加拿大数学竞赛历年试题(滑铁卢大学)

合集下载

2005滑铁卢竞赛试题答案

1.(a)Answer:a=5Since(a,a)lies on the line3x−y=10,then3a−a=10or2a=10or a=5.(b)Answer:(6,2)Solution1To get from A to B,we move2units to the right and1unit up.xSince C lies on the same straight line as A and B,then to get from B to C we move2 units to the right and1unit up twice,or4units to the right and2units up.Thus,the coordinates of C are(6,2).Solution2Label the origin as O and drop a perpendicular from C to P on the x-axis.xThen AOB is similar to CP B since both are right-angled and they have equal angles at B.Since BC=2AB,then CP=2AO=2(1)=2and BP=2BO=2(2)=4.Therefore,the coordinates of C are(2+4,0+2)=(6,2).(c)By the Pythagorean Theorem,AO2=AB2−OB2=502−402=900,so AO=30.Therefore,the coordinates of A are(0,30).By the Pythagorean Theorem,CD2=CB2−BD2=502−482=196,so CD=14.x Therefore,the coordinates of C are(40+48,14)=(88,14). Since M is the midpoint of AC,then the coordinates of M are1 2(0+88),12(30+14)=(44,22)2.(a)Answer:x=−2Solution1Since y=2x+3,then4y=4(2x+3)=8x+12.Since4y=8x+12and4y=5x+6,then8x+12=5x+6or3x=−6or x=−2.Solution2Since4y=5x+6,then y=54x+64=54x+32.Since y=2x+3and y=54x+32,then2x+3=54x+32or34x=−32or x=−2.Solution3Since the second equation contains a“5x”,we multiply theﬁrst equation by52to obtaina5x term,and obtain52y=5x+152.Subtracting this from4y=5x+6,we obtain32y=−32or y=−1.Since y=−1,then−1=2x+3or2x=−4or x=−2.(b)Answer:a=6Solution1Adding the three equations together,we obtain a−3b+b+2b+7c−2c−5c=−10+3+13 or a=6.Solution2Multiplying the second equation by3,we obtain3b−6c=9.Adding this new equation to theﬁrst equation,we obtain c=−1.Substituting this back into the original second equation,we obtain b=3+2c=1.Substituting into the third equation,a=−2b+5c+13=−2−5+13=6.(c)Solution1Let J be John’s score and M be Mary’s score.Since two times John’s score was60more than Mary’s score,then2J=M+60.Since two times Mary’s score was90more than John’s score,then2M=J+90.Adding these two equations,we obtain2J+2M=M+J+150or J+M=150orJ+M2=75.Therefore,the average of their two scores was75.(Note that we didn’t have to solve for their individual scores.)Solution2Let J be John’s score and M be Mary’s score.Since two times John’s score was60more than Mary’s score,then2J=M+60,so M=2J−60.Since two times Mary’s score was90more than John’s score,then2M=J+90.Substituting theﬁrst equation into the second,we obtain2(2J−60)=J+904J−120=J+903J=210J=70Substituting into M=2J−60gives M=80.Therefore,the average of their scores(ie.the average of70and80)is75.3.(a)Answer:x=50Simplifying using exponent rules,2(1612)+2(816)=2((24)12)+2((23)16)=2(248)+2(248)=4(248)=22(248)=250Therefore,since2x=2(1612)+2(816)=250,then x=50.(b)Solution1We factor the given equation(f(x))2−3f(x)+2=0as(f(x)−1)(f(x)−2)=0.Therefore,f(x)=1or f(x)=2.If f(x)=1,then2x−1=1or2x=2or x=1.If f(x)=2,then2x−1=2or2x=3or x=3..Therefore,the values of x are x=1or x=32Solution2Since f(x)=2x−1and(f(x))2−3f(x)+2=0,then(2x−1)2−3(2x−1)+2=04x2−4x+1−6x+3+2=04x2−10x+6=02x2−5x+3=0(x−1)(2x−3)=0Therfore,x=1or x=3.24.(a)Answer:1415Solution1The possible pairs of numbers on the tickets are(listed as ordered pairs):(1,2),(1,3), (1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),and(5,6).There areﬁfteen such pairs.(We treat the pair of tickets numbered2and4as being the same as the pair numbered4and2.)The pairs for which the smaller of the two numbers is less than or equal to4are(1,2), (1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),and(4,6).There are fourteen such pairs.Therefore,the probability of selecting such a pair of tickets is14.15Solution 2We ﬁnd the probability that the smaller number on the two tickets is NOT less than or equal to 4.Therefore,the smaller number on the two tickets is at least 5.Thus,the pair of numbers must be 5and 6,since two distinct numbers less than or equal to 6are being chosen.As in Solution 1,we can determine that there are ﬁfteen possible pairs that we can se-lected.Therefore,the probability that the smaller number on the two tickets is NOT less than orequal to 4is 115,so the probability that the smaller number on the two tickets IS less than or equal to 4is 1−115=1415.(b)Solution 1Since ∠HLP =60◦and ∠BLP =30◦,then ∠HLB =∠HLP −∠BLP =30◦.Also,since ∠HLP =60◦and ∠HP L =90◦,then ∠LHP =180◦−90◦−60◦=30◦.Therefore, HBL is isosceles and BL =HB =400m.In BLP ,BL =400m and ∠BLP =30◦,so LP =BL cos(30◦)=400 √32 =200√3m.Therefore,the distance between L and P is 200√3m.Solution 2Since ∠HLP =60◦and ∠BLP =30◦,then ∠HLB =∠HLP −∠BLP =30◦.Also,since ∠HLP =60◦and ∠HP L =90◦,then ∠LHP =180◦−90◦−60◦=30◦.Also,∠LBP =60◦.Let LP =x .HSince BLP is 30◦-60◦-90◦,then BP :LP =1:√3,so BP =1√3LP =1√3x .Since HLP is 30◦-60◦-90◦,then HP :LP =√3:1,so HP =√3LP =√3x .But HP =HB +BP so√3x =400+1√3x3x =400√3+x2x =400√3x =200√3Therefore,the distance from L to P is 200√3m.5.(a)Answer:(6,5)After 2moves,the goat has travelled 1+2=3units.After 3moves,the goat has travelled 1+2+3=6units.Similarly,after n moves,the goat has travelled a total of 1+2+3+···+n units.For what value of n is 1+2+3+···+n equal to 55?The fastest way to determine the value of n is by adding the ﬁrst few integers until we obtain a sum of 55.This will be n =10.(We could also do this by remembering that 1+2+3+···+n =12n (n +1)and solving for n this way.)So we must determine the coordinates of the goat after 10moves.We consider ﬁrst the x -coordinate.Since starting at (0,0)the goat has moved 2units in the positive x direction,4units in the negative x direction,6units in the positive x direction,8units in the negative x direction and 10units in the positive x direction,so its x coordinate should be 2−4+6−8+10=6.Similarly,its y -coordinate should be 1−3+5−7+9=5.Therefore,after having travelled a distance of 55units,the goat is at the point (6,5).(b)Solution 1Since the sequence 4,4r ,4r 2is also arithmetic,then the diﬀerence between 4r 2and 4r equals the diﬀerence between 4r and 4,or4r 2−4r =4r −44r 2−8r +4=0r 2−2r +1=0(r −1)2=Therefore,the only value of r is r =1.Solution 2Since the sequence 4,4r ,4r 2is also arithmetic,then we can write 4r =4+d and 4r 2=4+2d for some real number d .(Here,d is the common diﬀerence in this arithmetic sequence.)Then d =4r −4and 2d =4r 2−4or d =2r 2−2.Therefore,equating the two expressions for d ,we obtain 2r 2−2=4r −4or 2r 2−4r +2=0or r 2−2r +1=0or (r −1)2=0.Therefore,the only value of r is r =1.6.(a)Answer:4πFirst,we notice that whenever an equilateral triangle of side length 3is placed inside acircle of radius3with two of its vertices on the circle,then the third vertex will be at the centre of the circle.This is because if we place XY Z with Y and Z on the circle and connect Y and Z to the centre O,then OY=OZ=3,so OY Z is equilateral(since all three sides have length3).Thus XY Z and OY Z must be the same,so X is at the same point as O. Thus,in the starting position,A is at the centre of the circle.As the triangle is rotated about C,the point B traces out an arc of a circle of radius3. What fraction of the circle is traced out?When point A reaches point A1on the circle,we have AC=3and CA1=3.Since A is at the centre of the circle,then AA1=3as well,so AA1C is equilateral,and∠A1CA=60◦, so the triangle has rotated through60◦.1Therefore,B has traced out60◦360◦=16of a circle of radius3.Notice that A has also traced out an arc of the same length.When A reaches the circle, we have A and C on the circle,so B must be at the centre of the circle.Thus,on the next rotation,B again rotates through16of a circle of radius3as it movesto the circle.On the third rotation,the triangle rotates about B,so B does not move.After three rotations,the triangle will have A at the centre and B and C on the circle,with the net result that the triangle has rotated180◦about the centre of the circle.Thus,to return to its original position,the triangle must undergo three more of these rotations,and B will behave in the same way as it did for theﬁrst three rotations.Thus,in total,B moves four times along an arc equal to16of a circle of radius3.Therefore,the distance travelled by B is4(16)(2π(3))=4π.(b)In order to determine CD,we must determine one of the angles(or at least some infor-mation about one of the angles)in BCD.To do this,we look at∠A use the fact that∠A+∠C=180◦.ADCB 5674Using the cosine law in ABD ,we obtain72=52+62−2(5)(6)cos(∠A )49=61−60cos(∠A )cos(∠A )=15Since cos(∠A )=15and ∠A +∠C =180◦,then cos(∠C )=−cos(180◦−∠A )=−15.(We could have calculated the actual size of ∠A using cos(∠A )=15and then used this to calculate the size of ∠C ,but we would introduce the possibility of rounding error by doing this.)Then,using the cosine law in BCD ,we obtain72=42+CD 2−2(4)(CD )cos(∠C )49=16+CD 2−8(CD ) −150=5CD 2+8CD −1650=(5CD +33)(CD −5)So CD =−335or CD =5.(We could have also determined these roots using the quadratic formula.)Since CD is a length,it must be positive,so CD =5.(We could have also proceeded by using the sine law in BCD to determine ∠BDC and then found the size of ∠DBC ,which would have allowed us to calculate CD using the sine law.However,this would again introduce the potential of rounding error.)7.(a)Answer:Maximum =5,Minimum =1We rewrite by completing the square as f (x )=sin 2x −2sin x +2=(sin x −1)2+1.Therefore,since (sin x −1)2≥0,then f (x )≥1,and in fact f (x )=1when sin x =1(which occurs for instance when x =90◦).Thus,the minimum value of f (x )is 1.To maximize f (x ),we must maximize (sin x −1)2.Since −1≤sin x ≤1,then (sin x −1)2is maximized when sin x =−1(for instance,when x =270◦).In this case,(sin x −1)2=4,so f (x )=5.Thus,the maximum value of f (x )is 5.(b)From the diagram,the x -intercepts of the parabola are x =−k and x =3k .xy Since we are given that y =−14(x −r )(x −s ),then the x -intercepts are r and s ,so r and s equal −k and 3k in some order.Therefore,we can rewrite the parabola as y =−14(x −(−k ))(x −3k ).Since the point (0,3k )lies on the parabola,then 3k =−14(0+k )(0−3k )or 12k =3k 2or k 2−4k =0or k (k −4)=0.Thus,k =0or k =4.Since the two roots are distinct,then we cannot have k =0(otherwise both x -intercepts would be 0).Thus,k =4.This tells us that the equation of the parabola is y =−14(x +4)(x −12)or y =−14x 2+2x +12.We still have to determine the coordinates of the vertex,V .Since the x -intercepts of the parabola are −4and 12,then the x -coordinate of the vertex is the average of these intercepts,or 4.(We could have also used the fact that the x -coordinate is −b2a =−22(−14).)Therefore,the y -coordinate of the vertex is y =−14(42)+2(4)+12=16.Thus,the coordinates of the vertex are (4,16).8.(a)We look at the three pieces separately.If x <−4,f (x )=4so g (x )= 25−[f (x )]2=√25−42=√9=3.So g (x )is the horizontal line y =3when x <−4.If x >5,f (x )=−5so g (x )= 25−[f (x )]2= 25−(−5)2=√0=0.So g (x )is the horizontal line y =0when x >5.So far,our graph looks like this:If−4≤x≤5,f(x)=−x so g(x)=25−[f(x)]2=25−(−x)2=√25−x2.What is this shape?If y=g(x),then we have y=√25−x2or y2=25−x2or x2+y2=25.Therefore,this shape is a section of the upper half(since y is a positive square-root)of the circle x2+y2=25,ie.the circle with centre(0,0)and radius5.We must check the endpoints. When x=−4,we have g(−4)=25−(−4))2=3.When x=5,we have g(5)=√25−52=0.Therefore,the section of the circle connects up with the other two sections of our graph already in place.Thus,ourﬁnal graph is:(b)Solution 1Let the centres of the two circles be O 1and O 2.Join A and B to O 1and B and C to O 2.Designate two points W and X on either side of A on one tangent line,and two points Y and Z oneither side of C on the other tangent line.ZLet ∠XAB =θ.Since W X is tangent to the circle with centre O 1at A ,then O 1A is perpendicular to W X ,so ∠O 1AB =90◦−θ.Since O 1A =O 1B because both are radii,then AO 1B is isosceles,so ∠O 1BA =∠O 1AB =90◦−θ.Since the two circles are tangent at B ,then the line segment joining O 1and O 2passes through B ,ie.O 1BO 2is a straight line segment.Thus,∠O 2BC =∠O 1BA =90◦−θ,by opposite angles.Since O 2B =O 2C ,then similarly to above,∠O 2CB =∠O 2BC =90◦−θ.Since Y Z is tangent to the circle with centre O 2at C ,then O 2C is perpendicular to Y Z .Thus,∠Y CB =90◦−∠O 2CB =θ.Since ∠XAB =∠Y CB ,then W X is parallel to Y Z ,by alternate angles,as required.Solution2Let the centres of the two circles be O1and O2.Join A and B to O1and B and C to O2.Since AO1and BO1are radii of the same circle,AO1=BO1so AO1B is isosceles,so ∠O1AB=∠O1BA.Since BO2and CO2are radii of the same circle,BO2=CO2so BO2C is isosceles,so ∠O2BC=∠O2CB.Since the two circles are tangent at B,then O1BO2is a line segment(ie.the line segment joining O1and O2passes through the point of tangency of the two circles).Since O1BO2is straight,then∠O1BA=∠O2BC,by opposite angles.Thus,∠O1AB=∠O1BA=∠O2BC=∠O2CB.This tells us that AO1B is similar to BO2C,so∠AO1B=∠BO2C or∠AO1O2=∠CO2O1.Therefore,AO1is parallel to CO2,by alternate angles.But A and C are points of tangency,AO1is perpendicular to the tangent line at A and CO2is perpendicular to the tangent line at C.Since AO1and CO2are parallel,then the two tangent lines must be parallel.9.(a)Solution1We have(x−p)2+y2=r2and x2+(y−p)2=r2,so at the points of intersection,(x−p)2+y2=x2+(y−p)2x2−2px+p2+y2=x2+y2−2py+p2−2px=−2pyand so x=y(since we may assume that p=0otherwise the two circles would coincide).Therefore,a and b are the two solutions of the equation(x−p)2+x2=r2or2x2−2px+(p2−r2)=0or x2−px+12(p2−r2)=0.Using the relationship between the sum and product of roots of a quadratic equation andits coeﬃcients,we obtain that a+b=p and ab=12(p2−r2).(We could have solved for a and b using the quadratic formula and calculated these di-rectly.)So we know that a+b=p.Lastly,a2+b2=(a+b)2−2ab=p2−2 12(p2−r2)=r2,as required.Solution2Since the circles are reﬂections of one another in the line y=x,then the two points of intersection must both lie on the line y=x,ie.A has coordinates(a,a)and B has coordinates(b,b).Therefore,(a−p)2+a2=r2and(b−p)2+b2=r2,since these points lie on both circles.Subtracting the two equations,we get(b −p )2−(a −p )2+b 2−a 2=0((b −p )−(a −p ))((b −p )+(a −p ))+(b −a )(b +a )=0(b −a )(a +b −2p )+(b −a )(b +a )=0(b −a )(a +b −2p +b +a )=02(b −a )(a +b −p )=0Since a =b ,then we must have a +b =p ,as required.Since a +b =p ,then a −p =−b ,so substituting back into (a −p )2+a 2=r 2gives (−b )2+a 2=r 2,or a 2+b 2=r 2,as required.(b)We ﬁrst draw a diagram.yxWe know that C has coordinates (p,0)and D has coordinates (0,p ).Thus,the slope of line segment CD is −1.Since the points A and B both lie on the line y =x ,then the slope of line segment AB is 1.Therefore,AB is perpendicular to CD ,so CADB is a kite,and so its area is equal to 12(AB )(CD ).(We could derive this by breaking quadrilateral CADB into CAB and DAB .)Since C has coordinates (p,0)and D has coordinates (0,p ),then CD = p 2+(−p )2= 2p 2.(We do not know if p is positive,so this is not necessarily equal to √2p .)We know that A has coordinates (a,a )and B has coordinates (b,b ),soAB = (a −b )2+(a −b )2=√2a 2−4ab +2b 2= 2(a 2+b 2)−4ab = 2r 2−4 12(p 2−r 2) = 4r 2−2p 2Therefore,the area of quadrilateral CADB is 12(AB )(CD )=124r 2−2p 2 2p 2= 2r 2p 2−p 4To maximize this area,we must maximize 2r 2p 2−p 4=2r 2(p 2)−(p 2)2.Since r is ﬁxed,we can consider this as a quadratic polynomial in p 2.Since the coeﬃcient of (p 2)2is negative,then this is a parabola opening downwards,so we ﬁnd its maximum value by ﬁnding its vertex.The vertex of 2r 2(p 2)−(p 2)2is at p 2=−2r 22(−1)=r 2.So the maximum area of the quadrilateral occurs when p is chosen so that p 2=r 2.Since p 2=r 2,then (a +b )2=p 2=r 2so a 2+2ab +b 2=r 2.Since a 2+b 2=r 2,then 2ab =0so either a =0or b =0,and so either A has coordinates (0,0)or B has coordinates (0,0),ie.either A is the origin or B is the origin.(c)In (b),we calculated that AB = 4r 2−2p 2=√2 2r 2−p 2.Since r and p are integers (and we assume that neither r nor p is 0),then 2r 2−p 2=0,so the minimum possible non-negative value for 2r 2−p 2is 1,since 2r 2−p 2must be an integer.Therefore,the minimum possible distance between A and B should be √2√1=√2.Can we ﬁnd positive integers p and r that give us this value?Yes –if r =5and p =7,then 2r 2−p 2=1,so AB =√2.(There are in fact an inﬁnite number of positive integer solutions to the equation 2r 2−p 2=1or equivalently p 2−2r 2=−1.This type of equation is called Pell’s Equation.)10.(a)We proceed directly.On the ﬁrst pass from left to right,Josephine closes all of the even numbered lockers,leaving the odd ones open.The second pass proceeds from right to left.Before the pass,the lockers which are open are 1,3,...,47,49.On the second pass,she shuts lockers 47,43,39, (3)The third pass proceeds from left to right.Before the pass,the lockers which are open are 1,5,...,45,49.On the third pass,she shuts lockers 5,13, (45)This leaves lockers 1,9,17,25,33,41,49open.On the fourth pass,from right to left,lockers 41,25and 9are shut,leaving 1,17,33,49.On the ﬁfth pass,from left to right,lockers 17and 49are shut,leaving 1and 33open.On the sixth pass,from right to left,locker 1is shut,leaving 33open.Thus,f (50)=33.(b)&(c)Solution 1First,we note that if n =2k is even,then f (n )=f (2k )=f (2k −1)=f (n −1).See Solution 2for this justiﬁcation.Therefore,we only need to look for odd values of n in parts (b)and (c).Suppose that there was an n so that f (n )=2005,ie.2005is the last locker left open.On the ﬁrst pass,Josephine closes every other locker starting at the beginning,so she closes all lockers numbered m with m ≡0(mod 2).This leaves only odd-numbered lockers open,ie.only lockers m with m ≡1or 3(mod 4).On her second pass,she closes every other open locker,starting from the right-hand end.Thus,she will close every fourth locker from the original row.Since we want 2005to be left open and 2005≡1(mod 4),then she must close all lockers numbered m with m ≡3(mod 4).This leaves open only the lockers m with m ≡1(mod 4),or equivalently lockers with m ≡1or 5(mod 8).On her third pass,she closes every other open locker,starting from the left-hand end.Thus,she will close every eighth locker from the original row.Since locker1is still open,then she starts by closing locker5,and so closes all lockers mwith m≡5(mod8).But since2005≡5(mod8),then she closes locker2005on this pass,a contradiction.Therefore,there can be no integer n with f(n)=2005.Next,we show that there are inﬁnitely many positive integers n such that f(n)=f(2005).To do this,weﬁrst make a table of what happens when there are2005lockers in the row.We record the pass#,the direction of the pass,the leftmost locker that is open,therightmost locker that is open,all open lockers before the pass,which lockers will be closedon the pass,and which lockers will be left open after the pass:Pass#Dir.L Open R Open Open To close Leaves Open 1L to R12005All≡0(mod2)≡1(mod2) 2R to L12005≡1,3(mod4)≡3(mod4)≡1(mod4) 3L to R12005≡1,5(mod8)≡5(mod8)≡1(mod8) 4R to L12001≡1,9(mod16)≡9(mod16)≡1(mod16) 5L to R12001≡1,17(mod32)≡17(mod32)≡1(mod32) 6R to L11985≡1,33(mod64)≡33(mod64)≡1(mod64) 7L to R11985≡1,65(mod128)≡65(mod128)≡1(mod128) 8R to L11921≡1,129(mod256)≡1(mod256)≡129(mod256) 9L to R1291921≡129,385(mod512)≡385(mod512)≡129(mod512) 10R to L1291665≡129,641(mod1024)≡129(mod1024)≡641(mod1024) 11L to R6411665≡641,1665(mod2048)≡1665(mod2048)≡641(mod2048) Since there is only one integer between1and2005congruent to641(mod2048),thenthere is only one locker left open:locker641.Notice also that on any pass s,the“class”of lockers which are closed depends on what thenumber of the leftmost(on an odd-numbered pass)or rightmost(on an even-numberedpass)open locker number is congruent to mod2s.Consider n=2005+22a,where22a>2005,ie.a≥6.We show that f(n)=f(2005)=641.(See Solution2for a justiﬁcation of why we mighttry these values of n.)Suppose we were to try to make a table as above to calculate f(n).Then theﬁrst11passes in the table would be identical to the table above,except for therightmost open number;this number in the new table would be the number above plus22a.What will happen after pass11?After pass11,the lockers which are open are lockers with numbers≡641(mod2048).Thus,the leftmost open locker is641and the rightmost is22a+641.As the12th pass starts,the lockers which are still open are those with numbers≡641or2689(mod212).Since the rightmost open locker number(22a+641)is congruent to641(mod212),then alllockers with numbers≡2689(mod212)are closed,leaving open only those lockers withnumbers≡641(mod212).So after this12th pass,the lockers which are open are641,641+212,641+2(212),641+3(212),...,641+22a−12(212)=641+22a.The number of open lockers is22a−12+1.If we can now show that whenever we start with a number of lockers of the form22c+1,thelast locker remaining open is the leftmost locker,then we will be done,since of the lockersleft open above(22a−12+1of them,ie.2to an even power plus1),then the last locker re-maining open will be the leftmost one,that is locker641,so f(22a+2005)=641=f(2005).So consider a row of22c+1lockers.Notice that on any pass,if the number of lockers is odd,then the number of lockers whichwill be closed is one-half of one less than the total number of lockers,and theﬁrst andlast lockers will be left open.So on theﬁrst pass,there are22c−1lockers closed,leaving22c+1−22c−1=22c−1+1lockersopen,ie.an odd number of lockers open.On the next pass,there are22c−2lockers closed(since there are an odd number of lockersopen to begin),leaving22c−2+1lockers open.This continues,until there are21+1=3lockers open just before an even-numbered(ie.right to left)pass.Thus,the middle of these three lockers will be closed,leaving only theoriginal leftmost and rightmost lockers open.On the last pass(an odd-numbered pass from left to right),the rightmost locker will beclosed,leaving only the leftmost locker open.Therefore,starting with a row of22c+1open lockers,the leftmost locker will be the lastremaining open.Translating this to the above,we see that the leftmost locker of the22a−12+1still openis the last left open,ie.f(22a+2005)=641=f(2005)if a≥6.Therefore,there are inﬁnitely many positive integers n for which f(n)=f (2005).Solution2First,we calculate f(n)for n from1to32,to get a feeling for what happens.We obtain 1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11,1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11. This will help us to establish some patterns.Next,we establish two recursive formulas for f(n).First,from our pattern,it looks like f(2m)=f(2m−1).Why is this true in general?Consider a row of2m lockers.On theﬁrst pass through,Josephine shuts all of the even numbered lockers,leaving open lockers1,3,...,2m−1.These are exactly the same open lockers as if she had started with2m−1lockers in total. Thus,as she starts her second pass from right to left,the process will be the same now whether she started with2m lockers or2m−1lockers.Therefore,f(2m)=f(2m−1).This tells us that we need only focus on the values of f(n)where n is odd.Secondly,we show that f(2m−1)=2m+1−2f(m).(It is helpful to connect n=2m−1to a smaller case.)Why is this formula true?Starting with2m−1lockers,the lockers left open after theﬁrst pass are1,3,...,2m−1, ie.m lockers in total.Suppose f(m)=p.As Josephine begins her second pass,which is from right to left,we can think of this as being like theﬁrst pass through a row of m lockers.Thus,the last open locker will be the p th locker,counting from the right hand end,from the list1,3,...,2m−1.Theﬁrst locker from the right is2m−1=2m+1−2(1),the second is2m−3=2m+1−2(2), and so on,so the p th locker is2m+1−2p.Therefore,theﬁnal open locker is2m+1−2p,ie.f(2m−1)=2m+1−2p=2m+1−2f(m). Using these two formulae repeatedly,f(4k+1)=f(2(2k+1)−1)=2(2k+1)+1−2f(2k+1)=4k+3−2f(2(k+1)−1)=4k+3−2(2(k+1)+1−2f(k+1))=4k+3−2(2k+3−2f(k+1))=4f(k+1)−3andf(4k+3)=f(2(2k+2)−1)=2(2k+2)+1−2f(2k+2)=4k+5−2f(2k+1)=4k+5−2f(2(k+1)−1)=4k+5−2(2(k+1)+1−2f(k+1))=4k+5−2(2k+3−2f(k+1))=4f(k+1)−1From our initial list of values of f(n),it appears as if f(n)cannot leave a remainder of5 or7when divided by8.So we use these recursive relations once more to try to establish this:f(8l+1)=4f(2l+1)−3(since8l+1=4(2l)+1)=4(2l+3−2f(l+1))−3=8l+9−8f(l+1)=8(l−f(l+1))+9f(8l+3)=4f(2l+1)−1(since8l+3=4(2l)+3)=4(2l+3−2f(l+1))−1=8l+11−8f(l+1)=8(l−f(l+1))+11Similarly,f(8l+5)=8l+9−8f(l+1)and f(8l+7)=8l+11−8f(l+1). Therefore,since any odd positive integer n can be written as8l+1,8l+3,8l+5or8l+7, then for any odd positive integer n,f(n)is either9more or11more than a multiple of8. Therefore,for any odd positive integer n,f(n)cannot be2005,since2005is not9more or11more than a multiple of8.Thus,for every positive integer n,f(n)=2005,since we only need to consider odd values of n.Next,we show that there are inﬁnitely many positive integers n such that f(n)=f(2005). We do this by looking at the pattern we initially created and conjecturing thatf(2005)=f(2005+22a)if22a>2005.(We might guess this by looking at the connection between f(1)and f(3) with f(5)and f(7)and then f(1)through f(15)with f(17)through f(31).In fact,it appears to be true that f(m+22a)=f(m)if22a>m.)Using our formulae from above,f(2005+22a)=4f(502+22a−2)−3(2005+22a=4(501+22a−2)+1) =4f(501+22a−2)−3=4(4f(126+22a−4)−3)−3(501+22a−2=4(125+22a−4)+1)=16f(126+22a−4)−15=16f(125+22a−4)−15=16(4f(32+22a−6)−3)−15(125+22a−4=4(31+22a−6)+1)=64f(32+22a−6)−63=64f(31+22a−6)−63=64(4f(8+22a−8)−1)−63(31+22a−6=4(7+22a−8)+3)=256f(8+22a−8)−127=256f(7+22a−8)−127=256(4f(2+22a−10)−1)−127(7+22a−8=4(1+22a−10)+3)=1024f(2+22a−10)−383=1024f(1+22a−10)−383(Notice that we could have removed the powers of2from inside the functions and used this same approach to show that f(2005)=1024f(1)−383=641.)But,f(22b+1)=1for every positive integer b.Why is this true?We can prove this quickly by induction.For b=1,we know f(5)=1.Assume that the result is true for b=B−1,for some positive integer B≥2.Then f(22B+1)=f(4(22B−2)+1)=4f(22B−2+1)−3=4(1)−3=1by our induction hypothesis.Therefore,if a≥6,then f(1+22a−10)=f(1+22(a−5))=1sof(2005+22a)=1024(1)−383=641=f(2005)so there are inﬁnitely many integers n for which f(n)=f(2005).。

