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2005滑铁卢竞赛试题答案

1.(a)Answer:a=5

Since(a,a)lies on the line3x?y=10,then3a?a=10or2a=10or a=5.

(b)Answer:(6,2)

Solution1

To get from A to B,we move2units to the right and1unit up.

Since 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).

Solution2

Label the origin as O and drop a perpendicular from C to P on the x-axis.

Then 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 are

1 2(0+88),

(30+14)

=(44,22)

2.(a)Answer:x=?2

Solution1

Since y=2x+3,then4y=4(2x+3)=8x+12.

Since4y=8x+12and4y=5x+6,then8x+12=5x+6or3x=?6or x=?2.

Solution2

Since4y=5x+6,then y=5

4x+6

x+3

Since y=2x+3and y=5

4x+3

,then2x+3=5

x+3

or3

x=?3

or x=?2.

Solution3

Since the second equation contains a“5x”,we multiply the?rst equation by5

2to obtain

a5x term,and obtain5

2y=5x+15

Subtracting this from4y=5x+6,we obtain3

2y=?3

or y=?1.

Since y=?1,then?1=2x+3or2x=?4or x=?2.

(b)Answer:a=6

Solution1

Adding the three equations together,we obtain a?3b+b+2b+7c?2c?5c=?10+3+13 or a=6.

Solution2

Multiplying the second equation by3,we obtain3b?6c=9.

Adding this new equation to the?rst 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)Solution1

Let 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=150or

J+M

2=75.

Therefore,the average of their two scores was75.

(Note that we didn’t have to solve for their individual scores.)

Solution2

Let 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 the?rst equation into the second,we obtain

2(2J?60)=J+90

4J?120=J+90

3J=210

J=70

Substituting into M=2J?60gives M=80.

Therefore,the average of their scores(ie.the average of70and80)is75.

3.(a)Answer:x=50

Simplifying using exponent rules,

2(1612)+2(816)=2((24)12)+2((23)16)=2(248)+2(248)=4(248)=22(248)=250

Therefore,since2x=2(1612)+2(816)=250,then x=50.

(b)Solution1

We 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=3

Solution2

Since f(x)=2x?1and(f(x))2?3f(x)+2=0,then

(2x?1)2?3(2x?1)+2=0

4x2?4x+1?6x+3+2=0

4x2?10x+6=0

2x2?5x+3=0

(x?1)(2x?3)=0

Therfore,x=1or x=3

4.(a)Answer:14

Solution1

The 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 are?fteen 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

Solution 2

We ?nd 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 ?fteen possible pairs that we can se-lected.

Therefore,the probability that the smaller number on the two tickets is NOT less than or

equal 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 1

Since ∠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 2

Since ∠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 .

Since BLP is 30?-60?-90?,then BP :LP =1:√3,so BP =1√3LP =1√3

x .

Since HLP is 30?-60?-90?,then HP :LP =√3:1,so HP =√3LP =√3x .But HP =HB +BP so √3x =400+1√3x 3x =400√3+x 2x =400√3x =200√3

Therefore,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 ?rst 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 =1

2n (n +1)and solving for n this way.)

So we must determine the coordinates of the goat after 10moves.

We consider ?rst 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 1

Since the sequence 4,4r ,4r 2is also arithmetic,then the di?erence between 4r 2and 4r equals the di?erence between 4r and 4,or

4r 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 2

Since 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 di?erence 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:4π

First,we notice that whenever an equilateral triangle of side length 3is placed inside a https://www.sodocs.net/doc/05827416.html,

circle 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?.

Therefore,B has traced out60?

360?=1

of 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 through1

6of a circle of radius3as it moves

to 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 the?rst three rotations.

Thus,in total,B moves four times along an arc equal to1

6of a circle of radius3.

Therefore,the distance travelled by B is4(1

)(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?.

D C

674

Using the cosine law in ABD ,we obtain

72=52+62?2(5)(6)cos(∠A )

49=61?60cos(∠A )

cos(∠A )=15

Since cos(∠A )=1

5and ∠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 obtain

72=42+CD 2?2(4)(CD )cos(∠C )49=16+CD 2?8(CD ) ?15

0=5CD 2+8CD ?165

0=(5CD +33)(CD ?5)

So CD =?33

5or 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 =1

We 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 .

