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1998AIME 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10:26

1998AIME 真题及答案解析

答案解析请参考文末

Problem 1

For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ and $8^8$ , and $k$ ?

Problem 2

Find the number of ordered pairs $(x,y)$ of positive integers that satisfy $x le 2y le 60$ and $y le 2x le 60$ .

Problem 3

The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions. What is the area of the bounded region?

Problem 4

Nine tiles are numbered $1, 2, 3, cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Problem 5

Given that $A_k = frac {k(k - 1)}2cosfrac {k(k - 1)pi}2,$ find $|A_{19} + A_{20} + cdots + A_{98}|.$

Problem 6

Let $ABCD$ be a parallelogram. Extend $overline{DA}$ through $A$ to a point $P,$ and let $overline{PC}$ meet $overline{AB}$ at $Q$ and $overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$

Problem 7

Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $sum_{i = 1}^4 x_i = 98.$ Find $frac n{100}.$

Problem 8

Except for the first two terms, each term of the sequence $1000, x, 1000 - x,ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encountered. What positive integer $x$ produces a sequence of maximum length?

Problem 9

Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly $m$ minutes. The probability that either one arrives while the other is in the cafeteria is $40 %,$ and $m = a - bsqrt {c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a + b + c.$

Problem 10

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +bsqrt {c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a + b + c$ .

Problem 11

Three of the edges of a cube are $overline{AB}, overline{BC},$ and $overline{CD},$ and $overline{AD}$ is an interior diagonal. Points $P, Q,$ and $R$ are on $overline{AB}, overline{BC},$ and $overline{CD},$ respectively, so that $AP = 5, PB = 15, BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is the intersection of plane $PQR$ and the cube?

Problem 12

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $overline{BC}, overline{CA},$ and $overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $overline{DE}, overline{EF},$ and $overline{FD},$ respectively, with the property that $P$ is on $overline{CQ}, Q$ is on $overline{AR},$ and $R$ is on $overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + bsqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2} + b^{2} + c^{2}$ ?

Problem 13

If ${a_1,a_2,a_3,ldots,a_n}$ is a set of real numbers, indexed so that $a_1 < a_2 < a_3 < cdots < a_n,$ its complex power sum is defined to be $a_1i + a_2i^2+ a_3i^3 + cdots + a_ni^n,$ where $i^2 = - 1.$ Let $S_n$ be the sum of the complex power sums of all nonempty subsets of ${1,2,ldots,n}.$ Given that $S_8 = - 176 - 64i$ and $S_9 = p + qi,$ where $p$ and $q$ are integers, find $|p| + |q|.$

Problem 14

An $mtimes ntimes p$ rectangular box has half the volume of an $(m + 2)times(n + 2)times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $mle nle p.$ What is the largest possible value of $p$ ?

Problem 15

Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $j$ . Let $D_{40}$ be the set of all dominos whose coordinates are no larger than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}.$

1998AIME 详细解析

It is evident that has only 2s and 3s in its prime factorization, or .
- $6^6 = 2^6cdot3^6$
- $8^8 = 2^{24}$
- $12^{12} = 2^{24}cdot3^{12}$
The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$ . Therefore $12^{12} = 2^{24}cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{max(24,a)}3^{max(6,b)}$ , and $b = 12$ . Since $0 le a le 24$ , there are $boxed{25}$ values of $k$ .
Pick's theorem states that:
$A = I + frac B2 - 1$

The conditions give us four inequalities: $x le 30$ , $yle 30$ , $x le 2y$ , $y le 2x$ . These create a quadrilateral, whose area is $frac 12$ of the 30 by 30 square it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square.

So $A = frac 12 cdot 30^2 = 450$ . $B$ we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 lattice points, and the two diagonals each have 14 lattice points (for the top diagonal, every value of $x$ corresponds with an integer value of $y$ as $y = 2x$ and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders.

