3) –
1
1
–
1
1
1
2
–
=2–
1
3
=
3
2
3
2
=
1 1
– . What is the value of (1 * 2) * 3 ?
a b
1
2
–2–
1
3
=
3
2
–
7
3
=
9 – 14
6
=–
5
6
2
11 Two circles, one of radius 3 and the other of radius 5, are externally tangent and are
internally tangent to a larger circle. What is the area of the shaded region inside
the larger circle that is not contained in either of the two smaller circles?
R=
5+5+3+3
2
= 8 = radius of the large circle
Area = p 82 – p 52 – p 32 = p(64 – 25 – 9) = 30
12) Find all values of k so that the curves y = x2 – 3 k x –1 and y = 9 k x – 17 intersect in exactly one point.
x2 – 3 k x–1 = 9 k x – 17
x2 – 3 k x
9 k x – 1 + 17 = 0
x2 – 12 k x + 16 = 0
x=
12 k ≤
12 k 2 –4 16
2
For a single solution
12 k 2 –4 16 = 0 ï 12 k
4–x +
13) Find all real numbers x such that
2
16 – x2 = 4 x2
16 – x2 = 4 x2 – 8 ï
16 – x2 = 2 x2 – 4
16 – x2 = 4 x4 – 16 x2 + 16
4 x4 – 15 x2 = 0
x2 4 x2 – 15 = 0 ï x = 0 or ±
15
2
= 4(16) ï 12k = ± 8 ï k = ±
4 + x = 2x.
Square both sides:
4–x+4+x+2
2
2
3
15
2
Checking yields the only solution is x =
14) Let x and y be two positive real numbers such that logx y + log y x = 3. Find the value of logx y
logx y + log y x
logx y
logx y
2
2
2
2
+ log y x
2
.
= 32
+ log y x
+ log y x
2
2
+ 2 logx y log y x = 9
+2 =9
ï logx y
Using logx y =
2
+ log y x
2
1
log y x
=7
15) Let g be a function such that 3g(x) + 2g(1 – x) = 9 + 2x. Find the value of g(2).
Substituting x = 2 and x = – 1
3g(2) + 2g(–1) = 9 + 4
3g(–1) + 2g(2) = 9 – 2
3g(2) + 2g(–1) = 13
2g(2) + 3g(–1) = 7
Multiplying by 3 and 2
9g(2) + 6g(–1) = 39
4g(2) + 6g(–1) = 14
Subtract
5g(2) = 25 ï g(2) = 5
16) Find the exact value of cos3 (15°) sin(15°) – cos(15°) sin3 15 ° .
Factoring cos3 (15°) sin(15°) – cos(15°) sin3 15 ° = cos(15°) sin(15°)[cos2 (15°) – sin2 15 ° ]
Using the identities sin(2a) = 2cos(a)sin(a) and cos(2a) = cos2 a – sin2 a
cos3 (15°) sin(15°) – cos(15°) sin3 15 ° =
17) What integer n satisfies n <
5+2
1
2
sin(30°) cos(30°) =
2 +
5–2
1
2
1
2
3
2
=
3
8
2 < n + 1?
Squaring
n2 < 5 + 2
n2 < 10 + 2
Since 4 <
2 + 5–2
2 +2
17 < n + 1
17 < 5 ,
17 < n + 1
2
2
n2 < 18 < 20 < n + 1
2
ï n=4
18) In a random arrangement of the letters of GREENMOUNTAINS, what is the probability that the vowels are in alphabetical
order within the arrangement? Express your answer as a rational number in lowest terms.
There are 14 letters including 2 E’s and 3 N’s. 6 of the 14 letters are vowels.
Let T = the total number of permutations of the 14 letters =
14!
2! 3!
14
6
The vowels must appear in the order AEEIOU and can be placed in
Arrange the remaining 8 letters in
p=
14 8!
6 3!
14!
=
14! 8! 2! 3!
6! 8! 3! 14!
=
8!
3!
2!
6!
different positions.
ways in the remaining positions.
=
2
720
=
1
360
2! × 3!
19) If q is an acute angle such that tan(q) = 2, find the value of cos(4q). Express your answer as a rational number in lowest terms.
2
θ
1
2
sin(q) =
1
and cos(q) =
5
sin(2q) = 2sin(q)cos(q) = 2
5
cos(4q) = cos(2(2q)) = 1 – 2 sin2 2 q = 1 – 2
4 2
5
=1–
32
25
=–
2
1
5
5
=
4
5
7
25
20) Find the area of the region in the plane consisting of all points (x , y) that satisfy | 4x – 8 | § y § 12 .
Find points of intersection of
y = | 4x – 8 | and y = 12
x § 2 ï 8 – 4x = 12 ï 4x = – 4 ï x = – 1
x ¥ 2 ï 4x – 8 = 12 ï 4x = 20 ï x = 5
15
–1,12
5,12
10
5
−2
2
1
4
6
1
Area = 2 (base)(height) = 2 (5 + 1)(12) = 36
21) Find all positive real numbers x such that log2 x + log2 x – 12 = 6 .
log2 x + log2 x – 12 = 6
log2 x x–12 = 6
x(x – 12) = 26 = 64
x2 – 12 x – 64 = 0
(x – 16)(x + 4) = 0 ï x = 16
22) Let f be a function such that f
f(5w) = f(
25 w
5
= 25 w
2
x
5
= x2 + x + 2 . Find the sum of all values of w such that f (5w) = 2012.
+ 25 w + 2 = 2012
625 w2 + 25 w – 2010 = 0 ï w2 +
1
25
w–
2012
625
=0
Suppose the roots are a and b, then (w – a)(w – b) = 0 ï w2 – a + b w + ab = 0
Comparing coefficients, a + b = –
1
25
23) Suppose that 9a = 12, 12b = 15, 15c = 18, 18d = 21, 21e = 24 and 24 f = 27.
