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1996AIME 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10:41

1996AIME 真题及答案解析

答案解析请参考文末

Problem 1

In a magic square, the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ .

Problem 2

For each real number $x$ , let $lfloor x rfloor$ denote the greatest integer that does not exceed $x$ . For how many positive integers $n$ is it true that $n<1000$ and that $lfloor log_{2} n rfloor$ is a positive even integer?

Problem 3

Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms.

Problem 4

A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of a shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed $1000x$ .

Problem 5

Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a$ , $b$ , and $c$ , and that the roots of $x^3+rx^2+sx+t=0$ are $a+b$ , $b+c$ , and $c+a$ . Find $t$ .

Problem 6

In a five-team tournament, each team plays one game with every other team. Each team has a $50%$ chance of winning any game it plays. (There are no ties.) Let $dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$ .

Problem 7

Two squares of a $7times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible?

Problem 8

The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$ ?

Problem 9

A bored student walks down a hall that contains a row of closed lockers, numbered 1 to 1024. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?

Problem 10

Find the smallest positive integer solution to $tan{19x^{circ}}=dfrac{cos{96^{circ}}+sin{96^{circ}}}{cos{96^{circ}}-sin{96^{circ}}}$ .

Problem 11

Let $mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $mathrm {P}=r(cos{theta^{circ}}+isin{theta^{circ}})$ , where $0<r$ and $0leq theta <360$ . Find $theta$ .

Problem 12

For each permutation $a_1,a_2,a_3,cdots,a_{10}$ of the integers $1,2,3,cdots,10$ , form the sum

$|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|$ .

The average value of all such sums can be written in the form $dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Problem 13

In triangle $ABC$ , $AB=sqrt{30}$ , $AC=sqrt{6}$ , and $BC=sqrt{15}$ . There is a point $D$ for which $overline{AD}$ bisects $overline{BC}$ , and $angle ADB$ is a right angle. The ratio

$[dfrac{text{Area}(triangle ADB)}{text{Area}(triangle ABC)}]$

can be written in the form $dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Problem 14

A $150times 324times 375$ rectangular solid is made by gluing together $1times 1times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1times 1times 1$ cubes?

Problem 15

In parallelogram $ABCD,$ let $O$ be the intersection of diagonals $overline{AC}$ and $overline{BD}$ . Angles $CAB$ and $DBC$ are each twice as large as angle $DBA,$ and angle $ACB$ is $r$ times as large as angle $AOB$ . Find the greatest integer that does not exceed $1000r$ .

1996AIME 详细解析

Let's make a table. $[begin{array}{|c|c|c|} multicolumn{3}{c}{text{Table}}\hline x&19&96\hline 1&a&b\hline c&d&e\hline end{array}]$ $begin{eqnarray*} x+19+96=x+1+cRightarrow c=19+96-1=114,\ 114+96+a=x+1+114Rightarrow a=x-95 end{eqnarray*}$
$[begin{array}{|c|c|c|} multicolumn{3}{c}{text{Table in progress}}\hline x&19&96\hline 1&x-95&b\hline 114&d&e\hline end{array}]$

$begin{eqnarray*}19+x-95+d=x+d-76=115+xRightarrow d=191,\ 114+191+e=x+115Rightarrow e=x-190 end{eqnarray*}$ $[begin{array}{|c|c|c|} multicolumn{3}{c}{text{Table in progress}}\hline x&19&96\hline 1&x-95&b\hline 114&191&x-190\hline end{array}]$

$[3x-285=x+115Rightarrow 2x=400Rightarrow x=boxed{200}]$
For integers $k$ , we want $lfloor log_2 nrfloor = 2k$ , or $2k le log_2 n < 2k+1 Longrightarrow 2^{2k} le n < 2^{2k+1}$ . Thus, $n$ must satisfy these inequalities (since $n < 1000$ ):