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 ?b 2a =?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

So g (x )is the horizontal line y =3when 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,our?nal graph is:

(b)Solution 1

Let 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 on

either side of C on the other tangent line.

Let ∠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.

Solution2

Let 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)Solution1

We have(x?p)2+y2=r2and x2+(y?p)2=r2,so at the points of intersection,

(x?p)2+y2=x2+(y?p)2

x2?2px+p2+y2=x2+y2?2py+p2

?2px=?2py

and 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+1

2(p2?r2)=0.

Using the relationship between the sum and product of roots of a quadratic equation and

its coe?cients,we obtain that a+b=p and ab=1

2(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 1

(p2?r2)

=r2,as required.

Solution2

Since the circles are re?ections 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 )=0

Since 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 ?rst draw a diagram.

We 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 ),so

AB = (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 2

Therefore,the area of quadrilateral CADB is 12(AB )(CD )=12

4r 2?2p 2 2p 2= 2r 2p 2?p 4

To maximize this area,we must maximize 2r 2p 2?p 4=2r 2(p 2)?(p 2)2.

Since r is ?xed,we can consider this as a quadratic polynomial in p 2.Since the coe?cient of (p 2)2is negative,then this is a parabola opening downwards,so we ?nd its maximum value by ?nding 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 ?nd 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 in?nite 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 ?rst 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 ?fth 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 1

First,we note that if n =2k is even,then f (n )=f (2k )=f (2k ?1)=f (n ?1).See Solution 2for this justi?cation.

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 ?rst 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 m

with 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 in?nitely many positive integers n such that f(n)=f(2005).

To do this,we?rst 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,the

rightmost locker that is open,all open lockers before the pass,which lockers will be closed

on 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),then

there is only one locker left open:locker641.

Notice also that on any pass s,the“class”of lockers which are closed depends on what the

number of the leftmost(on an odd-numbered pass)or rightmost(on an even-numbered

pass)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 justi?cation of why we might

try these values of n.)

Suppose we were to try to make a table as above to calculate f(n).

Then the?rst11passes in the table would be identical to the table above,except for the

rightmost open number;this number in the new table would be the number above plus

22a.

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 all

lockers with numbers≡2689(mod212)are closed,leaving open only those lockers with

numbers≡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.

https://www.sodocs.net/doc/05827416.html,

If we can now show that whenever we start with a number of lockers of the form22c+1,the

last locker remaining open is the leftmost locker,then we will be done,since of the lockers

left 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 which

will be closed is one-half of one less than the total number of lockers,and the?rst and

last lockers will be left open.

So on the?rst pass,there are22c?1lockers closed,leaving22c+1?22c?1=22c?1+1lockers

open,ie.an odd number of lockers open.

On the next pass,there are22c?2lockers closed(since there are an odd number of lockers

open 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 the

original leftmost and rightmost lockers open.

On the last pass(an odd-numbered pass from left to right),the rightmost locker will be

closed,leaving only the leftmost locker open.

Therefore,starting with a row of22c+1open lockers,the leftmost locker will be the last

remaining open.

Translating this to the above,we see that the leftmost locker of the22a?12+1still open

is the last left open,ie.f(22a+2005)=641=f(2005)if a≥6.

Therefore,there are in?nitely many positive integers n for which f(n)=f (2005).

Solution2

First,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 the?rst 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 the?rst 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 the?rst 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.

The?rst 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,the?nal 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)?3

and

f(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)?1

From 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))+9

f(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))+11

Similarly,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 in?nitely many positive integers n such that f(n)=f(2005). We do this by looking at the pattern we initially created and conjecturing that

f(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))=1so

f(2005+22a)=1024(1)?383=641=f(2005)

so there are in?nitely many integers n for which f(n)=f(2005).

Solution3

We conjecture a formula for f(n)and prove this formula by induction,using the formulae that we proved in Solution2.

