1981年美国大学生数学竞赛试题

Problem A1

Let the largest power of 5 dividing 112233

... n n

be 5f(n)

. What is lim n→∞ f(n)/n 2

?

Solution Answer: 1/8.

Let 5k

be the largest power of 5 not exceeding n.

We proceed by looking first at the contribution of the single power of 5 in all multiples of 5, then at the additional contribution from the second power of 5 in

all multiples of 25 and so on. A single power of 5 in m appears m times in m m

, so the first power of 5 in 5, 10, 15, 20, ... contributes to f(n): 5 + 10 + 20 + ...

+ 5[n/5] = 5( 1 + 2 + ... + [n/5] ) = 1/2 5 ( [n/5]2

+ [n/5] ). Similarly, the second

power of 5 in 25, 50, 75, ... contributes an additional 1/2 52 ( [n/52]2 + [n/52

] ). And so on. Hence

2 f(n) = 5 ( [n/5]2

+ [n/5] ) + 52

( [n/52]2

+ [n/52

] ) + ... + 5k

( [n/5k ]2

+ [n/5k

]). Let [n/5i

] = n/5i

+ δi . Then 5i

( [n/5i ]2

+ [n/5i

] ) = 5i

( (n/5i

- δi )2

+ (n/5i

- δi ) = n 2/5i - 2n δi + 5i δi 2 + n - 5i

δi .

Hence 2 f(n) = n 2∑1/5i

- 2n ∑ δi + ∑ 5i

(δi 2

- δi ) + nk.

Now k = [log 5n], so k/n → 0 as n → ∞. Also |δi | and |δi 2

- δi | < 1. Hence lim 1/n 2 |- 2n ∑ δi + ∑ 5i (δi 2 - δi ) + nk| <= lim 2k/n + lim k/n + lim 1/n 2 ∑ 5i

=

0 + 0 + lim 1/n 2 (1/4 (5k+1

+ 1) = 0 (since the last term is less than 5/4n).

Hence lim f(n)/n 2

= 1/2 (1/5 + 1/52

+ ... ) = 1/2 1/5 1/(1 - 1/5) = 1/8.

Problem A2

We can label the squares of an 8 x 8 chess board from from 1 to 64 in 64! different ways. For each way we find D, the largest difference between the labels of two squares which are adjacent (orthogonally or diagonally). What is the smallest possible D? Solution Answer: 9.

Consider the straightforward ordering 1, 2, 3, 4, 5, 6, 7, 8 for the first row, 9, 10, ... , 16 for the second row, ... , 57, 58, ... , 64 for the last row. Adjacent squares in the same row have difference 1, adjacent squares in different rows have difference 7, 8 or 9. So for this ordering D = 9.

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If we take any two squares on a chessboard there is a path from one to the other of length at most 7, where each step of the path is to an adjacent square [if the squares are at opposite corners of an m x n rectangle with m >= n, then take m - 1 steps, n - 1 of them diagonal and the rest along the longest side.] So there is a path of at most 7 steps from 1 to 64. At least one step of that path must have a difference of at least 9 (since 7 x 9 = 64 - 1). So we always have D ≥ 9. Hence the minimal D is 9. Problem A3

Evaluate: lim k→∞ e -k

∫R (e x

- e y

) / (x - y) dx dy, where R is the rectangle 0 ≤ x, y, ≤ k. Solution

Answer: It diverges to plus infinity.

We use L'H?pital's rule. Let f(k) = ∫R (e x

- e y

) / (x - y) dx dy, then the limit

is the same as the limit of f'(k)/e k

. To work out f'(k), note that f(k + δk) - f(k)

= ∫δR (e x - e y

) / (x - y) dx dy, where δR is the strip from x = 0 to x = k at y = k, width δk and the strip from y = 0 to k at x = k width δk. The integral over

the two strips is obviously the same, so we get f'(k) = 2 ∫0k (e k - e x

) / (k - x)

dx. Hence f'(k)/e k = 2 ∫0k (1 - e

-(k-x)

) / (k - x) dx. Changing the integration variable to z = k - x, we get 2 ∫0k (1 - e -z

) / z dz. The integrand is always positive and

for z >= 1 is at least 1 - 1/e > 1/2. So the integral is at least ∫1k

1/z dz = ln k. This tends to infinity with k.

