Đề thi kèm đáp án chính thức kỳ thi toán học hoa kỳ AIME

MAA American Mathematics Competitions 40th Annual AIME II American Invitational Mathematics Examination II Wednesday, February 16, 2022 INSTRUCTIONS DO NOT OPEN THIS BOOKLET UNTIL YOUR COMPETITION MANAGER TELLS YOU TO BEGIN This is a 15-question competition All answers are integers ranging from 000 to 999, inclusive 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 SCORING: You will receive point for each correct answer, points for each problem left unanswered, and points for each incorrect answer Only blank scratch paper, rulers, compasses, and erasers are allowed as aids Prohibited materials include calculators, smartwatches, phones, computing devices, protractors, and graph paper Figures are not necessarily drawn to scale Before beginning the competition, your competition manager will ask you to record your name on the answer sheet You will have hours to complete the competition once your competition manager tells you to begin 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 competition 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 Bài Trong buổi hòa nhạc có 12 người tham dự người trưởng thành Sau xe buýt 11 chở thêm 50 người tới tham dự buổi hòa nhạc Khi 25 số người tham dự buổi hòa nhạc người trưởng thành Hỏi sau xe buýt tới nơi, buổi hòa nhạc có người trưởng thành tham dự? Bài Azar, Carl, Jon Sergey bốn vận động viên lọt vào vòng bán kết giải đấu tennis Họ chọn ngẫu nhiên thành hai cặp để thi đấu Những người chiến thắng cặp thi đấu với để định người vơ địch Khi Azar thi đấu với Carl xác suất Azar chiến thắng Khi Azar Carl thi đấu với Jon Sergey Azar Carl chiến thắng với xác suất Giả sử kết trận đấu độc lập với Gọi xác suất để Carl vô địch 𝑝 giải đấu 𝑞, với 𝑝 𝑞 số nguyên dương nguyên tố Tính 𝑝 + 𝑞 Bài Cho hình chóp tứ giác tích 54, độ dài cạnh đáy Các đỉnh hình chóp 𝑚 nằm hình cầu có bán kính 𝑛 , với 𝑚 𝑛 số nguyên dương nguyên tố Tính 𝑚 + 𝑛 1 Bài Cho số thực dương 𝑥 khác 20 thỏa mãn log 20𝑥 (22𝑥) = log 2𝑥 (202𝑥) 𝑚 Giá trị log 20𝑥 (22𝑥) viết dạng log10 ( 𝑛 ), với 𝑚 𝑛 số nguyên dương nguyên tố Tính 𝑚 + 𝑛 Bài Đánh dấu 20 điểm phân biệt đường tròn số từ đến 20 theo chiều kim đồng hồ Mỗi cặp điểm có hiệu hai số số nguyên tố nối đoạn thẳng Tìm số tam giác có cạnh đoạn thẳng đỉnh điểm cho Bài Cho số thực 𝑥1 ≤ 𝑥2 ≤ ∙∙∙ ≤ 𝑥100 thỏa mãn |𝑥1 | + |𝑥2 | + ∙∙∙ + |𝑥100 | = 𝑥1 + 𝑚 𝑥2 + ∙∙∙ + 𝑥100 = Giả sử giá trị lớn hiệu 𝑥76 − 𝑥16 nhận 𝑛 , với 𝑚 𝑛 số nguyên dương nguyên tố Tính 𝑚 + 𝑛 Bài Cho đường trịn bán kính đơn vị tiếp xúc ngồi đường trịn bán kính 24 đơn vị Tìm diện tích tam giác giới hạn ba đường tiếp tuyến chung hai đường tròn cho Bài Tìm số số nguyên dương 𝑛 ≤ 600 cho 𝑛 xác định biết trước 𝑛 𝑛 𝑛 giá trị ⌊ 4⌋, ⌊ 5⌋ ⌊ 6⌋ Trong đó, kí hiệu ⌊𝑥⌋ số nguyên lớn nhỏ số thực 𝑥 Bài Cho hai đường thẳng phân biệt 𝑙𝐴 𝑙𝐵 song song với Với số nguyên dương 𝑚 𝑛, ta lấy điểm phân biệt 𝐴1 , 𝐴2 , 𝐴3 ,…, 𝐴𝑚 nằm đường thẳng 𝑙𝐴 điểm phân biệt 𝐵1, 𝐵2, 𝐵3,…, 𝐵𝑛 nằm đường thẳng 𝑙𝐵 Đồng thời vẽ đoạn thẳng 𝐴𝑖 𝐵𝑗 với 𝑖 = 1, 2, 3, … , 𝑚 𝑗 = 1, 2, 3, … , 𝑛 khơng điểm hai đường thẳng 𝑙𝐴 𝑙𝐵 nằm nhiều hai đoạn thẳng vẽ Hỏi mặt phẳng bị chia thành phần có diện tích hữu hạn 𝑚 = 𝑛 = 5? Hình mặt phẳng bị chia thành phần có diện tích hữu hạn 𝑚 = 𝑛 = 40 𝑎 Bài 10 Tìm số dư chia ((2)) + ((2)) + ⋯ + (( )) cho 1000 Trong ( ) tổ hợp 2 2 chập a Bài 11 Cho tứ giác lồi ABCD với AB = 2, AD = CD = cho tia phân giác góc ̂ 𝐴𝐷𝐶 ̂ cắt trung điểm BC Tìm bình phương diện tích tứ giác ABCD nhọn 𝐷𝐴𝐵 𝑥2 𝑦2 Bài 12 Cho số thực a, b, x y