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2004AIME I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5:42

2004AIME I真题及答案解析

Problem 1

The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by 37?

Problem 2

Set $A$ consists of $m$ consecutive integers whose sum is $2m,$ and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is 99. Find $m.$

Problem 3

A convex polyhedron $P$ has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have?

Problem 4

A square has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k.$ Find $100k.$

Problem 5

Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

Problem 6

An integer is called snakelike if its decimal representation $a_1a_2a_3cdots a_k$ satisfies $a_i<a_{i+1}$ if $i$ is odd and $a_i>a_{i+1}$ if $i$ is even. How many snakelike integers between 1000 and 9999 have four distinct digits?

Problem 7

Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)cdots(1 + 14x)(1 - 15x).$ Find $|C|.$

Problem 8

Define a regular $n$ -pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,ldots, P_nP_1$ such that

the points $P_1, P_2,ldots, P_n$ are coplanar and no three of them are collinear,
each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint,
all of the angles at $P_1, P_2,ldots, P_n$ are congruent,
all of the $n$ line segments $P_2P_3,ldots, P_nP_1$ are congruent, and
the path $P_1P_2, P_2P_3,ldots, P_nP_1$ turns counterclockwise at an angle of less than 180 degrees at each vertex.

There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?

Problem 9

Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2.$ The minimum value of the area of $U_1$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Problem 10

A circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Problem 11

A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum-shaped solid $F,$ in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k.$ Given that $k=m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Problem 12

Let $S$ be the set of ordered pairs $(x, y)$ such that $0 < x le 1, 0<yle 1,$ and $left lfloor{log_2{left(frac 1xright)}}right rfloor$ and $left lfloor{log_5{left(frac 1yright)}}right rfloor$ are both even. Given that the area of the graph of $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ The notation $left lfloor{z}right rfloor$ denotes the greatest integer that is less than or equal to $z.$

Problem 13

The polynomial $P(x)=(1+x+x^2+cdots+x^{17})^2-x^{17}$ has 34 complex roots of the form $z_k = r_k[cos(2pi a_k)+isin(2pi a_k)], k=1, 2, 3,ldots, 34,$ with $0 < a_1 le a_2 le a_3 le cdots le a_{34} < 1$ and $r_k>0.$ Given that $a_1 + a_2 + a_3 + a_4 + a_5 = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Problem 14

A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower is $frac{a-sqrt{b}}c$ feet, where $a, b,$ and $c$ are positive integers, and $c$ is prime. Find $a+b+c.$

Problem 15

For all positive integers $x$ , let $[f(x)=begin{cases}1 &mbox{if }x = 1\ frac x{10} &mbox{if }xmbox{ is divisible by 10}\ x+1 &mbox{otherwise}end{cases}]$ and define a sequence as follows: $x_1=x$ and $x_{n+1}=f(x_n)$ for all positive integers $n$ . Let $d(x)$ be the smallest $n$ such that $x_n=1$ . (For example, $d(100)=3$ and $d(87)=7$ .) Let $m$ be the number of positive integers $x$ such that $d(x)=20$ . Find the sum of the distinct prime factors of $m$ .

