2010PascalSolution滑铁卢竞赛题答案

2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1Mary’s top book shelf holds five books with the follow ing widths, incentimeters: , , , , and . What is the average book width, in centimeters?SolutionTo find the average, we add up the widths , , , , and , to get a total sum of . Since there are books, the average book width isThe answer is .Problem 2Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is .Thus, the length of the rectangle is times large as the width. The answer is .Problem 3Tyrone had marbles and Eric had marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?SolutionLet be the number of marbles Tyrone gave to Eric. Then,. Solving for yields and . The answer is .Problem 4A book that is to be recorded onto compact discs takes minutes to read aloud. Each disc can hold up to minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?SolutionAssuming that there were fractions of compact discs, it would take CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have minutes on each of the 8 discs. The answer is .Problem 5The area of a circle whose circumference is is . What is the value of ?SolutionIf the circumference of a circle is , the radius would be . Since the area of a circle is , the area is . The answer is . Problem 6For positive numbers and the operation is defined asWhat is ?Solution. Then, is The answer isProblem 7Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to , which is equal to . The answer isTony works hours a day and is paid $per hour for each full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period?SolutionTony worked hours a day and is paid dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars, we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he was and . . Then isand we know that he was for days, and for days. Thus, the answer is .Problem 9A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?Solutionis at most , so is at most . The minimum value ofis . However, the only palindrome between and is , which means that must be .It follows that is , so the sum of the digits is .Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on a Saturday?Solution(E) 2017There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.For example: 5/27/08 Tue 5/27/09 WedHowever, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.For example: 5/27/11 Fri 5/27/12 SunYou can keep count forward to find that the first time this date falls on a Saturday is in 2017:5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11The length of the interval of solutions of the inequality is . What is ?SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .Subtract from all of the quantities:Divide all of the quantities by .Since we have the range of the solutions, we can make them equal to .Multiply both sides by 2.Re-write without using parentheses.Simplify.We need to find for the problem, so the answer isProblem 12Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionThe water tower holds times more water than Logan's miniature. Therefore, Logan should make his towertimes shorter than the actual tower. This ismeters high, or choice .Problem 13Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which equation could be used to solve for the time in hours that she drove before her stop?SolutionThe answer is （）because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasn't driving for minutes, or hours. Multiplying by gives the total distance, which is kms. Therefore, the answer isProblem 14Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?SolutionLet .Since ,Problem 15In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?SolutionSolution 1We can begin by first looking at Chris and LeRoy.Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.Clearly, Chris and LeRoy are different species, and so we have at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.Solution 2Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.As Mike is a frog, his statement is false, hence there is at most one toad.As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and three frogs.Problem 16Nondegenerate has integer side lengths, is an angle bisector, , and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If we use the next lowest values (and ), the Triangle Inequality is satisfied. Therefore, our answer is , or choice .Problem 17A solid cube has side length inches. A -inch by -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?SolutionSolution 1Imagine making the cuts one at a time. The first cut removes a box . The second cut removes two boxes, each of dimensions, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is .Therefore the volume of the rest of the cube is.Solution 2We can use Principle of Inclusion-Exclusion to find the final volume of the cube.There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has cubic inches. However, we can not just sum their volumes, as the central cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.Hence the total volume of the cuts is.Therefore the volume of the rest of the cube is.Solution 3We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.Each edge can be seen as a box, and each corner can be seen as a box..Problem 18Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?SolutionWe can solve this by breaking the problem down into cases and adding up the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .Case : Bernardo does not pick . Since the chance of Bernardo picking is , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.We get this probability to beProbability of Bernardo's number being greater isFactoring the fact that Bernardo could've picked a but didn't:Adding up the two cases we getProblem 19Equiangular hexagon has side lengthsand . The area of is of the area of the hexagon. What is the sum of all possible values of ?SolutionSolution 1It is clear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of is .If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .Solution 2As above, we find that the area of is .We also find by the sine triangle area formula that, and thusThis simplifies to.Problem 20A fly trapped inside a cubical box with side length meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?SolutionThe distance of an interior diagonal in this cube is and the distance of a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is.Problem 21The polynomial has three positive integer zeros. What is the smallest possible value of ?SolutionBy Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, which means will be and the answer is .Problem 22Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point insidethe circle. How many triangles with all three vertices in the interior of the circle are created?SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28,Problem 23Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let be the probability that Isabella stops afterdrawing exactly marbles. What is the smallest value of for which ?SolutionThe probability of drawing a white marble from box is . Theprobability of drawing a red marble from box is .The probability of drawing a red marble at box is thereforeIt is then easy to see that the lowest integer value of that satisfies the inequality is .Problem 24The number obtained from the last two nonzero digits of is equal to . What is ?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is equal to. Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .If we divide by by taking out all the factors of in , we canwrite as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .The number can be grouped as follows:Using the identity at the beginning of the solution, we can reducetoUsing the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore.Finally, combining with the fact that yields.Problem 25Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with , then his sequence contains numbers:Let be the smallest number for which Jim’s sequence has numbers. What is the units digit of ?SolutionWe can find the answer by working backwards. We begin with on the bottom row, then the goes to the right of the equal's sign in the row above. We find the smallest value for whichand , which is .We repeat the same procedure except with for the next row and for the row after that. However, at the fourth row, wesee that solving yields , in which case it would be incorrect since is not the greatest perfect square less than or equal to . So we make it a and solve . We continue on using this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows, we get:Hence the solution is the last digit of , which is .。

