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GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
1. S1 : If an angle is right, then it measures 90 degrees.
S 2 : If a triangle is equilateral, then it is an acute triangle.
S 3 : If two lines are parallel then they do not intersect.
S 4 : If a triangle is scalene then none of its interior angles are congruent.
S 5 : If a triangle is right, then it has two acute angles.
A = the number of statements above (out of S1 through S 5 ) which are true.
B = the number of statements above (out of S1 through S 5 ) which have true converses.
C = the number of statements above (out of S1 through S 5 ) which have true inverses.
D = the number of statements above (out of S1 through S 5 ) which are not true, but their
converses are true.
----------------------------------------------------------------------------------------------------------------------------E
2. F, G and H are collinear;
J, K and L are collinear;
F
G 3x  1
H
M, N and P are collinear;
G, K and N are intersection points
of transversal EQ and the respective
J
K x  39 L
lines FG, JK and MN . mHGK  3x 1 ,
mLKN  x  39 and mKNP  y  12 ;
y  12
all measures are in degrees.
M
N
P
Q
A = the value of  x  y  if FG, JK and MN are all parallel.
B = 2  mJKN   mLKN if FG, JK and MN are all parallel.
C = mMNK  mGKL if FG, JK and MN are all parallel.
D = the value of y if mKNM  2  x  39 and FH JL .
----------------------------------------------------------------------------------------------------------------------------3. RST is isosceles and RS=ST. mS  3x and mT  x .
A = the measure of the largest interior angle of RST .
PQR is isosceles and PQ=QR. mP  3 y and mR  y  10 .
B = the measure of the largest interior angle of PQR .
FGH is isosceles and FH=GH. GF=10 and GH=12.
C = the perimeter of FGH .
D = the area of FGH .
----------------------------------------------------------------------------------------------------------------------------
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
4. P and Q are points of tangency for external tangent
PQ , in circle R and circle T as shown. The circles
R and T are externally tangent at S.
P
Q
RP=10, TQ=5.
R
A = the length of PQ .
B = the number of distinct external common tangent
lines that can be drawn to circles R and T.
C = the ratio of the circumference of circle T to the
area of the circle with diameter RS .
D = the distance from point S to PQ (the length of the perpendicular
S
T
segment from S to PQ )
---------------------------------------------------------------------------------------------------------------------------P
5. Triangles PST and
QRS are right as
Q
12
R
shown. P, R and S
are collinear. PR=RS.
S
T
QR=12. PS=10. PT= 5 13 .
A = the value of k if the distance from Q to T is
B = the length of ST .
tan(T )
C = the ratio of
in fraction form.
sin(Q)
k.
D = p  q if the perimeter of quadrilateral QPTS (not drawn) is p  q .
-----------------------------------------------------------------------------------------------------------------------------6. EFK ~ FHL ~ HJM .
E, F, H and J are collinear and
EF=6, FH=4, and HJ=3. EK=10.
A= the perimeter of FHL .
sin(KEF )
B = the ratio
.
sin(LFH )
C = FL – HM, in simplified improper fraction
form (no mixed numbers).
D = HL – JM in simplified improper fraction
form (no mixed numbers).
E
10
6
F
K
4
H
3
J
L
M
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
7. Right triangle PQR has right angle Q.
If QS is the altitude to the hypotenuse
P
of PQR , and S is on PR , then
PS
A= the ratio of
.
SR
8
If QT is the angle bisector of Q
in PQR , and T is on PR , then
PT
B= the ratio of
.
TR
Q
10
R
If QS is the altitude to the hypotenuse
of PQR , and S is on PR , then
C= the length of QS .
If PU is the median to side QR in PQR , with U on side QR ,
D=the perimeter of PUQ .
----------------------------------------------------------------------------------------------------------------------------8.  R is twice the degree measure of the complement of S .
mR  mS  150 in RST . TR=10.
A = the degree measure of angle T in RST .
B = the measure of an exterior angle at  R in RST .
C = TS + SR.
D = the length of the longest altitude of RST .
---------------------------------------------------------------------------------------------------------------------------9. Regular polygon P1 has n sides and one interior angle measures 178 degrees.
A = the number of degrees of one exterior angle of a regular polygon with (n  80) sides.
Regular polygon P2 has interior angles with a total of 9000 degrees and has m sides.
B = the number of distinct positive integer factors of m.
Regular polygon P3 has one interior angle with measure 100 degrees more than one
exterior angle.
C = the sum of the total degrees for the interior angles of polygon P3 .
Regular polygon P4 has one interior angle that is ( s  134) degrees and it has s sides
for s  12 . D is the value of s .
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
10. SUR has vertex S on the origin. U on point (8, 0), R on (3, 6) and T, the foot of
the altitude from R to SU is at (3, 0).
y

