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1990AIME 真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月12日上午11:33

1990AIME 真题及答案解析

答案解析请参考文末

Problem 1

The increasing sequence $2,3,5,6,7,10,11,ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence.

Problem 2

Find the value of $(52+6sqrt{43})^{3/2}-(52-6sqrt{43})^{3/2}$ .

Problem 3

Let $P_1^{}$ be a regular $r~mbox{gon}$ and $P_2^{}$ be a regular $s~mbox{gon}$ $(rgeq sgeq 3)$ such that each interior angle of $P_1^{}$ is $frac{59}{58}$ as large as each interior angle of $P_2^{}$ . What's the largest possible value of $s_{}^{}$ ?

Problem 4

Find the positive solution to

$frac 1{x^2-10x-29}+frac1{x^2-10x-45}-frac 2{x^2-10x-69}=0$

Problem 5

Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $frac{n}{75}$ .

Problem 6

A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that 25% of these fish are no longer in the lake on September 1 (because of death and emigrations), that 40% of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1?

Problem 7

A triangle has vertices $P_{}^{}=(-8,5)$ , $Q_{}^{}=(-15,-19)$ , and $R_{}^{}=(1,-7)$ . The equation of the bisector of $angle P$ can be written in the form $ax+2y+c=0_{}^{}$ . Find $a+c_{}^{}$ .

Problem 8

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:

1) The marksman first chooses a column from which a target is to be broken.

2) The marksman must then break the lowest remaining target in the chosen column.

If the rules are followed, in how many different orders can the eight targets be broken?

Problem 9

A fair coin is to be tossed $10_{}^{}$ times. Let $i/j^{}_{}$ , in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$ .

Problem 10

The sets $A = {z : z^{18} = 1}$ and $B = {w : w^{48} = 1}$ are both sets of complex roots of unity. The set $C = {zw : z in A ~ mbox{and} ~ w in B}$ is also a set of complex roots of unity. How many distinct elements are in $C^{}_{}$ ?

Problem 11

Someone observed that $6! = 8 cdot 9 cdot 10$ . Find the largest positive integer $n^{}_{}$ for which $n^{}_{}!$ can be expressed as the product of $n - 3_{}^{}$ consecutive positive integers.

Problem 12

A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form

$a + b sqrt{2} + c sqrt{3} + d sqrt{6},$ where $a^{}_{}$ , $b^{}_{}$ , $c^{}_{}$ , and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$ .

Problem 13

Let $T = {9^k : k ~ mbox{is an integer}, 0 le k le 4000}$ . Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit?

Problem 14

The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 sqrt{3}$ and $BC^{}_{} = 13 sqrt{3}$ . Diagonals $overline{AC}$ and $overline{BD}$ intersect at $P^{}_{}$ . If triangle $ABP^{}_{}$ is cut out and removed, edges $overline{AP}$ and $overline{BP}$ are joined, and the figure is then creased along segments $overline{CP}$ and $overline{DP}$ , we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.

AIME 1990 Problem 14.png

Problem 15

Find $a_{}^{}x^5 + b_{}y^5$ if the real numbers $a_{}^{}$ , $b_{}^{}$ , $x_{}^{}$ , and $y_{}^{}$ satisfy the equations $[ax + by = 3^{}_{},]$ $[ax^2 + by^2 = 7^{}_{},]$ $[ax^3 + by^3 = 16^{}_{},]$ $[ax^4 + by^4 = 42^{}_{}.]$

