41. Let ABCD be a trapezoid of area 48 with AB parallel to DC, AB = 9, and DC = 3. Diagonals AC and BD intersect in point E. What is the area of 4ABE ?
41. Answer: 27. The altitude h of ABCD must satisfy h·(9+3)/2 = 48 so h = 8. Since ∠AEB = ∠CED, ∠ABE = ∠CDE and ∠BAE = ∠DCE,4ABE is similar to 4DCE with similarity ratio AB/DC = 3. The altitudes h1 and h2 of 4ABE and 4DCE from AB and DC respectively must therefore satisfy h1 + h2 = 8 and h1/h2 = 3. Then h1 = 6 and the area of 4ABE is (1/2) · 6 · 9 = 27.
41. 42. 43. Ali Baba has a number of gold coins in a wooden chest. On Day 1 he transfers half of the coins to a steel chest. On Day 2 he transfers one third of the coins in the steel chest back to the wooden chest. On Day 3 he transfers one fourth of the coins in the wooden chest back to the steel chest. Continuing in this fashion, on Day 2009 he transfers 1/2010 of the coins in the wooden chest to the steel chest. What fraction of the coins is now in the wooden chest? (A) 1/2010
(B) 1/2009
(D) 2010/4022
(C) 2009/4020
(E) 1/2
41. 42. 43. Answer (E): Let wk and sk denote the fractions of the coins in the wooden and steel chests, respectively, at the end of Day k. Then (w1 , s1 ) = and
(w2 , s2 ) =
and
(w3 , s3 ) =
1 , 2 2
1 1 1 1 1 1 + · , − · 2 3 2 2 3 2
2 1 2 1 1 2 − · , + · 3 4 3 3 4 3
2 1 , 3 3
1 1 , 2 2
=
=
In general, if k is odd and (wk , sk ) = (1/2, 1/2), then 1 1 1 1 1 1 k+3 k+1 · , − · = , (wk+1 , sk+1 ) = + 2k + 4 2k + 4 2 k+2 2 2 k+2 2 and (wk+2 , sk+2 ) =
1 k+3 k+1 1 k+3 k+3 − · , + · 2k + 4 k + 3 2k + 4 2k + 4 k + 3 2k + 4
=
1 1 , 2 2
In particular, the fraction of the coins in the wooden chest at the end of Day 2009 is w2009 = 1/2.
.
44. 45. 46. 47. A coin falls onto a checkerboard. The center of the coin is equally likely to lie on any point of the board. The coin is 0.5 inches in diameter and the lines on the board are 2 inches apart. What is the probability that the coin lies on a side of at least one square? (A) 7/16
(B) 1/2
(C) 9/16
(D) 5/8
(E) 11/16
48. In a certain congressional district, 50% of the voters belong to party D, 40% belong to party R, and 10% belong to party I. Members of party D vote for candidates from party D 70% of the time and for candidates from party I 30% of the time. Members of party R vote for candidates from party R 80% of the time and for candidates from party I 20% of the time. Members of party I always vote for candidates from party I. In an election with one candidate from each party, which party will win, and by what percentage of the total votes over the runner-up? (A) D by 2% R by 1%
(B) R by 2%
(C) I by 2%
(D) D by 1%
(E)
49. A pizza restaurant offers a basic pizza with the optional addition of up to 3 of the following extra toppings: anchovies, salami, onions, pepperoni, sausage, mushrooms, roasted peppers, broccoli, and sun-dried tomatoes. Using anywhere from none to at most 3 extra toppings, how many different kinds of pizza are possible?
47. Answer (A): Consider the 2-inch checkerboard square that contains the center of the coin. The coin lies on at least one side of that square unless the center is more than 0.25 inches from each edge, that is, unless the center is within a 1.5-inch square concentric with the checkerboard square. The area of the smaller square is 2.25 in2 , and the area of the checkerboard square is 4 in2 , so the coin touches at least one edge with probability 1 − (2.25/4) = 7/16.
48. Answer (A): The candidate from party D will receive (0.5)(0.7) = 35% of the votes. The candidate from party R will receive (0.4)(0.8) = 32% of the votes. The candidate from party I will receive 100%−35%−32% = 33% of the vote. Therefore the candidate from party D will win by a margin of 35% − 33% = 2% of the votes over the candidate from party I. 49. Answer: 130. There are nine toppings to choose from, so the number of combinations with extra toppings with at most three toppings is 9 9 9 9 + + + = 1 + 9 + 36 + 84 = 130 0 1 2 3
54. 55. For how many integer values of k does the equation
(x − k)2 + kx − k 2 =
3 −k 4
have no real solutions for x?
55. Answer: 1.
The equation may be rewritten as 3 2 x − kx + k − = 0. 4
The equation has no real solutions precisely when the discriminant is negative, that is, when k 2 − 4k + 3 = (k − 1)(k − 3) < 0. This condition is satisfied by only one integer, k = 2. 56. The first term ofa sequen ce is a1 = 1. For n > 1, the nth term of the √ √ sequence is an = 12 + 23 i an−1 where i = −1. If Sn = a1 +a2 +· · ·+an , what is S2015 ? √ √ √ 1 3 1 3 1 3 (A) + i (B) − + i (C) − i 2 2 2 2 2 2 √ √ 1 3 3 1 (D) − − i (E) − i 2 2 2 2
56. Answer (B): First note a2 = √ 2014 3 1 + . 2 2 i
1 2
+
√ 3 2 i ·1,
a3 =
1 2
+
√ 2 3 2 i ,
. . . a2015 =
Then √ ! 3 1 + i + S2015 = 1 + 2 2 √ Let x = 21 + 23 i . Then
√ !2 1 3 + i + ··· + 2 2
S2015 = 1 + x + x2 + · · · + x2014 =
√ !2014 1 3 + i . 2 2
1 − x2015 . 1−x
In polar form √ ! π π 1 3 + i = 1 · cos + i sin 2 2 3 3
so π 2015 π + i sin x2015 = cos 3 3 2015π 2015π = cos + i sin 3 3 5π 5π = cos 670π + + i sin 670π + 3 3 5π 5π = cos + i sin 3 3 √ 1 3 = −i 2 2 Then 1−
1 2
−
1−
1 2
+
S2015 =
=
1 2
+
1 2
−
√ 3 2 i √ 3 2 i
√ 3 2 i √ 3 2 i
√ 3 1 =− + i. 2 2 OR Since a1 through a6 represent the vertices of a hexagon inscribed in unit circle, their sum is 0. Then a1 + a2 + · · · + a2010 = 0, leaving a2011 + · · · + a2015
√ ! 1 3 + i + =1+ 2 2 √ ! 1 3 = − + i . 2 2
√ ! 1 3 − + i + (−1) + 2 2
√ ! 1 3 − − i 2 2
59. A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles, and the remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths 15 and 25 meters. What fraction of the yard is occupied by the flower beds?
(A) 1/8
(B) 1/6
(C) 1/5
(D) 1/4
(E) 1/3
59. Answer (C): Each triangle has side length (1/2)(25 − 15) = 5 and area (1/2)(5)2 = 25/2. Thus the flower beds have a total area of 25. The entire yard has length 25 and width 5, so its area is 125. The fraction of the yard occupied by the flower beds is 25/125 = 1/5.