We start by writing the positive integer n in its binary representation,ie.we write

n=b0+b1·2+b2·22+···+b2p?1·22p?1+b2p·22p

where each of b0,b1,···,b2p is0or1with either b2p=1,or b2p=0and b2p?1=1. Thus,in binary,n is equal to either(b2p b2p?1···b1b0)2or(b2p?1···b1b0)2.

We then conjecture that if n is odd(which tells us that b0=1for sure),then

f(n)=b0+b1·2+b3·23+···+b2p?1·22p?1

In other words,we omit the even-numbered powers of2from n.Looking at a few exam-ples:7=4+2+1,so f(7)=2+1=3,13=8+4+1,so f(13)=8+1=9,and 27=16+8+2+1,so f(27)=8+2+1=11.

We already know that if n is even,then f(n)=f(n?1)(we proved this in Solution2). Let’s assume that we’ve proved this formula.(We’ll prove it at the end.)

We can now solve parts(b)and(c)very quickly using our formula.

Are then any values of n such that f(n)=2005?

Writing2005as a sum of powers of2(ie.in binary),we get

2005=1024+512+256+128+64+16+4+1

Since the representation of2005does not use only odd-numbered powers of2,then there is no n for which f(n)=2005.

Lastly,we need to prove that there are in?nitely many positive integers n for which f(n)=f(2005).

To do this,we note that if n=2005+22a for some a≥6,then the last11binary digits of n agree with those of2005and the only1s in the representation of n=2005+22a in positions corresponding to odd-numbered powers of2come from the2005portion(since the extra“1”from22a corresponds to an even-numbered power of2).

Therefore,since we calculate f(2005+22a)by looking at only the odd-numbered powers of2,then f(2005+22a)=f(2005)for all integers a≥6.

Therefore,there are in?nitely many positive integers n for which f(n)=f(2005).

We now must prove that this formula is true.We use strong induction.

Looking at the list in Solution2,we can quickly see that the result holds for all odd values of n with n≤31.(We only need to establish this for a couple of small values of n to serve as base cases.)

Assume that the result holds for all odd positive integers n up to n=N?2for some odd positive integer N.

Consider n=N.

Case1:N=4q+1

Here we can write

N=1+b2·22+···+b2p?1·22p?1+b2p·22p

滑铁卢大学精算专业排名

加拿大滑铁卢大学计算机科学与技术。加拿大计算机最好的专业非滑铁卢大学莫属，据360教育集团介绍，滑铁卢大学1959年成立，在成立之初便把重点放在新兴的计算机行业上，并罕见地成立了数学学院来推动计算机专业的发展。滑铁卢最为人称道的成就是它创立的co-op program让电脑系学生在学习的同时有机会在IBM、Nortel、Bell等着名公司获得工作经验，现在已经为美加大学相竞效仿。滑铁卢大学位於安省南部，被《麦克琳》杂志排2010年加拿大综合性大学排名第三名，带薪实习项目(Co-op Program)规模位居全球之首。除著名的电脑科学、工程、会计等专业之外，滑大的数学学院精算专业近年来也吸引了大量学生入学。成立於1967年的滑大数学院(University of Waterloo Faculty of Mathematics)是全世界最大的数学和电脑科学教育和研究机构。数学院共有6000多名大学生、600多名研究生和200多名全职教授，开设数学、统计学、电脑科学等500多门相关课程，以其学术上的发现与创新在海内外赢得了很高的声誉。滑大数学院下属的精算专业(Actuarial Science Program)是北美最大的精算院校，也是世界上规模最大的精算院校之一，大学生近千人、研究生逾150名。精算科学将数学和统计学应用於财务金融和保险行业，主要用於评估长期风险、制定投资方针。从事精算的专业人员被称为精算师，在华尔街日报的「最理想职业调查」中屡屡名列前茅。滑大精算专业提供带薪实习项目，即学校与公司合作提供学生在校期间实习的机会，每个学期交替学习和实习，直到学生毕业。与滑大精算专业合作的公司大部分是保险公司、银行等大型企业，也有政府部门等其他机构，例如满地可银行、宏利金融、普华永道会计事务所、联邦退休金计画投资委员会，其中也有脸书等新生代的网际网路公司。想走入精算专业，首先要被滑大数学院录取。高中毕业生只需按照正常的申请程序，在安省大学申请中心(OUAC)的网站上填写申请表格、递交成绩、发送申请即可。申请人高中12年级成绩6门科目平均分需高於78分，要求的必修课程包括12年级英语以及方程、微积分、向量等数学科目。当然，数学成绩越好，申请的成功率就越大。如果是留学生或四年以内的移民，还需要提供托福、雅思等英语考试成绩。除正常申请所需的要求之外，滑大强烈推荐学生参与滑大主办的欧几里德数学竞赛(Euclid Mathematics Contest)。该竞赛每年4月举行，主要面向12年级毕业生或有兴趣参与的低年级学生，以考察学生的数学知识和学习能力为主，被称为「数学领域的托福考试」。虽然在竞赛中取得成绩并非申请滑大数学院的必备标准，但较好的竞赛成绩将成为录取决定的参考。

