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1994AIME 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11:01

1994AIME 真题及答案解析

答案解析请参考文末

Problem 1

The increasing sequence $3, 15, 24, 48, ldots,$ consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when the 1994th term of the sequence is divided by 1000?

Problem 2

A circle with diameter $overline{PQ},$ of length 10 is internally tangent at $P^{}_{}$ to a circle of radius 20. Square $ABCD,$ is constructed with $A,$ and $B,$ on the larger circle, $overline{CD},$ tangent at $Q,$ to the smaller circle, and the smaller circle outside $ABCD,$ . The length of $overline{AB},$ can be written in the form $m + sqrt{n},$ , where $m,$ and $n,$ are integers. Find $m + n,$ .

Problem 3

The function $f_{}^{}$ has the property that, for each real number $x,,$

$f(x)+f(x-1) = x^2,$ .

If $f(19)=94,,$ what is the remainder when $f(94),$ is divided by 1000?

Problem 4

Find the positive integer $n,$ for which

$lfloor log_2{1}rfloor+lfloorlog_2{2}rfloor+lfloorlog_2{3}rfloor+cdots+lfloorlog_2{n}rfloor=1994$ .

(For real $x,$ , $lfloor xrfloor,$ is the greatest integer $le x.,$ )

Problem 5

Given a positive integer $n,$ , let $p(n),$ be the product of the non-zero digits of $n,$ . (If $n,$ has only one digit, then $p(n),$ is equal to that digit.) Let

$S=p(1)+p(2)+p(3)+cdots+p(999)$ .

What is the largest prime factor of $S,$ ?

Problem 6

The graphs of the equations

$y=k, qquad y=sqrt{3}x+2k, qquad y=-sqrt{3}x+2k,$ are drawn in the coordinate plane for $k=-10,-9,-8,ldots,9,10.,$ These 63 lines cut part of the plane into equilateral triangles of side $2/sqrt{3}$ . How many such triangles are formed?

Problem 7

For certain ordered pairs $(a,b),$ of real numbers, the system of equations

$ax+by=1,$ $x^2+y^2=50,$ has at least one solution, and each solution is an ordered pair $(x,y),$ of integers. How many such ordered pairs $(a,b),$ are there?

Problem 8

The points $(0,0),$ , $(a,11),$ , and $(b,37),$ are the vertices of an equilateral triangle. Find the value of $ab,$ .

Problem 9

A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty. The probability that the bag will be emptied is $p/q,,$ where $p,$ and $q,$ are relatively prime positive integers. Find $p+q.,$

Problem 10

In triangle $ABC,,$ angle $C$ is a right angle and the altitude from $C,$ meets $overline{AB},$ at $D.,$ The lengths of the sides of $triangle ABC,$ are integers, $BD=29^3,,$ and $cos B=m/n,$ , where $m,$ and $n,$ are relatively prime positive integers. Find $m+n.,$

Problem 11

Ninety-four bricks, each measuring $4''times10''times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes $4'',$ or $10'',$ or $19'',$ to the total height of the tower. How many different tower heights can be achieved using all 94 of the bricks?

Problem 12

A fenced, rectangular field measures 24 meters by 52 meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence?

Problem 13

The equation

$x^{10}+(13x-1)^{10}=0,$ has 10 complex roots $r_1, overline{r_1}, r_2, overline{r_2}, r_3, overline{r_3}, r_4, overline{r_4}, r_5, overline{r_5},,$ where the bar denotes complex conjugation. Find the value of

$frac 1{r_1overline{r_1}}+frac 1{r_2overline{r_2}}+frac 1{r_3overline{r_3}}+frac 1{r_4overline{r_4}}+frac 1{r_5overline{r_5}}.$

Problem 14

A beam of light strikes $overline{BC},$ at point $C,$ with angle of incidence $alpha=19.94^circ,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $overline{AB},$ and $overline{BC},$ according to the rule: angle of incidence equals angle of reflection. Given that $beta=alpha/10=1.994^circ,$ and $AB=BC,,$ determine the number of times the light beam will bounce off the two line segments. Include the first reflection at $C,$ in your count.