thỏa mãn a > 4; b > 𝑎2 + 𝑎2 −16 = Tìm giá trị nhỏ a + b (𝑥−20)2 𝑏 −1 + (𝑦−11)2 𝑏2 = Bài 13 Cho đa thức 𝑃(𝑥) với hệ số nguyên thỏa mãn 𝑃(𝑥) = (𝑥 2310 −1)6 (𝑥 105 −1)(𝑥 70 −1)(𝑥 42 −1)(𝑥 30 −1) với giá trị < x < Tìm hệ số 𝑥 2022 đa thức 𝑃(𝑥) Bài 14 Bộ sưu tập tem tập hợp tem có mệnh giá a, b c đồng, loại mệnh giá có tem; a, b c số nguyên dương thỏa mãn 𝑎 < 𝑏 < 𝑐 Một sưu tập gọi ĐẸP với số nguyên dương nhỏ 1000 đồng, ta lấy từ sưu tập số tem có tổng giá trị số nguyên dương Gọi f(a, b, c) số tem nhỏ sưu tập ĐẸP với số a, b c cho trước Tìm tổng ba giá trị nhỏ c cho tồn a, b để f(a, b, c) = 97 Bài 15 Cho hai đường tròn 𝜔1 𝜔2 tiếp xúc với tâm 𝑂1 𝑂2 Đường tròn thứ ba Ω qua 𝑂1 𝑂2, cắt 𝜔1 B C, cắt 𝜔2 A D hình vẽ Giả sử AB = 2, 𝑂1 𝑂2 = 15, CD = 16 𝐴𝐵𝑂1 𝐶𝐷𝑂2 lục giác lồi Tính diện tích lục giác MAA American Mathematics Competitions 40th Annual AIME II American Invitational Mathematics Examination II Wednesday, February 16, 2022 INSTRUCTIONS DO NOT OPEN THIS BOOKLET UNTIL YOUR COMPETITION MANAGER TELLS YOU TO BEGIN This is a 15-question competition All answers are integers ranging from 000 to 999, inclusive 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 SCORING: You will receive point for each correct answer, points for each problem left unanswered, and points for each incorrect answer Only blank scratch paper, rulers, compasses, and erasers are allowed as aids Prohibited materials include calculators, smartwatches, phones, computing devices, protractors, and graph paper Figures are not necessarily drawn to scale Before beginning the competition, your competition manager will ask you to record your name on the answer sheet You will have hours to complete the competition once your competition manager tells you to begin 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 competition 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 2022 AIME II Problems Problem 1: Adults made up 12 of the crowd of people at a concert After a bus carrying of the people at the concert Find 50 more people arrived, adults made up 11 25 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 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 The five , where m and n are vertices of the pyramid all lie on a sphere with radius m n relatively prime positive integers Find m C n Problem 4: There is a positive real number x not equal to either 20 or such that log20x 22x/ D log2x 202x/: The value log20x 22x/ can be written as log10 prime positive integers Find m C n m n , where m and n are relatively Problem 5: Twenty distinct points are marked on a circle and labeled 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 Problem 6: Let x1 Ä x2 Ä Ä x100 be real numbers such that jx1 j C jx2 j C C jx100 j D and x1 Cx2 C Cx100 D Among all such 100-tuples of numbers, the greatest value that x76 x16 can achieve is m , where m and n are relatively prime positive n integers Find m C n 2022 AIME II Problems Problem 7: A circle with radius 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: Find the number of positive integers ˘ n ˘n Ä 600n ˘whose value can be uniquely detern mined when the values of , , and 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 and n D The figure shows that there are regions when m D and n D `B B1 B2 `A A1 A2 A3 Problem 10: Find the remainder when 2 is divided by 1000: ! C 2 ! C C 40 2 ! 