The perimeter of SUR is 8  3 p  q
for prime numbers p and q .
A = pq.
R(3, 6)




B =the area of SUR .


x
An equation of line RM , for RM the median
to SU , is y  mx  C for slope m and
y-intercept C . Give the value of C .

S

T(3,0)






U(8, 0)


y

The centroid of SUR lies on RM .
D = the sum of the coordinates of the
centroid of SUR .
R





S
U




M


x





---------------------------------------------------------------------------------------------------------------------------not drawn to scale.
11.
P
Q
R
4
60
S
T
U
V
W
Four congruent circles have centers S, T, U, and V and are shown drawn with tangent
segments RX and PQ at points of tangency R, P and Q. RX=4, and radii are each 3.
Circles are each 0.5 apart (that is, UV=6.5). P, Q and R are on circles T, U and V
respectively.
A=
B=
C=
D=
the length of WX .
the area of trapezoid PQVT.
the area of the 60 degree sector shown in circle S.
the area of QTX (not shown).
X
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
F
12. HF and HL are secant segments as shown,
with G on HF and K on HL . Chords GL
12
and KF intersect at J. mFML  200
HG=10, GF=12, HK=8.
o
G
M
10
J
200 o
H
A = the length of KL .
8
B = the number of possible integer values
K
that length of KF may have, by the Triangle
L
Inequality Theorem.
C = 2(mH )  mGK .
D = mF  mL .
------------------------------------------------------------------------------------------------------------------------------13.
Figures are
not drawn
to scale.
Angles are
measured
in degrees.
4y
h
Polygon P
Polygon Q
p
2y
Polygon R
Polygon P has eight sides, four angles of measure 120 degrees, with the other four
angles congruent.
A = the degree measure of the largest angle of polygon P.
Polygon Q has six sides, four angles of measure 125 degrees, with the other two angles
congruent.
B = the degree measure of the smallest angle of polygon Q.
Polygon R has five sides, with angles measures 95o , 95o , (2 x  10)o , (2 x)o and (3x  10)o .
C = the difference of the degree measures of the largest and smallest angle of polygon R.
Polygon Q and polygon R share one vertex V as shown, but not labeled.
7
If the angles of each were to be changed so that h would be
the degree measure of
6
7
that of a regular hexagon, and p to
the degree measure of that of a regular pentagon,
6
D = the value of y, with angles shown above around vertex V measure 2y, 4y, p and h.
Write D in simplified improper fraction form (no mixed numbers).
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
14. A: Given that square RSTU (not shown) has area 100, A = the length of diagonal RT .
B: Equilateral triangle PQR (not shown) has perimeter 12. It is inscribed in circle K.
An equilateral triangle is circumscribed about the same circle. B = the area of the
circumscribed triangle.
M
C: Rhombus MPQN has perimeter 52. MQ=24.
C = the area of rhombus MPQN, shown.
D = tan(NMQ)  tan(MQN ) from rhombus MPQN shown.
N
P
Q
----------------------------------------------------------------------------------------------------------------------------May not be
drawn to scale.
R
15. Quadrilateral RSUT is inscribed in the circle shown.
Angles S, R, T and U have degree measures
mT  2x , mR  x  60 , mS  x  2 y
and mU  2x  30 .
A = the value of x.
B = the value of y.
C = the degree measure of minor arc RU .
D = 0 if SRT is acute;
D = 1 if SRT is obtuse;
D = 2 if SRT is right.
x  60
T
2x
x  2y
S
2 x  30
U
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
ANSWERS!!
1
2
3
Part A
5
129
108
Part B
2
301
150
Part C
2
62
34
4
10 2
2
5
754
15
6
16
1
7
16
or 0.64
25
30
4
or 0.8
5
60
6
10
18
or 3.6 or
5
3
3
5
66
2
or 0.4
5
26
15
5
3
40 41
41
10  10 3
1260
24
24
11
2
117
or 29.25
4
1
or 29
4
15
3