2011EuclidSolution

1.(a)Since (x +1)+(x +2)+(x +3)=8+9+10,then 3x +6=27or 3x =21and so x =7.(b)Since 25+√x =6,then squaring both sides gives 25+√x =36or √x =11.Since √x =11,then squaring both sides again,we obtain x =112=121.Checking, 25+√121=√25+11=√36=6,as required.(c)Since (a,2)is the point of intersection of the lines with equations y =2x −4and y =x +k ,then the coordinates of this point must satisfy both equations.Using the ﬁrst equation,2=2a −4or 2a =6or a =3.Since the coordinates of the point (3,2)satisfy the equation y =x +k ,then 2=3+k or k =−1.2.(a)Since the side length of the original square is 3and an equilateral triangle of side length 1is removed from the middle of each side,then each of the two remaining pieces of each side of the square has length 1.Also,each of the two sides of each of the equilateral triangles that are shown has length 1.1111Therefore,each of the 16line segments in the ﬁgure has length 1,and so the perimeter of the ﬁgure is 16.(b)Since DC =DB ,then CDB is isosceles and ∠DBC =∠DCB =15◦.Thus,∠CDB =180◦−∠DBC −∠DCB =150◦.Since the angles around a point add to 360◦,then∠ADC =360◦−∠ADB −∠CDB =360◦−130◦−150◦=80◦.(c)By the Pythagorean Theorem in EAD ,we have EA 2+AD 2=ED 2or 122+AD 2=132,and so AD =√169−144=5,since AD >0.By the Pythagorean Theorem in ACD ,we have AC 2+CD 2=AD 2or AC 2+42=52,and so AC =√25−16=3,since AC >0.(We could also have determined the lengths of AD and AC by recognizing 3-4-5and 5-12-13right-angled triangles.)By the Pythagorean Theorem in ABC ,we have AB 2+BC 2=AC 2or AB 2+22=32,and so AB =√9−4=√5,since AB >0.3.(a)Solution 1Since we want to make 15−y x as large as possible,then we want to subtract as little as possible from 15.In other words,we want to make y x as small as possible.To make a fraction with positive numerator and denominator as small as possible,wemake the numerator as small as possible and the denominator as large as possible.Since 2≤x ≤5and 10≤y ≤20,then we make x =5and y =10.Therefore,the maximum value of 15−y x is 15−105=13.Solution2Since y is positive and2≤x≤5,then15−yx≤15−y5for any x with2≤x≤5andpositive y.Since10≤y≤20,then15−y5≤15−105for any y with10≤y≤20.Therefore,for any x and y in these ranges,15−yx≤15−105=13,and so the maximumpossible value is13(which occurs when x=5and y=10).(b)Solution1First,we add the two given equations to obtain(f(x)+g(x))+(f(x)−g(x))=(3x+5)+(5x+7)or2f(x)=8x+12which gives f(x)=4x+6.Since f(x)+g(x)=3x+5,then g(x)=3x+5−f(x)=3x+5−(4x+6)=−x−1.(We could alsoﬁnd g(x)by subtracting the two given equations or by using the second of the given equations.)Since f(x)=4x+6,then f(2)=14.Since g(x)=−x−1,then g(2)=−3.Therefore,2f(2)g(2)=2×14×(−3)=−84.Solution2Since the two given equations are true for all values of x,then we can substitute x=2to obtainf(2)+g(2)=11f(2)−g(2)=17Next,we add these two equations to obtain2f(2)=28or f(2)=14.Since f(2)+g(2)=11,then g(2)=11−f(2)=11−14=−3.(We could alsoﬁnd g(2)by subtracting the two equations above or by using the second of these equations.)Therefore,2f(2)g(2)=2×14×(−3)=−84.4.(a)We consider choosing the three numbers all at once.We list the possible sets of three numbers that can be chosen:{1,2,3}{1,2,4}{1,2,5}{1,3,4}{1,3,5}{1,4,5}{2,3,4}{2,3,5}{2,4,5}{3,4,5} We have listed each in increasing order because once the numbers are chosen,we arrange them in increasing order.There are10sets of three numbers that can be chosen.Of these10,the4sequences1,2,3and1,3,5and2,3,4and3,4,5are arithmetic sequences.Therefore,the probability that the resulting sequence is an arithmetic sequence is410or25.(b)Solution 1Join B to D .AConsider CBD .Since CB =CD ,then ∠CBD =∠CDB =12(180◦−∠BCD )=12(180◦−60◦)=60◦.Therefore, BCD is equilateral,and so BD =BC =CD =6.Consider DBA .Note that ∠DBA =90◦−∠CBD =90◦−60◦=30◦.Since BD =BA =6,then ∠BDA =∠BAD =12(180◦−∠DBA )=12(180◦−30◦)=75◦.We calculate the length of AD .Method 1By the Sine Law in DBA ,we have AD sin(∠DBA )=BA sin(∠BDA ).Therefore,AD =6sin(30◦)sin(75◦)=6×12sin(75◦)=3sin(75◦).Method 2If we drop a perpendicular from B to P on AD ,then P is the midpoint of AD since BDA is isosceles.Thus,AD =2AP .Also,BP bisects ∠DBA ,so ∠ABP =15◦.Now,AP =BA sin(∠ABP )=6sin(15◦).Therefore,AD =2AP =12sin(15◦).Method 3By the Cosine Law in DBA ,AD 2=AB 2+BD 2−2(AB )(BD )cos(∠ABD )=62+62−2(6)(6)cos(30◦)=72−72(√32)=72−36√3Therefore,AD = 36(2−√3)=6 2−√3since AD >0.Solution 2Drop perpendiculars from D to Q on BC and from D to R on BA .AThen CQ =CD cos(∠DCQ )=6cos(60◦)=6×12=3.Also,DQ =CD sin(∠DCQ )=6sin(60◦)=6×√32=3√3.Since BC =6,then BQ =BC −CQ =6−3=3.Now quadrilateral BQDR has three right angles,so it must have a fourth right angle and so must be a rectangle.Thus,RD =BQ =3and RB =DQ =3√3.Since AB =6,then AR =AB −RB =6−3√3.Since ARD is right-angled at R ,then using the Pythagorean Theorem and the fact that AD >0,we obtain AD =√RD 2+AR 2= 32+(6−3√3)2= 9+36−36√3+27= 72−36√3which we can rewrite as AD = 36(2−√3)=6 2−√3.5.(a)Let n be the original number and N be the number when the digits are reversed.Sincewe are looking for the largest value of n ,we assume that n >0.Since we want N to be 75%larger than n ,then N should be 175%of n ,or N =74n .Suppose that the tens digit of n is a and the units digit of n is b .Then n =10a +b .Also,the tens digit of N is b and the units digit of N is a ,so N =10b +a .We want 10b +a =74(10a +b )or 4(10b +a )=7(10a +b )or 40b +4a =70a +7b or 33b =66a ,and so b =2a .This tells us that that any two-digit number n =10a +b with b =2a has the required property.Since both a and b are digits then b <10and so a <5,which means that the possible values of n are 12,24,36,and 48.The largest of these numbers is 48.(b)We “complete the rectangle”by drawing a horizontal line through C which meets they -axis at P and the vertical line through B at Q .x A (0,Since C has y -coordinate 5,then P has y -coordinate 5;thus the coordinates of P are (0,5).Since B has x -coordinate 4,then Q has x -coordinate 4.Since C has y -coordinate 5,then Q has y -coordinate 5.Therefore,the coordinates of Q are (4,5),and so rectangle OP QB is 4by 5and so has area 4×5=20.Now rectangle OP QB is made up of four smaller triangles,and so the sum of the areas of these triangles must be 20.Let us examine each of these triangles:• ABC has area 8(given information)• AOB is right-angled at O ,has height AO =3and base OB =4,and so has area 12×4×3=6.• AP C is right-angled at P ,has height AP =5−3=2and base P C =k −0=k ,and so has area 1×k ×2=k .• CQB is right-angled at Q ,has height QB =5−0=5and base CQ =4−k ,andso has area 12×(4−k )×5=10−52k .Since the sum of the areas of these triangles is 20,then 8+6+k +10−52k =20or 4=32k and so k =83.6.(a)Solution 1Suppose that the distance from point A to point B is d km.Suppose also that r c is the speed at which Serge travels while not paddling (i.e.being carried by just the current),that r p is the speed at which Serge travels with no current (i.e.just from his paddling),and r p +c his speed when being moved by both his paddling and the current.It takes Serge 18minutes to travel from A to B while paddling with the current.Thus,r p +c =d 18km/min.It takes Serge 30minutes to travel from A to B with just the current.Thus,r c =d 30km/min.But r p =r p +c −r c =d 18−d 30=5d 90−3d 90=2d 90=d 45km/min.Since Serge can paddle the d km from A to B at a speed of d 45km/min,then it takes him 45minutes to paddle from A to B with no current.Solution 2Suppose that the distance from point A to point B is d km,the speed of the current of the river is r km/h,and the speed that Serge can paddle is s km/h.Since the current can carry Serge from A to B in 30minutes (or 12h),then d r =12.When Serge paddles with the current,his speed equals his paddling speed plus the speed of the current,or (s +r )km/h.Since Serge can paddle with the current from A to B in 18minutes (or 310h),then d r +s =310.The time to paddle from A to B with no current would be d sh.Since d r =12,then r d =2.Since d r +s =310,then r +s d =103.Therefore,s d =r +s d −r d =103−2=43.Thus,d s =34,and so it would take Serge 34of an hour,or 45minutes,to paddle from A to B with no current.Solution 3Suppose that the distance from point A to point B is d km,the speed of the current of the river is r km/h,and the speed that Serge can paddle is s km/h.Since the current can carry Serge from A to B in 30minutes (or 12h),then d r =12or d =1r .When Serge paddles with the current,his speed equals his paddling speed plus the speed of the current,or (s +r )km/h.Since Serge can paddle with the current from A to B in 18minutes (or 310h),then d r +s =310or d =310(r +s ).Since d =12r and d =310(r +s ),then 12r =310(r +s )or 5r =3r +3s and so s =23r .To travel from A to B with no current,the time in hours that it takes is d s =12r 2r =34,or 45minutes.(b)First,we note that a =0.(If a =0,then the “parabola”y =a (x −2)(x −6)is actuallythe horizontal line y =0which intersects the square all along OR .)Second,we note that,regardless of the value of a =0,the parabola has x -intercepts 2and 6,and so intersects the x -axis at (2,0)and (6,0),which we call K (2,0)and L (6,0).This gives KL =4.Third,we note that since the x -intercepts of the parabola are 2and 6,then the axis ofsymmetry of the parabola has equation x =12(2+6)=4.Since the axis of symmetry of the parabola is a vertical line of symmetry,then if theparabola intersects the two vertical sides of the square,it will intersect these at the same height,and if the parabola intersects the top side of the square,it will intersect it at two points that are symmetrical about the vertical line x =4.Fourth,we recall that a trapezoid with parallel sides of lengths a and b and height h hasarea 12h (a +b ).We now examine three cases.Case1:a<0Here,the parabola opens downwards.Since the parabola intersects the square at four points,it must intersect P Q at points M and N.(The parabola cannot intersect the vertical sides of the square since it gets “narrower”towards the vertex.)xx =4Since the parabola opens downwards,then MN<KL=4.Since the height of the trapezoid equals the height of the square(or8),then the area of the trapezoid is1h(KL+MN)which is less than1(8)(4+4)=32.But the area of the trapezoid must be36,so this case is not possible.Case2:a>0;M and N on P QWe have the following conﬁguration:xx =4Here,the height of the trapezoid is8,KL=4,and M and N are symmetric about x=4.Since the area of the trapezoid is36,then12h(KL+MN)=36or12(8)(4+MN)=36or4+MN=9or MN=5.Thus,M and N are each52units from x=4,and so N has coordinates(32,8).Since this point lies on the parabola with equation y=a(x−2)(x−6),then8=a(32−2)(32−6)or8=a(−12)(−92)or8=94a or a=329.Case3:a>0;M and N on QR and P Oxx =4Here,KL=4,MN=8,and M and N have the same y-coordinate.Since the area of the trapezoid is36,then12h(KL+MN)=36or12h(4+8)=36or6h=36or h=6.Thus,N has coordinates(0,6).Since this point lies on the parabola with equation y=a(x−2)(x−6),then 6=a(0−2)(0−6)or6=12a or a=12.Therefore,the possible values of a are329and12.7.(a)Solution1Consider a population of100people,each of whom is75years old and who behave ac-cording to the probabilities given in the question.Each of the original100people has a50%chance of living at least another10years,so there will be50%×100=50of these people alive at age85.Each of the original100people has a20%chance of living at least another15years,so there will be20%×100=20of these people alive at age90.Since there is a25%(or14)chance that an80year old person will live at least another10years(that is,to age90),then there should be4times as many of these people alive at age80than at age90.Since there are20people alive at age90,then there are4×20=80of the original100 people alive at age80.In summary,of the initial100people of age75,there are80alive at age80,50alive at age85,and20people alive at age90.Because50of the80people alive at age80are still alive at age85,then the probability that an80year old person will live at least5more years(that is,to age85)is50=5,or 62.5%.Solution2Suppose that the probability that a75year old person lives to80is p,the probability that an80year old person lives to85is q,and the probability that an85year old person lives to90is r.We want to the determine the value of q.For a75year old person to live at least another10years,they must live another5years (to age80)and then another5years(to age85).The probability of this is equal to pq. We are told in the question that this is equal to50%or0.5.Therefore,pq=0.5.For a75year old person to live at least another15years,they must live another5years (to age80),then another5years(to age85),and then another5years(to age90).The probability of this is equal to pqr.We are told in the question that this is equal to20% or0.2.Therefore,pqr=0.2Similarly,since the probability that an80year old person will live another10years is25%,then qr=0.25.Since pqr=0.2and pq=0.5,then r=pqrpq=0.20.5=0.4.Since qr=0.25and r=0.4,then q=qrr=0.250.4=0.625.Therefore,the probability that an80year old man will live at least another5years is0.625,or62.5%.(b)Using logarithm rules,the given equation is equivalent to22log10x=3(2·2log10x)+16or(2log10x)2=6·2log10x+16.Set u=2log10x.Then the equation becomes u2=6u+16or u2−6u−16=0.Factoring,we obtain(u−8)(u+2)=0and so u=8or u=−2.Since2a>0for any real number a,then u>0and so we can reject the possibility that u=−2.Thus,u=2log10x=8which means that log10x=3.Therefore,x=1000.8.