2004AIME I详细解析

A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${underline{(n+3)}},{underline{(n+2)}},{underline{( n+1)}},{underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n in lbrace0, 1, 2, 3, 4, 5, 6rbrace$ .Now, note that $3cdot 37 = 111$ so $30 cdot 37 = 1110$ , and $90 cdot 37 = 3330$ so $87 cdot 37 = 3219$ . So the remainders are all congruent to $n - 9 pmod{37}$ . However, these numbers are negative for our choices of $n$ , so in fact the remainders must equal $n + 28$ .Adding these numbers up, we get $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7cdot28 = boxed{217}$ , our answer.
Look at the problem... consecutive integers. Now, since set $A$ has the properties of $m$ integers that sum to $2m$ , it's obvious that the middle integer is just $2$ (the average that "balances out" everything), and the largest is $2 + frac{m-1}{2}$ . That's because there are $m-1$ to go after taking $2$ , and they are "evenly balanced" on either side.From there, we see that set $B$ 's average is $0.5$ . How can that happen? Only if the middle TWO values are $0$ and $1$ . Since set $B$ has $2m$ integers, its maximum must be $99$ larger than the maximum of set $A$ unless $m=1$ , which is impossible.From there, the largest element of set $B$ is $1 + (m-1) = m$ .Solving, we get $[m - 2 - frac{m-1}{2} = 99]$ $[m-frac{m}{2}+frac{1}{2}=101]$ $[frac{m}{2}=100frac{1}{2}.]$ There we go- $m$ is equal to none other than $boxed{201}$ .
Every pair of vertices of the polyhedron determines either an edge, a face diagonal or a space diagonal. We have ${26 choose 2} = frac{26cdot25}2 = 325$ total line segments determined by the vertices. Of these, $60$ are edges. Each triangular face has $0$ face diagonals and each quadrilateral face has $2$ , so there are $2 cdot 12 = 24$ face diagonals. This leaves $325 - 60 - 24 = boxed{241}$ segments to be the space diagonals.
Without loss of generality, let $(0,0)$ , $(2,0)$ , $(0,2)$ , and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$ . Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$ . Because the segment has length 2, $x^2+y^2=4$ . Using the midpoint formula, we find that the midpoint of the segment has coordinates $left(frac{x}{2},frac{y}{2}right)$ . Let $d$ be the distance from $(0,0)$ to $left(frac{x}{2},frac{y}{2}right)$ . Using the distance formula we see that $d=sqrt{left(frac{x}{2}right)^2+left(frac{y}{2}right)^2}= sqrt{frac{1}{4}left(x^2+y^2right)}=sqrt{frac{1}{4}(4)}=1$ . Thus the midpoints lying on the sides determined by vertex $(0,0)$ form a quarter-circle with radius 1. $[asy] size(100); pointpen=black;pathpen = black+linewidth(0.7); pair A=(0,0),B=(2,0),C=(2,2),D=(0,2); D(A--B--C--D--A); picture p; draw(p,CR(A,1));draw(p,CR(B,1));draw(p,CR(C,1));draw(p,CR(D,1)); clip(p,A--B--C--D--cycle); add(p); [/asy]$ The set of all midpoints forms a quarter circle at each corner of the square. The area enclosed by all of the midpoints is $4-4cdot left(frac{pi}{4}right)=4-pi approx .86$ to the nearest hundredth. Thus $100cdot k=boxed{086}$ $[asy] size(100); pointpen=black;pathpen = black+linewidth(0.7); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw(arc((0,2),1,270,360)); draw((0,1)--(1.7,2)); draw((0,2)--(1.7,1)); draw((0,1)--(1.7,1)--(1.7,2)); [/asy]$ If we imagine an arbitrary line with length $2$ connecting two sides of the square, we can draw the rectangle formed by drawing a perpendicular from where that line touches the square.Drawing the other diagonal of the rectangle, it also has length two, and it bisects with the original line. Since their intersection is the midpoint of both lines, the distance from the corner to the midpoint is always $1$ , which forms a circle with radius $1$ centered at the corner of the square.The area of the shape then follows from simple calculations.
Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number it gets right. Let $b$ be the number it gets right on day 2.These inequalities follow: $[frac{a}{q} < frac{160}{300} = frac{8}{15}]$ $[frac{b}{500-q} < frac{140}{200} = frac{7}{10}]$ Solving for a and b and adding the two inequalities: $[a + b < frac{8}{15}q + (350 - frac{7}{10}q)]$ $[a + b < 350 - frac{1}{6}q]$ From here, we see the largest possible value of $a+b$ is $349$ .Checking our conditions, we know that $a$ must be positive so therefore $q$ must be positive. A quick check shows that for $2le q le 5$ , $q$ follows all the conditions and results in $a+b=349$ .This makes Beta's success ratio $frac{349}{500}$ . Thus, the answer is $m+n = 349 + 500 = boxed{849}$ .We see that we want Beta to have more points where there is a higher Alpha success rate (that way, the score is shifted more towards the higher Alpha score). With that in mind, .7 > 8/15. Thus, we might as well put as many points as possible into the second day.That said, we need Beta to have a positive integer score on either end per requirements. The best way is for Beta to go 1/2 on the first day. That way, we "waste" the least points- both in the numerator and denominator.Thereafter, we see that, letting $x$ be the number of points Beta gained on the second day, $x/498 < 7/10$ ; thus $max(x) = 348$ .Aha. 348+1 = 349.
