KETTERING XX MATHEMATICS OLYMPIAD Star Wars ... - Kettering University

KETTERING XX MATHEMATICS OLYMPIAD

Star Wars Continued

Problem 1. Darth Vader urgently needed a new Death Star battle station. He sent requests to four planets asking how much time they would need to buld it. The Mandalorians answered that they can build it in one year, the Sorganians in one and a half year, the Nevarroins in two years, and the Klatooinians in three years. To expedise the work Darth Vader decided to hire all of them to work together. The Rebels need to know when the Death Star is operational. Can you help the Rebels and find the number of days needed if all four planets work together? We assume that one year=365 days.

Solution of Problem 1. Denote by x the total amount of work. In one day

the

Mandalorians

complete

x 365

,

the

Sorganians

2 3

?

x 365

,

the

Nevarroins

1 2

?

x 365

,

and

the

Klatooinians

1 3

?

x 365

.

Thus,

together

in

one

day

they

complete

x 2x 1x 1x x

211

+ ? + ? + ? = ? 1+ + +

365 3 365 2 365 3 365 365

323

5x x =? = .

2 365 146

Therefore, working together they will build the battle station in 146 days.

Problem 2.

Solve the inequality:

(sin

12

)

1-x

>

(sin

12

)x

.

Solution of Problem 2.

0 < sin < 1 implies 1 - x < x

12

1

-

x

0

x>0

1-

x

<

x2

0 0

Answer:

5-1 2

,

1

.

x2 + x - 1 = 0

-1 ? 5 x=

2

0

-1+ 2

5

-1 + 5

< x 1.

2

Problem 3. Solve the equation: x2 + 4x + 4 = x2 + 3x - 6

Solution of Problem 3. x2 + 4x + 4 = (x + 2)2 = |x + 2|

Case I. x < -2

|x + 2| = x2 + 3x - 6

Case II. x -2

|x + 2| = -x - 2

-x - 2 = x2 + 3x - 6

x2 + 4x - 4 = 0

x = -2 ? 8 = -2 ? 2 2

x < -2 x = -2 ? 2 2

x = -2 - 2 2

|x + 2| = x + 2

x + 2 = x2 + 3x - 6

x2 + 2x - 8 = 0

x = -1 ? 9 = -1 ? 3

x -2 x = -1 ? 3

Answer: {-2 - 2 2, 2}.

x=2

Problem 4. Solve the system of inequalities on [0, 2]:

sin(2x) sin(x) cos(2x) cos(x)

.

Solution of Problem 4.

1. sin(2x) sin(x)

2 sin x cos x sin x

sin x(2 cos x - 1) 0

Let us consider three cases: sin x = 0, sin x > 0, sin x < 0. I case. sin x = 0. In [0, 2] there are three solutions x = 0, x = , x = 2. II case. sin x > 0, that is 0 < x < . Then

1 cos x

2

In

(0, )

this

implies

that

0

<

x

3

.

III case. sin x < 0, that is < x < 2. Then

1 cos x

2

In

(, 2)

this

implies

that

<

x

5 3

.

Combining all three cases, one gets

0

x

3

x

5 3

x = 2

2. cos(2x) cos(x) Denote cos x = t, then

cos2 x - sin2 x cos x cos2 x - 1 + cos2 x cos x

2 cos2 x - cos x - 1 0

2t2 - t - 1 0

1? 1+8 1?3

t=

=

4

4

1 t = - or t = 1

2 1 - t1 2 1 - cos x 1 2 2 4 0 x or x 2 33

0 /3

5/3 2

0

2/3

4/3

2

0 x /3 4/3 x 5/3

x = 2

Answer:

0,

3

4 3

,

5 3

{2}

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