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.。

39th United States of America Mathematical Olympiad 20101.Solution by Titu Andreescu:Let T be the foot of the perpendicular from Y to line AB .We note the P,Q,T are the feet of the perpendiculars from Y to the sides of triangle ABX .Because Y lies on the circumcircle of triangle ABX ,points P,Q,T are collinear,by Simson’s theorem.Likewise,points S,R,T arecollinear.We need to show that ∠XOZ =2∠P T S or∠P T S =∠XOZ 2= XZ 2= XY 2+Y Z 2=∠XAY +∠ZBY =∠P AY +∠SBY.Because ∠P T S =∠P T Y +∠ST Y ,it suffices to prove that∠P T Y =∠P AY and ∠ST Y =∠SBY ;that is,to show that quadrilaterals AP Y T and BSY T are cyclic,which is evident,because ∠AP Y =∠AT Y =90◦and ∠BT Y =∠BSY =90◦.Alternate Solution from Lenny Ng and Richard Stong:Since Y Q,Y R are per-pendicular to BX,AZ respectively,∠RY Q is equal to the acute angle between lines BX and AZ ,which is 12( AX + BZ )=12(180◦− XZ )since X,Z lie on the circle with diameter AB .Also,∠AXB =∠AZB =90◦and so P XQY and SZRY are rectangles,whence ∠P QY =90◦−∠Y XB =90◦− Y B /2and ∠Y RS =90◦−∠AZY =90◦− AY /2.Finally,the angle between P Q and RS is∠P QY +∠Y RS −∠RY Q =(90◦− Y B /2)+(90◦− AY /2)−(90◦− XZ /2)= XZ /2=(∠XOZ )/2,as desired.This problem was proposed by Titu Andreescu.2.Solution from Kiran Kedlaya:Let h i also denote the student with height h i .We prove that for 1≤i <j ≤n ,h j can switch with h i at most j −i −1times.We proceed by induction on j −i ,the base case j −i =1being evident because h i is not allowed to switch with h i −1.For the inductive step,note that h i ,h j −1,h j can be positioned on the circle either in this order or in the order h i ,h j ,h j −1.Since h j −1and h j cannot switch,the only way to change the relative order of these three students is for h i to switch with either h j −1or h j .Consequently,any two switches of h i with h j must be separated by a switch of h i with h j −1.Since there are at most j −i −2of the latter,there are at most j −i −1of the former.The total number of switches is thus at mostn −1 i =1n j =i +1(j −i −1)=n −1 i =1n −i −1 j =0j =n −1 i =1 n −i 2=n −1 i =1 n −i +13− n −i 3 = n 3 .Note:One can also ask to prove that the number of switches before no more are possible depends only on the original ordering,or to find all initial positions for which n 3switches are possible (the only one is when the students are sorted in increasing order).Alternative Solution from Warut Suksompong:For i =1,2,...,n −1,let s i be the number of students with height no more than h i +1standing (possibly not directly)behind the student with height h i and (possibly not directly)in front of the one with height h i +1.Note that s i ≤i −1for all i .Now we take a look what happens when two students switch places.•If the student with height h n is involved in the switch,s n −1decreases by 1,while all the other s i ’s remain the same.•Otherwise,suppose the students with heights h a and h b are switched,with a +1<b <n ,then s b −1decreases by 1,while s b increases by 1.All the other s i ’s remain the same.Since s i ≤i −1for all i =1,2,...,n −1,the maximal number of switches is no more than the number of switches in the case where initially s i =i −1for all i .In that case,the number of switches is n −2i =1i (n −1−i )= n 3.Note:With this solution,it is also easy to see that the number of switches until no more are possible depends only on the original ordering.This problem was proposed by Kiran Kedlaya jointly with Travis Schedler and David Speyer.3.Solution from Gabriel Carroll:Multiplying together the inequalities a 2i −1a 2i ≤4i −1for i =1,2,...,1005,we geta 1a 2···a 2010≤3·7·11···4019.(1)The tricky part is to show that this bound can be attained.Leta 2008=4017·40184019,a 2009= 4019·40174018,a 2010= 4018·40194017,and define a i for i <2008by downward induction using the recursiona i =(2i +1)/a i +1.We then havea i a j =i +j whenever j =i +1or i =2008,j =2010.(2)We will show that (2)implies a i a j ≤i +j for all i <j ,so that this sequence satisfies the hypotheses of the problem.Since a 2i −1a 2i =4i −1for i =1,...,1005,the inequality (1)is an equality,so the bound is attained.We show that a i a j ≤i +j for i <j by downward induction on i +j .There are several cases:•If j =i +1,or i =2008,j =2010,then a i a j =i +j ,from (2).•If i=2007,j=2009,thena i a i+2=(a i a i+1)(a i+2a i+3)(a i+1a i+3)=(2i+1)(2i+5)2i+4<2i+2.Here the second equality comes from(2),and the inequality is checked by multiplying out:(2i+1)(2i+5)=4i2+12i+5<4i2+12i+8=(2i+2)(2i+4).•If i<2007and j=i+2,then we havea i a i+2=(a i a i+1)(a i+2a i+3)(a i+2a i+4)(a i+1a i+2)(a i+3a i+4)≤(2i+1)(2i+5)(2i+6)(2i+3)(2i+7)<2i+2.Thefirst inequality holds by applying the induction hypothesis for(i+2,i+4),and (2)for the other pairs.The second inequality can again be checked by multiplying out:(2i+1)(2i+5)(2i+6)=8i3+48i2+82i+30<8i3+48i2+82i+42= (2i+2)(2i+3)(2i+7).•If j−i>2,thena i a j=(a i a i+1)(a i+2a j)a i+1a i+2≤(2i+1)(i+2+j)2i+3<i+j.Here we have used the induction hypothesis for(i+2,j),and again we check the last inequality by multiplying out:(2i+1)(i+2+j)=2i2+5i+2+2ij+j< 2i2+3i+2ij+3j=(2i+3)(i+j).This covers all the cases and shows that a i a j≤i+j for all i<j,as required.Variant Solution by Paul Zeitz:It is possible to come up with a semi-alternative solution,after constructing the sequence,by observing that when the two indices differ by an even number,you can divide out precisely.For example,if you wanted to look at a3a8, you would use the fact that a3a4a5a6a7a8=(7)(11)(15)and a4a5a6a7=(9)(13).Hence we need to check that(7)(11)(15)/((9)(13))<11,which is easy AMGM/Symmetry.However,this attractive method requires much more subtlety when the indices differ by an odd number.It can be pulled off,but now you need,as far as I know,either to use the precise value of