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2005AIME II真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月9日下午5:30

2005AIME II真题及答案解析

答案解析请参考文末

Problem 1

A game uses a deck of $n$ different cards, where $n$ is an integer and $n geq 6.$ The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find $n.$

Problem 2

A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$

Problem 3

An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $frac mn$ where $m$ and $n$ are relatively prime integers. Find $m+n.$

Problem 4

Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$

Problem 5

Determine the number of ordered pairs $(a,b)$ of integers such that $log_a b + 6log_b a=5, 2 leq a leq 2005,$ and $2 leq b leq 2005.$

Problem 6

The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. Find the number of cards in the magical stack in which card number 131 retains its original position.

Not the actual problem statement, but to clarify: "retains its original position" means as we count from the top of the stack, the first number is position 1, the second is position 2, and so the 131st number down is position 131. This question is saying that the 131st number down is equal to 131. (From someone with interpretation issues, I thought it would make this question clearer.)

Problem 7

Let $x=frac{4}{(sqrt{5}+1)(sqrt[4]{5}+1)(sqrt[8]{5}+1)(sqrt[16]{5}+1)}.$ Find $(x+1)^{48}.$

Problem 8

Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $frac{msqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

Problem 9

For how many positive integers $n$ less than or equal to 1000 is $(sin t + i cos t)^n = sin nt + i cos nt$ true for all real $t$ ?

Problem 10

Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$

Problem 11

Let $m$ be a positive integer, and let $a_0, a_1,ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - frac 3{a_k}$ for $k = 1,2,ldots, m-1.$ Find $m.$

Problem 12

Square $ABCD$ has center $O, AB=900, E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, mangle EOF =45^circ,$ and $EF=400.$ Given that $BF=p+qsqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$

Problem 13

Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1cdot n_2.$

Problem 14

In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $overline{BC}$ with $CD=6.$ Point $E$ is on $overline{BC}$ such that $angle BAEcong angle CAD.$ Given that $BE=frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$

Problem 15

Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$

2005AIME II详细解析

The number of ways to draw six cards from $n$ is given by the binomial coefficient ${n choose 6} = frac{ncdot(n-1)cdot(n-2)cdot(n-3)cdot(n-4)cdot(n-5)}{6cdot5cdot4cdot3cdot2cdot1}$ .The number of ways to choose three cards from $n$ is ${nchoose 3} = frac{ncdot(n-1)cdot(n-2)}{3cdot2cdot1}$ .We are given that ${nchoose 6} = 6 {n choose 3}$ , so $frac{ncdot(n-1)cdot(n-2)cdot(n-3)cdot(n-4)cdot(n-5)}{6cdot5cdot4cdot3cdot2cdot1} = 6 frac{ncdot(n-1)cdot(n-2)}{3cdot2cdot1}$ .Cancelling like terms, we get $(n - 3)(n - 4)(n - 5) = 720$ .We must find a factorization of the left-hand side of this equation into three consecutive integers. Since 720 is close to $9^3=729$ , we try 8, 9, and 10, which works, so $n - 3 = 10$ and $n = boxed{013}$ .
Use construction. We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.
- Person 1: $frac{9 cdot 6 cdot 3}{9 cdot 8 cdot 7} = frac{9}{28}$
- Person 2: $frac{6 cdot 4 cdot 2}{6 cdot 5 cdot 4} = frac 25$
- Person 3: One roll of each type is left, so the probability here is $1$ .
Our answer is thus $frac{9}{28} cdot frac{2}{5} = frac{9}{70}$ , and $m + n = boxed{79}$ .

Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in $left(frac{3!}{1!1!1!}right)^3 = 6^3 = 216$ different manners. The total number of possible strings is $frac{9!}{3!3!3!} = 1680$ . The solution is therefore $frac{216}{1680} = frac{9}{70}$ , and $m + n = boxed{79}$ .
Let's call the first term of the original geometric series $a$ and the common ratio $r$ , so $2005 = a + ar + ar^2 + ldots$ . Using the sum formula for infinite geometric series, we have $;;frac a{1 -r} = 2005$ . Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + ldots$ . We know this series has sum $20050 = frac{a^2}{1 - r^2}$ . Dividing this equation by $frac{a}{1-r}$ , we get $10 = frac a{1 + r}$ . Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$ , $1995 = 2015r$ and finally $r = frac{1995}{2015} = frac{399}{403}$ , so the answer is $399 + 403 = boxed{802}$ .(We know this last fraction is fully reduced by the Euclidean algorithm -- because $4 = 403 - 399$ , $gcd(403, 399) | 4$ . But 403 is odd, so $gcd(403, 399) = 1$ .)We can write the sum of the original series as $a + aleft(dfrac{m}{n}right) + aleft(dfrac{m}{n}right)^2 + ldots = 2005$ , where the common ratio is equal to $dfrac{m}{n}$ . We can also write the sum of the second series as $a^2 + a^2left(dfrac{m}{n}right)^2 + a^2left(left(dfrac{m}{n}right)^2right)^2 + ldots = 20050$ . Using the formula for the sum of an infinite geometric series $S=dfrac{a}{1-r}$ , where $S$ is the sum of the sequence, $a$ is the first term of the sequence, and $r$ is the ratio of the sequence, the sum of the original series can be written as $dfrac{a}{1-frac{m}{n}}=dfrac{a}{frac{n-m}{n}}=dfrac{a cdot n}{n-m}=2005;text{(1)}$ , and the second sequence can be written as $dfrac{a^2}{1-frac{m^2}{n^2}}=dfrac{a^2}{frac{n^2-m^2}{n^2}}=dfrac{a^2cdot n^2}{(n+m)(n-m)}=20050;text{(2)}$ . Dividing $text{(2)}$ by $text{(1)}$ , we obtain $dfrac{acdot n}{m+n}=10$ , which can also be written as $acdot n=10(m+n)$ . Substitute this value for $acdot n$ back into $text{(1)}$ , we obtain $10cdot dfrac{n+m}{n-m}=2005$ . Dividing both sides by 10 yields $dfrac{n+m}{n-m}=dfrac{401}{2}$ we can now write a system of equations with $n+m=401$ and $n-m=2$ , but this does not output integer solutions. However, we can also write $dfrac{n+m}{n-m}=dfrac{401}{2}$ as $dfrac{n+m}{n-m}=dfrac{802}{4}$ . This gives the system of equations $m+n=802$ and $n-m=4$ , which does have integer solutions. Our answer is therefore $m+n=boxed{802}$ (Solving for $m$ and $n$ gives us $399$ and $403$ , respectively, which are co-prime).
$10^{10} = 2^{10}cdot 5^{10}$ so $10^{10}$ has $11cdot11 = 121$ divisors. $15^7 = 3^7cdot5^7$ so $15^7$ has $8cdot8 = 64$ divisors. $18^{11} = 2^{11}cdot3^{22}$ so $18^{11}$ has $12cdot23 = 276$ divisors.Now, we use the Principle of Inclusion-Exclusion. We have $121 + 64 + 276$ total potential divisors so far, but we've overcounted those factors which divide two or more of our three numbers. Thus, we must subtract off the divisors of their pair-wise greatest common divisors. $gcd(10^{10},15^7) = 5^7$ which has 8 divisors. $gcd(15^7, 18^{11}) = 3^7$ which has 8 divisors. $gcd(18^{11}, 10^{10}) = 2^{10}$ which has 11 divisors.So now we have $121 + 64 + 276 - 8 -8 -11$ potential divisors. However, we've now undercounted those factors which divide all three of our numbers. Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of $121 + 64 + 276 - 8 - 8 - 11 + 1 = boxed{435}$ .
The equation can be rewritten as Multiplying through by and factoring yields . Therefore, or , so either or .
- For the case $b=a^2$ , note that $44^2=1936$ and $45^2=2025$ . Thus, all values of $a$ from $2$ to $44$ will work.
- For the case $b=a^3$ , note that $12^3=1728$ while $13^3=2197$ . Therefore, for this case, all values of $a$ from $2$ to $12$ work.
There are $44-2+1=43$ possibilities for the square case and $12-2+1=11$ possibilities for the cube case. Thus, the answer is $43+11= boxed{054}$ .

Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs $(a,b)$ , and not for the number of possible values of $b$ . Were the problem to ask for the number of possible values of $b$ , the values of $b^6$ under $2005$ would have to be subtracted, which would just be $2$ values: $2^6$ and $3^6$ . However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)
Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly $131 - 1 = 130$ numbers must be in front of it. There are $frac{130}{2} = 65$ cards from each of piles A, B in front of card 131. This suggests that $n = 131 + 65 = 196$ ; the total number of cards is $196 cdot 2 = boxed{392}$ .
We note that in general, ${} (sqrt[2^n]{5} + 1)(sqrt[2^n]{5} - 1) = (sqrt[2^n]{5})^2 - 1^2 = sqrt[2^{n-1}]{5} - 1$ .It now becomes apparent that if we multiply the numerator and denominator of $frac{4}{ (sqrt{5}+1) (sqrt[4]{5}+1) (sqrt[8]{5}+1) (sqrt[16]{5}+1) }$ by $(sqrt[16]{5} - 1)$ , the denominator will telescope to $sqrt[1]{5} - 1 = 4$ , so $x = frac{4(sqrt[16]{5} - 1)}{4} = sqrt[16]{5} - 1$ .It follows that $(x + 1)^{48} = (sqrt[16]5)^{48} = 5^3 = boxed{125}$ .
Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$ . Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$ , respectively, and let the extension of $overline{T_1T_2}$ intersect the extension of $overline{O_1O_2}$ at a point $H$ . Let the endpoints of the chord/tangent be $A,B$ , and the foot of the perpendicular from $O_3$ to $overline{AB}$ be $T$ . From the similar right triangles $triangle HO_1T_1 sim triangle HO_2T_2 sim triangle HO_3T$ , $[frac{HO_1}{4} = frac{HO_1+14}{10} = frac{HO_1+10}{O_3T}.]$ It follows that $HO_1 = frac{28}{3}$ , and that $O_3T = frac{58}{7}$ . By the Pythagorean Theorem on $triangle ATO_3$ , we find that $[AB = 2AT = 2left(sqrt{r_3^2 - O_3T^2}right) = 2sqrt{14^2 - frac{58^2}{7^2}} = frac{8sqrt{390}}{7}]$ and the answer is $m+n+p=boxed{405}$ .
We know by De Moivre's Theorem that $(cos t + i sin t)^n = cos nt + i sin nt$ for all real numbers $t$ and all integers $n$ . So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.Recall the trigonometric identities $cos left(frac{pi}2 - uright) = sin u$ and $sin left(frac{pi}2 - uright) = cos u$ hold for all real $u$ . If our original equation holds for all $t$ , it must certainly hold for $t = frac{pi}2 - u$ . Thus, the question is equivalent to asking for how many positive integers $n leq 1000$ we have that $left(sinleft(fracpi2 - uright) + i cosleft(fracpi 2 - uright)right)^n = sin n left(fracpi2 -u right) + icos n left(fracpi2 - uright)$ holds for all real $u$ . $left(sinleft(fracpi2 - uright) + i cosleft(fracpi 2 - uright)right)^n = left(cos u + i sin uright)^n = cos nu + isin nu$ . We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all $n$ such that $cos n u = sin nleft(fracpi2 - uright)$ and $sin nu = cos nleft(fracpi2 - uright)$ hold for all real $u$ . $sin x = cos y$ if and only if either $x + y = frac pi 2 + 2pi cdot k$ or $x - y = fracpi2 + 2picdot k$ for some integer $k$ . So from the equality of the real parts we need either $nu + nleft(fracpi2 - uright) = fracpi 2 + 2pi cdot k$ , in which case $n = 1 + 4k$ , or we need $-nu + nleft(fracpi2 - uright) = fracpi 2 + 2pi cdot k$ , in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$ . Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 pmod 4$ work. There are $boxed{250}$ of them in the given range.
Let the side of the octahedron be of length $s$ . Let the vertices of the octahedron be $A, B, C, D, E, F$ so that $A$ and $F$ are opposite each other and $AF = ssqrt2$ . The height of the square pyramid $ABCDE$ is $frac{AF}2 = frac s{sqrt2}$ and so it has volume $frac 13 s^2 cdot frac s{sqrt2} = frac {s^3}{3sqrt2}$ and the whole octahedron has volume $frac {s^3sqrt2}3$ .Let $M$ be the midpoint of $BC$ , $N$ be the midpoint of $DE$ , $G$ be the centroid of $triangle ABC$ and $H$ be the centroid of $triangle ADE$ . Then $triangle AMN sim triangle AGH$ and the symmetry ratio is $frac 23$ (because the medians of a triangle are trisected by the centroid), so $GH = frac{2}{3}MN = frac{2s}3$ . $GH$ is also a diagonal of the cube, so the cube has side-length $frac{ssqrt2}3$ and volume $frac{2s^3sqrt2}{27}$ . The ratio of the volumes is then $left(frac{2s^3sqrt2}{27}right)big/left(frac{s^3sqrt2}{3}right) = frac29$ and so the answer is $boxed{011}$ .
