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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 sets.
Given a five-element subset $S$ of $\{1,2,\dots,14\}$ in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the $i$ -th character is $A$ if $i\in S$ and $B$ otherwise. Now we got a string with 5 $A$ s and 9 $B$ s. As no two numbers were consecutive, we know that in our string no two $A$ s are consecutive. We can now remove exactly one $B$ from between each pair of $A$ s to get a string with 5 $A$ s and 5 $B$ s. And clearly this is a bijection, as from each string with 5 $A$ s and 5 $B$ s we can reconstruct one original set by reversing the construction.
Hence we have $m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750$ , and the answer is $1750 \bmod 1000 = \boxed{750}$
| 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 and $5$ bars, where the stars represent the integers NOT in the subset while the bars ARE the integers in the chosen subset. We want the amount of combinations where there is at least one star between each bar (meaning that none of the integers in the subset are consecutive). To do this, we remove $4$ stars from the total $9$ stars ( $4$ because that is the amount that is between each bar). Now, we have $5$ stars and $5$ bars with no restrictions, so we calculate this as $\binom{10}{5}$ , which is $252$ . Hence, $\binom{14}{15} - \binom{10}{5} = 2002 - 252 = 1750$ , which becomes $\boxed{750}$
| 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 that $\dfrac{(2i)!}{(i!)^2}$ is simply ${2i \choose i}$ , hence this fraction is $\dfrac{{2i\choose i}}{2^{2i}}$ , and our sum turns into $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$
Let $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ .
Obviously $c$ is an integer, and $S$ can be written as $\dfrac{c}{2^{2\cdot 2009}}$ .
Hence if $S$ is expressed as a fraction in lowest terms, its denominator will be of the form $2^a$ for some $a\leq 2\cdot 2009$
In other words, we just showed that $b=1$ .
To determine $a$ , we need to determine the largest power of $2$ that divides $c$
Let $p(i)$ be the largest $x$ such that $2^x$ that divides $i$
We can now return to the observation that $(2i)! = (2i)!! \cdot (2i-1)!! = 2^i \cdot i! \cdot (2i-1)!!$ . Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$
It immediately follows that $p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)$ ,
and hence $p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)$
Obviously, for $i\in\{1,2,\dots,2009\}$ the function $f(i)=2\cdot 2009 - i - p(i!)$ is is a strictly decreasing function.
Therefore $p(c) = p\left( {2\cdot 2009\choose 2009} \right) = 2009 - p(2009!)$
We can now compute $p(2009!) = \sum_{k=1}^{\infty} \left\lfloor \dfrac{2009}{2^k} \right\rfloor = 1004 + 502 + \cdots + 3 + 1 = 2001$ .
Hence $p(c)=2009-2001=8$
And thus we have $a=2\cdot 2009 - p(c) = 4010$ , and the answer is $\dfrac{ab}{10} = \dfrac{4010\cdot 1}{10} = \boxed{401}$
| 401
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6,204
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https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7
| 3
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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 the numerator, a factor of $3$ occurs every $3$ terms, and starting out on the third term of the denominator, a factor of $3$ appears also every $3$ terms. Thus, the factors of $3$ on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form $1 \cdot 2^a$ . Since there will be no odd factors in the denominators, all the denominators will be in the form $2^a$ where $a$ is the number of factors of $2$ in $(2009 \cdot 2)! = 4018!$ . This is simply $\sum_{n=1}^{11} \left \lfloor{\frac{4018}{2^n}}\right \rfloor = 4010$ . Therefore, our answer is $\boxed{401}$
| 401
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6,205
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https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7
| 4
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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 common denominator.
Taking the p-adic valuations of each term, the powers of 2 in the denominator for $i$ is $(2i) - v_2(\binom{2i}{i}).$ We can use Kummers theorem to see that $v_2(\binom{2i}{i})$ is the number of digits carried over when $i$ is added to $i$ in base $2$ . This is simply the number of $1$ 's in the binary representation of $i$
Looking at the binary representations of some of the larger $i$ we see $2009 = 11111011001_2$ having eight $1$ 's. So the power of two is $2 \cdot 2009 - 8 = 4010$ . Experimenting with $2008, 2007, 2006$ we see that the power of two are all $< 4010$ , and under $2005$ the power of two $2i - v_2(\binom{2i}{i}) < 2i < 4010$ . Therefore $a = 4010$ and $\frac{ab}{10} = \boxed{401}.$
| 401
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6,206
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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 one of the number of times Linda rolls her die. Find $m+n$
|
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 combined probability is $\frac{11}{216}$
Let's do the same with the probability that Linda rolls twice and getting a six. This time it is $\frac{5}{6} * \frac{1}{6}$ . The probability that Dave meets the requirements set is $\frac{1}{6} + \frac{5}{6} * \frac{1}{6} + \frac{5}{6} * \frac{5}{6} * \frac{1}{6}$ . Combine the probabilities again to get $\frac{455}{7776}$ . (or not, because you can simplify without calculating later)
It's clear that as the number of rolls before getting a six increases, the probability that Dave meets the requirements is multiplied by $\frac{5}{6} * \frac{5}{6}$ . We can use this pattern to solve for the sum of an infinite geometric series.
First, set the case where Linda rolls only once aside. It doesn't fit the same pattern as the rest, so we'll add it separately at the end. Next, let $a = (\frac{5}{6} * \frac{1}{6}) * (\frac{1}{6} + \frac{5}{6} * \frac{1}{6} + \frac{5}{6} * \frac{5}{6} * \frac{1}{6}) = \frac{455}{7776}$ as written above. Each probability where the number of tosses Linda makes increases by one will be $a * (\frac{25}{36})^{n+1}$ . Let $S$ be the sum of all these probabilities.
