March Regional
Geometry Individual Solutions
Individual Solutions:
a 3
a2 3
1. Height of the equilateral triangle is
. So base*height/2 is
=> C
2
4
2. The third statement lets us know that if life as we know it isn’t real, then the Basic Math Otter
does a dance. The contrapositive of the second statement makes B true also. => A&B are true , so
E
3. The area of the smaller circle is 9π, the area of the larger circle is 25π. The probability of
randomly picking a point inside the smaller circle is
9π
9
=> B
=
25π 25
4. The points are (0,13), (13, 0), (-13,0), (0,-13), (5,12), (12,5), (-5,-12), (-12,-5), (-5,12), (12,-5),
(5,-12), (-12,5)=> B
5. Let’s call the midpoint of AB to be point P. Treat triangle CBP as a right triangle with right
angle CBP. To find M, the length of CP will be M. CB =
2 ; PB =
1
. Using the Pythagorean
2
3
2
theorem we find CP=M= . To find N, we treat the two faces Eli must cross to be a rectangle 1
unit tall and 2 units wide. CB=2; PB =
17
.
2
M+N =
1
. Using Pythagorean theorem gives the answer CP to be
2
3 + 17
=>C
2
6. Only (12, 83, 85) is not a Pythagorean triple. This can be confirmed with either the
Pythagorean theorem, or (2–m2, 2mn, n2+m2). => D
7. B
8. If p, then q. The converse is “If not p, then not q.” => B
9. For D to be a two-digit number, the radius of this circle must be an integer that falls in the
interval [5,9]. Testing each of these integers, we find the radius of the circle to be 9, so A is 81
and D is 18. 8+1 = 9 => B
10. Polygon ABCDE is a concave polygon, so the areas of ABCF and CDEF must be solved for
separately. Applying shoelace theorem, the area of ABCF is found to be 10 and the area of CDEF
is found to be 6. 10+6=16 => D
11. (CP)2+(PT)2=(PU)2+(PE)2
PE= 7 + 24 − 20 =15=>D.
12. Draw a perpendicular to PC from point O, the foot of this is X. PO = 4, PX=2, and so by
Pythagorean theorem OX = 2 3 . This is a 30-60-90 triangle, so the major arc BC is of measure
2
2
2
240 degrees, two thirds of a circle. So the measure of the area of major sector BC is
2
(9π )= 6π .
3
Minor arc AD has length 120, because PB and OA are parallel, and PC and OD are parallel, so
1
π
π = . The area remaining is the sum of two trapezoids:
3
3
1
19π
2 • • (1+ 3) • (2 3) = 8 3 , making the total area
+ 8 3 => B
2
3
the area of minor sector AD has area
13. The perimeter of the figure is found by summing two arc lengths with the two external
tangents. The arc length of the small circle is one third of the total perimeter of the circle (see
March Regional
Geometry Individual Solutions
solution to question 12) and the arc length of the larger circle is two thirds of the total perimeter
of the larger circle (see solution to question 12). The length of the two external tangents are both
2 3 (see solution to question 12).
1
2
14 π
(2π ) + (6π ) + 2 • 2 3 =
+ 4 3 =>C
3
3
3
14. Draw the octagon in the square so that four of the sides of the octagon lie on sides of the
square. Assign the side length of the octagon to be x.
8− x
x
. So x = 8 2 − 8 . The area of the
=
2
2
octagon is the area of the square less the area of the four triangles in the corners. The total area of
8 − (8 2 − 8) 2
) = 192 −128 2 . The area of the square is 64, so the area
2
of the octagon is 64 − (192 −128 2) = 128 2 −128 => A.
these four triangles is 2(
15. Let a, b, and c represent the dimensions of this rectangular prism. a + b + c = 10 .
2
2
2
2
2
2
2
So a + b + c = 10 . 2ab + 2bc + 2ac = 30 . (a + b + c) = a + b + c + 2ab + 2bc + 2ac . So
(a + b + c) 2 = 10 + 30 = 40 => B.
2
16. MN =
2
2
(AM • DC + MD • AB)
(AM(2AB) + 2AM(AB)) 4 AB
=
=> 8 =
. The length of AB
AD
3AM
3
is 6, so the length of DC is 12. 6+12 = 18 => B
17. Let the point of intersection of the diagonals be E, and the length of DE to be x. Using angle
bisector theorem,
x 15 − x
6
9
. So x=6. Using angle bisector theorem again,
=
= . So DC =
8
12
DC 10
20
=> A
3
18. Lisa needs to cut the pizza so that if five OR six people show up, each can receive the same
amount of pizza. She can do this by cutting six pieces, and dividing the 6th slice into fifths. It
takes three cuts to make the six pieces, then four more to divide the 6th piece into fifths. Thus,
seven cuts overall => A.
19. A cube has 9 planes of symmetry => D
20. There are 4 ways Niral-bug can climb up only crawling along two edges, 8 ways it can climb
up only crawling along three edges, 8 ways it can climb up only crawling along four edges, 8
ways it can climb up only crawling along five edges, and 8 ways it can climb up only crawling
along six edges. Overall = 36 => B.
21. Sum of the interior angles of a convex n-gon is (n − 2) •180 . (n − 2) •180 => n = 9 => C
22. Euclid is often referred to as the “Father of Geometry” and wrote Elements. => A
36
144
BD 9
36 DE BD DE
23. Observe similar triangles.
.
.
=> C.
= . BD =
=
= 5 . DE=
25
12 15
5 12
15 12
15
March Regional
Geometry Individual Solutions
24. The exterior angle must be a factor of 360 that is less than or equal to 120. There are 21
factors of 360 that are less than or equal to 120. => B.
25. The circle is centered at (0,0) and has a radius 9. Let the center be point O, the point
tangent to the circle be point A and point (15,0) be point B. OAB forms a right triangle
with one leg of length 9 and the hypotenuse of length 15. Thus, the second leg is of
length 12. Draw a line from A to line OB perpendicular to point OB and let that point of
intersection be point C. The triangles OCA and OAB are similar. So the tangent of angle
−1 −3
4
OAC is . The slope of the line tangent to the circle will thus have a slope of
= .
4 4
3
3
−3
−3
45
Using point-slope formula the line is found to be y − 0 = (x −15) => y =
=>
x+
4
4
4
D.
26. They are all constructable. => E
27. The quadrilateral is cyclic because the diagonal BD splits the quadrilateral into two right
triangles. Using Ptolemy’s theorem, we find that the second diagonal is 4 + 3 3 => C
28. Diagram is shown below. Triangles MAR and PAQ are similar.
MR 5
65
= . So MR = .
13 8
8
The altitude of triangle PQR to side PR is 12. Drop an altitude of triangle PMR to side PR.
65
15
MY MY
Because triangle RMY and triangle PQX are similar,
=
= 8 . So MY = . The area
QX
12
13
2
15
1 75
•10 • = => A
of PMR is then just
2
2 2
29. The two circles and tangent are shown below. Notice there is a right triangle with legs 11 and
60. The hypotenuse of this triangle is 61. 61-13-24=24 => D
30. The cross section of this figure shows that the altitudes of the triangular faces and the base
form an isosceles triangle that is inscribed around a circle. The sides of this isosceles triangle
March Regional
Geometry Individual Solutions
3
3
2
,
, and 1. The altitude of this triangle is
, so the area of this triangle is
2
2
2
r 3 r 3 r
+
+ =
If the radius of the sphere is r, the area of the circle can also be written as
4
4
2
6− 2
Solving, r =
=> D.
4
measure
2
.
4
2
.
4