id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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6,401 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_3 | 1 | 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? | Every pair of vertices of the polyhedron determines either an edge , a face diagonal or a space diagonal. We have ${26 \choose 2} = \frac{26\cdot25}2 = 325$ 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 t... | 241 |
6,402 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_4 | 1 | Square $ABCD$ 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 $100k$ | 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, $... | 86 |
6,403 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_5 | 1 | 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 atte... | Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he 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: ... | 849 |
6,404 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_6 | 1 | An integer is called snakelike if its decimal representation $a_1a_2a_3\cdots a_k$ 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? | 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_... | 882 |
6,405 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_6 | 2 | An integer is called snakelike if its decimal representation $a_1a_2a_3\cdots a_k$ 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? | 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 $5\cdot{10 \choose 4}$ But, this over-counts since it counts numbers like 0213. We can corr... | 882 |
6,406 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_6 | 3 | An integer is called snakelike if its decimal representation $a_1a_2a_3\cdots a_k$ 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? | We will first decide the order of the 4 digits, greatest to least. To do this, we will pretend that we have selected the digits 1,2,3,4, and we need to arrange them to create a snakelike number. By testing all permutations, there are only 5 ways to make a snakelike number: (1,3,2,4),(1,4,2,3),(2,3,1,4),(2,4,1,3),(3,4,1... | 882 |
6,407 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_7 | 1 | 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 $|C|.$ | 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 pol... | 588 |
6,408 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_7 | 2 | 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 $|C|.$ | 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_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) +... | 588 |
6,409 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_7 | 3 | 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 $|C|.$ | Let set $N$ be $\{-1, -3, \ldots -15\}$ and set $P$ be $\{2, 4, \ldots 14\}$ . The sum of the negative $x^2$ coefficients is the sum of the products of the elements in all two element sets such that one element is from $N$ and the other is from $P$ . Each summand is a term in the expansion of \[(-1 - 3 - \ldots - 15)(2... | 588 |
6,410 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_7 | 4 | 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 $|C|.$ | This is just another way of summing the subsets of 2 from $[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]$ . Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us $-15 * 7$ . Doing this for 14 gives us $14... | 588 |
6,411 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_8 | 1 | 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
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... | 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... | 199 |
6,412 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_10 | 1 | 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 $m + n.$ | 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 ince... | 817 |
6,413 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_10 | 2 | 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 $m + n.$ | 2004 I AIME-10.png
Let the bisector of $\angle CAD$ be $AE$ , with $E$ on $CD$ . By the angle bisector theorem, $DE = 36/5$ . Since $\triangle AOR \sim \triangle AED$ $O$ is the center of the circle), we find that $AR = 5$ since $OR = 1$ . Also $AT = 35$ so $RT = OQ = 30$
We can apply the same principle again to find t... | 817 |
6,414 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_10 | 3 | 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 $m + n.$ | 2004 I AIME-10b.png
Again, 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 $A$ be at the origin, $B (36,0)$ $C (36,15)$ $D (0,... | 817 |
6,415 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_11 | 1 | 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 th... | Our original solid has volume equal to $V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi$ 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 $A = 24\pi$
Let $x$ denote the radius of the small cone. Let $A_c$ and $... | 512 |
6,416 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_11 | 2 | 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 th... | 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 $A_\ell = 15\pi$ . The area of the base is $A_B = 3^2\pi = 9\pi$
$V$ and $C$ are similar cones, becaus... | 512 |
6,417 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_13 | 1 | The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$ 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 positi... | 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... | 482 |
6,418 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_14 | 1 | 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 towe... | 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 com... | 813 |
6,419 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_14 | 2 | 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 towe... | Note that by Power of a Point, the point the unicorn is at has power $4 \cdot 20 = 80$ which implies that the tangent from that point to the tower is of length $\sqrt{80}=4\sqrt{5},$ however this is length of the rope projected into 2-D. If we let $\theta$ be the angle between the horizontal and the rope, we have that ... | 813 |
6,420 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_15 | 1 | For all positive integers $x$ , let \[f(x)=\begin{cases}1 & \text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}\] 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 e... | 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... | 511 |
6,421 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_1 | 1 | chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}},$ where $a, b, c, d, e,$ and $f$ are positive integers $a$ and $... | Let $r$ be the length of the radius of the circle. A right triangle is formed by half of the chord, half of the radius (since the chord bisects it), and the radius. Thus, it is a $30^\circ$ $60^\circ$ $90^\circ$ triangle , and the area of two such triangles is $2 \cdot \frac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqrt{... | 592 |
6,422 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_2 | 1 | A jar has $10$ red candies and $10$ blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $m/n,$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$ | The probability that Terry picks two red candies is $\frac{10 \cdot 9}{20 \cdot 19} = \frac{9}{38}$ , and the probability that Mary picks two red candies after Terry chooses two red candies is $\frac{7\cdot8}{18\cdot17} = \frac{28}{153}$ . So the probability that they both pick two red candies is $\frac{9}{38} \cdot \... | 441 |
6,423 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_3 | 1 | A solid rectangular block is formed by gluing together $N$ congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the 1-cm cubes cannot be seen. Find the smallest possible value of $N.$ | The $231$ cubes which are not visible must lie below exactly one layer of cubes. Thus, they form a rectangular solid which is one unit shorter in each dimension. If the original block has dimensions $l \times m \times n$ , we must have $(l - 1)\times(m-1) \times(n - 1) = 231$ . The prime factorization of $231 = 3\cd... | 384 |
6,424 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_4 | 1 | How many positive integers less than 10,000 have at most two different digits | First, let's count numbers with only a single digit. We have nine of these for each length, and four lengths, so 36 total numbers.