$4leq n <8$
$16leq n<32$
$64leq n<128$

$256leq n<512$

There are $4$ for the first inequality, $16$ for the second, $64$ for the third, and $256$ for the fourth, so the answer is $4+16+64+256=boxed{340}$ .
Using Simon's Favorite Factoring Trick, we rewrite as $[(x+7)(y-3)]^n = (x+7)^n(y-3)^n$ . Both binomial expansions will contain $n+1$ non-like terms; their product will contain $(n+1)^2$ terms, as each term will have an unique power of $x$ or $y$ and so none of the terms will need to be collected. Hence $(n+1)^2 ge 1996$ , the smallest square after $1996$ is $2025 = 45^2$ , so our answer is $45 - 1 = boxed{044}$ .Alternatively, when $n = k$ , the exponents of $x$ or $y$ in $x^i y^i$ can be any integer between $0$ and $k$ inclusive. Thus, when $n=1$ , there are $(2)(2)$ terms and, when $n = k$ , there are $(k+1)^2$ terms. Therefore, we need to find the smallest perfect square that is greater than $1996$ . From trial and error, we get $44^2 = 1936$ and $45^2 = 2025$ . Thus, $k = 45rightarrow n = boxed{044}$ .
$[asy] import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5; real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw((0,unit,0)--(unit,unit,0)--(unit,unit,unit)--(0,unit,unit)); draw(P--(r*unit,0,0)--(r*unit,r*unit,0)--(0,r*unit,0)--P); draw(P--(r*unit,r*unit,0)); draw((r*unit,0,0)--(0,0,0)--(0,r*unit,0)); draw(P--(0,0,unit)--(unit,0,unit)--(unit,0,0)--(r*unit,0,0)--P,dashed+blue+linewidth(0.8)); label("$x$",(0,0,unit+unit/(r-1)/2),WSW); label("$1$",(unit/2,0,unit),N); label("$1$",(unit,0,unit/2),W); label("$1$",(unit/2,0,0),N); label("$6$",(unit*(r+1)/2,0,0),N); label("$7$",(unit*r,unit*r/2,0),SW); [/asy]$ (Figure not to scale) The area of the square shadow base is $48 + 1 = 49$ , and so the sides of the shadow are $7$ . Using the similar triangles in blue, $frac {x}{1} = frac {1}{6}$ , and $leftlfloor 1000x rightrfloor = boxed{166}$ .
By Vieta's formulas on the polynomial $P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0$ , we have $a + b + c = s = -3$ , $ab + bc + ca = 4$ , and $abc = 11$ . Then $t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)$ This is just the definition for $-P(-3) = boxed{023}$ .Alternatively, we can expand the expression to get $begin{align*} t &= -(-3-a)(-3-b)(-3-c)\ &= (a+3)(b+3)(c+3)\ &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\ t &= 11 + 3(4) + 9(-3) + 27 = 23end{align*}$
We can use complementary counting: finding the probability that at least one team wins all games or at least one team loses all games.No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games.Now we use PIE:
The probability that one team wins all games is $5cdot left(frac{1}{2}right)^4=frac{5}{16}$ .

Similarity, the probability that one team loses all games is $frac{5}{16}$ .

The probability that one team wins all games and another team loses all games is $left(5cdot left(frac{1}{2}right)^4right)left(4cdot left(frac{1}{2}right)^3right)=frac{5}{32}$ .

$frac{5}{16}+frac{5}{16}-frac{5}{32}=frac{15}{32}$

Since this is the opposite of the probability we want, we subtract that from 1 to get $frac{17}{32}$ .

$17+32=boxed{049}$
There are ${49 choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.

$[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5)); fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ $[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$

For most pairs, there will be
three other equivalent boards. For those symmetric about the center,
there is only one other.

Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $frac{49-1}{2}=24$ such pairs. There are then ${49 choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 choose 2}-24$ pairs, there are three other pairs that yield an equivalent board.

