2020年北欧数学竞赛考试试题目(无答案)
2020年北欧数学竞赛试题(含答案)
2020年北欧国家数学竞赛
1.对任息正整数，我们记g(x)为所有合{1,2,3,…,n}递增的三元组的个数。
求最小的正整数n,使得g(x)可以写成三个不同素数之积，且这三个素数都在一个公差为336的等差数列中(并不一定是连续的三项)。
2.卡特琳手上有2n+1张卡片,其中一张上写了0,剩下的卡片中写
有1,2，…n的个两张。
卡特琳想将这些卡片排成一排。
使得写了0的卡片正好在这一排卡片的中间,且对任意k=1,2，…n。
写有k的两张卡片之间的距离好为k(即这两张卡片之间一共有k-1张)
若1≤n≤10,当n为哪些数时，卡特琳可以排出这样的排列方式？
3.将凸四边形ABCD的边AB和CD分别三等分：E，F分别为AB三等分点，且AE=EF=FB,E，F分别为CD三等分点，且DP=PQ=QC.点M为四边形AEPD的两条对角线的交点。
点N为四边形FBCQ的两条对角线的交点。
证明△AMD和△BNC的面积之和等于△EPM和△FNQ的面积之和。
4.求所有函数:R R f →:.使得对任意满足(x+1)(y+1)(xy+1)≠0的数x ，y ∈R ，都有：
()⎪⎪⎭⎫ ⎝⎛++=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛+-111xy y x f y y f f x f 成立。
加拿大欧几里得数学竞赛真题解析2020EuclidSolution
1.(a)Solution1If x=−2,then 3x+6x+2=3(x+2)x+2=3.In other words,for every x=−2,the expression is equal to3.Therefore,when x=11,we get 3x+6x+2=3.Solution2When x=11,we obtain 3x+6x+2=3(11)+611+2=3913=3.(b)Solution1The point at which a line crosses the y-axis has x-coordinate0.Because A has x-coordinate−1and B has x-coordinate1,then the midpoint of AB is on the y-axis and is on the line through A and B,so is the point at which this line crossesthe x-axis.The midpoint of A(−1,5)and B(1,7)is 12(−1+1),12(5+7)or(0,6).Therefore,the line that passes through A(−1,5)and B(1,7)has y-intercept6.Solution2The line through A(−1,5)and B(1,7)has slope7−51−(−1)=22=1.Since the line passes through B(1,7),its equation can be written as y−7=1(x−1)or y=x+6.The line with equation y=x+6has y-intercept6.(c)First,wefind the coordinates of the point at which the lines with equations y=3x+7and y=x+9intersect.Equating values of y,we obtain3x+7=x+9and so2x=2or x=1.When x=1,we get y=x+9=10.Thus,these two lines intersect at(1,10).Since all three lines pass through the same point,the line with equation y=mx+17 passes through(1,10).Therefore,10=m·1+17which gives m=10−17=−7.2.(a)Suppose that m has hundreds digit a,tens digit b,and ones(units)digit c.From the given information,a,b and c are distinct,each of a,b and c is less than10, a=bc,and c is odd(since m is odd).The integer m=623satisfies all of these conditions.Since we are told there is only one such number,then623must be the only answer.Why is this the only possible value of m?We note that we cannot have b=1or c=1,otherwise a=c or a=b.Thus,b≥2and c≥2.Since c≥2and c is odd,then c can equal3,5,7,or9.Since b≥2and a=bc,then if c equals5,7or9,a would be larger than10,which is not possible.Thus,c=3.Since b≥2and b=c,then b=2or b≥4.If b≥4and c=3,then a>10,which is not possible.Therefore,we must have c=3and b=2,which gives a=6.(b)Since Eleanor has 100marbles which are black and gold in the ratio 1:4,then 15of her marbles are black,which means that she has 15·100=20black marbles.When more gold marbles are added,the ratio of black to gold is 1:6,which means thatshe has 6·20=120gold marbles.Eleanor now has 20+120=140marbles,which means that she added 140−100=40gold marbles.(c)First,we see that n 2+n +15n =n 2n +n n +15n =n +1+15n .This means that n 2+n +15n is an integer exactly when n +1+15nis an integer.Since n +1is an integer,then n 2+n +15n is an integer exactly when 15n is an integer.The expression 15nis an integer exactly when n is a divisor of 15.Since n is a positive integer,then the possible values of n are 1,3,5,and 15.3.(a)First,we note that a triangle with one right angle and one angle with measure 45◦isisosceles.This is because the measure of the third angle equals 180◦−90◦−45◦=45◦which means that the triangle has two equal angles.In particular, CDE is isosceles with CD =DE and EF G is isosceles with EF =F G .Since DE =EF =1m,then CD =F G =1m.Join C to G .ACB D FG HE 45°45°45°1 m 1 m 1 m 1 m Consider quadrilateral CDF G .Since the angles at D and F are right angles and since CD =GF ,it must be the case that CDF G is a rectangle.This means that CG =DF =2m and that the angles at C and G are right angles.Since ∠CGF =90◦and ∠DCG =90◦,then ∠BGC =180◦−90◦−45◦=45◦and ∠BCG =90◦.This means that BCG is also isosceles with BC =CG =2m.Finally,BD =BC +CD =2m +1m =3m.