We divide the problem into two cases: one in which zero is one of the digits and one in which it is not. In the latter case, suppose we pick digits $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ . There are five arrangements of these digits that satisfy the condition of being snakelike: $x_1x_3x_2x_4$ , $x_1x_4x_2x_3$ , $x_2x_3x_1x_4$ , $x_2x_4x_1x_3$ , $x_3x_4x_1x_2$ . Thus there are $5cdot {9choose 4}=630$ snakelike numbers which do not contain the digit zero.In the second case we choose zero and three other digits such that $0<x_2<x_3<x_4$ . There are three arrangements of these digits that satisfy the condition of being snakelike: $x_2x_30x_4$ , $x_2x_40x_3$ , $x_3x_40x_2$ . Because we know that zero is a digit, there are $3cdot{9choose 3}=252$ snakelike numbers which contain the digit zero. Thus there are $630+252=boxed{882}$ snakelike numbers.Let's create the snakelike number from digits $a < b < c < d$ , and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of $5cdot{10 choose 4}$ But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits $a < b < c$ . There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is $2cdot{9 choose 3}$ . Thus our answer is $5cdot{10 choose 4} - 2cdot{9 choose 3} = boxed{882}$
Let our polynomial be $P(x)$ .It is clear that the coefficient of $x$ in $P(x)$ is $-1 + 2 - 3 + ldots + 14 - 15 = -8$ , so $P(x) = 1 -8x + Cx^2 + Q(x)$ , where $Q(x)$ is some polynomial divisible by $x^3$ .Then $P(-x) = 1 + 8x + Cx^2 + Q(-x)$ and so $P(x)cdot P(-x) = 1 + (2C - 64)x^2 + R(x)$ , where $R(x)$ is some polynomial divisible by $x^3$ .However, we also know $P(x)cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) cdots (1 - 15x)(1 + 15x)$ $= (1 - x^2)(1 - 4x^2)cdots(1 - 225x^2)$ $= 1 - (1 + 4 + ldots + 225)x^2 + R(x)$ .Equating coefficients, we have $2C - 64 = -(1 + 4 + ldots + 225) = -1240$ , so $-2C = 1176$ and $|C| = boxed{588}$ .Let $S$ be the set of integers ${-1,2,-3,ldots,14,-15}$ . The coefficient of $x^2$ in the expansion is equal to the sum of the product of each pair of distinct terms, or $C = sum_{1 le i neq j}^{15} S_iS_j$ . Also, we know that $begin{align*}left(sum_{i=1}^{n} S_iright)^2 &= left(sum_{i=1}^{n} S_i^2right) + 2left(sum_{1 le i neq j}^{15} S_iS_jright)\ (-8)^2 &= frac{15(15+1)(2cdot 15+1)}{6} + 2Cend{align*}$ where the left-hand sum can be computed from: $sum_{i=1}^{15} S_i = S_{15} + left(sum_{i=1}^{7} S_{2i-1} + S_{2i}right) = -15 + 7 = -8$ and the right-hand sum comes from the formula for the sum of the first $n$ perfect squares. Therefore, $|C| = left|frac{64-1240}{2}right| = boxed{588}$ .
We use the Principle of Inclusion-Exclusion (PIE).If we join the adjacent vertices of the regular $n$ -star, we get a regular $n$ -gon. We number the vertices of this $n$ -gon in a counterclockwise direction: $0, 1, 2, 3, ldots, n-1.$ A regular $n$ -star will be formed if we choose a vertex number $m$ , where $0 le m le n-1$ , and then form the line segments by joining the following pairs of vertex numbers: $(0 mod{n}, m mod{n}),$ $(m mod{n}, 2m mod{n}),$ $(2m mod{n}, 3m mod{n}),$ $cdots,$ $((n-2)m mod{n}, (n-1)m mod{n}),$ $((n-1)m mod{n}, 0 mod{n}).$ If $gcd(m,n) > 1$ , then the star degenerates into a regular $frac{n}{gcd(m,n)}$ -gon or a (2-vertex) line segment if $frac{n}{gcd(m,n)}= 2$ . Therefore, we need to find all $m$ such that $gcd(m,n) = 1$ .Note that $n = 1000 = 2^{3}5^{3}.$ Let $S = {1,2,3,ldots, 1000}$ , and $A_{i}= {i in S mid i, textrm{ divides },1000}$ . The number of $m$ 's that are not relatively prime to $1000$ is: $mid A_{2}cup A_{5}mid = mid A_{2}mid+mid A_{5}mid-mid A_{2}cap A_{5}mid$ $= leftlfloor frac{1000}{2}rightrfloor+leftlfloor frac{1000}{5}rightrfloor-leftlfloor frac{1000}{2 cdot 5}rightrfloor$ $= 500+200-100 = 600.$ Vertex numbers $1$ and $n-1=999$ must be excluded as values for $m$ since otherwise a regular n-gon, instead of an n-star, is formed.The cases of a 1st line segment of (0, m) and (0, n-m) give the same star. Therefore we should halve the count to get non-similar stars.
Therefore, the number of non-similar 1000-pointed stars is $frac{1000-600-2}{2}= boxed{199}.$