AIME 1994 Problem 14.png

Problem 15

Given a point $P^{}_{}$ on a triangular piece of paper $ABC,,$ consider the creases that are formed in the paper when $A, B,,$ and $C,$ are folded onto $P.,$ Let us call $P_{}^{}$ a fold point of $triangle ABC,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppose that $AB=36, AC=72,,$ and $angle B=90^circ.,$ Then the area of the set of all fold points of $triangle ABC,$ can be written in the form $qpi-rsqrt{s},,$ where $q, r,,$ and $s,$ are positive integers and $s,$ is not divisible by the square of any prime. What is $q+r+s,$ ?

1994AIME 详细解析

One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$ . Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n equiv -1, 1 equiv 2, 1 pmod{3}$ . Since 1994 is even, $n$ must $equiv 1 pmod{3}$ . It will be the $frac{1994}{2} = 997$ th such term, so $n = 4 + (997-1) cdot 3 = 2992$ . The value of $n^2 - 1 = 2992^2 - 1 pmod{1000}$ is $063$ .
Call the center of the larger circle $O$ . Extend the diameter $overline{PQ}$ to the other side of the square (at point $E$ ), and draw $overline{AO}$ . We now have a right triangle, with hypotenuse of length $20$ . Since $OQ = OP - PQ = 20 - 10 = 10$ , we know that $OE = AB - OQ = AB - 10$ . The other leg, $AE$ , is just $frac 12 AB$ .Apply the Pythagorean Theorem:

$(AB - 10)^2 + left(frac 12 ABright)^2 = 20^2$

$AB^2 - 20 AB + 100 + frac 14 AB^2 - 400 = 0$

$AB^2 - 16 AB - 240 = 0$

The quadratic formula shows that the answer is $frac{16 pm sqrt{16^2 + 4 cdot 240}}{2} = 8 pm sqrt{304}$ . Discard the negative root, so our answer is $8 + 304 = 312$ .
$begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=cdots \ &= (94^2-93^2) + (92^2-91^2) +cdots+ (22^2-21^2)+ 20^2-f(19) \ &= 94+93+cdots+21+400-94 \ &= 4561 end{align*}$ So, the remainder is $boxed{561}$ .
Note that if $2^x le a<2^{x+1}$ for some $xinmathbb{Z}$ , then $lfloorlog_2{a}rfloor=log_2{2^{x}}=x$ .Thus, there are $2^{x+1}-2^{x}=2^{x}$ integers $a$ such that $lfloorlog_2{a}rfloor=x$ . So the sum of $lfloorlog_2{a}rfloor$ for all such $a$ is $xcdot2^x$ .Let $k$ be the integer such that $2^k le n<2^{k+1}$ . So for each integer $j<k$ , there are $2^j$ integers $ale n$ such that $lfloorlog_2{a}rfloor=j$ , and there are $n-2^k+1$ such integers such that $lfloorlog_2{a}rfloor=k$ .Therefore, $lfloorlog_2{1}rfloor+lfloorlog_2{2}rfloor+lfloorlog_2{3}rfloor+cdots+lfloorlog_2{n}rfloor= sum_{j=0}^{k-1}(jcdot2^j) + k(n-2^k+1) = 1994$ .Through computation: $sum_{j=0}^{7}(jcdot2^j)=1538<1994$ and $sum_{j=0}^{8}(jcdot2^j)=3586>1994$ . Thus, $k=8$ .
So, $sum_{j=0}^{k-1}(jcdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 Rightarrow n = boxed{312}$ .
Suppose we write each number in the form of a three-digit number (so $5 equiv 005$ ), and since our $p(n)$ ignores all of the zero-digits, replace all of the $0$ s with $1$ s. Now note that in the expansion of $(1+ 1 +2+3+4+5+6+7+8+9) (1+ 1 +2+3+cdots+9) (1+ 1 +2+3+cdots+9)$ we cover every permutation of every product of $3$ digits, including the case where that first $1$ represents the replaced $0$ s. However, since our list does not include $000$ , we have to subtract $1$ . Thus, our answer is the largest prime factor of $(1+1+2+3+cdots +9)^3 - 1 = 46^3 - 1 = (46-1)(46^2 + 46 + 1) = 3^3 cdot 5 cdot 7 cdot boxed{103}$ .
We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon. $[asy] size(200); picture pica, picb, picc; int i; for(i=-10;i<=10;++i){ if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i),black+0.7);} else{draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i));} } picb = rotate(120,origin)*pica; picc = rotate(240,origin)*pica; add(pica);add(picb);add(picc); [/asy]$ Solving the above equations for $k=pm 10$ , we see that the hexagon in question is regular, with side length $frac{20}{sqrt{3}}$ . Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just $left(frac{20/sqrt{3}}{2/sqrt{3}}right)^2 = 100$ . Thus, the total number of unit triangles is $6 times 100 = 600$ .There are $6 cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = boxed{660}$ .
$x^2+y^2=50$ is the equation of a circle of radius $sqrt{50}$ , centered at the origin. The lattice points on this circle are $(pm1,pm7)$ , $(pm5,pm5)$ , and $(pm7,pm1)$ . $ax+by=1$ is the equation of a line that does not pass through the origin. (Since $(x,y)=(0,0)$ yields $a(0)+b(0)=0 neq 1$ ).So, we are looking for the number of lines which pass through either one or two of the $12$ lattice points on the circle, but do not pass through the origin.It is clear that if a line passes through two opposite points, then it passes through the origin, and if a line passes through two non-opposite points, the it does not pass through the origin.There are $binom{12}{2}=66$ ways to pick two distinct lattice points, and thus $66$ distinct lines which pass through two lattice points on the circle. However, $frac{12}{2}=6$ of these lines pass through the origin.
Since there is a unique tangent line to the circle at each of these lattice points, there are $12$ distinct lines which pass through exactly one lattice point on the circle.