(a)First,we determine theﬁrst entry in the50th row.Since theﬁrst column is an arithmetic sequence with common diﬀerence3,then the50th entry in theﬁrst column(theﬁrst entry in the50th row)is4+49(3)=4+147=151.Second,we determine the common diﬀerence in the50th row by determining the second entry in the50th row.Since the second column is an arithmetic sequence with common diﬀerence5,then the 50th entry in the second column(that is,the second entry in the50th row)is7+49(5) or7+245=252.Therefore,the common diﬀerence in the50th row must be252−151=101.Thus,the40th entry in the50th row(that is,the number in the50th row and the40th column)is151+39(101)=151+3939=4090.(b)We follow the same procedure as in(a).First,we determine theﬁrst entry in the R th row.Since theﬁrst column is an arithmetic sequence with common diﬀerence3,then the R th entry in theﬁrst column(that is,theﬁrst entry in the R th row)is4+(R−1)(3)or 4+3R−3=3R+1.Second,we determine the common diﬀerence in the R th row by determining the second entry in the R th row.Since the second column is an arithmetic sequence with common diﬀerence5,then the R th entry in the second column(that is,the second entry in the R th row)is7+(R−1)(5) or7+5R−5=5R+2.Therefore,the common diﬀerence in the R th row must be(5R+2)−(3R+1)=2R+1.Thus,the C th entry in the R th row(that is,the number in the R th row and the C th column)is3R+1+(C−1)(2R+1)=3R+1+2RC+C−2R−1=2RC+R+C(c)Suppose that N is an entry in the table,say in the R th row and C th column.From(b),then N=2RC+R+C and so2N+1=4RC+2R+2C+1.Now4RC+2R+2C+1=2R(2C+1)+2C+1=(2R+1)(2C+1).Since R and C are integers with R≥1and C≥1,then2R+1and2C+1are each integers that are at least3.Therefore,2N+1=(2R+1)(2C+1)must be composite,since it is the product of two integers that are each greater than1.9.(a)If n=2011,then8n−7=16081and so √8n−7≈126.81.Thus,1+√8n−72≈1+126.812≈63.9.Therefore,g(2011)=2(2011)+1+8(2011)−72=4022+ 63.9 =4022+63=4085.(b)To determine a value of n for which f(n)=100,we need to solve the equation2n−1+√8n−72=100(∗)Weﬁrst solve the equation2x−1+√8x−72=100(∗∗)because the left sides of(∗)and(∗∗)do not diﬀer by much and so the solutions are likely close together.We will try integers n in(∗)that are close to the solutions to(∗∗). Manipulating(∗∗),we obtain4x−(1+√8x−7)=2004x−201=√8x−7(4x−201)2=8x−716x2−1608x+40401=8x−716x2−1616x+40408=02x2−202x+5051=0By the quadratic formula,x=202±2022−4(2)(5051)2(2)=202±√3964=101±√992and so x≈55.47or x≈45.53.We try n=55,which is close to55.47:f(55)=2(55)−1+8(55)−72=110−1+√4332Since √433≈20.8,then1+√4332≈10.9,which gives1+√4332=10.Thus,f(55)=110−10=100.Therefore,a value of n for which f(n)=100is n=55.(c)We want to show that each positive integer m is in the range of f or the range of g ,butnot both.To do this,we ﬁrst try to better understand the “complicated”term of each of the func-tions –that is,the term involving the greatest integer function.In particular,we start witha positive integer k ≥1and try to determine the positive integers n that give 1+√8n −72 =k .By deﬁnition of the greatest integer function,the equation 1+√8n −72 =k is equiv-alent to the inequality k ≤1+√8n −72<k +1,from which we obtain the following set of equivalent inequalities 2k ≤1+√8n −7<2k +22k −1≤√8n −7<2k +14k 2−4k +1≤8n −7<4k 2+4k +14k 2−4k +8≤8n <4k 2+4k +812(k 2−k )+1≤n <12(k 2+k )+1If we deﬁne T k =1k (k +1)=1(k 2+k )to be the k th triangular number for k ≥0,thenT k −1=12(k −1)(k )=12(k 2−k ).Therefore, 1+√8n −72 =k for T k −1+1≤n <T k +1.Since n is an integer,then 1+√8n −72=k is true for T k −1+1≤n ≤T k .When k =1,this interval is T 0+1≤n ≤T 1(or 1≤n ≤1).When k =2,this interval is T 1+1≤n ≤T 2(or 2≤n ≤3).When k =3,this interval is T 2+1≤n ≤T 3(or 4≤n ≤6).As k ranges over all positive integers,these intervals include every positive integer n and do not overlap.Therefore,we can determine the range of each of the functions f and g by examining the values f (n )and g (n )when n is in these intervals.For each non-negative integer k ,deﬁne R k to be the set of integers greater than k 2and less than or equal to (k +1)2.Thus,R k ={k 2+1,k 2+2,...,k 2+2k,k 2+2k +1}.For example,R 0={1},R 1={2,3,4},R 2={5,6,7,8,9},and so on.Every positive integer occurs in exactly one of these sets.Also,for each non-negative integer k deﬁne S k ={k 2+2,k 2+4,...,k 2+2k }and deﬁne Q k ={k 2+1,k 2+3,...,k 2+2k +1}.For example,S 0={},S 1={3},S 2={6,8},Q 0={1},Q 1={2,4},Q 2={5,7,9},and so on.Note that R k =Q k ∪S k so every positive integer occurs in exactly one Q k or in exactly one S k ,and that these sets do not overlap since no two S k ’s overlap and no two Q k ’s overlap and no Q k overlaps with an S k .We determine the range of the function g ﬁrst.For T k −1+1≤n ≤T k ,we have 1+√8n −72=k and so 2T k −1+2≤2n ≤2T k 2T k −1+2+k ≤2n + 1+√8n −72 ≤2T k +k k 2−k +2+k ≤g (n )≤k 2+k +k k 2+2≤g (n )≤k 2+2kNote that when n is in this interval and increases by 1,then the 2n term causes the value of g (n )to increase by 2.Therefore,for the values of n in this interval,g (n )takes precisely the values k 2+2,k 2+4,k 2+6,...,k 2+2k .In other words,the range of g over this interval of its domain is precisely the set S k .As k ranges over all positive integers (that is,as these intervals cover the domain of g ),this tells us that the range of g is precisely the integers in the sets S 1,S 2,S 3,....(We could also include S 0in this list since it is the empty set.)We note next that f (1)=2− 1+√8−72 =1,the only element of Q 0.For k ≥1and T k +1≤n ≤T k +1,we have 1+√8n −72=k +1and so 2T k +2≤2n ≤2T k +12T k +2−(k +1)≤2n − 1+√8n −72 ≤2T k +1−(k +1)k 2+k +2−k −1≤f (n )≤(k +1)(k +2)−k −1k 2+1≤f (n )≤k 2+2k +1Note that when n is in this interval and increases by 1,then the 2n term causes the value of f (n )to increase by 2.Therefore,for the values of n in this interval,f (n )takes precisely the values k 2+1,k 2+3,k 2+5,...,k 2+2k +1.In other words,the range of f over this interval of its domain is precisely the set Q k .As k ranges over all positive integers (that is,as these intervals cover the domain of f ),this tells us that the range of f is precisely the integers in the sets Q 0,Q 1,Q 2,....Therefore,the range of f is the set of elements in the sets Q 0,Q 1,Q 2,...and the range of g is the set of elements in the sets S 0,S 1,S 2,....These ranges include every positive integer and do not overlap.10.(a)Suppose that ∠KAB =θ.Since ∠KAC =2∠KAB ,then ∠KAC =2θand ∠BAC =∠KAC +∠KAB =3θ.Since 3∠ABC =2∠BAC ,then ∠ABC =23×3θ=2θ.Since ∠AKC is exterior to AKB ,then ∠AKC =∠KAB +∠ABC =3θ.This gives the following conﬁguration:BNow CAK is similar to CBA since the triangles have a common angle at C and ∠CAK =∠CBA .Therefore,AKBA=CACBordc=baand so d=bca.Also,CKCA=CACBora−xb=baand so a−x=b2aor x=a−b2a=a2−b2a,as required.(b)From(a),bc=ad and a2−b2=ax and so we obtainLS=(a2−b2)(a2−b2+ac)=(ax)(ax+ac)=a2x(x+c) andRS=b2c2=(bc)2=(ad)2=a2d2In order to show that LS=RS,we need to show that x(x+c)=d2(since a>0).Method1:Use the Sine LawFirst,we derive a formula for sin3θwhich we will need in this solution:sin3θ=sin(2θ+θ)=sin2θcosθ+cos2θsinθ=2sinθcos2θ+(1−2sin2θ)sinθ=2sinθ(1−sin2θ)+(1−2sin2θ)sinθ=3sinθ−4sin3θSince∠AKB=180◦−∠KAB−∠KBA=180◦−3θ,then using the Sine Law in AKB givesx sinθ=dsin2θ=csin(180◦−3θ)Since sin(180◦−X)=sin X,then sin(180◦−3θ)=sin3θ,and so x=d sinθsin2θandc=d sin3θsin2θ.This givesx(x+c)=d sinθsin2θd sinθsin2θ+d sin3θsin2θ=d2sinθsin22θ(sinθ+sin3θ)=d2sinθsin22θ(sinθ+3sinθ−4sin3θ)=d2sinθsin22θ(4sinθ−4sin3θ)=4d2sin2θsin22θ(1−sin2θ)=4d2sin2θcos2θsin22θ=4d2sin2θcos2θ(2sinθcosθ)2=4d2sin2θcos2θ4sin2θcos2θ=d2as required.We could have instead used the formula sin A +sin B =2sinA +B 2 cos A −B 2 toshow that sin 3θ+sin θ=2sin 2θcos θ,from which sin θ(sin 3θ+sin θ)=sin θ(2sin 2θcos θ)=2sin θcos θsin 2θ=sin 22θMethod 2:Extend ABExtend AB to E so that BE =BK =x and join KE .ENow KBE is isosceles with ∠BKE =∠KEB .Since ∠KBA is the exterior angle of KBE ,then ∠KBA =2∠KEB =2θ.Thus,∠KEB =∠BKE =θ.But this also tells us that ∠KAE =∠KEA =θ.Thus, KAE is isosceles and so KE =KA =d.ESo KAE is similar to BKE ,since each has two angles equal to θ.Thus,KA BK =AE KE or d x =c +x dand so d 2=x (x +c ),as required.Method 3:Use the Cosine Law and the Sine LawWe apply the Cosine Law in AKB to obtainAK 2=BK 2+BA 2−2(BA )(BK )cos(∠KBA )d 2=x 2+c 2−2cx cos(2θ)d 2=x 2+c 2−2cx (2cos 2θ−1)Using the Sine Law in AKB ,we get x sin θ=d sin 2θor sin 2θsin θ=d x or 2sin θcos θsin θ=d x and so cos θ=d 2x.Combining these two equations,d2=x2+c2−2cx2d24x2−1d2=x2+c2−cd2x+2cxd2+cd2x=x2+2cx+c2d2+cd2x=(x+c)2xd2+cd2=x(x+c)2d2(x+c)=x(x+c)2d2=x(x+c)as required(since x+c=0).(c)Solution1Our goal is toﬁnd a triple of positive integers that satisfy the equation in(b)and are the side lengths of a triangle.First,we note that if(A,B,C)is a triple of real numbers that satisﬁes the equation in(b)and k is another real number,then the triple(kA,kB,kC)also satisﬁes the equationfrom(b),since(k2A2−k2B2)(k2A2−k2B2+kAkC)=k4(A2−B2)(A2−B2+AC)=k4(B2C2)=(kB)2(kC)2 Therefore,we start by trying toﬁnd a triple(a,b,c)of rational numbers that satisﬁes the equation in(b)and forms a triangle,and then“scale up”this triple to form a triple (ka,kb,kc)of integers.To do this,we rewrite the equation from(b)as a quadratic equation in c and solve for c using the quadratic formula.Partially expanding the left side from(b),we obtain(a2−b2)(a2−b2)+ac(a2−b2)=b2c2which we rearrange to obtainb2c2−c(a(a2−b2))−(a2−b2)2=0By the quadratic formula,c=a(a2−b2)±a2(a2−b2)2+4b2(a2−b2)22b2=a(a2−b2)±(a2−b2)2(a2+4b2)2b2Since∠BAC>∠ABC,then a>b and so a2−b2>0,which givesc=a(a2−b2)±(a2−b2)√a2+4b22b2=(a2−b2)2b2(a±√a2+4b2)Since a2+4b2>0,then √a2+4b2>a,so the positive root isc=(a2−b2)2b2(a+a2+(2b)2)We try to ﬁnd integers a and b that give a rational value for c .We will then check to see if this triple (a,b,c )forms the side lengths of a triangle,and then eventually scale these up to get integer values.One way for the value of c to be rational (and in fact the only way)is for a 2+(2b )2to be an integer,or for a and 2b to be the legs of a Pythagorean triple.Since √32+42is an integer,then we try a =3and b =2,which givesc =(32−22)2·22(3+√32+42)=5and so (a,b,c )=(3,2,5).Unfortunately,these lengths do not form a triangle,since 3+2=5.(The Triangle Inequality tells us that three positive real numbers a ,b and c form a triangle if and only if a +b >c and a +c >b and b +c >a .)We can continue to try small Pythagorean triples.Now 152+82=172,but a =15and b =4do not give a value of c that forms a triangle with a and b .However,162+302=342,so we can try a =16and b =15which givesc =(162−152)2·152(16+√162+302)=31450(16+34)=319Now the lengths (a,b,c )=(16,15,319)do form the sides of a triangle since a +b >c and a +c >b and b +c >a .Since these values satisfy the equation from (b),then we can scale them up by a factor of k =9to obtain the triple (144,135,31)which satisﬁes the equation from (b)and are the side lengths of a triangle.(Using other Pythagorean triples,we could obtain other triples of integers that work.)Solution 2We note that the equation in (b)involves only a ,b and c and so appears to depend only on the relationship between the angles ∠CAB and ∠CBA in ABC .Using this premise,we use ABC ,remove the line segment AK and draw the altitude CF .CBA 3θ2θb aa c os 2θbc os 3θF Because we are only looking for one triple that works,we can make a number of assump-tions that may or may not be true in general for such a triangle,but which will help us ﬁnd an example.We assume that 3θand 2θare both acute angles;that is,we assume that θ<30◦.In ABC ,we have AF =b cos 3θ,BF =a cos 2θ,and CF =b sin 3θ=a sin 2θ.Note also that c =b cos 3θ+a cos 2θ.One way to ﬁnd the integers a,b,c that we require is to look for integers a and b and an angle θwith the properties that b cos 3θand a cos 2θare integers and b sin 3θ=a sin 2θ.Using trigonometric formulae,sin 2θ=2sin θcos θcos 2θ=2cos 2θ−1sin 3θ=3sin θ−4sin 3θ(from the calculation in (a),Solution 1,Method 1)cos 3θ=cos(2θ+θ)=cos 2θcos θ−sin 2θsin θ=(2cos 2θ−1)cos θ−2sin 2θcos θ=(2cos 2θ−1)cos θ−2(1−cos 2θ)cos θ=4cos 3θ−3cos θSo we can try to ﬁnd an angle θ<30◦with cos θa rational number and then integers a and b that make b sin 3θ=a sin 2θand ensure that b cos 3θand a cos 2θare integers.Since we are assuming that θ<30◦,then cos θ>√32≈0.866.The rational number with smallest denominator that is larger than √32is 78,so we try the acute angle θwith cos θ=7.In this case,sin θ=√1−cos 2θ=√158,and sosin 2θ=2sin θcos θ=2×78×√158=7√1532cos 2θ=2cos 2θ−1=2×4964−1=1732sin 3θ=3sin θ−4sin 3θ=3×√158−4×15√15512=33√15128cos 3θ=4cos 3θ−3cos θ=4×343512−3×78=7128To have b sin 3θ=a sin 2θ,we need 33√15128b =7√1532a or 33b =28a .To ensure that b cos 3θand a cos 2θare integers,we need 7128b and 1732a to be integers,andso a must be divisible by 32and b must be divisible by 128.The integers a =33and b =28satisfy the equation 33b =28a .Multiplying each by 32gives a =1056and b =896which satisfy the equation 33b =28a and now have the property that b is divisible by 128(with quotient 7)and a is divisible by 32(with quotient 33).With these values of a and b ,we obtain c =b cos 3θ+a cos 2θ=896×7128+1056×1732=610.We can then check that the triple (a,b,c )=(1056,896,610)satisﬁes the equation from(b),as required.As in our discussion in Solution 1,each element of this triple can be divided by 2to obtain the “smaller”triple (a,b,c )=(528,448,305)that satisﬁes the equation too.Using other values for cos θand integers a and b ,we could obtain other triples (a,b,c )of integers that work.。