Now, let's count those with two distinct digits. We handle the cases "0 included" and "0 not included" separately.
There are ${9 \choose 2}$ ways to choose two digits, $A$ and $B$ . Give... | 927 |
6,425 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_4 | 2 | How many positive integers less than 10,000 have at most two different digits | We use casework on the number of digits for this problem.
If the number has a single digit, namely the number $n \in [1,9],$ we can clearly all such $n$ work.
If the number has two digits, or the number $n \in [10,99]$ we can clearly see all such $n$ work.
If the number $n$ has three digits, there are a total of $900$ ... | 927 |
6,426 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_5 | 1 | In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was co... | A train is traveling at $1000$ miles per hour and has one hour to reach its destination $1000$ miles away. After $15$ minutes and $250$ miles it slows to $900$ mph, and thus takes $\frac{250}{900}(60)=\frac{50}{3}$ minutes to travel the next $250$ miles. Then it slows to $800$ mph, so the next quarter takes $\frac{250}... | 766 |
6,427 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_5 | 2 | In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was co... | Let each worker's speed be $w$ , the entire time be $t$ , and the total work be $1$
From the initial problem statement, we have $1000w\cdot\frac{1}{4}t=\frac{1}{4}$
We need to find the time the next quarter takes to complete the same amount of work, which is $\frac{1000}{900}\cdot\frac{1}{4}\cdot t=\frac{5}{18}t$
Simil... | 766 |
6,428 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_5 | 3 | In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was co... | Suppose $1000$ workers can complete one quarter of the job in one day. After the first day, there were $900$ workers remaining so the second quarter was completed in $\frac{10}{9}$ days. Now there are only $800$ workers remaining so the third quarter can be completed in $\frac{10}{8}$ days. It has been $1+\frac{10}{9}+... | 766 |
6,429 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_6 | 1 | Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third m... | Denote the number of bananas the first monkey took from the pile as $b_1$ , the second $b_2$ , and the third $b_3$ ; the total is $b_1 + b_2 + b_3$ . Thus, the first monkey got $\frac{3}{4}b_1 + \frac{3}{8}b_2 + \frac{11}{24}b_3$ , the second monkey got $\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3$ , and the thi... | 408 |
6,430 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_6 | 2 | Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third m... | Let the first monkey take $8x$ bananas, the second monkey take $8y$ , and the third monkey take $24z$ . I chose these numbers to make it so, when each monkey splits his bananas, they will get an integer amount of each variable. So, when the first monkey distributes his bananas, he gets $6x$ , and the other monkeys get ... | 408 |
6,431 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_6 | 3 | Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third m... | In this solution, you start out the same as solution one. Convert everything into the fractions of largest denominator terms (this is necessary) until you get
$27x=11z$ $27y=13z$
While solving, make sure to leave a list of numbers that must divide $x$ $y$ , and $z$ . For example, just by looking at the basic fractions ... | 408 |
6,432 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_6 | 4 | Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third m... | Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations \[\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=\frac{1}{2}\] \[\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=\frac{1}{3}\] \[\frac{1}{8}A+\frac{3}{8}B+\frac{2}{24}C=\frac{1}{6}.\] Solve this yo... | 408 |
6,433 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | 1 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | Since $EF$ is the perpendicular bisector of $\overline{BB'}$ , it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem , we have $AB' = 15$ . Similarly, from $BF = B'F$ , we have \begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC &= \frac{70}{3} \end{align*} Thus the ... | 293 |
6,434 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | 2 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | Let $A = (0,0), B=(0,25)$ , so $E = (0,8)$ and $F = (l,22)$ , and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \frac{14}{l}x$ . We know that $EF$ is perpendicular to and bisects $BB'$ . The slope of $BB'$ is thus $\frac{-l}{14}$ , and so the equ... | 293 |
6,435 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | 3 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | Firstly, observe that if we are given that $AE=8$ and $BE=17$ , the length of the triangle is given and the height depends solely on the length of $CF$ . Let Point $A = (0,0)$ . Since $AE=8$ , point E is at (8,0). Next, point $B$ is at $(25,0)$ since $BE=17$ and point $B'$ is at $(0,-15)$ since $BE=AE$ by symmetry. Dra... | 293 |
6,436 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | 4 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | Firstly, note that $B'E=BE=17$ , so $AB'=\sqrt{17^2-8^2}=15$ . Then let $\angle BEF=\angle B'EF=\theta$ , so $\angle B'EA = \pi-2\theta$ . Then $\tan(\pi-2\theta)=\frac{15}{8}$ , or
\[\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}\] using supplementary and double angle identities. Multiplying though and factoring ... | 293 |
6,437 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | 5 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | Use the prepared diagram for this solution.
