Download Varsity Meet 4 – March 6, 2013 ANSWERS

WORCESTER COUNTY MATHEMATICS LEAGUE
Varsity Meet 4 – March 6, 2013
ANSWERS
ROUND 1
(St. John’s, QSC, QSC)
ROUND 5
(Shepherd Hill, Quaboag, Hudson)
1. 1092
1. −5/2
2. 17/35
2. 25/72
3. 12158
ROUND 2
(Quaboag, [unknown school], Auburn)
1. 13
2. 21
3. $56 or $56.00
ROUND 3
(QSC, Algonquin, Shrewsbury)
3. π/4, 2π/3 (need both, either order)
TEAM ROUND
(Worc Acad, Hudson, St. Peter-Marian,
Algonquin, Algonquin, Assabet Valley, Doherty, Shepherd Hill, Assabet Valley)
−a + 1
a2
2. 12
1.
1. 25π
3. 264
2. 4
√
3. 3 39
4. 5
ROUND 4
(QSC, Algonquin, Worc Acad)
5. 30
6. 25
1. 5
7. (4, −2, 1/16) ≈ (4, −2, 0.063)
2. 2
8. 853
3. 2
9. 371
WORCESTER COUNTY MATHEMATICS LEAGUE
Varsity Meet 4 – March 6, 2013
FULL SOLUTIONS
ROUND 1
1. The prime factorization is 396 = 22 · 32 · 11. The sum of the divisors is
(20 + 21 + 22 )(30 + 31 + 32 )(110 + 111 )
because expanding that product will give each divisor as a term in the summation.
Therefore, the answer is 7 · 13 · 12 = 1092 .
[Notice that each term in the product is the sum of a geometric series. Recalling that
the sum of a geometric series is
a + ar + ar2 + . . . + arn =
arn+1 − a
,
r−1
the above product could be rewritten as
8 − 1 27 − 1 121 − 1
×
×
.
2−1
3−1
11 − 1
While not terribly useful in this case, it can be easier if the exponents are large.]
2. In total, there are 640+231+100+91+1003 = 2065 women. Therefore, our (unreduced)
answer is 1003/2065, and the problem is asking us to find the greatest common divisor
(gcd) of 1003 and 2065.
METHOD I: The prime factorizations are 1003 = 17·59 and 2065 = 5·7·59. Therefore,
17
17
the gcd is 59 the answer is
=
.
5·7
35
METHOD II: To find the gcd of two numbers, especially if they are large, the preferred
method is the Euclidean algorithm, first described by Euclid in 300 bc. In the
Euclidean algorithm, the smaller number is divided into the larger number and the
integer remainder taken. This process is repeated, and the last nonzero number is the
gcd. For this problem:
2065 = 2 × 1003 + 59
1003 = 17 × 59 + 0
[← the gcd]
The advantage to this method is that the two numbers need not be factored. For large
numbers (≈100 base 10 digits), prime factorization in a reasonable amount of time is not
possible even with supercomputers, but the number of steps required in the Euclidean
algorithm is at most 5 times the number of base 10 digits in the smaller number.
WORCESTER COUNTY MATHEMATICS LEAGUE
3. Since 43 = 64 = 82 , each group of three base 4 digits corresponds to a set of two base 8
digits. Split the given number into groups of three: 220 0314 . Working from right to left,
0314 = 1310 = 158 and 224 = 1010 = 128 . Putting it together, the answer is 12158 .
ROUND 2
1. Since 2x + 1 and 5 − 4x are opposites, 2x + 1 + 5 − 4x = 0. Therefore, 2x = 6 so x = 3.
We are asked to find the value of 7x − 8, so this is 21 − 8 = 13 .
2. Express the original fraction as
3x
. Then, we have
4x
2
3x − 5
= .
4x − 4
3
Cross-multiply to find 9x − 15 = 8x − 8 so x = 7. The numerator of the original fraction
is 3x, so 3x = 21 .
3. Keep track of how much money each person has at each stage:
Oscar Alfred
3
A
4
A
5
A
4
1
A
2
A
3
A
4
At the end, we know that Alfred has $8 less than Oscar, so
started with $24 + $32 = $56 .
