Rocket City Math League
Gemini Solutions
2014-2015
Round 2
5.
Answer: 207
Set x equal to the smallest angle and remember that there are 540 degrees in a
pentagon.
x  ( x  3)  ( x  6)  ( x  9)  ( x  12)  540
1.
2.
3.
Answer: 146
x + x +20 + x + 58 = 180
3x + 78 = 180
3x = 102
x = 34
The two biggest angles will be x +20 and x + 58.
x + 58 + x + 20
34+ 58 + 34 + 20 = 146
5 x  30  540
5 x  510
x  102
The two smallest angels are 102 and 105. The sum is 207.
6.
WZ has length 6 and is the hypotenuse of a 45-45-90 triangle with legs of ZV
and
Answer: 6
Area = (width)(length)
54 = 9(length)
54/9 = (length)
6 = (length)
7.
9
14
2
 3  8
9
 
 8
2


V1=
 V1= 4
V1= 9  2 V1= 18
V1=Bh V1=  r
The volume above (V1) is the volume of the entire cylindrical time capsule.
To get the volume of what each alien gets to put his/her item (V 2) into, divide
the volume by the number of aliens.
V2= 18
28
9
V2= 
14
2
 72.
Answer: y  3 x  14
The perpendicular slope to -1/3 is the negative reciprocal so the slope is 3. Using
point-slope form, simplify to slope-intercept form.
y  5  3( x  3)
y  5  3x  9
y  3 x  14
8.
2
 
square VOID is 2(3 2 )  6 2 . The area of the square is 6 2
Answer: 8 3
We are given that s=8 in the 30-60-90 triangle where s=the hypotenuse. The
short leg is 8/2=4. The height is 4* 3  4 3. The area of the triangle is thus
Answer:
VW . The lengths of these legs are 6  3 2. The length of the sides of
2
1
(4)( 4 3 )  8 3.
2
4.
Answer: 72
Answer:
432 3
125
The base of Prism H is made up of 12 equilateral triangles with side lengths of 12
 12 2 3 
3  1296 3. The volume of Prism


4


so the volume of Prism H is 12
T is the area of the base trapezoid times its height h. The area of the base of Prism
15
21  29  375. Set the volumes equal to each other to solve for h.
2
1296 3  375h
T is
h
1296 3 432 3

375
125
9.
Answer: 112
To find the volume of the frustum of the regular square pyramid, one must solve
all three clues. To solve the first clue, one must either know that the diagonal of a
cube is
s 3 or use the Pythagorean Theorem twice to find the diagonal. The first
use of the Pythagorean Theorem uses the length and the width. This is written as
10. Answer:
36   3
60
A regular icosahedron is a twenty sided 3D figure with faces of equilateral
triangles with side lengths of 12 3. The radii of the inscribed circles are


42  42
 1 
12 3
12 3 
. The area of each face is
4
2 3
(4 2 ) 2  42 or 4 3 . This gives the diagonal of the cube. Thus, the solution
each inscribed circle is
or 4 2 . The second use of the Pythagorean Theorem uses the depth
of the cube and the value of the first Pythagorean theorem. This is written as
for Clue 1 is 4 3 m. To solve the second clue, factor the quadratic and solve for
x, using the positive value as the height of the frustum.
2 x 2  5 x  12  0  (2 x  3)( x  4)  0  x   3 2 ,4
2
3
 108 3 and the area of
62   36 . Follow the table to find the amount of blue
surface area on the icosahedron.
Fraction of faces
Number of faces/20
Blue surface area
2/5
8/20
8(108 3  36 )
3/10
6/20
6(36 )
1/5
4/20
4(108 3  18 )
1/10
2/20
2(18 )
Thus the answer to Clue 2 is 4. For Clue 3, the altitude of an equilateral triangle is
equivalent to
s 3
4 3
 2 3. Now that all of the
so the value for Clue 3 is
2
2
clues are solved, one can find the volume of the frustum of the regular square
pyramid. To find the volume of a frustum, one can use the formula
v
h( B  B1  BB1 )
3
(where B = area of one base,
B1 = area of the other base,
and h = height). Because the bases are squares, one must multiply the side lengths
of the bases to the second power to find the areas of each base. One base has a side
Add the area together to get:
(864 3  432 3 )   (288  216  72  36)  1296 3  108
of length 4 3 m, so the area of the base is 48. The other base has a side length of
The total surface area is 20 times the area of each triangle so the total surface area
2 3 m, so the area of this base is 12. The height, which we found in clue 3, is
is
4m. Now one is able to apply the volume of a frustum formula.
Simplify the ratio to have the correct answer:
20(108 3 )  2160 3.
1296 3  108 12 3   36   3


60
2160 3
20 3
h( B  B1  BB1 )
3
4(48  12  48  12 )
v
3
4(60  24) 4(84)
v

 112
3
3
v
3
Thus, the volume of the frustum of the regular square pyramid is 112m .