$S = a + a * \frac{25}{36} + a * (\frac{25}{36})^2...$
$S * \frac{25}{36} = a * \frac{25}{36} + a * (\frac{25}{36})^2 + a * (\frac{25}{36})^3...$
Subtract the second equation from the first to get
$S * \frac{11}{36} = a$
$S = a * \frac{36}{11}$
$S = \frac{455}{2376}$
Don't forget to add the first case where Linda rolls once.
$\frac{455}{2376} + \frac{11}{216} = \frac{8}{33}$
$8 + 33 = \boxed{41}$
| 41
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6,207
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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 follows that the $50$ solutions for $k$ must be the integers $mn-1, mn-2, \dots, mn-50$ .
This will happen if and only if the lower bound on $k$ is in a suitable range -- we must have $mn-51 \leq \frac mn < mn-50$
Obviously there is no solution for $n=1$ . For $n>1$ the left inequality can be rewritten as $m\leq\dfrac{51n}{n^2-1}$ , and the right one as $m > \dfrac{50n}{n^2-1}$
Remember that we must have $m\geq n$ . However, for $n\geq 8$ we have $\dfrac{51n}{n^2-1} < n$ , and hence $m<n$ , which is a contradiction.
This only leaves us with the cases $n\in\{2,3,4,5,6,7\}$
Therefore the answer is $34\cdot 2 + 19\cdot 3 = 68 + 57 = \boxed{125}$
| 125
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6,208
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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 $2009 + (2009-1) + \cdots + (2009-(k-1)) = \frac{k(4019-k)}{2}$
Hence we get a necessary condition on $k$ : For a solution to exist, we must have $\frac{k(4019-k)}{2} \geq k(2k+1)$ . As $k$ is positive, this simplifies to $\frac{4019-k}{2} \geq 2k+1$ , whence $5k\leq 4017$ , and as $k$ is an integer, we have $k\leq \lfloor 4017/5\rfloor = 803$
If we now find a solution with $k=803$ , we can be sure that it is optimal.
From the proof it is clear that we don't have much "maneuvering space", if we want to construct a solution with $k=803$ .
We can try to use the $2k$ smallest numbers: $1$ to $2\cdot 803 = 1606$ .
When using these numbers, the average sum will be $1607$ . Hence we can try looking for a nice systematic solution that achieves all sums between $1607-401=1206$ and $1607+401=2008$ , inclusive.
Such a solution indeed does exist, here is one:
Partition the numbers $1$ to $1606$ into four sequences:
Sequences $A$ and $B$ have $402$ elements each, and the sums of their corresponding elements are $1206,1207,1208,1209,\dots,1606,1607$ .
Sequences $C$ and $D$ have $401$ elements each, and the sums of their corresponding elements are $1608,1609,1610,1611,\dots,2007,2008$
Thus we have shown that there is a solution for $k=\boxed{803}$ and that for larger $k$ no solution exists.
| 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 remainder when $n$ is divided by $1000$
|
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 $\omega=\text{cis}\frac{360^{\circ}}{14}$ ; then $C_1,\ldots, C_6$ correspond to $\omega,\ldots, \omega^6$ . Let $C_1',\ldots, C_6'$ be their reflections across the diameter. These points correspond to $\omega^8\ldots, \omega^{13}$ . Then the lengths of the segments are $|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|$ . Noting that $B$ represents 1 in the complex plane, the desired product is \begin{align*} BC_1\cdots BC_6 \cdot AC_1\cdots AC_6&= BC_1\cdots BC_6 \cdot BC_1'\cdots BC_6'\\ &= |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| \end{align*}
for $x=1$ .
However, the polynomial $(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})$ has as its zeros all 14th roots of unity except for $-1$ and $1$ . Hence \[(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})=\frac{x^{14}-1}{(x-1)(x+1)}=x^{12}+x^{10}+\cdots +x^2+1.\] Thus the product is $|x^{12}+\cdots +x^2+1|=7$ when the radius is 1, and the product is $2^{12}\cdot 7=28672$ . Thus the answer is $\boxed{672}$
| 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 remainder when $n$ is divided by $1000$
|
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 $n$ $(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {2\pi}{7})\dots(1 - \cos \frac{6\pi}{7})$ . It can be rearranged to form
$n$ $(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {6\pi}{7})\dots(1 - \cos \frac {3\pi}{7})(1 - \cos \frac {4\pi}{7})$
Since $\cos a = - \cos (\pi - a)$ , we have
$n$ $(8^6)(1 - \cos \frac {\pi}{7})(1 + \cos \frac {\pi}{7}) \dots (1 - \cos \frac {3\pi}{7})(1 + \cos \frac {3\pi}{7})$
$(8^6)(1 - \cos^2 \frac {\pi}{7})(1 - \cos^2 \frac {2\pi}{7})(1 - \cos^2 \frac {3\pi}{7})$
$(8^6)(\sin^2 \frac {\pi}{7})(\sin^2 \frac {2\pi}{7})(\sin^2 \frac {3\pi}{7})$
It can be shown that $\sin \frac {\pi}{7} \sin \frac {2\pi}{7} \sin \frac {3\pi}{7}$ $\frac {\sqrt {7}}{8}$ , so $n$ $8^6(\frac {\sqrt {7}}{8})^2$ $7(8^4)$ $28672$ , so the answer is $\boxed{672}$
| 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{25} = \frac{117}{125} \\ b_4 & = \frac 45\cdot \frac {117}{125} + \frac 35 \sqrt{1 - \left(\frac {117}{125}\right)^2} = \frac{468}{625} + \frac 35\cdot\frac {44}{125} = \frac{600}{625} = \frac{24}{25} \end{align*}
We now discovered that $b_4=b_2$ . And as each $b_{i+1}$ is uniquely determined by $b_i$ , the sequence becomes periodic. In other words, we have $b_3=b_5=b_7=\cdots=\frac{117}{125}$ , and $b_2=b_4=\cdots=b_{10}=\cdots=\frac{24}{25}$
Therefore the answer is
\begin{align*} \lfloor a_{10} \rfloor & = \left\lfloor 2^{10} b_{10} \right\rfloor = \left\lfloor \dfrac{1024\cdot 24}{25} \right\rfloor = \left\lfloor \dfrac{1025\cdot 24}{25} - \dfrac{24}{25} \right\rfloor \\ & = \left\lfloor 41\cdot 24 - \dfrac{24}{25} \right\rfloor = 41\cdot 24 - 1 = \boxed{983}
| 983
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6,212
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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 and the cosine. Both observations suggest that the formula may have a neat geometric interpretation.