2022 AIME II Problems Problem 11: Let ABCD be a convex quadrilateral with AB D 2, AD D 7, and CD D 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 > and b > such that y2 x2 x 20/2 y 11/2 C D 1: D C a2 a 16 b2 b2 Find the least possible value for a C b Problem 13: There is a polynomial P x/ with integer coefficients such that P x/ D x 105 x 2310 1/6 1/.x 70 1/.x 42 1/.x 30 1/ holds for every < x < 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 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 15 O1 16 !2 O2 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-inChief Jonathan Kane and Sergey Levin The problems appearing on this competition 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) AIME Câu 154 Câu 841 Câu 11 180 Câu 125 Câu 192 Câu 12 023 Câu 021 Câu 080 Câu 13 220 Câu 112 Câu 244 Câu 14 188 Câu 072 Câu 10 004 Câu 15 140 2022 AIME II Solutions Problem 1: Adults made up 12 of the crowd of people at a concert After a bus carrying 50 more people arrived, adults made up 11 of the people at the concert Find the minimum number of adults who could have been at the concert after the bus 25 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 , and the number of adults at the concert after the bus 12 arrived is 5m C n The fraction of adults at the concert after the bus arrived is 12 5m Cn 11 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 C 29 D 154 corresponds to m D 300 The requested minimum number of adults is 300 12 OR 11 m and 25 m C 50/ are integers, it follows that 300 divides m Setting m D 300 Let m be defined as above Because 12 300 C 50/ D 154, as above shows that the minimum number of adults at the concert is 11 25 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 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 C 4  1 C 4 à D 29 : 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 Problem 3: A right square pyramid with volume 54 has a base with side length The five vertices of the pyramid all lie on a sphere with radius m , where m and n are relatively prime positive integers Find m C n 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 ˇ ˇ p M is the center of square ABCD with side length 6, AM D Setting r D OP D OA gives OM D ˇ 29 r ˇ Applying the Pythagorean Theorem to 4OAM yields  from which it follows that r D 17 Ã2 r p Á2 C D r 2; The requested sum is 17 C D 21 P D O A C 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 P C h M Q ps A 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 D QM ; so PM AM ps 2r D h h ps : Solving for r gives rD s C 2h2 : 4h As in the first solution, h D 92 Thus the required radius is 62 C s C 2h2 D 4h 92 2 D 17 ; 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 D and height PM D 92 , so each of its legs has length s AP D P C D p AM C PM D p 2 p  Ã2 r 153 153 C D D : Therefore the area of 4AP C is given by AP P C CA D 4r 153 p p 459 D : 4r 8r However, the area of 4AP C is also p 1 p 27 AC PM D D : 2 2 Setting these equal and solving for r yields p 17 459 2 rD ; p D 27 as above Problem 4: There is a positive real number x not equal to either 20 or 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 Solution: A NSWER (112): Let y D log20x 22x/ Then the given equations imply 20x/y D 22x 2x/y D 202x: Thus 10y D 20x/y 11 D : y 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 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 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 to 19 Hence there are 2.15 C 13 C C 1/ D 72 triangles satisfying the given conditions Problem 6: Let x1 Ä x2 Ä Ä x100 be real numbers such that jx1 j C jx2 j C C jx100 j D and x1 C x2 C C x100 D Among all such 100-tuples of numbers, the