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2013美国女子数学大奖赛试题及解答(英文)


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2013 Solutions
Problem 1 The ?gure below shows two equilateral triangles each with area 1.

The intersection of the two triangles is a regular hexagon. What is the area of the union of the two triangles? Express your answer as a fraction in simplest form. 4 Answer: 3 Solution: The exterior angles of the regular hexagon are each 60? . So the 6 little triangles sticking out are equilateral triangles. The side length of these little equilateral triangles is 1 that of the 2 big equilateral triangles. Hence 3 1 . the area of each little equilateral triangle is 9 The union of the 2 big triangles is the disjoint union of 1 big triangle and 3 little triangles. So the total area is 1+3· 1 1 4 =1+ = . 9 3 3

Problem 2 When the binomial coe?cient how many zeros are at the rightmost end? Answer: 0

125 64

is written out in base 10,

1

Math Prize for Girls 2013

Solutions

Solution: The number of zeros at the end of 125 is the number of factors 64 125 of 10 in 64 . In turn, that number is the smaller of the number of factors of 2 in 125 and the number of factors of 5 in 125 . 64 64 125 The number of factors of 2 in 64 can be found by looking at the base-2 (binary) representation of 64 and 125 ? 64 = 61. The binary representation of 64 is (1000000)2 . The binary representation of 61 is (0111101)2 . By Kummer’s theorem, the number of factors of 2 in 125 is the number of 64 carries when (1000000)2 and (0111101)2 are added. But (1000000)2 and (0111101)2 have no 1’s in common, so there are no carries in their binary is 0. addition. Hence the number of factors of 2 in 125 64 125 Thus the number of factors of 10 in 64 is also 0 . Problem 3 Let S1 , S2 , . . . , S125 be 125 sets of 5 numbers each, comprising 625 distinct numbers. Let mi be the median of Si . Let M be the median of m1 , m2 , . . . , m125 . What is the greatest possible number of the 625 numbers that are less than M ? Answer: 436 Solution: Without loss of generality, assume that m1 > m2 > · · · > m125 . So M = m63 . Let the set Si be {ki , i , mi , ni , oi }, where ki > i > mi > ni > oi . The numbers ki , i , and mi for i ≤ 63 are each greater than or equal to M . All of the other numbers could be less than M . So the least possible number of the 625 numbers that are greater than or equal to M is 3 · 63 = 189. Hence the greatest possible number of the 625 numbers that are less than M is 625 ? 189, which is 436 . Problem 4 The MathMatters competition consists of 10 players P1 , P2 , . . . , P10 competing in a ladder-style tournament. Player P10 plays a game with P9 : the loser is ranked 10th, while the winner plays P8 . The loser of that game is ranked 9th, while the winner plays P7 . They keep repeating this process until someone plays P1 : the loser of that ?nal game is ranked 2nd, while the winner is ranked 1st. How many di?erent rankings of the players are possible? Answer: 512 Solution: The MathMatters tournament has 9 games. Each game has 2 possible winners. So the tournament has 29 possible sequences of winners. Each sequence of winners leads to a unique ranking. So the number of possible rankings is 29 , which is 512 . 2

Math Prize for Girls 2013

Solutions

Problem 5 Say that a 4-digit positive integer is mixed if it has 4 distinct digits, its leftmost digit is neither the biggest nor the smallest of the 4 digits, and its rightmost digit is not the smallest of the 4 digits. For example, 2013 is mixed. How many 4-digit positive integers are mixed? Answer: 1680 Solution: The number of sets of 4 digits is
10 4