1990AIME 详细解析

Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than $500$ . This happens to be $23^2=529$ . Notice that there are $23$ squares and $8$ cubes less than or equal to $529$ , but $1$ and $2^6$ are both squares and cubes. Thus, there are $529-23-8+2=500$ numbers in our sequence less than $529$ . Magically, we want the $500th$ term, so our answer is the smallest non-square and non-cube less than $529$ , which is $boxed{528}$ .
Suppose that $52+6sqrt{43}$ is in the form of $(a + bsqrt{43})^2$ . FOILing yields that $52 + 6sqrt{43} = a^2 + 43b^2 + 2absqrt{43}$ . This implies that $a$ and $b$ equal one of $pm1, pm3$ . The possible sets are $(3,1)$ and $(-3,-1)$ ; the latter can be discarded since the square root must be positive. This means that $52 + 6sqrt{43} = (sqrt{43} + 3)^2$ . Repeating this for $52-6sqrt{43}$ , the only feasible possibility is $(sqrt{43} - 3)^2$ .Rewriting, we get $(sqrt{43} + 3)^3 - (sqrt{43} - 3)^3$ . Using the difference of cubes, we get that $[sqrt{43} + 3 - sqrt{43} + 3] [(43 + 6sqrt{43} + 9) + (43 - 9) + (43 - 6sqrt{43} + 9)]$ $= (6)(3 cdot 43 + 9) = boxed{828}$
The formula for the interior angle of a regular sided polygon is $frac{(n-2)180}{n}$ .Thus, $frac{frac{(r-2)180}{r}}{frac{(s-2)180}{s}} = frac{59}{58}$ . Cross multiplying and simplifying, we get $frac{58(r-2)}{r} = frac{59(s-2)}{s}$ . Cross multiply and combine like terms again to yield $58rs - 58 cdot 2s = 59rs - 59 cdot 2r Longrightarrow 118r - 116s = rs$ . Solving for $r$ , we get $r = frac{116s}{118 - s}$ . $r ge 0$ and $s ge 0$ , making the numerator of the fraction positive. To make the denominator positive, $s < 118$ ; the largest possible value of $s$ is $117$ .This is achievable because the denominator is $1$ , making $r$ a positive number $116 cdot 117$ and $s = boxed{117}$ .
We could clear out the denominators by multiplying, though that would be unnecessarily tedious.To simplify the equation, substitute $a = x^2 - 10x - 29$ (the denominator of the first fraction). We can rewrite the equation as $frac{1}{a} + frac{1}{a - 16} - frac{2}{a - 40} = 0$ . Multiplying out the denominators now, we get: $[(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0]$ Simplifying, $-64a + 40 times 16 = 0$ , so $a = 10$ . Re-substituting, $10 = x^2 - 10x - 29 Longleftrightarrow 0 = (x - 13)(x + 3)$ . The positive root is $boxed{013}$ .
The prime factorization of $75 = 3^15^2 = (2+1)(4+1)(4+1)$ . For $n$ to have exactly $75$ integral divisors, we need to have $n = p_1^{e_1-1}p_2^{e_2-1}cdots$ such that $e_1e_2 cdots = 75$ . Since $75|n$ , two of the prime factors must be $3$ and $5$ . To minimize $n$ , we can introduce a third prime factor, $2$ . Also to minimize $n$ , we want $5$ , the greatest of all the factors, to be raised to the least power. Therefore, $n = 2^43^45^2$ and $frac{n}{75} = frac{2^43^45^2}{3 cdot 5^2} = 16 cdot 27 = boxed{432}$ .
Of the $70$ fish caught in September, $40%$ were not there in May, so $42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $frac{3}{42} = frac{60}{x} Longrightarrow boxed{x = 840}$ .(Note the 25% death rate does not affect the answer because both tagged and nontagged fish die.)
Use the distance formula to determine the lengths of each of the sides of the triangle. We find that it has lengths of side $15, 20, 25$ , indicating that it is a $3-4-5$ right triangle. At this point, we just need to find another point that lies on the bisector of $angle P$ . $[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]$ Use the angle bisector theorem to find that the angle bisector of $angle P$ divides $QR$ into segments of length $frac{25}{x} = frac{15}{20 -x} Longrightarrow x = frac{25}{2}, frac{15}{2}$ . It follows that $frac{QP'}{RP'} = frac{5}{3}$ , and so $P' = left(frac{5x_R + 3x_Q}{8},frac{5y_R + 3y_Q}{8}right) = (-5,-23/2)$ .The desired answer is the equation of the line $PP'$ . $PP'$ has slope $frac{-11}{2}$ , from which we find the equation to be $11x + 2y + 78 = 0$ . Therefore, $a+c = boxed{089}$ .
Suppose that the columns are labeled $A$ , $B$ , and $C$ . Consider the string $AAABBBCC$ . Since the arrangements of the strings is bijective to the order of shooting, the answer is the number of ways to arrange the letters which is $frac{8!}{3! cdot 3! cdot 2!} = boxed{560}$ .
Clearly, at least $5$ tails must be flipped; any less, then by the Pigeonhole Principle there will be heads that appear on consecutive tosses.Consider the case when $5$ tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled $(H)$ :

$(H) T (H) T (H) T (H) T (H) T (H)$

There are six slots for the heads to be placed, but only $5$ heads remaining. Thus, using stars-and-bars there are ${6choose5}$ possible combinations of 6 heads. Continuing this pattern, we find that there are $sum_{i=6}^{11} {ichoose{11-i}} = {6choose5} + {7choose4} + {8choose3} + {9choose2} + {{10}choose1} + {{11}choose0} = 144$ . There are a total of $2^{10}$ possible flips of $10$ coins, making the probability $frac{144}{1024} = frac{9}{64}$ . Thus, our solution is $9 + 64 = boxed{073}$ .