Thus, the number of inequivalent boards is $frac{{49 choose 2} - 24}{4} + frac{24}{2} = boxed{300}$ . For a $(2n+1) times (2n+1)$ board, this argument generalizes to $n(n+1)(2n^2+2n+1)$ inequivalent configurations.
The harmonic mean of $x$ and $y$ is equal to $frac{1}{frac{frac{1}{x}+frac{1}{y}}2} = frac{2xy}{x+y}$ , so we have $xy=(x+y)(3^{20}cdot2^{19})$ , and by SFFT, $(x-3^{20}cdot2^{19})(y-3^{20}cdot2^{19})=3^{40}cdot2^{38}$ . Now, $3^{40}cdot2^{38}$ has $41cdot39=1599$ factors, one of which is the square root ( $3^{20}2^{19}$ ). Since $x<y$ , the answer is half of the remaining number of factors, which is $frac{1599-1}{2}= boxed{799}$ .
On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$ , leaving only lockers $2 mod{8}$ and $6 mod{8}$ . Then he goes ahead and opens all lockers $2 mod {8}$ , leaving lockers either $6 mod {16}$ or $14 mod {16}$ . He then goes ahead and opens all lockers $14 mod {16}$ , leaving the lockers either $6 mod {32}$ or $22 mod {32}$ . He then goes ahead and opens all lockers $6 mod {32}$ , leaving $22 mod {64}$ or $54 mod {64}$ . He then opens $54 mod {64}$ , leaving $22 mod {128}$ or $86 mod {128}$ . He then opens $22 mod {128}$ and leaves $86 mod {256}$ and $214 mod {256}$ . He then opens all $214 mod {256}$ , so we have $86 mod {512}$ and $342 mod {512}$ , leaving lockers $86, 342, 598$ , and $854$ , and he is at where he started again. He then opens $86$ and $598$ , and then goes back and opens locker number $854$ , leaving locker number $boxed{342}$ untouched. He opens that locker.
$dfrac{cos{96^{circ}}+sin{96^{circ}}}{cos{96^{circ}}-sin{96^{circ}}} =$ $dfrac{sin{186^{circ}}+sin{96^{circ}}}{sin{186^{circ}}-sin{96^{circ}}} =$ $dfrac{sin{(141^{circ}+45^{circ})}+sin{(141^{circ}-45^{circ})}}{sin{(141^{circ}+45^{circ})}-sin{(141^{circ}-45^{circ})}} =$ $dfrac{2sin{141^{circ}}cos{45^{circ}}}{2cos{141^{circ}}sin{45^{circ}}} = tan{141^{circ}}$ .The period of the tangent function is $180^circ$ , and the tangent function is one-to-one over each period of its domain.Thus, $19x equiv 141 pmod{180}$ .
Since $19^2 equiv 361 equiv 1 pmod{180}$ , multiplying both sides by $19$ yields $x equiv 141 cdot 19 equiv (140+1)(18+1) equiv 0+140+18+1 equiv 159 pmod{180}$ .

Therefore, the smallest positive solution is $x = boxed{159}$ .
$begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + frac{z^5-1}{z-1}\ 0 &=& frac{(z^5 - 1)(z(z-1)+1)}{z-1} = frac{(z^2-z+1)(z^5-1)}{z-1} end{eqnarray*}$ Thus $z^5 = 1, z neq 1 Longrightarrow z = mathrm{cis} 72, 144, 216, 288$ ,or $z^2 - z + 1 = 0 Longrightarrow z = frac{1 pm sqrt{-3}}{2} = mathrm{cis} 60, 300$
(see cis).

Discarding the roots with negative imaginary parts (leaving us with $mathrm{cis} theta, 0 < theta < 180$ ), we are left with $mathrm{cis} 60, 72, 144$ ; their product is $P = mathrm{cis} (60 + 72 + 144) = mathrm{cis} boxed{276}$ .
Because of symmetry, we may find all the possible values for $|a_n - a_{n - 1}|$ and multiply by the number of times this value appears. Each occurs $5 cdot 8!$ , because if you fix $a_n$ and $a_{n + 1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ places $a_n$ and $a_{n + 1}$ can be.To find all possible values for $|a_n - a_{n - 1}|$ we have to compute $begin{eqnarray*} |1 - 10| + |1 - 9| + ldots + |1 - 2|\ + |2 - 10| + ldots + |2 - 3| + |2 - 1|\ + ldots\ + |10 - 9| end{eqnarray*}$ This is equivalent to
$[2sumlimits_{k = 1}^{9}sumlimits_{j = 1}^{k}j = 330]$

The total number of permutations is $10!$ , so the average value is $frac {330 cdot 8! cdot 5}{10!} = frac {55}{3}$ , and $m+n = boxed{058}$ .
Let $E$ be the midpoint of $overline{BC}$ . Since $BE = EC$ , then $triangle ABE$ and $triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$ , so $frac{[BDE]}{[BAE]} = frac{DE}{AE}$ (see area ratios). Now, $dfrac{[ADB]}{[ABC]} = frac{[ABE] + [BDE]}{2[ABE]} = frac{1}{2} + frac{DE}{2AE}.$ By Stewart's Theorem, $AE = frac{sqrt{2(AB^2 + AC^2) - BC^2}}2 = frac{sqrt {57}}{2}$ , and by the Pythagorean Theorem on $triangle ABD, triangle EBD$ , $begin{align*} BD^2 + left(DE + frac {sqrt{57}}2right)^2 &= 30 \ BD^2 + DE^2 &= frac{15}{4} \ end{align*}$
Subtracting the two equations yields $DEsqrt{57} + frac{57}{4} = frac{105}{4} Longrightarrow DE = frac{12}{sqrt{57}}$ . Then $frac mn = frac{1}{2} + frac{DE}{2AE} = frac{1}{2} + frac{frac{12}{sqrt{57}}}{2 cdot frac{sqrt{57}}{2}} = frac{27}{38}$ , and $m+n = boxed{065}$ .
Place one corner of the solid at and let be the general coordinates of the diagonally opposite corner of the rectangle, where .We consider the vector equation of the diagonal segment represented by: , where .Consider a point on the diagonal with coordinates . We have 3 key observations as this point moves from towards :
1. When exactly one of $ma$ , $mb$ , $mc$ becomes a positive integer, the point crosses one of the faces of a $1 times 1 times 1$ cube from the outside to the inside of the cube.
1. When exactly two of $ma$ , $mb$ , $mc$ become positive integers, the point enters a new cube through one of the edges of the cube from the outside to the inside of the cube.
1. When all three $ma$ , $mb$ , $mc$ become positive integers, the point enters a cube through one of its vertices from the outside to the inside of the cube.
The number of cubes the diagonal passes is equal to the number of points on the diagonal that has one or more positive integers as coordinates.