, which is

10 · 9 · 8 · 7 = 210. 4! Let’s ?x one choice for the set of 4 digits. The leftmost digit is neither the biggest nor the smallest, so there are 2 choices for it. (We don’t have to worry about starting with a leading zero, because the leftmost digit is not the smallest.) Given that choice, the rightmost digit is neither the smallest nor the leftmost, so there are 2 choices for it. Given those choices, the second digit from the left is neither the leftmost nor the rightmost, so there are 2 choices for it. Finally, given those choices, the third digit from the left has only 1 choice, the remaining digit. Hence, given a set of 4 digits, there are 2 · 2 · 2 · 1 = 8 ways to order the digits to form a mixed integer. So the number of mixed integers is 10 · 8, which is 210 · 8, or 1680 . 4 Problem 6 Three distinct real numbers form (in some order) a 3-term arithmetic sequence, and also form (in possibly a di?erent order) a 3-term geometric sequence. Compute the greatest possible value of the common ratio of this geometric sequence. Express your answer as a fraction in simplest form. 1 ?1 Answer: or ? 2 2 Solution: The geometric sequence has the form a, ar, ar2 , where a and r are real numbers. The three terms are distinct, so a = 0 and r = 1. Because the three numbers form an arithmetic sequence (in some order), one of them is the average of the other two. There are three cases, depending on which number is the average. Case 1: a is the average of ar and ar2 . In equation form, 2a = ar + ar2 . So r2 + r = 2. Hence r is ?2 or 1. But r can’t be 1, so r is ?2. Case 2: ar is the average of a and ar2 . In equation form, 2ar = a + ar2 . So r2 + 1 = 2r. Hence r is 1. But r can’t be 1, so Case 2 is impossible. 3

Math Prize for Girls 2013

Solutions

Case 3: ar2 is the average of a and ar. In equation form, 2ar2 = a + ar. So 2r2 = 1 + r. Hence r is ? 1 or 1. But r can’t be 1, so r is ? 1 . 2 2 Looking at all three cases, we conclude that r is ?2 or ? 1 . We can 2 1 1 1 achieve r = ? 2 , since 1, ? 2 , 4 is a geometric sequence with common ratio and ? 1 , 1 , 1 is an arithmetic sequence. So the greatest possible value ?1 2 2 4 1 of r is ? . 2 Problem 7 In the ?gure below, y A(1, 1) B ABC is an equilateral triangle.

C

x

Point A has coordinates (1, 1), point B is on the positive y -axis, and point C is on the positive √ x-axis. What is the area of ABC ? Express your answer in the form a n ? b, where a and b are positive integers and n is a square-free positive integer. √ Answer: 2 3 ? 3 Solution: We will use complex numbers. For example, the point A(1, 1) in complex form is 1 + i. ? → The vector AC in complex form is ? → AC = C ? A = C ? (1 + i) = C ? 1 ? i. ? → ? → The vector AB is a 60? clockwise rotation of AC . So we have √ ? → ? → 1 3 ? i (C ? 1 ? i). AB = cis(?60? )AC = 2 2 The complex multiplication on the right expands as follows: √ √ ? → C ? 1 ? 3 (C ? 1) 3 + 1 AB = ? i. 2 2 4

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Solutions

? → The vector AB has real part ?1. So we get the equation √ C ?1? 3 = ?1. 2 √ Solving for C gives C = 3 ? 1. ? → We can calculate the squared length of AC as follows: √ √ √ ? → |AC |2 = |C ? 1 ? i|2 = | 3 ? 2 ? i|2 = ( 3 ? 2)2 + 1 = 8 ? 4 3 . √ So the squared side length of equilateral triangle ABC is 8 ? 4 3. Hence the area of ABC is √ √ √ √ √ (8 ? 4 3) 3 = (2 ? 3) 3 = 2 3 ? 3 . 4 Problem 8 Let R be the set of points (x, y ) such that x and y are positive, x + y is at most 2013, and x y = x y . Compute the area of set R. Express your answer as a fraction in simplest form. Recall that a is the greatest integer that is less than or equal to a, and a is the least integer that is greater than or equal to a. 2013 Answer: 2 Solution: We may ignore the points when x or y is an integer, since the set of such points has area 0. Because x and y are non-integers, x = x + 1 and y = y + 1. So the given equation x y = x y becomes ( x + 1) y = x ( y + 1) . The displayed equation simpli?es to y = x . Let n be the ?oor of x (and y ). Because x + y ≤ 2013, the value of n is either 0, 1, . . . , 1005, or 1006. If n ≤ 1005, then (x, y ) is any point in the unit square n < x < n + 1 and n < y < n + 1. If n = 1006, then (x, y ) is any point in the right isosceles triangle given by 1006 < x < 1007, 1006 < y < 1007, and x + y ≤ 2013. The cases n = 0, 1, . . . , 1005 give 1006 unit squares, for an area of 1006. 1 The case n = 1006 gives a triangle of area 1 . So the total area is 1006 2 , 2 2013 which is . 2 5