滑铁卢竞赛数学题

滑铁卢竞赛数学题概述
滑铁卢竞赛数学题通常比较难，涉及的知识点广泛，包括代数、几何、数论、组合数学等多个领域。

以下是一些滑铁卢竞赛数学题的示例：
1. 有100个球，其中有一个与其他99个重量不同，但外观相同。

用一个天平，最少需要称多少次才能确定这个重量不同的球？
2. 一个正方形的面积为1，将其四边中点连接起来，形成另一个正方形。

如此重复，得到第五、第六个正方形，求第五个正方形的面积。

3. 一个圆被分成n个相等的扇形，其中一个是空心的，其他n-1个是实心的。

求空心扇形的圆心角是多少度？
4. 有100个人站成一排，从第1个人开始报数，每次报到奇数的人离开队伍。

经过若干轮后，只剩下一个人。

求这个人最初站在第几位？
5. 有5个不同质因数的最小正整数是多少？
以上仅是滑铁卢竞赛数学题的一些示例，实际上还有更多难题和技巧题。

如果想要深入了解滑铁卢竞赛数学题的解题技巧和策略，建议参考相关的竞赛书籍和资料，或者参加专业的数学竞赛培训课程。

欧几里得滑铁卢数学竞赛_2010EuclidSolution

Since Bea ﬂies at a constant speed, then the ratio of the two distances equals the ratio of
the corresponding times.
HF 60 minutes 4
Therefore, =
=.
GF 45 minutes 3
(b) Solution 1
Since ∠OP B = 90◦, then OP and P B are perpendicular, so the product of their slopes
is −1.
4−0 4
4−0
4
The slope of OP is
= and the slope of P B is
Since F GH is right-angled at F , then F GH must be similar to a 3-4-5 triangle, and
HG 5
so = .
GF 3
In
particular,
this
means
that
the
ratio
of
the
times
ﬂying
H
to
Wednesday, April 7, 2010
Solutions
©2010 Centre for Education in Mathematics and Computing
2010 Euclid Contest Solutions
Page 2
1. (a) Solution 1 Since 3x = 27, then 3x+2 = 3x32 = 27 · 9 = 243.

