2000 Solutions Euclid Contest

Transcription

1 Canadian Mathematics Competition n activit of The Centre for Education in Mathematics and Computing, Universit of Waterloo, Waterloo, Ontario 000 s Euclid Contest (Grade) for The CENTRE for EDUCTION in MTHEMTICS and COMPUTING wards 000 Waterloo Mathematics Foundation

2 000 Euclid s (a) If + 7 = 5, what is the value of? 5 = 5, 7 = Therefore, = 5 = (b) The line = a+ c is parallel to the line = and passes through the point (, 5) What is the value of c? Since the two given lines are parallel, the line = a+ c has slope and is of the form, = + c Since (, 5) is on the line, 5 = ()+ c c = (c) The parabola with equation = ( ) 6 has its verte at and intersects the -ais at B, as shown Determine the equation for the line passing through and B O B For = 0, 6 0 ( ) = [( ) ]( )+ 4 [ 4]= 0 Therefore = 6 or = Thus, the -intercepts of the parabola are and 6, and B has coordinates 6, 0 The verte of the parabola is at (, 6) Equation of line containing ( 6, 0) and (, 6) has slope 6 = 4 6 Thus the line has equation, 0 = 4 = ( ) (a) Si identical pieces are cut from a board, as shown in the diagram The angle of each cut is The pieces are assembled to form a heagonal picture frame as shown What is the value of?

3 000 Euclid s Each interior angle of a regular heagon is 0 Putting the frame together we would have the following = 0 (in degrees) = (b) If log = + log, what is the value of? 0 0 log 0 log0 = log0 = = 0 = 000 (c) If + = 6, determine all values of + Squaring both sides + = ; squaring = = = = 6 Creating a quadratic equation and solving = = 6 + 6= 0 0 ( )( )=

4 000 Euclid s 4 = or = For =, + = = = 6 97 = 6 For =, 9 = + 4 = (a) circle, with diameter B as shown, intersects the positive -ais at point D( 0, d) Determine d D 0, d, 0 O B 8, 0 The centre of the circle is, 0 Thus d + = 5 d = 5 d = 6 Therefore d = 4, since d > 0 ( ) and the circle has a radius of 5 Since B is a diameter of the circle, DB = 90 and OD = 90 DO ~ DBO Therefore, OD BO = O OD and d 8 = () d 6 = d = 4, since d > 0

5 000 Euclid s 5 DB = OD = BOD = 90 In OD, D = 4 + d In BOD, DB = 64 + d In DB, ( 4 + d )+( 64 + d )= 00 d = d = 4, d > 0 (b) square PQRS with side of length is subdivided into four triangular regions as shown so that area + area B = area C If PT = and RU = 5, determine the value of P T S C U B 5 Q R Since the side length of the square is, TS = and VS = 5 rea of triangle = ()() rea of triangle B= ( )( ) 5 rea of triangle C= ( 5)( ) From the given information, 5 ( )+ ( )= ( 5)( ) + 5 = = 0 ( )( ) = 5 0 Thus = 5 or = Therefore = 5 since = is inadmissible Labelled diagram P T S C 5 U B 5 Q R 4 (a) die, with the numbers,,, 4, 6, and 8 on its si faces, is rolled fter this roll, if an odd number appears on the top face, all odd numbers on the die are doubled If an even number appears on the top face, all the even numbers are halved If the given die changes in this wa, what is the probabilit that a will appear on the second roll of the die?

