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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6,201 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_6 | 1 | Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ | We can use complementary counting. We can choose a five-element subset in ${14\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$ s and 5 $B$ s, thereby showing that there are ${10\choose 5}$ such... | 750 |
6,202 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_6 | 3 | Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ | We will proceed by complementary counting. We can easily see that there are $\binom{14}{5}$ ways to select $5$ element subsets from the original set. Now, we must find the number of subsets part of that group that do NOT have any consecutive integers.
We can think of this situation as stars and bars. We have $9$ stars ... | 750 |
6,203 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7 | 1 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ | First, note that $(2n)!! = 2^n \cdot n!$ , and that $(2n)!! \cdot (2n-1)!! = (2n)!$
We can now take the fraction $\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$ . We get that this fraction is equal to $\dfrac{(2i)!}{(2i)!!^2} = \dfrac{(2i)!}{2^{2i}(i!)^2}$
Now we can recognize... | 401 |
6,204 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7 | 3 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ | We can logically deduce that the $b$ value will be 1. Listing out the first few values of odd and even integers, we have: $1, 3, 5, 7, 9, 11, 13...$ and $2, 4, 6, 8, 10, 12, 14, 16...$ . Obviously, none of the factors of $2$ in the denominator will cancel out, since the numerator is odd. Starting on the second term of ... | 401 |
6,205 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7 | 4 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ | Using the initial steps from Solution 1, $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$ . Clearly $b = 1$ as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find $a$ . Note that $2^a$ would be the largest denominator in all the summands, so when they are summed it is the c... | 401 |
6,206 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_8 | 3 | Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\dfrac mn$ is the probability that the number of times Dave rolls his die is equal to or within... | Let's write out the probabilities with a set number of throws that Linda rolls before getting a 6. The probability of Linda rolling once and gets 6 right away is $\frac{1}{6}$ . The probability that Dave will get a six in the same, one less, or one more throw is $\frac{1}{6} + \frac{5}{6} * \frac{5}{6}$ . Thus the comb... | 41 |
6,207 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_11 | 1 | For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$ . Find the sum of all possible values of the product $mn$ | We have $\log m - \log k = \log \left( \frac mk \right)$ , hence we can rewrite the inequality as follows: \[- \log n < \log \left( \frac mk \right) < \log n\] We can now get rid of the logarithms, obtaining: \[\frac 1n < \frac mk < n\] And this can be rewritten in terms of $k$ as \[\frac mn < k < mn\]
From $k<mn$ it f... | 125 |
6,208 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_12 | 1 | From the set of integers $\{1,2,3,\dots,2009\}$ , choose $k$ pairs $\{a_i,b_i\}$ with $a_i<b_i$ so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$ . Find the maximum possible value of $k$ | Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\cdots+2k=k(2k+1)$ . On the other hand, as the sum of each pair is distinct and at most equal to $2009$ , the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \cdots + (2009-(k-1)) = \frac{k(4... | 803 |
6,209 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_13 | 1 | Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$ . The arc is divided into seven congruent arcs by six equally spaced points $C_1$ $C_2$ $\dots$ $C_6$ . All chords of the form $\overline {AC_i}$ or $\overline {BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the r... | Let the radius be 1 instead. All lengths will be halved so we will multiply by $2^{12}$ at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then $C_1,\ldots, C_6$ are 6 of the 14th roots of unity. Let $\omega=\text{cis}\frac{360^{\circ}}{14}... | 672 |
6,210 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_13 | 2 | Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$ . The arc is divided into seven congruent arcs by six equally spaced points $C_1$ $C_2$ $\dots$ $C_6$ . All chords of the form $\overline {AC_i}$ or $\overline {BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the r... | Let $O$ be the midpoint of $A$ and $B$ . Assume $C_1$ is closer to $A$ instead of $B$ $\angle AOC_1$ $\frac {\pi}{7}$ . Using the Law of Cosines
$\overline {AC_1}^2$ $8 - 8 \cos \frac {\pi}{7}$ $\overline {AC_2}^2$ $8 - 8 \cos \frac {2\pi}{7}$ ,
.
.
. $\overline {AC_6}^2$ $8 - 8 \cos \frac {6\pi}{7}$
So $n$ $(8^6)(1 -... | 672 |
6,211 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_14 | 1 | The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac85a_n + \frac65\sqrt {4^n - a_n^2}$ for $n\geq 0$ . Find the greatest integer less than or equal to $a_{10}$ | We can now simply start to compute the values $b_i$ by hand:
\begin{align*} b_1 & = \frac 35 \\ b_2 & = \frac 45\cdot \frac 35 + \frac 35 \sqrt{1 - \left(\frac 35\right)^2} = \frac{24}{25} \\ b_3 & = \frac 45\cdot \frac {24}{25} + \frac 35 \sqrt{1 - \left(\frac {24}{25}\right)^2} = \frac{96}{125} + \frac 35\cdot\frac 7... | 983 |
6,212 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_14 | 2 | The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac85a_n + \frac65\sqrt {4^n - a_n^2}$ for $n\geq 0$ . Find the greatest integer less than or equal to $a_{10}$ | After we do the substitution, we can notice the fact that $\left( \frac 35 \right)^2 + \left( \frac 45 \right)^2 = 1$ , which may suggest that the formula may have something to do with the unit circle. Also, the expression $\sqrt{1-x^2}$ often appears in trigonometry, for example in the relationship between the sine an... | 983 |
6,213 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_1 | 1 | Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance? | Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance.