What is the value of a · b · c · d · e · f ?
Using the given equations from last to first,
27 = 24 f = 21ef = 18def = 15cdef = 12bcdef = 9abcdef
33 = 32 abcdef ï 2abcdef = 3 ï abcdef =
3
2
24) If x, y and z are positive real numbers such that x y = 40, x z = 60 and y z = 96, what is the value of x + y + z ?
(xz)(yz) = 60(96) ï xy z2 = 60 96 ï 40 z2 = 60 96 ï z2 =
60 96
40
= 144 ï z = 12
12 x = 60 ï x = 5 and 12 y = 96 ï y = 8
x + y + z = 5 + 8 + 12 = 25
25) Let S = 5, 52 , 53 , ÿ ÿ ÿ , 510 . Suppose that a and b are distinct integers chosen from S.
For how many ordered pairs a, b is loga b an integer?
Let a = 5k and b = 5n with k and n in { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
loga b œ Z + ï log5k 5n œ Z + ï k divides n
k
n
number
1
2, 3, 4, 5, 6, 7, 8, 9, 10
9
2
2, 4, 6, 8
4
3
6, 9
2
4
8
1
5
10
1
17
D
C
T
E
A
B
26 Square ABCD has side length 6. A semicircle whose diameter is AB is drawn
internally. The line that is tangent to the semicircle at T and passes through
D
C
vertex C intersects side AD at E. Find ET.
T
E
A
B
Since CB and CT are tangents to the circle from the same point, CB = CT = 6. Similarly, ET = EA = x.
From triangle EDC, CD2 + DE2 = CE2
Substituting, 62 + 6 – x
2
= 6+x
2
36 + 36 – 12 x + x2 = 36 + 12 x + x2 ï 36 = 24x ï x
3
2
27) Let S be the set of all three-digit positive integers, each of whose digits is 1 , 3 , 5 , 7 or 9. For example, 197 and 331 are two
elements in S. How many integers in S are divisible by 3?
An integer is divisible by 3 in and only if the sum of its digits is divisible by 3. So considering the possible sets of three allowed digits
and number of different arrangements:
Digits
1, 3, 5
1, 5, 9
3, 5, 7
5, 7, 9
1, 1, 7
3, 3, 9
7, 7, 1
9, 9, 3
1, 1, 1
3, 3, 3
5, 5, 5
7, 7, 7
9, 9, 9
Arrangements
6
6
6
6
3
3
3
3
1
1
1
1
1
Total = 41
Number = 41
28) Consider the binary sequence 01001000100001000001··· , where each block of 0’s contains one more 0 that the
preceding block of 0’s. Note that the first 1 appears in position 2. In what position does the 200th 1 appear?
Position
Digit
1
0
2
1
3
0
Let pk = position of k th 1
p1 = 2
p2 = 2 + 3
4
0
5
1
6
0
7
0
8
0
9
1
10
0
11
0
12
0
13
0
14
1
15
0
16
0
17
0
18
0
19
0
20
1
⋅ ⋅ ⋅
p3 = 2 + 3 + 4
···
pk = 2 + 3 + 4 + · · · k + 1 = 1 + 2 + 3 + 4 + · · · k + 1 – 1 =
p200 =
2002 +3 (200)
2
200 200+3
2
=
k+1 k+2
2
–1=
k 2 +3 k
2
= 100(203) = 20300
1
29) Define the sequence a n by a1 = 2012 and a n = 1 – a
for n ¥ 2. Find a 2012 .
n–1
a1 = 2012
a2 =
a3 =
a4 =
1
1 – 2012
1
1+
1
=
2011
1
1–
1
2011
2011
2012
=–
2011
=
2012
2012 – 2011
= 2012
2012
a3 k+1 = 2012
1
2011
2011
2012
a3 k+2 = –
a3 k =
2012 = 3(67) + 2 ï a2012
1
2011
–
30) Let a1 , a2 , a3 and a4 be positive real numbers. If a0 = 20, a5 = 12 and ak = ak–1 ak+1 for k = 1, 2, 3 and 4, find a4 .
a1
a2
a3
a4
= 20 a2
= a1 a3
= a2 a4
= a3 12
(1)
(2)
(3)
(4)
(1) & (2) ï a2 = 20 a2 a3 ï 1 = 20 a3 ï a3 =
1
20
(3) & (4) ï a3 = a2 12 a3 ï 1 = 12 a2 ï a2 =
(2) ï
1
12
= a1
(3) ï
1
20
=
a4 =
1
12
1
20
ï a1 =
a4 ï a4 =
20
12
12
20
=
=
1
12
5
3
3
5
3
5
31) Let S = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }. Let T be a subset consisting of three distinct elements chosen randomly from S.
What is the probability that the elements of T can be arranged to form an arithmetic progression with positive common
difference? Express your answer as a rational number in lowest terms.
Number 3 element subsets =
diff
1
2
3
4
9
3
=
9ÿ8ÿ7
3ÿ2
= 84
subsets
123,234,345,456,567,678,789
135,246,357,468,579
147,258,369
159
Number
7
5
3
1
16
p=
16
84
=
4
21
32) Determine the integer m so that the equation x4 – 3 m + 2 x2 + m2 = 0 has four real roots in arithmetic progression.