Problem A4

A particle moves in a straight line inside a square side 1. It is reflected from the sides, but absorbed by the four corners. It starts from an arbitrary point P inside the square. Let c(k) be the number of possible starting directions from which it reaches a corner after traveling a distance k or less. Find the smallest constant

a 2, such that from some constants a 1 and a 0, c(k) ≤ a 2k 2

+ a 1k + a 0 for all P and all k.

Solution Answer: π.

By a series of reflections we can transform the path into a straight line. The particle reaches a corner iff the transformed line passes through a lattice point. So the number of directions which lead to absorption in a distance k or less is the number of lattice points in a circle radius k. At least that would be true except for the shadowing problem - the starting point may lie on a line of lattice points

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in which case it could never reach the ones further out. Shadowing reduces the number of lattice points in the circle that we should count, so the number of lattice points is still an upper bound.

Evidently what we need are some estimates for the number of lattice points in the circle. They do not need to be good estimates - we only need the highest power to be correct. We can regard each lattice point as the centre of a unit square. The

resulting unit squares tile the plane. The circle radius k has area πk 2

, so it

contains around πk 2

of the unit squares and hence around πk 2 lattice points. However, the squares around the edge may be incomplete and the incomplete squares may or may not include their centres. The maximum distance between two points of a unit square is √2, so the circle radius k + √2 contains the entirety of any square which overlaps the circle radius k. Hence the number of lattice points in the circle radius k is

at most π(k + √2)2

. Thus if we take a 2 = π, a 1 = 2π√2, a 0 = 2π, then c(k) ≤ a 2k 2

+ a 1k + a 0 for all P and k.

Similarly, any unit square overlapping the circle radius k - √2 is entirely

contained in the circle radius k. There are at least π(k - √2)2

such squares, so

the circle radius k contains at least π(k - √2)2

lattice points. Take a coordinate system with axes parallel to the sides of the squares. Then all lines joining lattice points have rational slope, so if we take the point P to have one coordinate rational and the other irrational, it cannot lie on any such lines. Thus for these points

there is no shadowing problem and the number of directions is at least π(k - √2)2

. This shows that π is the minimal value of a 2 (because for any smaller value and

any a 1, a 0 we would have a 2k 2 + a 1k + a 0 < π(k - √2)2

for sufficiently large k).

Problem A5

p(x) is a real polynomial with at least n distinct real roots greater than 1. [To be precise we can find at least n distinct values a i > 1 such that p(a i ) = 0. It is possible that one or more of the a i is a multiple root, and it is possible that there

are other roots.] Put q(x) = (x 2 + 1) p(x) p'(x) + x p(x)2 + x p'(x)2

. Must q(x) have at least 2n - 1 distinct real roots?

Solution Answer: yes.

Notice that q(x) = (x p(x) + p'(x) )( p(x) + x p'(x) ). The second factor is (x p(x) )'. Now x p(x) has at least n + 1 distinct roots (those of p(x) plus the root x = 0), so its derivative has at least n distinct zeros. To be precise assume that p(x) has exactly m ≥ n values a i > 1 at which it is zero: 1 < a 1 < a 2 < ... < a m . Then (x p(x) )'

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has at least m zeros, one in the interval (0, a 1) and one in each of the intervals (a i , a i+1).

We would like to make a similar argument for x p(x) + p'(x). This needs a trick.

p(x) exp(x 2

/2) also has

zeros at a i . Hence its derivative, exp(x 2/2) ( x p(x) + p'(x) ) has at least m-1 roots, one in each interval (a i , a i+1). But exp(x 2

/2) is positive for all real x, so (x p(x) + p'(x) ) has a root in each interval (a i , a i+1). So we are done unless a root k of (x p(x) )' is also a root of (x p(x) + p'(x) ). But in this case k belongs to some open interval (a i , a i+1) and we have k p(k) + p'(k) = k p'(k) + p(k). Hence (k - 1)( p(k) - p'(k) ) = 0. But k > 1, so p'(k) = p(k). Hence kp(k) + p(k) = 0, so p(k) = 0. But this contradicts the assumption that the a i are the only roots greater than 1. Hence there are at least 2m - 1 (which is ≥ 2n - 1) roots. Problem A6

A, B, C are lattice points in the plane. The triangle ABC contains exactly one lattice point, X, in its interior. The line AX meets BC at E. What is the largest possible value of AX/XE? Solution Answer: 5.