6 000 Euclid s 6 There are onl two possibilities on the first roll - it can either be even or odd Possibilit The first roll is odd The probabilit of an odd outcome on the first roll is fter doubling all the numbers, the possible outcomes on the second roll would now be,, 6, 4, 6, 8 with the probabilit of a being Thus the probabilit of a on the second roll would be = 9 Possibilit The first is even The probabilit of an even outcome on the first roll is fter halving all the numbers, the possible outcomes on the second roll would be,,,,, 8 The probabilit of a on the second die would now be 6 Thus the probabilit of a on the second roll is 6 = 9 The probabilit of a appear on the top face is = 9 b) The table below gives the final standings for seven of the teams in the English Cricket League in 998 t the end of the ear, each team had plaed 7 matches and had obtained the total number of points shown in the last column Each win W, each draw D, each bonus bowling point, and each bonus batting point B received w, d, a and b points respectivel, where w, d, a and b are positive integers No points are given for a loss Determine the values of w, d, a and b if total points awarded are given b the formula: Points = w W + d D+ a + b B Final Standings W Losses D B Points Susse Warks Som Derbs Kent Worcs Glam

7 000 Euclid s 7 There are a variet of was to find the unknowns The most efficient wa is to choose equations that have like coefficients Here is one wa to solve the problem using this method For Susse: 6w+ 4d+ 0a+ 6b= 0 For Som: 6w+ 4d+ 0a+ 54b= 9 Subtracting, 9b = 9 b = If b = For Derbs: 6w+ 4d+ 8a+ 55 = 9 6w+ 4d+ 8a= 6 () For Susse: 6w+ 4d+ 0a+ 6 = 0 6w+ 4d+ 0a= 8 () Subtracting, () () a = a = We can now calculate d and w b substituting a =, b = into a pair of equations n efficient wa of doing this is b substituting a =, b = into Som and Worcs For Som: 6w+ 4d+ 84 = 9 6w+ 4d = 08 () For Worcs: 6w+ d+ 85 = 00 6w+ d = 05 (4) Subtracting, () (4) d = Substituting d = in either () or (4), 6w + 4()= 08 (substituting in ()) 6w = 96 w = 6 Therefore w = 6, d =, a= b= 5 (a) In the diagram, D = DC, sin DBC = 0 6 and CB = 90 What is the value of tan BC? D B C Let DB = 0 Therefore, DC = D =6 B the theorem of Pthagoras, BC = 0 6 = 64 Therefore, BC = 8

8 000 Euclid s 8 Thus, tan BC = = 8 (b) On a cross-sectional diagram of the Earth, the and -aes are placed so that O( 00, ) is the centre of the Earth and C( 640, 000 ) is the location of Cape Canaveral space shuttle is forced to land on an island at ( 549, ), as shown Each unit represents 000 km Determine the distance from Cape Canaveral to the island, measured on the surface of the earth, to the nearest 0 km O 0, 0 54, 9 C 640, 000 Calculating OC Calculating arc length Distance tan OC = 9 54 OC = tan 9 = The arc length C = [ ( π )( 6 40 ) ]= 57 units 60 The distance is approimatel 570 km 6 (a) Let represent the greatest integer which is less than or equal to For eample, =, 6 = If is positive and = 7, what is the value of? We deduce that 4 < < 5 Otherwise, if 4, 6, and if 5, 5 Therefore = 4 Since = 7 4 = 7 = 45

9 000 Euclid s 9 (b) The parabola = + 4 has verte P and intersects the -ais at and B The parabola is translated from its original position so that its verte moves along the line = +4 to the point Q In this position, the parabola intersects the -ais at B and C Determine the coordinates of C P Q 4 O B C The parabola = + 4 has verte P( 04, ) and intersects the -ais at ( 0, ) and B( 0, ) The intercept B( 0, ) has its pre-image, B on the parabola = + 4 To find B, we find the point of intersection of the line passing through B( 0, ), with slope, and the parabola = + 4 The equation of the line is = Intersection points, = = 0 ( + )( )= 0 Therefore, = or = For =, = = 5 Thus B has coordinates (, 5) If (, 5) (, 0) then the required general translation mapping = + 4 onto the parabola with verte Q is (, ) ( + 5, + 5 ) Possibilit Using the general translation, we find the coordinates of Q to be, P( 04, ) Q( 0+ 54, + 5)= Q( 59, ) If C is the reflection of B in the ais of smmetr of the parabola, ie = 5, C has coordinates ( 80, ) Possibilit If B has coordinates (, 5) then C is the reflection of B in the -ais Thus C has coordinates (, 5) If we appl the general translation then C has coordinates ( + 5, 5 + 5) or ( 8, 0) Thus C has coordinates ( 8, 0) Possibilit Using the general translation, we find the coordinates of Q to be, P( 04, ) Q( 0+ 54, + 5)= Q( 59, ) The equation of the image parabola is = ( 5) + 9

