Theta Circles & Polygons
MA National Convention 2012
Answers:
1. B
2. D
3. D
4. A
5. D
6. A
7. C
8. B
9. A
10. B
11. D
12. D
13. C
14. A
15. C
16. B
17. D
18. C
19. A
20. B changed to A at the convention
21. C
22. D
23. B
24. D
25. A
26. D
27. B
28. C
29. B
30. A
1
Theta Circles & Polygons
MA National Convention 2012
Solutions:
1.
180 12  2 
 150
12
2. A central angle with measure 1 radian would subtend an arc the same length of the radius.
Since 17.5  7  2.5 , the angle would be 2.5 radians.
3. Since the chord divides the diameter into lengths in the ratio 1:1, the chord must pass
through the center, making the chord a diameter also. Thus the length of the chord is 12.
4.
20  20  3
 170
2
5. The inscribed angle intercepts an arc of 256 , so the angle must be half of that, so it’s 128 .
6. 12  18  27x , so x  8 . Therefore the radius is

122  182  272  82
144  324  729  64

2
2
1261
.
2
7. Since two of the vertices of the triangle are already given (the ends of the diameter), the
third point must lie in one of the semicircles, so basically this triangle lies in a semicircle. If the
diameter is regarded as the base, the largest altitude would occur when that altitude is the
radius of the semicircle. Therefore, the ratio of the area enclosed by the triangle to the area
1
enclosed by the circle is  2r  r :  r 2  r 2 :  r 2  1:  .
2
8.
180  8  2   124  151  114  155 536

 134
4
4
9. 252  x 2  292   36  x   625  x 2  841  1296  72 x  x 2  72 x  1080  x  15 , so half of
2
the segment is of length
252  152  20 , making the segment have length 40.
10. The longest side of the triangle is a diameter, so the triangle is a right triangle, meaning the
hypotenuse has length
242  702  576  4900  5476  74 , so the radius is of length 37,
making the area   372  1369 .
2
Theta Circles & Polygons
MA National Convention 2012
11. 50  50  x   802  50  x  128  x  78 , so the total length of the secant is 50  78  128 .
2 179999
1
2
 2

  n  360000 . Therefore the
12. 180  1    179.999  1  
n
n 180000 180000 n

smallest number of sides is 360001.
13. All of the angles have measure 120 , so drawing a
perpendicular length between the two sides provides two right
triangles and a rectangle between the that perpendicular length
and the two sides of length 4 and 6. The sides of length 4 and 6
are the hypotenuses of the two right triangles, which are both
30  60  90 . The perpendicular length is made up of the two
sides opposite the two 60 angles, so the perpendicular length
is 2 3  3 3  5 3 .
14.
1
1
1
1 1 1


   x4
x
9
36 3 6 2
15. The distance from the point to the center of the circle is
 2  10    5  11
2
2
 400
 20 , and the radius of the circle is 7, so the point is 13 units away from the circle.
2
2


1 1
16. Let O have radius 2. The total enclosure then of the Oi 's is   12        ... 


2 4


 1  4
   , and the total enclosure of O is   22  4 , so the fraction of area enclosed
 
1 1  3

4
4
4   8 
3  3 2
by O not enclosed by any of the Oi 's is
4
4
3
3  122 3
 216 3 .
2
Then remove the rhombus eliminated when the one vertex is moved
to the center of the hexagon; that rhombus encloses an area of
1
 12  12 3  72 3 . Thus the enclosed area is 216 3  72 3  144 3 .
2
17. The area enclosed by the original hexagon is
3
Theta Circles & Polygons
MA National Convention 2012
18. Apeirogon is the name of such a degenerate shape. Myriagon – 10000 sides, chiliagon –
1000 sides, infinigon – made up.
19. Side-side-angle is not a valid postulate, but the other three are.
20. By analogue of Kepler Conjecture, the tiling that wastes the
least space is the one where three circles are all tangent and their
centers are the vertices of an equilateral triangle. Since equilateral
triangles would tile the plane in this arrangement, we need only
compute the wasted space in one triangle. If the circle’s radius had
22 3
1