Take A = (2, 4), B = (0, 0), C = (2, 1). This contains a single lattice point X (1, 1). B is a distance 2 from the line AC, and AC = 3, so ABC has area 3. B is a distance 1 from the line CX, and CX = 1, so XBC has area 1/2. So A is 6x the distance of X from BC. Hence AX = 5 XE. So the value 5 can certainly be achieved.

Let L, M, N be the midpoints of BC, CA, AB. Let B have the coordinates (b 1, b 2) and similarly for the other points. Take B' to be such that X is the midpoint of BB'. Then it is also a lattice point, since its coordinates are (2x 1 - b 1, 2x 2 - b 2). So it cannot lie in the interior of the triangle ABC (since X is the only such lattice point). So X cannot lie in the interior of the triangle BNL (which has dimensions half that of BAC and is similar to it). Similarly, X cannot lie in the interior of CLM.

Now consider the point L' on the ray LX such that LL' = 3LX. It is also a lattice point, since its coordinates are (3x 1 - (b 1 + c 1), 3x 2 - (b 2 + c 2) ). So it cannot lie in the interior of ABC. A fortiori, it cannot lie in the interior of LMN. Take M' on LM such that LM' = 1/3 LM, and N' on LN such that LN' = 1/3 LN. Then LM'N'

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is similar to LMN and 1/3 the dimensions. So X cannot lie inside LM'N' (otherwise L' would lie within LMN).

It follows that X must lie above the extended line M'N' (which is parallel to BC). But the distance of this line above BC is 1/6 the distance of A above BC. So if the line AX cuts it at F (and BC at E), then AF/FE = 5. Hence AX/XE ≤ 5. Problem B1

Evaluate lim n→∞ 1/n 5

∑ (5 r 4

- 18 r 2

s 2

+ 5 s 4

), where the sum is over all r, s satisfying 0 < r, s ≤ n.

Solution

Answer: - 1.

This seems curiously difficult for a qu 1, which makes me think I am missing something!

We need: ∑1n r 2 = 1/3 n 3 + 1/2 n 2

+ O(n), and ∑1n

r 4

= 1/5 n 5

+ 1/2 n 4

+ O(n 3

).

Applying these we get that the sum given is 5 (1/5 n 5

+ 1/2 n 4

+ O(n 3

) ) n - 18 (1/3 n 3 + 1/2 n 2 + O(n) )2 + 5 (1/5 n 5 + 1/2 n 4 + O(n 3) ) n = (n 6 + 5/2 n 5 + O(n 4) ) - (2n 6

+ 6n 5 + O(n 4) ) + (n 6 + 5/2 n 5 + O(n 4) ) = - n 5 + O(n 4

). So the required limit is - 1.

The tricky part is proving the relation for ∑ r 4

. A fairly trivial induction gives

that ∑ r(r+1)(r+2)...(r+s) = 1/(s+2) r(r+1)...(r+s+1). Hence ∑ r k = 1/(k+1) n k+1

+ O(n k ). Also ∑(r 4 + 6r 3 + O(r 2) ) = 1/5 (n 5 + 10 n 4 +O(n 3) ). Hence ∑ r 4 = 1/5 n 5

+ 2n 4 + O(n 3) - 6(1/4 n 4 + O(n 3) ) - ∑ O(n 2) = 1/5 n 5 + 1/2 n 4 + O(n 3

). The relation

for ∑ r 2

can be obtained similarly, although most people probably remember the exact formula 1/6 n(n+1)(2n+1), which is easy to prove by induction.

Problem B2

What is the minimum value of (a - 1)2

+ (b/a - 1)2

+ (c/b - 1)2

+ (4/c - 1)2

, over all real numbers a, b, c satisfying 1 ≤ a ≤ b ≤ c ≤ 4. Solution

Answer: 12 - 8√2.

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Let A = a, B = b/a, C = c/b, D = 4/c. Then we require the minimum of (A-1)2 + (B-1)2

+ (C-1)2 + (D-1)2

subject to ABCD = 4 and A, B, C, D >= 1.

We consider first the simpler case of minimising (A-1)2

+ (B-1)2

subject to AB = k 2

and A, B > 1. We show that it is achieved by taking A = B. This is routine. Set f(x)

= (x - 1)2 + (k 2/x - 1)2 and set f'(x) = 0, leading to a quartic (x - k)(x + k)(x 2

- x + k 2

) = 0 with real roots k, -k. Note that f'(x) < 0 for x just less than k, and > 0 for x just greater than k, so x = k is a minimum.

