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1、2002 Solutions Euclid Contest Grade 12 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 2002 Waterloo Mathematics Foundation 2002 Euclid Solut
2、ions2 1 a Solution 1 Midpoint Formula Since M is the midpoint of the line segment joining R and S then looking at the x coordinate of M 7 1 2 141 13 a a a Solution 2 Slopes Since the slope of RM is equal to the slope of MS then 3 6 3 7 76 13 a a a Solution 3 Distances SinceRMMS or RMMS 22 then 3637
3、01413 0131 22 2 a aa aa Therefore a 13 or a 1 but we reject a 1 since 110 does not lie on the line Thus a 13 Answer a 13 b The base of PQR has length 8 and the height has length k 2 since k 0 Since the area of PQR is 24 then 1 2 8224 4824 432 8 k k k k Answer k 8 c We first determine the point of in
4、tersection of lines yx 23 and yx 815 and then substitute this point into the line yxb 5 since it lies on all three lines So we set the first two equations equal to each other 2002 Euclid Solutions3 23815 126 2 xx x x Substitutingx 2 into the first equation we obtain y 2231 so the point of intersecti
5、on is 21 which must lie on the third line Thus 152 9 b b Therefore the value of b is 9 2 a Solution 1 Sincex 4 is a root then 43 40 2 c or c 4 Therefore the quadratic equation is xx 2 340 which we can factor as xx 410 This factorization is made easier since we already know one of the roots Therefore
6、 the second root is x 1 Solution 2 The sum of the roots of xxc 2 30 is 3 1 3 so since one root is 4 the second root must be x 1 Answer x 1 b Solution 1 Since the two expressions are the same then they must have the same value when we substitute any value for x In particular substitute x 2 and so we
7、get 2 21 23 2 23 92 7 2 22 A A A Solution 2 We compare the two expressions 21 3 2 3 23 33 26 3 2 22 2 22 2 2 x x A x x x A x xA x 2002 Euclid Solutions4 Since the expressions are the same the numerators must be the same and so 61A orA 7 Solution 3 21 3 267 3 237 3 2 7 3 2 2 2 2 2 2 2 x x x x x x x T
8、herefore A 7 Answer A 7 c Solution 1 The original parabola can be written as yxx 31 which means its roots are x 3 and x 1 When this parabola is shifted 5 units to the right the parabola obtained must thus have rootsx 358 and x 156 Therefore the new parabola is yxx xx 86 1448 2 and so d 48 Solution 2
9、 The original parabola yxx 2 43 can be written as yx 21 2 and so its vertex has coordinates 21 To get the vertex of the new parabola we shift the vertex of the original parabola 5 units to the right to the point 71 Substituting this point into the new parabola we obtain 1714 7 14998 48 2 d d d An ea
10、sier version of this solution is to recognize that if the original parabola passes through10 then 6 0 must be on the translated parabola Thus 03614 6 d or d 48 as above 2002 Euclid Solutions5 Solution 3 To carry out a translation of 5 units to the right we can define new coordinates X and Y withx yX
11、Y 5 So in these new coordinates the parabola will have equation YXX XXX XX 5453 10254203 1448 2 2 2 Comparing this with the given equation we see that d 48 3 a We make a table of the possible selections of balls a b c that give abc abc 211 312 321 413 422 431 Therefore since there are 6 ways to get
12、the required sum then the probability that he wins the prize is 6 64 3 32 Answer 3 32 b Since the product of the three integers is 216 then a ar ar a r ar ar 2 3 3 33 216 216 6 6 Now we are given that a is a positive integer but r is not necessarily an integer However we do know that the sequence is
13、 increasing so r 1 and thus a 0 bc 0 and cd 0 Adding these three inequalities abbccd 0 or abcd 220 We are going to show that this statement contradicts the facts that are known about the sequence We are told that the sum of any three consecutive terms is negative ie abc 0 and bcd 0 Adding these two
14、inequalities abcbcd 0 orabcd 220 and abcd 0 since ab 0 Then since abc 0 we must have that c 0 so d 0 This means that we have b 0 and cd 0 ie bcd 0 But from the conditions on the sequence bcd 0 In this case it is not immediately clear whether b has to be positive or negative However we do know that a
15、b 0 and abc 0 so it must be true that c 0 and cd 0 we must have both b 0andd 0 But then bcdbcd 0 since cd 0 and b 0 This is again a contradiction Therefore no such sequence exists with a 0 2002 Euclid Solutions16 9 a Let BAC Then by parallel lines DJHBDE Thus BED90o and so NEM since DEF90o SinceDGu
16、and HGv thenDHuv Similarly ENuw A D J B C M E KLGF N H wv w u w 90 u v v Looking at DHJ and MNE we see that tan uv v and tan w uw Therefore uv v w uw uv uwvw uuvuwvwvw u uvw 2 0 and since u 0 we must have uvw 0 or uvw Note If u 0 then the height of rectangle DEFG is 0 ie D coincides with point A and E coincides with point C which says that we must also have vw 0 ie the squares have no place to go b Consider the cross section of the sphere in the plane defined by the triangle This cross section w