10 000 Euclid s 0 ( ) + = To find its intercepts, ( ) = 5 9 5=± Therefore = 8 or = Thus C has coordinates ( 8, 0) The translation moving the parabola with equation = + 4 onto the parabola with verte Q is Ttt (, ) because the slope of the line = +4 is The pre-image of B is ( t, t) Since B is on the parabola with verte P, we have t = ( t) + 4 t = 4+ 4t t + 4 t 5t = 0 tt ( 5)= 0 Therefore, t = 0 or t = 5 Thus B is (, 5) Let C have coordinates ( c,0) The pre-image of C is ( c 5, 5) Therefore, 5= ( c 5) + 4 ( ) = Or, c 5 9 Therefore c 5= or c 5= c = 8 or c = Thus C has coordinates ( 8, 0) The translation moving the parabola with equation = + 4 onto the parabola with verte Q is T( p, p) because the slope of the line = +4 is Q will have coordinates p, p+ 4 ( ) Thus the equation of the image parabola is p p Since, 0 ( ) is on the parabola, 0 p p 4 p 5p= 0 pp ( 5)= 0 Therefore p = 0 or p = 5 The coordinates of Q are ( 5, 9) = ( ) + + = ( ) + + s in solution, we can use either reflection properties or the equation of the parabola to find that C has coordinates ( 8, 0) 4

11 000 Euclid s 7 (a) cube has edges of length n, where n is an integer Three faces, meeting at a corner, are painted red The cube is then cut into n smaller cubes of unit length If eactl 5 of these cubes have no faces painted red, determine the value of n If we remove the cubes which have red paint, we are left with a smaller cube with measurements, ( n ) ( n ) ( n ) Thus, ( n ) = 5 n = 6 (b) In the isosceles trapezoid BCD, B = CD = The area of the trapezoid is 80 and the circle with centre O and radius 4 is tangent to the four sides of the trapezoid Determine the value of B O C D Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram rea of trapezoid BCD = ()( BC + D) 8 = 4 ( b+ b) = 8 Thus, 8 = 80 Therefore, = 0 Bb b C b b 4 b O 4 b b b D 8 In parallelogram BCD, B = a and BC = b, where a> b The points of intersection of the angle bisectors are the vertices of quadrilateral PQRS (a) Prove that PQRS is a rectangle (b) Prove that PR = a b D E P S M N Q R F C B (a) In a parallelogram opposite angles are equal Since DF and BE bisect the two angles, let DF = CDF = BE = CBE = (in degrees) lso CDF = FD = (alternate angles) Let DM = BM = DCN = BCN = (in degrees)

12 000 Euclid s For an parallelogram, an two consecutive angles add to 80, + = 80 or, + =90 D E P S M N Q R F C B Therefore in PF, PF = 90 F P Using similar reasoning and properties of parallel lines we get right angles at Q, R and S Thus PQRS is a rectangle (b) Since M is a bisector of DB, let DM = BM = lso, DM = (alternate angles) This implies that DM is isosceles Using the same reasoning in CBN, we see that it is also isosceles and so the diagram ma now be labelled as: D N F a b b Q P R S b E M C b a B N = a b Thus DM and CBN are identical isosceles triangles lso, M NC (corresponding angles) or, P NR B using properties of isosceles triangles (or congruenc), P = NR impling that PRN is a parallelogram Thus N = PR and since N = a b, PR = a b (as required)