Thus, $\dfrac{3k}{5k + 20} = \dfrac{29}{50}$ , solving gives $k = 116$ . Thus, the number of people that like to dance is $2k + 20 = \boxed{252}$ | 252 |
6,214 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_1 | 2 | Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance? | Let the number of girls be $g$ . Let the number of total people originally be $t$
We know that $\frac{g}{t}=\frac{3}{5}$ from the problem.
We also know that $\frac{g}{t+20}=\frac{29}{50}$ from the problem.
We now have a system and we can solve.
The first equation becomes:
$3t=5g$
The second equation becomes:
$50g=29t+5... | 252 |
6,215 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_1 | 3 | Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance? | Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: \[0.4p+20 = 0.42(p+20)\] Solving for p gives us $p=580$ , so the solution is $0.4p+20 = \boxed{252}$ | 252 |
6,216 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3 | 1 | Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $... | Let the biking rate be $b$ , swimming rate be $s$ , jogging rate be $j$ , all in km/h.
We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$ . Subtracting the second from twice the first gives $4j + 5s = 57$ . Mod 4, we need $s\equiv1\pmod{4}$ . Thus, $(j,s) = (13,1),(8,5),(3,9)$
$(13,1)$ and $(3,9)$ give non-integral $b$ , bu... | 314 |
6,217 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3 | 2 | Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $... | Let $b$ $j$ , and $s$ be the biking, jogging, and swimming rates of the two people. Hence, $2b + 3j + 4s = 74$ and $4b + 2j + 3s = 91$ . Subtracting gives us that $2b - j - s = 17$ . Adding three times this to the first equation gives that $8b + s = 125\implies b\le 15$ . Adding four times the previous equation to ... | 314 |
6,218 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3 | 3 | Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $... | Creating two systems, we get $2x+3y+4z=74$ , and $2y+3z+4x=91$ . Subtracting the two expressions we get $y+z-2x=-17$ . Note that $-17$ is odd, so one of $x,y,z$ is odd. We see from our second expression that $z$ must be odd, because $91$ is also odd and $2y$ and $4x$ are odd. Thus, with this information, we can test ca... | 314 |
6,219 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3 | 4 | Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $... | Building on top of Solution 3, we can add $j+s-2b=17$ and $2b+3j+4s=74$ (sorry, I used different variables) to get $4j+5s=57$ . Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug ... | 314 |
6,220 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_7 | 1 | Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$ . For example, $S_4$ is the set ${400,401,402,\ldots,499}$ . How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square? | The difference between consecutive squares is $(x + 1)^2 - x^2 = 2x + 1$ , which means that all squares above $50^2 = 2500$ are more than $100$ apart.
Then the first $26$ sets ( $S_0,\cdots S_{25}$ ) each have at least one perfect square because the differences between consecutive squares in them are all less than $100... | 708 |
6,221 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_8 | 1 | Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\] | Since we are dealing with acute angles, $\tan(\arctan{a}) = a$
Note that $\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}$ , by tangent addition. Thus, $\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}$
Applying this to the first two terms, we get $\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\... | 47 |
6,222 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_8 | 2 | Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\] | Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\arctan\frac{1}{n}$ , is the argument of $n+i$ . The sum of these angles is then just the argument of the product \[(3+i)(4+i)(5+i)(n+i)\] and expansion give us $(48n-46)+(48+46n)i$ . Since the argument of th... | 47 |
6,223 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9 | 1 | Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ . The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probabi... | Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
\begin{align*}3a + 4b + 6c &= 41\\ a + b + c &= 10\end{align*}
Subtracting 3 times the second from the first gives $b + 3c = 11$ , or $(b,c) = (2,3),(5,2),(8,1),(11,0)$ . The last doesn't work, obviousl... | 190 |
6,224 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9 | 2 | Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ . The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probabi... | It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that $3$ and $6$ are both divisible by $3$ , so the number of $4$ -crates must be congruent to $41\bmod{3}$ , which is also congruent to $2\bmod{3}$ . Our solutions for the number of $4$ -crates will repeat mo... | 190 |
6,225 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9 | 3 | Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ . The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probabi... | Let's make two observations. We are trying to find the number of ways we can add $3\text{s}, 4\text{s}$ , and $6\text{s}$ to get $41$ , and the total number of (non-distinct) sums possible is $3^{10}$ .