2
200
0
110
70
47
3
5
or 0.83
6
2
8
9
12
13
39
2
1
or 19
2
150
19.5 or
14
10 2
16 3
120
15
30
45
120
Part D
0
50
5 119
20
3
366
4
3
13  89
Notes
Part C must be in
reduced fraction
form
Part C, D must
be in reduced
fraction form
5 3
36
2
17
or 5.6 or 5
3
3
27
Part D must be in
reduced fraction
form
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
SOLUTIONS
1. S1 : If an angle is right, then it measures 90 degrees. TRUE
S 2 : If a triangle is equilateral, then it is an acute triangle. TRUE
S 3 : If two lines are parallel then they do not intersect. TRUE
S 4 : If a triangle is scalene then none of its interior angles are congruent. TRUE
S 5 : If a triangle is right, then it has two acute angles. TRUE
Converses: S1 TRUE, S 2 FALSE (consider the triangle with angles 70, 80 and 30),
S 3 FALSE (skew lines), S 4 TRUE, S 5 FALSE (all triangles have two acute angles)
Inverses: S1 TRUE, S 2 FALSE (consider the triangle with angles 70, 80 and 30),
S 3 FALSE (skew lines), S 4 TRUE, S 5 FALSE (all triangles have two acute angles).
A = 5. B = 2. C = 2. D = 0, since all originals are true.
2. A, B, C: 3x 1  x  39 . x=20. Acute angles are 59 degrees. Obtuse angles 121. y+12=121
so y=109.
A = the value of x  y , 20+109=129.
B = 2  mJKN   mLKN = 121(2)+59=180+121=301.
C = mMNK  mGKL = acute-obtuse=59 – 121 = -62.
D = the value of y if mKNM  2  x  39 and FH JL . x is still 20 so
mKNM  2  59  118 . y+12=180-118=62. y=50.
3. A: S is the vertex angle, so 3x+x+x=180. 5x=180. 10x=360. x=36.
Largest angle is 3(36)=108.
B: Q is the vertex angle. 3y=y+10. y=5. Angles are 15, 15, 150. Largest is 150.
C: H is the vertex angle. GH=FH=12. Perimeter is 12+12+10=34.
1
D: height is 144  25  119 . Area is (10) 119 .
12
12
2
Area = 5 119
P
5
Q
5
5
5
5
15
4. A: 225  25  10 2
B: 2 external common tangents are possible to
two externally tangent circles.
2
C: 10 : 25 
5
D: extend PQ and RT to meet at U.
P
V
10
Q
x
5
R 10 S 5 T
U
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
Since TQ is half of PR and TR=15, TU=15.
x
5

Now put x on desired segment:
20 15
using similarity in triangles TQU and
SVU in the diagram. x=100/15=20/3
5. A: Set R at the origin. Q is (-12, 0) and T is (15, -5). Using the distance formula
A= 52  27 2 =754.
B: 102  ( ST )2  (5 13)2 . ST= 25(13) 100  225  15 . B=15.
10 5 2 13 26
 

C=
.
15 13 3 5 15
D: 13+13+ 5 13 +15= 41  5 13 . D=41+25(13)=366.
6
24
6. A: Since FK = 8 (triple), 
, perimeter of FHL is 16.
4 perim
B: Since the angles are congruent, due to similarity, the ratio of the sines is 1.
6 10

C:
so FL=20/3. Since the top triangle is twice the bottom, HM=5.
4 FL
C=20/3-15/3=5/3
6
8

D:
. HL=16/3. JM=4, again, half of FK. HL-JM= 16/3-12/3=4/3.
4 HL
7. A: Using Geometric Means, 64  PS ( PR) . 100  RS ( PR) .
B:
PS ( PR) PS 64 16