2002滑铁卢竞赛试题

Time: 2 1 2 hours
© 2002 Waterloo Mathematics Foundation
Calculators are permitted, provided they are non-programmable and without graphic displays. Do not open this booklet until instructed to do so. The paper consists of 10 questions, each worth 10 marks. Parts of each question can be of two types. SHORT ANSWER parts are worth 2 marks each (questions 1-2) or 3 marks each (questions 3-7). FULL SOLUTION parts are worth the remainder of the 10 marks for the question. Instructions for SHORT ANSWER parts: 1. SHORT ANSWER parts are indicated like this: .
Manulife Financial
Canadian Institute of Actuaries
Equitable Life of Canada
Great West Life and London Life
Chartered Accountants
Sybase Inc. (Waterloo)
iAnywhere Solutions
1 . f ( x)
(ii)

Fryer滑铁卢数学竞赛(Grade 9)-数学Mathematics-2009-试题 exam

2009Fryer Contest(Grade9)Wednesday,April8,20091.Emily sets up a lemonade stand.She has set-up costs of$12.00and each cup of lemonade costsher$0.15to make.She sells each cup of lemonade for$0.75.(a)What is the total cost,including the set-up,for her to make100cups of lemonade?(b)What is her proﬁt(money earned minus total cost)if she sells100cups of lemonade?(c)What is the number of cups that she must sell to break even(that is,to have a proﬁtof$0)?(d)Why is it not possible for her to make a proﬁt of exactly$17.00?2.If a>0and b>0,a new operation∇is deﬁned as follows:a∇b=a+b 1+ab.For example,3∇6=3+61+3×6=919.(a)Calculate2∇5.(b)Calculate(1∇2)∇3.(c)If2∇x=57,what is the value of x?(d)For some values of x and y,the value of x∇y is equal to x+y17.Determine all possibleordered pairs of positive integers x and y for which this is true.3.In the diagram,K,O and M are the centres of the three semi-circles.Also,OC=32and CB=36.(a)What is the length of AC?(b)What is the area of the semi-circle with centre K?(c)What is the area of the shadedregion?(d)Line l is drawn to touch the smaller semi-circlesat points S and E so that KS and ME are both perpendicular to l.Determine the area of quadrilateral KSEM.2009Fryer Contest Page24.The addition shown below,representing2+22+222+2222+···,has101rows and the lastterm consists of1012’s:2222222222...22 (2222)+222 (2222)···C B A(a)Determine the value of the ones digit A.(b)Determine the value of the tens digit B and the value of the hundreds digit C.(c)Determine the middle digit of the sum.。

滑铁卢数学竞赛

滑铁卢数学竞赛1、21.|x|>3表示的区间是（）[单选题] *A.（-∞，3）B.(-3,3)C. [-3,3]D. （-∞，-3）∪（3，+ ∞）(正确答案)2、15.下列数中，是无理数的为（）[单选题] *A.-3.14B.6/11C.√3(正确答案)D.03、9.一棵树在离地5米处断裂，树顶落在离树根12米处，问树断之前有多高（）[单选题] *A. 17(正确答案)B. 17.5C. 18D. 204、若tan（π-α）>0且cosα>0，则角α的终边在（）[单选题] *A.第一象限B.第二象限C.第三象限D.第四象限(正确答案)5、下列说法正确的是[单选题] *A.带“+”号和带“-”号的数互为相反数B.数轴上原点两侧的两个点表示的数是相反数C.和一个点距离相等的两个点所表示的数一定互为相反数D.一个数前面添上“-”号即为原数的相反数(正确答案)6、16、在中,则( ). [单选题] *A. AB<2AC (正确答案)B. AB=2ACC. AB>2ACD. AB与2AC关系不确定7、260°是第（）象限角？[单选题] *第一象限第二象限第三象限(正确答案)第四象限8、4.已知两圆的半径分别为3㎝和4㎝，两个圆的圆心距为10㎝,则两圆的位置关系是（）[单选题] *Ａ.内切Ｂ.相交Ｃ.外切Ｄ.外离(正确答案)9、下列各角中，是界限角的是（）[单选题] *A. 1200°B. -1140°C. -1350°(正确答案)D. 1850°10、下列各对象可以组成集合的是（）[单选题] *A、与1非常接近的全体实数B、与2非常接近的全体实数(正确答案)C、高一年级视力比较好的同学D、与无理数相差很小的全体实数11、若2?＝a2＝4 ?，则a?等于( ) [单选题] *A. 43B. 82C. 83(正确答案)D. 4?12、47、若△ABC≌△DEF，AB＝2，AC＝4，且△DEF的周长为奇数，则EF的值为（）[单选题] *A．3B．4C．1或3D．3或5(正确答案)13、8. 下列事件中，不可能发生的事件是(? ? )．[单选题] *A．明天气温为30℃B．学校新调进一位女教师C．大伟身长丈八(正确答案)D．打开电视机，就看到广告14、3．中国是最早采用正负数表示相反意义的量，并进行负数运算的国家．若零上10℃记作+10℃，则零下10℃可记作（）[单选题] *A．10℃B．0℃C.-10 ℃(正确答案)D．-20℃15、19．对于实数a、b、c,“a>b”是“ac2(c平方)>bc2(c平方) ; ”的（）[单选题] * A．充分不必要条件B．必要不充分条件(正确答案)C．充要条件D．既不充分也不必要条件16、22．如图棋盘上有黑、白两色棋子若干，找出所有使三颗颜色相同的棋在同一直线上的直线，满足这种条件的直线共有（）[单选题] *A．5条(正确答案)B．4条C．3条D．2条17、f（x）=-2x+5在x=1处的函数值为（）[单选题] *A、-3B、-4C、5D、3(正确答案)18、10. 如图所示，小明周末到外婆家，走到十字路口处，记不清哪条路通往外婆家，那么他一次选对路的概率是(? ? ?)．[单选题] *A．1/2B．1/3(正确答案)C．1/4D．119、15．如图所示，下列数轴的画法正确的是（）[单选题] *A．B．C．(正确答案)D．20、-950°是（）[单选题] *A. 第一象限角B. 第二象限角(正确答案)C. 第三象限角D. 第四象限角21、1．如图，∠AOB＝120°，∠AOC＝∠BOC，OM平分∠BOC，则∠AOM的度数为（）[单选题] *A．45°B．65°C．75°(正确答案)D．80°22、27．下列计算正确的是（）[单选题] *A．（﹣a3）2＝a6(正确答案)B．3a+2b＝5abC．a6÷a3＝a2D．（a+b）2＝a2+b223、16．“x2（x平方）－4x－5=0”是“x=5”的( ) [单选题] *A．充分不必要条件B．必要不充分条件(正确答案)C．充要条件D．既不充分也不必要条件24、19．下列两个数互为相反数的是（）[单选题] *A．（﹣）和﹣（﹣）B．﹣5和(正确答案)C．π和﹣14D．+20和﹣（﹣20）25、13．在数轴上，下列四个数中离原点最近的数是（）[单选题] *A．﹣4(正确答案)B．3C．﹣2D．626、14．命题“?x∈R，?n∈N*，使得n≥x2（x平方）”的否定形式是（）[单选题] * A．?x∈R，?n∈N*，使得n<x2B．?x∈R，?x∈N*，使得n<x2C．?x∈R，?n∈N*，使得n<x2D．?x∈R，?n∈N*，使得n<x2(正确答案)27、8．(2020·课标Ⅱ)已知集合U={－2，－1,0,1,2,3}，A={－1,0,1}，B={1,2}，则?U(A∪B)=( ) [单选题] *A．{－2,3}(正确答案)B．{－2,2,3}C．{－2，－1,0,3}D．{－2，－1,0,2,3}28、7人小组选出2名同学作正副组长，共有选法（）种。

Cayley滑铁卢数学竞赛(Grade 10)-数学Mathematics-1998-试题 exam

Chartered Accountants SybaseInc. (Waterloo) IBMCanada Ltd.Canadian Institute of ActuariesDo not open the contest booklet until you are told to do so.You may use rulers, compasses and paper for rough work.Calculators are permitted, providing they are non-programmable and without graphic displays.Part A: Each question is worth 5 credits.1.The value of 03012..()+ is(A ) 0.7(B ) 1(C ) 0.1(D ) 0.19(E ) 0.1092.The pie chart shows a percentage breakdown of 1000 votesin a student election. How many votes did Sue receive?(A ) 550(B ) 350(C ) 330(D ) 450(E ) 9353.The expression a a a 9153× is equal to(A ) a 45(B ) a 8(C ) a 18(D ) a 14(E ) a 214.The product of two positive integers p and q is 100. What is the largest possible value of p q +?(A ) 52(B ) 101(C ) 20(D ) 29(E ) 255.In the diagram, ABCD is a rectangle with DC =12. If the area of triangle BDC is 30, what is the perimeter ofrectangle ABCD ?(A ) 34(B ) 44(C ) 30(D ) 29(E ) 606.If x =2 is a solution of the equation qx –311=, the value of q is (A ) 4(B ) 7(C ) 14(D ) –7(E ) –47.In the diagram, AB is parallel to CD . What is the value ofy ?(A ) 75(B ) 40(C ) 35(D ) 55(E ) 508.The vertices of a triangle have coordinates 11,(), 71,() and 53,(). What is the area of this triangle?(A ) 12(B ) 8(C ) 6(D ) 7(E ) 99.The number in an unshaded square is obtained by adding thenumbers connected to it from the row above. (The ‘11’ is one such number.) The value of x must be (A ) 4(B ) 6(C ) 9(D ) 15(E) 10Scoring:There is no penalty for an incorrect answer.Each unanswered question is worth 2 credits, to a maximum of 20 credits.A BCD DAC B10.The sum of the digits of a five-digit positive integer is 2. (A five-digit integer cannot start with zero.)The number of such integers is(A ) 1(B ) 2(C ) 3(D ) 4(E ) 5Part B: Each question is worth 6 credits.11.If x y z ++=25, x y +=19 and y z +=18, then y equals(A ) 13(B ) 17(C ) 12(D ) 6(E ) –612. A regular pentagon with centre C is shown. The value of xis(A ) 144(B ) 150(C ) 120(D ) 108(E ) 7213.If the surface area of a cube is 54, what is its volume?(A ) 36(B ) 9(C ) 8138(D ) 27(E ) 162614.The number of solutions x y ,() of the equation 3100x y +=, where x and y are positive integers, is(A ) 33(B ) 35(C ) 100(D ) 101(E ) 9715.If y –55= and 28x =, then x y + equals(A ) 13(B ) 28(C ) 3316.Rectangle ABCDhas length 9 and width 5. Diagonal is divided into 5 equal parts at W , X , Y , and Z area of the shaded region.(A ) 36(B ) 365(C ) 18(D ) 41065(E ) 2106517.If N p q =()()()+75243 is a perfect cube, where p and q are positive integers, the smallest possible valueof p q + is(A ) 5(B ) 2(C ) 8(D ) 6(E ) 1218.Q is the point of intersection of the diagonals of one face ofa cube whose edges have length 2 units. The length of QRis(A ) 2(B ) 8(C ) 5(D ) 12(E ) 619.Mr. Anderson has more than 25 students in his class. He has more than 2 but fewer than 10 boys andmore than 14 but fewer than 23 girls in his class. How many different class sizes would satisfy these conditions?(A ) 5(B ) 6(C ) 7(D ) 3(E ) 420.Each side of square ABCD is 8. A circle is drawn through A and D so that it is tangent to BC . What is the radius of thiscircle?(A ) 4(B ) 5(C ) 6(D ) 42(E ) 5.25Part C: Each question is worth 8 credits.21.When Betty substitutes x =1 into the expression ax x c 32–+ its value is –5. When she substitutesx =4 the expression has value 52. One value of x that makes the expression equal to zero is(A ) 2(B ) 52(C ) 3(D ) 72(E ) 422. A wheel of radius 8 rolls along the diameter of a semicircleof radius 25 until it bumps into this semicircle. What is thelength of the portion of the diameter that cannot be touchedby the wheel?(A ) 8(B ) 12(C ) 15(D ) 17(E ) 2023.There are four unequal, positive integers a , b , c , and N such that N a b c =++535. It is also true thatN a b c =++454 and N is between 131 and 150. What is the value of a b c ++?(A ) 13(B ) 17(C ) 22(D ) 33(E ) 3624.Three rugs have a combined area of 2002m . By overlapping the rugs to cover a floor area of 1402m ,the area which is covered by exactly two layers of rug is 242m . What area of floor is covered by three layers of rug?(A ) 122m (B ) 182m (C ) 242m (D) 362m (E ) 422m 25.One way to pack a 100 by 100 square with 10000 circles, each of diameter 1, is to put them in 100rows with 100 circles in each row. If the circles are repacked so that the centres of any three tangent circles form an equilateral triangle, what is the maximum number of additional circles that can be packed?(A ) 647(B ) 1442(C ) 1343(D) 1443(E ) 1344。

1992 年加拿大数学竞赛试题

所有 U 及 V 的位置, 使得四邊形 P UQV 之面積為最大。
4. 5.
試解方程式 x2 +
x2 = 3。 (x + 1)2
假定一副牌由一張百搭 (Joker) 及其他 2n 張牌組成, 而對 1, 2, . . . , n 中之每一個數 k
均有兩張牌的號碼是 k。今欲將此 2n + 1 張牌排成一列, 滿足下述條件: 百搭在當中, 而對任意整數 k (1 ≤ k ≤ n), 在兩張號碼為 k 的牌中間恰好有 k − 1 張牌。 (a) 試求出所有使這種排列可能之 n (n ≤ 10) 之值。 (b) 對那些 n 之值所求之排列為不可能? —本文譯者任教於加拿大滑鐵盧之 Wilfrid Laurier 大學並為加拿大數學競賽委員會之會員—
1992 年加拿大數學競賽試題
王子俠譯
1. 試證 1 × 2 × . . . × n 能被 1 + 2 + . . . + n 整除之充分且必要條件為: n + 1 不為奇質數。 2. 設 x, y, z 為非負實數。試證不等式 x(x − z)2 + y(y − z)2 ≥ (x − z)(y − z)(x + y − z) 並決定出等號成立பைடு நூலகம்情形。 3. 在附圖中,ABCD 為一正方形。U及V 分別表在邊 AB 及 CD 上之任意內點。試決定出
1

2014CIMC滑铁卢国际加拿大数学竞赛

Canadian Intermediate Mathematics Contest NOTE:1.Please read the instructions on the front cover of this booklet.2.Write solutions in the answer booklet provided.3.It is expected that all calculations and answers will be expressed as exact numbers such as 4π,2+√7,etc.,rather than as 12.566...or4.646....4.While calculators may be used for numerical calculations,other mathematical steps must be shown and justiﬁed in your written solutions and speciﬁc marks may be allocated for these steps.For example,while your calculator might be able to ﬁnd the x -intercepts of the graph of an equation like y =x 3−x ,you should show the algebraic steps that you used to ﬁnd these numbers,rather than simply writing these numbers down.5.Diagrams are not drawn to scale.They are intended as aids only.6.No student may write both the Canadian Senior Mathematics Contest and the Canadian Intermediate Mathematics Contest in the same year.PART AFor each question in Part A,full marks will be given for a correct answer which is placed in the box.Part marks will be awarded only if relevant work is shown in the space provided in the answer booklet.1.There are 200people at the beach,and 65%of these people are children.If 40%of the children are swimming,how many children are swimming?2.If x +2y =14and y =3,what is the value of 2x +3y ?3.In the diagram,ABCD is a rectangle with points Pand Q on AD so that AB =AP =P Q =QD .Also,point R is on DC with DR =RC .If BC =24,whatis the area of P QR ?A B CD P Q R 4.At a given time,the depth of snow in Kingston is 12.1cm and the depth of snow in Hamilton is 18.6cm.Over the next thirteen hours,it snows at a constant rate of2.6cm per hour in Kingston and at a constant rate of x cm per hour in Hamilton.At the end of these thirteen hours,the depth of snow in Kingston is the same as the depth of snow in Hamilton.What is the value of x ?5.Scott stacks golfballs to make a pyramid.The ﬁrst layer,or base,of the pyramid is a square of golfballs and rests on a ﬂat table.Each golfball,above the ﬁrst layer,rests in a pocket formed by four golfballs in the layer below (as shown in Figure 1).Each layer,including the ﬁrst layer,is completely ﬁlled.For example,golfballs can be stacked into a pyramid with 3levels,as shown in Figure 2.The four triangular faces of the pyramid in Figure 2include a total of exactly 13diﬀerent golfballs.Scott makes a pyramid in which the four triangular faces include a total of exactly 145diﬀerent golfballs.How many layers does this pyramid have?Figure 1Figure 26.A positive integer is a prime number if it is greater than1and has no positive divisorsother than1and itself.For example,the number5is a prime number because its only two positive divisors are1and5.The integer43797satisﬁes the following conditions:•each pair of neighbouring digits(read from left to right)forms a two-digit primenumber,and•all of the prime numbers formed by these pairs are diﬀerent,because43,37,79,and97are all diﬀerent prime numbers.There are many integers with more thanﬁve digits that satisfy both of these conditions.What is the largest positive integer that satisﬁes both of these conditions?PART BFor each question in Part B,your solution must be well organized and contain words of explanation or justiﬁcation.Marks are awarded for completeness,clarity,and style of presentation.A correct solution,poorly presented,will not earn full marks.1.(a)Determine the average of the six integers22,23,23,25,26,31.(b)The average of the three numbers y+7,2y−9,8y+6is27.What is the valueof y?(c)Four positive integers,not necessarily diﬀerent and each less than100,have anaverage of94.Determine,with explanation,the minimum possible value forone of these integers.2.(a)In the diagram, P QR is right-angled at R.If P Q=25and RQ=24,determine the perimeterand area of P QR.PQR(b)In the diagram, ABC is right-angled at C withAB=c,AC=b,and BC=a.Also, ABC has perimeter144and area504.Determine all possible values of c.(You may use the facts that,for any numbers x and y, (x+y)2=x2+2xy+y2and(x−y)2=x2−2xy+y2.)AB C ab cCanadian Intermediate Mathematics Contest(English)20143.Vicky starts with a list(a,b,c,d)of four digits.Each digit is0,1,2,or3.Vickyenters the list into a machine to produce a new list(w,x,y,z).In the new list,w is the number of0s in the original list,while x,y and z are the numbers of1s,2s and3s,respectively,in the original list.For example,if Vicky enters(1,3,0,1),the machine produces(1,2,0,1).(a)What does the machine produce when Vicky enters(2,3,3,0)?(b)Vicky enters(a,b,c,d)and the machine produces the identical list(a,b,c,d).Determine all possible values of b+2c+3d.(c)Determine all possible lists(a,b,c,d)with the property that when Vicky enters(a,b,c,d),the machine produces the identical list(a,b,c,d).(d)Vicky buys a new machine into which she can enter a list of ten digits.Eachdigit is0,1,2,3,4,5,6,7,8,or9.The machine produces a new list whoseentries are,in order,the numbers of0s,1s,2s,3s,4s,5s,6s,7s,8s,and9s inthe original list.Determine all possible lists,L,of ten digits with the propertythat when Vicky enters L,the machine produces the identical list L.。

