2011 Northwest High School Math Championship
Individual Test Solutions
1. Rewrite the given equation as 19(x + 3y) = 19 · 5, so that x + 3y = 5. Thus 3x + 9y =
3(x + 3y) = 15, so D is the correct answer.
2. Since x and y are real, it follows that x = y. Thus x2010 = y 2010 , which means that the
given expression equals 0, thus C is the correct answer. Note that the denominator equals
(x670 + y 670 )3 = (2x670 )3 != 0 (since xy != 0).
2
3. We can rewrite the given equation as x(log x) = 1027 . Taking the log of both sides of this
equation, we get (log x)3 = 27, so log x = 3. Thus the only solution is x = 103 = 1000, so D
is the correct answer.
4. Factoring the numerator and denominator we have:
!
"
x3 + 7x2 + 7x − 15 = (x + 5) x2 + 2x − 3 = (x + 5) (x − 1) (x + 3)
!
"
x3 + 3x2 − 9x + 5 = (x + 5) x2 − 2x + 1 = (x + 5) (x − 1)2 .
Thus
x3 + 7x2 + 7x − 15
x+3
=
, so C is the correct answer.
3
2
x + 3x − 9x + 5
x−1
5. Substituting x = 5 − ky into the second equation, we have: ky 2 + y (k + 1) + 1 = 0. Since
k > 0, the system has a unique solution if and only if (k + 1)2 − 4k = 0, or k = 1. Therefore
A is the correct answer.
6. We consider the following cases:
1st case: x ≤ 1
The given expression becomes 15 − 3x ≥ 12 (since −3x ≥ −3). The equality occurs when
x = 1. Thus in this case, the minimum value is 12.
2nd case: 1 < x ≤ 5
The given expression becomes 13 − x ≥ 8 (since −1 ≥ −x ≥ −5). The equality occurs when
x = 5. Thus in this case, the minimum value is 8.
3rd case: 5 < x ≤ 9
The expression becomes x + 3 > 8 (since 5 < x ≤ 9). Equality does not occur since x > 5.
Thus in this case, there’s no minimum value.
4th case: 9 < x
The expression becomes 3x − 15 > 12. Equality does not occur since x > 9. Thus in this
case, there’s no minimum value.
From the four cases above, we conclude that the minimum value of the given expression is 8,
and is achieved when x = 5. Thus B is the correct answer.
1
1
1
= −
, we have:
n (n + 1)
n n+1
#
$ #
$ #
$
#
$
1
1 1
1 1
1
1
T = 1−
+
−
+
−
+ ··· +
−
.
2
2 3
3 4
20 21
7. Since each term of the sum can be rewritten as
Therefore T = 1 −
20
1
= , which means a + b = 41, so C is the correct answer.
21
21
8. Using the formula sin a = cos (90◦ − a) with a = 1◦ , 2◦ , . . . , 44◦ , we can rewrite S as follows:
"
!
" !
"
!
S = sin2 1◦ + cos2 1◦ + sin2 2◦ + cos2 2◦ + . . . + sin2 44◦ + cos2 44◦ + sin2 45◦ .
Since sin2 a + cos2 a = 1 for any a, we have S = 44 · 1 +
1
2
=
89
,
2
so B is the correct answer.
%
9. Note that 4 ≥ x ≥ −1. Squaring both sides of this equation yields (4 − x) (x + 1) = 2.
Squaring both sides again, we obtain −x2 + 3x = 0, so x = 0 or x = 3. Both of these solutions
satisfy the restriction 4 ≥ x ≥ −1, so they are the only solutions to the given equation. Thus
the sum of all solutions is 3, so A is the correct answer.
10. We can rewrite the given expression as follows:
#
$2
1
1
2
= 1 + 2 sin x cos x − sin2 2x − sin 2x −
(sin x + cos x) − sin 2x +
2
4
= 1 + sin 2x − sin 2x − sin2 2x −
1
3
3 − 4 sin2 2x
2 cos 4x + 1
= − sin2 2x =
=
.
4
4
4
4
Thus E is the correct answer.
2
log ab
log a2 − log b
2 log a − log b
log a
201
11. We have:
=
=
= 201 − 1 = 200 (since
=
). Thus
log b
log b
log b
log b
2
E is the correct answer.
12. We see that 12 = 3 · 4 and 30 = 2 · 3 · 5, which means N must be divisible by 3 · 4 · 5 = 60. Now,
since 60 has exactly 12 different positive integral factors and N = 60k (k is a positive integer),
N must be 60 (otherwise, N would have at least 13 different positive integral factors). Thus
A is the correct answer.
!" !" !"