$450 = I + frac {60}2 - 1$
$I = 421$

Since the inequalities also include the equals case, we include the boundaries, which gives us $421 + 60 = 481$ ordered pairs. However, the question asks us for positive integers, so $(0,0)$ doesn't count; hence, the answer is $480$ .
The equation given can be rewritten as:

$40|x| = - y^2 - 2xy + 400$

We can split the equation into a piecewise equation by breaking up the absolute value:

$40x = -y^2 - 2xy + 400quadquad xge 0$

$40x = y^2 + 2xy - 400 quad quad x < 0$

Factoring the first one: (alternatively, it is also possible to complete the square)

$40x + 2xy = -y^2 + 400$

$2x(20 + y)= (20 - y)(20 + y)$

Hence, either $y = -20$ , or $2x = 20 - y Longrightarrow y = -2x + 20$ .

Similarily, for the second one, we get $y = 20$ or $y = -2x - 20$ . If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is $boxed{800}$ .

The equation can be rewritten as: $(x+y)^2=(|x|-20)^2$ . Do casework as above.
In order for a player to have an odd sum, he must have an odd number of odd tiles: that is, he can either have three odd tiles, or two even tiles and an odd tile. Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ even tiles and $1$ odd tile. We count the number of ways this can happen. (We will count assuming that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.) $dbinom{5}{3} = 10$ choices for the tiles that he gets. The other two odd tiles can be distributed to the other two players in $2$ ways, and the even tiles can be distributed between them in $dbinom{4}{2} cdot dbinom{2}{2} = 6$ ways. This gives us a total of $10 cdot 2 cdot 6 = 120$ possibilities in which all three people get odd sums.In order to calculate the probability, we need to know the total number of possible distributions for the tiles. The first player needs three tiles which we can give him in $dbinom{9}{3} = 84$ ways, and the second player needs three of the remaining six, which we can give him in $dbinom{6}{3} = 20$ ways. Finally, the third player will simply take the remaining tiles in $1$ way. So, there are $dbinom{9}{3} cdot dbinom{6}{3} cdot 1 = 84 cdot 20 = 1680$ ways total to distribute the tiles.We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is $frac{360}{1680} = frac{3}{14},$ so the answer is $3 + 14 = boxed{017}$ .
Though the problem may appear to be quite daunting, it is actually not that difficult. $frac {k(k-1)}2$ always evaluates to an integer (triangular number), and the cosine of $npi$ where $n in mathbb{Z}$ is 1 if $n$ is even and -1 if $n$ is odd. $frac {k(k-1)}2$ will be even if $4|k$ or $4|k-1$ , and odd otherwise.So our sum looks something like: $left|sum_{i=19}^{98} A_iright| =- frac{19(18)}{2} + frac{20(19)}{2} + frac{21(20)}{2} - frac{22(21)}{2} - frac{23(22)}{2} + frac{24(23)}{2} cdots - frac{98(97)}{2}$ If we group the terms in pairs, we see that we need a formula for $-frac{(n)(n-1)}{2} + frac{(n+1)(n)}{2} = left(frac n2right)(n+1 - (n-1)) = n$ . So the first two fractions add up to $19$ , the next two to $-21$ , and so forth.If we pair the terms again now, each pair adds up to $-2$ . There are $frac{98-19+1}{2 cdot 2} = 20$ such pairs, so our answer is $|-2 cdot 20| = boxed{040}$
There are several similar triangles. $triangle PAQsim triangle PDC$ , so we can write the proportion:

$frac{AQ}{CD} = frac{PQ}{PC} = frac{735}{112 + 735 + RC} = frac{735}{847 + RC}$

Also, $triangle BRQsim DRC$ , so:

$frac{QR}{RC} = frac{QB}{CD} = frac{112}{RC} = frac{CD - AQ}{CD} = 1 - frac{AQ}{CD}$

$frac{AQ}{CD} = 1 - frac{112}{RC} = frac{RC - 112}{RC}$

Substituting,

$frac{AQ}{CD} = frac{735}{847 + RC} = frac{RC - 112}{RC}$

$735RC = (RC + 847)(RC - 112)$
$0 = RC^2 - 112cdot847$

Thus, $RC = sqrt{112*847} = 308$ .