Consider the equation: \[y = \frac45 x + \frac 35\sqrt{1 - x^2}\]
Note that for $t=\sin^{-1} \frac 35$ we have $\sin t=\frac 35$ and $\cos t = \frac 45$ . Now suppose that we have $x=\sin s$ for some $s$ . Then our equation becomes:
\[y=\cos t \cdot \sin s + \sin t \cdot |\cos s|\]
Depending on the sign of $\cos s$ , this is either the angle addition, or the angle subtraction formula for sine. In other words, if $\cos s \geq 0$ , then $y=\sin(s+t)$ , otherwise $y=\sin(s-t)$
We have $b_0=0=\sin 0$ . Therefore $b_1 = \sin(0+t) = \sin t$ $b_2 = \sin(t+t) = \sin (2t)$ , and so on. (Remember that $t$ is the constant defined as $t=\sin^{-1} \frac 35$ .)
This process stops at the first $b_k = \sin (kt)$ , where $kt$ exceeds $\frac{\pi}2$ . Then we'll have $b_{k+1} = \sin(kt - t) = \sin ((k-1)t) = b_{k-1}$ and the sequence will start to oscillate.
Note that $\sin \frac{\pi}6 = \frac 12 < \frac 35$ , and $\sin \frac{\pi}4 = \frac{\sqrt 2}2 > \frac 35$ , hence $t$ is strictly between $\frac{\pi}6$ and $\frac{\pi}4$ . Then $2t\in\left(\frac{\pi}3,\frac{\pi}2 \right)$ , and $3t\in\left( \frac{\pi}2, \frac{3\pi}4 \right)$ . Therefore surely $2t < \frac{\pi}2 < 3t$
Hence the process stops with $b_3 = \sin (3t)$ , we then have $b_4 = \sin (2t) = b_2$ . As in the previous solution, we conclude that $b_{10}=b_2$ , and that the answer is $\lfloor a_{10} \rfloor = \left\lfloor 2^{10} b_{10} \right\rfloor = \boxed{983}$
| 983
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6,213
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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
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6,214
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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:
$50g=29t+580$
Now we can sub in $30t=50g$ by multiplying the first equation by $10$ . We can plug this into our second equation.
$30t=29t+580$
$t=580$
We know that there were originally $580$ people. Of those, $\frac{2}{5}*580=232$ like to dance.
We also know that with these people, $20$ boys joined, all of whom like to dance. We just simply need to add $20$ to get $232+20=\boxed{252}$
| 252
|
6,215
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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
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3
| 1
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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 $3$ hours, and biking for $4$ hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.
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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 $b$ , but $(8,5)$ gives $b = 15$ . Thus, our answer is $15^{2} + 8^{2} + 5^{2} = \boxed{314}$
| 314
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6,217
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3
| 2
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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 $3$ hours, and biking for $4$ hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.
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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 the first given one gives us that $10b - j = 142\implies b > 14\implies b\ge 15$ . This gives us that $b = 15$ , and then $j = 8$ and $s = 5$ . Therefore, $b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}$
| 314
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6,218
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3
| 3
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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 $3$ hours, and biking for $4$ hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.
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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 cases quickly:
When readdressing the first equation, we see that if $2x+3y$ will be a multiple of $6$ $4z \equiv 2 \pmod{6} = 5$ , we get that $x=15$ and $y=8$ , which works because of integer values. Therefore, $225+64+25=\boxed{314}$
| 314
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6,219
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3
| 4
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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 $3$ hours, and biking for $4$ hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.
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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 it into the equation to get $4j+25=57\implies j=8$ . This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into $2b+3j+4s=74$ , we get $b=15$ . 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get $5^2+8^2+15^2=\boxed{314}$
| 314
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6,220
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_7
| 1
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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?
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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 $100$ . Also, since $316$ is the largest $x$ such that $x^2 < 100000$ $100000$ is the upper bound which all numbers in $S_{999}$ must be less than), there are $316 - 50 = 266$ other sets after $S_{25}$ that have a perfect square.
There are $1000 - 266 - 26 = \boxed{708}$ sets without a perfect square.
| 708
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6,221
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_8
| 1
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Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]
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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 $\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}$
Now, $\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}$
We now have $\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}$ . Thus, $\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1$ ; and simplifying, $23n + 24 = 24n - 23 \Longrightarrow n = \boxed{47}$
| 47
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6,222
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_8
| 2
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Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]
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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 this complex number is $\frac{\pi}{4}$ , its real and imaginary parts must be equal; then, we can we set them equal to get \[48n - 46 = 48 + 46n.\] Therefore, $n=\boxed{47}$
| 47
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6,223
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9
| 1
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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 probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers. Find $m$
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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, obviously. This gives the three solutions $(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$ . In terms of choosing which goes where, the first two solutions are analogous.
For $(5,2,3),(3,5,2)$ , we see that there are $2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7$ ways to stack the crates. For $(1,8,1)$ , there are $\dfrac{10!}{8!1!1!} = 90$ . Also, there are $3^{10}$ total ways to stack the crates to any height.