Call the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3.
Add all the sides together to ge... | 293 |
6,438 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | 6 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | Call the intersection of $B'C'$ $BC$ , and $EF$ $G$ . Since $FCBE$ and $FC'B'E$ are congruent, we know that the three lines intersect.
We already know $AB$ so we just need to find $CB$ , call it $x$ . Drop an altitude from $F$ to $AB$ and call it $H$ $EH=EB-FC=14$ . Using Pythagorean Theorem, we have $EF=\sqrt{x^2+14^2... | 293 |
6,439 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | 7 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | Let the endpoint of the intersection of the fold near $F$ be $G$ . Since trapezoid $BCFE$ is folded, it is congruent to trapezoid $B'C'FE$ . Therefore, $BE=B'E=17$ . Since $\triangle AB'E$ is a right triangle, $AB'=15$ from the pythagorean theorem. From here, we can see that triangles $\triangle AEB \sim \triangle DGB'... | 293 |
6,440 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_8 | 1 | How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers? | The prime factorization of 2004 is $2^2\cdot 3\cdot 167$ . Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$
We can count the number of divisors of a number by multiplying together one more than each of the exponents of the prime factors in its prime factorization. For example,... | 54 |
6,441 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_8 | 2 | How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers? | Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only $167 * 2^2 * 3$
167, 2, 2, 3
4, 3, 167
12, 167
4, 501
2, 1002
2, 3, 334
2, 2, 501*
6, 2, 167
3, 668
6, 334
2004*
To begin, the first multiple doesn't work because there are only 3 prime divisors of... | 54 |
6,442 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_9 | 1 | sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression , the second, third, and fourth terms are in arithmetic progression , and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms ... | Let $x = a_2$ ; then solving for the next several terms, we find that $a_3 = x^2,\ a_4 = x(2x-1),\ a_5$ $= (2x-1)^2,\ a_6$ $= (2x-1)(3x-2)$ , and in general, $a_{2n} = f(n-1)f(n)$ $a_{2n+1} = f(n)^2$ , where $f(n) = nx - (n-1)$
From \[a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\cdot 17 \cdot 19\] , we fin... | 973 |
6,443 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_9 | 2 | sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression , the second, third, and fourth terms are in arithmetic progression , and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms ... | Let $x = a_2$ . It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what $x$ can be. Finding that $x = 5$ works, after bashing out the rest of the terms we find that $a_{16} = 957$ and $a_{17} = 1089$ , hence our answer is $957 + 16 = \boxed{973}$ | 973 |
6,444 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_9 | 3 | sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression , the second, third, and fourth terms are in arithmetic progression , and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms ... | We can find the value of $a_{9}$ by its bounds using three conditions:
Rearranging conditions 1 and 2, we get:
\[\frac{646}{3} < a_{9} < \frac{646}{2}\]
trying all the terms from the third condition, it is clear that $a_9 = 289$ is the only solution.