13 000 Euclid s 9 permutation of the integers,,, n is a listing of these integers in some order For eample, (,, ) and (,, ) are two different permutations of the integers,, permutation ( a, a,, a n ) of the integers,,, n is said to be fantastic if a + a + + a k is divisible b k, for each k from to n For eample, (,, ) is a fantastic permutation of,, because is divisible b, + is divisible b, and + + is divisible b However, (,, ) is not fantastic because + is not divisible b (a) Show that no fantastic permutation eists for n = 000 (b) Does a fantastic permutation eist for n = 00? Eplain (a) (b) In our consideration of whether there is a fantastic permutation for n = 000, we start b looking at the 000th position Using our definition of fantastic permutation, it is necessar that 000 ( L + 000) Since L = ( )( ) = ( 000)( 00), it is required that ( 00) This is not possible and so no fantastic permutation eists for n = 000 ( 00)( 00) The sum of the integers from to 00 is = ( 00 )( 00 ) which is divisible b 00 If t, t,, t 00 is a fantastic permutation, when we remove t 00 from the above sum, and what remains must be divisible b 000 We now consider t + t + L + t000 and determine what integer is not included in the permutation t + t + L + t000 = ( )( ) t t = ( )( ) 00 ( )( )=, t 00 must be a number of the form k00 where k is odd Since The onl integer less than or equal to 00 with this propert is 00 Therefore t 00 = 00 So the sum up to t 000 is = When we remove t 000 we must get a multiple of 999 The largest multiple of 999 less than is ( 999)( 00)= This would make t 000 = = 00 which is impossible since t000 t00 If we choose lesser multiples of 999 to subtract from we will get values of t 000 which are greater than 00, which is also not possible Thus, a fantastic permutation is not possible for n = 00

14 000 Euclid s 4 0 n equilateral triangle BC has side length square, PQRS, is such that P lies on B, Q lies on BC, and R and S lie on C as shown The points P, Q, R, and S move so that P, Q and R alwas remain on the sides of the triangle and S moves from C to B through the interior of the triangle If the points P, Q, R and S alwas form the vertices of a square, show that the path traced out b S is a straight line parallel to BC S P R B Q C In essence, this solution establishes that the perpendicular distance from S to BC is s( sin θ+ cos θ) and then showing that this is a constant b finding s sin θ+ cos θ the base which is itself a constant length Let RQC = θ and from S draw a line perpendicular to the base at P Then TQB = 80 ( 90 + θ)= 90 θ Let s be the length of the side of the square From R draw a line perpendicular to BC at D and then through S draw a line parallel to BC From R draw a line perpendicular to this line at E T S ( ) as part of E R 0 B F P Q D C 90 From RQD, RD = s sin θ Since QRD = 90 θ then SRE = θ From SER, ER = s cos θ The perpendicular distance from S to BC is RD + ER = ssin θ+ s cos θ which we must now show is a constant We can now take each of the lengths DC, DQ, PF, FB and epress them in terms of s From RDC which is a triangle, DC RD = Since RD = s sin θ (from above) DC = ( s sin θ)= ssin θ 60

15 000 Euclid s 5 From RDQ, QD RQ = cos θ, QD = s cos θ From TFQ, sin θ= FQ s and cos θ= TF s or, FQ From TFB, BF TF = s sin θ and TF = s sin θ =, BF = TF = s cos θ= s cos θ Since DC + QD + FQ + BF =, ssin θ+ scos θ+ ssin θ+ scos θ= ( scos θ+ ssin θ)+ ( scos θ+ ssin θ)= scos θ+ ssin θ= + Thus scos θ+ ssin θ is a constant and the path traced out b S is a straight line parallel to BC Note: number of enquiries have been made about this question Several individuals have made the comment that it is not possible to do this under the given conditions What is not mentioned, and what is not realized, is that the size of the square changes This makes it possible for the square to eist under the given conditions