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2005AIME 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5:15

2005AIME I真题及答案解析

答案解析请参考文末

Problem 1

Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $lfloor K rfloor.$

Problem 2

For each positive integer $k,$ let $S_k$ denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is $k.$ For example, $S_3$ is the sequence $1,4,7,10,ldots.$ For how many values of $k$ does $S_k$ contain the term 2005?

Problem 3

How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50?

Problem 4

The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

Problem 5

Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.

Problem 6

Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $lfloor Prfloor.$

Problem 7

In quadrilateral $ABCD, BC=8, CD=12, AD=10,$ and $mangle A= mangle B = 60^circ.$ Given that $AB = p + sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$

Problem 8

The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots. Given that their sum is $frac mn$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Problem 9

Twenty-seven unit cubes are painted orange on a set of four faces so that the two unpainted faces share an edge. The 27 cubes are then randomly arranged to form a $3times 3 times 3$ cube. Given that the probability that the entire surface of the larger cube is orange is $frac{p^a}{q^br^c},$ where $p,q,$ and $r$ are distinct primes and $a,b,$ and $c$ are positive integers, find $a+b+c+p+q+r.$

Problem 10

Triangle $ABC$ lies in the Cartesian Plane and has an area of 70. The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$

Problem 11

A semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - sqrt{n},$ find $m+n.$

Problem 12

For positive integers $n,$ let $tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $tau (1)=1$ and $tau(6) =4.$ Define $S(n)$ by $S(n)=tau(1)+ tau(2) + cdots + tau(n).$ Let $a$ denote the number of positive integers $n leq 2005$ with $S(n)$ odd, and let $b$ denote the number of positive integers $n leq 2005$ with $S(n)$ even. Find $|a-b|.$

Problem 13

A particle moves in the Cartesian Plane according to the following rules:

From any lattice point $(a,b),$ the particle may only move to $(a+1,b), (a,b+1),$ or $(a+1,b+1).$
There are no right angle turns in the particle's path.

How many different paths can the particle take from $(0,0)$ to $(5,5)$ ?

Problem 14

Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by 1000.

