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1993AIME 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11:07

1993AIME 真题及答案解析

答案解析请参考文末

Problem 1

How many even integers between 4000 and 7000 have four different digits?

Problem 2

During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $frac{n^{2}}{2}$ miles on the $n^{mbox{th}}_{}$ day of this tour, how many miles was he from his starting point at the end of the $40^{mbox{th}}_{}$ day?

Problem 3

The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n,$ fish for various values of $n,$ .

$begin{array}{|c|c|c|c|c|c|c|c|c|} hline n & 0 & 1 & 2 & 3 & dots & 13 & 14 & 15 \ hline text{number of contestants who caught} n text{fish} & 9 & 5 & 7 & 23 & dots & 5 & 2 & 1 \ hline end{array}$ In the newspaper story covering the event, it was reported that

(a) the winner caught $15$ fish;

(b) those who caught $3$ or more fish averaged $6$ fish each;

What was the total number of fish caught during the festival?

Problem 4

How many ordered four-tuples of integers $(a,b,c,d),$ with $0 < a < b < c < d < 500,$ satisfy $a + d = b + c,$ and $bc - ad = 93,$ ?

Problem 5

Let $P_0(x) = x^3 + 313x^2 - 77x - 8,$ . For integers $n ge 1,$ , define $P_n(x) = P_{n - 1}(x - n),$ . What is the coefficient of $x,$ in $P_{20}(x),$ ?

Problem 6

What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

Problem 7

Three numbers, $a_1,$ , $a_2,$ , $a_3,$ , are drawn randomly and without replacement from the set ${1, 2, 3, dots, 1000},$ . Three other numbers, $b_1,$ , $b_2,$ , $b_3,$ , are then drawn randomly and without replacement from the remaining set of 997 numbers. Let $p,$ be the probability that, after a suitable rotation, a brick of dimensions $a_1 times a_2 times a_3,$ can be enclosed in a box of dimensions $b_1 times b_2 times b_3,$ , with the sides of the brick parallel to the sides of the box. If $p,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Problem 8

Let $S,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S,$ so that the union of the two subsets is $S,$ ? The order of selection does not matter; for example, the pair of subsets ${a, c},$ , ${b, c, d, e, f},$ represents the same selection as the pair ${b, c, d, e, f},$ , ${a, c},$ .

Problem 9

Two thousand points are given on a circle. Label one of the points 1. From this point, count 2 points in the clockwise direction and label this point 2. From the point labeled 2, count 3 points in the clockwise direction and label this point 3. (See figure.) Continue this process until the labels $1,2,3dots,1993,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as 1993?

Problem 10

Euler's formula states that for a convex polyhedron with $V,$ vertices, $E,$ edges, and $F,$ faces, $V-E+F=2,$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its $V,$ vertices, $T,$ triangular faces and $P^{}_{}$ pentagonal faces meet. What is the value of $100P+10T+V,$ ?

Problem 11

Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n,$ , where $m,$ and $n,$ are relatively prime positive integers. What are the last three digits of $m+n,$ ?

Problem 12

The vertices of $triangle ABC$ are $A = (0,0),$ , $B = (0,420),$ , and $C = (560,0),$ . The six faces of a die are labeled with two $A,$ 's, two $B,$ 's, and two $C,$ 's. Point $P_1 = (k,m),$ is chosen in the interior of $triangle ABC$ , and points $P_2,$ , $P_3,$ , $P_4, dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L,$ , where $L in {A, B, C}$ , and $P_n,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $overline{P_n L}$ . Given that $P_7 = (14,92),$ , what is $k + m,$ ?

Problem 13

Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Problem 14

A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form $sqrt{N},$ , for a positive integer $N,$ . Find $N,$ .

Problem 15

Let $overline{CH}$ be an altitude of $triangle ABC$ . Let $R,$ and $S,$ be the points where the circles inscribed in the triangles $ACH,$ and $BCH^{}_{}$ are tangent to $overline{CH}$ . If $AB = 1995,$ , $AC = 1994,$ , and $BC = 1993,$ , then $RS,$ can be expressed as $m/n,$ , where $m,$ and $n,$ are relatively prime integers. Find $m + n,$ .

