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March Regional
Geometry Team Solutions
1. The star’s angles trisect of the pentagon’s angles; since the sum of a pentagon’s angles is 540 , the sum of
the angles of the five-point star must be A  1
3
540   180 . Drawing the six-point star reveals to us that
it is formed by two overlapping equilateral triangles, and so each angle is 60 ; thus, B  360 . Therefore,
B
A
 360
180
 2.
2. The smallest integral length that allows the given triangle to exist is 6, and since 182  132  62 this
triangle is obtuse; thus, k  6 . The first integer to reverse this inequality is 13, as 182  132  132 but
182  132  122 ; thus, j  13 . If the given triangle is right, the missing leg can either be the hypotenuse
(where l 2  324  169 ) or one of the legs (where m2  324  169 ). Therefore, j  k  l 2  m2
 13  6   324  169   324  169  19  648  667
3. A is the area of a circle with radius 10, so Area  100 and
A  100 . For B and C, refer to the picture at right, where B is the
larger area outside the rectangle while C is the area within it. For B,
three-fourths of the area of A is not obstructed by the cage, so we
have an initial area of
3
100   75 . A remaining radius of 5 is
4
yielded since the cage’s side is shorter than the length of the panda’s
rope, so the panda can roam an additional quarter circle with radius
1
25
25 300 325


; thus, Area 
and B  325 . For C, starting from the
 25  
4
4
4
4
4
longer side it is attached to, the circular sector it can roam is obstructed by the other side when it makes a 30
1
25
100 
angle with the originating side, as shown in the picture. The area of the sector is
. The panda
12
3
25 25 3
25 1
  5 5 3 

can also roam the triangle enclosed by the sector, thus Area 
3
2
3
2
50  75 3

, and C  50 and D  75 . Therefore, A  B  C  D  100  325  50  75  550
6
5, or otherwise

4. The circle whose equation is given has radius

25  5 and center C  0,1 , so A  5  0 1  6 . The
 0 1 5 0  4  7 
,
   2, 1 and so B  2  1  1 . For C,
3
 3

centroid of the given triangle has coordinates 
C
5  2  5 1
2
2
 25  5 . Therefore, A  B  C  6 1  5  12 .
5. The sum of the exterior angles of a regular polygon, regardless of how many sides it has, is 360 , so
A  3  6  0  9 . The largest regular polygon described in part B is a hexagon (since it is made up for
equilateral triangles, which can also be tessellated without the use of any other shape), so B  6 . For C, the
answer has to be greater than one distinct angle, and the three lines can all intersect to form both 60 and
120 angles, so thus C  2 . The equation that relates vertices, faces and edges is V  F  E  2 (e.g., for a
cube, 8  6  12  2 ), and so if we have eight edges and six faces we must have V  E  2  F
 8  2  6  4 vertices. Therefore, the answer to the question is 9624 .
March Regional
Geometry Team Solutions
6. First we need to find the length of the slope the wheel will travel. Using trigonometry and if x is the length
   100 x and x  200 . Since the wheel travels one length of its circumference in
of the slope, then sin 30
    10 . Therefore, since 20010  20 , the
one revolution, we need to find its circumference; C  2 5
wheel will make exactly 20 revolutions before reaching the bottom of the hill.
7. For x, each side must be 15 units, and so x  152  225 . For y, a rhombus must have the same side
lengths, so each side must be 15 units; its diagonals must also be perpendicular to each other. Dividing the
rhombus into four right triangles, the legs must also be in the ratio 4: 3 since the rhombus’s diagonals are of
this ratio; the leg lengths that match a hypotenuse of 15 are 9 and 12, and so the rhombus’s area is
1
18 24    9  24   216 . For z, the rectangle’s side lengths that match the ratio 7 :8 and have
2
perimeter of 60 are 14 and 16, and so z  1416   224 . Therefore, x  y  z  225  216  224  665
y
8. From the information given, the radius of the circle is 4 units. Drawing this radius to the circle’s tangency
with the triangle, and drawing a line segment from the center to the triangle’s vertex, we can determine that
half the triangle’s side length must be 4 3 by the 30-60-90 triangle. Thus, this triangle’s perimeter is 24 3 .
9. Since the perimeters (and consequently the side lengths) of the similar triangles are at ratio of 4 : 5 , their
areas must be at ratio of 16 : 25 . We can set up two proportions; solving the first,
Solving the second,
16 400