Problem 15

Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

2005AIME I详细解析

Define the radii of the six congruent circles as $r$ . If we draw all of the radii to the points of external tangency, we get a regular hexagon. If we connect the vertices of the hexagon to the center of the circle $C$ , we form several equilateral triangles. The length of each side of the triangle is $2r$ . Notice that the radius of circle $C$ is equal to the length of the side of the triangle plus $r$ . Thus, the radius of $C$ has a length of $3r = 30$ , and so $r = 10$ . $K = 30^2pi - 6(10^2pi) = 300pi$ , so $lfloor 300pi rfloor = boxed{942}$ .
Suppose that the $n$ th term of the sequence $S_k$ is $2005$ . Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2cdot 3cdot 167$ . The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$ , $(2,1002)$ , $(3,668)$ , $(4,501)$ , $(6,334)$ , $(12,167)$ , $(167,12)$ , $(334,6)$ , $(501,4)$ , $(668,3)$ , $(1002,2)$ and $(2004,1)$ , and each of these gives a possible value of $k$ . Thus the requested number of values is $12$ , and the answer is $boxed{012}$ .Alternatively, notice that the formula for the number of divisors states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^2cdot 3^1cdot 167^1$ .
Suppose $n$ is such an integer. Then one of its factors is $1$ , so $n$ must be in the form $n=pcdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$ .In the first case, the three proper divisors of $n$ are $1$ , $p$ and $q$ . Thus, we need to pick two prime numbers less than $50$ . There are fifteen of these ( $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$ ) so there are ${15 choose 2} =105$ numbers of the first type.In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$ . Thus we need to pick a prime number whose square is less than $50$ . There are four of these ( $2, 3, 5$ and $7$ ) and so four numbers of the second type.Thus there are $105+4=boxed{109}$ integers that meet the given conditions.
If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ . If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ . Thus, if the number of columns is $n$ , the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n leq 14$ . In fact, when $n = 14$ we have $n(n + 7) = 14cdot 21 = 294 = 17^2 + 5$ , so this number works and no larger number can. Thus, the answer is $boxed{294}$ .
There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.There are ${8choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for $9$ total configurations. Thus, the answer is $70 cdot 9 = boxed{630}$ .
The left-hand side of that equation is nearly equal to $(x - 1)^4$ . Thus, we add 1 to each side in order to complete the fourth power and get $(x - 1)^4 = 2006$ .Let $r = sqrt[4]{2006}$ be the positive real fourth root of 2006. Then the roots of the above equation are $x = 1 + i^n r$ for $n = 0, 1, 2, 3$ . The two non-real members of this set are $1 + ir$ and $1 - ir$ . Their product is $P = 1 + r^2 = 1 + sqrt{2006}$ . $44^2 = 1936 < 2006 < 2025 = 45^2$ so $lfloor P rfloor = 1 + 44 = boxed{045}$ .
Draw the perpendiculars from $C$ and $D$ to $AB$ , labeling the intersection points as $E$ and $F$ . This forms 2 $30-60-90$ right triangles, so $AE = 5$ and $BF = 4$ . Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$ , we find another right triangle $triangle DGC$ . $DG = DE - CF = 5sqrt{3} - 4sqrt{3} = sqrt{3}$ . The Pythagorean Theorem yields that $GC^2 = 12^2 - sqrt{3}^2 = 141$ , so $EF = GC = sqrt{141}$ . Therefore, $AB = 5 + 4 + sqrt{141} = 9 + sqrt{141}$ , and $p + q = boxed{150}$ .
Let $y = 2^{111x}$ . Then our equation reads $frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ . Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1cdot r_2cdot r_3 = 4$ . Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ . Then the previous statement says that $2^{111cdot(x_1 + x_2 + x_3)} = 4$ so taking a logarithm of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = frac{2}{111}$ . Thus the answer is $111 + 2 = boxed{113}$ .
The unit cube at the center of our large cube has no exterior faces, so all of its orientations work.For the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face. This happens in $frac{4}{6} = frac{2}{3}$ of all orientations, so from these cubes we gain a factor of $left(frac{2}{3}right)^6$ .The twelve unit cubes along the edges of the large cube have two faces showing, and these two faces are joined along an edge. Thus, we need to know the number of such pairs that are both painted orange. We have a pair for each edge, and 7 edges border one of the unpainted faces while only 5 border two painted faces. Thus, the probability that two orange faces show for one of these cubes is $frac{5}{12}$ , so from all of these cubes we gain a factor of $left(frac{5}{12}right)^{12} = frac{5^{12}}{2^{24}3^{12}}$ .Finally, we need to orient the eight corner cubes. Each such cube has 3 faces showing, and these three faces share a common vertex. Thus, we need to know the number of vertices for which all three adjacent faces are painted orange. There are six vertices which are a vertex of one of the unpainted faces and two vertices which have our desired property, so each corner cube contributes a probability of $frac{2}{8} = frac{1}{4}$ and all the corner cubes together contribute a probability of $left(frac{1}{4}right)^8 = frac{1}{2^{16}}$ Since these probabilities are independent, the overall probability is just their product, $frac{2^6}{3^6} cdot frac{5^{12}}{2^{24}3^{12}} cdot frac{1}{2^{16}} = frac{5^{12}}{2^{34}cdot 3^{18}}$ and so the answer is $2 + 3 + 5 + 12 + 34 + 18 = boxed{074}$ .
The midpoint $M$ of line segment $overline{BC}$ is $left(frac{35}{2}, frac{39}{2}right)$ . The equation of the median can be found by $-5 = frac{q - frac{39}{2}}{p - frac{35}{2}}$ . Cross multiply and simplify to yield that $-5p + frac{35 cdot 5}{2} = q - frac{39}{2}$ , so $q = -5p + 107$ .Use determinants to find that the area of $triangle ABC$ is $frac{1}{2} begin{vmatrix}p & 12 & 23 \ q & 19 & 20 \ 1 & 1 & 1end{vmatrix} = 70$ (note that there is a missing absolute value; we will assume that the other solution for the triangle will give a smaller value of $p+q$ , which is provable by following these steps over again). We can calculate this determinant to become $140 = begin{vmatrix} 12 & 23 \ 19 & 20 end{vmatrix} - begin{vmatrix} p & q \ 23 & 20 end{vmatrix} + begin{vmatrix} p & q \ 12 & 19 end{vmatrix}$ $Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q$ $= -197 - p + 11q$ . Thus, $q = frac{1}{11}p - frac{337}{11}$ .Setting this equation equal to the equation of the median, we get that $frac{1}{11}p - frac{337}{11} = -5p + 107$ , so $frac{56}{11}p = frac{107 cdot 11 + 337}{11}$ . Solving produces that $p = 15$ . Substituting backwards yields that $q = 32$ ; the solution is $p + q = boxed{047}$ .
We note that aligning the base of the semicircle with a side of the square is certainly non-optimal. If the semicircle is tangent to only one side of the square, we will have "wiggle-room" to increase its size. Once it is tangent to two adjacent sides of the square, we will maximize its size when it touches both other sides of the square. This can happen only when it is arranged so that the center of the semicircle lies on one diagonal of the square.Now, let the square be $ABCD$ , and let $E in AB$ and $F in DA$ be the points at which the "corners" of the semicircle touch the square. Let $O$ be the center of the semicircle.We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, rsqrt{2}$ right triangle. This means that we have $r + rsqrt{2} = 8sqrt{2}$ so $r = frac{8sqrt{2}}{1+sqrt{2}} = 16 - 8sqrt{2}$ . Then the diameter is $32 - sqrt{512}$ giving us $32 + 512 = boxed{544}$
It is well-known that $tau(n)$ is odd if and only if $n$ is a perfect square. (Otherwise, we can group divisors into pairs whose product is $n$ .) Thus, $S(n)$ is odd if and only if there are an odd number of perfect squares less than $n$ . So $S(1), S(2)$ and $S(3)$ are odd, while $S(4), S(5), ldots, S(8)$ are even, and $S(9), ldots, S(15)$ are odd, and so on.So, for a given $n$ , if we choose the positive integer $m$ such that $m^2 leq n < (m + 1)^2$ we see that $S(n)$ has the same parity as $m$ .It follows that the numbers between $1^2$ and $2^2$ , between $3^2$ and $4^2$ , and so on, all the way up to the numbers between $43^2$ and $44^2 = 1936$ have $S(n)$ odd. These are the only such numbers less than $2005$ (because $45^2 = 2025 > 2005$ ).Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$ , and so on. The number of numbers from $n^2$ to $(n + 1)^2$ is $(n + 1 - n)(n + 1 + n) = 2n + 1$ . Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a = [2(1) + 1] + [2(3) + 1] ldots [2(43) + 1] = 3 + 7 + 11 ldots 87$ . $b = [2(2) + 1] + [2(4) + 1] ldots [2(42) + 1] + 70 = 5 + 9 ldots 85 + 70$ , the $70$ accounting for the difference between $2005$ and $44^2 = 1936$ , inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$ . Thus, the solution is $|a - b| = |b - a| = |2 cdot 21 + 70 - 87| = boxed{025}$ .Similarly, $b = (3^2 - 2^2) + (5^2 - 4^2) + ldots + (45^2 - 44^2) - 19$ , where the $-19$ accounts for those numbers between $2005$ and $2024$ .Thus $a = (2^2 - 1^2) + (4^2 - 3^2) + ldots + (44^2 - 43^2)$ .Then, $|a - b| = |2(2^2 + 4^2 + ldots + 44^2) - 2(1^2 + 3^2 + 5^2 + ldots 43^2) + 1^2 - 45^2 + 19|$ . We can apply the formula $1^2 + 2^2 + ldots + n^2 = frac{n(n + 1)(2n + 1)}{6}$ . From this formula, it follows that $2^2 + 4^2 + ldots + (2n)^2 = frac{2n(n + 1)(2n + 1)}{3}$ and so that