(b)We apply the process two more times:x y Before Step 1243After Step 1273After Step 2813After Step 3814x y Before Step 1814After Step 1854After Step 23404After Step 33405Therefore,the final value of x is 340.(c)The parabola with equation y =kx 2+6x +k has two distinct x -intercepts exactly whenthe discriminant of the quadratic equation kx 2+6x +k =0is positive.Here,the disciminant equals ∆=62−4·k ·k =36−4k 2.The inequality 36−4k 2>0is equivalent to k 2<9.Since k is an integer and k =0,then k can equal −2,−1,1,2.(If k ≥3or k ≤−3,we get k 2≥9so no values of k in these ranges give the desired result.)4.(a)Since a b <47and 47<1,then a b<1.Since a and b are positive integers,then a <b .Since the difference between a and b is 15and a <b ,then b =a +15.Therefore,we have 59<a a +15<47.We multiply both sides of the left inequality by 9(a +15)(which is positive)to obtain 5(a +15)<9a from which we get 5a +75<9a and so 4a >75.From this,we see that a >754=18.75.Since a is an integer,then a ≥19.We multiply both sides of the right inequality by 7(a +15)(which is positive)to obtain 7a <4(a +15)from which we get 7a <4a +60and so 3a <60.From this,we see that a <20.Since a is an integer,then a ≤19.Since a ≥19and a ≤19,then a =19,which means that a b =1934.(b)The first 6terms of a geometric sequence with first term 10and common ratio 12are 10,5,52,54,58,516.Here,the ratio of its 6th term to its 4th term is 5/165/4which equals 14.(We could have determined this without writing out the sequence,since moving from the 4th term to the 6th involves multiplying by 12twice.)The first 6terms of an arithmetic sequence with first term 10and common difference d are 10,10+d,10+2d,10+3d,10+4d,10+5d .Here,the ratio of the 6th term to the 4th term is 10+5d 10+3d .Since these ratios are equal,then 10+5d 10+3d =14,which gives 4(10+5d )=10+3d and so 40+20d =10+3d or 17d =−30and so d =−3017.5.(a)Let a=f(20).Then f(f(20))=f(a).To calculate f(f(20)),we determine the value of a and then the value of f(a).By definition,a=f(20)is the number of prime numbers p that satisfy20≤p≤30.The prime numbers between20and30,inclusive,are23and29,so a=f(20)=2.Thus,f(f(20))=f(a)=f(2).By definition,f(2)is the number of prime numbers p that satisfy2≤p≤12.The prime numbers between2and12,inclusive,are2,3,5,7,11,of which there are5.Therefore,f(f(20))=5.(b)Since(x−1)(y−2)=0,then x=1or y=2.Suppose that x=1.In this case,the remaining equations become:(1−3)(z+2)=01+yz=9or−2(z+2)=0yz=8From thefirst of these equations,z=−2.From the second of these equations,y(−2)=8and so y=−4.Therefore,if x=1,the only solution is(x,y,z)=(1,−4,−2).Suppose that y=2.In this case,the remaining equations become:(x−3)(z+2)=0x+2z=9From thefirst equation x=3or z=−2.If x=3,then3+2z=9and so z=3.If z=−2,then x+2(−2)=9and so x=13.Therefore,if y=2,the solutions are(x,y,z)=(3,2,3)and(x,y,z)=(13,2,−2).In summary,the solutions to the system of equations are(x,y,z)=(1,−4,−2),(3,2,3),(13,2,−2)We can check by substitution that each of these triples does indeed satisfy each of the equations.6.(a)Draw a perpendicular from S to V on BC .Since ASV B is a quadrilateral with three right angles,then it has four right angles and so is a rectangle.Therefore,BV =AS =r ,since AS is a radius of the top semi-circle,and SV =AB =4.Join S and T to P .Since the two semi-circles are tangent at P ,then SP T is a straight line,which means that ST =SP +P T =r +r =2r .64Consider right-angled SV T .We have SV =4and ST =2r .Also,V T =BC −BV −T C =6−r −r =6−2r .By the Pythagorean Theorem,SV 2+V T 2=ST 242+(6−2r )2=(2r )216+36−24r +4r 2=4r 252=24rThus,r =5224=136.(b)Since ABE is right-angled at A and is isosceles with AB =AE =7√2,then ABE is a 45◦-45◦-90◦triangle,which means that ∠ABE =45◦and BE =√2AB =√2·7√2=14.Since BCD is right-angled at C with DB DC =8x 4x =2,then BCD is a 30◦-60◦-90◦triangle,which means that ∠DBC =30◦.Since ∠ABC =135◦,then ∠EBD =∠ABC −∠ABE −∠DBC =135◦−45◦−30◦=60◦.Now consider EBD .We have EB =14,BD =8x ,DE =8x −6,and ∠EBD =60◦.Using the cosine law,we obtain the following equivalent equations:DE 2=EB 2+BD 2−2·EB ·BD ·cos(∠EBD )(8x −6)2=142+(8x )2−2(14)(8x )cos(60◦)64x 2−96x +36=196+64x 2−2(14)(8x )·12−96x =160−14(8x )112x −96x =16016x =160x =10Therefore,the only possible value of x is x =10.7.