1993AIME详细解析

The thousands digit is $in {4,5,6}$ . If the thousands digit is even ( $4, 6$ , 2 possibilities), then there are only $frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 cdot 4 cdot 8 cdot 7 = 448$ .If the thousands digit is odd ( $5$ , one possibility), then there are $5$ choices for the units digit, with $8$ digits for the hundreds and $7$ for the tens place. This gives $1 cdot 5 cdot 8 cdot 7 = 280$ possibilities. Together, the solution is $448 + 280 = boxed{728}$ .
On the first day, the candidate moves $[4(0) + 1]^2/2 text{east},, [4(0) + 2]^2/2 text{north},, [4(0) + 3]^2/2 text{west},, [4(0) + 4]^2/2 text{south}$ , and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 ldots +37^2 - 39^2 = left|sum_{i=0}^9 frac{(4i+1)^2}{2} - sum_{i=0}^9 frac{(4i+3)^2}{2}right|$ . Applying difference of squares, we see that $begin{align*} left|sum_{i=0}^9 frac{(4i+1)^2 - (4i+3)^2}{2}right| &= left|sum_{i=0}^9 frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}right|\ &= left|sum_{i=0}^9 -(8i+4) right|. end{align*}$ The N/S displacement is $[left|sum_{i=0}^9 frac{(4i+2)^2}{2} - sum_{i=0}^9 frac{(4i+4)^2}{2}right| = left|sum_{i=0}^9 -(8i+6) right|.]$ Since $sum_{i=0}^{9} i = frac{9(10)}{2} = 45$ , the two distances evaluate to $8(45) + 10cdot 4 = 400$ and $8(45) + 10cdot 6 = 420$ . By the Pythagorean Theorem, the answer is $sqrt{400^2 + 420^2} = 29 cdot 20 = boxed{580}$ .
Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-21$ fishers that caught 3 or more fish caught a total of $x - left(0(9) + 1(5) + 2(7)right) = x - 19$ fish. Since they averaged 6 fish, $6 = frac{x - 19}{y - 21} Longrightarrow x - 19 = 6y - 126.$ Similarily, those whom caught 12 or fewer fish averaged 5 fish per person, so $5 = frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = frac{x - 108}{y - 8} Longrightarrow x - 108 = 5y - 40.$ Solving the two equation system, we find that $y = 175$ and $x = boxed{943}$ , the answer.
Let $k = a + d = b + c$ so $d = k-a, b=k-c$ . It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$ . Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$ .Solve them in tems of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$ . The last two solutions don't follow $a < b < c < d$ , so we only need to consider the first two solutions.The first solution gives us $c - 93geq 1$ and $c + 1leq 499$ $implies 94leq cleq 498$ , and the second one gives us $32leq cleq 496$ .So the total number of such four-tuples is $405 + 465 = huge{boxed{870}}$ .
Notice thatUsing the formula for the sum of the first numbers, . Therefore,Substituting into the function definition, we get . We only need the coefficients of the linear terms, which we can find by the binomial theorem.
- $(x-210)^3$ will have a linear term of ${3choose1}210^2x = 630 cdot 210x$ .
- $313(x-210)^2$ will have a linear term of $-313 cdot {2choose1}210x = -626 cdot 210x$ .
- $-77(x-210)$ will have a linear term of $-77x$ .
Adding up the coefficients, we get $630 cdot 210 - 626 cdot 210 - 77 = boxed{763}$ .
Denote the first of each of the series of consecutive integers as $a, b, c$ . Therefore, $n = a + (a + 1) ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55$ . Simplifying, $9a = 10b + 9 = 11c + 19$ . The relationship between $a, b$ suggests that $b$ is divisible by $9$ . Also, $10b -10 = 10(b-1) = 11c$ , so $b-1$ is divisible by $11$ . We find that the least possible value of $b = 45$ , so the answer is $10(45) + 45 = 495$ .
Call the six numbers selected . Clearly, must be a dimension of the box, and must be a dimension of the brick.
- If $x_2$ is a dimension of the box, then any of the other three remaining dimensions will work as a dimension of the box. That gives us $3$ possibilities.
- If $x_2$ is not a dimension of the box but $x_3$ is, then both remaining dimensions will work as a dimension of the box. That gives us $2$ possibilities.
- If $x_4$ is a dimension of the box but $x_2, x_3$ aren’t, there are no possibilities (same for $x_5$ ).
The total number of arrangements is ${6choose3} = 20$ ; therefore, $p = frac{3 + 2}{20} = frac{1}{4}$ , and the answer is $1 + 4 = boxed{005}$ .