2005滑铁卢竞赛试题答案

1.(a)Answer:a=5 Since(a,a)lies on the line3x?y=10,then3a?a=10or2a=10or a=5. (b)Answer:(6,2) Solution1 To get from A to B,we move2units to the right and1unit up. x Since 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). Solution2 Label the origin as O and drop a perpendicular from C to P on the x-axis. x Then 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.

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Hypatia滑铁卢数学竞赛(Grade 11)-数学Mathematics-2008-试题 exam

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加拿大大学排名最新加拿大大学排名 USNews世界大学排名【】10月，U.S. News公布了202X年全球最佳大学排名，加拿大共有33所大学入榜。下面是关于加拿大大学在本次排名中的具体表现，供大家选校参考。 The U.S. News Best Global Universities rankings by country show the top institutions in 42 countries with five or more schools that are globally ranked in the top 1,250.The methodology for the country rankings is based entirely on how a school ranked in the overall Best Global Universities rankings covering the top 1,250 schools worldwide. These Canadian universities have been numerically ranked based on their positions in the overall Best Global Universities rankings. Schools were evaluated based on their research performance and their ratings by members of the academic community around the world and within North America. These are the top global universities in Canada. U.S. News全球最佳大学国家排名列出了42个国家的大学排名。这些国家进入全球排名的大学数量都在五所或五所以上。全球最佳大学排名评出了

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滑铁卢大学是一所以研究为主的中等大小的公立大学，创建于1957年。该校位于安大略省的西南面的滑铁卢市，以学习与实习并重的合作教育（co-operative education）而闻名。2011年到2013年，该校一直稳居麦克林杂志评选的加拿大综合性大学排名的第三位，是北美地区最优大学之一，其数学，计算机科学和工程学科教学水平居世界前列。滑铁卢大学是一所综合性公立大学，位于加拿大安大略省滑铁卢。学校于1957年由格里·哈格博士和艾拉·湾·尼德尔斯共同创立。学校有很高的声誉，特别是做为北美地区第一个经认可建立数学系的大学，以及拥有世界上最大的合作办学项目。滑铁卢大学共有六个学院。学校共授予100多个本科学位专业，28种硕士及博士学位专业。滑铁卢大学以数学、电脑、工程科学等学科闻名，学校的代表队曾多次获得ACM 国际大学生程序设计竞赛的冠军。据360教育集团介绍，滑铁卢大学教学设施完善，科研设备先进，学术力量雄厚，是一所著名的综合性公立大学。下设文学院、理学院、工程学院、数学院、环境学院、应用健康科学学院。滑铁卢大学设有北美唯一一所数学院，这也是全世界最大的数学和计算机的教育及研究中心。同时，在北美洲，该校是最早采用计算机教学的学府，而这所大学的计算机学科在整个北美洲也是极有名气的，加拿大其他大学在计算机科学领域上难以望其项背。除此之外，滑铁卢大学工程院拥有加拿大最好的软件工程，电子工程及机械工程专业。滑铁卢大学世界排名在1997年加拿大《Maclean‘s》杂志的加拿大大学排行榜中，滑铁卢大学荣获声誉排行及综合排名冠军，在本科综合教育性类别排名榜中排行第三位。 2000、2001、2002和2005年时ACM（美国电脑协会）将该校列举为北美地区最优大学之一。 2000年~2010年，该校在加拿大麦克林杂志（Maclean’s）排名中连续十年名列综合类大学第一名，全国知名度排名连续七年第一 2009年英国TIMES世界大学排名中滑铁卢大学位列113位。 2011年~2013年，滑铁卢大学在麦克林杂志（Maclean‘s）评选的加拿大综合性大学排名第3位，它还形容University of Waterloo是一所位居加拿大大学最具领导地位、最具有创新力的学校，比起其他的大学学生来，本校的学生有相当杰出的表现。在15次的排名里，University of Waterloo有13次排名第一位。Maclean’s形容这所大学“最强是数学、工程学和计算机科学方面”，并且描述大学“国际上被认定为空前未有的成功”。 2013年usnews计算机科学专业世界大学排名中，滑铁卢大学名列第36位。截止目前，留学360更新海外名校录取86789枚，中哈佛大学43人、耶鲁大学56人、斯坦福大学43人、麻省理工学院25人、牛津大学38人、剑桥大学35人、多伦多大学 290人、麦吉尔大学353人、悉尼大学1874人、墨尔本大学1286人、澳洲国立大学 1100人、香港大学120人、新加坡国立大学150人、南洋理工大学227人、新西兰奥克兰大学1241人。