$1^2 + 3^2 + ldots +(2n + 1)^2 = (1^2 + 2^2 + ldots +(2n + 1)^2) - (2^2 + 4^2 + ldots + (2n)^2)$

$= frac{(2n + 1)(2n + 2)(4n + 3)}{6} - frac{2n(n + 1)(2n + 1)}{3} = frac{(n + 1)(2n + 1)(2n + 3)}{3}$ . Thus,

$|a - b| = left| 2cdot frac{44cdot23cdot45}{3} - 2cdot frac{22 cdot 43 cdot 45}{3} - 45^2 + 20right| = |-25| =boxed{025}$ .
The length of the path (the number of times the particle moves) can range from to ; notice that gives the number of diagonals. Let represent a move to the right, represent a move upwards, and to be a move that is diagonal. Casework upon the number of diagonal moves:
- Case $d = 1$ : It is easy to see only $2$ cases.
- Case $d = 2$ : There are two diagonals. We need to generate a string with $3$ $R$ 's, $3$ $U$ 's, and $2$ $D$ 's such that no two $R$ 's or $U$ 's are adjacent. The $D$ 's split the string into three sections ( $-D-D-$ ): by the Pigeonhole principle all of at least one of the two letters must be all together (i.e., stay in a row).
If both $R$ and $U$ stay together, then there are $3 cdot 2=6$ ways.

If either $R$ or $U$ splits, then there are $3$ places to put the letter that splits, which has $2$ possibilities. The remaining letter must divide into $2$ in one section and $1$ in the next, giving $2$ ways. This totals $6 + 3cdot 2cdot 2 = 18$ ways.
- Case $d = 3$ : Now $2$ $R$ 's, $2$ $U$ 's, and $3$ $D$ 's, so the string is divided into $4$ partitions ( $-D-D-D-$ ).
If the $R$ 's and $U$ 's stay together, then there are $4 cdot 3 = 12$ places to put them.

