Đề thi chính thức kỳ thi Toán học Hoa Kỳ - AIME (có đáp án)

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MAA American Mathematics Competitions 40th Annual AIME II American Invitational Mathematics Examination II Wednesday, February 16, 2022 INSTRUCTIONS 1. DO NOT OPEN THIS BOOKLET UNTIL YOUR COMPETITION MANAGER TELLS YOU TO BEGIN. 2. This is a 15-question competition. All answers are integers ranging from 000 to 999, inclusive. 3. Mark your answer to each problem on the answer sheet with a #2 pencil. Check blackened answers for accuracy and erase errors completely. Only answers that are properly marked on the answer sheet will be scored. 4. SCORING: You will receive 1 point for each correct answer, 0 points for each problem left unanswered, and 0 points for each incorrect answer. 5. Only blank scratch paper, rulers, compasses, and erasers are allowed as aids. Prohibited materials include calculators, smartwatches, phones, computing devices, protractors, and graph paper. 6. Figures are not necessarily drawn to scale. 7. Before beginning the competition, your competition manager will ask you to record your name on the answer sheet. 8. You will have 3 hours to complete the competition once your competition manager tells you to begin. 9. When you finish the competition, sign your name in the space provided on the answer sheet. The MAA AMC Office reserves the right to disqualify scores from a school if it determines that the rules or the required security procedures were not followed. The publication, reproduction, or communication of the problems or solutions of this competi- tion during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via phone, email, or digital media of any type during this period is a violation of the competition rules. A combination of your AIME score and your AMC 10/12 score is used to determine eligibility for participation in the USA (Junior) Mathematical Olympiad. © 2022 Mathematical Association of America
5 Bài 1. Trong một buổi hòa nhạc có người tham dự là người trưởng thành. Sau đó một xe buýt 12 11 chở thêm 50 người tới tham dự buổi hòa nhạc. Khi ấy 25 số người tham dự buổi hòa nhạc là người trưởng thành. Hỏi sau khi xe buýt tới nơi, buổi hòa nhạc đó có ít nhất bao nhiêu người trưởng thành tham dự? Bài 2. Azar, Carl, Jon và Sergey là bốn vận động viên lọt vào vòng bán kết của một giải đấu tennis. Họ sẽ được chọn ngẫu nhiên thành hai cặp để thi đấu. Những người chiến thắng ở mỗi cặp sẽ thi đấu với nhau để quyết định người vô địch. Khi Azar thi đấu với Carl thì xác suất Azar chiến thắng 2 là 3. Khi Azar hoặc Carl thi đấu với Jon hoặc Sergey thì Azar hoặc Carl sẽ chiến thắng với xác 3 suất là . Giả sử rằng kết quả của mỗi trận đấu là độc lập với nhau. Gọi xác suất để Carl vô địch 4 𝑝 giải đấu là 𝑞, với 𝑝 và 𝑞 là các số nguyên dương nguyên tố cùng nhau. Tính 𝑝 + 𝑞. Bài 3. Cho hình chóp tứ giác đều có thể tích là 54, độ dài cạnh đáy là 6. Các đỉnh của hình chóp 𝑚 nằm trên một hình cầu có bán kính là 𝑛 , với 𝑚 và 𝑛 là các số nguyên dương nguyên tố cùng nhau. Tính 𝑚 + 𝑛. 1 1 Bài 4. Cho một số thực dương 𝑥 khác 20 và 2 thỏa mãn log 20𝑥 (22𝑥) = log 2𝑥 (202𝑥). 𝑚 Giá trị của log 20𝑥 (22𝑥) có thể viết dưới dạng log10 ( 𝑛 ), với 𝑚 và 𝑛 là các số nguyên dương nguyên tố cùng nhau. Tính 𝑚 + 𝑛. Bài 5. Đánh dấu 20 điểm phân biệt trên một đường tròn bằng các số từ 1 đến 20 theo chiều kim đồng hồ. Mỗi cặp điểm có hiệu hai số là một số nguyên tố được nối bởi một đoạn thẳng. Tìm số tam giác có cạnh là các đoạn thẳng trên và các đỉnh là các điểm đã cho. Bài 6. Cho các số thực 𝑥1 ≤ 𝑥2 ≤ ∙∙∙ ≤ 𝑥100 thỏa mãn |𝑥1 | + |𝑥2 | + ∙∙∙ + |𝑥100 | = 1 và 𝑥1 + 𝑚 𝑥2 + ∙∙∙ + 𝑥100 = 0. Giả sử giá trị lớn nhất của hiệu 𝑥76 − 𝑥16 có thể nhận được là 𝑛 , với 𝑚 và 𝑛 là các số nguyên dương nguyên tố cùng nhau. Tính 𝑚 + 𝑛. Bài 7. Cho đường tròn bán kính 6 đơn vị tiếp xúc ngoài một đường tròn bán kính 24 đơn vị. Tìm diện tích tam giác được giới hạn bởi ba đường tiếp tuyến chung của hai đường tròn đã cho.