a2010or establish inequalities for(a k)2for all values of k.It is ugly,but it may be attempted.This problem was suggested by Gabriel Carroll.4.Solution from Zuming Feng:The answer is no,it is not possible for segments AB,BC,BI,ID,CI,IE to all have integer lengths.Assume on the contrary that these segments do have integer side lengths.We set α=∠ABD =∠DBC and β=∠ACE =∠ECB .Note that I is the incenter of triangle ABC ,and so ∠BAI =∠CAI =45◦.Applying the Law of Sines to triangle ABI yields AB BI =sin(45◦+α)sin 45◦=sin α+cos α,by the addition formula (for the sine function).In particular,we conclude that s =sin α+cos αis rational.It is clear that α+β=45◦.By the subtraction formulas,we haves =sin(45◦−β)+cos(45◦−β)=√2cos β,from which it follows that cos βis not rational.On the other hand,from right triangle ACE ,we have cos β=AC/EC ,which is rational by assumption.Because cos βcannot not be both rational and irrational,our assumption was wrong and not all the segments AB ,BC ,BI ,ID ,CI ,IE can have integer lengths.Alternate Solution from Jacek Fabrykowski:Using notations as introduced in the problem,let BD =m ,AD =x ,DC =y ,AB =c ,BC =a and AC =b .The angle bisector theorem implies x b −x =c aand the Pythagorean Theorem yields m 2=x 2+c 2.Both equations imply that2ac =(bc )2m 2−c 2−a 2−c 2and since a 2=b 2+c 2is rational,a is rational too (observe that to reach this conclusion,we only need to assume that b ,c ,and m are integers).Therefore,x =bca +c is also rational,and so is y .Let now (similarly to the notations above from the solution by Zuming Feng)∠ABD =αand ∠ACE =βwhere α+β=π/4.It is obvious that cos αand cos βare both rational and the above shows that also sin α=x/m is rational.On the other hand,cos β=cos(π/4−α)=(√2/2)(sin α+sin β),which is a contradiction.The solution shows that a stronger statement holds true:There is no right triangle with both legs and bisectors of acute angles all having integer lengths.Alternate Solution from Zuming Feng:Prove an even stronger result:there is no such right triangle with AB,AC,IB,IC having rational side lengths.Assume on the contrary,that AB,AC,IB,IC have rational side lengths.Then BC 2=AB 2+AC 2is rational.On the other hand,in triangle BIC ,∠BIC =135◦.Applying the law of cosines to triangle BIC yieldsBC 2=BI 2+CI 2−√·CIwhich is irrational.Because BC2cannot be both rational and irrational,we conclude that our assumption was wrong and that not all of the segments AB,AC,IB,IC can have rational lengths.This problem was proposed by Zuming Feng.5.Solution by Titu Andreescu:We have2k(k+1)(k+2)=(k+2)−kk(k+1)(k+2)=1k(k+1)−1(k+1)(k+2) =1k−1k+1−1k+1−1k+2=1k+1k+1+1k+2−3k+1.Hence2S q=12+13+14+...+1q+1q+1+1q+2−313+16+...+1q+1=12+13+...+13p−12−1+12+...+1p−12,and so1−mn=1+2S q−1p=1p+12+...+1p−1+1p+1+...+13p−12 =1p+12+13p−12+...+1p−1+1p+1=pp+123p−12+...+p(p−1)(p+1).Because all denominators are relatively prime with p,it follows that n−m is divisible by p and we are done.This problem was suggested by Titu Andreescu.6.Solution by Zuming Feng and Paul Zeitz:The answer is43.Wefirst show that we can always get43points.Without loss of generality,we assume that the value of x is positive for every pair of the form(x,x)(otherwise,replace every occurrence of x on the blackboard by−x,and every occurrence of−x by x).Consider the ordered n-tuple(a1,a2,...,a n)where a1,a2,...,a n denote all the distinct absolute values of the integers written on the board.Letφ=√5−12,which is the positive root ofφ2+φ=1.We consider2n possible underliningstrategies:Every strategy corresponds to an ordered n-tuple s=(s1,...,s n)with s i=φor s i =1−φ(1≤i ≤n ).If s i =φ,then we underline all occurrences of a i on the blackboard.If s i =1−φ,then we underline all occurrences of −a i on the blackboard.The weight w (s )of strategy s equals the product ni =1s i .It is easy to see that the sum of weights of all 2n strategies is equal to s w (s )= ni =1[φ+(1−φ)]=1.For every pair p on the blackboard and every strategy s ,we define a corresponding cost coefficient c (p,s ):If s scores a point on p ,then c (p,s )equals the weight w (s ).If s does not score on p ,then c (p,s )equals 0.Let c (p )denote the the sum of of coefficients c (p,s )taken over all s .Now consider a fixed pair p =(x,y ):(a)In this case,we assume that x =y =a j .Then every strategy that underlines a j scores a point on this pair.Then c (p )=φ ni =j [φ+(1−φ)]=φ.(b)In this case,we assume that x =y .We have c (p )= φ2+φ(1−φ)+(1−φ)φ=3φ−1,(x,y )=(a k ,a );φ(1−φ)+(1−φ)φ+(1−φ)2=φ,(x,y )=(−a k ,−a );φ2+φ(1−φ)+(1−φ)2=2−2φ,(x,y )=(±a k ,∓a ).By noting that φ≈0.618,we can easily conclude that c (p )≥φ.We let C denote the sum of the coefficients c (p,s )taken over all p and s .These observations yield thatC =p,s c (p,s )= p c (p )≥ p φ=68φ>42.Suppose for the sake of contradiction that every strategy s scores at most 42points.Then every s contributes at most 42w (s )to C ,and we get C ≤42 s w (s )=42,which contradicts C >42.To complete our proof,we now show that we cannot always get 44points.Consider the blackboard contains the following 68pairs:For each of m =1,...,8,five pairs of (m,m )(for a total of 40pairs of type (a));For every 1≤m <n ≤8,one pair of (−m,−n )(for a total of 82=28pairs of type (b)).We claim that we cannot get 44points from this initial stage.Indeed,assume that exactly k of the integers 1,2,...,8are underlined.Then we get at most 5k points on the pairs of type (a),and at most 28− k 2 points on the pairs of type (b).We can get at most 5k +28− k 2 points.Note that the quadratic function 5k +28− k 2 =−k 22+11k 2+28obtains its maximum 43(for integers k )at k =5or k =6.Thus,we can get at most 43points with this initial distribution,establishing our claim and completing our solution.This problem was suggested by Zuming Feng.Copyright c Mathematical Association of America。