Thus, our probability is $\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}$ . Our answer is the numerator, $\boxed{190}$
| 190
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6,224
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9
| 2
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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 probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers. Find $m$
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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 mod $3$ , so if $x$ is a solution, so is $x+3$ . By inspection, we have that $2$ is solution, and so are $5$ and $8$ . Each solution splits into its own case.We must solve the equation $41-4z=6x+3y$ , simultaneously with $x+y=10-z$ . Note that we already know the possible values of $z$ . Solving these (it's AIME $9$ , you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution sets $\{8,1,1\},\{5,2,3\},$ and $\{2,3,5\}$ . We can count the number of possible arrangements for each solution by taking $\dbinom{10}{z}$ and then multiplying by $\dbinom{10-z}{x}$ (the solution sets, for the sake of consistency, are in the form $z,x,y$ ). Summing the results for all the solutions gives us $5130$ . Finally, to calculate the probability we must determine our denominator. Since we have $3$ ways to arrange each block, our denominator is $3^{10}$ $\frac{5130}{3^{10}}=\frac{190}{3^7}$ . The answer is $m=\boxed{190}$
| 190
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6,225
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9
| 3
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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 probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers. Find $m$
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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 is produced with $10$ threes, so WLOG we can solve for the number of ways to get $11$ with $0\text{s}, 1\text{s}$ , and $3\text{s}$
Case I: $0$ zeroes, $0$ threes, $11$ ones
Impossible, because there are only ten available spots.
Case II: $1$ zero, $1$ three, $8$ ones
This is just $\frac{10!}{8!}$ , so there are $90$ possibilities.
Case III: $3$ zeroes, $2$ threes, $5$ ones
This is just $\frac{10!}{3!2!5!}$ . This gives $2520$ possibilities.
Case IV: 5 zeroes, 3 threes, and 2 ones.
This is the same as case $3$ , so also $2520$ possibilities.
$90+2520+2520=5130$
$5130$ has three powers of $3$ , so $5130$ divided by $27$ is $\boxed{190}$
| 190
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9
| 4
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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 probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers. Find $m$
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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 want the coefficient of $x^{11}$ in $\left(1+x+x^3\right)^{10}.$ From multinomial theorem, this expansion is \[\sum_{a+b+c=10}\binom{10}{a,b,c}1^ax^bx^{3c}\] If we want the coefficient of $x^{11}$ then we need $b + 3c = 11.$ with $b + c \le 10$ (from the multinomial expansion).
This has the solutions $(b, c) = \{8, 1\}, \{5, 2\}, \{2, 3\}.$ Note that the denominator of the answer is just $3^{10}$ since there are 3 ways to orientate every crate and there are 10 creates. Thus, our answer is \[\frac{\binom{10}{8,1,1} + \binom{10}{5,2,3} + \binom{10}{2,3,5}}{3^{10}} = \frac{190}{3^7} \rightarrow \boxed{190}\]
| 190
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6,227
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_11
| 1
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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 length 14?
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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 string of length $n-1$ ending in an $A$ , or by appending two $B$ s to a string of length $n-2$ ending in a $B$ . Thus, we have the recursions \begin{align*} a_n &= a_{n-2} + b_{n-2}\\ b_n &= a_{n-1} + b_{n-2} \end{align*} By counting, we find that $a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0$ \[\begin{array}{|r||r|r|||r||r|r|} \hline n & a_n & b_n & n & a_n & b_n\\ \hline 1&0&1& 8&6&10\\ 2&1&0& 9&11&11\\ 3&1&2& 10&16&21\\ 4&1&1& 11&22&27\\ 5&3&3& 12&37&43\\ 6&2&4& 13&49&64\\ 7&6&5& 14&80&92\\ \hline \end{array}\] Therefore, the number of such strings of length $14$ is $a_{14} + b_{14} = \boxed{172}$
| 172
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| 2
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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 length 14?
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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 of length $n$ ending in $B$ \begin{align*} a_n &= a_{n-2} + b_{n-2}\\ b_n &= a_{n-1} + b_{n-2} \end{align*} The recursion for $a_n$ tells us that $a_n=a_{n-2}+b_{n-2}$ . However, this is also the definition for $t_{n-2}$ . Therefore, $a_n=t_{n-2}$
We also know from our recursion for $b_n$ that $b_n=a_{n-1}+b_{n-2}$ . Substituting for $a_n$ and $b_n$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+a_{n-1}+b_{n-2}$
Furthermore, note that since $a_n=t_{n-2}$ $a_{n-1}=t_{n-3}$ . Furthermore, using our definition for $t_{n-2}$ , we can rewrite $b_{n-2}$ as $t_{n-2}-a_{n-2}$ . Substituting for $a_{n-1}$ and $b_{n-2}$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+t_{n-3}+t_{n-2}-a_{n-2}$
Finally, note that since $a_n=t_{n-2}$ $a_{n-2}=t_{n-4}$ . Substituting for $a_{n-2}$ into our recursion for $t_n$ gives us $t_n=2t_{n-2}+t_{n-3}-t_{n-4}$ . We now have a recursion only in terms of $t$
By counting, we find that $t_1=1$ $t_2=1$ $t_3=3$ , and $t_4=2$ \[\begin{array}{|r|r||r|r|} \hline n & t_n & n & t_n\\ \hline 1&1&8&16\\ 2&1&9&22\\ 3&3&10&37\\ 4&2&11&49\\ 5&6&12&80\\ 6&6&13&113\\ 7&11&14&172\\ \hline \end{array}\]
Therefore, the number of such sequences of length 14 is $\boxed{172}$
| 172
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| 3
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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 length 14?
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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 amount of letters there can be.