Bài 8. Tìm số các số nguyên dương 𝑛 ≤ 600 sao cho 𝑛 được xác định duy nhất nếu biết trước các 𝑛 𝑛 𝑛 giá trị ⌊ 4⌋, ⌊ 5⌋ và ⌊ 6⌋. Trong đó, kí hiệu ⌊𝑥⌋ là số nguyên lớn nhất nhỏ hơn hoặc bằng số thực 𝑥. Bài 9. Cho hai đường thẳng phân biệt 𝑙𝐴 và 𝑙𝐵 song song với nhau. Với các số nguyên dương 𝑚 và 𝑛, ta lấy các điểm phân biệt 𝐴1 , 𝐴2 , 𝐴3 ,…, 𝐴𝑚 nằm trên đường thẳng 𝑙𝐴 và các điểm phân biệt 𝐵1, 𝐵2, 𝐵3,…, 𝐵𝑛 nằm trên đường thẳng 𝑙𝐵 . Đồng thời khi vẽ các đoạn thẳng 𝐴𝑖 𝐵𝑗 với mọi 𝑖 = 1, 2, 3, … , 𝑚 và 𝑗 = 1, 2, 3, … , 𝑛 thì không điểm nào ở giữa hai đường thẳng 𝑙𝐴 và 𝑙𝐵 nằm trên nhiều hơn hai đoạn thẳng đã vẽ. Hỏi mặt phẳng bị chia thành bao nhiêu phần có diện tích hữu hạn khi 𝑚 = 7 và 𝑛 = 5? Hình dưới đây chỉ ra rằng mặt phẳng bị chia thành 8 phần có diện tích hữu hạn khi 𝑚 = 3 và 𝑛 = 2. 3 4 40 𝑎 Bài 10. Tìm số dư khi chia ((2)) + ((2)) + ⋯ + (( 2 )) cho 1000. Trong đó ( ) là tổ hợp 2 2 2 2 chập 2 của a. Bài 11. Cho tứ giác lồi ABCD với AB = 2, AD = 7 và CD = 3 sao cho tia phân giác của các góc ̂ và 𝐴𝐷𝐶 nhọn 𝐷𝐴𝐵 ̂ cắt nhau tại trung điểm của BC. Tìm bình phương của diện tích tứ giác ABCD. 𝑥2 𝑦2 (𝑥−20)2 (𝑦−11)2 Bài 12. Cho các số thực a, b, x và y thỏa mãn a > 4; b > 1 và 𝑎2 + 𝑎2 −16 = + = 1. 𝑏 2 −1 𝑏2 Tìm giá trị nhỏ nhất của a + b.