If we slice the solid up by the $x$ -planes defined by $x=1,2,3,4, ldots, a$ , the diagonal will cut these $x$ -planes exactly $a$ times (plane of $x=0$ is not considered since $m ne 0$ ). Similar arguments for slices along $y$ -planes and $z$ -planes give diagonal cuts of $b$ , and $c$ times respectively. The total cuts by the diagonal is therefore $a+b+c$ , if we can ensure that no more than $1$ positive integer is present in the x, y, or z coordinate in all points $(ma,mb,mc)$ on the diagonal. Note that point $(a,b,c)$ is already one such exception.

But for each diagonal point $(ma,mb,mc)$ with 2 (or more) positive integers occurring at the same time, $a+b+c$ counts the number of cube passes as $2$ instead of $1$ for each such point. There are $gcd(a,b)+gcd(b,c)+gcd(c,a)$ points in such over-counting. We therefore subtract one time such over-counting from $a+b+c$ .

And for each diagonal point $(ma,mb,mc)$ with exactly $3$ integers occurring at the same time, $a+b+c$ counts the number of cube passes as $3$ instead of $1$ ; ie $a+b+c$ over-counts each of such points by $2$ . But since $gcd(a,b)+gcd(b,c)+gcd(c,a)$ already subtracted three times for the case of $3$ integers occurring at the same time (since there are $3$ of these gcd terms that represent all combinations of 3 edges of a cube meeting at a vertex), we have the final count for each such point as $1 = 3-3+k Rightarrow k=1$ , where $k$ is our correction term. That is, we need to add $k=1$ time $gcd(a,b,c)$ back to account for the case of 3 simultaneous integers.

Therefore, the total diagonal cube passes is: $D = a+b+c-left[ gcd(a,b)+gcd(b,c)+gcd(c,a) right]+gcd(a,b,c)$ .

For $(a,b,c) = (150, 324, 375)$ , we have: $gcd(150,324)=6$ , $gcd(324,375)=3$ , $gcd(150,375)=75$ , $gcd(150,324,375)=3$ .

Therefore $D = 150+324+375-(6+3+75)+3 = boxed{768}$ .
Let $theta = angle DBA$ . Then $angle CAB = angle DBC = 2theta$ , $angle AOB = 180 - 3theta$ , and $angle ACB = 180 - 5theta$ . Since $ABCD$ is a parallelogram, it follows that $OA = OC$ . By the Law of Sines on $triangle ABO,, triangle BCO$ , $frac{sin angle CBO}{OC} = frac{sin angle ACB}{OB} quad text{and} quad frac{sin angle DBA}{OC} = frac{sin angle BAC}{OB}.$ Dividing the two equalities yields $[frac{sin 2theta}{sin theta} = frac{sin (180 - 5theta)}{sin 2theta} Longrightarrow sin^2 2theta = sin 5theta sin theta.]$
Pythagorean and product-to-sum identities yield

$[1 - cos^2 2 theta = frac{cos 4theta - cos 6 theta}{2},]$

and the double and triple angle ( $cos 3x = 4cos^3 x - 3cos x$ ) formulas further simplify this to

$[4cos^3 2theta - 4cos^2 2theta - 3cos 2theta + 3 = (4cos^2 2theta - 3)(cos 2theta - 1) = 0]$

The only value of $theta$ that fits in this context comes from $4 cos^2 2theta - 3 = 0 Longrightarrow cos 2theta = frac{sqrt{3}}{2} Longrightarrow theta = 15^{circ}$ . The answer is $lfloor 1000r rfloor = leftlfloor 1000 cdot frac{180 - 5theta}{180 - 3theta} rightrfloor = left lfloor frac{7000}{9} right rfloor = boxed{777}$ .