For $0 < k < m$ , we have $a_{k}a_{k+1} = a_{k-1}a_{k} - 3$ .Thus the product $a_{k}a_{k+1}$ is a monovariant: it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37cdot 72$ , so when $k = frac{37 cdot 72}{3} = 888$ , $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = boxed{889}$ , our answer.
Let $G$ be the foot of the perpendicular from $O$ to $AB$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). Then $tan angle EOG = frac{x}{450}$ , and $tan angle FOG = frac{y}{450}$ .By the tangent addition rule $left( tan (a + b) = frac{tan a + tan b}{1 - tan a tan b} right)$ , we see that $[tan 45 = tan (EOG + FOG) = frac{frac{x}{450} + frac{y}{450}}{1 - frac{x}{450} cdot frac{y}{450}}.]$ Since $tan 45 = 1$ , this simplifies to $1 - frac{xy}{450^2} = frac{x + y}{450}$ . We know that $x + y = 400$ , so we can substitute this to find that $1 - frac{xy}{450^2} = frac 89 Longrightarrow xy = 150^2$ .Substituting $x = 400 - y$ again, we know have $xy = (400 - y)y = 150^2$ . This is a quadratic with roots $200 pm 50sqrt{7}$ . Since $y < x$ , use the smaller root, $200 - 50sqrt{7}$ .Now, $BF = BG - FG = 450 - (200 - 50sqrt{7}) = 250 + 50sqrt{7}$ . The answer is $250 + 50 + 7 = boxed{307}$ .
We define $Q(x)=P(x)-x+7$ , noting that it has roots at $17$ and $24$ . Hence $P(x)-x+7=A(x-17)(x-24)$ . In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$ . Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$ , where $A$ , $(x-17)$ , and $(x-24)$ are integers. This cannot occur if $xle 17$ or $xge 24$ because the product $(x-17)(x-24)$ will either be too large or not be a divisor of $10$ . We find that $x=19$ and $x=22$ are the only values that allow $(x-17)(x-24)$ to be a factor of $10$ . Hence the answer is $19cdot 22=boxed{418}$ .
By the Law of Sines and since $angle BAE = angle CAD, angle BAD = angle CAE$ , we have $begin{align*} frac{CD cdot CE}{AC^2} &= frac{sin CAD}{sin ADC} cdot frac{sin CAE}{sin AEC} \ &= frac{sin BAE sin BAD}{sin ADB sin AEB} \ &= frac{sin BAE}{sin AEB} cdot frac{sin BAD}{sin ADB}\ &= frac{BE cdot BD}{AB^2} end{align*}$ Substituting our knowns, we have $frac{CE}{BE} = frac{3 cdot 14^2}{2 cdot 13^2} = frac{BC - BE}{BE} = frac{15}{BE} - 1 Longrightarrow BE = frac{13^2 cdot 15}{463}$ . The answer is $q = boxed{463}$ .
Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$ .Let $w_3$ have center $(x,y)$ and radius $r$ . Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$ , and if they are internally tangent, it is $|r_1 - r_2|$ . So we have $begin{align*} r + 4 &= sqrt{(x-5)^2 + (y-12)^2} \ 16 - r &= sqrt{(x+5)^2 + (y-12)^2} end{align*}$ Solving for $r$ in both equations and setting them equal, then simplifying, yields $begin{align*} 20 - sqrt{(x+5)^2 + (y-12)^2} &= sqrt{(x-5)^2 + (y-12)^2} \ 20+x &= 2sqrt{(x+5)^2 + (y-12)^2} end{align*}$ Squaring again and canceling yields $1 = frac{x^2}{100} + frac{(y-12)^2}{75}.$ So the locus of points that can be the center of the circle with the desired properties is an ellipse. $[asy] size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d; pair A = (-5, 12), B = (5, 12), C = (0, 0); D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red); D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow(4)); void bluecirc (real x) { pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue); D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype("4 4")); } bluecirc(-9.2); bluecirc(-4); bluecirc(3); [/asy]$ Since the center lies on the line $y = ax$ , we substitute for $y$ and expand: $[1 = frac{x^2}{100} + frac{(ax-12)^2}{75} Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.]$
We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$ , so $(-96a)^2 - 4(3+4a^2)(276) = 0$ .

Solving yields $a^2 = frac{69}{100}$ , so the answer is $boxed{169}$ .