Call the number of ways of flipping $n$ coins and not receiving any consecutive heads $S_n$ . Notice that tails must be received in at least one of the first two flips.

If the first coin flipped is a T, then the remaining $n-1$ flips must fall under one of the configurations of $S_{n-1}$ .

If the first coin flipped is a H, then the second coin must be a T. There are then $S_{n-2}$ configurations.

Thus, $S_n = S_{n-1} + S_{n-2}$ . By counting, we can establish that $S_1 = 2$ and $S_2 = 3$ . Therefore, $S_3 = 5, S_4 = 8$ , forming the Fibonacci sequence. Listing them out, we get $2,3,5,8,13,21,34,55,89,144$ , and the 10th number is $144$ . Putting this over $2^{10}$ to find the probability, we get $frac{9}{64}$ . Our solution is $9+64=boxed{073}$ .
The least common multiple of $18$ and $48$ is $144$ , so define $n = e^{2pi i/144}$ . We can write the numbers of set $A$ as ${n^8, n^{16}, ldots n^{144}}$ and of set $B$ as ${n^3, n^6, ldots n^{144}}$ . $n^x$ can yield at most $144$ different values. All solutions for $zw$ will be in the form of $n^{8k_1 + 3k_2}$ . $8$ and $3$ are relatively prime, and by the Chicken McNugget Theorem, for two relatively prime integers $a,b$ , the largest number that cannot be expressed as the sum of multiples of $a,b$ is $a cdot b - a - b$ . For $3,8$ , this is $13$ ; however, we can easily see that the numbers $145$ to $157$ can be written in terms of $3,8$ . Since the exponents are of roots of unities, they reduce $mod{144}$ , so all numbers in the range are covered. Thus the answer is $boxed{144}$ .
The product of $n - 3$ consecutive integers can be written as $frac{(n - 3 + a)!}{a!}$ for some integer $a$ . Thus, $n! = frac{(n - 3 + a)!}{a!}$ , from which it becomes evident that $a ge 3$ . Since $(n - 3 + a)! > n!$ , we can rewrite this as $frac{n!(n+1)(n+2) ldots (n-3+a)}{a!} = n! Longrightarrow (n+1)(n+2) ldots (n-3+a) = a!$ . For $a = 4$ , we get $n + 1 = 4!$ so $n = 23$ . For greater values of $a$ , we need to find the product of $a-3$ consecutive integers that equals $a!$ . $n$ can be approximated as $^{a-3}sqrt{a!}$ , which decreases as $a$ increases. Thus, $n = 23$ is the greatest possible value to satisfy the given conditions.
The easiest way to do this seems to be to find the length of each of the sides and diagonals. To do such, draw the radii that meet the endpoints of the sides/diagonals; this will form isosceles triangles. Drawing the altitude of those triangles and then solving will yield the respective lengths.
- The length of each of the 12 sides is $2 cdot 12sin 15$ . $24sin 15 = 24sin (45 - 30) = 24frac{sqrt{6} - sqrt{2}}{4} = 6(sqrt{6} - sqrt{2})$ .
- The length of each of the 12 diagonals that span across 2 edges is $2 cdot 12sin 30 = 12$ (or notice that the triangle formed is equilateral).
- The length of each of the 12 diagonals that span across 3 edges is $2 cdot 12sin 45 = 12sqrt{2}$ (or notice that the triangle formed is a $45 - 45 - 90$ right triangle).
- The length of each of the 12 diagonals that span across 4 edges is $2 cdot 12sin 60 = 12sqrt{3}$ .
- The length of each of the 12 diagonals that span across 5 edges is $2 cdot 12sin 75 = 24sin (45 + 30) = 24frac{sqrt{6}+sqrt{2}}{4} = 6(sqrt{6}+sqrt{2})$ .
- The length of each of the 6 diameters is $2 cdot 12 = 24$ .
Adding all of these up, we get $12[6(sqrt{6} - sqrt{2}) + 12 + 12sqrt{2} + 12sqrt{3} + 6(sqrt{6}+sqrt{2})] + 6 cdot 24$

