Download Round 1 - Mu Alpha Theta

Mu Alpha Theta National Convention 2004
Theta State Bowl
Round 1
1. Simplify for all x: 2x  2x  2x  2x .
C
2. In the given triangle, AB = BC. If the area of triangle ABE is x,
what is the area of triangle ACD?
B
118Þ
A
62Þ
E
D
Round 2
3. The sides of a triangle are in an arithmetic progression with the middle term = 2 and the
angle opposite the side of 2 is 60 . Find the area of the triangle.
4. The perimeter of trapezoid ABCD is 50. If the two bases BC = 9 and AD = 21, what is
the length of the diagonal AC?
Round 3
5. A square and a hexagon have the same perimeter. If the area of the square is 2.25, what is
the area of the hexagon?
6. In the figure, rectangle ABCD is inscribed in a circle. If the radius
of the circle is 1 and AB = 1, what is the area of the shaded area.
A
B
O
D
C
Round 4
7. A regular pyramid is composed of a square base of area 12 and four equilateral triangles.
What is the volume of the pyramid?
 1 1  1 
8. What is the sum of the infinite geometric series 2       ...
 2  8  32
Round 5
9. If 42x2  64, then x 
2
2
10. Consider the ellipse with the equation 9x  4y  36x  32y  64  0 . Find the equation of
the circle, in graphing form, with its center at the center of the ellipse and its area the same
as the ellipse.
Round 6
11. Change 3.2513513513513… to a fraction. At lowest terms
12. Solve over the real numbers: x  x  20  0
Round 7
13. A goat is tethered by a 100 ft. rope attached to an outside corner of an 80 ft. by 80 ft.
square barn. How much grazing area outside the barn can the goat reach?
14. A triangle has sides of lengths 1, 2, and
triangle.
3 . Find the radius of a circle inscribed in the
Round 8
15. Evaluate log
7
9
3
 log
 2log where logs are to base 2.
8
14
4
16. A snowman is made using three balls of snow with diameters of 20 cm, 30 cm, and 40 cm.
If the head of the snowman weighs 1 kg, what is the total weight of the snowman? (the
head is the 20 cm snowball)
Round 9
17. Find the positive integer x for which
2x
4
2
8
.
5x  8
18. Find the equations of the asymptotes for 16x2 – 9y2 – 32x – 54y – 209 = 0 in slope
intercept form.
Round 10
19.
44776389827563405877634
Simplify: i
20. In an arithmetic progression of positive numbers, the common difference is three times the
first term, and the sum of the first five terms is equal to the square of the first term. Find
the first term.
Mu Alpha Theta National Convention 2004
Theta State Bowl Answers
Answer
#
Answer
#
1-1
2x+2
6-11
1203
370
1-2
4x
6-12
25
7-13
7700
2-3
3
3 1
2
2-4
17
7-14
3-5
3 3
2
8-15
0
8-16
3
12 kg
8
9-17
12
3-6

3
4-7

3
4
4 6
4-8
8
5
9-18
4
13
x
and
3
3
4
5
y
x
3
3
5-9
1
2
10-19
-1
10-20
35
5-10
(x – 2)2+(y + 4)2=6
y
Round 1
1. Simplify for all x: 2x  2x  2x  2x .
x2
Answer: 2
Solution: 4 2x  22 2x  2x2
 
C
2. In the given triangle, AB = BC. If the area of triangle ABE is x,
B
what is the area of triangle ACD?
118Þ
A
E
62Þ
D
Answer: 4x
Solution:
Since 118o  62o  180o, BE PCD. Triangle ACD and triangle ABE are similar
with sides in ratio of 2:1. Their areas are in ratio of 4:1. Area of triangle ACD = 4x
Round 2
3. The sides of a triangle are in an arithmetic progression with the middle term = 2 and the
angle opposite
the side of 2 is 60 . Find the area of the triangle.
3
Answer:
C
Solution:
2
2 +d
Y
Z
O
6 0Þ
A
X
2 -d
B
2  d  2  2  d 
CY  AY  2
the perimeter = 6
CZ  AX  2
AngleOBX  30
the semiperimeter = 3
AB  BC  4
BZ  BX
triangle OB is a 30, 60, 90 triangle
 both = 1
OX = the inradius =
3
3
 3 
Area  inradius semiperimeter  
 3 
3  3
 
4. The perimeter of trapezoid ABCD is 50. If the two bases BC = 9 and AD = 21, what is the
length of
the diagonal AC?
Answer: 17
Solution:
B
9
C
Since the perimeter is 50 and the legs are
10
A
10
8
congruent, the legs =
50 - 21+ 9
2
Dropping ' s from B and C forms rt
6
9
6
D
 10
triangles 6, 8, & 10. The diagonal forms
a rt triangle of 8, 15, & hypotenuse.
AC =
152  82
 17
Round 3
5. A square and an hexagon have the same perimeter. If the area of the square is 2.25, what is
the area of
the hexagon?
3 3
Answer:
2
Solution:
Area of square = 2.25 means side of square = 1.5. Perimeter of square = 6  Perimeter
s2 3
12 3 3 3
6