(a)Solution 1Since the function g is linear and has positive slope,then it is one-to-one and so invertible.This means that g −1(g (a ))=a for every real number a and g (g −1(b ))=b for every real number b .Therefore,g (f (g −1(g (a ))))=g (f (a ))for every real number a .This means thatg (f (a ))=g (f (g −1(g (a ))))=2(g (a ))2+16g (a )+26=2(2a −4)2+16(2a −4)+26=2(4a 2−16a +16)+32a −64+26=8a 2−6Furthermore,if b =f (a ),then g −1(g (f (a )))=g −1(g (b ))=b =f (a ).Therefore,f (a )=g −1(g (f (a )))=g −1(8a 2−6)Since g (x )=2x −4,then y =2g −1(y )−4and so g −1(y )=12y +2.Therefore,f (a )=12(8a 2−6)+2=4a 2−1and so f (π)=4π2−1.Solution 2Since the function g is linear and has positive slope,then it is one-to-one and so invertible.To find a formula for g −1(y ),we start with the equation g (x )=2x −4,convert to y =2g −1(y )−4and then solve for g −1(y )to obtain 2g −1(y )=y +4and so g −1(y )=y +42.We are given that g (f (g −1(x )))=2x 2+16x +26.We can apply the function g −1to both sides to obtain successively:f (g −1(x ))=g −1(2x 2+16x +26)f (g −1(x ))=(2x 2+16x +26)+42(knowing a formula for g −1)f (g −1(x ))=x 2+8x +15f x +42 =x 2+8x +15(knowing a formula for g −1)f x +42 =x 2+8x +16−1f x +42=(x +4)2−1We want to determine the value of f (π).Thus,we can replace x +42with π,which is equivalent to replacing x +4with 2π.Thus,f (π)=(2π)2−1=4π2−1.(b)Solution 1Using logarithm laws,the given equations are equivalent tolog 2(sin x )+log 2(cos y )=−32log 2(sin x )−log 2(cos y )=12Adding these two equations,we obtain 2log 2(sin x )=−1which gives log 2(sin x )=−12and so sin x =2−1/2=121/2=1√2.Since 0◦≤x <180◦,then x =45◦or x =135◦.Since log 2(sin x )+log 2(cos y )=−32and log 2(sin x )=−12,then log 2(cos y )=−1,which gives cos y =2−1=12.Since 0◦≤y <180◦,then y =60◦.Therefore,(x,y )=(45◦,60◦)or (x,y )=(135◦,60◦).Solution 2First,we note that 21/2=√2and 2−3/2=123/2=12121/2=12√2.From the given equations,we obtainsin x cos y =2−3/2=12√2sin x cos y=21/2=√2Multiplying these two equations together,we obtain (sin x )2=12which gives sin x =±1√2.Since 0◦≤x <180◦,it must be the case that sin x ≥0and so sin x =1√2.Since 0◦≤x <180◦,we obtain x =45◦or x =135◦.Since sin x cos y =12√2and sin x =1√2,we obtain cos y =12.Since 0◦≤y <180◦,then y =60◦.Therefore,(x,y )=(45◦,60◦)or (x,y )=(135◦,60◦).8.(a)Solution1Let x be the probability that Bianca wins the tournament.Because Alain,Bianca and Chen are equally matched and because their roles in the tour-nament are identical,then the probability that each of them wins will be the same.Thus,the probability that Alain wins the tournament is x and the probability that Chen wins the tournament is x.Let y be the probability that Dave wins the tournament.Since exactly one of Alain,Bianca,Chen,and Dave wins the tournament,then3x+y=1and so x=1−y3.We can calculate y in terms of p.In order for Dave to win the tournament,he needs to win two matches.No matter who Dave plays,his probability of winning each match is p.Thus,the probability that he wins his two consecutive matches is p2and so the probability that he wins the tournament is y=p2.Thus,the probability that Bianca wins the tournament is 1−p23.(We could rewrite this as −p2+0p+13to match the desired form.)Solution2Let x be the probability that Bianca wins the tournament. There are three possible pairings for thefirst two matches: (i)Bianca versus Alain,and Chen versus Dave(ii)Bianca versus Chen,and Alain versus Dave(iii)Bianca versus Dave,and Alain versus ChenEach of these three pairings occurs with probability13.In(i),Bianca wins either if Bianca beats Alain,Chen beats Dave,and Bianca beats Chen, or if Bianca beats Alain,Dave beats Chen,and Bianca beats Dave.Since Bianca beats Alain with probability12,Chen beats Dave with probability1−p,andBianca beats Chen with probability12,then thefirst possibility has probability12·(1−p)·12.Since Bianca beats Alain with probability12,Dave beats Chen with probability p,andBianca beats Dave with probability1−p,then