greatest value that x76 x16 can achieve is m , where m and n are relatively n 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 equal to 50 , it follows that x76 Ä 50 Similarly, because there cannot be more than 16 numbers less than or equal to 1 1 41 , it follows that x Thus x76 x16 Ä 50 C 32 D 800 : 16 32 32 41 1 To see that the bound 800 can be achieved, let xi D 32 for i Ä 16, let xi D for 17 Ä i Ä 75, and let xi D 50 for 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 2022 AIME II Solutions Problem 7: A circle with radius 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 B C d r s D Because 4CBD and 4CAG are similar, E G F d d CsCr D ; s r from which  d Ds r Cs r s à : 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 gives p 24 24 D 192: 24 Problem 8: Find the number of positive integers n Ä 600 whose value can be uniquely determined when the values of ˘ and n6 are given, where bxc denotes the greatest integer less than or equal to the real number x n ˘ , n ˘ , 2022 AIME II Solutions 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 ; n ; n are not identical, so they must differ in at least one coordinate ˘ ˘ ˘ ˘ nC1 ˘ nC1 ˘ This implies that n is a multiple of 4, 5, or Similarly, n4 ; n5 ; n6 and nC1 ; ; are not identical, so n C is a multiple of 4, 5, or Because it is impossible for both n and n C to be even, one of these must be a multiple of These conditions are both necessary and sufficient Assume first that n is a multiple of and n C is a multiple of or or both Then n Á mod 5/ and n is congruent to modulo either or 6, implying that n is 3, 5, 7, or 11 modulo 12 In this case, the Chinese Remainder Theorem implies that there are good values of n from through 12 D 60 They are 5, 15, 35, and 55 Next assume that n C is a multiple of and n is a multiple of or or both Then n Á mod 5/ and n Á 0, 4, 6, or mod 12/, and again there are good values of n from through 60 They are 4, 24, 44, and 54 Hence there are C D good integers from through 60, so there are 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 ˘the given values n D is an interval of Fix a positive integer n between and 60 D lcm.4; 5; 6/ The set of integers m such that m 4 four˘ consecutive integers, where the least of these integers is divisible by Similarly, the set of integers m such that ˘ m n D is an interval of five consecutive integers, where the least is divisible by 5; and the set of integers m such ˘ n˘ D is an interval of six consecutive integers, where the least is divisible by Under this reformulation that m 6 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 and 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 locations for the interval of length 4; then, by the even length condition, there are ways to place the interval of length so that all three intervals intersect at a single point Thus there are D 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 values in the set f1; 2; : : : ; 60g are good, so 10 D 80 values in the set f1; 2; : : : ; 600g are good 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 and n D The figure shows that there are regions when m D and n D `B B1 B2 `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 , respectively, 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 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 plus 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 points fAp ; Aq ; Br ; Bs g with Ä p < q Ä m and Ä r < s Ä n, the total number of regions is ! ! m n C m n 2/ C : 2 Substituting m D and n D gives C ! ! 