Then we can calculate the next few terms from there since we have $a... | 973 |
6,445 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_10 | 1 | Let $S$ be the set of integers between $1$ and $2^{40}$ whose binary expansions have exactly two $1$ 's. If a number is chosen at random from $S,$ the probability that it is divisible by $9$ is $p/q,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Any number from 1 to $2^{40}$ can be represented in binary with 40 digits (because $2^{40}$ has 41) with leading zeroes if necessary. Therefore the number of sets where there are exactly two 1’s in this binary representation is just $\binom {40}{2}$ because we’re choosing 2 1s to go in 40 digit slots. This is equal to ... | 913 |
6,446 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_10 | 2 | Let $S$ be the set of integers between $1$ and $2^{40}$ whose binary expansions have exactly two $1$ 's. If a number is chosen at random from $S,$ the probability that it is divisible by $9$ is $p/q,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | A positive integer $n$ has exactly two 1s in its binary representation exactly when $n = 2^j + 2^k$ for $j \neq k$ nonnegative integers. Thus, the set $S$ is equal to the set $\{n \in \mathbb{Z} \mid n = 2^j + 2^k \,\mathrm{ and }\, 0 \leq j < k \leq 39\}$ . (The second condition ensures simultaneously that $j \neq k... | 913 |
6,447 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_10 | 3 | Let $S$ be the set of integers between $1$ and $2^{40}$ whose binary expansions have exactly two $1$ 's. If a number is chosen at random from $S,$ the probability that it is divisible by $9$ is $p/q,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | As mentioned above, there are 780 possible combinations. Since we are essentially adding two powers of two together, thinking about the properties of this sum organizes our solution. All powers of two are even except for $2^0$ , so we begin with labeling an entire group "where one of the 1s is in the rightmost spot". I... | 913 |
6,448 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_11 | 1 | right circular cone has a base with radius $600$ and height $200\sqrt{7}.$ A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is $125$ , and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}.$ ... | The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$ -axis and the angle $\theta$ going counterclockwise. The circumference of the base is $C=1200\pi$ . The sector's radius (cone's sweep) is $R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt... | 625 |
6,449 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_12 | 1 | Let $ABCD$ be an isosceles trapezoid , whose dimensions are $AB = 6, BC=5=DA,$ and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m... | Let the radius of the center circle be $r$ and its center be denoted as $O$
Clearly line $AO$ passes through the point of tangency of circle $A$ and circle $O$ . Let $y$ be the height from the base of the trapezoid to $O$ . From the Pythagorean Theorem \[3^2 + y^2 = (r + 3)^2 \Longrightarrow y = \sqrt {r^2 + 6r}.\]
We ... | 134 |
6,450 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_13 | 1 | Let $ABCDE$ be a convex pentagon with $AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$ . Since $AB \parallel CE, BC \parallel AD,$ it follows that $ABCF$ is a parallelogram , and so $\triangle ABC \cong \triangle CFA$ . Also, as $AC \parallel DE$ , it follows that $\triangle ABC \sim \triangle EFD$
By the Law of Cosines $AC^2 = 3^2 + 5^2 ... | 484 |
6,451 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_14 | 1 | Consider a string of $n$ $7$ 's, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression . For example, $7+77+777+7+7=875$ could be obtained from eight $7$ 's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$ | Suppose we require $a$ $7$ s, $b$ $77$ s, and $c$ $777$ s to sum up to $7000$ $a,b,c \ge 0$ ). Then $7a + 77b + 777c = 7000$ , or dividing by $7$ $a + 11b + 111c = 1000$ . Then the question is asking for the number of values of $n = a + 2b + 3c$
Manipulating our equation, we have $a + 2b + 3c = n = 1000 - 9(b + 12c) \L... | 108 |
6,452 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_14 | 2 | Consider a string of $n$ $7$ 's, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression . For example, $7+77+777+7+7=875$ could be obtained from eight $7$ 's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$ | To simplify, replace all the $7$ ’s with $1$ ’s.