Let r1 , r2 , r3 and r4 be the roots. Expanding x–r1 x–r2 x–r3 x–r4 :
x4 r1 + r2 + r3 + r4 x3 + r1 r2 + r1 r3 + r1 r4 + r2 r3 + r2 r4 + r3 r4 x2
r1 r2 r3
r1 r2 r4
r1 r3 r4
r2 r3 r4 x + r1 r2 r3 r4
Since the roots are in arithmetic progression, r1 = r , r2 = r + d , r3 = r + 2 d and r4 = r + 3 d
Equating coefficients gives :
x3 : r + r + d + r + 2 d + r + 3 d = 0
x0 : r r + d r + 2 d r + 3 d = m2
x2 : r r + d + r r + 2 d + r r + 3 d + r + d r + 2 d + r + d r + 3 d + r + 2 d r + 3 d =– 3 m + 2
(1) ï 4 r + 6 d = 0 ï r = –
(2) ï
–
–
(3) ï –
5
2
3
2
3
2
3
2
d –
d –
1
2
d –
3
2
1
2
d +d –
d
1
2
d + –
d
3
2
d
3
2
(3)
d
3
3
d +2d – 2 d +3d
2
3
9
d = m2 ï 16 d 4
2
1
2
(1)
(2)
3
2
d + –
d
3
2
= m2
4
= m2 ï d 2 = ≤ 3 m
1
1
2
d + –2 d
d + –
(A)
1
2
3
2
d
d +
d2 = 3 m + 2
1
2
3
2
d
d =– 3m+2
(B)
Using (A) & (B) :
d2 =
4
3
4
m ï
d2 = – 3 m ï
5
2
5
2
4
3
m = 3m+2 ï
4
10
3
m=3m+2 ï
–3 m = 3m+2 ï –
10
3
10
3
m–3 m = 2 ï
m=3m+2 ï –
10
3
1
3
m=2 ï m=6
m–3 m = 2 ï –
19
3
6
m = 2 ï m = – 19
m=6
33 Right triangle ABC has AB = 24, AC = 32 and BC = 40. Points D and E are
chosen on sides BC and AC respectively so that triangle ADE is similar to
triangle ABC and AE > AD > DE. Find the area of triangle ADE.
C
α
E
β
a
D
32
40
b
α
A
From DABC, tan(b) =
32
24
=
4
3
β
h
24 – z β
z
F
B
From DAFD, tan(b) =
h
z
From DFBD, tan(b) =
h
24 – z
4
3
4
3
z=
h=
4
3
=
4
3
ï h=
=
4
3
4
3
z
ï h=
4
3
24 – z
24 – z ï z = 24 – z ï 2z = 24 ï z = 12
z ï h=
4
3
12 = 16
From DAFD, b2 = 122 + 162 = 400 ï b = 20
From DABC, tan(a) =
24
32
From DADE, tan(a) =
a
b
Area =
1
2
ab =
1
2
=
=
3
4
3
4
ï a=
3
4
b =
3
4
20 = 15
15 20 = 150
34) In how many ways can the squares in a 2 by 5 array be colored red (R), green (G) or blue (B) so that no two
squares with a common edge are the same color? One such example is
RBGBR
BGBRB
.
Let Nk be the number of colorings of a 2 by k grid.
Consider any coloring of a 2 by k grid and look at the last two squares. For example, suppose the are colored R and G.
R
G
For the 2 by k + 1 grid, the added top square could possibly be G or B and bottom added square could possibly be R or B.
R G
G R
R G
G B
R B
G R
R B
G B
One of these leads to matching colors, B and B. Thus for this “ending” choice of the 2 by k grid there are 3 possible new colorings.
Hence Nk+1 = 3 * Nk .
For k =1 there are 3 · 2 choices. So N1 = 2 · 3 , N2 = 2 · 3 · 3 = 2 · 32 · · · Nk = 2 · 3k
N5 = 2 · 3 5 = 2 · 243 = 486
35 Let S1 be a square with side length
2
8
. A sequence of smaller squares is
constructed by joining the midpoints of the sides of each previous square to
form a new square. That is, the corners of square S2 are the midpoints of the
sides of square S1 ; the corners of square S3 are the midpoints of square S2 ;
and so on. The first 5 such squares are shown in the figure. Let Pi be the
¶
perimeter of square Si . Find
Pk .
k= 1
Consider consecutive squares, one with side s and the next with side z.
s 2
z
s 2
2
z2 =
s 2
2
+
s2
4
=
s2
4
+
=
s2
2
s 2
s
ï z=
2
Starting with square of side s,
P1 = 4 s
s
P2 = 4
2
s
P3 = 4
2
2
ÿÿÿ
s
Pk = 4
2
k–1
¶
1
Pk = 4s 1 +
2
k= 1
1
+
+
2
2
1
2
3
1
+ ÿ ÿ ÿ =4s
1
1–
2
¶
With s =
2
8
,
Pk = 4
k= 1
¶
36) Express
S=
k= 1
S=
1
11
+
2
112
+
3
113
k
11k
+
2
8
2
2 –1
=
1
2 –1
=
2 +1
2 –1
2 +1
as a rational number in lowest terms.
4
114
+
5
115
+ ÿ ÿÿ
=
2 +1
2–1
=
2
1
1
1
+
11
112
1
112
S=
+
+
+
1
1
+
113
1
+
113
1
115
1
+
114
1
+
115
1
+
114
1
+
113
1
+
114
1
115
1
+
114
115
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅ ⋅ ⋅ ⋅⋅⋅
S=
1
11
1 +
1
11
1
11
+
S= 1 +
+
1
112
1
112
+
+
1
113
1
113
+ ÿÿ ÿ +
+ ÿ ÿÿ
1
11
1
112
+
1
112
1 +
+
1
11
1
113
+
+
1
112
1
114
+
1
113
+ ÿ ÿÿ +
1
112
1 +
+ ÿ ÿ ÿ
1
1
S=
1–
11
1
1–
11
1
=
11
1
11 – 1 11 – 1
=
11
100
11
37) If x + y + z = 2, xyz = 12 and xy + xz + yz = 5, find the value of x3 + y3 + z3 .
x + y + z 3 = x3 + y3 + z3 + 3 xy2 + 3 xz2 + 3 yx2 + 3 yz2 + 3 zx2 + 3 zy2 + 6xyz
x + y + z 3 = x3 + y3 + z3 + 3 xy x + y + z + 3 xz x + y + z + 3 yz x + y + z – 3xyz
x + y + z 3 = x3 + y3 + z3 + 3 x + y + z xy + xz + yz – 3xyz
23 = x3 + y3 + z3 + 3 2 5 – 3(8)
x3 + y3 + z3 = 8 – 3 2 5 + 3(12) = 14
20122
2012 k . Find the remainder when s is divided by 7.