Then we just use casework to easily and directly solve for the number of ways to get $41$ . To begin, the minimum sum... | 190 |
6,226 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9 | 4 | Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ . The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probabi... | Note we are placing 10 crates where each "height" is 3, 4, 6 and we want all the heights to sum to 41.
We can model this as the generating function \[\left(x^3+x^4+x^6\right)^{10}\] where we want the coefficient of $x^{41}.$ First off, factor this to get \[{x^3}^{10}\left(1+x+x^3\right)^{10}\] and then see that we wan... | 190 |
6,227 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_11 | 1 | Consider sequences that consist entirely of $A$ 's and $B$ 's and that have the property that every run of consecutive $A$ 's has even length, and every run of consecutive $B$ 's has odd length. Examples of such sequences are $AA$ $B$ , and $AABAA$ , while $BBAB$ is not such a sequence. How many such sequences have len... | Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$ . If a sequence ends in an $A$ , then it must have been formed by appending two $A$ s to the end of a string of length $n-2$ . If a sequence ends in a $B,$ it must have either been formed by appending one $B$ to a stri... | 172 |
6,228 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_11 | 2 | Consider sequences that consist entirely of $A$ 's and $B$ 's and that have the property that every run of consecutive $A$ 's has even length, and every run of consecutive $B$ 's has odd length. Examples of such sequences are $AA$ $B$ , and $AABAA$ , while $BBAB$ is not such a sequence. How many such sequences have len... | Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$
Additionally, let $t_n$ denote the total number of sequences of length $n$ . Then, $t_n=a_n+b_n$ , as the total amount of sequences of length $n$ consists of the sequences of length $n$ ending in $A$ and the sequences ... | 172 |
6,229 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_11 | 3 | Consider sequences that consist entirely of $A$ 's and $B$ 's and that have the property that every run of consecutive $A$ 's has even length, and every run of consecutive $B$ 's has odd length. Examples of such sequences are $AA$ $B$ , and $AABAA$ , while $BBAB$ is not such a sequence. How many such sequences have len... | There must be an even amount of runs of consecutive $B$ s due to parity. Thus, we can split this sequence into the following cases: $A$ $BAAB$ $AABAAB$ $BAABAA$ $AABAABAA$ $BAABAABAAB$ $AABAABAABAAB$ $BAABAABAABAA$ , and $AABAABAABAABAA$ , in which the amount of letters in one run does not necessarily represent the amo... | 172 |
6,230 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_12 | 1 | On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling ... | Let $n$ be the number of car lengths that separates each car (it is easy to see that this should be the same between each pair of consecutive cars.) Then their speed is at most $15n$ . Let a unit be the distance between the cars (front to front). Then the length of each unit is $4(n + 1)$ . To maximize, in a unit, the ... | 375 |
6,231 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_12 | 2 | On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling ... | Disclaimer: This is for the people who may not understand calculus, and is also how I did it.
First, we assume several things. First, we assume the speeds must be multiples of 15 to maximize cars, because any less will be a waste. Second, we start with one car in front of the photoelectric eye.
We first set the speed o... | 375 |
6,232 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14 | 2 | Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $\overline{AB} = 18$ , and let $m$ denote the maximum possible length of seg... | From the diagram, we see that $BQ = OT + BO \sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)$ , and that $QP = BA\cos\theta = 18\cos\theta$
\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \sin\theta)^2 + 18^2\cos^2\theta\\ &= 9^2[1 + 2\sin\theta + \sin^2\theta + 4(1 - \sin^2\theta)]\\ BP^2 &= 9^2[5 + 2\sin\theta - 3\sin^2\... | 432 |
6,233 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14 | 3 | Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $\overline{AB} = 18$ , and let $m$ denote the maximum possible length of seg... | (Diagram credit goes to Solution 2)
We let $AC=x$ . From similar triangles, we have that $PC=\frac{x\sqrt{x^2+18x}}{x+9}$ (Use Pythagorean on $\triangle\omega TC$ and then using $\triangle\omega CT\sim\triangle ACP$ ). Similarly, $TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}$ . Using the Pythagorean Theorem again and $\triangle C... | 432 |
6,234 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14 | 4 | Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $\overline{AB} = 18$ , and let $m$ denote the maximum possible length of seg... | Let $h$ be the distance from $A$ to $CT$ . Observe that $h$ takes any value from $0$ to $2r$ , where $r$ is the radius of the circle.
Let $Q$ be the foot of the altitude from $B$ to $CT$ . It is clear that $T$ is the midpoint of $PQ$ , and so the length $OT$ is the average of $AP$ and $BQ$ . It follows thus that $BQ = ... | 432 |
6,235 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_15 | 1 | A square piece of paper has sides of length $100$ . From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up alon... | In the original picture, let $P$ be the corner, and $M$ and $N$ be the two points whose distance is $\sqrt{17}$ from $P$ . Also, let $R$ be the point where the two cuts intersect.