加拿大大学排名

2016-2017最新加拿大大学排名最新公布的一版QS世界大学排名中，我们可以看到有26所加拿大大学进入排名，在这些大学中，有3所加拿大顶尖大学位于世界前50名，另外10所加拿大大学位于世界前300名。下面请看出国留学网为大家独家翻译的今年加拿大顶尖大学排名，先从加拿大排名前10的大学开始吧，供大家在选择加拿大留学学校的时候进行参考。第一名、麦吉尔大学麦吉尔大学连续第二年位列加拿大顶尖大学之首，不过在今年的世界大学排名中下降了6个名次，位列第30名。麦吉尔大学位于蒙特利尔，有大约40000名学生就读，其中25%的学生是国际留学生。麦吉尔大学有比任何其他加拿大大学更多的罗德奖学金学者和诺贝尔奖获得者。麦吉尔大学成立于1821年，是加拿大最古老的大学，几百年以来一直在国际上享有盛誉，被称为“加拿大的哈佛大学”。麦吉尔大学一直都以严谨的学风而闻名，想要申请麦吉尔大学，申请条件和录取分数都是加拿大所有大学之首，麦吉尔大学在校学生获得的国家级和国际级奖项也远远多过其他加拿大大学，名列加拿大所有大学的第一位。第二名、多伦多大学多伦多大学今年更接近于其竞争对手麦吉尔大学，在今年的世界大学排名中上升了

两个名次，位于第32位。多伦多大学位于加拿大最大的城市多伦多，多伦多大学每年为加拿大的经济贡献157亿加元，并且多伦多大学凭借其研究和创新成果而闻名，包括胰岛素和干细胞的发现。多伦多大学始建于1827年，是加拿大传统四大名校之一，多伦多大学的经费、捐款、国家教授奖项、研究出版规模和藏书量都位于加拿大大学的首位，在学术及研究方面也一直处于领先地位，是美国大学协会中仅有的两所非美国大学之一。第三名、英属哥伦比亚大学英属哥伦比亚大学今年上升了五个名次，位于世界大学排名第45位，与此同时也保持了加拿大大学排名第三位的位置。英属哥伦比亚大学位于温哥华和基洛纳(加拿大西部大不列颠哥伦比亚省的一个城市)，有61,100名学生就读，其中包括来自155个国家的13,200名国际留学生。英属哥伦比亚大学与麦吉尔大学、多伦多大学并称加拿大大学“三强”，在加拿大国内的排名中始终保持在前三名之内，是加拿大最难申请的大学之一，同时也是加拿大淘汰率最高的大学之一。英属哥伦比亚大学是一所世界一流的研究型大学，不仅教学水平一流，其温哥华校区还被称为整个北美最漂亮的校园。第四名、阿尔伯塔大学