38) Let s =
k= 0
2012 mod 7 = 3
2012k mod 7 = 1 , 3 , 2 , 6 , 4 , 5 , 1 , 3 , 2 , · · · = 0 , 1 , 2 , 3 , 4 , · · ·
Each 6 add to 0 mod 7. (20122 + 1) mod 6 = 2 · 2 + 1 = 5
20122
2012 k mod 7 = (1 + 3 + 2 + 6 + 4) mod 7 = 16 mod 7 = 2
k= 0
39) How many positive perfect cubes divide the product 1! ä 2! ä 3! ä · · · ä 10! ?
k!
2!
3!
4!
5!
6!
7!
8!
9!
10!
Expanded
2
3⋅2
2⋅2⋅3⋅2
5⋅2⋅2⋅3⋅2
3⋅2⋅5⋅2⋅2⋅3⋅2
7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
2⋅2⋅2⋅7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
3⋅3⋅2⋅2⋅2⋅7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
5⋅2⋅3⋅3⋅2⋅2⋅2⋅7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
2's
1
1
3
3
4
4
7
7
8
38
3's
5's
7's
1
1
1
2
2
2
4
4
17
1
1
1
1
1
2
7
1
1
1
1
4
P = 1! ä 2! ä 3! ä · · · ä 10! = 238 ÿ317 ÿ57 ÿ 74
If the positive perfect cube n divides P, it must have the form n = 23 s ÿ33 t ÿ 53 u ÿ 73 v
1
11
+
1
112
+
1
113
+ ÿ ÿ ÿ +· · ·
So must pick
0 § s § 12, 0 § t § 5, 0 § u § 2 and 0 § v § 1
Number of n = 13(6)(3)(2) = 468
40 Let S be the square with vertices 0, 0 , 3, 0 , 3, 3 and 0, 3 .
Each side of this square is trisected and the points of trisection
are joined as indicated in the sketch to form two overlapping
squares. Find the area of the region common to these two
squares.
Line (0,1) to (2,0) y = 1 –
1
2
x
Line (0,2) to (1,0) y = – 2(x – 1)
Intersect 1 –
1
2
x = – 2(x – 1) ï 1 –
1
2
x = – 2x + 2 ï
3
x
2
=1 ïx=
2
3
ï y=
2
3
Line (0,1) to (1,3) y = 2x + 1
Line (0,2) to (1,0) y = – 2(x – 1)
Intersect 2x + 1 = – 2(x – 1) ï 2x + 1 = – 2x + 2 ï 4x = 1 ï x =
Side if inscribed square =
1 2 + 22 =
1
4
ï y=
3
2
5 Let A = area of square = 5.
Length (0,1) to (1/4,3/2) =
1 2
4
+
1 2
2
=
1
16
Length (0,1) to (2/3,2/3) =
2 2
3
+
1 2
3
=
4
9
+
+
Let T = area triangle ((0,1) - (1/4,3/2) - (2/3,2/3) =
Shaded area = area of square – 4 T = 5 – 4
5
24
1
4
=
1
9
=
1
2
5
4
=5–
5
6
5
4
5
3
=
5
3
=
5
24
25
6
C
A
D
E
B
41 Two circles of radius 16 have their centers on the
circumference of each other. AB is the diameter
of the right–hand circle that passes through the
centers of the two circles. A smaller circle is constructed
tangent to AB and the two given circles as in the sketch.
Find the radius of the smaller circle.
C
A
Let the radius of each of the large circles be R and the radius of the small circle be r.
AD = R, AC = R + r DC = R – r
AE2 = (R + r 2 – r2 = R2 + 2Rr DE = (R – r 2 – r2 = R2 – 2Rr
AD = AE – DE
R2 + 2 Rr –
R=
R2 –2 Rr
R2 = R2 + 2Rr + R2 – 2Rr – 2
R2 = 2
R4 – 4 R2 r2
R4 = 4 R4 – 16 R2 r2
R2 = 4 R2 – 16 r2
r2 =
3
16
R2
r=
3
4
R
r=
3
4
· 16 = 4
3
R4 – 4 R2 r2
D
E
B
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-FIFTH ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 14, 2012
1) What is the smallest positive integer n such that 2n + 1 and 3n + 1 are both primes?
n
2n+1
1
3
2
5
3n+1
4
7
n=2
2) Every student in Ms. Math’s class took the AP Calculus Exam and received a score of either a 4 or a 5. If 32 of the students earned
a 5 and that was 80% of the class, how many students earned a 4 on the AP Exam?
Let n = number of students.
0.8 n = 32 ï n =
32
0.8
=
320
8
= 40
Number scoring 4 = n – 32
1
3) Express
2+
40 – 32 = 8
as a rational number in lowest terms.
3
5
4+ 6
1
=
3
2+
5
4+ 6
1
3
2 + 29
=
1
2+
18
29
=
1
76
=
29
76
29
6
4 M , A, B and N are collinear with MA = AB = BN. Arcs MA , AN , NB and MB
are all semicircles. If MN = 9, what is the perimeter of the shaded region ?
P =2p
3
2
+ 2p(3) = 3p + 6p = 9
5) If p and q are the roots of the equation x2 – 6 x + 2 = 0, find the value of
1
1
+
.
p
q
(x – p)(x – q) = x2 – p + q x + pq = x2 – 6 x + 2 ï pq = 2 and p + q = 6
1
1
+
=
p
q
p+q
pq
=
6
2
=3
6) Among the 41 students in a class,
12 have a cat,
5 have a dog,
8 have a robot,
2 have a cat and a dog,
6 have a cat and a robot,
3 have a dog and a robot, and
1 has a cat, a dog and a robot.