, and x  625 .
25
x
16
y

, and y  256 . Thus, x  y  625  256  881 .
25 400
10. If we draw line segments between these three points, all of the perpendicular bisectors of these segments
will intersect at the circle’s center. Therefore, if we find where two of these bisectors meet, we can determine
the center. The perpendicular bisector of the last two points has slope of 1 and passes through  3,3 , and the
perpendicular bisector of the first and last point has slope 2 and passes through  2,1 . Therefore, the
equations of these lines are x  y  0 and 2 x  y  5 , respectively. By solving the system their intersection

is at 5 , 5
 , and if we measure the radius from the center and the point  0, 0 we can conclude the radius
3 3
5 2
must be
.
3
11. This triangle cannot exist since its two shorter sides are not longer than the third, and so it therefore
cannot have a perimeter of 21, and it cannot be scalene or obtuse (though it would if the triangle could exist);
therefore, A  0 . B is true unless the two lines lie on different planes, in which case they would be
skew and not parallel; therefore, B  0 . For C, we can use deductive reasoning to determine a
formula to give us our answer, since it may be hard to accurately draw the six lines; 1 line makes 2
regions, 2 lines make four regions, 3 lines make seven regions, and four lines make 11 regions.
Continuing this pattern, five lines make 16 regions and six lines make 22 regions; therefore, C  1 .
D is true; the altitude to any base divides the triangle into two smaller and equal triangles, and is
therefore both the median, perpendicular and angle bisector at the same time; therefore D  1 . For E,
consider that the exterior angle of is 180  144  36 . Since 360 10  36 , a ten-sided polygon must
March Regional
Geometry Team Solutions
have interior angle measure of 144 ; therefore E  1 . F is true, since a rectangle does not have the
property of having perpendicular diagonals unless its sides are of equal length; therefore F  1 .
Finally, for G, in the case described, two consecutive angles are supplementary, not two opposite
angles; therefore, G  0 . Thus, A  B  C  D  E  F  G  4 .
12. Consider a rectangle of side lengths 3 and 4—or simply two right triangles with legs 3 and 4 connected
by their hypotenuses (which we know to each be 5). The diagonals of the rectangle (the hypotenuses in this
case) bisect each other, and so if we look at the case of one of the triangles, the median must therefore be
equal to half of the hypotenuse, so median  5 .
2
13. A kite is composed of four right triangles, whose sides must be of integral length as required by the
problem; its shorter diagonal must also be bisected by the larger diagonal. When the diagonals of the kite are
16 and 21 the kite’s area is 168, and by dividing the longer diagonal into 15 and 6, we can form four right
triangles (two having lengths of 6, 8 and 10; two having lengths of 8, 15 and 17); when the diagonals of the
kite are 24 and 14 the kite the kite’s area is also 168, and by dividing the shorter diagonal into 9 and 5, we can
form four right triangles (two having lengths of 5, 12, and 13; two having lengths of 9, 12, and 15). Therefore
our conditions are satisfied; the first kite’s perimeter is 54 (the sum of the four triangles’ hypotenuses), while
the second kite’s perimeter is 56; thus, our answer is 54  56  110 .
14. It is easiest to approach this problem in 2D, since the circumference of the sphere’s great circle is the
circumference of the sphere’s cross section. Given the circumference in A, the radius of the circle must be 4
units; by the 2D cross-section the square’s length must be 8, and so the volume of the cube must be 83  512
cubic units. For B, a cube with an eighth of the cube in part A will have volume of 512  64 cubic units,
8
and so each side must be 4 units. Using the cross section once more, the radius of the circle must be 2, and so
3
the volume of this sphere is 4   2   32
3
3
, so B  32 and A  B  512  32  544
n  n  3
, while the second series can be described
2
 n  2  2n  2   n  2 n  1 . Therefore, A  100 100  3  5150 and
by the formula

 
2
2
B  100  2100  1  10302 , and so A  B  5150 10302  15452 .
15. The first series can be described by the formula