Math Prize for Girls 2013

Solutions

Problem 9 Let A and B be distinct positive integers such that each has the same number of positive divisors that 2013 has. Compute the least possible value of |A ? B |. Answer: 1 Solution: The prime factorization of 2013 is 3 · 11 · 61. So 2013 has 8 positive divisors. If A and B are distinct integers, then |A ? B | ≥ 1. Can we ?nd two consecutive integers with exactly 8 positive divisors each? A positive integer with exactly 8 positive divisors has the form p7 or p3 q or pqr, where p, q , and r are distinct primes. In particular, if q is a prime bigger than 2, then every number of the form 8q has exactly 8 positive divisors. Numbers of that form include 24, 40, 56, 88, 104, 136. Let’s factorize the numbers that are 1 away from these numbers. We see that 105 = 3 · 5 · 7. So both 104 and 105 have exactly 8 positive divisors. Hence, we can set A = 104 and B = 105 to get |A ? B | = 1 . Problem 10 The following ?gure shows a walk of length 6:

O This walk has three interesting properties: 1. It starts at the origin, labelled O. 2. Each step is 1 unit north, east, or west. There are no south steps. 3. The walk never comes back to a point it has been to. Let’s call a walk with these three properties a northern walk. There are 3 northern walks of length 1 and 7 northern walks of length 2. How many northern walks of length 6 are there? Answer: 239 or 236 6

Math Prize for Girls 2013

Solutions

Solution: If k is a nonnegative integer, let N (k ) be the number of northern walks of length k that end in a north step. (For convenience, say that the empty walk of length 0 ends in a north step.) Let E (k ) be the number of northern walks of length k that end in an east step. By symmetry, E (k ) is also the number of northern walks of length k that end in a west step. For initial conditions, we have N (0) = 1 and E (0) = 0. We claim that E (k + 1) = N (k ) + E (k ). That’s because every northern walk of length k + 1 that ends in an east step consists of a northern walk of length k that ends in a north or west step followed by a ?nal east step. Similarly, we claim that N (k + 1) = N (k ) + 2E (k ). That’s because every northern walk of length k + 1 that ends in a north step consists of a northern walk of length k (which ends with a north, east, or west step) followed by a ?nal north step. Using these recurrence relations, we can compute the ?rst few values of N and E . 0 1 2 3 4 5 6 k N (k ) 1 1 3 7 17 41 99 E (k ) 0 1 2 5 12 29 70 From the table above, we see that N (6) = 99 and E (6) = 70. The number of northern walks of length 6 is N (6) + 2E (6), because every northern walk of length 6 ends in a north, east, or west step. Hence the number of northern walks of length 6 is N (6) + 2E (6) = 99 + 2 · 70 = 99 + 140 = 239. Some students interpreted the problem as excluding walks that went beyond the displayed grid. Under that interpretation, three walks counted above should be excluded: the straight walks of length 6 going in the purely north, east, or west direction. So, under that interpretation, the answer is 239 ? 3 = 236. Our intended answer was 239, but we accepted either 239 or 236 . Problem 11 Alice throws two standard dice, with A being the number on her ?rst die and B being the number on her second die. She then draws the line Ax + By = 2013. Boris also throws two standard dice, with C being the number on his ?rst die and D being the number on his second die. He then draws the line Cx + Dy = 2014. Compute the probability that these two lines are parallel. Express your answer as a fraction in simplest form. 7

Math Prize for Girls 2013
Answer: 43 648

Solutions

A Solution: The slope of the line Ax + By = 2013 is ? B . The slope of the C line Cx + Dy = 2014 is ? D . The lines are parallel if and only if their slopes A C ?B and ? D are the same, which is equivalent to AD = BC . Because A and D are from the set {1, 2, 3, 4, 5, 6}, the possible values of AD are given by the multiplication table below.