For the first case and the last case, there is only one possible sequence of letters.
For all other cases, we can insert two of the same letter at a time into a run that has the exact same letter. For example, for the second case, we can insert two $A$ s and make the sequence $BAAAAB$ . There are three "slots" in which we can insert two additional letters in, and we must insert five groups of new letters. By stars and bars , the number of ways for the second case is $\binom{7}{2}=21$
Applying this logic to all of the other cases gives us $\binom{7}{3}$ $\binom{7}{3}$ $\binom{7}{4}$ $\binom{8}{6}$ $\binom{8}{1}$ , and $\binom{8}{1}$ . Adding 1+ $\binom{7}{2}$ $\binom{7}{3}$ $\binom{7}{3}$ $\binom{7}{4}$ $\binom{8}{6}$ $\binom{8}{1}$ $\binom{8}{1}$ gives us the answer $\boxed{172}$
| 172
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6,230
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_12
| 1
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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 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let $M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $M$ is divided by $10$
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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 CAR comes first, THEN the empty space. So at time zero, the car is right at the eye.
Hence, we count the number of units that pass the eye in an hour: $\frac {15,000n\frac{\text{meters}}{\text{hour}}}{4(n + 1)\frac{\text{meters}}{\text{unit}}} = \frac {15,000n}{4(n + 1)}\frac{\text{units}}{\text{hour}}$ . We wish to maximize this.
Observe that as $n$ gets larger, the $+ 1$ gets less and less significant, so we take the limit as $n$ approaches infinity
Now, as the speeds are clearly finite, we can never actually reach $3750$ full UNITs. However, we only need to find the number of CARS. We can increase their speed so that the camera stops (one hour goes by) after the car part of the $3750$ th unit has passed, but not all of the space behind it. Hence, $3750$ cars is possible, and the answer is $\boxed{375}$
| 375
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6,231
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_12
| 2
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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 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let $M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $M$ is divided by $10$
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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 of the cars as $15k$ . Then, the distance between them is $\frac{4}{1000} \times k\text{km}$ . Therefore, it takes the car closest to the eye not on the eye $\frac{\frac{k}{250}}{15k}$ hours to get to the eye. There is one hour, so the amount of cars that can pass is $\frac{1}{\frac{\frac{k}{250}}{15k}}$ , or $3750$ cars. When divided by ten, you get the quotient of $\boxed{375}$
| 375
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6,232
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14
| 2
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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 segment $BP$ . Find $m^{2}$
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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\theta]\end{align*}
This is a quadratic equation , maximized when $\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}$ . Thus, $m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}$
| 432
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6,233
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14
| 3
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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 segment $BP$ . Find $m^{2}$
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(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 $\triangle CAP\sim\triangle CBQ$ $BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}$ . Using the Pythagorean Theorem $\bold{again}$ $BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}$ . After a large bashful simplification, $BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}$ . The fraction is equivalent to $729\frac{2x-9}{(x+9)^2}$ . Taking the derivative of the fraction and solving for x, we get that $x=18$ . Plugging $x=18$ back into the expression for $BP$ yields $\sqrt{432}$ , so the answer is $(\sqrt{432})^2=\boxed{432}$
| 432
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6,234
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14
| 4
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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 segment $BP$ . Find $m^{2}$
|
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 $BQ = 2r - h$
We compute $PT = \sqrt{r^2 - (r - h)^2} = \sqrt{h(2r - h)},$ and so $BP^2 = PQ^2 + BQ^2 = 4PT^2 + BQ^2 = 4h(2r - h) + (2r-h)^2 = (2r-h)(2r + 3h)$ .
This is $\frac{1}{3}(6r - 3h)(2r + 3h) \le \frac{1}{3} \cdot \left( \frac{8r}{2} \right)^2$ . Equality is attained, so thus we extract the answer of $\frac{16 \cdot 9^2}{3} = 27 \cdot 16 = \boxed{432}.$
| 432
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6,235
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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 along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $\sqrt[n]{m}$ , where $m$ and $n$ are positive integers, $m<1000$ , and $m$ is not divisible by the $n$ th power of any prime. Find $m+n$
|
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 $MR = NR = \sqrt{34}$ . (Alternatively, we could find this by the Law of Sines .)
The length of the perpendicular from $P$ to $MN$ in $\triangle{MNP}$ is $\frac{\sqrt{17}}{\sqrt{2}}$ , and the length of the perpendicular from $R$ to $MN$ in $\triangle{MNR}$ is $\frac{\sqrt{51}}{\sqrt{2}}$ . Adding those two lengths, $PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}$ . (Alternatively, we could have used that $\sin 75^{\circ} = \sin (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}$ .)
Drop a perpendicular from $R$ to the side of the square containing $M$ and let the intersection be $G$
\begin{align*}PG&=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\ MG=PG-PM&=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}
Let $A'B'C'D'$ be the smaller square base of the tray and let $ABCD$ be the larger square, such that $AA'$ , etc, are edges. Let $F$ be the foot of the perpendicular from $A$ to plane $A'B'C'D'$
We know $AA'=MR=\sqrt{34}$ and $A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}$ . Now, use the Pythagorean Theorem on triangle $AFA'$ to find $AF$
\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&=34\\AF&=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&=\sqrt[4]{867}\end{align*}
The answer is $867 + 4 = \boxed{871}$
| 871
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6,236
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_15
| 2
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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 along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $\sqrt[n]{m}$ , where $m$ and $n$ are positive integers, $m<1000$ , and $m$ is not divisible by the $n$ th power of any prime. Find $m+n$
|
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 $z$ coordinate. Let $\alpha$ be the angle made with the positive $x$ axis . Define $\beta$ and $\gamma$ analogously.