Bài 13. Cho đa thức 𝑃(𝑥) với các hệ số nguyên thỏa mãn (𝑥 2310 −1)6 𝑃(𝑥) = với mọi giá trị 0 < x < 1. (𝑥 105 −1)(𝑥 70 −1)(𝑥 42 −1)(𝑥 30 −1) Tìm hệ số của 𝑥 2022 trong đa thức 𝑃(𝑥). Bài 14. Bộ sưu tập tem là tập hợp các con tem có mệnh giá a, b hoặc c đồng, mỗi loại mệnh giá có ít nhất một con tem; trong đó a, b và c là các số nguyên dương thỏa mãn 𝑎 < 𝑏 < 𝑐. Một bộ sưu tập được gọi là ĐẸP nếu với mọi số nguyên dương nhỏ hơn hoặc bằng 1000 đồng, ta có thể lấy ra từ bộ sưu tập một số con tem có tổng giá trị là số nguyên dương đó. Gọi f(a, b, c) là số tem nhỏ nhất có thể của một bộ sưu tập ĐẸP với các bộ số a, b và c cho trước. Tìm tổng ba giá trị nhỏ nhất của c sao cho tồn tại a, b để f(a, b, c) = 97. Bài 15. Cho hai đường tròn 𝜔1 và 𝜔2 tiếp xúc ngoài với các tâm lần lượt là 𝑂1 và 𝑂2. Đường tròn thứ ba Ω đi qua 𝑂1 và 𝑂2, cắt 𝜔1 tại B và C, cắt 𝜔2 tại A và D như hình vẽ dưới đây. Giả sử rằng AB = 2, 𝑂1 𝑂2 = 15, CD = 16 và 𝐴𝐵𝑂1 𝐶𝐷𝑂2 là một lục giác lồi. Tính diện tích của lục giác đó.
MAA American Mathematics Competitions 40th Annual AIME II American Invitational Mathematics Examination II Wednesday, February 16, 2022 INSTRUCTIONS 1. DO NOT OPEN THIS BOOKLET UNTIL YOUR COMPETITION MANAGER TELLS YOU TO BEGIN. 2. This is a 15-question competition. All answers are integers ranging from 000 to 999, inclusive. 3. Mark your answer to each problem on the answer sheet with a #2 pencil. Check blackened answers for accuracy and erase errors completely. Only answers that are properly marked on the answer sheet will be scored. 4. SCORING: You will receive 1 point for each correct answer, 0 points for each problem left unanswered, and 0 points for each incorrect answer. 5. Only blank scratch paper, rulers, compasses, and erasers are allowed as aids. Prohibited materials include calculators, smartwatches, phones, computing devices, protractors, and graph paper. 6. Figures are not necessarily drawn to scale. 7. Before beginning the competition, your competition manager will ask you to record your name on the answer sheet. 8. You will have 3 hours to complete the competition once your competition manager tells you to begin. 9. When you finish the competition, sign your name in the space provided on the answer sheet. The MAA AMC Office reserves the right to disqualify scores from a school if it determines that the rules or the required security procedures were not followed. The publication, reproduction, or communication of the problems or solutions of this competi- tion during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via phone, email, or digital media of any type during this period is a violation of the competition rules. A combination of your AIME score and your AMC 10/12 score is used to determine eligibility for participation in the USA (Junior) Mathematical Olympiad. © 2022 Mathematical Association of America
2 2022 AIME II Problems Problem 1: 5 Adults made up 12 of the crowd of people at a concert. After a bus carrying 50 more people arrived, adults made up 11 25 of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. Problem 2: Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability 2 3 . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability 43 . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is pq , where p and q are relatively prime positive integers. Find p C q. Problem 3: A right square pyramid with volume 54 has a base with side length 6. The five vertices of the pyramid all lie on a sphere with radius m n , where m and n are relatively prime positive integers. Find m C n. Problem 4: 1 1 There is a positive real number x not equal to either 20 or 2 such that log20x .22x/ D log2x .202x/: The value log20x .22x/ can be written as log10 m n , where m and n are relatively prime positive integers. Find m C n. Problem 5: Twenty distinct points are marked on a circle and labeled 1 through 20 in clock- wise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles whose sides are three of these line segments and whose vertices are three distinct points from among the original 20 points. Problem 6: Let x1 x2 x100 be real numbers such that jx1 j C jx2 j C C jx100 j D 1 and x1 Cx2 C Cx100 D 0. Among all such 100-tuples of numbers, the greatest value that x76 x16 can achieve is m n , where m and n are relatively prime positive integers. Find m C n.