Note that in general, the number of $n$ -pointed stars is given by $frac{phi(n)}{2} - 1$ (dividing by $2$ to remove the reflectional symmetry, subtracting $1$ to get rid of the $1$ -step case), where $phi(n)$ is the Euler's totient function. It is well-known that $phi(n) = nleft(1-frac{1}{p_1}right)left(1-frac{1}{p_2}right)cdots left(1-frac{1}{p_n}right)$ , where $p_1,,p_2,ldots,,p_n$ are the distinct prime factors of $n$ . Thus $phi(1000) = 1000left(1 - frac 12right)left(1 - frac 15right) = 400$ , and the answer is $frac{400}{2} - 1 = 199$ .
We let for the purpose of labeling. Clearly, the dividing segment in must go through one of its vertices, without loss of generality . The other endpoint () of the segment can either lie on or . is a trapezoid with a right angle then, from which it follows that contains one of the right angles of , and so is similar to . Thus , and hence , are s.Suppose we find the ratio of the smaller base to the larger base for , which consequently is the same ratio for . By similar triangles, it follows that by the same ratio , and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, it follows that .

$[asy] defaultpen(linewidth(0.7)); pair A=(0,0),B=(0,3),C=(4,0); draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); draw((9*4/14,0)--(9*4/14,5*3/14),dashed); label("(V_1)",(1,1.2)); label("(U_1)",(3,0.3)); [/asy]$ $[asy] defaultpen(linewidth(0.7)); pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(4.5,6); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,N)),dashed); label("(U_2)",(1,3)); label("(V_2)",(5,3)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("9/2",(E+H)/2,N); [/asy]$