Using $\triangle{MNP}$ (a 45-45-90 triangle), $MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}$ $\triangle{MNR}$ is equilateral , so $MR... | 871 |
6,236 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_15 | 2 | A square piece of paper has sides of length $100$ . From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up alon... | In the final pyramid, let $ABCD$ be the smaller square and let $A'B'C'D'$ be the larger square such that $AA'$ , etc. are edges.
It is obvious from the diagram that $\angle A'AB = \angle A'AD = 105^\circ$
Let $AB$ and $AD$ be the positive $x$ and $y$ axes in a 3-d coordinate system such that $A'$ has a positive $z$ coo... | 871 |
6,237 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_1 | 1 | Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$ | Rewriting this sequence with more terms, we have
Factoring this expression yields
Next, we get
Then,
Dividing $10100$ by $1000$ yields a remainder of $\boxed{100}$ | 100 |
6,238 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_1 | 2 | Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$ | By observation, we realize that the sequence \[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2\] alternates every 4 terms. Simplifying, we get \[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2 = 8a + 12\] , turning $N$ into a arithmetic sequence with 25 terms, them being $1, 5, 9, \dots ,97$ , as the series $8a + 12$ alternates every 4 terms.
App... | 100 |
6,239 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_2 | 1 | Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $... | Let $r$ be the time Rudolph takes disregarding breaks and $\frac{4}{3}r$ be the time Jennifer takes disregarding breaks. We have the equation \[r+5\left(49\right)=\frac{4}{3}r+5\left(24\right)\] \[125=\frac13r\] \[r=375.\] Thus, the total time they take is $375 + 5(49) = \boxed{620}$ minutes. | 620 |
6,240 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_2 | 2 | Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $... | Let the total time that Jennifer and Rudolph bike, including rests, to be $t$ minutes. Furthermore, let Rudolph's biking rate be $r$ so Jennifer's biking rate is $\frac{3}{4}r$ . Note that Rudolf takes 49 breaks, taking $49\cdot 5$ minutes, and Jennifer takes 24 breaks, taking $24\cdot 5$ minutes. Since they both rea... | 620 |
6,241 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_3 | 2 | A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic... | A more intuitive way to solve it is by seeing that to keep the volume of the rectangular cheese the greatest, we must slice the cheese off to decrease the greatest length of the cheese (this is easy to check). Here are the ten slices:
${10, 13, 14} \rightarrow {10, 13, 13} \rightarrow {10, 12, 13} \rightarrow {10, 12, ... | 729 |
6,242 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5 | 1 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ | Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$ . Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$
As $\angle AED = 90^{\circ}$ , note that the midpoint of $\overline{AD}$ $N$ , is the center of the circumcircle of $\triangle AED$ . We can do the same with the circumcircle about $\triangle BEC$ and $M$ ... | 504 |
6,243 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5 | 2 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ | Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$ , respectively. Let $x = NH$ , so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$ . Also, let $h = BF = CG = HM$
By AA~, we have that $\triangle AFB \sim \triangle CGD$ , and so \[\frac{BF}{AF} = \frac {DG}{CG} \Longle... | 504 |
6,244 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5 | 3 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ | Plot the trapezoid such that $B=\left(1000\cos 37^\circ, 0\right)$ $C=\left(0, 1000\sin 37^\circ\right)$ $A=\left(2008\cos 37^\circ, 0\right)$ , and $D=\left(0, 2008\sin 37^\circ\right)$
The midpoints of the requested sides are $\left(500\cos 37^\circ, 500\sin 37^\circ\right)$ and $\left(1004\cos 37^\circ, 1004\sin 37^... | 504 |
6,245 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5 | 5 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ | Let the height be h. Note that if $\overline{NH} = x$ then if we draw perpendiculars like in solution 1, $\overline{FN} = 500 - x, \overline{AF} = 504 + x, \overline{HG} = 500, \overline{GD} = 504 - x.$ Note that we wish to find $\overline{MN} = \sqrt{x^2 + h^2}.$ Let's find $\tan(53)$ in two ways. Finding $\tan(53)$ f... | 504 |
6,246 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5 | 6 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ | Rotate trapezoid $MNCD$ 180 degrees around point $N$ so that $AN$ coincides with $ND$ . Let the image of trapezoid $MNCD$ be $ANM'C'$ . Since angles are preserved during rotations, $\angle BAC' = 37^{\circ} + 53 ^{\circ} =90 ^{\circ}$ . Since $BM=CM=C'M'$ and $BM || C'M'$ $BMM'C'$ is a parallelogram. Thus, $MM'=BC'$ .