Now it follows that all four of A, B, C, D must be equal. For if any two were unequal, we could keep the others fixed and reduce the sum of squares by equalising the unequal pair whilst keeping their product fixed. Thus the minimum value is achieved with A = B = C = D = √2, which gives value 12 - 8√2. Problem B3

Prove that infinitely many positive integers n have the property that for any prime

p dividing n 2 + 3, we can find an integer m such that (1) p divides m 2

+ 3, and (2) m 2

< n. Solution

I started by finding some small n with the required property. This led to: (1) 52

+ 3 = 227, with 7 | 22 + 3, 2 | 12 + 3; (2) 122 + 1 = 723, with 3 | 32 + 3; (3) 232

+ 1 = 227 19 with 192 | 42

+ 3.

That suggested as the first line of attack looking at special n such as 2a 7b

. That led nowhere.

My second line of attack was to look for relations of the type (a 2

+ 3)(b 2

+ 3) =

f(a, b)2 + 3. After some playing around, I obtained (m 2 + 3)( (m+1)2 + 3) = M 2

+ 3,

where M = (m 2 + m + 3). We are almost there. Any prime factor of M 2

+ 3 which divides m 2 + 3 has the required property since m 2

< M. But that is not true for prime factors

of (m+1)2

+ 3. However, all we have to do is to iterate. We need to split the larger factor:

(m 2

+ 3)( (m+1)2 + 3)( (m 2+m+2)2

+ 3) = ( (m 2

+m+2)2

+ (m 2

+m+2) + 3)2

+ 3.

This shows that any n of the form ( (m 2

+m+2)2

+ (m 2

+m+2) + 3) has the required property. Problem B4

A is a set of 5 x 7 real matrices closed under scalar multiplication and addition. It contains matrices of ranks 0, 1, 2, 4 and 5. Does it necessarily contain a matrix of rank 3?

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Answer: no.

The 5 x 7 is something of a red herring. Note that we would expect the answer to be no, because addition and scalar multiplication do not impose any mixing on the matrix elements.

Consider the 5 x 5 matrix with a in the first 4 positions on the diagonal, c is the last position, b at positions (4,5) and (5,4) and zeros elsewhere. Taking (a, b, c) = (1, 0, 1), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0) gives matrices of rank 5, 4, 2, 1, 0 respectively. If a = 0, then all the entries in rows 1, 2 and 3 are zero, so the rank is at most 2. If a is non-zero, then there is certainly a 4 x 4 unit submatrix, so the rank is at least 4. Thus no member of the set has rank 3. If we add two columns of zeros to every member of the set, then we get a counter-example for the 5 x 7 case. Problem B5

f(n) is the number of 1s in the base 2 representation of n. Let k = ∑ f(n) / (n + n 2), where the sum is taken over all positive integers. Is e k

rational?

Solution Answer: yes.

Obviously f(2n+1) = f(2n) + 1 (because the base 2 representation for 2n+1 is the same as that for 2n with an additional 1 at the end). Note also that f(2n) = f(n).

Let us look at the adjacent terms f(2n) /( 2n(2n + 1) ), and f(2n + 1) / ( (2n+1)(2n+2) ). They sum to 1/ ( (2n+1)(2n+2) ) + 2 f(2n) / ( 2n(2n+2) ) = 1/ ( (2n+1)(2n+2) ) +

1/2 f(n) / ( n(n+1) ). Thus we get k = f(1)/2 + ∑1∞

( 1/ ( (2n+1)(2n+2) ) + 1/2 f(n) / ( n(n+1) ) = (1 - 1/2 + 1/3 - 1/4 + ... ) + 1/2 k [since 1/( (2n+1)(2n+2) )

= 1/(2n+1) - 1/(2n+2) ]. Thus k = 2 ln 2 = ln 4. So e k

= 4.

Problem B6

Let P be a convex polygon each of whose sides touches a circle C of radius 1. Let A be the set of points which are a distance 1 or less from P. If (x, y) is a point of A, let f(x, y) be the number of points in which a unit circle center (x, y) intersects P (so certainly f(x, y) ≥ 1). What is sup 1/|A| ∫A f(x, y) dx dy, where the sup is taken over all possible polygons P?

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Answer: 8/3.