the second possibility has probability1 2·p·(1−p).Therefore,the probability of Bianca winning,given that possibility(i)occurs,is12·(1−p)·12+12·p·(1−p).In(ii),Bianca wins either if Bianca beats Chen,Alain beats Dave,and Bianca beats Alain, or if Bianca beats Alain,Dave beats Alain,and Bianca beats Dave.The combined probability of these is12·(1−p)·12+12·p·(1−p).In(iii),Bianca wins either if Bianca beats Dave,Alain beats Chen,and Bianca beats Alain,or if Bianca beats Dave,Chen beats Alain,and Bianca beats Chen.The combined probability of these is(1−p)·12·12+(1−p)·12·12.Therefore,x=13 14(1−p)+12p(1−p)+14(1−p)+12p(1−p)+14(1−p)+14(1−p)=13(p(1−p)+(1−p))=13(p−p2+1−p)Thus,the probability that Bianca wins the tournament is 1−p23.(b)Throughout this solution,we will mostly not include units,but will assume that all lengthsare in kilometres,all times are in seconds,and all speeds are in kilometres per second.We place the points in the coordinate plane with B at(0,0),A on the negative x-axis, and C on the positive x-axis.We put A at(−1,0)and C at(2,0).Suppose that P has coordinates(x,y)and that the distance from P to B is d km.Since the sound arrives at A12s after arriving at B and sound travels at13km/s,then Ais(12s)·(13km/s)=16km farther from P than B is.Thus,the distance from P to A is(d+16)km.Since the sound arrives at C an additional1second later,then C is an additional13kmfarther,and so is(d+16)km+(13km)=(d+12)km from P.Since the distance from P to B is d km,then(x−0)2+(y−0)2=d2.Since the distance from P to A is(d+16)km,then(x+1)2+(y−0)2=(d+16)2.Since the distance from P to C is(d+12)km,then(x−2)2+(y−0)2=(d+12)2.When these equations are expanded and simplified,we obtainx2+y2=d2x2+2x+1+y2=d2+13d+136x2−4x+4+y2=d2+d+14 Subtracting thefirst equation from the second,we obtain2x+1=13d+136Subtracting thefirst equation from the third,we obtain−4x+4=d+14 Therefore,2(2x+1)+(−4x+4)=2(13d+136)+(d+14)6=23d+118+d+14216=24d+2+36d+9(multiplying by36)205=60dd=4112Therefore,the distance from B to P is4112km.9.(a)After each round,each L shape is divided into4smaller L shapes.This means that the number of L shapes increases by a factor of4after each round.After1round,there are4L shapes.After2rounds,there are42=16L’s of the smallest size.After3rounds,there are43=64L’s of the smallest size.(b)There are four orientations of L shapes of a given size:,,,.When an L of each orientation is subdivided,the followingfigures are obtained:From thesefigures,we can see that after each subsequent round,•Each produces2,0,1,and1of the smallest size.•Each produces0,2,1,and1.•Each produces1,1,2,and0.•Each produces1,1,0,and2.After1round,there are2,0,1,and1.After2rounds,the number of L’s of each orientation are as follows:•:2·2+0·0+1·1+1·1=6•:2·0+0·2+1·1+1·1=2•:2·1+0·1+1·2+1·0=4•:2·1+0·1+1·0+1·2=4After3rounds,the number of L’s of each orientation are as follows:•:6·2+2·0+4·1+4·1=20•:6·0+2·2+4·1+4·1=12•:6·1+2·1+4·2+4·0=16•:6·1+2·1+4·0+4·2=16Where do these numbers come from?For example,to determine the number of after2rounds,we look at the number of L’s of each orientation after round1(2,0,1,1)and ask how many each of these produces at the next level.Since the four types each produce2,0,1,and1,then the total number of after2rounds equals2·2+0·0+1·1+1·1which equals6.As a second example,to determine the number of after3rounds,we note that after2 rounds the number of L’s of the four different orientations are6,2,4,4and that each L of each of the four types produces0,2,1,1.This means that the total number ofafter3rounds is6·0+2·2+4·1+4·1=12.Putting all of this together,the number of L’s of the smallest size in the same orientation as the original L is20.