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March Statewide Invitational
Geometry Team
Question
Part A
Part B
Part C
Part D
1
.
x=15
Scalene and
acute
250
200
2.
(-3,-2)
(𝑥 + 3)2 +
(𝑦 + 2)2=85
A=85π
3.
94
66
12
16
4.
congruent and
parallel
Perpendicular
bisectors, with a
ratio of 2 to 1
5.
60
a=134
b=14
240
6.
70
120
Alternate
interior angles
2
7.
5
10
75
161
8.
150
135
45
60
9.
30
150
88
34
10.
Reflexive Prop.
SAS
CPCTC
Def. of
Seg.Bisector.
11.
X’(1,3),
Y’(-3,5) Z’(-4,2)
X’’(-1,3),
Y’’(3,5),
Z’’(4,-2)
D’(-1,4)
E’(-3,-2)
F’(2,-4)
D’’(4,1)
E’’(-2, 3)
F’’(-4,-2)
12.
1/49 or about
2%
25/49 or about
51%
24/49 or about
49%
second scoring
zone
13.
4
1
1
3
14.
5
7
8
578
2√85
2√13
rhombus
Solutions:
̂ = 𝑚𝑄𝑅
̂ + 𝑚𝑅𝑃
̂ =
1. A. 6x+10x+10+8x-10=360; x=15 B. scalene and acute C. 𝑚𝑄𝑅𝑃
̂ = 𝑚𝑄𝑃
̂ + 𝑚𝑄𝑅
̂ = 110 + 90 = 𝟐𝟎𝟎
160 + 90 = 𝟐𝟓𝟎 D. 𝑚𝑄𝑃𝑅
(−10+4) (−8+4)
2. A. 2 , 2 ; center (-3,-2) B. √(−10 − 4)2 + (−8 − 4)2; d=𝟐√𝟖𝟓 C. (𝒙 +
𝟑)𝟐 + (𝒚 + 𝟐)𝟐 =85 D. A=πr2 A=85π
March Statewide Invitational
Geometry Team
̂ − 𝐵𝐷
̂ =226, 226-x=y, 226̂ ) =m<A, (134-x)=40 x=94; B. 360-134= 𝑚𝐵𝐷𝐶
3. A. 1/2(𝐵𝐶
2
94 =132, ½(132)=y y=66 C. 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = 𝑜𝑢𝑡𝑠𝑖𝑑𝑒(𝑖𝑛𝑠𝑖𝑑𝑒 + 𝑜𝑢𝑡𝑠𝑖𝑑𝑒) = (2𝑝)2 =
(2𝑝 − 3)(𝑝 + 1 + 2𝑝 − 3); 4𝑝2 = 6𝑝2 -13p+6 after solving p=6, because p is an integer.
AB=12 D. AC=AD+CD=16
4. A. AB= √65r, CD=√65, mAB=-7/4, mCD=-7/4; AB and CD are congruent and
parallel B. BD=√(−9 − 3)2 + (6 + 2)2 = 4√13;CA=√(−1 + 5)2 + (5 + 1)2 = 2√13
because Quadrilateral ABCD is a rhombus segments BD and CA are perpendicular
bisectors, with a ratio of 2 to 1 C.CA =𝟐√𝟏𝟑 D. segments BC and AD are parallel and
measure rad65, therefore triangle ABCD is a rhombus
5. A. Major arc NM=240, Minor arc NM=120, 1/2(240-120) =60 m<P=60, B. 1/2(â + 𝑚𝐴𝐵
̂ =240
b)=60, (a-b)=120, a+b+212=360 solve the system a=134 C. b=14 D. 𝑚𝐴𝑋
6. A In the following diagram, two additional (dotted) lines are drawn to illustrate the
properties of the transversal line intersecting a pair of parallel lines. With AY and LO as
the transversal line, alternate interior angles are congruent (at 30° and 50°). Since , the
remaining upper part of this angle will be 90 – 50 = 40°. Since alternate interior angles
are congruent with transversal lines (LY), the measure of angle AYL is 30 + 40 = 70°.