How many students have neither a cat nor a dog nor a robot?
Dog
Cat
1
1
5
1
2
5
0
Robot
None = 41 – (1 + 1 + 1 + 2 + 5 + 5) = 41 – 15 = 26
7) The total cost of tickets to the school play for one adult and 3 children is $27. If the cost of a ticket for an adult is $1 more than
the cost of a ticket for a child, what is the cost of an adult ticket?
A = cost of adult ticket and C = cost of child ticket.
A + 3C = 27 ï A + 3(A – 1) = 27 ï 4A = 30 ï A =
30
4
= 7.50
8) A line passing through the points (3 , – 6) and (9 , 3) intersects the x-axis at the point (a , 0). What is the value of a ?
9,3
a,0
3,−6
3
9–a
=
6
a–3
9) Let f (n) =
ï 3a – 9 = 54 – 6a ï 9a = 63 ï a = 7
f 4
3 n!
. Find
.
3n !
f 5
3 ÿ 4!
f 4
3 ÿ 4!
15!
15 ÿ 14 ÿ 13
12!
= 3 ÿ 5! =
·
=
= 3 ÿ 14 ÿ 13 = 546
f 5
12!
3 ÿ 5!
5
15!
10) For non-zero real numbers a and b, define a * b = a b –
1 * 2 = 1(2) –
1
2
1
* 3 = 2 (3) –
1
1
–
1
1
1
2
–
=2–
1
3
=
3
2
3
2
=
1 1
– . What is the value of (1 * 2) * 3 ?
a b
1
2
–2–
1
3
=
3
2
–
7
3
=
9 – 14
6
=–
5
6
2
11 Two circles, one of radius 3 and the other of radius 5, are externally tangent and are
internally tangent to a larger circle. What is the area of the shaded region inside
the larger circle that is not contained in either of the two smaller circles?
R=
5+5+3+3
2
= 8 = radius of the large circle
Area = p 82 – p 52 – p 32 = p(64 – 25 – 9) = 30
12) Find all values of k so that the curves y = x2 – 3 k x –1 and y = 9 k x – 17 intersect in exactly one point.
x2 – 3 k x–1 = 9 k x – 17
x2 – 3 k x
9 k x – 1 + 17 = 0
x2 – 12 k x + 16 = 0
x=
12 k ≤
12 k 2 –4 16
2
For a single solution
12 k 2 –4 16 = 0 ï 12 k
4–x +
13) Find all real numbers x such that
2
16 – x2 = 4 x2
16 – x2 = 4 x2 – 8 ï
16 – x2 = 2 x2 – 4
16 – x2 = 4 x4 – 16 x2 + 16
4 x4 – 15 x2 = 0
x2 4 x2 – 15 = 0 ï x = 0 or ±
15
2
= 4(16) ï 12k = ± 8 ï k = ±
4 + x = 2x.
Square both sides:
4–x+4+x+2
2
2
3
15
2
Checking yields the only solution is x =
14) Let x and y be two positive real numbers such that logx y + log y x = 3. Find the value of logx y
logx y + log y x
logx y
logx y
2
2
2
2
+ log y x
2
.
= 32
+ log y x
+ log y x
2
2
+ 2 logx y log y x = 9
+2 =9
ï logx y
Using logx y =
2
+ log y x
2
1
log y x
=7
15) Let g be a function such that 3g(x) + 2g(1 – x) = 9 + 2x. Find the value of g(2).
Substituting x = 2 and x = – 1
3g(2) + 2g(–1) = 9 + 4
3g(–1) + 2g(2) = 9 – 2
3g(2) + 2g(–1) = 13
2g(2) + 3g(–1) = 7
Multiplying by 3 and 2
9g(2) + 6g(–1) = 39
4g(2) + 6g(–1) = 14
Subtract
5g(2) = 25 ï g(2) = 5
16) Find the exact value of cos3 (15°) sin(15°) – cos(15°) sin3 15 ° .
Factoring cos3 (15°) sin(15°) – cos(15°) sin3 15 ° = cos(15°) sin(15°)[cos2 (15°) – sin2 15 ° ]
Using the identities sin(2a) = 2cos(a)sin(a) and cos(2a) = cos2 a – sin2 a
cos3 (15°) sin(15°) – cos(15°) sin3 15 ° =
17) What integer n satisfies n <
5+2
1
2
sin(30°) cos(30°) =
2 +
5–2
1
2
1
2
3
2
=
3
8
2 < n + 1?
Squaring
n2 < 5 + 2
n2 < 10 + 2
Since 4 <
2 + 5–2
2 +2
17 < n + 1
17 < 5 ,
17 < n + 1
2
2
n2 < 18 < 20 < n + 1
2
ï n=4
18) In a random arrangement of the letters of GREENMOUNTAINS, what is the probability that the vowels are in alphabetical
order within the arrangement? Express your answer as a rational number in lowest terms.
There are 14 letters including 2 E’s and 3 N’s. 6 of the 14 letters are vowels.
Let T = the total number of permutations of the 14 letters =
14!
2! 3!
14
6
The vowels must appear in the order AEEIOU and can be placed in
Arrange the remaining 8 letters in
p=
14 8!
6 3!
14!
=
14! 8! 2! 3!
6! 8! 3! 14!
=
8!
3!
2!
6!
different positions.
ways in the remaining positions.
=
2
720
=
1
360
2! × 3!
19) If q is an acute angle such that tan(q) = 2, find the value of cos(4q). Express your answer as a rational number in lowest terms.
2
θ
1
2
sin(q) =
1
and cos(q) =
5
sin(2q) = 2sin(q)cos(q) = 2
5
cos(4q) = cos(2(2q)) = 1 – 2 sin2 2 q = 1 – 2
4 2
5
=1–
32
25
=–
2
1
5
5
=
4
5
7
25
20) Find the area of the region in the plane consisting of all points (x , y) that satisfy | 4x – 8 | § y § 12 .