(c)In(b),we determined the number of L’s of the smallest size in each orientation after1,2and3rounds.We continue to determine the number of L’s of the smallest size after4rounds.After4rounds,the number of L’s of each orientation are as follows:•:20·2+12·0+16·1+16·1=72•:20·0+12·2+16·1+16·1=56•:20·1+12·1+16·2+16·0=64•:20·1+12·1+16·0+16·2=64This gives us the following tables of the numbers of L’s of the smallest size in each orien-tation after1,2,3,and4rounds:After Round1201126244320121616472566464We re-write these numbers in the third row as16+4,16−4,16,16and the numbers in the fourth row as64+8,64−8,64,64.Based on this,we might guess that the numbers of L’s of the smallest size in each orien-tation after n rounds are4n−1+2n−1,4n−1−2n−1,4n−1,4n−1.If this guess is correct,then,after2020rounds,the number of L’s of the smallest size in the same orientation as the original L is42019+22019.We prove that these guesses are right by using an inductive process.First,we note that the table above shows that our guess is correct when n=1,2,3,4.Next,if we can show that our guess being correct after a given number of rounds implies that it is correct after the next round,then it will be correct after every round.This is because being correct after4rounds will mean that it is correct after5rounds,being correct after5rounds will mean that it is correct after6rounds,and so on to be correct after any number of rounds.Suppose,then,that after k rounds the numbers of L’s of the smallest size in each orienta-tion are4k−1+2k−1,4k−1−2k−1,4k−1,4k−1.After k+1rounds(that is,after the next round),the number of L’s of each orientation is:•:(4k−1+2k−1)·2+(4k−1−2k−1)·0+4k−1·1+4k−1·1=4·4k−1+2·2k−1=4k+2k•:(4k−1+2k−1)·0+(4k−1−2k−1)·2+4k−1·1+4k−1·1=4·4k−1−2·2k−1=4k−2k•:(4k−1+2k−1)·1+(4k−1−2k−1)·1+4k−1·2+4k−1·0=4·4k−1=4k•:(4k−1+2k−1)·1+(4k−1−2k−1)·1+4k−1·0+4k−1·2=4·4k−1=4kSince k=(k+1)−1,these expressions match our guess.This means that our guess is correct after every number of rounds.Therefore,after2020rounds,the number of L’s of the smallest size in the same orientation as the original L is42019+22019.10.(a)Here,the pairwise sums of the numbers a1≤a2≤a3≤a4are s1≤s2≤s3≤s4≤s5≤s6.The six pairwise sums of the numbers in the list can be expressed asa1+a2,a1+a3,a1+a4,a2+a3,a2+a4,a3+a4Since a1≤a2≤a3≤a4,then the smallest sum must be the sum of the two smallest numbers.Thus,s1=a1+a2.Similarly,the largest sum must be the sum of the two largest numbers,and so s6=a3+a4.Since a1≤a2≤a3≤a4,then the second smallest sum is a1+a3.This is because a1+a3 is no greater than each of the four sums a1+a4,a2+a3,a2+a4,and a3+a4: Since a3≤a4,then a1+a3≤a1+a4.Since a1≤a2,then a1+a3≤a2+a3.Since a1≤a2and a3≤a4,then a1+a3≤a2+a4.Since a1≤a4,then a1+a3≤a3+a4.Thus,s2=a1+a3.Using a similar argument,s5=a2+a4.So far,we have s1=a1+a2and s2=a1+a3and s5=a2+a4and s6=a3+a4.This means that s3and s4equal a1+a4and a2+a3in some order.It turns out that either order is possible.Case1:s3=a1+a4and s4=a2+a3Here,a1+a2=8and a1+a3=104and a2+a3=110.Adding these three equations gives(a1+a2)+(a1+a3)+(a2+a3)=8+104+110and so2a1+2a2+2a3=222or a1+a2+a3=111.Since a2+a3=110,then a1=(a1+a2+a3)−(a2+a3)=111−110=1.Since a1=1and a1+a2=8,then a2=7.Since a1=1and a1+a3=104,then a3=103.Since a3=103and a3+a4=208,then a4=105.Thus,(a1,a2,a3,a4)=(1,7,103,105).Case2:s3=a2+a3and s4=a1+a4Here,a1+a2=8and a1+a3=104and a2+a3=106.Using the same process,a1+a2+a3=109.From this,we obtain(a1,a2,a3,a4)=(3,5,101,107).Therefore,Kerry’s two possible lists are1,7,103,105and3,5,101,107.(b)Suppose that the values of s1,s2,s3,s4,s5,s6,s7,s8,s9,s10arefixed,but unknown.In terms of the numbers a1≤a2≤a3≤a4≤a5,the ten pairwise sums are a1+a2,a1+a3,a1+a4,a1+a5,a2+a3,a2+a4,a2+a5,a3+a4,a3+a5,a4+a5 These will equal s1,s2,s3,s4,s5,s6,s7,s8,s9,s10in some order.Using a similar analysis to that in(a),the smallest sum is a1+a2and the largest sum is a4+a5.Thus,s1=a1+a2and s10=s4+s5.Also,the second smallest sum will be s2=a1+a3and the second largest sum will be s9=a3+a5.We letS=s1+s2+s3+s4+s5+s6+s7+s8+s9+s10Note that S has afixed,but unknown,value.Even though we do not know the order in which these pairwise sums are assigned to s1 through s10,the value of S will equal the sum of these ten pairwise expressions.In other words,S=4a1+4a2+4a3+4a4+4a5,since each of the numbers in the list occurs in four sums.Thus,a1+a2+a3+a4+a5=14S and so(a1+a2)+a3+(a4+a5)=14S.This means that s1+a3+s10=14S and so a3=14S−s1−s10.Since the values of s1,s10and S arefixed,then we are able to determine the value of a3 from the list of sums s1through s10.Using the value of a3,the facts that s2=a1+a3and s9=a3+a5,and that s2and s9are known,we can determine a1and a5.Finally,using s1=a1+a2and s10=a4+a5and the values of a1and a5,we can determine a2and a4.Therefore,given the ten sums s1through s10,we can determine the values of a3,a1,a5, a2,a4and so there is only one possibility for the list a1,a2,a3,a4,a5.