× 1 2 3 4 5 6

1 2 3 1 2 3 2 4 6 3 6 9 4 8 12 5 10 15 6 12 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

Looking at the table above, we see 5 numbers (1, 9, 16, 25, and 36) that each appear 1 time, 10 numbers (2, 3, 5, 8, 10, 15, 18, 20, 24, and 30) that each appear 2 times, 1 number (4) that appears 3 times, and 2 numbers (6 and 12) that each appear 4 times. By symmetry, BC has the same distribution. So the probability that AD = BC is 5· 1 36
2

+ 10 ·

2 36

2

+1·

3 36

2

+2·

4 36

2

,

which simpli?es to 86 86 43 = = . 2 36 1296 648 Problem 12 The rectangular parallelepiped (box) P has some special properties. If one dimension of P were doubled and another dimension were halved, then the surface area of P would stay the same. If instead one dimension of P were tripled and another dimension were divided by 3, then the surface area of P would still stay the same. If the middle (by length) dimension of P is 1, compute the least possible volume of P . Express your answer as a fraction in simplest form. 2 Answer: 3 8

Math Prize for Girls 2013

Solutions

Solution: Let the dimensions of the parallelepiped P be a, b, and c. The surface area of P is 2(ab + ac + bc). If a were doubled and b were halved, then the surface area would be 2(ab + 2ac + bc/2). Equating the old and new surface area, we get 2(ab + ac + bc) = 2(ab + 2ac + bc/2). Simplifying gives b = 2a. So the preservation of surface area under a doubling/halving means that one dimension is double another. If a were tripled and b were divided by 3, then the surface area of P would be 2(ab + 3ac + bc/3). Equating the old and new surface area, we get 2(ab + bc + ac) = 2(ab + 3ac + bc/3). Simplifying gives b = 3a. So the preservation of surface area under a tripling/dividing by 3 means that one dimension is triple another. We may assume that a ≤ b ≤ c. Because the middle dimension is 1, we have b = 1. Because one dimension is triple another, we have c ≥ 3a. So the 1 or c = 2. doubling involves b. Hence either a = 2 1 Suppose that a = 2 . Because one dimension is triple another, c is either 3 or 3. Either way, the volume of P is at least 3 . 2 4 Suppose that c = 2. Because one dimension is triple another, a is either 1 or 2 . Either way, the volume of P is at least 2 . 3 3 3 Combining the two cases, we see that the volume of P is at least 2 . We 3 1 can achieve that bound by setting a = 3 , b = 1, and c = 2. So the least 2 possible volume of P is . 3 Problem 13 Each of n boys and n girls chooses a random number from the set {1, 2, 3, 4, 5}, uniformly and independently. Let pn be the probability that every boy chooses a di?erent number than every girl. As n approaches √ in?nity, what value does n pn approach? Express your answer as a fraction in simplest form. 6 Answer: 25 Solution: The probability that each of the n boys chooses a number from the set {1, 2, 3} and each of the n girls chooses a number from {4, 5} is 3 5
n

·

2 5

n

= 9

6 25

n

.

Math Prize for Girls 2013

Solutions

So pn ≥ (6/25)n . More generally, let S be a subset of {1, 2, 3, 4, 5}. Let AS be the event that each of the n boys chooses a number from S and each of the n girls chooses a number not from S . The probability of AS is |S | 5
n

·

5 ? |S | 5

n

=

|S |(5 ? |S |) 25

n

.