2008年全国大学生英语竞赛样题(C级)Part IListening Comprehension (25 minutes，30 marks)Section A（5 marks)Directions：In this section, you will hear 5 short conversations。

At the end of each conversation, a question will be asked about what was said。

Both the conversation and the question will be read only once. After each question, there will be a pause。

During the pause, you must read the three choices marked A, B and C, and decide which is the best answer。

Then mark the corresponding letter on the Answer Sheet with a single line through the centre。

1。

A. The man is not suitable for the position.B。

The job has been given to someone else.C. She hadn't received the man’s application。

2. A. He is going to see his section chief.B. He is going to have a job interview。

C。

He is going to see his girlfriend.3. A。

Ask to see the man’s ID card。

2010年全国高中数学联合竞赛加试 试题参考答案及评分标准（B 卷）说明：1. 评阅试卷时，请严格按照本评分标准的评分档次给分.2. 如果考生的解答方法和本解答不同，只要思路合理、步骤正确，在评卷时可参考本评分标准适当划分档次评分，10分为一个档次，不要增加其他中间档次。

一、（本题满分40分）如图，锐角三角形ABC 的外心为O ，K 是边BC 上一点（不是边BC 的中点），D 是线段AK 延长线上一点，直线BD 与AC 交于点N ，直线CD 与AB 交于点M ．求证：若OK ⊥MN ，则A ，B ，D ，C 四点共圆．证明：用反证法．若A ，B ，D ，C 不四点共圆，设三角形ABC 的外接圆与AD 交于点E ，连接BE 并延长交直线AN 于点Q ，连接CE 并延长交直线AM 于点P ，连接PQ ．因为2PK =P 的幂（关于⊙O ）+K 的幂（关于⊙O ） ()()2222PO rKOr=-+-，同理 ()()22222QK QO r KO r =-+-，所以 2222P O P K Q O Q K -=-，故 OK ⊥PQ ． （10分）由题设，OK ⊥MN ，所以PQ ∥MN ，于是AQ APQN PM=． ① 由梅内劳斯（Menelaus ）定理，得1NB DE AQBD EA QN⋅⋅=， ② 1MC DE APCD EA PM⋅⋅=． ③ 由①，②，③可得NB MCBD CD=， （30分） 所以ND MDBD DC=，故△DMN ∽ △DCB ，于是DMN DCB ∠=∠，所以BC ∥MN ，故OK ⊥BC ，即K 为BC 的中点，矛盾！从而,,,A B D C 四点共圆. （40分）M注1：“2PK =P 的幂（关于⊙O ）+K 的幂（关于⊙O ）”的证明：延长PK 至点F ，使得PK KF AK KE ⋅=⋅， ④则P ，E ，F ，A 四点共圆，故PFE PAE BCE ∠=∠=∠，从而E ，C ，F ，K 四点共圆，于是PK PF PE PC ⋅=⋅， ⑤⑤-④，得 2PK PE PC AK KE =⋅-⋅=P 的幂（关于⊙O ）+K 的幂（关于⊙O ）．注2：若点E 在线段AD 的延长线上，完全类似．二、（本题满分40分）设m 和n 是大于1的整数，求证：11111112(1)().1m m n mmmk k jj m m k j i n n C n C i m -+===⎧⎫+++=+-⎨⎬+⎩⎭∑∑∑ 证明：11101)m m j jm j q C q +++=+=∑由（得到 1110(1),mm m j jm j q qC q +++=+-=∑ 1,2,,q n =分别将代入上式得：11021,mm jm j C ++=-=∑111322,mm m j jm j C +++=-=∑F E QP O NM K DCBA1110(1)(1),mm m j jm j n n C n +++=--=-∑ 1110(1).m m m j j m j n nC n +++=+-=∑ n 将上面个等式两边分别相加得到：1101(1)1(),mnm jjm j i n Ci++==+-=∑∑ （20分）11111(1)(1)1(1),m nnmj j mm j i i n n n Ci m i -+===++-=+++∑∑∑（）11111112(1)().1m m nm mmk k j j m m k j i n n C n C i m -+===⎧⎫+++=+-⎨⎬+⎩⎭∑∑∑ （40分） 三、（本题满分50分）设,,x y z 为非负实数, 求证：22232222223()()()()()32xy yz zx x y z x xy y y yz z z zx x ++++≤-+-+-+≤.证明：首先证明左边不等式.因为 2222211[()3()]()44x xy y x y x y x y -+=++-≥+, 同理,有2221()4y yz z y z -+≥+, 2221()4z zx x z x -+≥+; （10分） 于是22222221()()()[()()()]64x xy y y yz z z zx xx y y z z x -+-+-+≥+++21[()()]64x y z xy yz zx xyz =++++-; （20分） 由算术-几何平均不等式, 得 1()()9xyz x y z xy yz zx ≤++++,所以222222221()()()()()81x xy y y yz z z zx x x y z xy yz zx -+-+-+≥++++ 22221(222)()81x y z xy yz zx xy yz zx =+++++++3()3xy yz zx ++≥. 左边不等式获证, 其中等号当且仅当x y z ==时成立. （30分）下面证明右边不等式.根据欲证不等式关于,,x y z 对称, 不妨设x y z ≥≥, 于是 22222()()z z x x y y z z xy -+-+≤, 所以222222222()()()()x x y y y y z z z z x x xx y y x y-+-+-+≤-+. （40分）运用算术-几何平均不等式, 得222222222()()()2x xy y xy x xy y x y x xy y xy xy xy -++-+=-+⋅⋅≤⋅ 22222()()22x xy y xy x y -+++≤⋅2222233()()22x y x y z +++=≤. 右边不等式获证, 其中等号当且仅当,,x y z 中有一个为0,且另外两个相等时成立. （50分）四、（本题满分50分）设k 是给定的正整数，12r k =+．记(1)()()f r f r r r ==⎡⎤⎢⎥，()()l f r = (1)(()),2l f f r l -≥．证明：存在正整数m ，使得()()m f r 为一个整数．这里，x ⎡⎤⎢⎥表示不小于实数x的最小整数，例如：112⎡⎤=⎢⎥⎢⎥，11=⎡⎤⎢⎥． 证明：记2()v n 表示正整数n 所含的2的幂次．则当2()1m v k =+时，()()m f r 为整数． 下面我们对2()v k v =用数学归纳法．当0v =时，k 为奇数，1k +为偶数，此时()111()1222f r k k k k ⎛⎫⎡⎤⎛⎫=++=++ ⎪ ⎪⎢⎥⎝⎭⎢⎥⎝⎭为整数． （10分)假设命题对1(1)v v -≥成立．对于1v ≥，设k 的二进制表示具有形式1212222v v v v v k αα++++=+⋅+⋅+，这里，0i α=或者1，1,2,i v v =++． （20分)于是 ()111()1222f r k k k k ⎛⎫⎡⎤⎛⎫=++=++ ⎪ ⎪⎢⎥⎝⎭⎢⎥⎝⎭2122kk k =+++ 11211212(1)2()222v v v vv v v ααα-++++=+++⋅++⋅+++12k '=+， ① （40分）这里1121122(1)2()22v v v v v v v k ααα-++++'=++⋅++⋅+++．显然k '中所含的2的幂次为1v -．故由归纳假设知，12r k ''=+经过f 的v 次迭代得到整数，由①知，(1)()v f r +是一个整数，这就完成了归纳证明． （50分）出师表两汉：诸葛亮先帝创业未半而中道崩殂，今天下三分，益州疲弊，此诚危急存亡之秋也。