It is easy to see that if $P: = (x,y,z)$ , then $x = AA'\cdot \cos\alpha$ . Furthermore, this means that $\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
We have that $\alpha = \beta = 105^\circ$ , so $\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt [4]{\frac {3}{4}}$
It is easy to see from the Law of Sines that $\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}$
Now, $z = AA'\cdot \cos\gamma = \sqrt [4]{34^2\cdot \frac {3}{4}} = \sqrt [4]{867}$
It follows that the answer is $867 + 4 = \boxed{871}$
| 871
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6,237
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_1
| 1
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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
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6,238
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_1
| 2
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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.
Applying the sum of arithmetic sequence formula, we get
So the answer would be \[\frac{10100}{1000} = \boxed{100}\]
| 100
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6,239
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_2
| 1
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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 $50$ -mile mark at exactly the same time. How many minutes has it taken them?
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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
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6,240
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_2
| 2
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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 $50$ -mile mark at exactly the same time. How many minutes has it taken them?
|
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 reach the 50 mile mark, then by $d=rt$ , the rate times time taken for Rudolph and Jennifer must equal. Hence, we disregard the breaks from the total time taken and get the equation \[r(t-49\cdot 5)=\frac{3}{4}r(t-24\cdot 5),\] yielding $t=\boxed{620}$
| 620
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6,241
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_3
| 2
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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 cm of the remaining block of cheese after ten slices have been cut off?
|
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:
${10, 13, 14} \rightarrow {10, 13, 13} \rightarrow {10, 12, 13} \rightarrow {10, 12, 12} \rightarrow {10, 11, 12} \rightarrow {10, 11, 11} \rightarrow {10, 10, 11} \rightarrow {10, 10, 10} \rightarrow {9, 10, 10} \rightarrow {9, 9, 10} \rightarrow {9, 9, 9}.$
So the greatest possible volume is thus $9 \times 9 \times 9 = \boxed{729}$
| 729
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6,242
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5
| 1
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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 $M$ (or we could apply the homothety to find $ME$ in terms of $NE$ ). It follows that
\[NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500.\]
Thus $MN = NE - ME = \boxed{504}$
| 504
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6,243
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5
| 2
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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 \[\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow x^2 + h^2 = 504^2.\]
By the Pythagorean Theorem on $\triangle MHN$ \[MN^{2} = x^2 + h^2 = 504^2,\] so $MN = \boxed{504}$
| 504
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6,244
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5
| 3
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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 $\left(1004\cos 37^\circ, 1004\sin 37^\circ\right)$
To find the distance from $M$ to $N$ , we simply apply the distance formula and the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to get $MN=\boxed{504}$
| 504
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6,245
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5
| 5
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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 $\tan(53)$ from $\triangle BAF$ yields $\tan(53) = \frac{504+x}{h}.$ Finding it from $\triangle CDG$ yields $\frac{h}{504-x}.$ Setting these equal yields \[\frac{504+x}{h}=\frac{h}{504-x} \rightarrow h^2 = 504^2-x^2 \rightarrow \sqrt{x^2+h^2} = \sqrt{504^2} = \boxed{504}\]
| 504
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6,246
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5
| 6
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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, $MM'=BC'$ .
Let the point where $BC'$ intersects $AD$ be $E$ . Since $BMNE$ is a parallelogram, $AE=AN-BM=1004-500-504$ . Since $BE=EC$ and $\angle BAC= 90^{\circ}$ $AE$ is a median to the hypotenuse of $BAC'$ . Therefore, $BC'=2 AE= 1008$ , and $BE=MN=\boxed{504}.$
| 504
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6,247
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_6
| 1
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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 $\frac{b_n}{b_{n-1}} = (n-1) + \frac{b_{1}}{b_0} = n+2$ . These recursions $a_{n} = na_{n-1}$ and $b_{n} = (n+2)b_{n-1}$ , respectively, correspond to the explicit functions $a_n = n!$ and $b_n = \frac{(n+2)!}{2}$ (after applying our initial conditions). It follows that $\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = \boxed{561}$
| 561
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6,248
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_8
| 1
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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(n+1)a)\\ &= -\sin(0) + \sin(n(n+1)a) = \sin(n(n+1)a) \end{align*}
Thus, we need $\sin \left(\frac{n(n+1)\pi}{2008}\right)$ to be an integer; this can be only $\{-1,0,1\}$ , which occur when $2 \cdot \frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1$ . We know that $n$ cannot be $250$ as $250$ isn't divisible by $4$ , so 1004 doesn't divide $n(n+1) = 250 \cdot 251$ . Therefore, it is clear that $n = \boxed{251}$ is the smallest such integer.