2022 AIME II Problems 3 Problem 7: A circle with radius 6 is externally tangent to a circle with radius 24. Find the area of the triangular region bounded by the three common tangent lines of these two circles. Problem 8: n ˘ n ˘n 600 Find the number of positive integers n ˘whose value can be uniquely deter- mined when the values of 4 , 5 , and 6 are given, where bxc denotes the greatest integer less than or equal to the real number x. Problem 9: Let `A and `B be two distinct parallel lines. For positive integers m and n, distinct points A1 ; A2 ; A3 ; : : : ; Am lie on `A , and distinct points B1 ; B2 ; B3 ; : : : ; Bn lie on `B . Additionally, when segments Ai Bj are drawn for all i D 1; 2; 3; : : : ; m and j D 1; 2; 3; : : : ; n, no point strictly between `A and `B lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when m D 7 and n D 5. The figure shows that there are 8 regions when m D 3 and n D 2. B1 B2 `B 3 2 6 1 8 5 4 7 `A A1 A2 A3 Problem 10: Find the remainder when 3 ! 4 ! 40 ! 2 C 2 C C 2 2 2 2 is divided by 1000:
4 2022 AIME II Problems Problem 11: Let ABCD be a convex quadrilateral with AB D 2, AD D 7, and CD D 3 such that the bisectors of acute angles ∠DAB and ∠ADC intersect at the midpoint of BC . Find the square of the area of ABCD. Problem 12: Let a; b; x; and y be real numbers with a > 4 and b > 1 such that x2 y2 .x 20/2 .y 11/2 C 2 D C D 1: a2 a 16 b2 1 b2 Find the least possible value for a C b. Problem 13: There is a polynomial P .x/ with integer coefficients such that .x 2310 1/6 P .x/ D .x 105 1/.x 70 1/.x 42 1/.x 30 1/ holds for every 0 < x < 1. Find the coefficient of x 2022 in P .x/. Problem 14: For positive integers a, b, and c with a < b < c, consider collections of postage stamps in denominations a, b, and c cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to 1000 cents, let f .a; b; c/ be the minimum number of stamps in such a collection. Find the sum of the three least values of c such that f .a; b; c/ D 97 for some choice of a and b.
2022 AIME II Problems 5 Problem 15: Two externally tangent circles !1 and !2 have centers O1 and O2 , respectively. A third circle passing through O1 and O2 intersects !1 at B and C and !2 at A and D, as shown. Suppose that AB D 2, O1 O2 D 15, CD D 16, and ABO1 CDO2 is a convex hexagon. Find the area of this hexagon. !1 B A !2 2 15 O1 O2 16 D C The problems and solutions for the American Invitational Mathematics Exams are selected and edited by the AIME Editorial Board of the MAA, with Co-Editors-in- Chief Jonathan Kane and Sergey Levin. The problems appearing on this competi- tion were authored by Chris Jeuell, David Altizio, David Wells, Evan Chen, Ivan Borsenco, Jerrold Grossman, Jonathan Kane, Michael Tang, and Zachary Franco. We thank them all for their contributions.
AIME 2022 ĐÁP ÁN KỲ THI TOÁN HỌC HOA KỲ - AIME II NĂM 2022 (American Invitational Mathematics Exemination AIME II) Câu 1. 154 Câu 6. 841 Câu 11. 180 Câu 2. 125 Câu 7. 192 Câu 12. 023 AIME Câu 3. 021 Câu 8. 080 Câu 13. 220 Câu 4. 112 Câu 9. 244 Câu 14. 188 Câu 5. 072 Câu 10. 004 Câu 15. 140
Official Solutions MAA American Mathematics Competitions 40th Annual AIME II American Invitational Mathematics Examination II Wednesday, February 16, 2022 This official solutions booklet gives at least one solution for each problem on this year’s competition and shows that all problems can be solved without the use of a calculator. When more than one solution is provided, this is done to illustrate a significant contrast in methods. These solutions are by no means the only ones possible, nor are they superior to others the reader may devise. Teachers are encouraged to share copies of the problem booklet and official solutions with their students for educational purposes. All problems should be credited to the MAA AMC (for example, “2017 AIME II, Problem #2”). The publication, reproduction, or communication of the competition’s problems or solutions for revenue-generating purposes requires written permission from the Mathematical Association of America (MAA). Questions and comments about this competition should be sent to: amcinfo@maa.org or MAA American Mathematics Competitions P.O. Box 471 Annapolis Junction, MD 20701 The problems and solutions for this AIME were prepared by the MAA AIME Editorial Board under the direction of: Jonathan Kane and Sergey Levin, co-Editors-in-Chief © 2022 Mathematical Association of America