1.For real numbers a and b with a≥0and b≥0,the operation is defined bya b=√a+4b.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 justification,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 beenfilled 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 exactlyfive 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 thefirst 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 thefirst 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,find 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.。

1.Calculating,2×9−√36+1=18−6+1=13.Answer:(D)2.On Saturday,Deepit worked6hours.On Sunday,he worked4hours.Therefore,he worked6+4=10hours in total on Saturday and Sunday.Answer:(E) 3.Since1piece of gum costs1cent,then1000pieces of gum cost1000cents.Since there are100cents in a dollar,the total cost is$10.00.Answer:(D) 4.Since each of the18classes has28students,then there are18×28=504students who attendthe school.On Monday,there were496students present,so504−496=8students were absent.Answer:(A) 5.The sum of the angles around any point is360◦.Therefore,5x◦+4x◦+x◦+2x◦=360◦or12x=360or x=30.Answer:(D) 6.When−1is raised to an even exponent,the result is1.When−1is raised to an odd exponent,the result is−1.Thus,(−1)5−(−1)4=−1−1=−2.Answer:(A) 7.Since P Q is horizontal and the y-coordinate of P is1,then the y-coordinate of Q is1.Since QR is vertical and the x-coordinate of R is5,then the x-coordinate of Q is5.Therefore,the coordinates of Q are(5,1).Answer:(C)8.When y=3,we have y3+yy2−y=33+332−3=27+39−3=306=5.Answer:(D)9.Since there are4♣’s in each of thefirst two columns,then at least1♣must be moved out ofeach of these columns to make sure that each column contains exactly three♣’s.Therefore,we need to move at least2♣’s in total.If we move the♣from the top left corner to the bottom right corner♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣and the♣from the second row,second column to the third row,fifth column♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣then we have exactly three♣’s in each row and each column.Therefore,since we must move at least2♣’s and we can achieve the configuration that we want by moving2♣’s,then2is the smallest number.(There are also other combinations of moves that will give the required result.)Answer:(B) 10.Solution1Since z=4and x+y=7,then x+y+z=(x+y)+z=7+4=11.Solution2Since z=4and x+z=8,then x+4=8or x=4.Since x=4and x+y=7,then4+y=7or y=3.Therefore,x+y+z=4+3+4=11.Answer:(C) 11.We write out thefive numbers to5decimal places each,without doing any rounding:5.076=5.07666...5.076=5.07676...5.07=5.070005.076=5.076005.076=5.07607...We can use these representations to order the numbers as5.07000,5.07600,5.07607...,5.07666...,5.07676...so the number in the middle is5.076.Answer:(E) 12.Solution1Since there are24hours in a day,Francis spends13×24=8hours sleeping.Also,he spends14×24=6hours studying,and18×24=3hours eating.The number of hours that he has left is24−8−6−3=7hours. Solution2Francis spends13+14+18=8+6+324=1724of a day either sleeping,studying or eating.This leaves him1−1724=724of his day.Since there are24hours in a full day,then he has7hours left.Answer:(D) 13.Solution1Since the sum of the angles in a triangle is180◦,then∠QP S=180◦−∠P QS−∠P SQ=180◦−48◦−38◦=94◦Therefore,∠RP S=∠QP S−∠QP R=94◦−67◦=27◦.Solution2Since the sum of the angles in a triangle is180◦,then∠QRP=180◦−∠P QR−∠QP R=180◦−48◦−67◦=65◦Therefore,∠P RS=180◦−∠P RQ=180◦−65◦=115◦.Using P RS,∠RP S=180◦−∠P RS−∠P SR=180◦−115◦−38◦=27◦Answer:(A)14.The perimeter of the shaded region equals the sum of the lengths of OP and OQ plus the lengthof arc P Q.Each of OP and OQ has length5.Arc P Q forms34of the circle with centre O and radius5,because the missing portion corre-sponds to a central angle of90◦,and so is14of the total circle.Thus,the length of arc P Q is34of the circumference of this circle,or34(2π(5))=152π.Therefore,the perimeter is5+5+152π≈33.56which,of the given answers,is closest to34.Answer:(A)15.After some trial and error,we obtain the two lists{4,5,7,8,9}and{3,6,7,8,9}.Why are these the only two?If the largest number of thefive integers was8,then the largest that the sum could be would be8+7+6+5+4=30,which is too small.This tells us that we must include one9in the list.(We cannot include any number larger than9,since each number must be a single-digit number.)Therefore,the sum of the remaining four numbers is33−9=24.If the largest of the four remaining numbers is7,then their largest possible sum would be 7+6+5+4=22,which is too small.Therefore,we also need to include an8in the list.Thus,the sum of the remaining three numbers is24−8=16.If the largest of the three remaining numbers is6,then their largest possible sum would be 6+5+4=15,which is too small.Therefore,we also need to include an7in the list.Thus,the sum of the remaining two numbers is16−7=9.This tells us that we need two different positive integers,each less than7,that add to9.These must be3and6or4and5.This gives us the two lists above,and shows that they are the only two such lists.Answer:(B) 16.The area of the entire grid is4×9=36.The area of P QR is12(QR)(P Q)=12(3)(4)=6.The area of ST U is12(ST)(UT)=12(4)(3)=6.The area of the rectangle with base