Note that the $1000$ in the problem, is not used, and is cleverly bypassed in the solution, because we can call our six numbers $x_1,x_2,x_3,x_4,x_5,x_6$ whether they may be $1,2,3,4,5,6$ or $999,5,3,998,997,891$ .
Call the two subsets $m$ and $n$ . For each of the elements in $S$ , we can assign it to either $m$ , $n$ , or both. This gives us $3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $m$ and $n$ contain all $6$ elements of $S$ . So our final answer is then $frac {3^6 - 1}{2} + 1 = boxed{365}$
The label $1993$ will occur on the $frac12(1993)(1994) pmod{2000}$ th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $frac12(n)(n + 1)equiv frac12(1993)(1994) pmod{2000}$ .Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)equiv 0pmod{2000}$ . Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$ .For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$ , $n equiv 118 pmod {125}$ and $nequiv 6 pmod {16}$ . The smallest $n$ for this case is $118$ .In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$ , $nequiv 9pmod{16}$ and $nequiv 6pmod{125}$ . The smallest $n$ for this case is larger than $118$ , so $boxed{118}$ is our answer.
Note: One can just substitute $1993equiv-7pmod{2000}$ and $1994equiv-6pmod{2000}$ to simplify calculations.
The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=frac{3t+5p}{2}$ , (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$ . Finally, using Euler's formula we have $V=E-30=30$ .In summary, the solution to the problem is $100P+10T+V=boxed{250}$ .
The probability that the $n$ th flip in each game occurs and is a head is $frac{1}{2^n}$ . The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is $frac{1}{2}+frac{1}{8}+frac{1}{32}+cdots = frac{frac{1}{2}}{1-frac{1}{4}}=frac{2}{3}$ , and the probability of the second person winning is $frac{1}{3}$ .Let $a_n$ be the probability that Alfred wins the $n$ th game, and let $b_n$ be the probability that Bonnie wins the $n$ th game.If Alfred wins the $n$ th game, then the probability that Alfred wins the $n+1$ th game is $frac{1}{3}$ . If Bonnie wins the $n$ th game, then the probability that Alfred wins the $n+1$ th game is $frac{2}{3}$ .Thus, $a_{n+1}=frac{1}{3}a_n+frac{2}{3}b_n$ .
Similarly, $b_{n+1}=frac{2}{3}a_n+frac{1}{3}b_n$ .

Since Alfred goes first in the $1$ st game, $(a_1,b_1)=left(frac{2}{3}, frac{1}{3}right)$ .

Using these recursive equations:

$(a_2,b_2)=left(frac{4}{9}, frac{5}{9}right)$

$(a_3,b_3)=left(frac{14}{27}, frac{13}{27}right)$

$(a_4,b_4)=left(frac{40}{81}, frac{41}{81}right)$

$(a_5,b_5)=left(frac{122}{243}, frac{121}{243}right)$

$(a_6,b_6)=left(frac{364}{729}, frac{365}{729}right)$

Since $a_6=frac{364}{729}$ , $m+n = 1093 equiv boxed{093} pmod{1000}$ .
If we have points (p,q) and (r,s) and we want to find (u,v) so (r,s) is the midpoint of (u,v) and (p,q), then u=2r-p and v=2s-q. So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have: $P_7=(14,92) P_6=(2cdot14-0, 2cdot92-0)=(28,184) P_5=(2cdot28-0, 2cdot 184-0)=(56,368) P_4=(2cdot56-0, 2cdot368-420)=(112,316) P_3=(2cdot112-0, 2cdot316-420)=(224,212) P_2=(2cdot224-0, 2cdot212-420)=(448,4) P_1=(2cdot448-560, 2cdot4-0)=(336,8)$ So the answer is $boxed{344}$ .
Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$ , they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are $begin{align}y&=-frac{100}{t}x+200-frac{5000}{t}\50^2&=x^2+y^2end{align}.$ When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get $[-frac{x}{y}=-frac{100}{t}]$ or $xt=100y.$ Now substitute $[y= frac{xt}{100}]$ into $(2)$ and get $[x=frac{5000}{sqrt{100^2+t^2}}.]$ Now substitute this and $[y=frac{xt}{100}]$ into $(1)$ and solve for $t$ to get $[t=frac{160}{3}.]$ Finally, the sum of the numerator and denominator is $160+3=boxed{163}.$
Answer: 448.Solution: Put the rectangle on the coordinate plane so its vertices are at $(pm4,pm3)$ , for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$ .Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be $A(4,y)$ , $B(-x,3)$ , $C(-4,-y)$ and $D(x,-3)$ for nonnegative $x,y$ . Then this is a rectangle, so $OA=OB$ , or $16+y^2=9+x^2$ , so $x^2=y^2+7$ .

Reflect $D$ across the side of the rectangle containing $C$ to $D'(-8-x,-3)$ . Then $BD'=sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10$ is constant, and the perimeter of the rectangle is equal to $2(BC+CD')$ . The midpoint of $overline{BD'}$ is $(-4-x,0)$ , and since $-4>-4-x$ and $-yle0$ , $C$ always lies below $overline{BD'}$ .

If $y$ is positive, it can be decreased to $y'<y$ . This causes $x$ to decrease as well, to $x'$ , where $x'^2=y'^2+7$ and $x'$ is still positive. If $B$ and $D'$ are held in place as everything else moves, then $C$ moves $(y-y')$ units up and $(x-x')$ units left to $C'$ , which must lie within $triangle BCD'$ . Then we must have $BC'+C'D'<BC+CD'$ , and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with $y=0$ , so $x=sqrt7$ .

By the distance formula, this minimum perimeter is $[2left(sqrt{(4-sqrt7)^2+3^2}+sqrt{(4+sqrt7)^2+3^2}right)=4left(sqrt{8-2sqrt7}+sqrt{8+2sqrt7}right)]$ $[=4(sqrt7-1+sqrt7+1)=8sqrt7=sqrt{448}.]$
$[asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label("$R$",y[0],NE); label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S); [/asy]$ From the Pythagorean Theorem, $AH^2+CH^2=1994^2$ , and $(1995-AH)^2+CH^2=1993^2$ . Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$ . After simplification, we see that $2*1995AH-1995^2=3987$ , or $AH=frac{1995}{2}+frac{3987}{2*1995}$ . Note that $AH+BH=1995$ . Therefore we have that $BH=frac{1995}{2}-frac{3987}{2*1995}$ . Therefore $AH-BH=frac{3987}{1995}$ .Now note that $RS=|HR-HS|$ , $RH=frac{AH+CH-AC}{2}$ , and $HS=frac{CH+BH-BC}{2}$ . Therefore we have $RS=left| frac{AH+CH-AC-CH-BH+BC}{2} right|=frac{|AH-BH-1994+1993|}{2}$ .
Plugging in $AH-BH$ and simplifying, we have $RS=frac{1992}{1995*2}=frac{332}{665} rightarrow 332+665=boxed{997}$ .

Edit by GameMaster402: It can be shown that in any triangle with side lengths $n-1, n, n+1$ , if you draw an altitude from the vertex to the side of $n+1$ , and draw the incircles of the two right triangles, the distance between the two tangency points is simply $frac{n-2}{2n+2}$ Plugging in $n=1994$ yields that the answer is $1992/1995*2$ , which simplifies to $332/665$