We have $triangle BRQsim triangle DRC$ so $frac{112}{RC} = frac{BR}{DR}$ . We also have $triangle BRC sim triangle DRP$ so $frac{ RC}{847} = frac {BR}{DR}$ . Equating the two results gives $frac{112}{RC} = frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=boxed{308}$
Define $x_i = 2y_i - 1$ . Then $2left(sum_{i = 1}^4 y_iright) - 4 = 98$ , so $sum_{i = 1}^4 y_i = 51$ .So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is $n = {50choose3} = frac{50 * 49 * 48}{3 * 2} = 19600$ , and $frac n{100} = boxed{196}$ .
The best way to start is to just write out some terms.

0 1 2 3 4 5 6

$quad 1000 quad$ aa $quad x quad$ aaa $1000 - x$ $2x - 1000$ a $2000 - 3x$ $5x - 3000$ $5000 - 8x$

It is now apparent that each term can be written as
- $n equiv 0 pmod{2}quadquad F_{n-1}cdot 1000-F_ncdot x$
- $n equiv 1 pmod{2}quadquad F_{n}cdot 1000-F_{n-1}cdot x$
where the $F_{n}$ are Fibonacci numbers. This can be proven through induction.
Let the two mathematicians be $M_1$ and $M_2$ . Consider plotting the times that they are on break on a coordinate plane with one axis being the time $M_1$ arrives and the second axis being the time $M_2$ arrives (in minutes past 9 a.m.). The two mathematicians meet each other when $|M_1-M_2| leq m$ . Also because the mathematicians arrive between 9 and 10, $0 leq M_1,M_2 leq 60$ . Therefore, $60times 60$ square represents the possible arrival times of the mathematicians, while the shaded region represents the arrival times where they meet. $[asy] import graph; size(180); real m=60-12*sqrt(15); draw((0,0)--(60,0)--(60,60)--(0,60)--cycle); fill((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle,lightgray); draw((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle); xaxis("$M_1$",-10,80); yaxis("$M_2$",-10,80); label(rotate(45)*"$M_1-M_2le m$",((m+60)/2,(60-m)/2),NW,fontsize(9)); label(rotate(45)*"$M_1-M_2ge -m$",((60-m)/2,(m+60)/2),SE,fontsize(9)); label("$m$",(m,0),S); label("$m$",(0,m),W); label("$60$",(60,0),S); label("$60$",(0,60),W); [/asy]$ It's easier to compute the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet:

$frac{(60-m)^2}{60^2} = .6$
$(60-m)^2 = 36cdot 60$
$60 - m = 12sqrt{15}$
$Rightarrow m = 60-12sqrt{15}$

So the answer is $60 + 12 + 15 = 087$ .

Case 1:

Case 2:

We draw a number line representing the time interval. If mathematician $M_1$ comes in at the center of the time period, then the two mathematicions will meet if $M_2$ comes in somewhere between $m$ minutes before and after $M_1$ comes (a total range of $2m$ minutes). However, if $M_1$ comes into the cafeteria in the first or last $m$ minutes, then the range in which $M_2$ is reduced to somewhere in between $m$ and $2m$ .

We know try to find the weighted average of the chance that the two meet. In the central $60-2m$ minutes, $M_1$ and $M_2$ have to enter the cafeteria within $m$ minutes of each other; so if we fix point $M_1$ then $M_2$ has a $frac{2m}{60} = frac{m}{30}$ probability of meeting.

In the first and last $2m$ minutes, the probability that the two meet ranges from $frac{m}{60}$ to $frac{2m}{60}$ , depending upon the location of $M_1$ with respect to the endpoints. Intuitively, the average probability will occur at $frac{frac{3}{2}m}{60} = frac{m}{40}$ .