Find points of intersection of
y = | 4x – 8 | and y = 12
x § 2 ï 8 – 4x = 12 ï 4x = – 4 ï x = – 1
x ¥ 2 ï 4x – 8 = 12 ï 4x = 20 ï x = 5
15
–1,12
5,12
10
5
−2
2
1
4
6
1
Area = 2 (base)(height) = 2 (5 + 1)(12) = 36
21) Find all positive real numbers x such that log2 x + log2 x – 12 = 6 .
log2 x + log2 x – 12 = 6
log2 x x–12 = 6
x(x – 12) = 26 = 64
x2 – 12 x – 64 = 0
(x – 16)(x + 4) = 0 ï x = 16
22) Let f be a function such that f
f(5w) = f(
25 w
5
= 25 w
2
x
5
= x2 + x + 2 . Find the sum of all values of w such that f (5w) = 2012.
+ 25 w + 2 = 2012
625 w2 + 25 w – 2010 = 0 ï w2 +
1
25
w–
2012
625
=0
Suppose the roots are a and b, then (w – a)(w – b) = 0 ï w2 – a + b w + ab = 0
Comparing coefficients, a + b = –
1
25
23) Suppose that 9a = 12, 12b = 15, 15c = 18, 18d = 21, 21e = 24 and 24 f = 27.
What is the value of a · b · c · d · e · f ?
Using the given equations from last to first,
27 = 24 f = 21ef = 18def = 15cdef = 12bcdef = 9abcdef
33 = 32 abcdef ï 2abcdef = 3 ï abcdef =
3
2
24) If x, y and z are positive real numbers such that x y = 40, x z = 60 and y z = 96, what is the value of x + y + z ?
(xz)(yz) = 60(96) ï xy z2 = 60 96 ï 40 z2 = 60 96 ï z2 =
60 96
40
= 144 ï z = 12
12 x = 60 ï x = 5 and 12 y = 96 ï y = 8
x + y + z = 5 + 8 + 12 = 25
25) Let S = 5, 52 , 53 , ÿ ÿ ÿ , 510 . Suppose that a and b are distinct integers chosen from S.
For how many ordered pairs a, b is loga b an integer?
Let a = 5k and b = 5n with k and n in { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
loga b œ Z + ï log5k 5n œ Z + ï k divides n
k
n
number
1
2, 3, 4, 5, 6, 7, 8, 9, 10
9
2
2, 4, 6, 8
4
3
6, 9
2
4
8
1
5
10
1
17
D
C
T
E
A
B
26 Square ABCD has side length 6. A semicircle whose diameter is AB is drawn
internally. The line that is tangent to the semicircle at T and passes through
D
C
vertex C intersects side AD at E. Find ET.
T
E
A
B
Since CB and CT are tangents to the circle from the same point, CB = CT = 6. Similarly, ET = EA = x.
From triangle EDC, CD2 + DE2 = CE2
Substituting, 62 + 6 – x
2
= 6+x
2
36 + 36 – 12 x + x2 = 36 + 12 x + x2 ï 36 = 24x ï x
3
2
27) Let S be the set of all three-digit positive integers, each of whose digits is 1 , 3 , 5 , 7 or 9. For example, 197 and 331 are two
elements in S. How many integers in S are divisible by 3?
An integer is divisible by 3 in and only if the sum of its digits is divisible by 3. So considering the possible sets of three allowed digits
and number of different arrangements:
Digits
1, 3, 5
1, 5, 9
3, 5, 7
5, 7, 9
1, 1, 7
3, 3, 9
7, 7, 1
9, 9, 3
1, 1, 1
3, 3, 3
5, 5, 5
7, 7, 7
9, 9, 9
Arrangements
6
6
6
6
3
3
3
3
1
1
1
1
1
Total = 41
Number = 41
28) Consider the binary sequence 01001000100001000001··· , where each block of 0’s contains one more 0 that the
preceding block of 0’s. Note that the first 1 appears in position 2. In what position does the 200th 1 appear?
Position
Digit
1
0
2
1
3
0
Let pk = position of k th 1
p1 = 2
p2 = 2 + 3
4
0
5
1
6
0
7
0
8
0
9
1
10
0
11
0
12
0
13
0
14
1
15
0
16
0
17
0
18
0
19
0
20
1
⋅ ⋅ ⋅
p3 = 2 + 3 + 4
···
pk = 2 + 3 + 4 + · · · k + 1 = 1 + 2 + 3 + 4 + · · · k + 1 – 1 =
p200 =
2002 +3 (200)
2
200 200+3
2
=
k+1 k+2
2
–1=
k 2 +3 k
2
= 100(203) = 20300
1
29) Define the sequence a n by a1 = 2012 and a n = 1 – a
for n ¥ 2. Find a 2012 .
n–1
a1 = 2012
a2 =
a3 =
a4 =
1
1 – 2012
1
1+
1
=
2011
1
1–
1
2011
2011
2012
=–
2011
=
2012
2012 – 2011
= 2012
2012
a3 k+1 = 2012
1
2011
2011
2012
a3 k+2 = –
a3 k =
2012 = 3(67) + 2 ï a2012
1
2011
–
30) Let a1 , a2 , a3 and a4 be positive real numbers. If a0 = 20, a5 = 12 and ak = ak–1 ak+1 for k = 1, 2, 3 and 4, find a4 .
a1
a2
a3
a4
= 20 a2
= a1 a3
= a2 a4
= a3 12
(1)
(2)
(3)
(4)
(1) & (2) ï a2 = 20 a2 a3 ï 1 = 20 a3 ï a3 =
1
20
(3) & (4) ï a3 = a2 12 a3 ï 1 = 12 a2 ï a2 =
(2) ï
1
12
= a1
(3) ï
1
20
=
a4 =
1
12
1
20
ï a1 =
a4 ï a4 =
20
12
12
20
=
=
1
12
5
3
3
5
3
5
31) Let S = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }. Let T be a subset consisting of three distinct elements chosen randomly from S.