Thus, there are a total of $66-6+12=boxed{072}$ distinct lines which pass through either one or two of the $12$ lattice points on the circle, but do not pass through the origin.
Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so: $[(a+11i)left(text{cis},60^{circ}right) = (a+11i)left(frac 12+frac{sqrt{3}i}2right)=b+37i.]$ Equating the real and imaginary parts, we have: $begin{align*}b&=frac{a}{2}-frac{11sqrt{3}}{2}\37&=frac{11}{2}+frac{asqrt{3}}{2} end{align*}$ Solving this system, we find that $a=21sqrt{3}, b=5sqrt{3}$ . Thus, the answer is $boxed{315}$ .
Note: There is another solution where the point $b+37i$ is a rotation of $-60$ degrees of $a+11i$ ; however, this triangle is just a reflection of the first triangle by the $y$ -axis, and the signs of $a$ and $b$ are flipped. However, the product $ab$ is unchanged.
Let be the probability of emptying the bag when it has pairs in it. Let's consider the possible draws for the first three cards:
- Case 1. We draw a pair on the first two cards. The second card is the same as the first with probability $frac {1}{2k - 1}$ , then we have $k - 1$ pairs left. So this contributes probability $frac {P_{k - 1}}{2k - 1}$ .
- Case 2. We draw a pair on the first and third cards. The second card is different from the first with probability $frac {2k - 2}{2k - 1}$ and the third is the same as the first with probability $frac {1}{2k - 2}$ . We are left with $k - 1$ pairs but one card already drawn. However, having drawn one card doesn't affect the game, so this also contributes probability $frac {P_{k - 1}}{2k - 1}$ .
- Case 3. We draw a pair on the second and third cards. This is pretty much the same as case 2, so we get $frac {P_{k - 1}}{2k - 1}$ .
Therefore, we obtain the recursion $P_k = frac {3}{2k - 1}P_{k - 1}$ . Iterating this for $k = 6,5,4,3,2$ (obviously $P_1 = 1$ ), we get $frac {3^5}{11*9*7*5*3} = frac {9}{385}$ , and $p+q=boxed{394}$ .
Since $triangle ABC sim triangle CBD$ , we have $frac{BC}{AB} = frac{29^3}{BC} Longrightarrow BC^2 = 29^3 AB$ . It follows that $29^2 | BC$ and $29 | AB$ , so $BC$ and $AB$ are in the form $29^2 x$ and $29 x^2$ , respectively, where x is an integer.By the Pythagorean Theorem, we find that $AC^2 + BC^2 = AB^2 Longrightarrow (29^2x)^2 + AC^2 = (29 x^2)^2$ , so $29x | AC$ . Letting $y = AC / 29x$ , we obtain after dividing through by $(29x)^2$ , $29^2 = x^2 - y^2 = (x-y)(x+y)$ . As $x,y in mathbb{Z}$ , the pairs of factors of $29^2$ are $(1,29^2)(29,29)$ ; clearly $y = frac{AC}{29x} neq 0$ , so $x-y = 1, x+y= 29^2$ . Then, $x = frac{1+29^2}{2} = 421$ .Thus, $cos B = frac{BC}{AB} = frac{29^2 x}{29x^2} = frac{29}{421}$ , and $m+n = boxed{450}$ .
We have the smallest stack, which has a height of $94 times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$ 's, $6$ 's, and $15$ 's. Because $0$ , $6$ , and $15$ are all multiples of $3$ , the change will always be a multiple of $3$ , so we just need to find the number of changes we can get from $0$ 's, $2$ 's, and $5$ 's.From here, we count what we can get: $[0, 2 = 2, 4 = 2+2, 5 = 5, 6 = 2+2+2, 7 = 5+2, 8 = 2+2+2+2, 9 = 5+2+2, ldots]$ It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem, which says that the greatest number that cannot be expressed in the form of $2m + 5n$ for $m,n$ being positive integers is $5 times 2 - 5 - 