Mathematics Contest Worksheet - Bluffton - 2009 Page 5
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611. Larry, Curly, and Moe divide a pile of n pennies like so: If n is even, Larry takes half the pile.
If n is odd, Moe takes one, then Curly takes half of those that remain. This process is repeated
until the pile is gone. If the original pile contains 2009 pennies, how many will Curly get?
Solution: The table on the right shows the entire
Pennies
Larry
Moe
Curly
process of dividing the pile of pennies. At the
2009
–
1
1004
1004
502
–
–
end, Curly has 1217 pennies.
502
251
–
–
251
–
1
125
Had they started with 2000 pennies, Curly would
125
–
1
62
only have ended with 88 pennies. Starting with
62
31
–
–
2007 pennies, Curly would have collected 1842
31
–
1
15
pennies.
15
–
1
7
7
–
1
3
3
–
1
1
1
–
1
–
784
8
1217
◦
12. The base of an isosceles triangle is 25 cm long, and the vertex angle measures 40
. Find the
length (in cm) of a leg. (If appropriate, round to two decimal places.)
Solution 1: Call this leg length L. Draw an altitude from the vertex (which bisects the vertex
◦
angle), creating a right triangle with an acute angle measuring 20
opposite a side of length
◦
=
12.5
12.5 cm. The hypotenuse of this right triangle has length L, so sin 40
, meaning that
L
L =
12.5
= 36.5475 . . . cm, which rounds to 36.55 cm
sin 20 ◦
◦
=
25
L
Solution 2: The base angles of the triangle measure 70
, so by the law of sines,
.
sin 40 ◦
sin 70 ◦
25 sin 70 ◦
Then L =
= 36.5475 . . . cm, as before.
sin 40 ◦
Section C. Each correct answer is worth 3 points.
− 14x
− 72x = 0. Express your answers
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3
13. Find all solutions (both real and complex) to x
either exactly or rounded to two decimal places.
√
Solution: The solution set is {0, ±2i, ±3
2}.
Clearly, x = 0 is a solution. Pulling out a factor of x, we have x(x
− 14x
− 72) = 0. Viewing
4
2
2
the fourth-degree factor as a quadratic expression in x
, we have
14 ±
− 4(1)(−72)
2
14
2
=
= 7 ± 11 = 18 or −4.
x
2
√
√
= −4 gives the solutions x = ±2i, and x
= 18 gives the solutions x = ±
18 = ±3
2
2
x
2.
14. The length of each side of an isosceles triangle is an integer. The legs have length x + 1, and
the base length is 3x − 2. Determine all possible values for this triangle’s perimeter.
Solution: In a triangle, the sum of any two side lengths must exceed the third, so 2x + 2 > 3x − 2
and 4x − 1 > x + 1. The first inequality implies that x < 4, while the second says that 3x > 2.
Combined with the fact that all side lengths are integers, we see that x must be 1, 2, or 3. The
perimeter will be 2(x + 1) + 3x − 2 = 5x, or 5, 10, 15.
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