RS is2×4=8.Therefore,the total shaded area is6+6+8=20and so the unshaded area is36−20=16.The ratio of the shaded area to the unshaded area is20:16=5:4.Answer:(E) 17.We can suppose that each test is worth100marks.Since the average of herfive test marks is73%,then the total number of marks that she received is5×73=365.Once her teacher removes a mark,her new average is76%so the sum of the remaining four marks is4×76=304.Since365−304=61,then the mark removed was61%.Answer:(B) 18.Solution1From December31,1988to December31,2008,a total of20years have elapsed.A time period of20years is the same asfive4year periods.Thus,the population of Arloe has doubled 5times over this period to its total of 3456.Doubling 5times is equivalent to multiplying by 25=32.Therefore,the population of Arloe on December 31,1988was 345632=108.Solution 2The population doubles every 4years going forward,so is halved every 4years going backwards in time.The population on December 31,2008was 3456.The population on December 31,2004was 3456÷2=1728.The population on December 31,2000was 1728÷2=864.The population on December 31,1996was 864÷2=432.The population on December 31,1992was 432÷2=216.The population on December 31,1988was 216÷2=108.Answer:(D)19.Since Pat drives 60km at 80km/h,this takes him 60km 80km/h =34h.Since Pat has 2hours in total to complete the trip,then he has 2−34=54hours left to complete the remaining 150−60=90km.Therefore,he must travel at 90km 54h =360km/h =72km/h.Answer:(C)20.Since the three numbers in each straight line must have a product of 3240and must include 45,then the other two numbers in each line must have a product of 3240=72.The possible pairs of positive integers are 1and 72,2and 36,3and 24,4and 18,6and 12,and 8and 9.The sums of the numbers in these pairs are 73,38,27,22,18,and 17.To maximize the sum of the eight numbers,we want to choose the pairs with the largest possible sums,so we choose the first four pairs.Thus,the largest possible sum of the eight numbers is 73+38+27+22=160.Answer:(E)21.Since each of Alice and Bob rolls one 6-sided die,then there are 6×6=36possible combinationsof rolls.Each of these 36possibilities is equally likely.Alice wins when the two values rolled differ by 1.The possible combinations that differ by 1are (1,2),(2,3),(3,4),(4,5),(5,6),(2,1),(3,2),(4,3),(5,4),and (6,5).Therefore,there are 10combinations when Alice wins.Thus her probability of winning is 1036=518.Answer:(C)22.Diameters P Q and RS cross at the centre of the circle,which we call O .The area of the shaded region is the sum of the areas of P OS and ROQ plus the sum of the areas of sectors P OR and SOQ .Each of P OS and ROQ is right-angled and has its two perpendicular sides of length 4(the radius of the circle).Therefore,the area of each of these triangles is 12(4)(4)=8.Each of sector P OR and sector SOQ has area 14of the total area of the circle,as each hascentral angle 90◦(that is,∠P OR =∠SOQ =90◦)and 90◦is one-quarter of the total central angle.Therefore,each sector has area 14(π(42))=14(16π)=4π.Thus,the total shaded area is 2(8)+2(4π)=16+8π.Answer:(E)23.The maximum possible mass of a given coin is 7×(1+0.0214)=7×1.0214=7.1498g.The minimum possible mass of a given coin is 7×(1−0.0214)=7×0.9786=6.8502g.What are the possible numbers of coins that could make up 1000g?To find the largest number of coins,we want the coins to be as light as possible.If all of the coins were as light as possible,we would have 10006.8502≈145.98coins.Now,we cannot have a non-integer number of coins.This means that we must have at most 145coins.(If we had 146coins,the total mass would have to be at least 146×6.8502=1000.1292g,which is too heavy.)Practically,we can get 145coins to have a total mass of 1000g by taking 145coins at the minimum possible mass and making each slightly heavier.To find the smallest number of coins,we want the coins to be as heavy as possible.If all of the coins were as heavy as possible,we would have 10007.1498≈139.86coins.Again,we cannot have a non-integer number of coins.This means that we must have at least 140coins.(If we had 139coins,the total mass would be at most 139×7.1498=993.8222g,which is too light.)Therefore,the difference between the largest possible number and smallest possible number of coins is 145−140=5.Answer:(B)24.Divide the large cube of side length 40into 8smaller cubes of side length 20,by making threecuts of the large cube through its centre using planes parallel to the pairs of faces.Each of these small cubes has the centre of the large cube as its vertex.Each of these small cubes also just encloses one of the large spheres,in the sense that the sphere just touches each of the faces of the small cube.We call the sphere that fits in the central space the inner sphere.To make this sphere as large possible,its centre will be at the centre of the large cube.(If this was not the case,the centre be outside one of the small cubes,and so would be farther away from one of the large spheres than from another.)To find the radius of the inner sphere,we must find the shortest distance from the centre of the large cube (that is,the centre of the inner sphere)to one of the large spheres.