Galois滑铁卢数学竞赛(Grade 10)-数学Mathematics-2007-试题 exam

2007Galois Contest (Grade 10) Wednesday,April 18,2007 1.Jim shops at a strange fruit store.Instead of putting prices on each item,the mathematical store owner will answer questions about combinations of items. (a)In Aisle 1,Jim receives the following answers to his questions: Jim’s Question Answer What is the sum of the prices of an Apple and a Cherry?62cents What is the sum of the prices of a Banana and a Cherry?66cents What is di?erence between the prices of an Apple and a Banana?Which has a higher price?Explain how you obtained your answer. (b)In Aisle 2,Jim receives the following answers to his questions: Jim’s Question Answer What is the sum of the prices of a Mango and a Nectarine?60cents What is the sum of the prices of a Pear and a Nectarine?60cents What is the sum of the prices of a Mango and a Pear?68cents What is the price of a Pear?Explain how you obtained your answer. (c)In Aisle 3,Jim receives the following answers to his questions: Jim’s Question Answer What is the sum of the prices of a Tangerine and a Lemon?60cents How much more does a Tangerine cost than a Grapefruit?6cents What is the sum of the prices of Grapefruit,a Tangerine and a Lemon?94cents What is the price of a Lemon?Explain how you obtained your answer.2.(a)In the diagram,what is the perimeter of the sector of the circle with radius 12?Explain how you obtained your answer. (b)Two sectors of a circle of radius 12are placed side by side,as shown.Determine the area of ?gure ABCD .Explain how you obtained your answer. A (c)In the diagram,AO B is a sector of a circle with ∠AOB =60?. OY is drawn perpendicular to AB and intersects AB at X .What is the length of XY ?Explain how you obtained your answer.A O B X Y 12 12 (d)See over...

世界500强大学排名

,. 2018年QS世界大学排名500强全球排名大学名称所在国家第1位麻省理工学院美国第2位斯坦福大学美国第3位哈佛大学美国第4位加州理工学院美国第5位剑桥大学英国第6位牛津大学英国第7位伦敦大学学院英国第8位伦敦帝国理工学院英国第9位芝加哥大学美国第10位苏黎世联邦理工学院瑞士第11位南洋理工大学新加坡第12位洛桑联邦理工学院瑞士第13位普林斯顿大学美国第14位康奈尔大学美国第15位新加坡国立大学新加坡第16位耶鲁大学美国第17位约翰斯霍普金斯大学美国

,. 第18位哥伦比亚大学美国第19位宾夕法尼亚大学美国第20位澳大利亚国立大学澳大利亚并列第21位杜克大学美国并列第21位密歇根大学美国并列第23位伦敦大学国王学院英国并列第23位爱丁堡大学英国第25位清华大学中国第26位香港大学中国香港第27位加州大学伯克利分校美国并列第28位西北大学美国并列第28位东京大学日本第30位香港科技大学中国香港第31位多伦多大学加拿大第32位麦吉尔大学加拿大第33位加州大学洛杉矶分校美国第34位曼彻斯特大学英国第35位伦敦政治经济学院英国并列第36位京都大学日本并列第36位首尔国立大学韩国并列第38位北京大学中国

,. 并列第38位加州大学圣地亚哥分校美国第40位复旦大学中国并列第41位韩国科学技术院韩国并列第41位墨尔本大学澳大利亚第43位巴黎高等师范学院法国第44位布里斯托大学英国第45位新南威尔士大学澳大利亚第46位香港中文大学中国香港并列第47位卡内基梅隆大学美国并列第47位昆士兰大学澳大利亚第49位香港城市大学中国香港第50位悉尼大学澳大利亚第51位英属哥伦比亚大学加拿大第52位纽约大学美国第53位布朗大学美国第54位代夫特理工大学荷兰第55位威斯康辛大学麦迪逊分校美国第56位东京工业大学日本第57位华威大学英国第58位阿姆斯特丹大学荷兰第59位巴黎综合理工学院法国

2015QS世界大学排名大全(最新版)