13. We see that there are 41 · 62 !· 82" ways to draw exactly 2 blue, 1 red, and 2 yellow marbles.
On the other hand, there are 18
ways to draw exactly 5 marbles, regardless of their colors.
5
(4)·(6)·(8)
,
Thus the probability of drawing exactly 2 blue, 1 red, and 2 yellow marbles is 1 182 2 = 10
51
(5)
which implies a + b = 61. Thus C is the correct answer.
14. Let b1 , w1 , b2 , w2 be the numbers of black and white marbles in jar 1 and jar 2, respectively.
From the given ratios, we have b1 = 7w1 and b2 = 9w2 . Since the numbers of marbles in these
two jars are the same, we have 8w1 = 10w2 or 4w5 1 = w2 . Thus 180 = 9w5 1 , or w1 = 100 and
w2 = 80. Thus the number of black marbles in jar 1 is 700, so B is the correct anwer.
!
"
!
"
!
"
15. We can rewrite the equation as sin x + π4 = sin π2 − x − π6 = sin π3 − x . Since x ∈ [0, 2π],
π
we must have either x + π4 = π3 − x or x + π4 = 7π
− x. The only solutions are x = 24
and
3
13π
25π
x = 24 , and their sum is 12 . Thus a + b = 25, so B is the correct answer.
16. Let x be the given expression. Since the fraction is infinite
√ with repeated pattern, we can write
1
1+ 3
x=1+
. Solving this equation, we get x =
(since x > 0). Thus a + b + c = 6 .
2
2 + x1
17. We see that P = 2 · 3 · 5 · 7 · . . . · 97. Since 2 · 5 = 10 has units digit 0, it follows that the units
digit of P is 0 .
18. Applying the Pythagorean Theorem, we have: 169 = a2 + b2 . Since (a, b, c) = (5, 12, 13) is the
only Pythagorean triple with c = 13, we conclude that the two sides of this rectangle must be
5 and 12. Thus the perimeter of this rectangle is 34 .
19. The first inequality implies either n ≥ 1 or n ≤ −1. The second inequality implies −1 ≤ n ≤
100. From these two inequalities, we see that either n = −1 or 1 ≤ n ≤ 100 will satisfy the
given inequalities. Thus there are 101 possible integral values for n.
20. Note that all points that lie inside or on the boundary of the circle of radius 1 centered at the
same point as the circle of radius 2 lie within 1 unit from that center. The area of the circle
of radius 1 is π, while the area of the circle of radius 2 is 4π. Therefore the probability that
π
a point inside the circle of radius 2 is farther than 1 unit from the center is 1 − 4π
= 34 .
21. We have: 4x − y + i (4y + x) = 17 + 17i. Because x and y are real, this equation is true if
and only if 4x − y = 17 and 4y + x = 17. Solving this system, we get x = 5 and y = 3. Thus
x2 + y 2 = 25 + 9 = 34 .
2
22. From the
√ triangle inequality, we have: 1 < x < 7 and 7 < x < 25. The second inequality
implies 7 < x < 5, which also satisfies the first inequality. Since x is a positive integer, the
possible values of x are 3 and 4, and the sum of their squares is 32 + 42 = 25 .
3
−53
11
23. We have: nn−4
= n2 + 4n + 16 + n−4
. This is an integer if and only if n − 4 is a factor of 11,
so n − 4 = ±1 or ±11. Thus the possible values of n are −7, 3, 5, or 15, and their sum is 16 .
24. For real x, we have x2 + 1 ≥ 1 and cos x ≤ 1. Thus the equation is true if and only if x = 0 .
25. Let x and y be the current ages of me and David, respectively. Let k be the number of years
it takes me to be as old as David is now. From the given hypothesis, we have x + k = y and
y + k = x + 6. Adding these two equations side by side, we get k = 3. Therefore David is 3
years older than I am.
26. We can rewrite the polynomial as n3 + 6n2 + 11n + 6 + 2 = (n + 1) (n2 + 5n + 6) + 2 =
(n + 1) (n + 2) (n + 3) + 2. Note that this expression is always divisible by 2, so it is only
prime when it equals 2. Hence n = −1, −2, or −3, and the sum of the squares of these integers
is 14 .
27. Note that there are 3 possibilities for the sexes of the two people leaving the room: 2 boys, or
2 girls, or 1 boy and 1 girl. Thus we examine the probability for each of these events to occur.