The numerator |S |(5 ? |S |) is at most 6, so the probability of AS is at most (6/25)n . The event “every boy chooses a di?erent number than every girl” is the union of the events AS over all subsets S of {1, 2, 3, 4, 5}. In other words, pn = Pr[
S

AS ].

Because the probability of a union is at most the sum of the individual probabilities, we have pn ≤ Pr[AS ].
S

In the previous paragraph, we showed that Pr[AS ] is at most (6/25)n , so pn ≤
S

6 25

n

.

Because S ranges over the 32 subsets of {1, 2, 3, 4, 5}, we have pn ≤ 32 6 25
n

.

Our previous work has shown that √ (6/25)n ≤ pn ≤ 32(6/25)n . Taking √ √ 6 6 n the nth root shows that 25 ≤ n pn ≤ 25 32. As n approaches in?nity, n 32 6 √ approaches 1. So n pn approaches . 25 Problem 14 How many positive integers n satisfy the inequality n n +1> ? 101 100 Recall that a is the least integer that is greater than or equal to a. Answer: 15,049 10

Math Prize for Girls 2013

Solutions

Solution: Every positive integer n can be written uniquely in the form 101q ? r, where q is a positive integer and r is a nonnegative integer such that r ≤ 100. Here q = n/101 . The given inequality becomes q+1> 101q ? r . 100

Solving for q gives q < 100 + r. Because q is an integer, that inequality is equivalent to q ≤ 99 + r. So for any given r, there are 99 + r choices for q . Hence the total number of choices is
100 100

(99 + r) = 101(99) +
r=0 r=0

r = 9999 + 5050 = 15049 .

Problem 15 Let ABC be a triangle with AB = 7, BC = 8, and AC = 9. 45? . What is the length Point D is on side AC such that CBD has measure √ √ of BD? Express your answer in the form m x ? n y , where m and n are positive integers x and y are square-free positive integers. √ and √ Answer: 40 2 ? 16 10 Solution: The diagram below shows the triangle ABC and the given information.

A
7 9 D 45? 8

B

C

By the Law of Cosines, we have 72 = 82 + 92 ? 2 · 8 · 9 cos C. Solving for cosine gives cos C = 82 + 92 ? 72 96 6 2 = = = . 2·8·9 2·8·9 9 3 11

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Hence the sine is sin C = √ 1 ? cos2 C = 1? 2 3
2

Solutions

√ 5 . = 3

Because the angles of triangle BCD add to 180? , we have BDC = 180? ? 45? ? C = 135? ? C. By the sine subtraction law, we have √ √ √ √ √ 2 2 2 5 10 + 2 2 · + · = . sin BDC = sin(135? ? C ) = 2 3 2 3 6 By the Law of Sines, we have √ √ 8 8·6 BD √ = 24 10 ? 48 2. = =√ sin C sin BDC 10 + 2 2 Solving for BD gives √ √ BD = (sin C )(24 10 ? 48 2 ) = √ √ √ √ 5 √ (24 10 ? 48 2 ) = 40 2 ? 16 10 . 3
3

and x = 1, de?ne C (x) = 1x . The real Problem 16 If ?3 ≤ x < 3 2 ?x 3 root of the cubic 2x + 3x ? 7 is of the form pC ?1 (q ), where p and q are rational numbers. What is the ordered pair (p, q )? Express your answer using fractions in simplest form. 7 27 Answer: , 3 98 Solution: Let x be the real root of 2x2 + 3x ? 7. By the Rational Root theorem, we see that x is irrational. According to the statement of the problem, x = pC ?1 (q ), where p and q are rational. Solving for q , we ?nd that q=C x p = (x/p)3 x3 7 ? 3x = 3 = . 1 ? x/p p ? p2 x 2p3 ? 2p2 x

Clearing fractions gives 2p3 q ? 2p2 qx = 7 ? 3x. 12

Math Prize for Girls 2013
Grouping the x terms together, we get (2p2 q ? 3)x = 2p3 q ? 7.