What is the probability that the elements of T can be arranged to form an arithmetic progression with positive common
difference? Express your answer as a rational number in lowest terms.
Number 3 element subsets =
diff
1
2
3
4
9
3
=
9ÿ8ÿ7
3ÿ2
= 84
subsets
123,234,345,456,567,678,789
135,246,357,468,579
147,258,369
159
Number
7
5
3
1
16
p=
16
84
=
4
21
32) Determine the integer m so that the equation x4 – 3 m + 2 x2 + m2 = 0 has four real roots in arithmetic progression.
Let r1 , r2 , r3 and r4 be the roots. Expanding x–r1 x–r2 x–r3 x–r4 :
x4 r1 + r2 + r3 + r4 x3 + r1 r2 + r1 r3 + r1 r4 + r2 r3 + r2 r4 + r3 r4 x2
r1 r2 r3
r1 r2 r4
r1 r3 r4
r2 r3 r4 x + r1 r2 r3 r4
Since the roots are in arithmetic progression, r1 = r , r2 = r + d , r3 = r + 2 d and r4 = r + 3 d
Equating coefficients gives :
x3 : r + r + d + r + 2 d + r + 3 d = 0
x0 : r r + d r + 2 d r + 3 d = m2
x2 : r r + d + r r + 2 d + r r + 3 d + r + d r + 2 d + r + d r + 3 d + r + 2 d r + 3 d =– 3 m + 2
(1) ï 4 r + 6 d = 0 ï r = –
(2) ï
–
–
(3) ï –
5
2
3
2
3
2
3
2
d –
d –
1
2
d –
3
2
1
2
d +d –
d
1
2
d + –
d
3
2
d
3
2
(3)
d
3
3
d +2d – 2 d +3d
2
3
9
d = m2 ï 16 d 4
2
1
2
(1)
(2)
3
2
d + –
d
3
2
= m2
4
= m2 ï d 2 = ≤ 3 m
1
1
2
d + –2 d
d + –
(A)
1
2
3
2
d
d +
d2 = 3 m + 2
1
2
3
2
d
d =– 3m+2
(B)
Using (A) & (B) :
d2 =
4
3
4
m ï
d2 = – 3 m ï
5
2
5
2
4
3
m = 3m+2 ï
4
10
3
m=3m+2 ï
–3 m = 3m+2 ï –
10
3
10
3
m–3 m = 2 ï
m=3m+2 ï –
10
3
1
3
m=2 ï m=6
m–3 m = 2 ï –
19
3
6
m = 2 ï m = – 19
m=6
33 Right triangle ABC has AB = 24, AC = 32 and BC = 40. Points D and E are
chosen on sides BC and AC respectively so that triangle ADE is similar to
triangle ABC and AE > AD > DE. Find the area of triangle ADE.
C
α
E
β
a
D
32
40
b
α
A
From DABC, tan(b) =
32
24
=
4
3
β
h
24 – z β
z
F
B
From DAFD, tan(b) =
h
z
From DFBD, tan(b) =
h
24 – z
4
3
4
3
z=
h=
4
3
=
4
3
ï h=
=
4
3
4
3
z
ï h=
4
3
24 – z
24 – z ï z = 24 – z ï 2z = 24 ï z = 12
z ï h=
4
3
12 = 16
From DAFD, b2 = 122 + 162 = 400 ï b = 20
From DABC, tan(a) =
24
32
From DADE, tan(a) =
a
b
Area =
1
2
ab =
1
2
=
=
3
4
3
4
ï a=
3
4
b =
3
4
20 = 15
15 20 = 150
34) In how many ways can the squares in a 2 by 5 array be colored red (R), green (G) or blue (B) so that no two
squares with a common edge are the same color? One such example is
RBGBR
BGBRB
.
Let Nk be the number of colorings of a 2 by k grid.
Consider any coloring of a 2 by k grid and look at the last two squares. For example, suppose the are colored R and G.
R
G
For the 2 by k + 1 grid, the added top square could possibly be G or B and bottom added square could possibly be R or B.
R G
G R
R G
G B
R B
G R
R B
G B
One of these leads to matching colors, B and B. Thus for this “ending” choice of the 2 by k grid there are 3 possible new colorings.
Hence Nk+1 = 3 * Nk .
For k =1 there are 3 · 2 choices. So N1 = 2 · 3 , N2 = 2 · 3 · 3 = 2 · 32 · · · Nk = 2 · 3k
N5 = 2 · 3 5 = 2 · 243 = 486
35 Let S1 be a square with side length
2
8
. A sequence of smaller squares is
constructed by joining the midpoints of the sides of each previous square to
form a new square. That is, the corners of square S2 are the midpoints of the
sides of square S1 ; the corners of square S3 are the midpoints of square S2 ;
and so on. The first 5 such squares are shown in the figure. Let Pi be the
¶
perimeter of square Si . Find
Pk .
k= 1
Consider consecutive squares, one with side s and the next with side z.
s 2
z
s 2
2
z2 =
s 2
2
+
s2
4
=
s2
4
+
=
s2
2
s 2
s
ï z=
2
Starting with square of side s,
P1 = 4 s
s
P2 = 4
2
s
P3 = 4
2
2
ÿÿÿ
s
Pk = 4
2
k–1
¶
1
Pk = 4s 1 +
2
k= 1
1
+
+
2
2
1
2
3
1
+ ÿ ÿ ÿ =4s
1
1–
2
¶
With s =
2
8
,
Pk = 4
k= 1
¶
36) Express
S=
k= 1
S=
1
11
+
2
112
+
3
113
k
11k
+
2
8
2
2 –1
=
1
2 –1
=
2 +1
2 –1
2 +1
as a rational number in lowest terms.