2015QS世界大学排名大全（最新版）最全的世界大学排名，前一百大学美国28所入选，英国19所，澳大利亚8所，加拿大5所，中国3所。其中英美院校占了近一半。更多留学问题可以联系我哦！QQ：449224440 排名学校名称学校英文名国家/ 地区评分 1 麻省理工学院Massachusetts Institute of Technology 美国100 2 剑桥大学University of Cambridge 英国99.4 2 帝国理工学院Imperial College London 英国99.4 4 哈佛大学Harvard University 美国99.3 5 牛津大学University of Oxford 英国99.2 5 伦敦大学学院UCL (University College London) 英国99.2 7 斯坦福大学Stanford University 美国98.3 8 加州理工学院California Institute of Technology (Caltech) 美国97.1 9 普林斯顿大学Princeton University 美国96.6 10 耶鲁大学Yale University 美国96.5 11 芝加哥大学University of Chicago 美国95.5 12 苏黎世联邦理工学院 ETH Zurich (Swiss Federal Institute of Technology) 瑞士95.3 13 宾夕法尼亚大学University of Pennsylvania 美国94.5 14 哥伦比亚大学Columbia University 美国94.1

世界500强大学排名

2018年QS世界大学排名500强全球排名大学名称所在国家第1位麻省理工学院美国第2位斯坦福大学美国第3位哈佛大学美国第4位加州理工学院美国第5位剑桥大学英国第6位牛津大学英国第7位伦敦大学学院英国第8位伦敦帝国理工学院英国第9位芝加哥大学美国第10位苏黎世联邦理工学院瑞士第11位南洋理工大学新加坡第12位洛桑联邦理工学院瑞士第13位普林斯顿大学美国第14位康奈尔大学美国第15位新加坡国立大学新加坡第16位耶鲁大学美国第17位约翰斯霍普金斯大学美国第18位哥伦比亚大学美国第19位宾夕法尼亚大学美国第20位澳大利亚国立大学澳大利亚并列第21位杜克大学美国并列第21位密歇根大学美国并列第23位伦敦大学国王学院英国并列第23位爱丁堡大学英国第25位清华大学中国第26位香港大学中国香港第27位加州大学伯克利分校美国

并列第28位西北大学美国并列第28位东京大学日本第30位香港科技大学中国香港第31位多伦多大学加拿大第32位麦吉尔大学加拿大第33位加州大学洛杉矶分校美国第34位曼彻斯特大学英国第35位伦敦政治经济学院英国并列第36位京都大学日本并列第36位首尔国立大学韩国并列第38位北京大学中国并列第38位加州大学圣地亚哥分校美国第40位复旦大学中国并列第41位韩国科学技术院韩国并列第41位墨尔本大学澳大利亚第43位巴黎高等师范学院法国第44位布里斯托大学英国第45位新南威尔士大学澳大利亚第46位香港中文大学中国香港并列第47位卡内基梅隆大学美国并列第47位昆士兰大学澳大利亚第49位香港城市大学中国香港第50位悉尼大学澳大利亚第51位英属哥伦比亚大学加拿大第52位纽约大学美国第53位布朗大学美国第54位代夫特理工大学荷兰第55位威斯康辛大学麦迪逊分校美国第56位东京工业大学日本第57位华威大学英国第58位阿姆斯特丹大学荷兰

Fermat滑铁卢数学竞赛(Grade 11)-数学Mathematics-2002-试题 exam

Canadian Institute of Actuaries Chartered Accountants Sybase i Anywhere Solutions

Scoring:There is no penalty for an incorrect answer.Each unanswered question is worth 2, to a maximum of 10 unanswered questions. Part A: Each correct answer is worth 5. 1.If x =3, the numerical value of 522–x is (A ) –1(B ) 27(C ) –13(D )–31(E ) 3 2.33 2232++ is equal to (A ) 3(B ) 6(C ) 2(D )32(E ) 5 3.If it is now 9:04 a.m., in 56 hours the time will be (A ) 9:04 a.m.(B ) 5:04 p.m.(C ) 5:04 a.m.(D ) 1:04 p.m.(E ) 1:04 a.m. 4.Which one of the following statements is not true? (A ) 25 is a perfect square. (B ) 31 is a prime number. (C ) 3 is the smallest prime number. (D ) 8 is a perfect cube. (E ) 15 is the product of two prime numbers. 5. A rectangular picture of Pierre de Fermat, measuring 20 cm by 40 cm, is positioned as shown on a rectangular poster measuring 50 cm by 100 cm. What percentage of the area of the poster is covered by the picture? (A ) 24%(B ) 16%(C ) 20% (D ) 25%(E ) 40% 6.Gisa is taller than Henry but shorter than Justina. Ivan is taller than Katie but shorter than Gisa. The tallest of these five people is (A ) Gisa (B ) Henry (C ) Ivan (D ) Justina (E ) Katie 7. A rectangle is divided into four smaller rectangles. The areas of three of these rectangles are 6, 15 and 25, as shown.The area of the shaded rectangle is (A ) 7(B ) 15(C ) 12 (D ) 16(E ) 10