Solutions

Because x is irrational, it follows that 2p2 q = 3 and 2p3 q = 7. Dividing one equation by the other shows that p= Hence q= 2p2 q 7 = . 2p3 q 3

3 3 33 27 27 = = = = . 2 2 2 2p 2(7/3) 2(7) 2 · 49 98

Therefore, the pair (p, q ) is

7 27 . , 3 98 Note: The function C ?1 is called the “curly root” function by Dan Kalman in his book Uncommon Mathematical Excursions: Polynomia and Related Realms. He uses curly roots as an alternative method for solving cubic equations.

Problem 17 Let f be the function de?ned by f (x) = ?2 sin(πx). How many values of x such that ?2 ≤ x ≤ 2 satisfy the equation f (f (f (x))) = f (x)? Answer: 61 Solution: The graph of f on the interval [?2, 2] is shown below. y

x

13

Math Prize for Girls 2013

Solutions

As we can see, the equation f (x) = 0 has 5 solutions in [?2, 2]. If y is a ?xed nonzero number such that ?2 < y < 2, then the equation f (x) = y has 4 solutions in [?2, 2]. Let’s enumerate the solutions of f (f (f (x))) = f (x) in [?2, 2]. Let y = f (x). Then the equation becomes f (f (y )) = y . So we need to enumerate the solutions of f (f (y )) = y in [?2, 2]. Let x = f (y ). Then y = f (x) and x = f (y ). The graph of y = f (x) is the graph of f , as shown above. The graph of x = f (y ) is the inverse graph, the re?ection of y = f (x) around the line y = x. The diagram below shows the two graphs, restricted to x and y in [?2, 2]. y

x

As we can see, ignoring the origin, the two graphs intersect at 4 points in the ?rst quadrant, 3 points in the second quadrant, 4 points in the third quadrant, and 3 points in the fourth quadrant. Altogether, there are 14 intersection points, not counting the origin. So there are 14 nonzero solutions to f (f (y )) = y in [?2, 2]. Neither 2 nor ?2 is a solution. As we just discovered, there are 14 nonzero solutions to f (f (y )) = y in the open interval (?2, 2). As we mentioned earlier, for every nonzero y in (?2, 2), there are 4 solutions to f (x) = y in [?2, 2]. So there are 14 · 4 = 56 solutions to f (f (f (x))) = f (x) and f (x) = 0 in [?2, 2]. Plus the 5 solutions to f (x) = 0 in [?2, 2] give 5 solutions to f (f (f (x))) = f (x) and f (x) = 0 in [?2, 2]. Altogether, the number of solutions to f (f (f (x))) = f (x) in [?2, 2] is 56 + 5, which is 61 . Problem 18 Ranu starts with one standard die on a table. At each step, she rolls all the dice on the table: if all of them show a 6 on top, then she places one more die on the table; otherwise, she does nothing more on this 14

Math Prize for Girls 2013

Solutions

step. After 2013 such steps, let D be the number of dice on the table. What is the expected value (average value) of 6D ? Answer: 10,071 Solution: For every nonnegative integer n, let Dn be the number of dice on the table after n steps. At the start, D0 = 1. Given Dn , we are told that Dn+1 is Dn + 1 with probability 6?Dn and is Dn otherwise. Hence the expected value of 6Dn+1 , given Dn , is E (6Dn+1 | Dn ) = 6?Dn · 6Dn +1 + 1 ? 6?Dn · 6Dn = 6 + 6Dn ? 1 = 6Dn + 5. Averaging over Dn shows that E (6Dn+1 ) = E (E (6Dn+1 | Dn )) = E (6Dn + 5) = E (6Dn ) + 5. Each expected value is 5 more than the previous one So we have E (6D2013 ) = E (6D0 ) + 2013 · 5 = E (61 ) + 10065 = 6 + 10065 = 10071 . Note: This problem is related to the “approximate counting” randomized algorithm in computer science that is used to count a large number of events with a small amount of memory. Problem 19 If n is a positive integer, let φ(n) be the number of positive integers less than or equal to n that are relatively prime to n. Compute the value of the in?nite sum ∞ φ(n)2n . n ? 2n 9 n=1 Express your answer as a fraction in simplest form. 18 Answer: 49
2 Solution: Let x = 9 . Then the given sum can be rewritten as ∞ n=1

φ(n)

xn . 1 ? xn

By geometric series, we have xn = 1 ? xn 15
∞ k=1

xkn .