RS ( PR) RS 100 25
PT 8

by theorem. B=0.8 or 4/5.
TR 10
C: See part I and get the hypotenuse is 2 41 . QS=
64 100
8 10 40 41

=
41
2 41 2 41 2 41
D: PU= 64  25  89 . Perimeter = 8+5+ 89 =13+ 89 .
8. A: R+S=150 and R=2(90-S). 180-2S+S=150. S=30. R=120. T=30.
B: see part A, and Exterior at R = 180-120=60.
10
R
10
C: see picture: 10  10 3
30
30
1
5 3 5 3
T
S
D: Area of the triangle is (10 3)(5)  25 3 .
2
1
Let base be the shortest side, h the longest altitude to that side: (10)h  25 3 . h= 5 3
2
9. A: One exterior angle is 180-178=2. n=360/2=180 sides. n-80=100 sides to the new
polygon, and one exterior angle to that polygon is 360/100 =3.6 degrees or 18/5 degrees.
B: 9000=180(n-2). n-2=9000/180=900/18=100/2=50. n=52. 52  22 131 . Using the
sequential counting principle, the number of factors is 3x2=6.
C: Exterior + interior = 100+e+e=180 gives the exterior angle is 40. 360/40=9 sides.
7(180)=total interior = 1260
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
360
 180  ( s  134) . 360   s 2  46s . s 2  46s  360  0 . ( s  36)(s  10)  0 . s  12 , s=36.
s
10. A: Use the Pythagorean Theorem to get SR= 9  36  3 5 and UR= 61 .
D:
Perimeter = 8  3 5  61 . p=5 and q=61. A=66.
1
B: Area  (8)(6)  24 .
2
C: Using (3, 6) and (4, 0), y  6 x  b;0  6(4)  b . So C=y-intercept=24.
D: Average the vertices of the triangle for get (11/3, 2). Sum = D = 17/3
11. A: Since the tangent to a radius is perpendicular at the point of tangency, VRX is a
right triangle. If radius is 3, then this is a 3-4-5 triangle. VX=5. WX=2.
1
1  13 26  117
B: Area = (3)  6.5  13  (3)    =
or 29.25
2
2 2 2  4
1
3
C:  9    .
6
2
1
D: (18)(3)  27
2
12. A: 10(22)  8(8  KL) . 19.5 or 39/2.
B: The sides of the triangle KFH are 8, 22 and x. x must be between 30 and 14.
Integer possibilities are 15 through 29, which are 15 in number.
1
C: mH  (200  mGK ) . 2(mH )  mGK =200.
2
D: Angles F and L intercept the same arc, so they are congruent. Difference =0.
6(180)  4(120)
 one missing angle.
4
=270-120=150. Largest angle is 150 degrees.
4(180)  4(125)
 one missing angle. 360-250=110. Smallest angle is 110 degrees.
B:
2
C: 540  95  95  (2 x  10)  2 x  (3 x  10) . 350=7x. x=50. Angles are 95, 95, 90, 100, 160.
C=160-90=70.
7
7
47
D: h  (120)  140 . p  (108)  7(18)  126 . 6 y  360  140  126 . 6 y  94 . y=
6
6
3
13. A:
GEOMETRY TEAM QUESTIONS
Sponsors' Copy
Sickles Invitational
February 27, 2016
14. A: sides of RSTU are 10 each. Diagonal is 10 2 .
4
B: The radius of circle K (see picture to right) is
3
4/ 3
shrunk
4/ 3
4
2
2
4
64
3  16 3
4
C: One side of the rhombus is 52/4=13. Each small right triangle in the rhombus is a
5-12-13 triangle. Area is 4 (1/2) 5(12)=120.
D: 5/12 + 5/12 = 10/12 = 5/6.
15. A: x  60  2x  30  180 . x=30.
B: x  2 y  2 x  180. Using part A, we get y=45.
The new triangle has side 8, and area
C: Since part A gave x=30, we have 2x= mT =60 and so measure of RU is 120.
D: Since angle R is 90 degrees, RST is right, and so D=2.