... | 504 |
6,247 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_6 | 1 | The sequence $\{a_n\}$ is defined by \[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\] The sequence $\{b_n\}$ is defined by \[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\] Find $\frac {b_{32}}{a_{32}}$ | Rearranging the definitions, we have \[\frac{a_n}{a_{n-1}} = \frac{a_{n-1}}{a_{n-2}} + 1,\quad \frac{b_n}{b_{n-1}} = \frac{b_{n-1}}{b_{n-2}} + 1\] from which it follows that $\frac{a_n}{a_{n-1}} = 1+ \frac{a_{n-1}}{a_{n-2}} = \cdots = (n-1) + \frac{a_{1}}{a_0} = n$ and $\frac{b_n}{b_{n-1}} = (n-1) + \frac{b_{1}}{b_0} =... | 561 |
6,248 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_8 | 1 | Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer. | By the product-to-sum identities , we have that $2\cos a \sin b = \sin (a+b) - \sin (a-b)$ . Therefore, this reduces to a telescope series: \begin{align*} \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(... | 251 |
6,249 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_8 | 2 | Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer. | We proceed with complex trigonometry. We know that for all $\theta$ , we have $\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)$ and $\sin \theta = \dfrac{1}{2i} \left( z - \dfrac{1}{z} \right)$ for some complex number $z$ on the unit circle. Similarly, we have $\cos n \theta = \dfrac{1}{2} \left( z^n + \dfra... | 251 |
6,250 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_10 | 1 | The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maximu... | We label our points using coordinates $0 \le x,y \le 3$ , with the bottom-left point being $(0,0)$ . By the Pythagorean Theorem , the distance between two points is $\sqrt{d_x^2 + d_y^2}$ where $0 \le d_x, d_y \le 3$ ; these yield the possible distances (in decreasing order) \[\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{... | 240 |
6,251 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_11 | 1 | In triangle $ABC$ $AB = AC = 100$ , and $BC = 56$ Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\triangle ABC$ . The radius of circle $Q$ can be ex... | Refer to the above diagram. Let the larger circle have center $O_1$ , the smaller have center $O_2$ , and the incenter be $I$ . We can easily calculate that the area of $\triangle ABC = 2688$ , and $s = 128$ and $R = 21$ , where $R$ is the inradius.
Now, Line $\overline{AI}$ is the perpendicular bisector of $\overline{... | 254 |
6,252 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_11 | 2 | In triangle $ABC$ $AB = AC = 100$ , and $BC = 56$ Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\triangle ABC$ . The radius of circle $Q$ can be ex... | Let the incenter be O and the altitude from A to $\overline{BC}$ be T. Note that by AA, $\triangle BQY \sim \triangle OBT$ and $\triangle PXC \sim \triangle OTC.$ Note that from $A = rs$ , the inradius of the big triangle is $21$ Using ravi substitution(or noticing that $\overline{AT}$ is an altitude), we then have tha... | 254 |
6,253 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15 | 1 | Find the largest integer $n$ satisfying the following conditions: | Write $n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1$ , or equivalently, $(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2$
Since $2n + 1$ and $2n - 1$ are both odd and their difference is $2$ , they are relatively prime . But since their product is three times a square, one of them must be a square and the other th... | 181 |
6,254 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15 | 2 | Find the largest integer $n$ satisfying the following conditions: | Suppose that the consecutive cubes are $m$ and $m + 1$ . We can use completing the square and the first condition to get: \[(2n)^2 - 3(2m + 1)^2 = 1\equiv a^2 - 3b^2\] where $a$ and $b$ are non-negative integers. Now this is a Pell equation , with solutions in the form $(2 + \sqrt {3})^k = a_k + \sqrt {3}b_k,$ $k = 0,1... | 181 |
6,255 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15 | 3 | Find the largest integer $n$ satisfying the following conditions: | Let us generate numbers $1$ to $1000$ for the second condition, for squares. We know for $N$ to be integer, the squares must be odd. So we generate $N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801$ $N$ cannot exceed $1000$ since it is AIME problem. Now take the first criterion, let $a$ be the smaller consecutive c... | 181 |
6,256 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_2 | 1 | A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the... | Clearly we have people moving at speeds of $6,8$ and $10$ feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there is som... | 52 |
6,257 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_4 | 1 | Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are 60, 84, and 140 years. The three planets and the star are currently collinear . What is the fewest number of years from now that they will all be collinear again? | Denote the planets $A, B, C$ respectively. Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years. These are given by $a(t) = \frac{t \pi}{30}$ $b(t) = \frac{t \pi}{42}$ $c(t) = \frac{t \pi}{70}$
In order for the planets and the central star to be ... | 105 |
6,258 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5 | 1 | The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer
For how many integer Fahrenhe... | Examine $F - 32$ modulo 9.
