《欧几里得滑铁卢数学竞赛_2000EuclidSolution》由会员分享，可在线阅读，更多相关《欧几里得滑铁卢数学竞赛_2000EuclidSolution（15页珍藏版）》请在金锄头文库上搜索。
1、2000 Solutions Euclid Contest Grade12 Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing University of Waterloo Waterloo Ontario for Awards The CENTRE for EDUCATION in MATHEMATICS and COMPUTING 2000 Waterloo Mathematics Foundation 2000 Euclid Soluti
2、ons2 1 a If x 27125 1 3 1 3 what is the value of x Solution 1255 1 3 273 1 3 Therefore x 532 b The line yaxc is parallel to the line yx 2 and passes through the point 1 5 What is the value of c Solution Since the two given lines are parallel the line yaxc has slope 2 and is of the form yxc 2 Since 1
3、 5 is on the line 52 1 c c 3 c The parabola with equation yx 216 2 has its vertex at A and intersects the x axis at B as shown Determine the equation for the line passing through A and B y x O A B Solution For y 0 x 2160 2 xx 24240 Therefore x 6 or x 2 Thus the x intercepts of the parabola are 2 and
4、 6 and B has coordinates 6 0 The vertex of the parabola is at A 216 Equation of line containing 6 0 and 216 has slope 16 26 4 Thus the line has equation y x yx 0 6 4424 2 a Six identical pieces are cut from a board as shown in the diagram The angle of each cut is x The pieces are assembled to form a
5、 hexagonal picture frame as shown What is the value of x 2000 Euclid Solutions3 x x x x x x x Solution Each interior angle of a regular hexagon is 120 Putting the frame together we would have the following 2120 x in degrees x 60 120 120 x x b If loglog 1010 3xy what is the value of x y Solution log
6、log log 1010 10 3 3 xy x y x y 101000 3 c If x x 113 6 determine all values of x x 2 2 1 Solution 1 Squaring both sides x x 113 6 22 squaring x x 2 2 2 1169 36 x x 2 2 1169 32 2 x x 2 2 1169 36 72 36 97 36 Solution 2 Creating a quadratic equation and solving 6 1 6 13 6 x x x x 6613 2 xx 61360 2 xx 3
7、2 230 xx 2000 Euclid Solutions4 x 2 3 or x 3 2 For x 2 3 x x 2 2 1 2 3 1 2 3 2 2 4 9 9 4 8116 36 97 36 For x 3 2 3 2 1 3 2 2 2 9 4 4 9 97 36 3 a A circle with diameter AB as shown intersects the positive y axis at point Dd0 Determine d D 0 d A 2 0B 8 0 y x O Solution 1 The centre of the circle is 3
8、0 and the circle has a radius of 5 Thus d2 2 35 d d 222 2 53 16 Therefore d 4 since d 0 Solution 2 Since AB is a diameter of the circle ADB90 and AOD90 ADODBO Therefore OD AO BO OD and d22 8 d216 d 4 since d 0 2000 Euclid Solutions5 Solution 3 ADBAODBOD90 In AOD ADd 22 4 In BOD DBd 22 64 In ADBdd 46
9、4100 22 232 2 d d 4 d 0 b A square PQRS with side of length x is subdivided into four triangular regions as shown so that area A area B area C If PT 3 and RU 5 determine the value of x P T S U RQx x 3 5 A B C Solution Since the side length of the square is x TSx 3 and VSx 5 Area of triangle Ax 1 2 3
10、 Area of triangle Bx 1 2 5 Area of triangle Cxx 1 2 53 From the given information 1 2 3 1 2 5 1 2 53xxxx 35815 2 xxxx xx 2 16150 xx 1510 Thus x 15 or x 1 Therefore x 15 since x 1 is inadmissible Labelled diagram P T S U RQx x 3 5 A B C x 3 x 5 4 a A die with the numbers 1 2 3 4 6 and 8 on its six fa
11、ces is rolled After 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 way what is the probability that a 2 will appear on the second roll of the die 2000
12、 Euclid Solutions6 Solution There are only two possibilities on the first roll it can either be even or odd Possibility 1 The first roll is odd The probability of an odd outcome on the first roll is 1 3 After doubling all the numbers the possible outcomes on the second roll would now be 2 2 6 4 6 8
13、with the probability of a 2 being 1 3 Thus the probability of a 2 on the second roll would be 1 3 1 3 1 9 Possibility 2 The first is even The probability of an even outcome on the first roll is 2 3 After halving all the numbers the possible outcomes on the second roll would be 1 1 3 2 3 8 The probab
14、ility of a 2 on the second die would now be 1 6 Thus the probability of a 2 on the second roll is 2 3 1 6 1 9 The probability of a 2 appear on the top face is 1 9 1 9 2 9 b The table below gives the final standings for seven of the teams in the English Cricket League in 1998 At the end of the year e
15、ach team had played 17 matches and had obtained the total number of points shown in the last column Each win W each draw D each bonus bowling point A and each bonus batting point B received w d a and b points respectively where w d a and b are positive integers No points are given for a loss Determi
16、ne the values of w d a and b if total points awarded are given by the formula Points wWdDaAbB Final Standings W LossesD A B Points Sussex 6743063201 Warks 6833560200 Som 6743054192 Derbys 6742855191 Kent 5571859178 Worcs 4673259176 Glam 4673655176 2000 Euclid Solutions7 Solution There are a variety of ways to find the unknowns The most efficient way is to choose equations that have like coefficients Here is one way to solve the problem using this method For Sussex 643063201wdab For Som 643054192