$= 12(12 + 12sqrt{2} + 12sqrt{3} + 12sqrt{6}) + 144 = 288 + 144sqrt{2} + 144sqrt{3} + 144sqrt{6}$ . Thus, the answer is $144 cdot 5 = boxed{720}$ .
Since $9^{4000}$ has 3816 digits more than $9^1$ , $4000 - 3816 = boxed{184}$ numbers have 9 as their leftmost digits.Let's divide all elements of $T$ into sections. Each section ranges from $10^{i}$ to $10^{i+1}$ And, each section must have 1 or 2 elements. So, let's consider both cases.If a section has 1 element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be $9^{k}$ and the section ranges from $10^{i}$ to $10^{i+1}$ . To the contrary, let's assume the number ( $9^{k}$ ) does have 9 as the leftmost digit. Thus, $9 cdot 10^i leq 9^k$ . But, if you divide both sides by 9, you get $10^i leq 9^{k-1}$ , and because $9^{k-1} < 9^{k} < 10^{i+1}$ , so we have another number ( $9^{k-1}$ ) in the same section ( $10^i leq 9^{k-1} < 10^{i+1}$ ). Which is a contradiction to our assumption that the section only has 1 element. So in this case, the number doesn't have 9 as the leftmost digit.If a section has 2 elements, we claim one has to have a 9 as the leftmost digit, one doesn't. Let the elements be $9^{k}$ and $9^{k+1}$ , and the section ranges from $10^{i}$ to $10^{i+1}$ . So We know $10^{i} leq 9^{k} < 9^{k+1} < 10^{i+1}$ . From $10^{i} leq 9^{k}$ , we know $9 cdot 10^{i} leq 9^{k+1}$ , and since $9^{k+1} < 10^{i+1}$ . The number( $9^{k+1}$ )'s leftmost digit must be 9, and the other number( $9^{k}$ )'s leftmost digit is 1.There are total 4001 elements in $T$ and 3817 sections that have 1 or 2 elements. And, no matter how many elements a section has, each section contains exactly one element that doesn't begin with 9. We can take 4001 elements, subtract 3817 elements that don't have 9 as the leftmost digit, and get $boxed{184}$ numbers that have 9 as the leftmost digit.
$[asy] import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, (99/133)^.5); Pa=(P.x,P.y,0); draw(P--Pa--A); draw(C--Pa--D); draw((C+D)/2--A--C--D--P--C--P--A--D); label("A", A, NE); label("P", P, N); label("C", C); label("D", D, S); label("$13sqrt{3}$", (A+D)/2, E, small); label("$13sqrt{3}$", (A+C)/2, N, small); label("$12sqrt{3}$", (C+D)/2, SW, small); [/asy]$ Our triangular pyramid has base $12sqrt{3} - 13sqrt{3} - 13sqrt{3} triangle$ . The area of this isosceles triangle is easy to find by $[ACD] = frac{1}{2}bh$ , where we can find $h_{ACD}$ to be $sqrt{399}$ by the Pythagorean Theorem. Thus $A = frac 12(12sqrt{3})sqrt{399} = 18sqrt{133}$ . $[asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label("A", A, NE); label("C", C, NW); label("D", D, S); label("P",P , N); label("P$'$", Pa, SW); label("$13sqrt{3}$", (A+D)/2, E, small); label("$13sqrt{3}$", (A+C)/2, NW, small); label("$12sqrt{3}$", (C+D)/2, SW, small); label("h", (P + Pa)/2, W); label("$frac{sqrt{939}}{2}$", (C+P)/2 ,NW); [/asy]$ To find the volume, we want to use the equation $frac 13Bh = 6sqrt{133}h$ , so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = frac{sqrt{939}}{2}$ . If we let $P$ be the center of a sphere with radius $frac{sqrt{939}}{2}$ , then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $triangle ACD$ .From here we just need to perform some brutish calculations. Using the formula $A = 18sqrt{133} = frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = frac{12sqrt{3} cdot (13sqrt{3})^2}{4cdot 18sqrt{133}} = frac{13^2sqrt{3}}{2sqrt{133}}$ (there are slightly simpler ways to calculate $R$ since we have an isosceles triangle). By the Pythagorean Theorem,
$begin{align*}h^2 &= PA^2 - R^2 \ &= left(frac{sqrt{939}}{2}right)^2 - left(frac{13^2sqrt{3}}{2sqrt{133}}right)^2\ &= frac{939 cdot 133 - 13^4 cdot 3}{4 cdot 133} = frac{13068 cdot 3}{4 cdot 133} = frac{99^2}{133}\ h &= frac{99}{sqrt{133}} end{align*}$

Finally, we substitute $h$ into the volume equation to find $V = 6sqrt{133}left(frac{99}{sqrt{133}}right) = boxed{594}$ .
Set $S = (x + y)$ and $P = xy$ . Then the relationship $[(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})]$ can be exploited: $begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)end{eqnarray*}$ Therefore:
$begin{eqnarray*}7S & = & 16 + 3P \ 16S & = & 42 + 7Pend{eqnarray*}$

Consequently, $S = - 14$ and $P = - 38$ . Finally:

$begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \ (42)(S) & = & (ax^5 + by^5) + (P)(16) \ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \ ax^5 + by^5 & = & boxed{020}end{eqnarray*}$