2=3$ .But we also have a maximum change ( $94 times 5$ ), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either $0$ 's, $3$ 's, or $5$ 's. The maximum we can't get is $5 times 3-5-3=7$ , so the numbers $94 times 5-8$ and below, except $3$ and $1$ , work. Now there might be ones that we haven't counted yet, so we check all numbers between $94 times 5-8$ and $94 times 5$ . $94 times 5-7$ obviously doesn't work, $94 times 5-6$ does since 6 is a multiple of 3, $94 times 5-5$ does because it is a multiple of $5$ (and $3$ ), $94 times 5-4$ doesn't since $4$ is not divisible by $5$ or $3$ , $94 times 5-3$ does since $3=3$ , and $94 times 5-2$ and $94 times 5-1$ don't, and $94 times 5$ does.
Thus the numbers $0$ , $2$ , $4$ all the way to $94 times 5-8$ , $94 times 5-6$ , $94 times 5-5$ , $94 times 5-3$ , and $94times 5$ work. That's $2+(94 times 5 - 8 - 4 +1)+4=boxed{465}$ numbers. That's the number of changes you can make to a stack of bricks with dimensions $4 times 10 times 19$ , including not changing it at all.
Suppose there are $n$ squares in every column of the grid, so there are $frac{52}{24}n = frac {13}6n$ squares in every row. Then $6|n$ , and our goal is to maximize the value of $n$ .Each vertical fence has length $24$ , and there are $frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$ , and there are $n-1$ such fences. Then the total length of the internal fencing is $24left(frac{13n}{6}-1right) + 52(n-1) = 104n - 76 le 1994 Longrightarrow n le frac{1035}{52} approx 19.9$ , so $n le 19$ . The largest multiple of $6$ that is $le 19$ is $n = 18$ , which we can easily verify works, and the answer is $frac{13}{6}n^2 = boxed{702}$ .
Let $t = 1/x$ . After multiplying the equation by $t^{10}$ , $1 + (13 - t)^{10} = 0Rightarrow (13 - t)^{10} = - 1$ .Using DeMoivre, $13 - t = text{cis}left(frac {(2k + 1)pi}{10}right)$ where $k$ is an integer between $0$ and $9$ . $t = 13 - text{cis}left(frac {(2k + 1)pi}{10}right) Rightarrow bar{t} = 13 - text{cis}left(-frac {(2k + 1)pi}{10}right)$ .Since $text{cis}(theta) + text{cis}(-theta) = 2cos(theta)$ , $tbar{t} = 170 - 26cos left(frac {(2k + 1)pi}{10}right)$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2pi$ are involved in the product.The expression to find is $sum tbar{t} = 850 - 26sum_{k = 0}^4 cos frac {(2k + 1)pi}{10}$ .
But $cos frac {pi}{10} + cos frac {9pi}{10} = cos frac {3pi}{10} + cos frac {7pi}{10} = cos frac {pi}{2} = 0$ so the sum is $boxed{850}$ .
At each point of reflection, we pretend instead that the light continues to travel straight. $[asy] pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); D(A--B--(1.5,0));D(r);D(anglemark(C,B,D(A)));D(anglemark((1.5,0),C,C+.4*expi(beta*pi/180)));MP("beta",B,(5,1.2),fontsize(9));MP("alpha",C,(4,1.2),fontsize(9)); for(int i = 0; i < 180/alpha; ++i){ path l = B -- (1+i/2)*expi(-i * alpha * pi / 180); D(l, linetype("4 4")); D(IP(l,r)); } D(B); [/asy]$ Note that after $k$ reflections (excluding the first one at $C$ ) the extended line will form an angle $k beta$ at point $B$ . For the $k$ th reflection to be just inside or at point $B$ , we must have $kbeta le 180 - 2alpha Longrightarrow k le frac{180 - 2alpha}{beta} = 70.27$ . Thus, our answer is, including the first intersection, $leftlfloor frac{180 - 2alpha}{beta} rightrfloor + 1 = boxed{071}$ .

Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $overline{PA}$ and $overline{PB}$ , and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $triangle PAB, PBC, PCA$ . According to the problem statement, the circumcenters of the triangles cannot lie within the interior of the respective triangles, since they are not on the paper. It follows that $angle APB, angle BPC, angle CPA > 90^{circ}$ ; the locus of each of the respective conditions for $P$ is the region inside the (semi)circles with diameters $overline{AB}, overline{BC}, overline{CA}$ .We note that the circle with diameter $AC$ covers the entire triangle because it is the circumcircle of $triangle ABC$ , so it suffices to take the intersection of the circles about $AB, BC$ . We note that their intersection lies entirely within $triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $triangle ABC$ from $B$ ). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = overline{AB}, overline{BC}$ and note that $triangle M_1BM_2 sim triangle ABC$ , we see that thse segments respectively cut a $120^{circ}$ arc in the circle with radius $18$ and $60^{circ}$ arc in the circle with radius $18sqrt{3}$ .

$[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype("2 3") + linewidth(0.7), dashes = linetype("8 6")+linewidth(0.7)+blue, bluedots = linetype("1 4") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP("P",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP("A",A)) -- D(MP("B",B)) -- D(MP("C",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]$

The diagram shows $P$ outside of the grayed locus; notice that the creases [the dotted blue] intersect within the triangle, which is against the problem conditions. The area of the locus is the sum of two segments of two circles; these segments cut out $120^{circ}, 60^{circ}$ angles by simple similarity relations and angle-chasing.

Hence, the answer is, using the $frac 12 absin C$ definition of triangle area, $left[frac{pi}{3} cdot 18^2 - frac{1}{2} cdot 18^2 sin frac{2pi}{3} right] + left[frac{pi}{6} cdot left(18sqrt{3}right)^2 - frac{1}{2} cdot (18sqrt{3})^2 sin frac{pi}{3}right] = 270pi - 324sqrt{3}$ , and $q+r+s = boxed{597}$ .