| 251
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6,249
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_8
| 2
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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 $\cos n \theta = \dfrac{1}{2} \left( z^n + \dfrac{1}{z^n} \right)$ and $\sin n \theta = \dfrac{1}{2i} \left(z^n - \dfrac{1}{z^n} \right)$ . Thus, we have $\cos n^2 a \sin n a = \dfrac{1}{4i} \left( z^{n^2} + \dfrac{1}{z^{n^2}} \right) \left( z^{n} - \dfrac{1}{z^n} \right)$
$= \dfrac{1}{4i} \left( z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \dfrac{1}{z^{n^2 - n}} \right)$
$= \dfrac{1}{2} \left( \dfrac{1}{2i} \left(z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} \right) - \dfrac{1}{2i} \left(z^{n^2 - n} - \dfrac{1}{z^{n^2 - n}} \right) \right)$
$= \dfrac{1}{2} \left( \sin ((n^2 + n)a) - \sin ((n^2 - n)a) \right)$
$= \dfrac{1}{2} \left( \sin(((n+1)^2 - (n+1))a) - \sin((n^2 - n)a) \right)$
which clearly telescopes! Since the $2$ outside the brackets cancels with the $\dfrac{1}{2}$ inside, we see that the sum up to $n$ terms is
$\sin ((2^2 - 2)a) - \sin ((1^2 - 1)a) + \sin ((3^3 - 3)a) - \sin ((2^2 - 2)a) \cdots + \sin (((n+1)^2 - (n+1))a) - \sin ((n^2 - n)a)$
$= \sin (((n+1)^2 - (n+1))a) - \sin(0)$
$= \sin ((n^2 + n)a) - 0$
$= \sin \left( \dfrac{n(n+1) \pi}{2008} \right)$
This expression takes on an integer value iff $\dfrac{2n(n+1)}{2008} = \dfrac{n(n+1)}{1004}$ is an integer; that is, $1004 \mid n(n+1)$ . Clearly, $1004 = 2^2 \cdot 251$ , implying that $251 \mid n(n+1)$ . Since we want the smallest possible value of $n$ , we see that we must have ${n,n+1} = 251$ . If $n+1 = 251 \rightarrow n=250$ , then we have $n(n+1) = 250(251)$ , which is clearly not divisible by $1004$ . However, if $n = 251$ , then $1004 \mid n(n+1)$ , so our answer is $\boxed{251}$
| 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 maximum possible number of points in a growing path, and let $r$ be the number of growing paths consisting of exactly $m$ points. Find $mr$
|
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) \[\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\ \sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2},\ \sqrt{1}\] As these define $9$ lengths, so the maximum value of $m$ is $10$ . For now, we assume that $m = 10$ is achievable. Because it is difficult to immediately impose restrictions on a path with increasing distances, we consider the paths in shrinking fashion. Note that the shrinking paths and growing paths are equivalent, but there are restrictions upon the locations of the first edges of the former.
The $\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$
The $\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$
From $(1,0)$ , there are two possible ways to move $\sqrt{10}$ away, either to $(0,3)$ or $(2,3)$ . However, from $(0,3)$ , there is no way to move $\sqrt{9}$ away, so we discard it as a possibility.
From $(2,3)$ , the lengths of $\sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2}$ fortunately are all determined, with the endpoint sequence being $(2,3)-(2,0)-(0,2)-(2,1)-(0,1)-(1,2)$
From $(1,2)$ , there are $3$ possible lengths of $\sqrt{1}$ (to either $(1,1),(2,2),(1,3)$ ). Thus, the number of paths is $r = 4 \cdot 2 \cdot 3 = 24$ , and the answer is $mr = 10 \cdot 24 = \boxed{240}$
| 240
|
6,251
|
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_11
| 1
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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 expressed in the form $m - n\sqrt {k}$ , where $m$ $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$
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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 $\overline{BC}$ , as $\triangle ABC$ is isosceles. Letting the point of intersection be $X$ , we get that $BX = 28$ and $IX = 21$ , and $B, O_2, I$ are collinear as $O_2$ is equidistant from $\overline{AB}$ and $\overline{BC}$ . By Pythagoras, $BI = 35$ , and we notice that $\triangle BIX$ is a 3-4-5 right triangle.
Letting $r$ be the desired radius and letting $Y$ be the projection of $O_2$ onto $\overline{BC}$ , we find that $BY = \frac{4r}{3}$ . Similarly, we find that the distance between the projection from $O_1$ onto $\overline{BC}$ $W$ , and $C$ , is $\frac{64}{3}$ . From there, we let the projection of $O_2$ onto $\overline{O_1W}$ be $Z$ , and we have $O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}$ $O_1Z = 16 - r$ , and $O_1O_2 = 16 + r$ . We finish with Pythagoras on $\triangle O_1O_2Z$ , whence we get the desired answer of $\boxed{254}$ . - Spacesam
| 254
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6,252
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_11
| 2
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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 expressed in the form $m - n\sqrt {k}$ , where $m$ $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$
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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 that $TB = TC = 28.$ From similar triangles, we can now find $\overline{BY}.$ We have \[\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r\] Now, note that as in solution 1, drawing the perpendicular from Q to $\overline{PX}$ (call it Z) yields $\overline{PZ} = 16 - r, \overline{ZX} = r.$ Then, from this, \[\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}\] Using similar similarity as was done to find $\overline{BY}$ we have $\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{3}$ .
Now adding all these up and equating them to $\overline{BC}$ yields \[\frac{4}{3}r + 8\sqrt{r}+ \frac{16}{3} = 56 \rightarrow r = 44 - 6\sqrt{35} \rightarrow 44 + 6\cdot 35 = \boxed{254}\]
| 254
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6,253
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15
| 1
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Find the largest integer $n$ satisfying the following conditions:
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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 three times a square. We cannot have $2n - 1$ be three times a square, for then $2n + 1$ would be a square congruent to $2$ modulo $3$ , which is impossible.
Thus $2n - 1$ is a square, say $b^2$ . But $2n + 79$ is also a square, say $a^2$ . Then $(a + b)(a - b) = a^2 - b^2 = 80$ . Since $a + b$ and $a - b$ have the same parity and their product is even, they are both even. To maximize $n$ , it suffices to maximize $2b = (a + b) - (a - b)$ and check that this yields an integral value for $m$ . This occurs when $a + b = 40$ and $a - b = 2$ , that is, when $a = 21$ and $b = 19$ . This yields $n = 181$ and $m = 104$ , so the answer is $\boxed{181}$
| 181
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6,254
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15
| 2
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Find the largest integer $n$ satisfying the following conditions:
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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 $(2 + \sqrt {3})^k = a_k + \sqrt {3}b_k,$ $k = 0,1,2,3,...$ . However, $a$ is even and $b$ is odd. It is easy to see that the parity of $a$ and $b$ switch each time (by induction). Hence all solutions to the first condition are in the form: \[(2 + \sqrt {3})^{2k + 1} = a_k + \sqrt {3}b_k\] where $k = 0,1,2,..$ . So we can (with very little effort) obtain the following: $(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)$ . It is an AIME problem so it is implicit that $n < 1000$ , so $2n < 2000$ . It is easy to see that $a_n$ is strictly increasing by induction. Checking $2n = 362\implies n =\boxed{181}$ in the second condition works (we know $b_k$ is odd so we don't need to find $m$ ). So we're done.