4
114
+
5
115
+ ÿ ÿÿ
=
2 +1
2–1
=
2
1
1
1
+
11
112
1
112
S=
+
+
+
1
1
+
113
1
+
113
1
115
1
+
114
1
+
115
1
+
114
1
+
113
1
+
114
1
115
1
+
114
115
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅ ⋅ ⋅ ⋅⋅⋅
S=
1
11
1 +
1
11
1
11
+
S= 1 +
+
1
112
1
112
+
+
1
113
1
113
+ ÿÿ ÿ +
+ ÿ ÿÿ
1
11
1
112
+
1
112
1 +
+
1
11
1
113
+
+
1
112
1
114
+
1
113
+ ÿ ÿÿ +
1
112
1 +
+ ÿ ÿ ÿ
1
1
S=
1–
11
1
1–
11
1
=
11
1
11 – 1 11 – 1
=
11
100
11
37) If x + y + z = 2, xyz = 12 and xy + xz + yz = 5, find the value of x3 + y3 + z3 .
x + y + z 3 = x3 + y3 + z3 + 3 xy2 + 3 xz2 + 3 yx2 + 3 yz2 + 3 zx2 + 3 zy2 + 6xyz
x + y + z 3 = x3 + y3 + z3 + 3 xy x + y + z + 3 xz x + y + z + 3 yz x + y + z – 3xyz
x + y + z 3 = x3 + y3 + z3 + 3 x + y + z xy + xz + yz – 3xyz
23 = x3 + y3 + z3 + 3 2 5 – 3(8)
x3 + y3 + z3 = 8 – 3 2 5 + 3(12) = 14
20122
2012 k . Find the remainder when s is divided by 7.
38) Let s =
k= 0
2012 mod 7 = 3
2012k mod 7 = 1 , 3 , 2 , 6 , 4 , 5 , 1 , 3 , 2 , · · · = 0 , 1 , 2 , 3 , 4 , · · ·
Each 6 add to 0 mod 7. (20122 + 1) mod 6 = 2 · 2 + 1 = 5
20122
2012 k mod 7 = (1 + 3 + 2 + 6 + 4) mod 7 = 16 mod 7 = 2
k= 0
39) How many positive perfect cubes divide the product 1! ä 2! ä 3! ä · · · ä 10! ?
k!
2!
3!
4!
5!
6!
7!
8!
9!
10!
Expanded
2
3⋅2
2⋅2⋅3⋅2
5⋅2⋅2⋅3⋅2
3⋅2⋅5⋅2⋅2⋅3⋅2
7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
2⋅2⋅2⋅7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
3⋅3⋅2⋅2⋅2⋅7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
5⋅2⋅3⋅3⋅2⋅2⋅2⋅7⋅3⋅2⋅5⋅2⋅2⋅3⋅2
2's
1
1
3
3
4
4
7
7
8
38
3's
5's
7's
1
1
1
2
2
2
4
4
17
1
1
1
1
1
2
7
1
1
1
1
4
P = 1! ä 2! ä 3! ä · · · ä 10! = 238 ÿ317 ÿ57 ÿ 74
If the positive perfect cube n divides P, it must have the form n = 23 s ÿ33 t ÿ 53 u ÿ 73 v
1
11
+
1
112
+
1
113
+ ÿ ÿ ÿ +· · ·
So must pick
0 § s § 12, 0 § t § 5, 0 § u § 2 and 0 § v § 1
Number of n = 13(6)(3)(2) = 468
40 Let S be the square with vertices 0, 0 , 3, 0 , 3, 3 and 0, 3 .
Each side of this square is trisected and the points of trisection
are joined as indicated in the sketch to form two overlapping
squares. Find the area of the region common to these two
squares.
Line (0,1) to (2,0) y = 1 –
1
2
x
Line (0,2) to (1,0) y = – 2(x – 1)
Intersect 1 –
1
2
x = – 2(x – 1) ï 1 –
1
2
x = – 2x + 2 ï
3
x
2
=1 ïx=
2
3
ï y=
2
3
Line (0,1) to (1,3) y = 2x + 1
Line (0,2) to (1,0) y = – 2(x – 1)
Intersect 2x + 1 = – 2(x – 1) ï 2x + 1 = – 2x + 2 ï 4x = 1 ï x =
Side if inscribed square =
1 2 + 22 =
1
4
ï y=
3
2
5 Let A = area of square = 5.
Length (0,1) to (1/4,3/2) =
1 2
4
+
1 2
2
=
1
16
Length (0,1) to (2/3,2/3) =
2 2
3
+
1 2
3
=
4
9
+
+
Let T = area triangle ((0,1) - (1/4,3/2) - (2/3,2/3) =
Shaded area = area of square – 4 T = 5 – 4
5
24
1
4
=
1
9
=
1
2
5
4
=5–
5
6
5
4
5
3
=
5
3
=
5
24
25
6
C
A
D
E
B
41 Two circles of radius 16 have their centers on the
circumference of each other. AB is the diameter
of the right–hand circle that passes through the
centers of the two circles. A smaller circle is constructed
tangent to AB and the two given circles as in the sketch.
Find the radius of the smaller circle.
C
A
Let the radius of each of the large circles be R and the radius of the small circle be r.
AD = R, AC = R + r DC = R – r
AE2 = (R + r 2 – r2 = R2 + 2Rr DE = (R – r 2 – r2 = R2 – 2Rr
AD = AE – DE
R2 + 2 Rr –
R=
R2 –2 Rr
R2 = R2 + 2Rr + R2 – 2Rr – 2
R2 = 2
R4 – 4 R2 r2
R4 = 4 R4 – 16 R2 r2
R2 = 4 R2 – 16 r2
r2 =
3
16
R2
r=
3
4
R
r=
3
4
· 16 = 4
3
R4 – 4 R2 r2
D
E
B