2/ C D 244: 2 2022 AIME II Solutions OR 2, let f n/ be the number of bounded regions for given values of n Then f 1/ D m To 1/, notice that adding the segment Ai Bn creates C m i /.n 1/ new regions, so adding For a fixed value of m obtain f n/ from f n Bn to `B creates C Œ1 C n 1/ C Œ1 C 2.n 1/ C C Œ1 C m 1/ D m C 1/.n n 1/m.m 1/ new regions as segments from the new point to the points on `A are drawn Therefore f n/ D f n 1/ C m C n 1/m.m 1/ : Computing recursively when m D 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 n line `A , n vertices on line `B , and p D m vertices between the two lines for a total of m C n C p vertices Vertices 2 A1 and Am each have degree n C 1, and each vertex Ai for < i < n has degree n C Similarly, vertices B1 and Bn each have degree m C 1, and each vertex Bj for < j < n has degree m C Each vertex between lines `A and `B has degree Thus the total of all the degrees of the vertices in the graph is m.n C 2/ C n.m C 2/ C 4p D mn C m C n C 2p/ : Hence the number of edges in the graph is half of this number which is mn C m C n C 2p Euler’s Formula gives the number of bounded regions for a planar graph as more than the number of edges minus the number of vertices, which, in this case, is mn C m C n C 2p/ When m D and n D 5, this equals C ! m C n C p/ C D mn C p 1: D 244 Problem 10: Find the remainder when C ! C C 40 ! is divided by 1000: Solution: A NSWER (004): Because k ! 1D it follows that k 2 ! D k k 2 k.k Á D 1/ 2 D k k C 1/k.k 2/.k C 1/ ; 1/.k 2/ ! kC1 D3 : 10 2022 AIME II Solutions By the Hockey-stick Identity, n X kD3 ! ! kC1 nC2 D3 : Then the given sum equals ! 42 41 40 39 38 42 D D 42 38/ 41 39/ 120 Á 596 599 Á 404/ 401/ Á mod 1000/: The requested remainder is OR Note that k ! is the number of ways of choosing two distinct sets of two integers from to k The two sets chosen can have either or element in common Choose two sets with no elements in common by choosing elements and then separating those elements into two sets of 2, and this can be done in k4 ways Choose two sets with one element in common by choosing the common element and then choosing two elements to pair with it, and this can be done in k k ways This shows that k 2 ! ! ! ! ! ! k k k k kC1 D3 Ck D3 C3 D3 : 4 Then the solution continues as above Problem 11: Let ABCD be a convex quadrilateral with AB D 2, AD D 7, and CD D such that the bisectors of acute angles ∠DAB and ∠ADC intersect at the midpoint of BC Find the square of the area of ABCD Solution: A NSWER (180): Let I be the midpoint of BC , and let R and S denote the reflections of C and B across DI and AI , respectively Note that, because DI and AI are angle bisectors, R and S lie on segment AD Then AS D and DR D 3, so RS D Furthermore, IR D IC D IB D IS , so ∠ABI D ∠ASI D ∠DRI D ∠DCI; and, if T is the foot of the perpendicular from I to AD, then RT D S T D C I B ˛ ˇ D R T S A 2022 AIME II Solutions 11 Now let ˛ D ∠BAI D ∠IAD, ˇ D ∠CDI D ∠IDA, and D ∠ABC D ∠BCD Because the sum of the measures of the interior angles of a quadrilateral is 360ı , it follows that ˛ C ˇ C D 180ı This implies ∠BIA D 180ı ˛ D ˇ; p p and likewise ∠CID D ˛ Thus 4ABI 4ICD, so BI D CI D AB CD D p Finally, applying the Pythagorean Theorem to 4IRT yields I T D 5, so Area.ABCD/ D Area.4AIS / C Area.4RIS / C Area.4DIR/ p Á p p Á p D 21 C2 C 12 D 5: p Á2 The square of the area of ABCD is D 180 OR Let X be the intersection of rays AB and DC Then the angle bisectors of ∠DAB and ∠CDA meet at the