Because the sum is congruent to $n \pmod 9$ and \[1000 \equiv 1 \pmod 9 \implies n \equiv 1 \pmod 9\] Also, $n \leq 1000$ . There are $\big\lfloor \tfrac{1000}{9} \big\rfloor + 1 = 112$ positive integers that satisfy both conditions i.e. $\{1, 10, 19, 28, 37, 46, . . . ... | 108 |
6,453 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_14 | 3 | Consider a string of $n$ $7$ 's, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression . For example, $7+77+777+7+7=875$ could be obtained from eight $7$ 's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$ | It's obvious that we cannot have any number $\ge 7777$ because $7777 > 7000$ so the max number that an occur is $777$
Let's say we have $a$ 777's , $b$ 77's and $c$ 7's
From here we get our required equation as $777a + 77b + 7c = 7000$
Now comes the main problem , one might think that if we find number of $(a,b,c)$ the... | 1 |
6,454 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_15 | 1 | A long thin strip of paper is $1024$ units in length, $1$ unit in width, and is divided into $1024$ unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a $512$ by $1$ strip of double thickness.... | Number the squares $0, 1, 2, 3, ... 2^{k} - 1$ . In this case $k = 10$ , but we will consider more generally to find an inductive solution. Call $s_{n, k}$ the number of squares below the $n$ square after the final fold in a strip of length $2^{k}$
Now, consider the strip of length $1024$ . The problem asks for $s_{... | 593 |
6,455 | https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_15 | 2 | A long thin strip of paper is $1024$ units in length, $1$ unit in width, and is divided into $1024$ unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a $512$ by $1$ strip of double thickness.... | We can keep track of the position of the square labeled 942 in each step. We use an $(x,y)$ coordinate system, so originally the 942 square is in the position $(942,1)$ . In general, suppose that we've folded the strip into an array $r=2^k$ squares wide and $c=1024/r=2^{10-k}$ squares tall (so we've made $10-k$ folds).... | 593 |
6,456 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_1 | 1 | Given that
where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$ | Note that \[{{\left((3!)!\right)!}\over{3!}}= {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.\] Because $120\cdot719!<720!$ , we can conclude that $n < 720$ . Thus, the maximum value of $n$ is $719$ . The requested value of $k+n$ is therefore $120+719=\boxed{839}$ | 839 |
6,457 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_2 | 1 | One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to t... | To get the green area, we can color all the circles of radius $100$ or below green, then color all those with radius $99$ or below red, then color all those with radius $98$ or below green, and so forth. This amounts to adding the area of the circle of radius $100$ , but subtracting the circle of radius $99$ , then ad... | 301 |
6,458 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_2 | 2 | One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to t... | We want to find $\frac{\sum\limits_{n=1}^{50} (4n-1)\pi}{10000\pi}=\frac{\sum\limits_{n=1}^{50} (4n-1)}{10000}=\frac{(\sum\limits_{n=1}^{50} (4n) )-50}{10000}=\frac{101}{200} \rightarrow 101+200=\boxed{301}$ | 301 |
6,459 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_2 | 3 | One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to t... | The sum of the areas of the green regions is
\[\left[(2^2-1^2)+(4^2-3^2)+(6^2-5^2)+\cdots+(100^2-99^2)\right]\pi\] \[=\left[(2+1)+(4+3)+(6+5)+\cdots+(100+99)\right]\pi\] \[={1\over2}\cdot100\cdot101\pi.\]
Thus the desired ratio is \[{1\over2}\cdot{{100\cdot101\pi}\over{100^2\pi}}={101\over200},\] and $m+n=\boxed{301}$ | 301 |
6,460 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_3 | 1 | Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list. | Order the numbers in the set from greatest to least to reduce error: $\{34, 21, 13, 8, 5, 3, 2, 1\}.$ Each element of the set will appear in $7$ two-element subsets , once with each other number.
Therefore the desired sum is $34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\boxed{484}$ | 484 |
6,461 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_3 | 2 | Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list. | Thinking of this problem algorithmically, one can "sort" the array to give: \[{1, 2, 3, 5, 8, 13, 21, 34}\]
Now, notice that when we consider different pairs, we are only going to fixate one element and look at the all of the next elements in the array, basically the whole $j = i + 1$ shebang. Then, we see that if we s... | 484 |
6,462 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_5 | 1 | Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures $3$ by $4$ by $5$ units. Given that the volume of this set is $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers , and $n$ and $p$ are relatively prime , find $m + n + p.$ | The set can be broken into several parts: the big $3\times 4 \times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$ , the $1/8$ spheres (one centered at each vertex of the large parallelepiped), and the $1/4$ cylinders connecting each adjacen... | 505 |
6,463 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_6 | 1 | The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers . Find $m + n + p.$ | Since there are $8$ vertices of a cube , there are ${8 \choose 3} = 56$ total triangles to consider. They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal ), those which lie in a plane perpendicular to one face of... | 348 |
6,464 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_7 | 1 | Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers . Let $s$ be the sum of all possible perimeters of $\triangle ACD$ . Find $s.$ | Denote the height of $\triangle ACD$ as $h$ $x = AD = CD$ , and $y = BD$ . Using the Pythagorean theorem , we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$ . Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$ . The LHS is difference of squares , so $(x + y)(x - y) = 189$ . As both $x,\ y$ are integers, $... | 380 |