2007PascalSolution滑铁卢竞赛题答案

1.Calculating,3×(7−5)−5=3×2−5=6−5=1.Answer:(B) 2.Since x is less than−1and greater than−2,then the best estimate of the given choices is−1.3.Answer:(B) 3.The shaded square has side length1so has area12=1.The rectangle has dimensions3by5so has area3×5=15.Thus,the fraction of the rectangle that is shaded is115.Answer:(A)4.Calculating,25−52=32−25=7.Answer:(E) 5.In3hours,Leona earns$24.75,so she makes$24.75÷3=$8.25per hour.Therefore,in a5hour shift,Leona earns5×$8.25=$41.25.Answer:(E)6.Calculating,√64+√36√64+36=8+6√100=1410=75.Answer:(A)7.Solution1In total Megan and Dan inherit$1010000.Since each donates10%,then the total donated is10%of$1010000,or$101000.Solution2Megan donates10%of$1000000,or$100000.Dan donates10%of$10000,or$1000.In total,they donate$100000+$1000=$101000.Answer:(A) 8.We think of BC as the base of ABC.Its length is12.Since the y-coordinate of A is9,then the height of ABC from base BC is9.Therefore,the area of ABC is12(12)(9)=54.Answer:(B)9.Calculating the given diﬀerence using a common denominator,we obtain58−116=916.Since916is larger than each of12=816and716,then neither(D)nor(E)is correct.Since916is less than each of34=1216,then(A)is not correct.As a decimal,9=0.5625.Since35=0.6and59=0.5,then916>59,so(C)is the correct answer.Answer:(C)10.Since M=2007÷3,then M=669.Since N=M÷3,then N=669÷3=223.Since X=M−N,then X=669−223=446.Answer:(E)11.The mean of6,9and18is 6+9+183=333=11.Thus the mean of12and y is11,so the sum of12and y is2(11)=22,so y=10.Answer:(C) 12.In P QR,since P R=RQ,then∠RP Q=∠P QR=48◦.Since∠MP N and∠RP Q are opposite angles,then∠MP N=∠RP Q=48◦.In P MN,P M=P N,so∠P MN=∠P NM.Therefore,∠P MN=12(180◦−∠MP N)=12(180◦−48◦)=12(132◦)=66◦.Answer:(D)13.The prime numbers smaller than10are2,3,5,and7.The two of these numbers which are diﬀerent and add to10are3and7.The product of3and7is3×7=21.Answer:(B) 14.Since there were21males writing and the ratio of males to females writing is3:7,then thereare73×21=49females writing.Therefore,the total number of students writing is49+21=70.Answer:(D) 15.Solution1Theﬁrst stack is made up of1+2+3+4+5=15blocks.The second stack is made up of1+2+3+4+5+6=21blocks.There are36blocks in total.We start building the new stack from the top.Since there are more than21blocks,we need at least6rows.For7rows,1+2+3+4+5+6+7=28blocks are needed.For8rows,1+2+3+4+5+6+7+8=36blocks are needed.Therefore,Clara can build a stack with0blocks leftover.Solution2Since the new stack will be larger than the second stack shown,let us think about adding new rows to this second stack using the blocks from theﬁrst stack.Theﬁrst stack contains1+2+3+4+5=15blocks in total.Theﬁrst two rows that we would add to the bottom of the second stack would have7and8 blocks in them,for a total of15blocks.This uses all of the blocks from theﬁrst stack,with none left over,and creates a similar stack.Therefore,there are0blocks left over.Answer:(A) 16.Solution1The sum of the numbers in the second row is10+16+22=48,so the sum of the numbers in any row,column or diagonal is48.In theﬁrst row,P+4+Q=48so P+Q=44.In the third row,R+28+S=48so R+S=20.Therefore,P+Q+R+S=44+20=64.Solution 2The sum of the numbers in the second row is 10+16+22=48,so the sum of the numbers in any row,column or diagonal is 48.From the ﬁrst row,P +4+Q =48so P +Q =44.From the ﬁrst column,P +10+R =48so P +R =38.Subtracting these two equations gives (P +Q )−(P +R )=44−38or Q −R =6.From one of the diagonals,R +16+Q =48or Q +R =32.Adding these last two equations,2Q =38or Q =19,so R =32−Q =13.Also,P =44−Q =25.From the last row,13+28+S =48,or S =7.Thus,P +Q +R +S =25+19+13+7=64.Solution 3The sum of all of the numbers in the grid isP +Q +R +S +10+16+22+28+4=P +Q +R +S +80But the sum of the three numbers in the second column is 4+16+28=48,so the sum of the three numbers in each column is 48.Thus,the total of the nine numbers in the grid is 3(48)=144,so P +Q +R +S +80=144or P +Q +R +S =64.Answer:(C)17.At present,the sum of Norine’s age and the number of years that she has worked is 50+19=69.This total must increase by 85−69=16before she can retire.As every year passes,this total increases by 2(as her age increases by 1and the number of years that she has worked increases by 1).Thus,it takes 8years for her total to increase from 69to 85,so she will be 50+8=58when she can retire.Answer:(C)18.By the Pythagorean Theorem in P QR ,P Q 2=P R 2−QR 2=132−52=144,so P Q =√144=12.By the Pythagorean Theorem in P QS ,QS 2=P S 2−P Q 2=372−122=1225,so QS =√1225=35.Therefore,the perimeter of P QS is 12+35+37=84.Answer:(D)19.Since the reciprocal of 310is 1x+1 ,then 1x +1=1031x =73x =37so x =37.Answer:(C)20.Draw a line from F to BC ,parallel to AB ,meeting BC at P .AD CB E F PSince EB is parallel to F P and ∠F EB =90◦,then EBP F is a rectangle.Since EB =40,then F P =40;since EF =30,then BP =30.Since AD =80,then BC =80,so P C =80−30=50.Therefore,the area of EBCF is sum of the areas of rectangle EBP F (which is 30×40=1200)and F P C (which is 12(40)(50)=1000),or 1200+1000=2200.Since the areas of AEF CD and EBCF are equal,then each is 2200,so the total area ofrectangle ABCD is 4400.Since AD =80,then AB =4400÷80=55.Therefore,AE =AB −EB =55−40=15.Answer:(D)21.Let us ﬁrst consider the possibilities for each integer separately:•The two-digit prime numbers are 11,13,17,19.The only one whose digits add up to a prime number is 11.Therefore,P =11.•Since Q is a multiple of 5between 2and 19,then the possible values of Q are 5,10,15.•The odd numbers between 2and 19that are not prime are 9and 15,so the possible values of R are 9and 15.•The squares between 2and 19are 4,9and 16.Only 4and 9are squares of prime numbers,so the possible values of S are 4and 9.•Since P =11,the possible value of Q are 5,10and 15,and T is the average of P and Q ,then T could be 8,10.5or 13.Since T is also a prime number,then T must be 13,so Q =15.We now know that P =11,Q =15and T =13.Since the ﬁve numbers are all diﬀerent,then R cannot be 15,so R =9.Since R =9,S cannot be 9,so S =4.Therefore,the largest of the ﬁve integers is Q =15.Answer:(B)22.By the Pythagorean Theorem,P R =QR 2+P Q 2=√152+82=√289=17km.Asafa runs a total distance of 8+15+7=30km at 21km/h in the same time that Florence runs a total distance of 17+7=24km.Therefore,Asafa’s speed is 30=5of Florence’s speed,so Florence’s speed is 4×21=84km/h.Asafa runs the last 7km in 721=13hour,or 20minutes.Florence runs the last7km in 7845=3584=512hour,or25minutes.Since Asafa and Florence arrive at S together,then Florence arrived at R5minutes before Asafa.Answer:(E) 23.The total area of the larger circle isπ(22)=4π,so the total area of the shaded regions mustbe512(4π)=53π.Suppose that∠ADC=x◦.The area of the unshaded portion of the inner circle is thusx360of the total area of the innercircle,orx360(π(12))=x360π(since∠ADC isx360of the largest possible central angle(360◦)).The area of the shaded portion of the inner circle is thusπ−x360π=360−x360π.The total area of the outer ring is the diﬀerence of the areas of the outer and inner circles,orπ(22)−π(12)=3π.The shaded area in the outer ring will bex360of this total area,since∠ADC isx360of thelargest possible central angle(360◦). So the shaded area in the outer ring isx360(3π)=3x360π.So the total shaded area(which must equal 53π)is,in term of x,3x360π+360−x360π=360+2x360π.Therefore,360+2x360=53=600360,so360+2x=600or x=120.Thus,∠ADC=120◦.Answer:(B) 24.First,we complete the next several spaces in the spiral to try to get a better sense of thepattern:1716151413185431219612112078910212223242526We notice from this extended spiral that the odd perfect squares lie on a diagonal extending down and to the right from the1,since1,9,25and so on will complete a square of numbers when they are written.(Try blocking out the numbers larger than each of these to see this.) This pattern does continue since when each of these odd perfect squares is reached,the number of spaces up to that point in the sequence actually does form a square.Theﬁrst odd perfect square larger than2007is452=2025.2025will lie18spaces to the left of2007in this row.(The row with2025will actually be long enough to be able to move18spaces to the left from2025.)The odd perfect square before2025is432=1849,so1850will be the number directly above 2025,as the row containing1849will continue one more space to the right before turning up.Since1850is directly above2025,then1832is directly above2007.The odd perfect square after2025is472=2209,so2208will be the number directly below 2025,since2209will be one space to the right and one down.Since2208is directly below2025,then2190is directly below2007.Therefore,the sum of the numbers directly above and below2007is1832+2190=4022.Answer:(E) 25.For x and3x to each have even digits only,x must be in one of the following forms.(Here,a,b,c represent digits that can each be0,2or8,and n is a digit that can only be2or8.)•nabc(2×3×3×3=54possibilities)•na68(2×3=6possibilities)•n68a(2×3=6possibilities)•68ab(3×3=9possibilities)•n668(2possibilities)•668a(3possibilities)•6668(1possibility)•6868(1possibility)In total,there are82possibilities for x.In general terms,these are the only forms that work,since digits of0,2and8in x pro-duce even digits with an even“carry”(0or2)thus keeping all digits in3x even,while a6may be used,but must be followed by8or68or668in order to give a carry of2.More precisely,why do these forms work,and why are they the only forms that work?First,we note that3×0=0,3×2=6,3×4=12,3×6=18and3×8=24.Thus,each even digit of x will produce an even digit in the corresponding position of3x,but may aﬀect the next digit to the left in3x through its“carry”.Note that a digit of0or2in x produces no carry,while a digit of8in x produces an even carry.Therefore,none of these three digits can possibly create an odd digit in3x(either directly or through carrying),as they each create an even digit in the corresponding position of3x and do not aﬀect whether the next digit to the left is even or odd.(We should note that the carry into any digit in3x can never be more than2,so we do not have to worry about creating a carry of1from a digit in x of0or2,or a carry of3from a digit of8in x through multiple carries.) So a digit of2or8can appear in any position of x and a digit of0can appear in any position of x except for theﬁrst position.A digit of4can never appear in x,as it will always produce a carry of1,and so will al-ways create an odd digit in3x.A digit of6can appear in x,as long as the carry from the previous digit is2to make thecarry forward from the6equal to2.(The carry into the6cannot be larger than2.)When this happens,we have3×6+Carry=20,and so a2is carried forward,which does not aﬀect whether next digit is even or odd.A carry of2can occur if the digit before the6is an8,or if the digit before the6is a6whichis preceded by8or by68.Combining the possible uses of0,2,6,and8gives us the list of possible forms above,and hence82possible values for x.Answer:(A)。

1999滑铁卢竞赛试题答案

( x – 2)( x + 1) = 0
∴ x = 2 or x = –1
Solution 2 1 6 x2 – x – 6 1– – 2 = x x x2 x 2 – x – 2) – 4 ( = x2 –4 2 2 (since x – x – 2 = 0 ) x 2 But x – x – 2 = ( x – 2)( x + 1) = 0 ∴ x = 2 or x = –1. x = –1, Substituting x = 2, or –4 –4 = 4 1 = –1. = –4 =
y
6
–3
–1 O
x
Solution 3 Let the equation of the parabola be y = a( x + 2)2 + c . Since (0, 6) is on parabola, 6 = 4 a + c , and ( –1, 0) is on parabola, 0 = a + c . Solving, a = 2, c = – 2. ∴ Equation is y = 2( x + 2)2 – 2 . 3. (a) How many equilateral triangles of side 1 cm, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side 10 cm?
(b)
If the point P( – 3, 2) is on the line 3 x + 7ky = 5 , what is the value of k? Solution Since P is on the line, its coordinates must satisfy the equation of the line. Thus, 3( – 3) + 7k (2) = 5 14 k = 14 k =1

滑铁卢数学竞赛

滑铁卢数学竞赛年1月10日更新这篇主要来介绍一下滑铁卢系列数学竞赛，下面是该项赛事的特点以及备考建议：【特点】（1）全年龄段，从7年级到12年级都有，建议参加12年级Euclid 竞赛；（2）全球统考，高含金量，对于申请有帮助；（2）有些比赛项目并不是选择题，是填空、简答题，对于英语表达有一定要求；（3）难度相对比较小，获个奖比较容易；【建议】做真题！做真题！做真题！真题网址：下面是Waterloo数学竞赛的详细介绍以及21年Euclid真题卷：滑铁卢大学始建于1957年，在加拿大最权威的教育杂志Maclean's （麦克林）的排名榜上，连续五年综合排名第一第二。

滑铁卢大学设有加拿大唯一一所数学学院，这也是北美乃至全世界最大的数学学院，因滑铁卢大学在数学领域的优良声誉及传统，以及欧几里德数学竞赛考察标准的严格性和专业性，该竞赛成绩在加拿大和美国大学中已经得到广泛认可，被誉为类似加拿大“数学托福”的考试。