( 5)
2
The probability that 2 boys leave is 152 = 21
. Similarly, the probability of 2 girls leaving the
(2)
(10)
= 10
room is 152 = 37 , and that of 1 girl and 1 boy leaving the room is 5·10
. Thus the most
21
(2)
(152)
likely ratio of girls remaining to boys remaining occurs when 1 boy and 1 girl leave the room,
and this ratio is 94 .
k
k
k
k
28. We
√ have |(1 − 2i)√| =k |5 |,kwhich implies |1 − 2i| = |5| . Note that |1 − 2i| =
5, so we have ( 5) = 5 , which is only true when k = 0 .
%
12 + (−2)2 =
29. Let x1 , x2 be the roots of x2 − ax + a + 5. Applying Vieta’s theorem, we have x1 + x2 = a
and x1 · x2 = a + 5. From the given hypothesis, we have (x1 − 1)(x2 − 1) = q. Thus
q = x1 x2 − (x1 + x2 ) + 1 = (a + 5) − a + 1 = 6 .
30. Since Batman doesn’t win, Superman beats Wonder Woman if and only if Superman wins the
P (Superman wins)
race on that day. Thus the conditional probability for this event to occur is P (Batman
=
doesn’t win)
1/2
1−1/6
=
3
5
.
31. The number of digits of 22011 is '2011 log10 2(+1 = 606 (since 2011 log10 2 ≈ 605.37). Similarly,
the number of digits of 32011 is 960 (since 2011 log10 3 ≈ 959.49). Thus d = 960 − 606 = 354 .
d
d
32. Let d be the distance that Alex travels. From the given hypothesis, we have dr = 20
−1 = 30
+1.
120
Solving for d from the last two equations, we get d = 120. Therefore r = 5 = 24 .
! 5"
33. We see that there are
5
subsets
that
have
exactly
1
element,
= 10 subsets that have
2 ! "
! 5"
5
exactly 2 elements, 3 = 10 subsets that have exactly 3 elements, 4 = 5 subsets that have
exactly 4 elements, and 1 subset that has exactly 5 elements. Thus the sum that Peter gets
is 5 + 20 + 30 + 20 + 5 = 80 .
34. Let d be the common difference in this arithmetic progression. From the given hypothesis, we
have: x3 + x5 = 2x1 + 6d = 14 and x1 + x2 + . . . + x11 + x12 = 12x1 + 66d = 129. Solving
this system of equation yields: d = 23 and x1 = 52 . Thus xn = 52 + 32 (n − 1) = 193 implies
n = 128 .
35. Let O be the center of the inscribed circle, and let H, K, M , and N be the feet of the perpendicular segments from O to AB, BC, CD, and DA, respectively. Without loss of generality,
assume AD is the fourth side. Now, since BH and BK are both tangent to circle O, we have
BH = BK. Similarly, we also have AH = AN . Adding these two equations side by side,
we get BK + AN = AB = 4. By symmetry, we also get DN + CK = DC = 16. Therefore
AD + BC = 20, which means AD = 20 − 9 = 11 .
36. If Ali wins in the first round, the probability is 21 . If Ali wins in the second round, the probability is 12 · 18 . If Ali wins in the third round, the probability is 12 · ( 18 )2 , and !so on. Therefore "the
probability that Ali wins is the sum of the following geometric series: 12 1 + 18 + 812 + . . . =
1
2
·
1
1− 18
=
4
7
.
37. Note that log2 p + log2 q = log2 pq. The product of the roots of 2x2 − 5x + 2 = 0 is
log2 pq = log2 1 = 0 .
2
2
= 1, thus
38. Adding these equations side by side and simplifying yields 2 (x + y + z)2 = 288 or x + y + z =
±12. If x + y + z = 12, then x + y = 12 − z. Substituting this into the first equation, we get
z = 2. Substituting y + z = 12 − x into the second equation, we get x = 4. Finally, since
y = 12 − x − z, we have y = 6. Thus in this case, (x, y, z) = (4, 6, 2). If x + y + z = −12,
then z = −2, x = −4, and y = −6. Thus (x, y, z) = (4, 6, 2) or (−4, −6, −2), and the largest
possible value of xyz is 48 .
39. Note that sin kπ = cos π7 = sin( π2 − π7 ) = sin 5π
. Because of principal values, kπ =
14
k=
5
14
5π
,
14
so
.
40. By the triangle inequality, the largest side cannot be greater than or equal to 6. Because the
triangle has perimeter 12, the largest side cannot be smaller than 4. If the largest side is 4,
then the only possibility is (x, y, z) = (4, 4, 4). If the largest side is 5, then the other two sides
can be either 5 and 2 or 4 and 3. If the triangle is 5-5-2, then there are 3 possible triples
(x, y, z). If the triangle is 5-4-3, then there are 6 possible triples (x, y, z), for a total of 10
possible triples.