 3     12  3   1.732  1.571  0.161 . Divide this by
4
6
2
0.161
 0.0929 , which is about 9.29%. Of
the total area enclosed by the triangle, 3 , to get
1.732
the answer choices, the closest one is 10%.
length 1, that wasted space is
21. Any tiles that cover the plane must have interior angles whose degree measure divides 360
evenly. A hexagon (interior angle of 120 ) would suffice, but the next larger factor of 360 is
180, which is not an interior angle of a regular polygon. Therefore, the answer is 6.
22. List the three angles in non-decreasing order. For example, if the smallest angle is 1 , then
the other two angles could be any pair from 1 ,178 to 89 ,90 . The table gives those values:
Therefore, if the smallest angle is odd, the
Smallest Small
Large Number
nd
total number of triangles is 89  86  83  ...
angle
2
2nd
of
30
angle
angle
triples
2   89  2   1365 , and if the smallest
2
pair
pair
angle is even, the total number of triangles
89
1
1 ,178 89 ,90
30
is 88  85  82  ...  1   88  1   1335 .
88
2
2 ,176 89 ,89
2
86
3
3 ,174 88 ,89
Therefore, the total number of triangles is
85
4
1365  1335  2700 .
4 ,172 88 ,88
83
5
5 ,170 87 ,88
23. This triangle contains three right angles,
82
6
6 ,168 87 ,87
so the sum of its interior angles is 270
24. The six points make a hexagon, so we
are looking for the product of the three
diagonals and two of the edges of the
4
…
…
…
58
58 ,64
61 ,61
…
4
59
59 ,62
60 ,61
2
60
60 ,60
60 ,60
1
Theta Circles & Polygons
MA National Convention 2012
hexagon. The longest diagonal is a diameter of the circle, so its length is 2. Both of the other
two diagonals are equal, and since the circle had radius 1, each side length of the hexagon has
3 . Therefore, the product of the lengths of
length 1. This makes the length of the diagonal
the five chords is 1  1  2  3  3  6 .
25. The region we want is shaded yellow. Notice that Region 1 + Region 2 = Region 4. The
areas of each region are as follows:
Region 1:
Region 4:

6
4 
2
42 3 8

4 3
4
3
42 3
4 3
4
Region 3:
Total circle  2  Region 2  7  Region 1  Region 4
 Total circle  5  Region 1  3  Region 4
8
2
 8

   4   5
4 33 4 3 
8 3
3
 3

Total desired area:


 8

3  Region 3  3  Region 1  Region 4  3 
8 3
 3

 8

3 
 4 3   4 3  8  24 3  8  12 3  4 3  16  16 3
 3

26. According to the picture, cos
180 y 2 y

 , so y 
n
x2 x
180
. Thus, for any of the n-gons , to get the side length
n
of the next n-gon , multiply the previous side length by
x cos
cos
180
. The sum of the perimeters, then, is
n
nx  nx cos
180
180
 nx cos2
 ... 
n
n
6x

1  cos30
6x
3
1
2
in parentheses.



nx
. Plugging in n  6 , we get
180
1  cos
n


2 3
12x

 12 2  3 x  12 , since x is the reciprocal of what is
2 3 2 3


5
Theta Circles & Polygons
MA National Convention 2012
27. Since these are equilateral triangles, each next polygon has side length half that of the
previous side length. Therefore, the series is geometric with ratio 1 , and the first triangle
4
 3
encloses an area of
4
2
4
3
3
3
3
4
 4 1.
 . Thus the sum of all enclosed areas is
3
4
1 1
4
4
28. From problem 26, this ratio with n  4 is cos
180
2
 cos45 
.
4
2
7
41
29. By Pick’s Theorem, the area enclosed by the heptagon is 18   1 
, and by the
2
2
Shoelace Method, the area enclosed is
3
1
12
0
2
0
2b
16  2b
2
3
1
3
0
2
a
2
1
1
4
3
1
0
b
1
2
12
3
1
a
3
 A  16  2b   14  2b  a   15  2b  . Setting these
2
2
0
2b
a
14  2b  a
equal, 30  4b  a  41 , meaning either 4b  a  11 or 4b  a  71 . Points on this second line
would violate the extra intersections hypothesis, so 4b  a  11 . The lattice points satisfying
a2  b2  200 are  11,0  ,  7,1 ,  3,2 , 1,3 ,  5,4  , and 9,5 . However,  11,0  would
have the edge connecting it to  2,0  pass through other lattice points (for example,  3,0  ).
 7,1 connected to 2,1 passes through  6,1 ; 1,3 connected to  2,0 passes through
 0,2 ; and  5,4  connected to 2,1 passes through 3,2 . The edge through the point  3,2
2
1
and
, and the edge
1
5
5
through the point  9,5 is OK since the unreduced slopes through the adjacent vertices are
11
7
and . Therefore, two points are possible.
4
is OK since the unreduced slopes through the adjacent vertices are
6
Theta Circles & Polygons
MA National Convention 2012
30. By problems 26 and 28 (which is essentially the same problem as this one), the ratio of the
sides is cos
 180 
180
, so the ratio of the areas is cos2 
.
n
n


7