6,465 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_8 | 1 | In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms. | Denote the first term as $a$ , and the common difference between the first three terms as $d$ . The four numbers thus are in the form $a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}$
Since the first and fourth terms differ by $30$ , we have that $\frac{(a + 2d)^2}{a + d} - a = 30$ . Multiplying out by the denominator, \[(a^... | 129 |
6,466 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_8 | 2 | In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms. | The sequence is of the form $a-d,$ $a,$ $a+d,$ $\frac{(a+d)^2}{a}$ . Since the first and last terms differ by 30, we have \[\frac{(a+d)^2}{a}-a+d=30\] \[d^2+3ad=30a\] \[d^2+3ad-30a=0\] \[d=\frac{-3a + \sqrt{9a^2+120a}}{2}.\] Let $9a^2+120a=x^2$ , where $x$ is an integer. This yields the following: \[9a^2+120a-x^2=0\] \... | 129 |
6,467 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_8 | 3 | In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms. | We represent the values as $a-d$ $a$ $a+d$ , and $\frac{(a+d)^2}{a}$ Take the difference between the first and last values \[\frac{(a+d)^2}{a}-a+d=30\] Manipulating the values by expanding and then long division we see \[\frac{a^2+2ad+d^2}{a}-a+d=30\] \[\frac{(a+2d)a+d^2}{a}-a+d=30\] \[a+2d+\frac{d^2}{a}-a+d=30\] Combi... | 129 |
6,468 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_9 | 1 | An integer between $1000$ and $9999$ , inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there? | If the common sum of the first two and last two digits is $n$ , such that $1 \leq n \leq 9$ , there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits (since zero may not be the first digit). This gives $\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers. If the common sum of the firs... | 615 |
6,469 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_9 | 2 | An integer between $1000$ and $9999$ , inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there? | Call the number $\overline{abcd}$ . Then $a+b=c+d$ . Set $a+b=x$
Clearly, $0\le x \le18$
If $x=0$ $0000$ is not acceptable.
If $x=1$ : The only case is $1001$ or $1010$ . 2 choices.
If $x=2$ : then since $a\neq0$ $a=1=b$ or $a=2, b=0$ . There are 3 choices for $(c,d)$ $(2,0), (0, 2), (1, 1)$ $2*3=6$ here.
If $x=3$ : Cl... | 615 |
6,470 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_9 | 3 | An integer between $1000$ and $9999$ , inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there? | We ignore the requirement that the first digit is non-zero, and do casework on the sum of the sum of the pairs of digits.
If two digits $a$ and $b$ sum to $0$ , we have $1$ possibility: $(a,b) = (0,0)$
If $a+b = 1$ , we have $2$ possibilities: $(a,b) = (0,1)$ and $(a,b) = (1,0)$
$a+b = 2$ $(0,2)$ $(2,0)$ , and $(1,1)$ ... | 615 |
6,471 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_10 | 1 | Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$ | Take point $N$ inside $\triangle ABC$ such that $\angle CBN = 7^\circ$ and $\angle BCN = 23^\circ$
$\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ$ . Also, since $\triangle AMC$ and $\triangle BNC$ are congruent (by ASA), $CM = CN$ . Hence $\triangle CMN$ is an equilateral triangle , so $\angle CNM = 60^\circ$
The... | 83 |
6,472 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_10 | 2 | Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$ | From the givens, we have the following angle measures $m\angle AMC = 150^\circ$ $m\angle MCB = 83^\circ$ . If we define $m\angle CMB = \theta$ then we also have $m\angle CBM = 97^\circ - \theta$ . Then apply the Law of Sines to triangles $\triangle AMC$ and $\triangle BMC$ to get
\[\frac{\sin 150^\circ}{\sin 7^\circ}... | 83 |
6,473 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_10 | 3 | Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$ | Without loss of generality , let $AC = BC = 1$ . Then, using the Law of Sines in triangle $AMC$ , we get $\frac {1}{\sin 150} = \frac {MC}{\sin 7}$ , and using the sine addition formula to evaluate $\sin 150 = \sin (90 + 60)$ , we get $MC = 2 \sin 7$
Then, using the Law of Cosines in triangle $MCB$ , we get $MB^2 = 4\... | 83 |
6,474 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_10 | 4 | Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$ | Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros
First, take point $E$ outside of $\triangle{ABC}$ so that $\triangle{CEB}$ is equilateral. Then, connect $A$ $C$ , and $M$ to $E$ . Also, let $ME$ intersect $AB$ at $F$ $\angle{MCE} ... | 83 |
6,475 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_12 | 1 | In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$ . Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$ | By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$ ),
\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\] \[(AD^2 - BC^2) = 360(AD - BC) \cos A\] \[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\] \[(AD + BC) = 360 \cos A\] We know tha... | 777 |
6,476 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_12 | 2 | In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$ . Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$ | Notice that $AB = CD$ , and $BD = DB$ , and $\angle{DAB} \cong \angle{BCD}$ , so we have side-side-angle matching on triangles $ABD$ and $CDB$ . Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$ , we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matchi... | 777 |
6,477 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_13 | 1 | Let $N$ be the number of positive integers that are less than or equal to $2003$ and whose base- $2$ representation has more $1$ 's than $0$ 's. Find the remainder when $N$ is divided by $1000$ | In base- $2$ representation, all positive numbers have a leftmost digit of $1$ . Thus there are ${n \choose k}$ numbers that have $n+1$ digits in base $2$ notation, with $k+1$ of the digits being $1$ 's.
