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March Regional
Geometry Individual Solutions
1. If we translate this philosophy as ~ p ~ q , the contrapositive of this is q  p and the converse of this
is p  q , or “If there is a question whose answer is E, I will place individually.” E.
2. Regardless how the shapes are fitted together, their area remains the same. Thus, we can fit these rhombi
 s2 3  3 2
 4   2 s 3 . A


to form a regular hexagon whose area with side s is A  6 
3. Refer to figure at right. Suppose the radius of the circle
is x. Because the triangle is right, the distance from the
point of tangency to the radius to the right angle must also
be x. From this, the other portions of the sides can be
represented by 20 – x and 21 – x, and because tangents
that meet at the same external point must be equal, the
hypotenuse must be composed of 20 – x and 21 – x, as
shown in the figure. So, 20  x  21  x  29 , and x  6 .
Therefore,
Atriangle  Acircle  .5  21 20     6   210  36 . B
2
4. SA   r 2   rl   r 2   r r 2  h2 , and since


r  h , SA   r 2   r 2r 2   r 2 1  2 . C
5. The longest length of DE occurs when the two triangles are on opposite sides of the line segment.
Incidentally, the sides of the two triangles form DE ; to prove this, mACD  mBCE  60 because of
equilateral triangles, and since they both lie on segment AB they become vertical angles and the segments
EC and DC must coincide on the same line as DE . Thus, DE = 6. D
52  102  102  225  15 . B
6.
7. I and III are false when the triangle inscribed is a right triangle. The larger circle’s center coincides with
the triangle’s circumcenter and not its incenter, so II is false. IV is true since the ratio of the circle’s radii is
2 :1 as proven by the 30-60-90 special triangle rules. Therefore, A
8. Given a set perimeter, the circle maximizes the enclosed area. A
9. For the moment we can divide the trapezoid into two triangles and one rectangle. Since the hypotenuse of
the triangle including the 45 angle is 4, each side must be 2 2 , which is also the rectangle’s height. Thus,

the area of this triangle is .5 2 2
 2 2   4 and the area of the rectangle is  2 2   4  8
2 . The height
of 2 2 is across the 30 angle in the other triangle, making the other leg of length 2 6 ; thus the area is



.5 2 6 2 2  2 12  4 3 , and the area of the trapezoid is 4  8 2  4 3 . C
10. Because the original triangle is equilateral, its tangent circles are of equal radius length and are
equidistant from each other—and thus the triangle formed by their centers is also equilateral. In fact, the
original triangle makes up a fourth of the larger triangle’s area, and its dimensions are half of the larger one’s
(in the case of an equilateral triangle); because the circles are equidistant, point A bisects the larger triangle’s
March Regional
Geometry Individual Solutions
side, as do its equivalents on the other sides. Therefore, we only need the radius of the circle to determine
AB . If we draw a perpendicular from the radius of a circle to its tangency with the triangle, a special 30-6090 triangle is formed, thus, the radius is
3 3
3 3 63 3

. Therefore, AB  3 
.A
2
2
2
11. The area of a triangle given side lengths a, b, and c is A  s  s  a  s  b  s  c  , where s is the
semiperimeter of the triangle. So, s 
567
 9 , and A  9  9  5 9  6  9  7   6 6 . C
2
12. The angle measure in a regular pentagon is 108 . The numbers 106, 107, 108, 109 and 110 are
consecutive natural numbers and have the same sum as 5 108 , and can therefore be the angles of a
pentagon; the smallest of these numbers is 106. D
13. Given the trapezoid is isosceles and since the midsegment m is parallel to its bases, m 
Since Atrapezoid 
1
 b1  b2   12 .
2
1
h  b1  b2  , 60  12h and h  5 . B
2
14. Solving for the cone’s radius,
6
1
  r 2 8 ,
 3
3
 18   1 
2
. Because the right cone’s
 2     r , and r 
2
   8 
circular cross sections’ radii are all proportional to the perpendicular distance to the cone’s apex (its height),
3
6
3
8
, and
 2 , where x is the radius of the sliced cone; 8x  , x 