官方网址：【比赛特点】（1）可选择的比赛种类非常多，有适合各种不同年龄段学生的比赛；（2）比赛难度相对较小，比较适合想参加数学类竞赛，但本身数学程度并不是特别出挑的学生；（3）在加拿大的高校中有比较大的影响力，尤其是Euclid的比赛成绩；（4）部分赛题以测试学生的分析能力、逻辑思维能力为主，在2021年MAT考试中出现了21年Euclid的类似试题。

【比赛形式】注：上述数学竞赛都可以使用计算器。

【2021-2022赛程】大家可以看到Waterloo数学竞赛各项赛事跨度比较大，主要集中在四、五月份，特别是Euclid数学竞赛，是一项含金量比较高的全球数学竞赛。

【报名方式】当地可提供报名的机构或以自己学校的名义注册报名注：Waterloo报名时需要填写一个学校代码，所以一般需要通过学校报名。

【奖项设置】（1）国际生全球排名前25%的学生将获得杰出荣誉证书，2021年大概68分；（2）在参赛学校中成绩最高的学生会得到一个校级冠军的奖牌；（3）每位参赛者都可以获得一个参赛证书；从上述奖项的设置可以看出，我们参加Waterloo数学竞赛获一个奖相对而言是比较容易的。

滑铁卢数学竞赛高中试题

滑铁卢数学竞赛高中试题一、选择题1. 已知函数$ f(x) = ax^2 + bx + c $，其中$ a, b, c $为实数，且$ f(1) = 2 $，$ f(-1) = 0 $，$ f(2) = 6 $。

求$ a $的值。

2. 一个圆的半径为5，圆心位于原点，求圆上点$ P(3,4) $到圆心的距离。

3. 若$ \sin(\alpha + \beta) = \frac{1}{2} $，$ \cos(\alpha + \beta) = \frac{\sqrt{3}}{2} $，且$ \alpha $在第二象限，$ \beta $在第一象限，求$ \sin(\alpha) $的值。

二、填空题1. 计算$ \int_{0}^{1} x^2 dx $。

2. 若$ \log_{2}8 = n $，则$ n $的值为______。

3. 一个等差数列的前三项分别为2，5，8，求该数列的第10项。

三、解答题1. 证明：对于任意正整数$ n $，$ 1^3 + 2^3 + ... + n^3 =\frac{n^2(n+1)^2}{4} $。

2. 一个矩形的长是宽的两倍，若矩形的周长为24，求矩形的面积。

3. 已知一个等比数列的前三项分别为3，9，27，求该数列的第5项。

四、应用题1. 一个工厂每天生产相同数量的零件，如果每天生产100个零件，工厂可以在30天内完成订单。

如果每天生产150个零件，工厂可以在20天内完成订单。

求工厂每天实际生产的零件数量。

2. 一个圆环的外圆半径是内圆半径的两倍，且圆环的面积为π。

求外圆的半径。

五、证明题1. 证明：对于任意实数$ x $，$ \cos(x) + \cos(2x) + \cos(3x) $可以表示为一个单一的余弦函数。

六、开放性问题1. 考虑一个无限大的棋盘，每个格子可以放置一个硬币。

2014-2012加拿大滑铁卢大学11年级数学竞赛试题

2014-2012加拿⼤滑铁卢⼤学11年级数学竞赛试题1.For real numbers a and b with a≥0and b≥0,the operation is de?ned bya b=√For example,5 1=5+4(1)=√9=3.(a)What is the value of8 7?(b)If16 n=10,what is the value of n?(c)Determine the value of(9 18) 10.(d)With justi?cation,determine all possible values of k such that k k=k.2.Each week,the MathTunes Music Store releases a list of the Top200songs.A newsong“Recursive Case”is released in time to make it onto the Week1list.The song’s position,P,on the list in a certain week,w,is given by the equation P=3w2?36w+110.The week number w is always a positive integer.(a)What position does the song have on week1?(b)Artists want their song to reach the best position possible.The closer that theposition of a song is to position#1,the better the position.(i)What is the best position that the song“Recursive Case”reaches?(ii)On what week does this song reach its best position?(c)What is the last week that“Recursive Case”appears on the Top200list?3.A pyramid ABCDE has a square base ABCD of side length 20.Vertex E lies on theline perpendicular to the base that passes through F ,the centre of the base ABCD .It is given that EA =EB =EC =ED = 18.(a)Determine the surface area of the pyramidABCDEincluding its base.(b)Determine the height EF of the pyramid.A B C D EF 2018(c)G and H are the midpoints of ED and EA ,respectively.Determine the area of thequadrilateral BCGH .4.The triple ofpositive integers (x,y,z )is called an Almost Pythagorean Triple (or APT)if x >1and y >1and x 2+y 2=z 2+1.For example, (5,5,7)is an APT.(a)Determine the values of y and z so that (4,y,z )is an APT.(b)Prove that for any triangle whose side lengths form an APT,the area of thetriangle is not an integer.(c)Determine two 5-tuples (b,c,p,q,r )of positive integers with p ≥100for which(5t +p,bt +q,ct +r )is an APT for all positive integers t .1.At the JK Mall grand opening,some lucky shoppers are able to participate in a moneygiveaway.A large box has been?lled with many$5,$10,$20,and$50bills.The lucky shopper reaches into the box and is allowed to pull out one handful of bills.(a)Rad pulls out at least two bills of each type and his total sum of money is$175.What is the total number of bills that Rad pulled out?(b)Sandy pulls out exactly?ve bills and notices that she has at least one bill of eachtype.What are the possible sums of money that Sandy could have?(c)Lino pulls out six or fewer bills and his total sum of money is$160.There areexactly four possibilities for the number of each type of bill that Lino could have.Determine these four possibilities.2.A parabola has equation y=(x?3)2+1.(a)What are the coordinates of the vertex of the parabola?(b)A new parabola is created by translating the original parabola3units to the leftand3units up.What is the equation of the translated parabola?(c)Determine the coordinates of the point of intersection of these two parabolas.(d)The parabola with equation y=ax2+4,a<0,touches the parabola withequation y=(x?3)2+1at exactly one point.Determine the value of a.3.A sequence of m P’s and n Q’s with m>n is called non-predictive if there is some pointin the sequence where the number of Q’s counted from the left is greater than or equal to the number of P’s counted from the left.For example,if m=5and n=2the sequence PPQQPPP is non-predictive because in counting the?rst four letters from the left,the number of Q’s is equal to the number of P’s.Also,the sequence QPPPQPP is non-predictive because in counting the? rst letterfrom the left,the number of Q’s is greater than the number ofP’s.(a)If m=7and n=2,determine the number of non-predictive sequences that beginwith P.(b)Suppose that n=2.Show that for every m>2,the number of non-predictivesequences that begin with P is equal to the number of non-predictive sequences that begin with Q.(c)Determine the number of non-predictive sequences with m=10and n=3.4.(a)Twenty cubes,each with edge length1cm,are placed together in4rows of5.What isthe surface area of this rectangularprism?(b)A number of cubes,each with edge length1cm,are arranged to form a rectangularprism having height1cm and a surface area of180cm2.Determine the number of cubes in the rectangular prism.(c)A number of cubes,each with edge length1cm,are arranged to form a rectangularprism having length l cm,width w cm,and thickness1cm.A frame is formed byremoving a rectangular prism with thickness 1cm located k cm from each of the sides of the original rectangular prism,as shown. Each of l,w and k is a positive integer.If the frame has surface area532cm2,determine all possible values for l and w such that l≥w.l cmw cmk cmk cmk cmk cm1 cm1.Quadrilateral P QRS is constructed with QR =51,as shown.The diagonals of P QRS intersect at 90?at point T ,such that P T =32and QT =24.322451P QRST (a)Calculate the length of P Q.(b)Calculate the area of P QR .(c)If QS :P R =12:11,determine the perimeter of quadrilateral P QRS .2.(a)Determine the value of (a +b )2,given that a 2+b 2=24and ab =6.(b)If (x +y )2=13and x 2+y 2=7,determine the value of xy .(c)If j +k =6and j 2+k 2=52,determine the value of jk .(d)If m 2+n 2=12and m 4+n 4=136,determine all possible values of mn .3.(a)Points M (12,14)and N (n,n 2)lie on theparabola with equation y =x 2,as shown.Determine the value of n such that∠MON =90?.yx(b)Points A (2,4)and B (b,b 2)are the endpointsofa chord of the parabola with equationy =x 2,as shown.Determine the value of bso that ∠ABO =90?.y x(c)Right-angled triangle P QR is inscribed inthe parabola with equation y =x 2,asshown.Points P,Q and R have coordinates(p,p 2),(q,q 2)and (r,r 2),respectively.If p ,qand r are integers,show that 2q +p +r =0.y x4.The positive divisors of 21are 1,3,7and 21.Let S (n )be the sum of the positive divisors of the positive integer n .For example,S (21)=1+3+7+21=32.(a)If p is an odd prime integer,?nd the value of p such that S (2p 2)=2613.(b)The consecutive integers 14and 15have the property that S (14)=S (15).Determine all pairs of consecutive integers m and n such that m =2p and n =9q for prime integers p,q >3,and S (m )=S (n ).(c)Determine the number of pairs of distinct prime integers p and q ,each less than 30,with the property that S (p 3q )is not divisible by 24.。

2008滑铁卢竞赛试题答案

Canadian Mathematics Competition
An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario
ADB ,
AB 2 = BD2 + DA2 = 92 + 122 = 81 + 144 = 225 √ so AB = 225 = 15 since AB > 0. By the Pythagorean Theorem in ADC , DC 2 = CA2 − AD2 = 202 − 122 = 400 − 144 = 256 √ so DC = 256 = 16 since AD > 0. The perimeter of ABC is AB + BC + CA = AB + (BD + DC ) + CA = 15 + (9 + 16) + 20 = 60 Solution 2 Since BD : DA = 9 : 12 = 3 : 4 and BDA is right-angled at B , then 5 BD = 15. to a 3-4-5 triangle. Thus, AB = 3 Since CA : DA = 20 : 12 = 5 : 3 and ADC is right-angled at D, then to a 3-4-5 triangle. Thus, DC = 4 CA = 16. 5 Therefore, the perimeter of ABC is
ADB is similar ADC is similar