(Think of starting the inner sphere as a point at the centre of the cube and inflating it until it just touches the large spheres.)Consider one of these small cubes and the sphere inside it.Join the centre of the small cube to one of its vertices.Since the small cube has side length 20,then this new segment has length √102+102+102or √300,since to get from the centre to a vertex,we must go over 10,down 10and across 10.(See below for an explanation of why this distance is thus √300.)The inner sphere will touch the large sphere along this segment.Thus,the radius of the inner sphere will be this distance (√300)minus the radius of the large sphere (10),and so is √300−10≈7.32.Of the given answers,this is closest to 7.3.(We need to justify why the distance from the centre of the small cube to its vertex is √102+102+102.Divide the small cube into 8tiny cubes of side length 10each.The distance from the centre of the small cube to its vertex is equal to the length of a diagonal of one of the tiny cubes.Consider a rectangular prism with edge lengths a ,b and c .What is the length,d ,of the diag-onal inside the prism?By the Pythagorean Theorem,a face with side lengths a and b has a diagonal of length √a 2+b 2.Consider the triangle formed by this diagonal,the diagonal of the prism and one of the vertical edges of the prism,of length c .cThis triangle is right-angled since the vertical edge is perpendicular to the top face.By the Pythagorean Theorem again,d 2=(√a 2+b 2)2+c 2,so d 2=a 2+b 2+c 2or d =√a 2+b 2+c 2.)Answer:(B)25.The three machines operate in a way such that if the two numbers in the output have a commonfactor larger than 1,then the two numbers in the input would have to have a common factor larger than 1.To see this,let us look at each machine separately.We use the fact that if two numbers are each multiples of d ,then their sum and difference are also multiples of d .Suppose that (m,n )is input into Machine A.The output is (n,m ).If n and m have a common factor larger than 1,then m and n do as well.Suppose that (m,n )is input into Machine B.The output is (m +3n,n ).If m +3n and n have a common factor d ,then (m +3n )−n −n −n =m has a factor of d as each part of the subtraction is a multiple of d .Therefore,m and n have a common factor of d .Suppose that (m,n )is input into Machine C.The output is (m −2n,n ).If m −2n and n have a common factor d ,then (m −2n )+n +n =m has a factor of d as each part of the addition is a multiple of d .Therefore,m and n have a common factor of d .In each case,any common factor that exists in the output is present in the input.Let us look at the numbers in the five candidates.After some work,we can find the prime factorizations of the six integers:2009=7(287)=7(7)(41)1016=8(127)=2(2)(2)(127)1004=4(251)=2(2)(251)1002=2(501)=2(3)(167)1008=8(126)=8(3)(42)=16(3)(3)(7)=2(2)(2)(2)(3)(3)(7)1032=8(129)=8(3)(43)=2(2)(2)(3)(43)Therefore,the only one of 1002,1004,1008,1016,1032that has a common factor larger than 1with 2009is 1008,which has a common factor of 7with 2009.How does this help?Since2009and1008have a common factor of7,then whatever pair was input to produce(2009,1008)must have also had a common factor of7.Also,the pair that was input to create this pair also had a common factor of7.This can be traced back through every step to say that the initial pair that produces the eventual output of(2009,1008)must have a common factor of7.Thus,(2009,1008)cannot have come from(0,1).Notes:•This does not tell us that the other four pairs necessarily work.It does tell us,though, that(2009,1008)cannot work.•We can trace the other four outputs back to(0,1)with some effort.(This process is easier to do than it is to describe!)To do this,we notice that if the output of Machine A was(a,b),then its input was(b,a), since Machine A switches the two entries.Also,if the output of Machine B was(a,b),then its input was(a−3b,b),since MachineB adds three times the second number to thefirst.Lastly,if the output of Machine C was(a,b),then its input was(a+2b,b),since MachineC subtracts two times the second number from thefirst.Consider(2009,1016)for example.We try tofind a way from(2009,1016)back to(0,1).We only need tofind one way that works,rather than looking for a specific way.We note before doing this that starting with an input of(m,n)and then applying MachineB then MachineC gives an output of((m+3n)−2n,n)=(m+n,n).Thus,if applyingMachine B then Machine C(we call this combination“Machine BC”)gives an output of (a,b),then its input must have been(a−b,b).We can use this combined machine to try to work backwards and arrive at(0,1).This will simplify the process and help us avoid negative numbers.We do this by making a chart and by attempting to make the larger number smaller wher-ever possible:Output Machine Input(2009,1016)BC(993,1016)(993,1016)A(1016,993)(1016,993)BC(23,993)(23,993)A(993,23)(993,23)BC,43times(4,23)(4,23)A(23,4)(23,4)BC,5times(3,4)(3,4)A(4,3)(4,3)BC(1,3)(1,3)A(3,1)(3,1)B(0,1)Therefore,by going up through this table,we can see a way to get from an initial input of(0,1)to afinal output of(2009,1016).In a similar way,we can show that we can obtainfinal outputs of each of(2009,1004), (2009,1002),and(2009,1032).Answer:(D)。