1
A = $8. Therefore, they
4
WORCESTER COUNTY MATHEMATICS LEAGUE
ROUND 3
1. A formula from geometry that relates the tangents and secants of a circle states that
AP 2 = (AB)(AE). Plugging in AP = 12 and AB = 8, we find that AE = 18 so
EB = 10. Also, EB is a diameter, so the area of the circle is π(5)2 = 25π .
P
E
A
B
O
2. Let O be the center of the circumscribed circle as shown:
A
E
B
O
D
C
Since O is the center of the circle, OA = OD. Additionally, since 4ABE is equilateral and the figure has a vertical line of symmetry, AO k ED. Draw segment AD.
Since AO k ED, m∠EDA = m∠OAD. However, triangles AED and AOD are isosceles (AE = ED because the equilateral triangle shares a side with the square), so
m∠EAD = m∠EDA = m∠OAD = m∠ODA.
Therefore, by ASA congruency, 4AED ∼
= 4AOD, and since corresponding parts of
congruent triangles are equal (“CPCTE”), the radius OD = ED = 4 .
WORCESTER COUNTY MATHEMATICS LEAGUE
3. The lateral faces of the pyramid have base 6 and height 4, so the lateral edge has length
5 (3-4-5 right triangle).
D
A
6
X
B
3
5
X
A
C
C
5
6
The left figure is the base and the right √
figure is the pyramid. By the Pythagorean
Theorem and 30-60-90 triangles, AX = 2 3 (alternatively, use the property that medians divideq
each other in 2 : 1 ratios at the centroid). Therefore, the height DX of the
√
√
pyramid is 52 − (2 3)2 = 13. The area of the base (equilateral triangle with side
√
√
1 √ √
length 6) is 9 3, so the volume of the pyramid is (9 3)( 13) = 3 39 .
3
ROUND 4
√
√
1. Since log10 N = 8.8, we have that N = 108.8√and N = 104.4 . Therefore, 104 < N <
105 . Since 104 = 10000 and 105 = 100000, N must have 5 digits to the left of the
decimal point.
2. The bases of all of the numbers in the fraction are powers of 2, so rewrite:
16x
4x+2
24x
22x+4
4x − (2x + 4)
2x − 4
6
2
= 32x−2
= 25x−10
=
=
=
=
5x − 10
5x − 10
3x
x
WORCESTER COUNTY MATHEMATICS LEAGUE
3. Simplify using the logarithm change-of-base formula and then cancel out common terms:
(log2x 3)(log2x 25)
= 3
(log2x 5)(log4x 3)
log 3 2 log 5
log 2x log 2x
log 5 log 3
log 2x log 4x
= 3
2 log 4x
= 3
log 2x
log2x [(4x)2 ] = 3
16x2 = (2x)3
2
= x
ROUND 5
1. Plug in the values:
2π
−2π
5π
π
tan
+ cot
tan
sin
3
3
3
3
√
√
√
3
1
=
·− 3+ √ ·− 3
2
3
3
= − −1
2
= −5/2 .
2. Write in terms of sines and cosines:
sin x cos x
144
+
=
cos x sin x
25
1
144
=
sin x cos x
25
by the Pythagorean identity sin2 x + cos2 x = 1. Since sin 2x = 2 sin x cos x, this is equal
25
to
.
72
WORCESTER COUNTY MATHEMATICS LEAGUE
3. First, recognize the Pythagorean identity that has been divided through by cos2 x:
tan2 x + 1 = sec2 x. Therefore, we have
tan x + 2(sin x − cos x) − 1 = 0.
Multiply through by cos x:
sin x + 2 sin x cos x − 2 cos2 x − cos x = 0
(sin x − cos x)(1 + 2 cos x) = 0
Therefore, we have either sin x = cos x and so x = π/4 or cos x = −1/2 so x = 2π/3 .
TEAM ROUND
1. We work our way from the bottom:
1
−1 =
1
1+
a+
1
a−1
1
−1
a−1
1+ 2
a −a+1
a2 − a + 1
−1
a2
−a + 1
.