加拿大大学世界排名

去加拿大留学之前，一定要选好合适的院校专业就读，加拿大作为北欧留学的热门国家，世界顶尖院校众多。今天就老师给大家介绍一下最新QS世界大学排名中加拿大排名前10的院校。 1.麦吉尔大学在世界名列第24位，麦吉尔大学已再次超过多伦多大学，回到其立之前作为加拿大大学最高排名的位置。成立于1821 年，有来自150 个国家的国际学生，已连续10 年被评为加拿大医学博士高校第一名。 2.多伦多大学世界排名第34 位，多伦多大学位于加拿大最大城市。其广受赞誉的领域包括文学批评、通信和理论和一系列的研究和创新成果，包括胰岛素和干细胞研究的诞生。 3.不列颠哥伦比亚大学位于西海岸，不列颠哥伦比亚大学在世界排名第50，有几乎60，000名学生(几乎一万二千人为国际学生) 在其温哥华和基洛纳校区。 4.阿尔伯塔大学今年与赫尔辛基大学一起在QS 世界大学排名中并列第96，阿尔伯塔大学是在艾伯塔省的主要经济驱动力，其艺术和人文学科的课程特别受人尊敬。 5.蒙特利尔大学蒙特利尔大学是加拿大的主要研究中心之一，从2011年开始分配5亿2400 万到超过150个研究中心进行研究。它今年世界排名第115，并且有超过7000 名加拿大留学生。

6.麦克马斯特大学今年麦克马斯特大学与天主教鲁汶大学一起并列149。它有300 英亩的校园，位于安大略省汉密尔顿，近年来因其日益增长的艺术场景已经吸引了越来越多的人来加拿大留学。 7.滑铁卢大学今年世界排名第152，滑铁卢大学相比前一年的排名已攀升17 个名次。大学是U15成员，一群加拿大研究型大学联盟，并声称要在加拿大建立头号心理学系，根据研究引文。 8.西安大略大学西安大略大学今年世界排名192，尤其以其医学、科学及商业的课程而闻名–是最大的社会科学学院。 9.卡尔加里大学卡尔加里大学与博洛尼亚大学一起并列世界204 ，是芯片重大发明的发源地，当前的加拿大总理哈珀也算是其校友。 10.皇后大学设在金斯敦，安大略省皇后大学这一年名列世界第206 名，有多个校园跨越1，400 公顷(3500 英亩) 整个安大略省的土地。皇后大学是西部沿海省份承认妇女，并形成一个学生政府第一所大学。其实，在最新的QS世界大学排名前800所院校中，加拿大院校占了26个位置，这里只是简单列举了前10个院校。

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 by a 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 new song“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 the position 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?

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

Chartered Accountants Sybase Inc. (Waterloo) IBM Canada Ltd. Canadian Institute of Actuaries Do 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 0301 2..()+ is (A ) 0.7(B ) 1(C ) 0.1(D ) 0.19(E ) 0.109 2.The pie chart shows a percentage breakdown of 1000 votes in 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 21 4.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 ) 25 5.In the diagram, ABCD is a rectangle with DC =12. If the area of triangle BDC is 30, what is the perimeter of rectangle 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 ) –4 7.In the diagram, AB is parallel to CD . What is the value of y ?(A ) 75(B ) 40(C ) 35 (D ) 55(E ) 50 8.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 ) 9 9.The number in an unshaded square is obtained by adding the numbers 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 B C D D A C B

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