So the weighted average is:

$frac{frac{m}{30}(60-2m) + frac{m}{40}(2m)}{60} = frac{40}{100}$

$0 = frac{m^2}{60} - 2m + 24$

$0 = m^2 - 120m + 1440$

Solving this quadratic, we get two roots, $60 pm 12sqrt{15}$ . However, $m < 60$ , so we discard the greater root; and thus our answer $60 + 12 + 15 = 087$ .
The key is to realize the significance that the figures are spheres, not circles. The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.Let us examine the relation between one of the outside 8 spheres and the center one (with radius $r$ ):If we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$ . Then by the Pythagorean Theorem: $[x^2 + (r-100)^2 = (r+100)^2 Longrightarrow x = 20sqrt{r}]$ $x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40sqrt{r}$ . We can draw another right triangle as shown above. One leg has a length of $200$ . The other can be found by partitioning the leg into three sections and using $45-45-90 triangle$ s to see that the leg is $100sqrt{2} + 200 + 100sqrt{2} = 200(sqrt{2} + 1)$ . Pythagorean Theorem: $begin{eqnarray*} (40sqrt{r})^2 &=& 200^2 + [200(sqrt{2}+1)]^2 1600r &=& 200^2[(1 + sqrt{2})^2 + 1] r &=& 100 + 50sqrt{2} end{eqnarray*}$ Thus $a + b + c = 100 + 50 + 2 = boxed{152}$ .
This approach uses analytical geometry. Let $A$ be at the origin, $B$ at $(20,0,0)$ , $C$ at $(20,0,20)$ , and $D$ at $(20,20,20)$ . Thus, $P$ is at $(5,0,0)$ , $Q$ is at $(20,0,15)$ , and $R$ is at $(20,10,20)$ .Let the plane $PQR$ have the equation $ax + by + cz = d$ . Using point $P$ , we get that $5a = d$ . Using point $Q$ , we get $20a + 15c = d Longrightarrow 4d + 15c = d Longrightarrow d = -5c$ . Using point $R$ , we get $20a + 10b + 20c = d Longrightarrow 4d + 10b - 4d = d Longrightarrow d = 10b$ . Thus plane $PQR$ ’s equation reduces to $frac{d}{5}x + frac{d}{10}y - frac{d}{5}z = d Longrightarrow 2x + y - 2z = 10$ .We know need to find the intersection of this plane with that of $z = 0$ , $z = 20$ , $x = 0$ , and $y = 20$ . After doing a little bit of algebra, the intersections are the lines $y = -2x + 10$ , $y = -2x + 50$ , $y = 2z + 10$ , and $z = x + 5$ . Thus, there are three more vertices on the polygon, which are at $(0,10,0)(0,20,5)(15,20,20)$ .We can find the lengths of the sides of the polygons now. There are 4 right triangles with legs of length 5 and 10, so their hypotenuses are $5sqrt{5}$ . The other two are of $45-45-90 triangle$ s with legs of length 15, so their hypotenuses are $15sqrt{2}$ . So we have a hexagon with sides $15sqrt{2},5sqrt{5}, 5sqrt{5},15sqrt{2}, 5sqrt{5},5sqrt{5}$ By symmetry, we know that opposite angles of the polygon are congruent. We can also calculate the length of the long diagonal by noting that it is of the same length of a face diagonal, making it $20sqrt{2}$ . $[asy] size(190); pointpen=black;pathpen=black; real s=2^.5; pair P=(0,0),Q=(7.5*s,2.5*s),R=Q+(0,15*s),Pa=(0,20*s),Qa=(-Q.x,Q.y),Ra=(-R.x,R.y); D(P--Q--R--Pa--Ra--Qa--cycle);D(R--Ra);D(Q--Qa);D(P--Pa); MP("15sqrt{2}",(Q+R)/2,E); MP("5sqrt{5}",(P+Q)/2,SE); MP("5sqrt{5}",(R+Pa)/2,NE); MP("20sqrt{2}",(P+Pa)/2,W); [/asy]$ The height of the triangles at the top/bottom is $frac{20sqrt{2} - 15sqrt{2}}{2} = frac{5}{2}sqrt{2}$ . The Pythagorean Theorem gives that half of the base of the triangles is $frac{15}{sqrt{2}}$ . We find that the middle rectangle is actually a square, so the total area is $(15sqrt{2})^2 + 4left(frac 12right)left(frac 52sqrt{2}right)left(frac{15}{sqrt{2}}right) = 525$ .
We let $x = EP = FQ$ , $y = EQ$ , $k = PQ$ . Since $AE = frac {1}{2}AB$ and $AD = frac {1}{2}AC$ , $triangle AED sim triangle ABC$ and $ED parallel BC$ .By alternate interior angles, we have $angle PEQ = angle BFQ$ and $angle EPQ = angle FBQ$ . By vertical angles, $angle EQP = angle FQB$ .Thus $triangle EQP sim triangle FQB$ , so $frac {EP}{EQ} = frac {FB}{FQ}Longrightarrowfrac {x}{y} = frac {1}{x}Longrightarrow x^{2} = y$ .Since $triangle EDF$ is equilateral, $EQ + FQ = EF = BF = 1Longrightarrow x + y = 1$ . Solving for $x$ and $y$ using $x^{2} = y$ and $x + y = 1$ gives $x = frac {sqrt {5} - 1}{2}$ and $y = frac {3 - sqrt {5}}{2}$ .Using the Law of Cosines, we get