Scores and official competition solutions will be sent to your competition manager who can share that information with you. For more information about the MAA American Mathematics Competitions program, please visit maa.org/amc. Questions and comments about this competition should be sent to: amcinfo@maa.org or MAA American Mathematics Competitions P.O. Box 471 Annapolis Junction, MD 20701 The problems and solutions for this AIME were prepared by the MAA AIME Editorial Board under the direction of: Jonathan Kane and Sergey Levin, co-Editors-in-Chief MAA Partner Organizations We acknowledge the generosity of the following organizations in supporting the MAA AMC and Invitational Competitions: Akamai Foundation Army Educational Outreach Program Art of Problem Solving AwesomeMathGirls.org Casualty Actuarial Society Jane Street Capital MathWorks Society for Industrial and Applied Mathematics TBL Foundation The D. E. Shaw Group Tudor Investment Corporation Two Sigma
2 2022 AIME II Solutions Problem 1: 5 Adults made up 12 of the crowd of people at a concert. After a bus carrying 50 more people arrived, adults made up 11 25 of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. Solution: A NSWER (154): Let m be the number of people at the concert before the bus arrived, and let n be the number of adults on the bus. Then the number of adults at the concert before the bus arrived is 5m 12 , and the number of adults at the concert after the bus arrived is 5m 12 C n. The fraction of adults at the concert after the bus arrived is 5m 11 Cn D 12 ; 25 m C 50 and this equation simplifies to 7m C 6600 D 300n. It follows that m must be a multiple of 300, and there is a positive integer k such that m D 300k and 7k C 22 D n. The minimum value of k is 1, corresponding to n D 29. This 5 corresponds to m D 300. The requested minimum number of adults is 300 12 C 29 D 154. OR 5 11 Let m be defined as above. Because 12 m and 25 .m C 50/ are integers, it follows that 300 divides m. Setting m D 300 shows that the minimum number of adults at the concert is 11 25 .300 C 50/ D 154, as above. Problem 2: Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability 32 . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability 34 . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is pq , where p and q are relatively prime positive integers. Find p C q. Solution: A NSWER (125): There are two cases, depending on whether Azar and Carl meet in the semifinals. If they do, which occurs with probability 13 , Carl will win the tournament if and only if he beats Azar and goes on to beat the winner of the other semifinal match, which occurs with probability 13 43 D 14 . If they do not, which occurs with probability 23 , Carl must beat Jon or Sergey in the semifinal match, which occurs with probability 34 , and go on to win the final match. If Azar wins her semifinal match, which occurs with probability 43 , Carl must beat Azar, which occurs with probability 13 . If Azar loses her semifinal match, which occurs with probability 41 , Carl must beat Azar’s opponent, which occurs with probability 43 . Thus Carl will win the tournament with probability 1 1 2 3 3 1 1 3 29 C C D : 3 4 3 4 4 3 4 4 96 The requested sum is 29 C 96 D 125. Note: This problem was inspired by an article entitled “Surprises in Knockout Tournaments” from Mathematics Magazine, 93:3 (June, 2020), 193–199.