incenter I of 4XAD Because XI is both an angle bisector and a median of 4BXC , it is the perpendicular bisector of BC Let T , E, and F be the projections of I onto lines AD, XA, and XD, respectively X r2 r2 F r E r C I B D T A Because 4BXI is congruent to 4CXI , it follows that CF D BE By equal tangents, DT D DF D C CF D C BE D C AE D C AT Because DT C AT D AD D 7, AT D and DT D from which CF D BE D AE AB D AT AB D D Let r be the inradius of 4ADX Because 4BEI 4IEX, it follows that 2 XE D IE D r BE Because the inradius of a triangle with sides a, b, and c and semiperimeter s is r s a/.s b/.s c/ rD ; s it follows that implying that r D p r r C C 4/ D r 4; 12 2022 AIME II Solutions Then Area.ABCD/ D Area.4ABI / C Area.4CDI / C Area.4AID/ D p r AB r CD r AD C C D 5; 2 as above Problem 12: Let a; b; x; and y be real numbers with a > and b > such that x2 y2 x 20/2 y 11/2 C D C D 1: a2 a2 16 b2 b2 Find the least possible value for a C b Solution: A NSWER (023): The graph of y2 x2 C D1 a2 a2 16 is an ellipse p centered at 0; 0/ with major axis parallel to the x-axis of length 2a: The distance from the center to the foci is a2 a2 16/ D 4, so the foci are F1 D 4; 0/ and F2 D 4; 0/ Similarly, the graph of y 11/2 x 20/2 C D1 b b2 is an ellipse pcentered at 20; 11/ with major axis parallel to the y-axis of length 2b: The distance from the center to the foci is b b 1/ D 1, so the foci are G1 D 20; 10/ and G2 D 20; 12/ G2 G1 F1 F2 Let P x; y/ be a point that lies on both ellipses, so by a property of ellipses, 2a D PF1 C PF2 and 2b D P G1 C P G2 : 2022 AIME II Solutions 13 Adding these equations and using the triangle inequality yields 2a C 2b D PF1 C PF2 C P G1 C P G2 D PF1 C P G1 / C PF2 C P G2 / F1 G1 C F2 G2 p p D 242 C 102 C 162 C 122 D 46: Thus a C b 23: Equality is achieved by running this argument in reverse: that is, by taking P to be the intersection of F1 G1 and F2 G2 ; and then setting a D 12 PF1 C PF2 / and b D 12 P G1 C P G2 /: In this case, a D 16, b D 7, and P D 14; 7:5/ Therefore the least possible value of a C b is 23 Problem 13: There is a polynomial P x/ with integer coefficients such that P x/ D x 105 x 2310 1/6 1/.x 42 1/.x 30 1/.x 70 1/ holds for every < x < Find the coefficient of x 2022 in P x/ Solution: A NSWER (220): For < x < 1, the given rational expression is equal to x 2310 1/2 D x 2310 x 2310 x 105 1/2 21 X x 2310 x 2310 x 2310 1 x 70 x 42 x 30 x 105a aD0 32 X x 70b 54 X x 42c cD0 bD0 76 X x 30d: d D0 The coefficient of x 2022 is therefore equal to the number of quadruples of nonnegative integers a; b; c; d / that satisfy 105a C 70b C 42c C 30d D 2022: Considering this equation modulo 2, 3, 5, yields aÁ0 mod 2/ bÁ0 mod 3/ c Á 42 2022 Á mod 5/ d Á 30 2022 Á mod 7/: Thus there are nonnegative integers w, x, y, and z satisfying a D 2w, b D 3x, c D 5y C 1, and d D 7z C The given equation becomes 2022 D 105.2w/ C 70.3x/ C 42.5y C 1/ C 30.7z C 3/ D 210.w C x C y C z/ C 132; from which w C x C y C z D By the sticks-and-stones method, the number of solutions to this equation in nonnegative integers is given by 9C3 D 220 14 2022 AIME II Solutions 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 subcollections 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 Solution: A NSWER (188): First note that a must be To make every positive integer number of cents up to c requires at most c stamps of denominations and b, and the number of c-cent stamps then required to make every positive integer number of cents up to 1000 is at most 1000 c 1/ 