4
4
2
A
B
6. In the figure, rectangle ABCD is inscribed in a circle. If the radius
of the circle is 1 and AB = 1, what is the area of the shaded area.
of hexagon = 6

side of hexagon = 1

Area = 6
O
D
C
Answer:

3

3
4
Triangle AOB is equilateral  mAOD = 120o
2
1
1

Area of sector =  r 2   1 
3
3
3
Area of segment = sector - triangle =

Solution:
 1  1
  
3  2  2

 3  3 
3
4
Round 4
7. A regular pyramid is composed of a square base of area 12 and four equilateral triangles.
What is the
volume of the pyramid?
Answer: 4 6
Square base with area = 12  side = 2 3
The right triangle with hypotenuse = a lateral
Solution:
edge of the pyramid = 2 3, one side =
1
diagonal of square = 6 and the 3rd
2
side is the height.
h 6

V
2 3 
2
2
h 
 6
2
1
1
Bh  12 6  4 6
3
3
 1 1  1 
8. What is the sum of the infinite geometric series 2       ...
 2  8  32
8
3
Answer:
1
5
5
Solution:
1

1
2
2 8
The first term a 1  2, and the common ratio r  2  
 Sum =

-1 5  5
2
4
1-  
4  4
Round 5
9. If 42x2  64, then x 
1
Answer: x 
2

64
1
 4  2x  1  x 
16
2
2
2
10. Consider the ellipse with the equation 9x  4y  36x  32y  64  0 . Find the equation of
the
circle, in graphing form, with its center at the center of the ellipse and its area the same as
the ellipse.
Solution: 42x 42  64
42x 

Answer: x  22  y  4  6
2
The ellipse has the equation
x  22 y  42
4
Solution: Area of ellipse = ab  23  6
For the circle


9
center at 2, - 4
1 
r2  6

r 6
x  22  y  42  6
Round 6
11. Change 3.2513513513513… to a fraction. At lowest terms
1203
370
Let N = 3.2513513513...
10000N = 32513.513513...
Solution: -10N = -32.513513513.....
Answer :
9990N = 32481

N=
32481 1203

9990
370
12. Solve over the real numbers: x  x  20  0
Answer: x=25
2
1
2
2
Factoring x  x  20  0
Solution:
1
2
x 5
 x  25
 1
 1

 2
 2

 x  5x  4  0 






1
x 2  4 impossible
Round 7
13. A goat is tethered by a 100 ft. rope attached to an outside corner of an 80 ft. by 80 ft.
square barn.
How much grazing area outside the barn can the goat reach?
Answer: 7700 sq units
3
of a circle with radius 100 and
4
Solution:
1

3
1
2  of circle of radius 20   A   10000   400  7700
4

4
2
14. A triangle has sides of lengths 1, 2, and 3 . Find the radius of a circle inscribed in the
triangle.
The grazing area consists of
3 1
2
Answer:
Solution:
Drop ' s from the incenter to the 3 sides and
draw segments from the incenter to the vertices.
The right angle is bisected and creates isosceles
A
triangles. Side a is divided into r and a  r, a and
likewise for side b.  since they are tangents from
an external point,
b-r
r
2r  a  b  c 
r
C
a  r   b  r   c
r
r
r
a-r
1 3  2

2

3 1
2
B
Round 8
15. Evaluate log
7
9
3
 log
 2log where logs are to base 2.
8
14
4
Answer: 0
7  9 
 9 
  
 
8 14 
16

log
 log 1  0
Solution: log
 9 
32
 
 
16
4
16. A snow man is made using three balls of snow with diameters of 20 cm, 30 cm, and 40 cm.
If the
head of the snow man weighs 1 kg, what is the total weight of the snow man? (the head is
the 20 cm
snowball)
Answer: 12
3
kg
8
Weight is proportional to volume, and volume is proportional
to the cube of the diameter.
20
30
3
Solution:
weight of 30 cm ball = x
3
30 1  27
x
 8
20
3
1

x
40 1  8

20

3
3
weight of 40 cm ball =
3

Total = 1
27
3
 8  12
8
8
Round 9
2x
17. Find the positive integer x for which
4
2
8
5x  8 .
Answer: 12
2x
Solution:
2

 23
5x
22
 
8
x  12x  2  0
18.
x2  10x  24


x2  10x  24  0

x  12
Find the equations of the asymptotes for
16x 2  9y2  32x  54y  209  0 in slope intercept form.
Answer: y 
4
13
x
and
3
3
x  1  y  3
2
9
Solution:
y
4
5
x
3
3
2
16
 1  The asymptotes are y+3=  (x-1)
In slope intercept form y 
Round 10
19.

44776389827563405877634
Simplify: i
Answer: -1
4
13
4
5
x
and y   x 
3
3
3
3
i1  i
i2  1
i3  i
i4  1
After 4 it repeats
Solution: You only need to divide the last two digits of the exponents by 4
34
 8 with remainder = 2 
4

value = -1
20. In an arithmetic progression of positive numbers, the common difference is three times the
first term,
and the sum of the first five terms is equal to the square of the first term. Find the first term.
Answer: 35
Solution: d  3a

a  4a  7a  10a  13a  a2

a2  35a
 a  35