Generalizing this, we define that $9x + k = F - 32$ . Thus, $F = \left[\frac{9}{5}\left[\frac{5}{9}(9x + k)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + \left[\frac{5}{9}k\right]) + 32\right] \Longrightarrow F = \left[\frac{9}{5} \left[\frac{5}{9}k \right] \right] + 9x + 32$ ... | 539 |
6,259 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5 | 3 | The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer
For how many integer Fahrenhe... | Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $\boxed{539}$ solutions. | 539 |
6,260 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6 | 1 | A frog is placed at the origin on the number line , and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A move sequence is ... | Another way would be to use a table representing the number of ways to reach a certain number
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\ \hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\ \end{tabular}$
How we came with ... | 169 |
6,261 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6 | 2 | A frog is placed at the origin on the number line , and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A move sequence is ... | Let $f(n)$ be the number of ways one can get to $39$ starting at position $n.$ We wish to compute $f(0).$ Now it's just a long simplifications until you get to $f(36) = 1.$ We have \[f(0) = f(3) + f(13) = f(6) +2f(13) + f(9) + 3f(12) + f(12) + 4f(12) + f(15) + 5f(12).\]
Most of these steps are valid since at any $n$ th... | 169 |
6,262 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_7 | 1 | Let $N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )$
Find the remainder when $N$ is divided by 1000. ( $\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$ , and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$ .) | The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer ); otherwise, it is equal to 1. Thus, we need to find when or not $\log_{\sqrt{2}} k$ is an integer.
The change of base formula shows that $\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}$ . For the $\log 2$ ... | 477 |
6,263 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9 | 1 | In right triangle $ABC$ with right angle $C$ $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii . The circle with center $O_1$ is tangent to the hypotenuse and to the extension ... | A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$ $A$ at $(0,30)$ , and $B$ at $(16,0)$ . We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance b... | 737 |
6,264 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9 | 2 | In right triangle $ABC$ with right angle $C$ $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii . The circle with center $O_1$ is tangent to the hypotenuse and to the extension ... | It is known that $O_1O_2$ is parallel to AB. Thus, extending $O_1F$ and $GO_2$ to intersect at H yields similar triangles $O_1O_2H$ and BAC, so that $O_1O_2 = 2r$ $O_1H = \frac{16r}{17}$ , and $HO_2 = \frac{30r}{17}$ . It should be noted that $O_2G = r$ . Also, FHGC is a rectangle, and so AF = $\frac{47r}{17} - 30$ and... | 737 |
6,265 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | 1 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | Consider the first column. There are ${6\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.
Now consider the 3x3 that is next to th... | 860 |
6,266 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | 2 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | We start by showing that every group of $6$ rows can be grouped into $3$ complementary pairs.
We proceed with proof by contradiction. Without loss of generality, assume that the first row has columns $1$ and $2$ shaded. Note how if there is no complement to this, then all the other five rows must have at least one squa... | 860 |
6,267 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | 3 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares.
There are ${6 \choose 3}$ ways to choose which rows have 1 shaded square (which we'll call ... | 860 |
6,268 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | 4 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | We can use generating functions. Suppose that the variables $a$ $b$ $c$ , and $d$ represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function $ab+ac+ad+bc+bd+cd$ , which we can wri... | 860 |
6,269 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11 | 1 | For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000. | $\left(k- \frac 12\right)^2=k^2-k+\frac 14$ and $\left(k+ \frac 12\right)^2=k^2+k+ \frac 14$ . Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k<p\leq k^2+k$ . There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$ $44<\sqrt{2007}<45$ , so all numbers $1$ to $44$ have... | 955 |
6,270 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11 | 2 | For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000. | Let $p$ be in the range of $a^2 \le p < (a+1)^2$ . Then, we need to find the point where the value of $b(p)$ flips from $k$ to $k+1$ . This will happen when $p$ exceeds $(a+\frac{1}{2})^2$ or $a(a+1)+\frac{1}{4}$ . Thus, if $a^2 \le p \le a(a+1)$ then $b(p)=a$ . For $a(a+1) < p < (a+1)^2$ , then $b(p)=a+1$ . There are ... | 955 |
6,271 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12 | 1 | In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ . Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the regi... | Redefine the points in the same manner as the last time ( $\triangle AB'C'$ , intersect at $D$ $E$ , and $F$ ). This time, notice that $[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])$
The area of $[\triangle AB'C'] = [\triangle ABC]$ . The altitude of $\triangle ABC$ is clearly $10 \tan 75 = 10 \tan... | 875 |
6,272 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12 | 2 | In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ . Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the regi... | Call the points of the intersections of the triangles $D$ $E$ , and $F$ as noted in the diagram (the points are different from those in the diagram for solution 1). $\overline{AD}$ bisects $\angle EDE'$
Through HL congruency, we can find that $\triangle AED$ is congruent to $\triangle AE'D$ . This divides the region $A... | 875 |
6,273 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12 | 3 | In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ . Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the regi... | From the given information, calculate the coordinates of all the points (by finding the equations of the lines, equating). Then, use the shoelace method to calculate the area of the intersection.
