Intermediate Math Circles Wednesday October Problem Set 1

Transcription

1 Intermediate Math ircles Wednesda October Problem Set 1 1. In the diagram, is parallel to. etermine the values of and. b 3 50 F a G 13 Let a = GF, b = F G. Since, we know that alternate interior angles are equal so a = 50. Observe that a and the angle 13 form a straight angle. Then, a + 13 = = 180 = 10 Similarl, using b, the 50 angle, and the angle 3 = 30, b = 180 b + 80 = 180 b = 100 Since is an eternal angle to F G, = a + b = 150. Therefore, = 10, = Triangle has a right angle at. is etended to so that =. The bisector of angle meets at. Prove that = 45. Since bisects, we can let = =. Since =, is isosceles so = =. In, b the sum of interior angles of a triangle, 2 + (90 + ) + = = = 90 + = 45 In, using the sum of interior angles, + (90 + ) + = ( + ) + = = 180 (from above) 45 + = 90 = 45 (as required) 1

2 3. In the diagram, is parallel to and = =. If = 52, determine the measure of. is isosceles since =. Therefore = = 52. Then in, = 180 = = 76 Since, we have = = 76. Since =, is isosceles. Therefore, = = 76 Therefore, b sum of interior angles of a triangle, = = The diagram shows three squares of the same size. What is the value of? In a square, the corner angles are 90. The triangle is equilateral (all sides equal), so we know all the angles are equal and hence must be 60 each. If we look at the place where the triangle and two squares meet (where is located), we notice it is made up of four angles; two corner angles of a square, one corner angle of a triangle, and. These four angles form a complete revolution, so the must sum up to 360. Then, = = 360 = 120 Therefore the measure of angle is

3 5. The diagram shows a rhombus F GHI and an isosceles triangle F GJ in which GF = GJ. ngle F JI equals 111. What is the measure of angle JF I? F J G I H Since F JI = 111 is part of a straight angle with F JG, we have that F JG = 69. We see that because GF = GJ, F GJ is isosceles, with equal base angles F JG and GF J, we get GF J = 69 and so F GJ = 42 ecause F G IH, F GI = GIH = 42. lso, IHG is isosceles since GH = HI, so IGH = GIH = 42 Since GH F I, F IG = IGH = 42. Using F JI, we see F JI + F IJ + JF I = JF I = 180 JF I = is a square. The point is outside the square so that is an equilateral triangle. etermine the measure of. Since is a square, =. Since is equilateral, = =. Therefore, = = = and so =. the properties of a square, = 90. the properties of equilateral triangles, = 60. Therefore = + = = 150. Since =, is isosceles. So = =. In this triangle, we have + + = = 180 = 15 So = = 15. Note that 60 = = + = Therefore, = = 45. 3

4 7. The diagram shows two isosceles triangles in which the four angles marked are equal. The two angles marked are also equal. Find an equation relating and. onsider the angles opposite to the angles marked. Since the are opposite angles, the are equal to. The quadrilateral formed in the overlap must have angle sum 360. We know two of the angles are. The other two angles are actuall the missing angle of the two isosceles triangles. In the left triangle, this angle is 180 2; for the triangle on the right, it is also These four angles have to sum to 360. Therefore, + + (180 2) + (180 2) = = = 4 = 2 = 2 is our desired relationship. 8. In the diagram, QSR is a straight line. P QP S = 12 and P Q = P S = RS. What is the measure of QP R? Let SP R =. Then, QP R = QP S + SP R = Q S R Since P S = SR, SP R is isosceles and so P RS = SP R =. Since P S = P Q, P QS is isosceles and so P QS = P SQ =. Then = = 168 = 84 Since QSR is a straight line, = P SQ is eternal to P SR, so 84 = = + = 2. Therefore, = 42. nd QP R = QP S + SP R = 12 + = = 54. 4

5 L 9. The three angle bisectors of triangle LMN meet at a point O as shown. ngle LNM is 68. What is the size of angle LOM? Since we are using angle bisectors, let LN O = ON M =, NLO = OLM =, and LMO = OMN = z. ut 68 = LNM = NLO + OLM = 2, so = 34. O N M We also have LON = 180 ( + ) = 146, LOM = 180 ( + z), and NOM = 180 ( + z) = 146 z. LON, NOM, and LOM form a complete revolution. So, LOM = 360 LON NOM = 360 (146 ) (146 z) = z Using the entire triangle, LNM + NLM + LMN = z = z = z = 56 Therefore, substituting back in, we get LOM = = In the figure shown, = F and, F, F, and are all straight lines. etermine an equation relating, and z. F Since = F, F is isosceles, so F = F = a. Since F and F are opposite angles, F = F = a. z is eternal to, so = + and a = + z follows. (1) is eternal to F, so = F + and = a + z follows. (2) Substituting (1) into (2) for a, we obtain = + z + z. Rearranging and simplifing we obtain + 2z = 0. This is the equation relating,, z. 5