$[asy] defaultpen(linewidth(0.7)); pair A=(0,0),B=(0,3),C=(4,0); draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); draw((3.5,0)--(3.5,3/8),dashed); label("(V_1)",(1.5,1)); label("(U_1)",(3.8,0.4)); [/asy]$ $[asy] defaultpen(linewidth(0.7)); pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(7,21/4); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,NW)),dashed); label("(V_2)",(2,3)); label("(U_2)",(5,1)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("21/4",(G+H)/2,(-1,0)); [/asy]$
- If $D'$ lies on $overline{EF}$ , then $ED' = frac{9}{2},, 8$ ; the latter can be discarded as extraneous. Therefore, $D'F = frac{5}{2}$ , and the ratio $r = frac{D'F}{DG} = frac{5}{14}$ . The area of $[U_1] = 6left(frac{5}{14}right)^2$ in this case.
- If $D'$ lies on $overline{FG}$ , then $GD' = frac{21}{4},, frac{28}{3}$ ; the latter can be discarded as extraneous. Therefore, $D'F = frac{3}{4}$ , and the ratio $r = frac{D'F}{DE} = frac{1}{8}$ . The area of $[U_1] = 6left(frac{1}{8}right)^2$ in this case.
Of the two cases, the second is smaller; the answer is $frac{3}{32}$ , and $m+n = boxed{035}$ .
The location of the center of the circle must be in the $34 times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $overline{AC}$ . Let this triangle be $A'B'C'$ .Notice that $ABC$ and $A'B'C'$ share the same incenter; this follows because the corresponding sides are parallel, and so the perpendicular inradii are concurrent, except that the inradii of $triangle ABC$ extend one unit farther than those of $triangle A'B'C'$ . From $A = rs$ , we note that $r_{ABC} = frac{[ABC]}{s_{ABC}} = frac{15 cdot 36 /2}{(15+36+39)/2} = 6$ . Thus $r_{A'B'C'} = r_{ABC} - 1 = 5$ , and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, $[A'B'C'] = [ABC] cdot left(frac{r_{A'B'C'}}{r_{ABC}}right)^2 = frac{15 times 36}{2} cdot frac{25}{36} = frac{375}{2}$ .The probability is $frac{2[A'B'C']}{34 times 13} = frac{375}{442}$ , and $m + n = boxed{817}$ .
Our original solid has volume equal to $V = frac13 pi r^2 h = frac13 pi 3^2cdot 4 = 12 pi$ and has surface area $A = pi r^2 + pi r ell$ , where $ell$ is the slant height of the cone. Using the Pythagorean Theorem, we get $ell = 5$ and $A = 24pi$ .Let $x$ denote the radius of the small cone. Let $A_c$ and $A_f$ denote the area of the painted surface on cone $C$ and frustum $F$ , respectively, and let $V_c$ and $V_f$ denote the volume of cone $C$ and frustum $F$ , respectively. Because the plane cut is parallel to the base of our solid, $C$ is similar to the uncut solid and so the height and slant height of cone $C$ are $frac{4}{3}x$ and $frac{5}{3}x$ , respectively. Using the formula for lateral surface area of a cone, we find that $A_c=frac{1}{2}ccdot ell=frac{1}{2}(2pi x)left(frac{5}{3}xright)=frac{5}{3}pi x^2$ . By subtracting $A_c$ from the surface area of the original solid, we find that $A_f=24pi - frac{5}{3}pi x^2$ .Next, we can calculate $V_c=frac{1}{3}pi r^2h=frac{1}{3}pi x^2 left(frac{4}{3}xright)=frac{4}{9}pi x^3$ . Finally, we subtract $V_c$ from the volume of the original cone to find that $V_f=12pi - frac{4}{9}pi x^3$ . We know that $frac{A_c}{A_f}=frac{V_c}{V_f}=k.