Now, we can equate the equations to find the intersections of all the points.
We take these points and tie them together by ... | 875 |
6,274 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | 1 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | Define a function $f(n)$ on the non-negative integers, as \[f(n) = \frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_{n}}\] We want $\left\lfloor f(2006) \right\rfloor$
Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Dividing through by $a_{n}a_{n-1}$ , we get \[.\phantom{------------}... | 224 |
6,275 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | 2 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | We are given that
$a_{n+1}a_{n-1}= a_{n}^{2}+2007$ $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$
Add these two equations to get
This is an invariant . Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$ , the above equation means
$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$
We can thus calculate that $b_{2007}+\frac{1... | 224 |
6,276 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | 3 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | The equation $a_{n+1}a_{n-1}-a_n^2=2007$ looks like the determinant \[\left|\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right|=2007.\] Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence $... | 224 |
6,277 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | 4 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | We will try to manipulate $\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\frac{a_1^2+a_2^2}{a_1a_2}$
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2... | 224 |
6,278 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | 5 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$ , and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$ , respectively. It will be clear why I decided to factor these exp... | 224 |
6,279 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_15 | 1 | Let $ABC$ be an equilateral triangle , and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ . Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ . The area of triangle $DEF$ is $14\sqrt{3}$ . The two possible values of the length of side $AB$ are $p \pm q \sqrt{... | AIME I 2007-15.png
Denote the length of a side of the triangle $x$ , and of $\overline{AE}$ as $y$ . The area of the entire equilateral triangle is $\frac{x^2\sqrt{3}}{4}$ . Add up the areas of the triangles using the $\frac{1}{2}ab\sin C$ formula (notice that for the three outside triangles, $\sin 60 = \frac{\sqrt{3}}... | 989 |
6,280 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_15 | 2 | Let $ABC$ be an equilateral triangle , and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ . Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ . The area of triangle $DEF$ is $14\sqrt{3}$ . The two possible values of the length of side $AB$ are $p \pm q \sqrt{... | First of all, assume $EC=x,BD=m, ED=a, EF=b$ , then we can find $BF=m-3, AE=2+m-x$ It is not hard to find $ab*sin60^{\circ}*\frac{1}{2}=14\sqrt{3}, ab=56$ , we apply LOC on $\triangle{DEF}, \triangle{BFD}$ , getting that $(m-3)^2+m^2-m(m-3)=a^2+b^2-ab$ , leads to $a^2+b^2=m^2-3m+65$ Apply LOC on $\triangle{CED}, \trian... | 989 |
6,281 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_1 | 1 | A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. ... | There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.
Thus, $N = 2520 + 1200 = 3720$ , and $\frac{N}{10} = \boxed{372}$ | 372 |
6,282 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_2 | 1 | Find the number of ordered triples $(a,b,c)$ where $a$ $b$ , and $c$ are positive integers $a$ is a factor of $b$ $a$ is a factor of $c$ , and $a+b+c=100$ | Denote $x = \frac{b}{a}$ and $y = \frac{c}{a}$ . The last condition reduces to $a(1 + x + y) = 100$ . Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$
Subtracting the one, we see that $x + y = \{0,1,3,4,9,19,24,49,99\}$ . There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y = n$ . ... | 200 |
6,283 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_3 | 1 | Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $EF^{2}$ [asy]unitsize(0.2 cm); pair A, B, C, D, E, F; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); draw(A--B--C--D--cycle); draw(... | Drawing $EF$ , it clearly passes through the center of $ABCD$ . Letting this point be $P$ , we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\frac{13}{\sqrt{2}}.$ Now, from Ptolemy's, $13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP=\frac{17\sqrt{2}}{2}$ . Since $EF=EP+FP=2\cdot... | 578 |
6,284 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_3 | 2 | Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $EF^{2}$ [asy]unitsize(0.2 cm); pair A, B, C, D, E, F; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); draw(A--B--C--D--cycle); draw(... | We first see that the whole figure is symmetrical and reflections across the center that we will denote as $O$ bring each half of the figure to the other half. Thus we consider a single part of the figure, namely $EO.$
First note that $\angle BAO = 45^{\circ}$ since $O$ is the center of square $ABCD.$ Also note that $\... | 578 |
6,285 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_4 | 1 | The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoo... | Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.
Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):
Solve the system of equations with the first two equations to find that $(x,y) = \left(... | 450 |
6,286 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_6 | 1 | An integer is called parity-monotonic if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $a_{i}$ is odd , and $a_{i}>a_{i+1}$ if $a_{i}$ is even . How many four-digit parity-monotonic integers are there? | This problem can be solved via recursion since we are "building a string" of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications( $0$ can't be at the front and no digit is less than $9$ ). There are $4$ options to add no matter what(try some examples ... | 640 |
6,287 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_8 | 1 | rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
Given that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic ... | Denote the number of horizontal lines drawn as $x$ , and the number of vertical lines drawn as $y$ . The number of basic rectangles is $(x - 1)(y - 1)$ $5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}$ . Substituting, we find that $(x - 1)\left(-\frac 54x + \frac{2003}4\right)$
FOIL this to get a quadratic, $-\f... | 896 |
6,288 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_9 | 1 | Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$ | Several Pythagorean triples exist amongst the numbers given. $BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105$ . Also, the length of $EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287$
Use the Two Tangent Theorem on $\triangle BEF$ . Since both circles are inscribed in congruent triangles, they a... | 259 |
6,289 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_10 | 1 | Let $S$ be a set with six elements . Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$ , not necessarily distinct, are chosen independently and at random from $\mathcal{P}$ . The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$ $n$ , and $r$ are posit... | Let $|S|$ denote the number of elements in a general set $S$ . We use complementary counting.
There is a total of $2^6$ elements in $P$ , so the total number of ways to choose $A$ and $B$ is $(2^6)^2 = 2^{12}$
Note that the number of $x$ -element subset of $S$ is $\binom{6}{x}$ . In general, for $0 \le |A| \le 6$ , in ... | 710 |
6,290 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_11 | 1 | Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface . The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$ . It rolls over the smaller tube and continues rolling along the flat surface until it comes to ... | 2007 AIME II-11.png
If it weren’t for the small tube, the larger tube would travel $144\pi$ . Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.
Drawing the radii as shown in the diagram, notice that the hypotenuse of the right... | 179 |
6,291 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_13 | 1 | triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$ th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in the given diagram). In each square of the eleventh row... | Label each of the bottom squares as $x_0, x_1 \ldots x_9, x_{10}$
Through induction , we can find that the top square is equal to ${10\choose0}x_0 + {10\choose1}x_1 + {10\choose2}x_2 + \ldots {10\choose10}x_{10}$ . (This also makes sense based on a combinatorial argument: the number of ways a number can "travel" to the... | 640 |
6,292 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14 | 1 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | If the leading term of $f(x)$ is $ax^m$ , then the leading term of $f(x)f(2x^2) = ax^m \cdot a(2x^2)^m = 2^ma^2x^{3m}$ , and the leading term of $f(2x^3 + x) = 2^max^{3m}$ . Hence $2^ma^2 = 2^ma$ , and $a = 1$ . Because $f(0) = 1$ , the product of all the roots
of $f(x)$ is $\pm 1$ . If $f(\lambda) = 0$ , then $f(2\lam... | 676 |
6,293 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14 | 2 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | Let $r$ be a root of $f(x)$ . Then we have $f(r)f(2r^2)=f(2r^3+r)$ ; since $r$ is a root, we have $f(r)=0$ ; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$ , so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no... | 676 |
6,294 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14 | 3 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | Let $r$ be a root of $f(x).$ This means that $f(r)f(2r^2)=f(2r^3+r).$ In other words, $2r^3+r$ is a root of $f(x)$ too. Since $f(x)$ can't have infinitely many roots, \[Q(x)=P(P(\dotsb P(P(r)) \dotsb))\] is cyclic, where $P(x)=2x^3+x.$ Now, we will do casework.
Case 1: $\deg f\geq1$
Subcase 1: $|r|>1$
This means that \... | 676 |
6,295 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | 1 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ an... | [asy] defaultpen(fontsize(12)+0.8); size(350); pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,... | 389 |
6,296 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | 2 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ an... | 2007 AIME II-15b.gif
Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$ , where $s$ is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, $u.$
The circumradius is $R=\frac{abc}{4rs}=\frac{1... | 389 |
6,297 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | 3 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ an... | Let $A'$ $B'$ $C'$ , and $O$ be the centers of circles $\omega_{A}$ $\omega_{B}$ $\omega_{C}$ $\omega$ , respectively, and let $x$ be their radius.
Now, triangles $ABC$ and $A'B'C'$ are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for $x$
Since $OA'=OB'... | 389 |
6,298 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | 4 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ an... | According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is $2r$ . Now denoting $AB=13;BC=14;AC=15$ , and centers of circles tangent to $AB,AC;AC,BC;AB,BC$ are relatively $M,N,O$ with $OJ,NK$ both perpendicular ... | 389 |
6,299 | https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15 | 5 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ an... | [asy] defaultpen(fontsize(12)+0.8); size(300); pair A,B,C,X,Y,Z,P,Q,R; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); draw(A--B--C--A); dra... | 389 |
6,300 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_2 | 1 | Let set $\mathcal{A}$ be a 90- element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$ | The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$ . The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$ . All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$
Alternatively, for ease of calculation, let set $\mathcal{B}$ be a 10-element... | 901 |
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