Download 2014 High School Interstellar Problems 2 31. The sum of 1 + 1 + 1 +

2014 High School Interstellar Problems
31. The sum of
is p + q ?
1
2
2
1
1
1
+ 61 + 12
+ 20
+ · · · 8010
is written as a common fraction, pq . What
32. A right hexagonal pyramid is cut by 2 planes parallel to the hexagonal base√to
form two √
hexagonal cross sections. The areas of the cross sections are 253.5 3
and 1014 3 square inches. If the two hexagonal planes are 10 inches apart,
what is the distance of the larger cross section from the apex of the pyramid?
33. Three darts land at random with uniform distribution in the 3 by 3 board shown.
What is the probability that they land in three distinct unit squares forming either a
horizontal, vertical, or diagonal line?
(A)
1
81
(B)
1
63
(C)
1
16
(D)
8
81
(E)
48
729
34. The Fort Worth Zoo has a number of birds which all have two legs and a number
of four-legged land animals in its care. On one visit to the zoo, Margie counts
200 heads and 522 legs. How many of the animals that Margie surveyed were
birds?
35. The two circles pictured have the same center C; Chord AD is tangent to the
inner circle at B; the length of radius AC is 13; and the length of chord AD is
24. The area of the ring between the two circles is
D
B
A
C
(A) 25π
(B) 64π
(C) 100π
(D) 144π
(E) 169π
38. In convex pentagon ABCDE, angles A, B, and D are right angles, AB = 2AE =
2BC, and DE = DC. If the area of ABCDE is 240, what is the area of 4DEC?
40. Positive integers 1, 2, 3, . . . , 2013 are spaced in that order around the perimeter of a
circular table. Then the numbers 1, 6, 11, . . , 2011 are removed and the remaining
numbers are evenly spaced around the circle with their cyclic ordering maintained.
What integer appears diametrically opposite 1000?
(A) 1009
(B) 1013
(C) 1014
(D) 2005
(E) 2007
2014 High School Interstellar Problems
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31. Answer: (179)
Observe that the denominators of each term are the product of consecutive
1
1
1
1
1
+ 2·3
+ 3·4
+ 4·5
+ · · · + n·(n+1)
.
integers. That is, the sum can be written as 1·2
1
1
1
n+1−n
The denominators are of the form n · (n + 1) and n·(n+1) = n − n+1 = n·(n+1) =
1
1
1
1
1
1
1
n·(n+1). We can therefore re-write the sum as 1 − 2 + 2 − 3 + 3 − 4 +
1
1
4 − 5 + · · · . Remove the parentheses and every term cancels except the first
and last. The last denominator is 8010. It is of the form n · (n + 1) so that
n2 + n = 8010 or n2 + n − 8010 = 0 or the root that applies is n = 89. Thus
1
) or 89
we’re looking at 89 and 90. All but the first and last terms leave (1 − 90
90 .
The sum of 89 and 90 is 179.
32. Answer: (20)
If s is the side of a regular hexagon, then since√the hexagon
consists of 6 equilat√
2
2
eral triangles, the area of the hexagon is 6 · s 4 3 = 3s 2 3 . Using this expression
with the two given cross sections and solving for the corresponding sides, we get
side lengths of 13 and 26. We set up a ratio as shown using the diagram below
x
x+10
= 13
to obtain x = 10 inches. Thus the distance from the larger cross
26
section to the apex is x + 10 = 20 inches.
2014 High School Interstellar Problems
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33. Answer (E): Each of the first, second and third darts may land on any one
of the 9 squares, so there are 93 possible dart landing possibilities, each equally
likely by the assumptions of the problem. Among these possibilities there are 3
that lie in a vertical line, 3 in a horizontal line, and 2 that fall along a diagonal.
Each of those lines can have any of the 6 arrangements of the first, second and
48
third darts. So the probability is 6·(3+3+2)
= 729
.
93
34. Answer: (139)
Let a be the number of 4-legged animals and b the number of birds. Then
a + b = 200 and 4a + 2b = 522. Solving gives a = 61 and b = 139.
OR
If one were to assume that all 200 heads had two feet, then 400 of the legs could
2014 High School Interstellar Problems
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be connected to the heads. The means additional 122 legs must be assigned to
four-legged animals. Since the four-legged animals would need two legs each to
complete their complement, 122
2 = 61 four-legged animals were at the zoo. So,
200 − 61 = 139 birds must be at the zoo.
35. Answer (D): In the right triangle ABC, the length of AB is 12, 122 + BC 2 =
132 so BC = 5.The area of the outer circle is 169π, the area of the inner circle
is 25π thus the area of the ring between the circles is 144π.
38. Answer: (80)
√
If BC = s, then AB = EC = 2s so that DC = DE = 2s Thus area
2
2
2
2
ABCDE
√ 2 = 22s + s = 3s = 240 so s = 80. Then the area of 4DEC =
1
2 ( 2s) = s = 80.
40. Answer (E): Label the remaining positive integers in clockwise order by
{a1 , a2 , a3 , . . . , a1610 }, with a1 = 1. Notice that a4k = 5k, so 1000 = a800 . Now
a800 is diametrically across a800+805 = a1605 . Since a1604 = 2005, a1605 = 2007.