$ Plugging in our values for $A_c$ , $A_f$ , $V_c$ , and $V_f$ , we obtain the equation $frac{frac{5}{3}pi x^2}{24pi - frac{5}{3}pi x^2}=frac{frac{4}{9}pi x^3}{12pi - frac{4}{9}pi x^3}$ . We can take reciprocals of both sides to simplify this equation to $frac{72}{5x^2} - 1 = frac{27}{x^3} - 1$ and so $x = frac{15}{8}$ . Then $k = frac{frac{5}{3}pi x^2}{24pi - frac{5}{3}pi x^2}= frac{125}{387} = frac mn$ so the answer is $m+n=125+387=boxed{512}$ .Our original solid $V$ has surface area $A_v = pi r^2 + pi r ell$ , where $ell$ is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain $ell = 5$ and lateral area $A_ell = 15pi$ . The area of the base is $A_B = 3^2pi = 9pi$ . $V$ and $C$ are similar cones, because the plane that cut out $C$ was parallel to the base of $V$ . Let $x$ be the scale factor between the original cone and the small cone $C$ in one dimension. Because the scale factor is uniform in all dimensions, $x^2$ relates corresponding areas of $C$ and $V$ , and $x^3$ relates corresponding volumes. Then, the ratio of the painted areas $frac{A_c}{A_f}$ is $frac{15pi x^2}{9pi + 15pi - 15pi x^2} = frac{5 x^2}{8 - 5 x^2} = k$ and the ratio of the volumes $frac{V_c}{V_f}$ is $frac{x^3}{1 - x^3} = k$ . Since both ratios are equal to $k$ , they are equal to each other. Therefore, $frac{5 x^2}{8 - 5 x^2} = frac{x^3}{1 - x^3}$ .Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives $5 x^2(1 - x^3) = x^3(8 - 5 x^2)$ . Dividing both sides by $x^2$ and distributing the $x$ on the right, we have $5 - 5 x^3 = 8 x - 5 x^3$ , and so $8 x = 5$ and $x = frac{5}{8}$ . Substituting back into the easier ratio, we have $frac{(frac{5}{8})^3}{1 - (frac{5}{8})^3} = frac{frac{125}{512}}{frac{387}{512}} = frac{125}{387}$ . And so we have $m + n = 125 + 387 = boxed{512}$ .
$leftlfloorlog_2left(frac{1}{x}right)rightrfloor$ is even when $[x in left(frac{1}{2},1right) cup left(frac{1}{8},frac{1}{4}right) cup left(frac{1}{32},frac{1}{16}right) cup cdots]$ Likewise: $leftlfloorlog_5left(frac{1}{y}right)rightrfloor$ is even when $[y in left(frac{1}{5},1right) cup left(frac{1}{125},frac{1}{25}right) cup left(frac{1}{3125},frac{1}{625}right) cup cdots]$ Graphing this yields a series of rectangles which become smaller as you move toward the origin. The $x$ interval of each box is given by the geometric sequence $frac{1}{2} , frac{1}{8}, frac{1}{32}, cdots$ , and the $y$ interval is given by $frac{4}{5} , frac{4}{125}, frac{4}{3125}, cdots$ Each box is the product of one term of each sequence. The sum of these boxes is simply the product of the sum of each sequence or: $[left(frac{1}{2} + frac{1}{8} + frac{1}{32} ldots right)left(frac{4}{5} + frac{4}{125} + frac{4}{3125} ldots right)=left(frac{frac{1}{2}}{1 - frac{1}{4}}right)left(frac{frac{4}{5}}{1-frac{1}{25}}right)= frac{2}{3} cdot frac{5}{6} = frac{5}{9},]$ and the answer is $m+n = 5 + 9 = boxed{014}$ .
We see that the expression for the polynomial $P$ is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series: $begin{align*} P(x) &= left(frac{x^{18} - 1}{x - 1}right)^2 - x^{17} = frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\ &= frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2} end{align*}$ This expression has roots at every $17$ th root and $19$ th roots of unity, other than $1$ . Since $17$ and $19$ are relatively prime, this means there are no duplicate roots. Thus, $a_1, a_2, a_3, a_4$ and $a_5$ are the five smallest fractions of the form $frac m{19}$ or $frac n {17}$ for $m, n > 0$ . $frac 3 {17}$ and $frac 4{19}$ can both be seen to be larger than any of $frac1{19}, frac2{19}, frac3{19}, frac 1{17}, frac2{17}$ , so these latter five are the numbers we want to add. $frac1{19}+ frac2{19}+ frac3{19}+ frac 1{17}+ frac2{17}= frac6{19} + frac 3{17} = frac{6cdot17 + 3cdot19}{17cdot19} = frac{159}{323}$ and so the answer is $159 + 323 = boxed{482}$ .