(Can you write out expressions for each of a1through a5in terms of s1through s10only?)(c)Suppose that the lists a1,a2,a3,a4and b1,b2,b3,b4produce the same list of sums s1,s2,s3,s4,s5,s6.(Examples of such lists can be found in(a).)Let x be a positive integer.Consider the following list with8entries:a1,a2,a3,a4,b1+x,b2+x,b3+x,b4+xFrom this list,there are three categories of pairwise sums:(i)a i+a j,1≤i<j≤4:these give the sums s1through s6(ii)(b i+x)+(b j+x),1≤i<j≤4:each of these is2x greater than the six sums s1 through s6because the pairwise sums b i+b j give the six sums s1through s6 (iii)a i+(b j+x),1≤i≤4and1≤j≤4Consider also the list with8entries:a1+x,a2+x,a3+x,a4+x,b1,b2,b3,b4From this list,there are again three categories of pairwise sums:(i)b i+b j,1≤i<j≤4:these give the sums s1through s6(ii)(a i+x)+(a j+x),1≤i<j≤4:each of these is2x greater than the six sums s1 through s6because the pairwise sums a i+a j give the six sums s1through s6 (iii)(a i+x)+b j,1≤i≤4and1≤j≤4Thus,the28pairwise sums in each case are the same.In each case,there are6sums in (i),6sums in(ii),and16sums in(iii).If we choose the initial lists to have the same pairwise sums and choose the value of x to be large enough so that a i+x is not equal to any b j and b i+x is not equal to any a j,we obtain two different lists of8numbers that each produce the same list of28sums.For example,if we choose a1,a2,a3,a4to be1,7,103,105and b1,b2,b3,b4to be3,5,101,107 and x=10000,we get the lists1,7,103,105,10003,10005,10101,10107and3,5,101,107,10001,10007,10103,10105Using a similar analysis to that above,if the lists a1,a2,a3,a4,a5,a6,a7,a8and b1,b2, b3,b4,b5,b6,b7,b8have the same set of pairwise sums,then the listsa1,a2,a3,a4,a5,a6,a7,a8,b1+y,b2+y,b3+y,b4+y,b5+y,b6+y,b7+y,b8+y anda1+y,a2+y,a3+y,a4+y,a5+y,a6+y,a7+y,a8+y,b1,b2,b3,b4,b5,b6,b7,b8will also have the same pairwise sums.Therefore,setting y=1000000,we see that the lists1,7,103,105,10003,10005,10101,10107,1000003,1000005,1000101,1000107,1010001,1010007,1010103,1010105and3,5,101,107,10001,10007,10103,10105,1000001,1000007,1000103,1000105,1010003,1010005,1010101,1010107have the same list of sums s1,s2,...,s120,as required.。
国际数学奥林匹克竞赛试题及解答
国际数学奥林匹克竞赛试题及解答国际数学奥林匹克竞赛是世界范围内最具影响力和声誉的数学竞赛之一。
每年，来自各个国家的数学高手们聚集在一起，参与这项激烈而充满挑战的竞赛。
本文将介绍一些历年的国际数学奥林匹克竞赛试题，并提供相应的解答。
试题一：证明：当n为正整数时，4^n + n^4不是素数。
解答一：我们可以通过反证法来证明这个命题。
假设4^n + n^4是一个素数，即不存在其他因子能够整除它。
考虑到任何正整数n都可以写成2k或2k+1的形式，其中k是整数。
当n为偶数时，可以将n表示为2k的形式。
那么我们有：4^n + n^4 = (2^2)^n + (2k)^4 = 2^(2n) + (2k)^4我们可以看出，2^(2n)是一个完全平方数，而(2k)^4也是一个完全平方数。
根据完全平方数的性质，它们的和2^(2n) + (2k)^4也是一个完全平方数。
因此，当n为偶数时，4^n + n^4不可能是素数。
当n为奇数时，可以将n表示为2k+1的形式。
那么我们有：4^n + n^4 = (2^2)^n + (2k+1)^4 = 2^(2n) + (2k+1)^4同样地，我们可以看出，2^(2n)是一个完全平方数，而(2k+1)^4也是一个完全平方数。
根据完全平方数的性质，它们的和2^(2n) + (2k+1)^4也是一个完全平方数。
因此，当n为奇数时，4^n + n^4同样不可能是素数。
综上所述，我们可以得出结论：当n为正整数时，4^n + n^4不是素数。
试题二：证明：对于任意正整数n，n^2 + 3n + 1不是完全平方数。
解答二：我们同样可以使用反证法来证明这个命题。
假设n^2 + 3n + 1是一个完全平方数，即存在另一个正整数m，使得m^2 = n^2 + 3n + 1。
根据完全平方数的性质，m^2必然是一个奇数，因为奇数的平方也是奇数。
我们可以将n^2 + 3n + 1拆分为两部分，即(n^2 + 2n + 1) + n。
imo数学竞赛试题及答案
imo数学竞赛试题及答案IMO数学竞赛试题及答案一、选择题1. 下列哪个数是最小的正整数？A. 0B. 1C. 2D. 3答案：B2. 如果一个数的立方等于它本身，那么这个数可以是：A. -1B. 0C. 1D. 2答案：ABC3. 一个长方体的长、宽、高分别是8cm、6cm和5cm，那么它的表面积是多少平方厘米？A. 236B. 284C. 312D. 376答案：B二、填空题4. 一个数的平方根是3，那么这个数是_________。
答案：95. 一个等差数列的前三项分别是2，4，6，那么它的第10项是_________。
答案：22三、解答题6. 证明：对于任意的正整数 \( n \)，\( n^5 - n \) 总是能被30整除。
解答：首先，我们可以将 \( n^5 - n \) 分解为 \( n(n^4 - 1) \)。
接下来，我们注意到 \( n^4 - 1 \) 可以表示为 \( (n^2 +1)(n^2 - 1) \)。
而 \( n^2 - 1 \) 可以进一步分解为 \( (n +1)(n - 1) \)。
因此，我们有：\( n^5 - n = n(n^2 + 1)(n + 1)(n - 1) \)。
由于 \( n \) 是正整数，\( n - 1 \) 和 \( n + 1 \) 也是整数。