2022 AIME II Solutions 3 Problem 3: A right square pyramid with volume 54 has a base with side length 6. The five vertices of the pyramid all lie on a sphere with radius m n , where m and n are relatively prime positive integers. Find m C n. Solution: A NSWER (021): Let ABCD be the base of the pyramid, P be its apex, and h be its height. Furthermore, let O be the center of the sphere that circumscribes the pyramid. Let M be the foot of the altitude from P . Then O, P , and M are collinear. Because the volume of a pyramid is 31 times the area of its base times its height, 13 62 h D 54, so MP D h D 92 . Because p M is the center of square ABCD with side length 6, AM D 3 2. Setting r D OP D OA gives OM D ˇ 29 r ˇ. ˇ ˇ Applying the Pythagorean Theorem to 4OAM yields 2 p 2 9 r C 3 2 D r 2; 2 17 from which it follows that r D 4 . The requested sum is 17 C 4 D 21. P D O C A M B OR More generally, let the pyramid have height h and base side length s. Let A and C be diagonally opposite vertices of the base of the pyramid, and let P be the apex of the pyramid. Let M be the center of squarepABCD, and let Q be the point on the sphere such that PQ is a diameter of the sphere. Then M is on PQ, AC D s 2, and AM D ps . 2 P h C A M ps 2 Q
4 2022 AIME II Solutions Let the sphere have radius r. Because 4PAQ is a right triangle with altitude AM , triangles 4PMA and 4AMQ are similar. Thus AM PM D QMAM ; so ps 2 2r h D : h ps 2 Solving for r gives s 2 C 2h2 rD : 4h As in the first solution, h D 92 . Thus the required radius is 2 s 2 C 2h2 62 C 2 92 17 D D ; 4 92 4h 4 as above. OR Define A, B, C , D, M , P , and r as in the first solution, where it was shown that PM D 29 . Note that the radius of the p p sphere is also the circumradius of 4AP C . The isosceles triangle 4AP C has base AC D AB 2 D 6 2 and height PM D 92 , so each of its legs has length s 2 r p p p 2 9 153 153 AP D P C D AM 2 C PM 2 D 3 2 C D D : 2 4 2 Therefore the area of 4AP C is given by 153 p p AP P C CA 4 6 2 459 2 D D : 4r 4r 8r However, the area of 4AP C is also p 1 1 p 9 27 2 AC PM D 6 2 D : 2 2 2 2 Setting these equal and solving for r yields p 459 2 2 17 rD p D ; 8 27 2 4 as above. Problem 4: 1 1 There is a positive real number x not equal to either 20 or 2 such that log20x .22x/ D log2x .202x/: The value log20x .22x/ can be written as log10 m n , where m and n are relatively prime positive integers. Find m C n.
2022 AIME II Solutions 5 Solution: A NSWER (112): Let y D log20x .22x/. Then the given equations imply .20x/y D 22x .2x/y D 202x: Thus .20x/y 11 10y D y D : .2x/ 101 11 Hence y D log10 101 . The requested sum is 11 C 101 D 112. The value of x that satisfies the original equation is approximately equal to 0:047630. Problem 5: Twenty distinct points are marked on a circle and labeled 1 through 20 in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles whose sides are three of these line segments and whose vertices are three distinct points from among the original 20 points. Solution: A NSWER (072): Suppose i , j , and k are the labels of the three vertices of a triangle with i > j > k. Note that .i j /C.j k/ D i k, so one of i j or j k must be 2, and furthermore, the other two primes must be twin primes. Thus .i j; j k; i k/ must be one of .2; 3; 5/; .3; 2; 5/; .2; 5; 7/; .5; 2; 7/; .2; 11; 13/; .11; 2; 13/; .2; 17; 19/; .17; 2; 19/: In particular, for any pairs of vertices .a; a C d /, where d 2 f5; 7; 13; 19g, there are exactly two locations for the middle vertex that yield a triangle. There are 20 d pairs of vertices .a; a C d / for every d from 1 to 19. Hence there are 2.15 C 13 C 7 C 1/ D 72 triangles satisfying the given conditions. Problem 6: Let x1 x2 x100 be real numbers such that jx1 j C jx2 j C C jx100 j D 1 and x1 C x2 C C x100 D 0. Among all such 100-tuples of numbers, the greatest value that x76 x16 can achieve is m n , where m and n are relatively prime positive integers. Find m C n. Solution: A NSWER (841): Let s be the sum of all the positive numbers in the list. Then the sum of the negative numbers in the list is s and the sum of all the absolute values is 2s. Hence s D 12 . Because there cannot be more than 25 numbers greater than or 1 1 equal to 50 , it follows that x76 50 . Similarly, because there cannot be more than 16 numbers less than or equal to 1 1 1 1 41 32 , it follows that x16 32 . Thus x76 x16 50 C 32 D 800 : 41 1 1 To see that the bound 800 can be achieved, let xi D 32 for i 16, let xi D 0 for 17 i 75, and let xi D 50 for 1 1 41 i 76. Then all the conditions in the problem are satisfied and x76 x16 D 50 C 32 D 800 . Hence the greatest value 41 that x76 x16 can achieve is 800 . The requested sum is 41 C 800 D 841.