Construct a rectangle on each side of P, with the other dimension of each rectangle 1. Between each adjacent pair of rectangles construct a sector radius 1, center the common vertex. Then A comprises the interior of P, the rectangles and the sectors. By joining each vertex of P to the center of C we divide the interior of P into triangles, each with area 1/2 x length of side. So the area of P = k/2, where k is the perimeter of P. Obviously, the area of the rectangles is k and since the sectors form a circle their area is π. So |A| = π + 3k/2.

The center of a unit circle which cuts a segment length δx must lie in one of two crescents centered on δx, which have total area 4δx. We may now integrate around P to get the area 4k. In this integration points are counted multiple times if the unit circles centered on them intersect P multiple times. In fact, the integral is exactly ∫A f(x, y) dx dy. Note that most unit circles meeting P will meet it twice, but those with centres near a vertex may meet it 4 times.

Thus 1/|A| ∫A f(x, y) dx dy = 4k/(π + 3k/2) = 8/(3 + 2π/k). So clearly 8/3 is an upper bound. But we can make k arbitrarily large (by taking one vertex of P off to infinity), so 8/3 is also the least upper bound.

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AMC／AIME美国数学竞赛 试题真题

AMC／AIME美国数学竞赛试题真题 考试信息 AMC最新考试时间： ●2010年第26届AMC8于 11月16日，星期二 ●2011第12届AMC10A，第62届AMC12A 于2月8日，星期二 ●2011第12届AMC10B，第62届AMC12B 于2月23日，星期三 ●2011第29届AIME－1于3月17日，星期四 2011第29届AIME－2于3月30日，星期三 ●2009年AMC8考试情况

●2008年考试情况 AMC/AIME中国历程： 1983第1届AIME上海有76名同学获得参赛资格 1984年第2届AIME有110人获得参赛资格 1985年第3届AIME北京有118名同学获得参赛资格 1986年第4届AIME上海有154名同学获得参赛资格，我国首次参加IMO的上海向明中学吴思皓就是在第四届AIME中获得满分 1992年第10届AIME上海有一千多名同学获得参赛资格，其中格致中学潘毅明，交大附中张觉，上海中学葛建庆均获满分1993年第11届AIME上海有一千多名同学获得参赛资格，其中华东师大二附中高一王海栋，格致中学高二（女）黄静，市西中学高二张

亮，复旦附中高三韩志刚四人获得满分，前三名总分排名复旦附中41分，华东师大二附中41分，上海中学40分。 北京地区参加2006年AMC的共有7所市重点学校的842名学生，有515名学生获得参加AIME资格，其中，清华附中有61名学生参加AMC，45名学生获得AIME资格，20名学生获得荣誉奖章 据悉中国大陆以下地区可以报名参加考试： 北京地区：中国数学会奥林匹克委员会负责组织实施 长春地区、哈尔滨地区也有参加考试 在华举办的美国人子弟学校也有参加考试广州地区：《数学奥林匹克报》负责组织实施。 在中国大陆报名者就在中国大陆考试。考题采用英文版。 2009年AMC中国地区参赛学校一览表

2010年美国大学生数学建模竞赛B题一等奖

Summary Faced with serial crimes,we usually estimate the possible location of next crime by narrowing search area.We build three models to determine the geographical profile of a suspected serial criminal based on the locations of the existing crimes.Model One assumes that the crime site only depends on the average distance between the anchor point and the crime site.To ground this model in reality,we incorporate the geographic features G,the decay function D and a normalization factor N.Then we can get the geographical profile by calculating the probability density.Model Two is Based on the assumption that the choice of crime site depends on ten factors which is specifically described in Table5in this paper.By using analytic hierarchy process (AHP)to generate the geographical profile.Take into account these two geographical profiles and the two most likely future crime sites.By using mathematical dynamic programming method,we further estimate the possible location of next crime to narrow the search area.To demonstrate how our model works,we apply it to Peter's case and make a prediction about some uncertainties which will affect the sensitivity of the program.Both Model One and Model Two have their own strengths and weaknesses.The former is quite rigorous while it lacks considerations of practical factors.The latter takes these into account while it is too subjective in application. Combined these two models with further analysis and actual conditions,our last method has both good precision and operability.We show that this strategy is not optimal but can be improved by finding out more links between Model One and Model Two to get a more comprehensive result with smaller deviation. Key words:geographic profiling,the probability density,anchor point, expected utility

2018年美国“数学大联盟杯赛”(中国赛区)初赛三年级试卷及答案

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