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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 . mHGK 3x 1 , mLKN x 39 and mKNP 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 mJKN mLKN if FG, JK and MN are all parallel. C = mMNK mGKL if FG, JK and MN are all parallel. D = the value of y if mKNM 2 x 39 and FH JL . ----------------------------------------------------------------------------------------------------------------------------3. RST is isosceles and RS=ST. mS 3x and mT x . A = the measure of the largest interior angle of RST . PQR is isosceles and PQ=QR. mP 3 y and mR 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 . mR mS 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 = pq. 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(mH ) mGK . D = mF mL . ------------------------------------------------------------------------------------------------------------------------------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 mT 2x , mR x 60 , mS x 2 y and mU 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 mJKN mLKN = 121(2)+59=180+121=301. C = mMNK mGKL = acute-obtuse=59 – 121 = -62. D = the value of y if mKNM 2 x 39 and FH JL . x is still 20 so mKNM 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: mH (200 mGK ) . 2(mH ) 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= mT =60 and so measure of RU is 120. D: Since angle R is 90 degrees, RST is right, and so D=2.