6 2022 AIME II Solutions Problem 7: A circle with radius 6 is externally tangent to a circle with radius 24. Find the area of the triangular region bounded by the three common tangent lines of these two circles. Solution: A NSWER (192): More generally, let the larger circle have radius r and center A and the smaller circle have radius s and center B. Let the two circles be tangent at E, let the common external tangents intersect at C , let one of those tangents be tangent to the larger circle at G and to the smaller circle at D, let that tangent intersect the common internal tangent at F , and let d D BC , as shown. A E B d r C s D F G Because 4CBD and 4CAG are similar, d d CsCr D ; s r from which r Cs d Ds : r s Becausep∠DFE and ∠EF G are supplementary, and BF and AF bisect these angles, 4AFB is a right triangle, so FE D rs. The required area is then p p 2rs rs .d C s/ rs D : r s Substituting r D 24 and s D 6 gives p 2 24 6 24 6 D 192: 24 6 Problem 8: n˘ n˘ Find the number of positive integers n 600 whose value can be uniquely determined when the values of 4 , 5 , and n6 are given, where bxc denotes the greatest integer less than or equal to the real number x. ˘
2022 AIME II Solutions 7 Solution: A NSWER (080): Call an integer n good if it is uniquely determined by the values of n4 , n5 , and n6 . If n is good, then the ordered ˘ ˘ ˘ triples n4 ; n5 ; n6 and n 4 1 ; n 5 1 ; n 6 1 are not identical, so they must differ in at least one coordinate. ˘ ˘ ˘ ˘ ˘ ˘ This implies that n is a multiple of 4, 5, or 6. Similarly, n4 ; n5 ; n6 and nC1 ˘ ˘ ˘ ˘ nC1 ˘ nC1 ˘ 4 ; 5 ; 6 are not identical, so n C 1 is a multiple of 4, 5, or 6. Because it is impossible for both n and n C 1 to be even, one of these must be a multiple of 5. These conditions are both necessary and sufficient. Assume first that n is a multiple of 5 and n C 1 is a multiple of 4 or 6 or both. Then n 0 .mod 5/ and n is congruent to 1 modulo either 4 or 6, implying that n is 3, 5, 7, or 11 modulo 12. In this case, the Chinese Remainder Theorem implies that there are 4 good values of n from 1 through 5 12 D 60. They are 5, 15, 35, and 55. Next assume that n C 1 is a multiple of 5 and n is a multiple of 4 or 6 or both. Then n 1 .mod 5/ and n 0, 4, 6, or 8 .mod 12/, and again there are 4 good values of n from 1 through 60. They are 4, 24, 44, and 54. Hence there are 4 C 4 D 8 good integers from 1 through 60, so there are 8 10 D 80 good positive integers less than or equal to 600. OR As in the first solution, use the term good to refer to an integer that can be uniquely determined ˘ from n ˘the given values. Fix a positive integer n between 1 and 60 D lcm.4; 5; 6/. The set of integers m such that m 4 D 4 is an interval of four consecutive m˘ n˘ integers, where the least of these integers is divisible by 4. Similarly, the set of integers m such that D 5 ˘5 ˘ is an interval of five consecutive integers, where the least is divisible by 5; and the set of integers m such that m 6 D n 6 is an interval of six consecutive integers, where the least is divisible by 6. Under this reformulation of the problem, n is good if and only if these three intervals intersect in exactly one point. There are two key observations about such intervals. First, these three intervals are guaranteed to have nonempty intersection because n lies in all three intervals. Second, the intervals of lengths 4 and 6 must intersect in an interval of even length because the leftmost numbers in both intervals have the same parity. From these two observations, all relative positions of the three intervals can be determined. Indeed, fixing the position of the interval of length 6, there are 4 locations for the interval of length 4; then, by the even length condition, there are 2 ways to place the interval of length 5 so that all three intervals intersect at a single point. Thus there are 4 2 D 8 ways to position the intervals relative to each other to obtain the desired condition. These eight configurations are displayed below. □□■□ □□□■ ■□□□ □□□■ ■□□□□□ □■□□□□ ■□□□□□ □□□■□□ □□□□■ ■□□□□ □□□□■ ■□□□□ ■□□□ □□□■ ■□□□ □■□□ □□■□□□ □□□□□■ □□□□■□ □□□□□■ □□□□■ ■□□□□ □□□□■ ■□□□□ Finally, by the Chinese Remainder Theorem, for every integer m, each configuration above can be achieved by exactly one integer n in fm C 1; m C 2; m C 3; : : : ; m C 60g. This means that 8 values in the set f1; 2; : : : ; 60g are good, so 8 10 D 80 values in the set f1; 2; : : : ; 600g are good.