6 11. What is the measure of the angle formed b the hands of a clock at 9:10? ver minute after the hour, the minute hand moves = 6 from 12 o clock. So after 10 minutes, it has moved 10 6 = 60 past 12 o clock. In one hour, the hour hand moves 360 = In ten minutes, it will have moved 1 of this, 6 so it has moved 1 30 = 6 5 closer to 12 o clock. 9 o clock is located 90 before 12 o clock, so the hour hand will be 85 before 12 o clock. Therefore, the total angle between the hour and minute hand will be = etermine the sum of the angles at,,,, and in the five-pointed star shown. Let a, b, c, d, and e be the angles at,,,, and, respectivel. Let P and Q be the points where intersects and, respectivel. Let p represent P Q and q represent QP. This information is marked on the above diagram. P Q is eterior to Q. Therefore, P Q = Q + Q. It follows that q = a + c. (1) QP is eterior to P. Therefore, QP = P + P. It follows that p = b + d. (2) In P Q, P Q + QP + P Q = 180. Substituting, we obtain e + p + q = 180. Further substituting for p from (2) and q from (1), we obtain e + b + d + a + c = 180. That is, the sum of the angles at,,,, and is

7 13. In P QR, P Q = P R. P Q is etended to S so that QS = QR. Prove that P RS = 3( QSR). P Since P Q = P R and QS = QR, we can label the diagram as shown. Q R Note that SP R = Using SP R, we see the angle sum gives us 180 = SP R + P SR + P RS 180 = (180 2) + + ( + ) 180 = = 2 S So P RS = + = + 2 = 3 = 3( QSR) as required. T 14. beam of light shines from point S, reflects off a reflector at point P, and reaches point T so that P T is perpendicular to RS. What is the value of? tend T P to RS, intersecting RS at the point Q as in the diagram. Then P QS is a right triangle. R P Q 26 S Since T P S is eterior to P QS, T P S = = 116. Since the reflector forms a straight line, the two angles marked and T P S form a straight angle. Then T P S + + = = = 64 = 32 7

8 R 15. In the diagram, P W is parallel to QX, S and T lie on QX, and U and V are the points of intersection of P W with SR and T R, respectivel. If SUV = 120 and V T X = 112, what is the measure of URV? P Q S U V T W X Since P W QX, we have SUV + T SU = T SU = 180 T SU = 60 RT X is eterior to RST. RT X = SRT + RST. (1) ut RT X = V T X = 112 (same angle, given info) and RST = T SU = 60 (same angle) substituting in (1), we have SRT + 60 = 112 SRT = 52 ut SRT and URV are the same angle. URV = In the diagram, let M be the point of intersection of the three altitudes of triangle. If = M, then what is in degrees? Let the three altitudes be, and F. In F and, we have F = = 90. lso, F and are the same angle, so F. = F =. ppling the same argument to F and, we get F = =. F M In M and, M = = M = = 90 M = M = M = and = So is right isosceles, hence = = 45. Therefore = 45, since =. 8

9 17. The diagram shows a regular nonagon with two sides etended to meet at point X. What is the size of the acute angle at X? X 140 o 140 o In a regular nonagon (9 sides), the sum of the interior angles is (9 2) 180 = Since the figure is regular, all the interior angles are equal. each angle is = 140. Using our diagram, the two etended sides each form a straight angle. One part of each straight angle is the interior angle, 140. The other part we will call must be 40. is part of a revolution; the other part of the revolution is one interior angle of the nonagon, 140. So = 220. The shape containing the angles X,, is a quadrilateral. The interior sum must therefore be 360. So, X = 360. Plugging in our values for,, we see Therefore, X = 60. X = = = The angles of a nonagon are nine consecutive numbers. What are these numbers? In problem 9, we determined that the sum of the interior angles of a nonagon is Order the angles from least to greatest, and let the middle angle (the 5th) be. Since the are consecutive numbers, the angles are { 4, 3, 2, 1,, + 1, + 2, + 3, + 4} Summing these angles should give us If ou add the nine angles, ou get 9. So 9 = = 140. This is the fifth angle. Therefore, the list of angles is {136, 137, 138, 139, 140, 141, 142, 143, 144 }. 9

10 19. regular pentagon is a five-sided figure which has all of its angles equal and all of its side lengths equal. In the diagram, T RN is a regular pentagon, P is an equilateral triangle, and OP N is a square. etermine the size of R. P R T O N Since P is equilateral, P = 60. Since OP N is a square, P N = 90. Since T RN is a regular pentagon, with interior angle sum is 540, each angle equals = 108. So NR = 108. t, the angles make a complete rotation, so R = 360 P P N NR = = 102 Since P is equilateral, = P. Since OP N is a square, P = N. Since T RN is a regular pentagon, N = R. Therefore = P = N = R and R is isosceles. It follows that R = R =. In R, we then have R + R + R = = = 78 = 39 Therefore, R = 39 10

11 20. Three regular polgons meet at a point and do not overlap. One has 3 sides and one has 42 sides. How man sides does the third polgon have? an ou find other sets of three polgons that have this propert? ach angle in a regular 3 sided polgon is 180 = (42 2) ach angle in a regular 42 sided polgon is (n 2) ach angle in a regular n-gon is. n The 3 angles form a complete revolution. = sides n sides 60 o 42 sides (n 2) + = 360 n 180(n 2) = n 7 180(n 2) = 900 n 7 (n 2) = 5 n 7 7n 14 = 5n 2n = 14 n = 7 it is a 7-sided figure. 11