1001 1000 D 1D : c c c Hence the possible values of c must satisfy the inequality c 1/ C 1000 c 97; p p 1401 or c 49 C 1401, and the integer solutions are c Ä 11 implying that c 98c C 1000 Thus c Ä 49 or c 87 Let the number of stamps of denominations 1, b, and c be na , nb , and nc , respectively, and consider two cases Case c Ä 11: Note that 97 10 D 970 < 1000, so no value of c less than 11 is possible, but 11 is a possible value of c if b D and na ; nb ; nc / D 6; 1; 90/ Case c 87: If c D 87, then nc Ä 11 and na C nb Ä 86, and 87 is a possible value of c only if equality holds in both cases However, if na C nb D 86, then the collection contains 85 1-cent stamps and 86-cent stamp The total value of those stamps in cents is 171, and because 1000 171 < 10 87, only 10 87-cent stamps are needed to make all of the required amounts If c D 88, then nc Ä 11 and na C nb Ä 87, so if 88 is a possible value of c, then na C nb ; nc / D 86; 11/ or na Cnb ; nc / D 87; 10/ In fact both 88 and 89 are possible values of c if b D 87 and na ; nb ; nc / D 86; 1; 10/ The sum of the three least possible values of c is therefore 11 C 88 C 89 D 188 2022 AIME II Solutions 15 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 15 O1 O2 16 D C Solution: A NSWER (140): First observe that AO2 D O2 D and BO1 D O1 C Let points A0 and B be the reflections of A and B, respectively, across the diameter of that is the perpendicular bisector of O1 O2 Thus A0 and B are on with A0 B D AB D Then quadrilaterals ABO1 O2 and A0 B O2 O1 are congruent, so hexagons ABO1 CDO2 and B A0 O1 CDO2 have the > > > > same area Furthermore, because A0 O1 D DO2 and O1 C D O2 B , it follows that B D D A0 C and quadrilateral 0 B A CD is an isosceles trapezoid A0 B0 ˛ R O1 O2 C D 16 2022 AIME II Solutions Because A0 O1 D DO2 , quadrilateral A0 O1 DO2 is an isosceles trapezoid In turn, A0 D D O1 O2 D 15, and similarly B C D 15 Thus Ptolemy’s Theorem applied to B A0 CD yields B D A0 C C 16 D 152 , whence B D D A0 C D p 193 Let ˛ D ∠B A0 D The Law of Cosines applied to 4B A0 D yields cos ˛ D 152 C 22 p 2 15 193 Á2 D ; and hence sin ˛ D 45 Let R be the point on line A0 B such that A0 R ? DR Because 4A0 RD is a right triangle whose hypotenuse has length 15 with cos.∠DA0 R/ D 35 , it follows that A0 R D and DR D 12 In particular, the distance from A0 B to CD is DR D 12, which implies that the area of B A0 CD is 12 12 C 16/ D 108 Now let O1 C D O2 B D r1 and O2 D D O1 A0 D r2 The tangency of circles !1 and !2 implies r1 C r2 D 15 Furthermore, ∠B O2 D and ∠B A0 D are opposite angles in cyclic quadrilateral A0 B O2 D, which implies that the measure of ∠B O2 D is 180ı ˛ Therefore the Law of Cosines applied to 4B O2 D yields  à 193 D r12 C r22 2r1 r2 D r12 C 2r1 r2 C r22 r1 r2 r1 r2 D r1 C r2 /2 r1 r2 : D 225 Thus r1 r2 D 40, so the area of 4B O2 D is 12 r1 r2 sin ˛ D 16 The area of the hexagon is 108 C 16 D 140 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 competition 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 ... 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) AIME Câu 154 Câu 841 Câu 11 180 Câu 125... vận động viên lọt vào vòng bán kết giải đấu tennis Họ chọn ngẫu nhiên thành hai cặp để thi đấu Những người chiến thắng cặp thi đấu với để định người vô địch Khi Azar thi đấu với Carl xác suất Azar... 40th Annual AIME II American Invitational Mathematics Examination II Wednesday, February 16, 2022 INSTRUCTIONS DO NOT OPEN THIS BOOKLET UNTIL YOUR COMPETITION MANAGER TELLS YOU TO BEGIN This is a