In order for there to be more $1$ 's than $0$ 's, we must have $k+1 > \frac{n+1}{2} \implies k > \frac{n-1}{2} \impl... | 155 |
6,478 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_13 | 2 | Let $N$ be the number of positive integers that are less than or equal to $2003$ and whose base- $2$ representation has more $1$ 's than $0$ 's. Find the remainder when $N$ is divided by $1000$ | We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$ . If there are even digits, $2n$ , then the leftmost digit is $1$ , the rest, $2n-1$ , has odd number of digits. In order for the base-2 representation to have more $1$ 's, we will n... | 155 |
6,479 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_15 | 1 | In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overl... | By the Angle Bisector Theorem, we know that $[CBD]=\frac{169}{289}[ABC]$ . Therefore, by finding the area of triangle $CBD$ , we see that \[\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].\] Solving for $BD$ yields \[BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.\] Furthermore, $\cos\frac{B}{2}=\frac{BD}{BF}$ , ... | 289 |
6,480 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_15 | 2 | In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overl... | Let $\angle{DBM}=\theta$ and $\angle{DBC}=\alpha$ . Then because $BM$ is a median we have $360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}$ . Now we know \[\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE... | 289 |
6,481 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_15 | 3 | In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overl... | Firstly, angle bisector theorem yields $\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}$ . We're given that $AM=MC$ . Therefore, the cross ratio
\[(A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120}\]
We need a fourth point for this cross ratio to be useful, so reflect point $F$ over angle bisector $BD$ to a point $F... | 289 |
6,482 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_15 | 4 | In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overl... | Extend $DF$ to intersect with the extension of $AB$ at $G$ . Notice that $\triangle{BDF} \cong \triangle{BDG}$ , so $GD=DF$ . We now use Menelaus on $\triangle{GBF}$ , as $A$ $D$ , and $C$ are collinear; this gives us $\frac{GA}{BA} \cdot \frac{BC}{FC} \cdot \frac{DF}{GD}=1$ . As $GD=DF$ , we have $\frac{GA}{AB}=\frac{... | 289 |
6,483 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_1 | 1 | The product $N$ of three positive integers is $6$ times their sum , and one of the integers is the sum of the other two. Find the sum of all possible values of $N$ | Let the three integers be $a, b, c$ $N = abc = 6(a + b + c)$ and $c = a + b$ . Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$ . Since $a$ and $b$ are positive, $ab = 12$ so $\{a, b\}$ is one of $\{1, 12\}, \{2, 6\}, \{3, 4\}$ so $a + b$ is one of $13, 8, 7$ so the sum of all possible values of $N$ is $12 \cdot (... | 336 |
6,484 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_2 | 1 | Let $N$ be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when $N$ is divided by 1000? | We want a number with no digits repeating, so we can only use the digits $0-9$ once in constructing our number. To make the greatest number, we want the greatest digit to occupy the leftmost side and the least digit to occupy the rightmost side. Therefore, the last three digits of the greatest number should be an arran... | 120 |
6,485 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_3 | 1 | Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$ $B$ , and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$ $B$ is never immediately followed by $C$ , and $C$ is never immediately followed by $A$ . How many seven... | There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is $3*2^6=192$
Therefore, there are $\boxed{192}$ seven-letter good words. | 192 |
6,486 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_3 | 2 | Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$ $B$ , and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$ $B$ is never immediately followed by $C$ , and $C$ is never immediately followed by $A$ . How many seven... | There are three choices for the first letter and two choices for each subsequent letter, so there are $3\cdot2^{n-1}\ n$ -letter good words. Substitute $n=7$ to find there are $3\cdot2^6=\boxed{192}$ seven-letter good words. ~ aopsav (Credit to AoPS Alcumus) | 192 |
6,487 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_4 | 1 | In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Embed the tetrahedron in 4-space to make calculations easier.