Looking from an overhead view, call the center of the circle $O$ , the tether point to the unicorn $A$ and the last point where the rope touches the tower $B$ . $triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$ . We use the Pythagorean Theorem to find the horizontal component of $AB$ has length $4sqrt{5}$ . $[asy] defaultpen(fontsize(10)+linewidth(0.62)); pair A=(-4*sqrt(5),4), B=(0,4*(8*sqrt(6)-4*sqrt(5))/(8*sqrt(6))), C=(8*sqrt(6)-4*sqrt(5),0), D=(-4*sqrt(5),0), E=(0,0); draw(A--C--D--A);draw(B--E); label("(A)",A,(-1,1));label("(B)",B,(1,1));label("(C)",C,(1,0));label("(D)",D,(-1,-1));label("(E)",E,(0,-1)); label("$4sqrt{5}$",D/2+E/2,(0,-1));label("$8sqrt{6}-4sqrt{5}$",C/2+E/2,(0,-1)); label("$4$",D/2+A/2,(-1,0));label("$x$",C/2+B/2,(1,0.5));label("$20-x$",0.7*A+0.3*B,(1,0.5)); dot(A^^B^^C^^D^^E); [/asy]$ Now look at a side view and "unroll" the cylinder to be a flat surface. Let $C$ be the bottom tether of the rope, let $D$ be the point on the ground below $A$ , and let $E$ be the point directly below $B$ . Triangles $triangle CDA$ and $triangle CEB$ are similar right triangles. By the Pythagorean Theorem $CD=8cdotsqrt{6}$ .Let $x$ be the length of $CB$ . $[frac{CA}{CD}=frac{CB}{CE}implies frac{20}{8sqrt{6}}=frac{x}{8sqrt{6}-4sqrt{5}}implies x=frac{60-sqrt{750}}{3}]$ Therefore $a=60, b=750, c=3, a+b+c=boxed{813}$ .
We backcount the number of ways. Namely, we start at $x_{20} = 1$ , which can only be reached if $x_{19} = 10$ , and then we perform $18$ operations that either consist of $A: (-1)$ or $B: (times 10)$ . We represent these operations in a string format, starting with the operation that sends $f(x_{18}) = x_{19}$ and so forth downwards. There are $2^9$ ways to pick the first $9$ operations; however, not all $9$ of them may be $A: (-1)$ otherwise we return back to $x_{10} = 1$ , contradicting our assumption that $20$ was the smallest value of $n$ . Using complementary counting, we see that there are only $2^9 - 1$ ways.Since we performed the operation $B: (times 10)$ at least once in the first $9$ operations, it follows that $x_{10} ge 20$ , so that we no longer have to worry about reaching $1$ again.But what can force us to subtract some of these operations? If there is just a group of 9 $A$ s right at the VERY BACK. This makes just one possibility. The remaining $9$ operations can be picked in $2^9$ ways, with a total of $2^9(2^9 - 1) = 2^{18} - 2^9$ strings so far.However, we must also account for a sequence of $10$ or more $A: (-1)$ s in a row, because that implies that at least one of those numbers was divisible by $10$ , yet the $frac{x}{10}$ was never used, contradiction. We must use complementary counting again by determining the number of strings of $A,B$ s of length $18$ such that there are $10$ $A$ s in a row. The first ten are not included since we already accounted for that scenario above, so our string of $10$ $A$ s must be preceded by a $B$ . There are no other restrictions on the remaining seven characters. Letting $square$ to denote either of the functions, and $^{[k]}$ to indicate that the character appears $k$ times in a row, our bad strings can take the forms: $begin{align*} &underbrace{BA^{[10]}}square square square square square square square\ &square underbrace{BA^{[10]}}square square square square square square \ &qquad quad cdots quad cdots \ &square square square square square square underbrace{BA^{[10]}} square \ &square square square square square square square underbrace{BA^{[10]}} \ end{align*}$ There are $2^7$ ways to select the operations for the $7$ $square$ s, and $8$ places to place our $BA^{[10]}$ block. Thus, our answer is $2^{18} - 2^9 - 8 cdot 2^7 = 2^9 times 509$ , and the answer is $boxed{511}$ .