这意味着 \( n^5 - n \) 中至少包含因子2和3（因为 \( n^2 + 1 \) 至少是奇数，从而至少包含一个2的因子）。
因此，\( n^5 - n \)可以被30整除。
7. 一个圆的半径是15厘米，求圆的面积。
解答：圆的面积可以通过公式 \( A = \pi r^2 \) 计算，其中\( A \) 是面积，\( r \) 是半径，\( \pi \) 是圆周率，约等于3.14159。
将给定的半径 \( r = 15 \) 厘米代入公式，我们得到：\( A = \pi \times 15^2 = \pi \times 225 \approx 706.86 \)平方厘米。
2020年国际数学奥林匹克(IMO)全部试题解答
2020年第61届国际数学奥林匹克（IMO）全部试题解答海亮高级中学高三康榕博高二陈昶旭第一天第1题. 考虑凸四边形ABCD. 设P 是ABCD 内部一点. 且以下比例等式成立:∠PAD:∠PBA:∠DPA＝1: 2 :3＝∠CBP:∠BAP:∠BPC.证明: ∠ADP 的内角平分线、∠PCB 的内角平分线和线段AB 的垂直平分线三线共点.证明：如图，设∠PAD＝α,∠PBC＝β,则∠ABP＝2α,∠BAP＝2β, ∠APD＝3α,∠BPC＝3β,取△ABP外心O, 则∠AOP＝4α＝π－∠ADP∴A, O, P, D共圆.∴∠ADO＝∠APO＝∠PAO＝∠PDO∴OD平分∠PDA.同理, OC平分∠PCB.而O为△ABP外心, 显然在AB中垂线上.故∠PDA平分线, ∠PCB平分线, AB中垂线均过点O.证毕.第2题. 设实数a, b, c, d 满足a ≥b ≥c ≥d > 0, 且 a + b + c + d = 1. 证明:(234)1a b c d a b c d a b c d +++<. 证明: 由加权AM －GM 不等式, 我们有2222a b c d a b c d a a b b c c d d a b c d <⋅+⋅+⋅+⋅=+++ 故只需证明22223(234)()()cyca b c d a b c d a ++++++<∑ (*)注意到332()36cyc cyc sym cyca a ab abc =++∑∑∑∑, 及32222cyca ab ad a a ++≥∑2232222222cyca b ab b bc bd b a ++++≥∑2222233333cyca cbc ac cd c a +++≥∑22234444cyc a d a b abd acd bcd d a ++++≥∑∴ (*)成立. 故原不等式成立.第3题. 有4n 枚小石子, 重量分别为1, 2, 3, . . . , 4n. 每一枚小石子都染了n 种颜色之一, 使得每种颜色的小石子恰有四枚. 证明: 我们可以把这些小石子分成两堆, 同时满足以下两个条件:• 两堆小石子有相同的总重量;• 每一堆恰有每种颜色的小石子各两枚.证明: 引理:将n 种颜色的点个4个两两分组, 则可取n 组使得每种颜色的点各2个.即证: n 阶4-正则图G(不一定简单)必有2-正则生成子图. n ＝1, G 为v 的2个自环, 成立.设0n n ≤成立, 则01n n =+时:若G 有点含两自环或有两点含4重边, 对其余部分用归纳假设,该部分取1自环或2重边即可.下设无这样的结构.若G 含三重边,设x,y 间有三条边, 且,(,)xu yv G u y v x ∈≠≠. 考虑将x,y 去掉, 并添入边uv 得到图G ’. 由归纳假设, 图G ’有2-正则生成子图, 若该图含添入的边 uv, 删去该边并加入ux, xy, yv 即可. 若不含, 加入xy, xy 即可.下设无三重边.显然G 有圈. 设最小圈为121,,...,t x x x x . 由G 无2自环,3重边知01t n <+, i x 有两边不指向12,,...t x x x . 设这两边指向,i i u v ，以下下标模t.在G 中删去点12,,...t x x x 并加入边1(1)i i i e u v i t +=≤≤得到G’. 由归纳假设, G ’有2-正则子图G 1.对1≤i ≤t, 若1i e G ∈, 则选择G 中的边11,i i i i x u x v ++, 若1i e G ∉, 则选自1i i x x +, 其余边按G 1中边选择, 则选出的边即为G 的2-正则生成子图的边集.结论成立.回到原题. 将重量为{,41}k n k +-的小石子分为一组.(12)k n ≤≤, 由引理可取n 组使每种颜色的小石子恰2个. 这2n 个分为一组, 其余分为一组, 此即满足条件的分法, 命题成立.第二天第4题. 给定整数n > 1. 在一座山上有n2个高度互不相同的缆车车站. 有两家缆车公司A和B, 各运营k辆缆车; 每辆从一个车站运行到某个更高的车站(中间不停留其他车站). A 公司的k辆缆车的k个起点互不相同，k个终点也互不相同, 并且起点较高的缆车，它的终点也较高. B公司的缆车也满足相同的条件. 我们称两个车站被某个公司连接，如果可以从其中较低的车站通过该公司的一辆或多辆缆车到达较高的车站(中间不允许在车站之间有其他移动). 确定最小的正整数k, 使得一定有两个车站被两个公司同时连接.解: 由题意得, 每个缆车与1或2个缆车相连. (否则有两辆缆车起点不同, 终点相同)∴A, B各自的缆车线路图可划分为若干个链.注意到每条链长度大于等于2, 且首尾两点不能作为终点和起点, 故恰有2n k-条链.若21k n n≥-+, 则A最多由n－1条链.由抽屉原理, 其中至少有一条链上有221nnn⎡⎤=+⎢⎥-⎢⎥个点, 设为P. 而B仅有n－1条链, 故P上一定有两个点同时在B 的一条链上, 则这两点可被两个公司同时连接.另一方面, 2k n n=-时, 记2n个车站高度排序为21,2,...n (从低到高)令A的2n n-辆缆车为2(1)i n i i n n→+≤≤-令B的2n n-辆缆车为21(11,|)i i i n n i→+≤≤-/易见此时任两个车站不能被两个公司同时相连.2 min 1k n n∴=-+.第5题. 有一叠n > 1张卡片. 在每张卡片上写有一个正整数. 这叠卡片具有如下性质：其中任意两张卡片上的数的算术平均值也等于这叠卡片中某一张或几张卡片上的数的几何平均值.确定所有的n, 使得可以推出这叠卡片上的数均相等? 解: 设这n 张卡片上的数为1212,,....(...)n n x x x x x x ≤≤. 若12gcd(,,...)1n x x x d =>, 用i x d 代替i x , 不影响结果. 故不妨设12gcd(,,...)1n x x x =.由题意得, 1,2i jx x i j n +∀≤≤≤为代数整数.则2|i j i x x x +⇒模2同余. 又12gcd(,,...)1n x x x =, 故i x 全为奇数.任取一个素数p, p ≥3.记{|1,|},{|1,|}i i i i A x i n p x B x i n p x =≤≤=≤≤/ 则对,,2x y x A y B +∀∈∈不为p 的倍数. 设121(...)2k k i i i x y x x x +=, 则121|(...)2k k i i i x y p x x x +=/ ∴对1,j i j k x B ∀≤≤∈.max 2i i x B x y x ∈+∴≤. 取max ,max i i i i x A x B x x y x ∈∈==, 则max max i i i i x A x B x x ∈∈≤若1n x ≠, 取n x 的奇素因子p, 由12gcd(,,...)1n x x x =知, i ∃, 使|i p x /.取0max{|1,|}i i i i n p x =≤≤/, 由上述结论知0n i x x ≤, 则o n i x x =. 又0|,|i n p x p x /, 矛盾!1n x ∴=. 则1,1i i n x ∀≤≤=.∴对任意n ≥2, 卡片上的所有数均相等.第6题. 证明: 存在正常数c 具有如下性质:对任意整数n > 1, 以及平面上n 个点的集合S, 若S 中任意两点之间的距离不小于1，则存在一条分离S 的直线ℓ, 使得S 中的每个点到直线ℓ 的距离不小于13cn -.(我们称直线ℓ分离点集S, 如果某条以S 中两点为端点的线段与ℓ 相交.)证明: 以每个点为圆心,12为半径作圆, 则这些圆两两公共部分面积为0.引理1: 对凸多边形P, 其内部最多由421s l π++个点在S 中,其中s,l 代表P 的面积和周长. 证明: 如图, 将P 的每条边往外侧平移12, 并以P 上每个点为圆心, 12为半径作圆, 拓展区域面积为124l π+. ∴P 内部最多1422414S l s l πππ+++=+个点. 现在对于一条直线l, 作S 中每个点在l 上的投影. 任取相邻两个投影点, 则这两点连线的中垂线分离点集S, 且所有的到该直线的距离≥12投影点距离.设S 的直径为D, 则可作一个以D 为边长的正方形覆盖S. 由引理1, 122481()D Dn D n π++≥⇒=Ω 设P,Q ∈S, PQ ＝D. 将PQ 作为上述l, 记我们所能做到的使每个点到一条直线的距离均不小于该数的最大值为d.由于仅与夹角有关, 故d 存在.而l 上除P,Q 外有n －2个投影点.2(1)2D D d n n∴≥>-. 又12()D n =Ω, 故12()d n -=Ω. 需证明13()d n -=Ω .取点集S 的凸包P. 若一直线过P 上一点且使得S 中所有点都在该线一侧, 我们认为其亦分离S. 称其为支撑边. 对于任一常数C, 作两条平行的距离为C 的直线, 满足这两条直线分离S. 作他们的垂线l, 设这个带状区域内有m 个S 中的点, 则11c c d m m d≥⇒≥-+. 不妨设(1)d o =, 则可以认为m 远远大于1. 为使m 尽量小, 应取两直线其中之一为支撑边.∴现在对于一条分离S 的直线l, 设l 与P 围成的区域内部有B 个点. P 中与l 距离最近的点到l 距离为0s , 则01s d B ≥+ (以下用≥代表数量级估计) 我们证明d≥从而311D d n D n ≥⋅= 则13()d n -=Ω. 如图, P 夹在这样一个区域里, 取XY 上一点Z, 使得0YZ s =. 过Z 作MN ⊥XY , 点M,N 在以X 为圆心, D 为半径的圆上. 则B ≤YMN 内S 中点的个数.不妨设XY 为x 轴, 对YMN 内任意两点1122(,),(,)x y x y , 221201212||,()()1x x s x x y y -≤-+-≥, 则12||1y y B -≥⇒≤+.而MN =02s d MN∴≥=+由于0(1)s =Θd ∴≥, 则13d n -≥, 即13()d n -=Ω证毕.。