8 2022 AIME II Solutions Problem 9: Let `A and `B be two distinct parallel lines. For positive integers m and n, distinct points A1 ; A2 ; A3 ; : : : ; Am lie on `A , and distinct points B1 ; B2 ; B3 ; : : : ; Bn lie on `B . Additionally, when segments Ai Bj are drawn for all i D 1; 2; 3; : : : ; m and j D 1; 2; 3; : : : ; n, no point strictly between `A and `B lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when m D 7 and n D 5. The figure shows that there are 8 regions when m D 3 and n D 2. B1 B2 `B 3 2 6 1 8 5 4 7 `A A1 A2 A3 Solution: A NSWER (244): Assume that the points A1 ; A2 ; A3 ; : : : ; Am and B1 ; B2 ; B3 ; : : : ; Bn lie in these orders on the lines `A and `B , respec- tively, so that `A and `B together with the segments A1 B1 and Am Bn bound one region in the plane. Consider adding the other line segments Ai Bj one at a time. One of these line segments divides each of k regions into two regions where k 1 is the number of times Ai Bj intersects another of these line segments at a point between `A and `B . It follows that the final number of regions must be 1 plus 2 less than the number of line segments drawn plus the number of intersection points of these segments between `A and `B . Because there is one intersection point for every set of 4 points fAp ; Aq ; Br ; Bs g with 1 p < q m and 1 r < s n, the total number of regions is ! ! m n 1 C .m n 2/ C : 2 2 Substituting m D 7 and n D 5 gives ! ! 7 5 1 C .7 5 2/ C D 244: 2 2
2022 AIME II Solutions 9 OR For a fixed value of m 2, let f .n/ be the number of bounded regions for given values of n. Then f .1/ D m 1. To obtain f .n/ from f .n 1/, notice that adding the segment Ai Bn creates 1 C .m i /.n 1/ new regions, so adding Bn to `B creates .n 1/m.m 1/ 1 C Œ1 C .n 1/ C Œ1 C 2.n 1/ C C Œ1 C .m 1/.n 1/ D m C 2 new regions as segments from the new point to the points on `A are drawn. Therefore .n 1/m.m 1/ f .n/ D f .n 1/ C m C : 2 Computing recursively when m D 7 gives f .1/ D 6, f .2/ D 34, f .3/ D 83, f .4/ D 153, and f .5/ D 244, which is the requested number of regions. OR Consider the graph whose vertices are A1 ; A2 ; A3 ; : : : ; Am , B1 ; B2 ; B3 ; : : : ; Bn , and the intersections of the Ai Bj line segments, and whose edges follow the Ai Bj , B1 Bn , or A1 Am line segments. Then this graph has m vertices on line `A , n vertices on line `B , and p D m n 2 2 vertices between the two lines for a total of m C n C p vertices. Vertices A1 and Am each have degree n C 1, and each vertex Ai for 1 < i < n has degree n C 2. Similarly, vertices B1 and Bn each have degree m C 1, and each vertex Bj for 1 < j < n has degree m C 2. Each vertex between lines `A and `B has degree 4. Thus the total of all the degrees of the vertices in the graph is m.n C 2/ 2 C n.m C 2/ 2 C 4p D 2 .mn C m C n 2 C 2p/ : Hence the number of edges in the graph is half of this number which is mn C m C n 2 C 2p. Euler’s Formula gives the number of bounded regions for a planar graph as 1 more than the number of edges minus the number of vertices, which, in this case, is .mn C m C n 2 C 2p/ .m C n C p/ C 1 D mn C p 1: 7 5 When m D 7 and n D 5, this equals 7 5 C 2 2 1 D 244. Problem 10: Find the remainder when ! ! ! 3 4 40 2 C 2 C C 2 2 2 2 is divided by 1000: Solution: A NSWER (004): Because ! k k.k 1/ 2 .k 2/.k C 1/ 1D D ; 2 2 2 it follows that k k ! 2 1 ! k 2 2 .k C 1/k.k 1/.k 2/ kC1 D D D3 : 2 2 8 4