Its vertices are $(1,0,0,0)$ $(0,1,0,0)$ $(0,0,1,0)$ $(0,0,0,1)$
To get the center of any face, we take the average of the three coordinates of that face. The vertices of the center of the faces are: $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 0)$ $(\frac{1}{3}... | 28 |
6,488 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_5 | 1 | A cylindrical log has diameter $12$ inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a $45^\circ$ angle with the plane of the first cut. The intersection of these two planes has e... | The volume of the wedge is half the volume of a cylinder with height $12$ and radius $6$ . (Imagine taking another identical wedge and sticking it to the existing one). Thus, $V=\dfrac{6^2\cdot 12\pi}{2}=216\pi$ , so $n=\boxed{216}$ | 216 |
6,489 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_6 | 1 | In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle "glued" together on the $12$ side, $[ABC]=\frac{1}{2}\cdot12\cdot14=84$
There are six points of intersection between $\Delta ABC$ and $\Delta A'B'C'$ . Connect each of these points to $G$
[asy] size(8cm); pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1... | 112 |
6,490 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_6 | 2 | In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | First, find the area of $\Delta ABC$ either like the first solution or by using Heron’s Formula. Then, draw the medians from $G$ to each of $A, B, C, A’, B’,$ and $C’$ . Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. ... | 112 |
6,491 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_7 | 1 | Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively. | The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$ . The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$
The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$ , where $a$ $b$ , and $c$ are the sides and $R$ is the circ... | 400 |
6,492 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_7 | 2 | Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively. | Let $\theta=\angle BDA$ . Let $AB=BC=CD=x$ . By the extended law of sines, \[\frac{x}{\sin\theta}=25\] Since $AC\perp BD$ $\angle CAD=90-\theta$ , so \[\frac{x}{\sin(90-\theta)=\cos\theta}=50\] Hence $x=25\sin\theta=50\cos\theta$ . Solving $\tan\theta=2$ $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\sqrt{5}}$ . ... | 400 |
6,493 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_8 | 1 | Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$ $f(2)=1716$ , and $f(3)=1848$ . Plugging in the values for x gives us a system of three equations:
$a+b+c=1440$
$4a+2b+c=1716$
$9a+3b+c=1848$
Solvin... | 348 |
6,494 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_8 | 2 | Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | Use the same rationale as in solution 1; instead of using terms $1,2,3$ , we use $-1,0,1$ and solve the $6$ th term.
$a-b+c=1440$
$c=1716$
$a+b+c=1848$
Accordingly we will solve
$a=-72, b=204, c=1716$
$36a+6b+c= \boxed{348}.$ | 348 |
6,495 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_8 | 3 | Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | Setting one of the sequences as $a+nr_1$ and the other as $b+nr_2$ , we can set up the following equalities
$ab = 1440$
$(a+r_1)(b+r_2)=1716$
$(a+2r_1)(b+2r_2)=1848$
We want to find $(a+7r_1)(b+7r_2)$
Foiling out the two above, we have
$ab + ar_2 + br_1 + r_1r_2 = 1716$ and $ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848$
Pluggin... | 348 |
6,496 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_9 | 2 | Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$ | Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0=4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]
Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P... | 6 |
6,497 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_10 | 1 | Two positive integers differ by $60$ . The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers? | Call the two integers $b$ and $b+60$ , so we have $\sqrt{b}+\sqrt{b+60}=\sqrt{c}$ . Square both sides to get $2b+60+2\sqrt{b^2+60b}=c$ . Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$ , and $(b+n+30)(b-n+30)=900$ . The sum of these two factors is $2b+60$ , so they must both be even. To maximize $b$ , we wan... | 156 |
6,498 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_11 | 1 | Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers... | We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$
Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$ $MN=BN-BM$ , and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$
From the third equation, we get $CN=\frac{168} {25}.$
By the Pythagorean Theorem in $\Delt... | 578 |
6,499 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_11 | 2 | Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers... | Let $E$ be the intersection of lines $BC$ and $DM$ . Since triangles $\Delta CME$ and $\Delta CMD$ share a side and height, the area of $\Delta CDM$ is equal to $\frac{DM}{EM}$ times the area of $\Delta CME$ .
By AA similarity, $\Delta EMB$ is similar to $\Delta ACB$ $\frac{EM}{AC}=\frac{MB}{CB}$ . Solving yields $EM=... | 578 |
6,500 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_12 | 1 | The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of ... | Let $v_i$ be the number of votes candidate $i$ received, and let $s=v_1+\cdots+v_{27}$ be the total number of votes cast. Our goal is to determine the smallest possible $s$
Candidate $i$ got $\frac{v_i}s$ of the votes, hence the percentage of votes they received is $\frac{100v_i}s$ . The condition in the problem statem... | 134 |
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