Math Prize for Girls 2013
By setting m = kn, we can rewrite the last sum as xn = xm , 1 ? xn m=1
n|m ∞

Solutions

where m ranges over all positive multiples of n. So our original sum becomes the double sum ∞ ∞ φ(n)xm .
n=1 m=1 n|m

By switching the order of the sums, we get
∞ ∞

φ(n)xm .

m=1 n=1 n |m

For every ?xed positive integer m, we claim that
∞ n=1 n|m

φ(n) = m.

If d is a positive divisor of m, then the number of positive integers x less than or equal to m such that gcd(x, m) = d is φ(m/d). Summing over all d shows that φ(m/d) = m.
d d|m

Setting n = m/d proves the claim. We can use the claim of the last paragraph to simplify the double sum before that: ∞ ∞ ∞ φ(n)xm = mxm .
m=1 n=1 n|m m=1

The sum on the right can be evaluated in standard ways:
∞ m=1

mx =

m



m

x =

m





x =

m

∞ k=1

m=1 k=1

k=1 m=k

x xk = . 1?x (1 ? x)2

16

Math Prize for Girls 2013
2 Plugging in x = 9 , we get

Solutions

2/9 18 2·9 . = 2 = 2 (1 ? 2/9) 7 49 Note: The in?nite sum in the statement of the problem is an example of a Lambert series. Problem 20 Let a0 , a1 , a2 , . . . be an in?nite sequence of real numbers such 4 and that a0 = 5 an = 2a2 n?1 ? 1 for every positive integer n. Let c be the smallest number such that for every positive integer n, the product of the ?rst n terms satis?es the inequality a0 a1 . . . an?1 ≤ c . 2n

What is the value of 100c, rounded to the nearest integer? Answer: 167 Solution: The recurrence an = 2a2 n?1 ? 1 resembles the double-angle identity 2 cos 2θ = 2 cos θ ? 1. With that in mind, we shall set θ = cos?1 a0 = cos?1 Then we have
2 a1 = 2a2 0 ? 1 = 2 cos θ ? 1 = cos 2θ.

4 . 5

In the same way, we ?nd that ak = cos 2k θ. We’re interested in the product cos θ cos 2θ cos 4θ . . . cos 2n?1 θ. Using the double-angle identity sin 2θ = 2 sin θ cos θ, we get the extended identity sin θ cos θ cos 2θ cos 4θ . . . cos 2n?1 θ = sin 2n θ. Because ak = cos 2k θ, we have 2n (sin θ)a0 a1 a2 . . . an?1 = sin 2n θ. 17

Math Prize for Girls 2013
4 3 Because cos θ = 5 , we have sin θ = 5 . Therefore

Solutions

a0 a1 a2 . . . an?1 = Because sin 2n θ ≤ 1, we have

5 sin 2n θ . 3 · 2n

a0 a1 a2 . . . an?1 ≤

5 . 3 · 2n

. Hence c ≤ 5 3 If sin 2n θ gets arbitrarily close to 1, then the analysis above would show 5 that c = 3 . Does sin 2n θ get arbitrarily close to 1? Probably, but I don’t know how to prove it. (It’s related to the binary representation of θ.) Nevertheless, we can check by computer that sin 236 θ > 0.999. Then the analysis above shows that c≥ 5 5 sin 236 θ > · 0.999 = 1.665. 3 3
500 3 2 = 166 3 . Hence 100c rounded

So 100c is greater than 166.5 and at most to the nearest integer is 167 .

18


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