B. 70+50=120
C. alternate interior angles
D. 2
7. A. first 2 square numbers are 1 and 4, 1+4=5, B. first 3 triangular numbers are 1,3,6,
1+3+6=10 C. first 5 pentagonal numbers are 1,5,12,22,35, 1+5+12+22+35=75 D. first 6
hexagonal numbers are 1,6,15,28,45,66, 1+6+15+28+45+66=161
8. The sides of a regular dodecagon divide its circumcircle into 12 equal arcs. Therefore
each of these small arcs has a measure of 360/12=30. A. Angle ABC is inscribed in the
arc that consists of 10 of these smaller arcs. Therefore the major arc AC has measure
10(30)=300, so angle ABC =300/2=150 B. angle ACD is inscribed in an arc that consists
of 9 of these little arcs, so it has a measure (9*30)/2=135 C. angle ADJ is inscribed in an
arc that consists of three of these little arcs JK, KL, LA, so it has a measure (3*30)/2=45
D. The acute angles between the chords each equal Arc (AB+FI)/2. Since arc AB =30
and FI consists of 3 of little arcs of measure 30 or angle has measure (30+3*30)/2=60
March Statewide Invitational
Geometry Team
̂ = 300, 𝑚𝑌𝑍
̂ ≅ 𝑚𝑍𝑊
̂ =
9. Given: 𝑚 < 𝑋 = 150, ̅̅̅̅
𝑌𝑍 ≅ ̅̅̅̅̅
𝑊𝑍, 𝑚 < 𝑌 = 92.; 𝑚𝑌𝑍𝑊
̂
̂ = 60, 𝑚𝑋𝑊𝑍
̂ = 184, 𝑚𝑋𝑌𝑍 = 176, 𝑚 < 𝑤 =
̂ =
150, 𝑚𝑊𝑌
88, 𝑚 < 𝑧 = 30, 𝑚𝑋𝑊
̂
34, 𝑚𝑋𝑌 = 26
A. 30
B. 150
C. 88
D. 34
10. A. Reflexive Property B. Side Angle Side (SAS) C. Corresponding Parts of
Congruent Triangles are Congruent (CPCTC) D. Definition of Segment Bisector
11. A. Translations move the pre-image in the direction of the vector 2 left and 1 up.
X’(1,3), Y’(-3,5) Z’(-4,-2) B. Reflection moves the image over the y-axis, switching the
sign of the x value of the ordered pair. X’’(-1,3), Y’’(3,5), Z’’(4,-2) C. Reflection across
y= -x of the pre-image switches the x and y values and changes the sign. D’(-1,4), E’(-3,2), F’(2,-4) D. rotation -270 of the image is the same as a 90 degree rotation. The x and y
values are switched and the new y value changes signs. D’’(4,1), E’’(-2, 3), F’’(-4,-2)
12. Probability is equal to area of the zone divided by total area. Area of target is 49π.
A. Area of first zone is 1π/49π =1/49 or about 2%.
B. Area of forth zone is 49π-25π(zones 1-3). Add to area of first zone=25π. Probability
=25/49 or about 51%.
C. Area of third zone is 25π. Subtract first zone =24π. Probability is 24/49 or about 49%.
D. D. Area of second zone is 9π minus area of first= 8π. Probability of hitting second
zone is 8/49 or about 16%.
13. A. (3,4,5), (25,312,313), (21,220,221), (5,12,13), = 4 B. (17,144,145) =1 C. (8,15,17)
=1 D. (16,63,65), (15,36,39), (24,45,51) =3
14.
A parallelogram is a rectangle. (s)
A parallelogram is a rhombus. (s)
A parallelogram is a square. (s)
A parallelogram is a trapezoid. (n)
A rectangle is a parallelogram. (a)
A rectangle is a rhombus. (s)
March Statewide Invitational
A rectangle is a square. (s)
A rectangle is a trapezoid. (n)
A rhombus is a parallelogram. (a)
A rhombus is a rectangle. (s)
A rhombus is a square. (s)
A rhombus is a trapezoid. (n)
A square is a parallelogram. (a)
A square is a rectangle. (a)
A square is a rhombus. (a)
A square is a trapezoid. (n)
A trapezoid is a parallelogram. (n)
A trapezoid is a rectangle. (n)
A trapezoid is a rhombus. (n)
A trapezoid is a square. (n)
100A + 10B + C = 100(5) + 10(7) + (8) = 578
A. 5
B. 7
C. 8
D. 578
Geometry Team