If one of them splits and the other stays together, then there are $4 cdot {3choose 2}$ places to put them, and $2$ ways to pick which splits, giving $4 cdot 3 cdot 2 = 24$ ways.

If both groups split, then there are ${4choose 2}=6$ ways to arrange them. These add up to $12 + 24 + 6 = 42$ ways.
- Case $d = 4$ : Now $1$ $R$ , $1$ $U$ , $4$ $D$ 's ( $-D-D-D-D-$ ). There are $5$ places to put $R$ , $4$ places to put $U$ , giving $20$ ways.
- Case $d = 5$ : It is easy to see only $1$ case.
Together, these add up to $2 + 18 + 42 + 20 + 1 = boxed{083}$ .
Consider a point $E$ such that $AE$ is perpendicular to $BD$ , $AE$ intersects $BD$ , and $AE = BD$ . E will be on the same side of the square as point $C$ .Let the coordinates of $E$ be $(x_E,y_E)$ . Since $AE$ is perpendicular to $BD$ , and $AE = BD$ , we have $9 - 7 = x_E - 0$ and $10 - ( - 4) = 12 - y_E$ The coordinates of $E$ are thus $(2, - 2)$ .Now, since $E$ and $C$ are on the same side, we find the slope of the sides going through $E$ and $C$ to be $frac { - 2 - 0}{2 - 8} = frac {1}{3}$ . Because the other two sides are perpendicular, the slope of the sides going through $B$ and $D$ are now $- 3$ .Let $A_1,B_1,C_1,D_1$ be the vertices of the square so that $A_1B_1$ contains point $A$ , $B_1C_1$ contains point $B$ , and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find $2$ vertices of the square, then apply the distance formula.We find the coordinates of $C_1$ to be $(12.5,1.5)$ and the coordinates of $D_1$ to be $( - 0.7, - 2.9)$ . Applying the distance formula, the side length of our square is $sqrt {left( frac {44}{10} right)^2 + left( frac {132}{10} right)^2} = frac {44}{sqrt {10}}$ .Hence, the area of the square is $K = frac {44^2}{10}$ . The remainder when $10K$ is divided by $1000$ is $936$ .Let $(a,b)$ denote a normal vector of the side containing $A$ . Note that $overline{AC}, overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$ , $ax+by=8a$ , $bx-ay=10b-9a$ , and $bx-ay=-4b-7a$ . The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$ , hence $8a+0b-12b=-4b-7a-10b+9a$ , or $-3a=b$ . We can take $a=-1$ and $b=3$ . So the side of the square is $frac{44}{sqrt{10}}$ , the area is $K=frac{1936}{10}$ , and the answer to the problem is $boxed{936}$ .
Let $E$ , $F$ and $G$ be the points of tangency of the incircle with $BC$ , $AC$ and $AB$ , respectively. Without loss of generality, let $AC < AB$ , so that $E$ is between $D$ and $C$ . Let the length of the median be $3m$ . Then by two applications of the Power of a Point Theorem, $DE^2 = 2m cdot m = AF^2$ , so $DE = AF$ . Now, $CE$ and $CF$ are two tangents to a circle from the same point, so $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$ . Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$ .Now, by Stewart's Theorem in triangle $triangle ABC$ with cevian $overline{AD}$ , we have $[(3m)^2cdot 20 + 20cdot10cdot10 = 10^2cdot10 + (30 - 2c)^2cdot 10.]$ Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$ , so we combine these two results to solve for $c$ and we get $[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 quad Longrightarrow quad c^2 - 12c + 20 = 0.]$ Thus $c = 2$ or $= 10$ . We discard the value $c = 10$ as extraneous (it gives us an equilateral triangle) and are left with $c = 2$ , so our triangle has area $sqrt{28 cdot 18 cdot 8 cdot 2} = 24sqrt{14}$ and so the answer is $24 + 14 = boxed{038}$ .Use Power of a point similar to the first solution to find that $AB = 30$ and that the side $AC = 2 cdot x sqrt{2}$ , where $x$ is one third of the median's length. Then use systems of law of cosines, creating two triangles, with $10-10-3x$ with angle $theta$ , and $10-20-2 cdot x sqrt{2}$ with the same angle. Solving the system yields $x = 4 sqrt{2}$ . Solving using Heron's Formula gets the answer $24 sqrt{14}$ , or $boxed{038}$ .