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)。

PASCAL竞赛辅导习题二问题１【问题】甲乙丙丁戊五个人在运动会上分获百米、二百米、跳高、跳远和铅球冠军，有四个人猜测比赛结果：Ａ说：乙获铅球冠军，丁获跳高冠军。

Ｂ说：甲获百米冠军，戊获跳远冠军。

Ｃ说：丙获跳远冠军，丁获二百米冠军。

Ｄ说：乙获跳高冠军，戊获铅球冠军。

其中每个人都只说对一句，说错一句。

求五人各获哪项冠军。

【算法】用1,2,3,4,5分别代表百米、二百米、跳高、跳远和铅球５个项目，用a,b,c,d,e 分别代表五人。

如b=3 表示乙获跳高冠军。

用多重循环穷举出来。

【答案】甲乙丙丁戊１２４３５【参考程序】var a,b,c,d,e:byte;beginfor a:=1 to 5 dofor b:=1 to 5 dofor c:=1 to 5 dofor d:=1 to 5 do begine:=15-a-b-c-d;if (ord(b=5)+ord(d=3)=1) and(ord(a=1)+ord(e=4)=1) and(ord(c=4)+ord(d=2)=1) and(ord(b=3)+ord(e=5)=1) and (a*b*c*d*e=120) thenwriteln('a:',a,'b:',b,'c:',c,'d:',d,'e:',e);end;end.问题２【问题】５家工厂的产品在一次评比中分获１，２，３，４，５，在公布结果前，已知Ｅ厂产品肯定不是第二、三名，五厂代表猜测评比结果，Ａ厂的代表说：Ｅ厂一定能获得第一名。

Ｂ厂的代表说：我厂的产品可能获第二名。

Ｃ厂的代表说：Ａ厂的产品质量最次。

Ｄ厂的代表说：Ｃ厂的产品不是最好的。

Ｅ厂的代表说：Ｄ厂的产品会获得第一名。

公布结果后，证明只有产品获第一名和第二名的两个厂的代表猜对了。

求５个厂产品各获第几名。

【答案】ＡＢＣＤＥ５２１３４【参考程序】var a,b,c,d,e:byte;beginfor a:=1 to 5 dofor b:=1 to 5 dofor c:=1 to 5 dofor d:=1 to 5 do begine:=15-a-b-c-d;if (e<>2) and (e<>3) and (a*b*c*d*e=120) thenif(ord(e=1)+ord(b=2)+ord(a=5)+ord(c<>1)+ ord(d=1)=2) and (ord((e=1) and ((a=1) or (a=2)))+ord((b=2) and ((b=1) or (b=2)))+ord((a=5) and ((c=1) or (c=2)))+ord((c<>1) and ((d=1) or (d=2)))+ord((d=1) and ((e=1) or (e=2)))=2) thenwriteln('a:',a,' b:',b,' c:',c,' d:',d,' e:',e);end;end.问题3逻辑判断\v谁是小偷\a问题３\a\v谁获冠军？\a问题１\a\v猜测产品质量评奖\a问题２\a问题３【问题】有Ａ、Ｂ、Ｃ、Ｄ四名偷窃嫌疑犯，其中一人是小偷，审问中，Ａ说：“我不是小偷”，Ｂ说：“Ｃ是小偷”，Ｃ说：“小偷肯定是Ｄ”，Ｄ说：“Ｃ在冤枉人”，有三人说真话，一人说假话，问到底谁是小偷？【参考程序】var thief:char;beginfor thief:='A' to 'D' doif ord(thief<>'A')+ord(thief='C')+ord(thief='D')+ord(thief<>'D')=3 then writeln('The Thief is : ',thief); end.问题4字母塔【问题】输出由字母组成的“字母塔”。

Pascal语言基础知识测验题第一套1、sqrt(4)的值为（2.0）（保留一位小数）。

2、’A’>’a’的值为（False）。

3、下列标识符哪个是合法的（A）。

A、X1B、a(c)C、varD、1234、下列程序段输出结果为（9）。

Var a,b,c:integer;Begin a:=5;b:=4; c:=a*a-b*b;writeln(c); End.5、下列程序段输出结果为（45）。

Var m,n:integer;Begin M:=0;For n:=2 to 3 Dom:=m+2; Write(m); End.第二套1、Pascal 语言中的赋值号为(:=)。

2、ord(37)的值为(37)。

3、用Pascal语言表达式表示X为偶数。

（odd）4、下列程序段输出结果为（40）。

Var A:integer;Begin A:=100;A:=A+20; A:=A Div 3;Writeln(A); End.5、下列程序段输出结果为（4）。

Var Y:integer;Begin Y:=2;While(Y<=32)DoY:=Y*Y; Writeln(Y); End.第三套1、Turbo Pascal7.0中，Integer类型的取值范围为(-32768)到32767。

2、已知chr(97)的值为字母a则chr(100)的值为字母(d)。

3、-15 mod 4的值为(-3)。

4、下列程序段输出结果为（-1）。

Var X,Y:integer;Begin X:=-100;If (X<0) Then Y:=-1 ElseY:=1; Writeln(Y); End.5、下列程序段输出结果为（200）。

Var X:integer;Begin X:=0;For I:=1 to 100 DoX:=X+2; Writeln(X); End.第四套1、odd(24)的值(False)。

2、16 Div 5的值为(3)。

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)。