2008PascalSolution滑铁卢竞赛题答案

1.Calculating,2+3+42×3×4=924=38.Answer:(E)2.Since3x−9=12,then3x=12+9=21.Since3x=21,then6x=2(3x)=2(21)=42.(Note that we did not need to determine the value of x.)Answer:(A)3.Calculating,√52−42=√25−16=√9=3.Answer:(B)4.Solution1Since JLMR is a rectangle and JR=2,then LM=2.Similarly,since JL=8,then RM=8.Since RM=8and RQ=3,then QM=8−3=5.Since KLMQ is a rectangle with QM=5and LM=2,its area is5(2)=10.Solution2Since JL=8and JR=2,then the area of rectangle JLMR is2(8)=16.Since RQ=3and JR=2,then the area of rectangle JKQR is2(3)=6.The area of rectangle KLMQ is the diﬀerence between these areas,or16−6=10.Answer:(C) 5.Since x=12and y=−6,then3x+y x−y =3(12)+(−6)12−(−6)=3018=53Answer:(C)6.Solution1Since∠P QS is an exterior angle of QRS,then∠P QS=∠QRS+∠QSR,so136◦=x◦+64◦or x=136−64=72.Solution2Since∠P QS=136◦,then∠RQS=180◦−∠P QS=180◦−136◦=44◦.Since the sum of the angles in QRS is180◦,then44◦+64◦+x◦=180◦or x=180−44−64=72.Answer:(A) 7.In total,there are5+6+7+8=26jelly beans in the bag.Since there are8blue jelly beans,the probability of selecting a blue jelly bean is826=413.Answer:(D)8.Since Olave sold108apples in6hours,then she sold108÷6=18apples in one hour.A time period of1hour and30minutes is equivalent to1.5hours.Therefore,Olave will sell1.5×18=27apples in1hour and30minutes.Answer:(A)9.Since the length of the rectangular grid is 10and the grid is 5squares wide,then the side length of each square in the grid is 10÷5=2.There are 4horizontal wires,each of length 10,which thus have a total length of 4×10=40.Since the side length of each square is 2and the rectangular grid is 3squares high,then the length of each vertical wire is 3×2=6.Since there are 6vertical wires,the total length of the vertical wires is 6×6=36.Therefore,the total length of wire is 40+36=76.Answer:(E)10.Solution 1Since Q is at 46and P is at −14,then the distance along the number line from P to Q is 46−(−14)=60.Since S is three-quarters of the way from P to Q ,then S is at −14+34(60)=−14+45=31.Since T is one-third of the way from P to Q ,then T is at −14+13(60)=−14+20=6.Thus,the distance along the number line from T to S is 31−6=25.Solution 2Since Q is at 46and P is at −14,then the distance along the number line from P to Q is 46−(−14)=60.Since S is three-quarters of the way from P to Q and T is one-third of the way from P to Q ,then the distance from T to S is 60 34−13 =60 912−412 =60 512 =25.Answer:(D)11.In total,30+20=50students wrote the Pascal Contest at Mathville Junior High.Since 30%(or 310)of the boys won certiﬁcates and 40%(or 410)of the girls won certiﬁcates,then the total number of certiﬁcates awarded was 310(30)+410(20)=9+8=17.Therefore,17of 50participating students won certiﬁcates.In other words,1750×100%=34%of the participating students won certiﬁcates.Answer:(A)12.Since the perimeter of the rectangle is 56,then2(x +4)+2(x −2)=562x +8+2x −4=564x +4=564x=52x =13Therefore,the rectangle is x +4=17by x −2=11,so has area 17(11)=187.Answer:(B)ing exponent rules,23×22×33×32=23+2×33+2=25×35=(2×3)5=65.Answer:(A)14.Solution 1The wording of the problem tells us that a +b +c +d +e +f must be the same no matter what numbers abc and def are chosen that satisfy the conditions.An example that works is 889+111=1000.In this case,a +b +c +d +e +f =8+8+9+1+1+1=28,so this must always be the value.Solution 2Consider performing this “long addition”by hand.Consider ﬁrst the units column.Since c +f ends in a 0,then c +f =0or c +f =10.The value of c +f cannot be 20or more,as c and f are digits.Since none of the digits is 0,we cannot have c +f =0+0so c +f =10.(This means that we “carry”a 1to the tens column.)Since the result in the tens column is 0and there is a 1carried into this column,then b +e ends in a 9,so we must have b +e =9.(Since b and e are digits,b +e cannot be 19or more.)In the tens column,we thus have b +e =9plus the carry of 1,so the resulting digit in the tens column is 0,with a 1carried to the hundreds column.Using a similar analysis in the hundreds column to that in the tens column,we must have a +d =9.Therefore,a +b +c +d +e +f =(a +d )+(b +e )+(c +f )=9+9+10=28.Answer:(D)15.Each of P SQ and RSQ is right-angled at S ,so we can use the Pythagorean Theorem inboth triangles.In RSQ ,we have QS 2=QR 2−SR 2=252−202=625−400=225,so QS =√225=15since QS >0.In P SQ ,we have P Q 2=P S 2+QS 2=82+225=64+225=289,so P Q =√289=17since P Q >0.Therefore,the perimeter of P QR is P Q +QR +RP =17+25+(20+8)=70.Answer:(E)16.Suppose the radius of the circle is r cm.Then the area M is πr 2cm 2and the circumference N is 2πr cm.Thus,πr 22πr =20or r 2=20or r =40.Answer:(C)17.Solution 1The large cube has a total surface area of 5400cm 2and its surface is made up of 6identical square faces.Thus,the area of each face,in square centimetres,is 5400÷6=900.Because each face is square,the side length of each face is √900=30cm.Therefore,each edge of the cube has length 30cm and so the large cube has a volume of 303=27000cm 3.Because the large cube is cut into small cubes each having volume 216cm 3,then the number of small cubes equals 27000÷216=125.Solution 2Since the large cube has 6square faces of equal area and the total surface area of the cube is 5400cm 2,then the surface area of each face is 5400÷6=900cm 2.Since each face is square,then the side length of each square face of the cube is √900=30cm,and so the edge length of the cube is 30cm.Since each smaller cube has a volume of 216cm 3,then the side length of each smaller cube is 3√216=6cm.Since the side length of the large cube is 30cm and the side length of each smaller cube is 6cm,then 30÷6=5smaller cubes ﬁt along each edge of the large cube.Thus,the large cube is made up of 53=125smaller cubes.Answer:(B)18.Solution1Alex has265cents in total.Since265is not divisible by10,Alex cannot have only dimes,so must have at least1quarter.If Alex has1quarter,then he has265−25=240cents in dimes,so24dimes.Alex cannot have2quarters,since265−2(25)=215is not divisible by10.If Alex has3quarters,then he has265−3(25)=190cents in dimes,so19dimes.Continuing this argument,we can see that Alex cannot have an even number of quarters,since the total value in cents of these quarters would end in a0,making the total value of the dimes end in a5,which is not possible.If Alex has5quarters,then he has265−5(25)=140cents in dimes,so14dimes.If Alex has7quarters,then he has265−7(25)=90cents in dimes,so9dimes.If Alex has9quarters,then he has265−9(25)=40cents in dimes,so4dimes.If Alex has more than9quarters,then he will have even fewer than4dimes,so we do not need to investigate any more possibilities since we are told that Alex has more dimes than quarters.So the possibilities for the total number of coins that Alex has are1+24=25,3+19=22, 5+14=19,and7+9=16.Therefore,the smallest number of coins that Alex could have is16.(Notice that each time we increase the number of quarters above,we are in eﬀect exchanging 2quarters(worth50cents)for5dimes(also worth50cents).)Solution2Suppose that Alex has d dimes and q quarters,where d and q are non-negative integers.Since Alex has$2.65,then10d+25q=265or2d+5q=53.Since the right side is odd,then the left side must be odd,so5q must be odd,so q must be odd.If q≥11,then5q≥55,which is too large.Therefore,q<11,leaving q=1,3,5,7,9which give d=24,19,14,9,4.The solution with d>q and d+q smallest is q=7and d=9,giving16coins in total.Answer:(B) 19.From the deﬁnition,theﬁrst and second digits of an upright integer automatically determinethe third digit,since it is the sum of theﬁrst two digits.Considerﬁrst those upright integers beginning with1.These are101,112,123,134,145,156,167,178,and189,since1+0=1,1+1=2,and so on.(The second digit cannot be9,otherwise the last“digit”would be1+9=10,which is impossible.)There are9such numbers.Beginning with2,the upright integers are202,213,224,235,246,257,268,and279.There are8of them.We can continue the pattern and determine the numbers of the upright integers beginning with 3,4,5,6,7,8,and9to be7,6,5,4,3,2,and1.Therefore,there are9+8+7+6+5+4+3+2+1=45positive3-digit upright integers.Answer:(D) 20.The sum of the six given integers is1867+1993+2019+2025+2109+2121=12134.The four of these integers that have a mean of2008must have a sum of4(2008)=8032.(We do not know which integers they are,but we do not actually need to know.)Thus,the sum of the remaining two integers must be12134−8032=4102.=2051.Therefore,the mean of the remaining two integers is41022(We can verify that1867,2019,2025and2121do actually have a mean of2008,and that1993 and2109have a mean of2051.)Answer:(D)21.The maximum possible value of pqis when p is as large as possible(that is,10)and q is as smallas possible(that is,12).Thus,the maximum possible value of pqis1012=56.The minimum possible value of pqis when p is as small as possible(that is,3)and q is as largeas possible(that is,21).Thus,the maximum possible value of pqis321=17.The diﬀerence between these two values is 56−17=3542−642=2942.Answer:(A)22.Suppose that the distance from Ginger’s home to her school is d km.Since there are60minutes in an hour,then33minutes(or15minutes)is15×1=1of anhour.Since Ginger walks at4km/h,then it takes her d4hours to walk to school.Since Ginger runs at6km/h,then it takes her d6hours to run to school.Since she saves116of an hour by running,then the diﬀerence between these times is116of anhour,sod 4−d6=1163d 12−2d12=116d12=116d=1216=34Therefore,the distance from Ginger’s home to her school is34km.Answer:(E)23.Suppose that the distance from line M to line L is d m.Therefore,the total length of piece W to the left of the cut is d m.Since piece X is3m from line M,then the length of piece X to the left of L is(d−3)m, because3of the d m to the left of L are empty.Similarly,the lengths of pieces Y and Z to the left of line L are(d−2)m and(d−1.5)m.Therefore,the total length of lumber to the left of line L isd+(d−3)+(d−2)+(d−1.5)=4d−6.5mSince the total length of lumber on each side of the cut is equal,then this total length is1 2(5+3+5+4)=8.5m.(We could insteadﬁnd the lengths of lumber to the right of line L to be5−d,6−d,7−d,and 5.5−d and equate the sum of these lengths to the sum of the lengths on the left side.) Therefore,4d−6.5=8.5or4d=15or d=3.75,so the length of the part of piece W to the left of L is3.75m.Answer:(D)24.We label the ﬁve circles as shown in the diagram.P QRS TWe note that there are 3possible colours and that no two adjacent circles can be coloured the same.Consider circle R .There are three possible colours for this circle.For each of these colours,there are 2possible colours for T (either of the two colours that R is not),since it cannot be the same colour as R .Circles Q and S are then either the same colour as each other,or are diﬀerent colours.Case 1:Q and S are the same colourIn this case,there are 2possible colours for Q (either of the colours that R is not)and 1pos-sibility for S (the same colour as Q ).For each of these possible colours for Q /S ,there are two possible colours for P (either of the colours that Q and S are not).P Q R S T3 choices2 choices2 choices1 choice2 choices In this case,there are thus 3×2×2×1×2=24possible ways of colouring the circles.Case 2:Q and S are diﬀerent coloursIn this case,there are 2possible colours for Q (either of the colours that R is not)and 1possibility for S (since it must be diﬀerent from R and diﬀerent from Q ).For each of these possible colourings of Q and S ,there is 1possible colour for P (since Q and S are diﬀerent colours,P is diﬀerent from these,and there are only 3colours in total).P Q R S T3 choices2 choices1 choice1 choice In this case,there are thus 3×2×2×1×1=12possible ways of colouring the circles.In total,there are thus 24+12=36possible ways to colour the circles.Answer:(D)25.Since P Q =2and M is the midpoint of P Q ,then P M =MQ =12(2)=1.Since P QR is right-angled at P ,then by the Pythagorean Theorem,RQ = P Q 2+P R 2= 22+(2√3)2=√4+12=√16=4(Note that we could say that P QR is a 30◦-60◦-90◦triangle,but we do not actually need this fact.)Since P L is an altitude,then ∠P LR =90◦,so RLP is similar to RP Q (these triangles have right angles at L and P respectively,and a common angle at R ).Therefore,P L QP =RP RQ or P L =(QP )(RP )RQ =2(2√3)4=√3.Similarly,RL RP =RP RQ so RL =(RP )(RP )RQ =(2√3)(2√3)4=3.Therefore,LQ =RQ −RL =4−3=1and P F =P L −F L =√3−F L .So we need to determine the length of F L .Drop a perpendicular from M to X on RQ .RP QM LFX Then MXQ is similar to P LQ ,since these triangles are each right-angled and they share a common angle at Q .Since MQ =12P Q ,then the corresponding sides of MXQ are half as long as those of P LQ .Therefore,QX =12QL =12(1)=12and MX =12P L =12(√3)=√32.Since QX =12,then RX =RQ −QX =4−12=72.Now RLF is similar to RXM(they are each right-angled and share a common angle at R).Therefore,F LMX=RLRXso F L=(MX)(RL)RX=√32(3)72=3√3.Thus,P F=√3−3√37=4√37.Answer:(C)。

2003滑铁卢竞赛试题

parts are indicated like this: .Enter the answer in the appropriate box in the answer booklet.be given for a correct answer which is placed in the box. Part marks will be awarded parts are indicated like this: .Finished solutions must be written in the appropriate location in the answer booklet.Sybasei Anywhere SolutionsCanadian Instituteof ActuariesChartered Accountants Great West Lifeand London LifeNOTE: 1.Please read the instructions on the front cover of this booklet.2.Place all answers in the answer booklet provided. 3.For questions marked “”, full marks will be given for a correct answer placed in theappropriate box in the answer booklet. Marks may be given for work shown . Students are strongly encouraged to show their work.4.It is expected that all calculations and answers will be expressed as exact numbers such as4π, 27+, etc., except where otherwise indicated.1.(a)In the diagram, the parabola cuts the y -axis at the point 08,(), cuts the x -axis at the points 20,() and 40,(),and passes through the point a ,8(). What is the value of a ?(b)The quadratic equation x x k 260++= has two equal roots. What is the value of k ?(c)The line y x =+22 intersects the parabola y x x c =+23– at two points. One of these points is 14,(). Determine the coordinates of the second point of intersection.2.(a)If 090o o <<x and 3150sin cos x ()−()=o , what is the value of x to the nearest tenth of a degree?(b)In the diagram, ∆ABC is right-angled at B and AC =20. If sin C =35, what is the length of sideBC ?(c) A helicopter is flying due west over level ground at a constant altitude of 222 m and at aconstant speed. A lazy, stationary goat, which is due west of the helicopter, takes two measurements of the angle between the ground and the helicopter. The first measurement the goat makes is 6° and the second measurement, which he makes 1 minute later, is 75°. If the helicopter has not yet passed over the goat, as shown, how fast is the helicopter travelling to the nearest kilometre per hour?A B CThe function f x () has the property that f x f x 2323+()=()+ for all x .If f 06()=, what is the value of f 9()?Suppose that the functions f x () and g x () satisfy the system of equations f x g x x x f x g x x ()+()=++()+()=+36242422for all x . Determine the values of x for which f x g x ()=().4.(a)In a short-track speed skating event, there are five finalists including two Canadians. The first three skaters to finish the race win a medal. If all finalists have the same chance of finishing in any position, what is the probability that neither Canadian wins a medal?(b)Determine the number of positive integers less than or equal to 300 that are multiples of 3or 5, but are not multiples of 10 or 15.5.(a)In the series of odd numbers 1357911131517192123+++++––––––... the signs alternate every three terms, as shown. What is the sum of the first 300 terms of the series?(b)A two-digit number has the property that the square of its tens digit plus ten times its units digit equals the square of its units digit plus ten times its tens digit. Determine all two-digit numbers which have this property, and are prime numbers.6.(a)A lead box contains samples of two radioactive isotopes of iron. Isotope A decays so that after every 6 minutes, the number of atoms remaining is halved. Initially, there are twice as many atoms of isotope A as of isotope B, and after 24 minutes there are the same number of atoms of each isotope. How long does it take the number of atoms of isotopeB to halve?(b)Solve the system of equations:log log log log 103102102103113x y x y ()+()=()−()=7.(a) A regular hexagon is a six-sided figure which has all of its angles equal and all of its side lengths equal. Inthe diagram, ABCDEF is a regular hexagon with anarea of 36. The region common to the equilateral triangles ACE and BDF is a hexagon, which isshaded as shown. What is the area of the shadedhexagon?(b)At the Big Top Circus, H erc theHuman Cannonball is fired out of the cannon at ground level. (For the safetyof the spectators, the cannon ispartially buried in the sand floor.)Herc ’s trajectory is a parabola until he catches the vertical safety net, on his way down, at point B . Point B is 64 mdirectly above point C on the floor ofthe tent. If Herc reaches a maximumheight of 100 m, directly above a point30 m from the cannon, determine thehorizontal distance from the cannon tothe net.8.(a) A circle with its centre on the y-axis intersects the graph of y x = at the origin, O , and exactly two otherdistinct points, A and B , as shown. Prove that the ratioof the area of triangle ABO to the area of the circle isalways 1 : π.(b)In the diagram, triangle ABC has a right angle at Band M is the midpoint of BC . A circle is drawn usingBC as its diameter. P is the point of intersection of thecircle with AC . The tangent to the circle at P cutsABat Q . Prove that QM is parallel to AC .9.Cyclic quadrilateral ABCD has AB AD ==1, CD ABC =∠cos , and cos –∠=BAD 13. Provethat BC is a diameter of the circumscribed circle.10. A positive integer n is called “savage” if the integers 12,,...,n{} can be partitioned into three sets A, B and C such thati)the sum of the elements in each of A, B, and C is the same,ii)A contains only odd numbers,iii)B contains only even numbers, andiv)C contains every multiple of 3 (and possibly other numbers).(a)Show that 8 is a savage integer.(b)Prove that if n is an even savage integer, then n+412is an integer.(c)Determine all even savage integers less than 100.。

2004滑铁卢竞赛试题答案

www.linstituteatics Competition
An activity of The Centre for Educati on in Ma the matics and Computing , University of W aterloo, Wa terloo, Ontario
[
]
square units.

Solution 2 Since ABCD is a square of side length 6 and each of AE : EB, BF : FC, CG : GD, and DH : HA is equal to 1 : 2, then AE = BF = CG = DH = 2 and EB = FC = GD = HA = 4 . Thus, each of the triangles HAE, EBF, FCG, and GDH is right-angled, with one leg of length 2 and the other of length 4. By the Pythagorean Theorem,
EF = FG = GH = HE = 2 2 + 4 2 = 20 . Since the two triangles HAE and EBF are congruent (we know the lengths of all three o sides of each), then ∠AHE = ∠BEF . But ∠AHE + ∠AEH = 90 , so o o ∠BEF + ∠AEH = 90 , so ∠HEF = 90 . In a similar way, we can show that each of the four angles of EFGH is a right-angle, and so EFGH is a square of side length 20 .

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

Each question is worth 10 marks
Calculating devices are allowed, provided that they do not have any of the following features: (i) internet access, (ii) the ability to communicate with other devices, (iii) information previously stored by students (such as formulas, programs, notes, etc.), (iv) a computer algebra system, (v) dynamic geometry software.
Parts of each question can be of two types: 1. SHORT ANSWER parts indicated by
• worth 3 marks each • full marks given for a correct answer which is placed in the box • part marks awarded only if relevant work is shown in the space provided
WRITE ALL ANSWERS IN THE ANSWER BOOKLET PROVIDED. • Extra paper for your ﬁnished solutions must be supplied by your supervising teacher and inserted into your answer booklet. Write your name, school name, and question number on any inserted pages. • Express answers as simpli√ﬁed exact numbers except where otherwise indicated. For example, π + 1 and 1 − 2 are simpliﬁed exact numbers.
4. For questions marked
, provide a well-organized solution in the answer booklet.

Use mathematical statements and words to explain all of the steps of your solution.
NOTE: 1. Please read the instructions on the front cover of this booklet. 2. Write all answers in the answer booklet provided.
3. For questions marked , place your answer in the appropriate box in the answer booklet and show your work.
Do not discuss the problems or solutions from this contest online for the next 48 hours.
The name, grade, school and location, and score range of some top-scoring students will be published on our website, cemc.uwaterloo.ca. In addition, the name, grade, school and location, and score of some top-scoring students may be shared with other mathematical organizations for other recognition opportunities.
2. FULL SOLUTION parts indicated by
• worth the remainder of the 10 marks for the question • must be written in the appropriate location in the answer booklet • marks awarded for completeness, clarity, and style of presentation • a correct solution poorly presented will not earn full marks
Work out some details in rough on a separate piece of paper before writing your ﬁnished
solution.
5. Diagrams are not drawn to scale. They are intended as aids only.
(outside of North America and South America)
Time:
2
1 2
hours
c 2019 University of Waterloo
Do not open this booklet until instructed to do so.
Number of questions: 10
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING
cemc.uwaterloo.ca
Euclid Contest
Wednesday, April 3, 2019
(in North America and South America)
Thursday, April 4, 2019