$k^{2} = x^{2} + y^{2} - 2xycos{frac {pi}{3}}$

$= left(frac {sqrt {5} - 1}{2}right)^{2} + left(frac {3 - sqrt {5}}{2}right)^{2} - 2left(frac {sqrt {5} - 1}{2}right)left(frac {3 - sqrt {5}}{2}right)cos{frac {pi}{3}}$

$= 7 - 3sqrt {5}$

We want the ratio of the squares of the sides, so $frac {(2)^{2}}{k^{2}} = frac {4}{7 - 3sqrt {5}} = 7 + 3sqrt {5}$ so $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = boxed{083}$ .
We note that the number of subsets (for now, including the empty subset, which we will just define to have a power sum of zero) with $9$ in it is equal to the number of subsets without a $9$ . To easily see this, take all possible subsets of ${1,2,ldots,8}$ . Since the sets are ordered, a $9$ must go at the end; hence we can just append a $9$ to any of those subsets to get a new one.Now that we have drawn that bijection, we can calculate the complex power sum recursively. Since appending a $9$ to a subset doesn't change anything about that subset's complex power sum besides adding an additional term, we have that $S_9 = 2S_8 + T_9$ , where $T_9$ refers to the sum of all of the $9i^x$ .It a subset of size 1 has a 9, then its power sum must be $9i$ , and there is only $1$ of these such subsets. There are ${8choose1}$ with $9cdot i^2$ , ${8choose2}$ with $9cdot i^3$ , and so forth. So $T_9 =sum_{k=0}^{8} 9{8choose{k}}i^{k+1}$ . This is exactly the binomial expansion of $9i cdot (1+i)^8$ . We can use De Moivre's Theorem to calculate the power: $(sqrt{2})^8cos{8cdot45} = 16$ . Hence $T_9 = 16cdot9i = 144i$ , and $S_9 = 2S_8 + 144i = 2(-176 -64i) + 144i = -352 + 16i$ . Thus, $|p| + |q| = |-352| + |16| = 368$ .
Let’s solve for :Clearly, we want to minimize the denominator, so we test . The possible pairs of factors of are . These give and respectively. Substituting into the numerator, we see that the first pair gives , while the second pair gives . We now check that is optimal, setting , in order to simplify calculations. SinceWe haveWhere we see gives us our maximum value of .
- Note that $0 le (a-1)(b-1)$ assumes $m,n ge 3$ , but this is clear as $frac{2m}{m+2} = frac{(n+2)(p+2)}{np} > 1$ and similarly for $n$ .
We can draw a comparison between the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes.You need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the point. (Reasoning for this: Everytime you go to a vertex, you have to leave the vertex, so every vertex reached is equivalent to adding 2 more segments. So the degree of each vertex must be even, with the exception of endpoints) Since there are 39 segments coming from each point it is impossible to touch every segment.But you can get up to 38 on each segment because you go in to the point then out on a different segment. Counting going out from the starting and ending at the ending point we have: $frac{38cdot 40 + 2}2 = boxed{761}$