2020年欧洲女子奥林匹克数学竞赛
2020年欧洲女子奥林匹克数学竞赛题1.一列正整数a 0,a 1,a 2，……a 3030满足对任意n=0,1,2,……,3028都有2a n+2=a n+1+4a n证明：a 0,a 1,a 2，……a 3030至少有一个数可以被22020整除。
题2.求所有满足下列三个条件的非负实数组(x 1,x 2，……，x 2020):1.x 1≤x 2≤……≤x 2020；2.x 2020≤x 1+1；3.存在(x 1,x 2，……，x 2020)的一个置换(y 1,y 2，……，y 2020)满足:题3.设 ABCDEF 是一个凸的六边形,满足∠A=∠C=∠E ，∠B=∠D=∠F并且∠A 、∠C 、∠E 的内角平分线交于一点。
证明:∠B,∠D,∠F 的内角平分线也交于一点。
()()[]∑∑===++202013202012811i i i i i x y x题4.如果数组（1,2,……，m ）的一个置换满足:1. 不存在任何一个正整数k<m 使得这个置换中的前k 个数字刚 好是按某种顺序排列的1,2,…,k 。
那么我们称这个置換为“新鲜的”,设m f 是所有（1,2,……，m ）的新鲜的置换的种数。
证明:对任何n ≥3都有1-•≥n n f n f题5 在△ABC 中,∠BCA>90°,△ABC 的外接圆的半径为R.在线段AB 中有一点P ，满足PB=PC,具PA=R.记D 、E 为PB 的垂直分线与△ABC 的外接圆的两个交点。
证明:P 是△CDE 的内心。
题6 设m>1是一个整数,数列a 1,a 2，a 3……定义为a 1=a 2=1，a 3=4且对任意n ≥4都有a n =m(a n-1+a n-2)-a n-3求所有使得这个数列中每个项都是平方数的m 。
国际数学奥林匹克竞赛试题及解答
国际数学奥林匹克竞赛试题及解答第一题：在一个正方形的边上选择10个点，然后连接相邻点之间得到一个多边形。
问这个多边形内部最多能够放置多少个相互不相交的小正方形？解答：这个问题可以通过找规律进行解答。
我们可以先考虑较小的正方形个数，再逐渐递增。
当只有1个小正方形时，我们可以把它放在正方形中心。
当有2个小正方形时，我们可以把它们放在相邻的两个顶点上。
当有3个小正方形时，我们可以放置两个在相邻的两个顶点上，另一个放在中心位置。
当有4个小正方形时，我们可以把它们分别放在四个顶点上。
当有5个小正方形时，我们可以把其中4个放在四个顶点上，然后将剩下的一个放在中心位置。
当有6个小正方形时，我们可以把其中4个放在四个顶点上，另外两个放在中点和中心位置。
...通过逐个增加小正方形的个数，我们可以得出规律：在一个正方形上最多可以放置 n(n+1)/2 个相互不相交的小正方形，其中 n 为偶数。
第二题：求方程组|y - x^2| = 3|y - x - 4| = 5的解。
解答：首先，对于第一个方程 |y - x^2| = 3，我们可以将其分为两部分进行讨论：1. y - x^2 = 3，解得 y = x^2 + 3；2. -(y - x^2) = 3，解得 y = -x^2 - 3。
然后，将得到的两个解代入第二个方程 |y - x - 4| = 5，得到：1. |(x^2 + 3) - x - 4| = 5，即 |x^2 - x - 1| = 5；2. |(-x^2 - 3) - x - 4| = 5，即 |-x^2 - x - 7| = 5。
我们分别解这两个方程：1. x^2 - x - 1 = 5，解得 x = -2 或 x = 3。
2. -x^2 - x - 7 = 5，解得 x = -3 或 x = 2。
将上述解代入方程 y = x^2 + 3 或 y = -x^2 - 3，则可求出相应的 y 值。
因此，该方程组的解为 (-2, 7)，(3, 12)，(-3, -6)，(2, -1)。
2020年摩尔多瓦IMO代表队选拔考试数学考试试题(word版,无答案)
2020 摩尔多瓦 IMO 代表队选拔考试第一天1. 有无穷项的等比数列{},1n b n ≥的每一项都是某个等差数列的项. 已知1b 为整数. 证明{}n b 的每一项都是整数.2. 证明对所有正实数 a,b,c, 均有不等式22233a b c b c a a b c +++≥+++成立.3. 设n ≥3为正整数. 集合A ＝{1,2,…n}, 我们将A 中所有元素随机排成数列{}12,,,n a a a ⋅⋅⋅. 若整数i, j 满足1,i j i j n a a ≤<≤>,就称 i,j 是一组“反转对”. 那么, 集合 A 中的元素有多少种排列方法, 使得其中恰有 3 对“反转对”?4. 锐角△ABC 中, H 为垂心, B 1和C 1分别为B 和C 到对边的垂足, M 为AH 中点. 在线段B 1C 1上取不在AH 上的一点K, 直线 AK 与 MB 1 和 MC 1 分别交于 E 和 F, 直线 BE 和 CF 交于点N. 求证 K 为△NBC 垂心.5. 设 n 为正整数. 在区间 [0, 4π)上,求方程组sin cos 22tan 23n x x x n ⎧+=⎪⎪⎨-⎪=⎪⎩的所有解.6. 设 n 为不小于 3 的自然数. 多项式()()()()20121121n n f x x x nx a a x a x a x =++⋅⋅⋅+=+++⋅⋅⋅+证明:3a 整除()22211223.n n n k C C C a ++=-7. 证明对所有正实数 a,b,c, 均有不等式2222222221777a b c a b c a b c ++≤++++++成立.8. △ABC 中, ∠ABC 和∠ACB 为锐角, AB 中点为 M. D 为 CB 延长线上一点, 且∠DAB ＝∠BCM, 过 B 作 CD 垂线与 AB 的垂直平分线交于点 E. 证明 DE ⊥AC.9. △ABC 内接于以 AP 为直径的圆 Ω. E, F 分别为 B 到 AC 和AP 的垂足. EF 中点为 M, CP 中点为 N.求证: BM ⊥MN.10. 设 n 为给定正整数. 正实数 a,b,c 满足1111a b c++=.求 ()212221222122,,n n nn n n n n n n n na b c E a b c a b c b c b c a c a c a b a b +++=+++⋅+⋅+⋅+⋅+⋅+⋅的最大值.11. 求所有函数[]:1,1f R -→,使得 f(sinx )+f (cosx ) =2020对所有实数 x 恒成立.12. 在一个象棋单循环赛中, 每个选手与其他选手各比赛一局. 已知所有的选手得分都不相同. 所有选手中的最后一名得了k 分. 那么, 第一名至少得了多少分? (对每一局棋, 胜者得 1 分, 败者得 0 分, 平局则双方各得 0.5 分)。
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2020年北欧国家数学竞赛
1.对任息正整数，我们记g(x)为所有合{1,2,3,…,n}递增的三元组的个数。
求最小的正整数n,使得g(x)可以写成三个不同素数之积，且这三个素数都在一个公差为336的等差数列中(并不一定是连续的三项)。
2.卡特琳手上有2n+1张卡片,其中一张上写了0,剩下的卡片中写 有1,2，…n 的个两张。
卡特琳想将这些卡片排成一排。
使得写了0的卡片正好在这一排卡片的中间,且对任意k=1,2，…n 。
写有k 的两张卡片之间的距离好为k(即这两张卡片之间一共有k-1张)
若1≤n ≤10,当n 为哪些数时，卡特琳可以排出这样的排列方式？
3.将凸四边形ABCD 的边AB 和CD 分别三等分：E ，F 分别为AB 三等分点，且AE=EF=FB,E ，F 分别为CD 三等分点，且DP=PQ=QC.点M 为四边形AEPD 的两条对角线的交点。
点N 为四边形FBCQ 的两条对角线的交点。
证明△AMD 和△BNC 的面积之和等于△EPM 和 △FNQ 的面积之和。
4.求所有函数:R R f →:.使得对任意满足(x+1)(y+1)(xy+1)≠0的数x ，y ∈R ，都有：
()⎪⎪⎭⎫ ⎝⎛++=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛+-111xy y x f y y f f x f 成立。