4
4
x
2
2
6 3
21
1  3 
3
1  3 
6

 
. Since Vlargecone   
.D
  4 
  8   , our answer is 
3  4 
4
3  2 

 4 4
we can create a proportion:
Vsmallcone
15. Since AB  AD , AE  AE , and BE  DE (because of the properties of the kite), we can say that
triangles BEA and DEA are congruent by the SSS theorem, and we can also say that the area of triangle BEA
is half that of BAD. Therefore, the ratio of the areas of triangle BCE and BEA is 5: 3 . Suppose BE  x and
4x
ABCE 5
12
4
AE  y . Then,
  2  , and y  . C
5
ABAE 3 xy
y
2
16. We can treat the handshakes as the number of diagonals and sides of a regular polygon, and add
Werther’s extra handshakes afterwards. By deductive reasoning, since 3 sides yields 3, 4 yields 6, 5 yields 10
and etc, we can describe this pattern as
n  n  1
15 15  1
 15  7   105 . Then,
for n  2 . With n  15 ,
2
2
to add Werther’s additional 14 handshakes (since he presumably does not give himself a handshake),
105 14  119 . C
March Regional
Geometry Individual Solutions
 
17. By the tangent-secant theorem,  7   3 JL , and JL 
2
49
40
40
3 
. Since C   d , C 
.B
3
3
3
1
18. mNLJ  110  46   32 . B
2
49
. Therefore, JM  JL  ML
3

19. If we bisect a 60 angle, we end up with angle measures of 30 , and then if we repeat this process we
get 15 . Similarly, bisecting a 45-degree angle gives us 22.5 , and repeating the process gives us 11.25 .
Because 26.25  15 11.25 , 33.75  22.5 11.25 , 37.5  22.5  15 and 67.5  45  22.5 , our answer is
E
20. Triangle ABC is proportional to triangle BDC , which is proportional to both triangle BED and
triangle DEC . We can say that
9
8
6
, and thus DC  . Using the Pythagorean Theorem to find

2
6 DC
2
3 7
9
the length of BC , we get BC   6     
. Finally, setting up one more proportion, we
2
2
2
BC DC
have
,

DC EC
3 7
9
2  2 , and EC   81   2   27 7 . D
 

9
14
EC
 4  3 7 
2
21. The intersection of the altitudes and the intersection of the perpendicular bisectors of a triangle, or
otherwise the orthocenter and the circumcenter respectively, can exist outside the perimeter of the triangle if
the given triangle is obtuse. Only the centroid and the incenter, or the intersection of the medians and angle
bisectors respectively, exist solely within the triangle for all cases. Therefore, B
22. Setting up the proportion between the man and the lamppost,
from the lamppost), 24  6x 12 and x  2 . B
12 x  2

(where x is how far he is
6
2
23. If the angle is not divisible by 180, then it cannot be the sum of the interior angles in a regular polygon.
Since
1080
3420
7380
6940
 6,
 19 and
 41 , but 38 
 39 , the answer is C
180
180
180
180
24. A triangular prism capped with tetrahedrons has faces composed of three rectangles and six equilateral
triangles. These rectangles make up an area of 3 10 4  120 square units. The six equilateral triangles
 42 3 
have area of 6 
  24 3 square units, and so the total surface area is 120  24 3 . B
4


25. The shape described is a rectangle, and rotated about the given line gives us a cylinder of radius 2 and
height of 6. Therefore, our volume is V   r 2 h    2   6   24 . B
2
March Regional
Geometry Individual Solutions
26. A is false, since two skew lines do not like in the same plane. C is false, as the vertices of a triangle prove
this. D can be false if one line lies on the xy-plane and the other on the xz-plane; both can cross the x-axis at
the same place and both can have the same slope, but they are not the same line. B is true, since all sets of
three points determine some plane, and thus three points must be coplanar. Therefore, B
27. By corresponding angles and by vertical angles, we can prove that angles 1, 2, and 3 form a 180 angle.
Therefore, our answer is 180  30  150 . B
28. Since mCAB  m2 , and since the other corresponding angles of the triangles are the same
(as a result of parallel lines), the two triangles are proportional by the AAA theorem. Thus,
4
AB 6 2
  , and the ratio between the areas is therefore . A
9
AE 9 3
29. Because the lines are parallel we can conclude that m1  mABC  30 . Drawing the height
perpendicular to base AB , we can say A  .5bh  .5  6  h  3 7sin  30   21 . A
2


30. Choice A gives us SAS, choice B gives us SSA, choice C gives us SAA, and choice D gives us
ASA. Of these, SSA does not prove congruency. B