=
a2
=
2. In base 10, 305b = 3b2 + 5 and 1321b = b3 + 3b2 + 2b + 1. We have (by long division, or
any preferred method)
( 3b − 4)
b
b+3
b − 12
b3 + 3b2 + 2b + 1
=
+1 + 2
=
+ 2
.
2
3b + 5
3
3b + 5
3
9b + 15
Since 1321b is a multiple of 305b , the right side must be an integer. If the RHS is an
integer, then 3 times it must also be an integer. Three times the RHS is
(b + 3) +
b − 12
.
3b2 + 5
The first term will always be an integer, and since 3b2 − 5 > b − 12 for all real b, the
only integer value the second term will take on is zero, when b = 12 . This is the only
possible answer; we must check that it is not extraneous. The original RHS with b = 12
gives (4 + 1) + 0 = 5, an integer, so the solution is good.
[In base 10, we have 132112 = 2185 and 30512 = 437, and 2185 = 437 · 5.]
WORCESTER COUNTY MATHEMATICS LEAGUE
3. Use the Binomial Theorem: the coefficient is
12
2
(2)2 (−1)10 = 66 · 4 · 1 = 264 .
4. Given that f (x) = px + q, we have
f (f (f (x))) = f (f (px + q))
= f (p2 x + pq + q)
= p3 x + p2 q + pq + q.
Since this is equal to 8x + 21, p = 2, giving 4q + 2q + q = 21 and q = 3. The sum is
p+q = 5.
5. Since BO and CO are angle bisectors, m∠DBO = m∠OBC and m∠ECO = m∠OCB.
Since DE k BC, m∠DOB = m∠OBC and m∠EOC = m∠OCB. By transitivity,
m∠DBO = m∠DOB and m∠EOC = m∠ECO, so triangles BOD and ECO are
isosceles. Therefore, the perimeter of 4ADE is AB + AC = 12 + 18 = 30 .
A
D
B
O
E
C
√
√
B2 C
B2 C
6. We are given that A ∝
, so
= k. Plugging in the given values, k =
DE
ADE
√
42 81
1
= .
3 · 8 · 12
2
√
√
1
42 C
Therefore, =
so C = 5 and C = 25 .
2
2 · 10 · 8
7. Combining the first and fourth statements, logB B −2 = E so E = −2. Then use the
1
second statement to find N = −2E
3 = 1/16 and the third to find that B = −2E = 4.
Therefore, the ordered triple is (B, E, N ) = (4, −2, 1/16) .
√
√
8. If x + 47 and x − 12 are both positive integers, then there must be two perfect
squares that differ by 47 − (−12) = 59. Let the two perfect squares be x2 and y 2 , for
x and y positive integers and x > y. Then, 59 = x2 − y 2 = (x − y)(x + y). Since 59 is
prime, we must have x − y = 1 and x + y = 59 so (x, y) = (30, 29). Therefore, the two
perfect squares in question are 302 = 900 and 292 = 841.
Since x + 47 = 900 and x − 12 = 841, pick one to solve and get x = 853 .
WORCESTER COUNTY MATHEMATICS LEAGUE
9. Draw AC to split the quadrilateral into two triangles.
D
A
18
C
The area of 4ABC is
128°
112°
20
13
B
1
· 13 · 20 sin 128◦ ≈ 102.4414. The law of cosines gives
2
AC 2 = 132 + 202 − 2 · 13 · 20 cos 128◦
so AC ≈ 29.8185.
Next, use the law of sines:
29.8185
13
=
◦
sin 128
sin m∠ACB
and m∠ACB ≈ 20.093◦ .
Therefore, m∠DCA = 112◦ − m∠ACB ≈ 91.907◦ .
1
The area of triangle ACD is therefore 18 · 29.8185 sin 91.907◦ ≈ 268.2181.
2
Hence the area of the quadrilateral is [ABC] + [ACD] ≈ 102.4414 + 268.2181 ≈ 371 .
[The same method could have been used but drawing BD instead of AC.]