| 181
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6,255
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https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15
| 3
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Find the largest integer $n$ satisfying the following conditions:
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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 cube. We then get:
$N^2 = (A + 1)^3 - A^3$
$N^2 - 1 = 3A^2 + 3A$
$(N + 1)(N - 1) = 3A(A + 1)$
Now we know either $N + 1$ or $N - 1$ must be factor of $3$ , hence $N = 1 \pmod 3$ or $N = 2 \pmod 3$ . Only $1, 73, 181, 721$ satisfy this criterion. Testing each of the numbers in the condition yields $181$ as the largest that fits both, thus answer $= \boxed{181}$
| 181
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6,256
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_2
| 1
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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 walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.
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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 some point when Al is halfway between Cy and Bob. At this time $s$ , we have that
$\frac{8(s-4)+10(s-2)}{2}=6s$ After solving, $s=\frac{26}{3}$ . At this time, Al has traveled $6\cdot\frac{26}{3}=52$ feet.
We could easily check that Al is in the middle by trying all three possible cases. $\frac{6s + 8(s-4)}{2} = 10(s-2)$ yields that $s = \frac 43$ , which can be disregarded since both Bob and Cy hadn't started yet. $\frac{6s + 10(s-2)}{2} = 8(s-4)$ yields that $-10=-32$ , a contradiction. Thus, the answer is $\boxed{52}$
| 52
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6,257
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_4
| 1
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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?
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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 collinear, $a(t)$ $b(t)$ , and $c(t)$ must differ by a multiple of $\pi$ . Note that $a(t) - b(t) = \frac{t \pi}{105}$ and $b(t) - c(t) = \frac{t \pi}{105}$ , so $a(t) - c(t) = \frac{ 2 t \pi}{105}$ . These are simultaneously multiples of $\pi$ exactly when $t$ is a multiple of $105$ , so the planets and the star will next be collinear in $\boxed{105}$ years.
| 105
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6,258
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5
| 1
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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 Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature?
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Examine $F - 32$ modulo 9.
Generalizing this, we define that $9x + k = F - 32$ . Thus, $F = \left[\frac{9}{5}\left[\frac{5}{9}(9x + k)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + \left[\frac{5}{9}k\right]) + 32\right] \Longrightarrow F = \left[\frac{9}{5} \left[\frac{5}{9}k \right] \right] + 9x + 32$ . We need to find all values $0 \le k \le 8$ that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$ . Testing every value of $k$ shows that $k = 0, 2, 4, 5, 7$ , so $5$ of every $9$ values of $k$ work.
There are $\lfloor \frac{1000 - 32}{9} \rfloor = 107$ cycles of $9$ , giving $5 \cdot 107 = 535$ numbers that work. Of the remaining $6$ numbers from $995$ onwards, $995,\ 997,\ 999,\ 1000$ work, giving us $535 + 4 = \boxed{539}$ as the solution.
| 539
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6,259
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5
| 3
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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 Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature?
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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
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6,260
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6
| 1
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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 a sequence of coordinates that correspond to valid moves, beginning with 0 and ending with 39. For example, $0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39$ is a move sequence. How many move sequences are possible for the frog?
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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 each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$ , we can reach it from $13, 15, 18, 21, 24$ , so we add all those values to get the value for $26$ . For $27$ , it is only reachable from $24$ or $26$ , so we have $29 + 6 = 35$
The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = \boxed{169}$
| 169
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6,261
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6
| 2
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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 a sequence of coordinates that correspond to valid moves, beginning with 0 and ending with 39. For example, $0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39$ is a move sequence. How many move sequences are possible for the frog?
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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 $n$ that is a multiple of $3$ we can either go to the next multiple of $3$ or we can skip to the next multiple of $13$ which is simply $13.$
From these equations we have deduced $f(0) = f(15) + 5f(12).$ Continuing we have \[f(15) + 5f(12) = f(15) + 5(f(26) +f(15) = 5f(26) + 6f(15) = 5f(26)+ 6f(26) + 6f(18) = 5f(26) + 12f(26) + 6f(21) = 5f(26) + 18f(26) + 6f(24) = 5f(26) + 24f(26) + 6f(27) = 29f(26) + 6f(27).\]
Finally, note that $f(26) = 1 + f(27) = 2 + f(30) = 3+f(33) = 4+f(36) = 5$ since at any point we can either go to the next multiple of $3$ or go to the next multiple of $13$ which happens to be $39.$ Therefore $f(26) = 5.$ Similarly we find $f(27) = 1+f(30) = 2 + f(33) = 3+f(36) = 4$ so the end answer is $5 \cdot 29 + 6 \cdot 4 = \boxed{169}.$
| 169
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6,262
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_7
| 1
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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 $k$ .)
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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 $\log 2$ term to cancel out, $k$ is a power of $2$ . Thus, $N$ is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from $2^0 = 1$ to $2^9 = 512$
The formula for the sum of an arithmetic sequence and the sum of a geometric sequence yields that our answer is $\left[\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9)\right] \mod{1000}$
Simplifying, we get $\left[1000\left(\frac{1000+1}{2}\right) -1023\right] \mod{1000} \equiv [500-23] \mod{1000} \equiv 477 \mod{1000}.$ The answer is $\boxed{477}$
| 477
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6,263
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9
| 1
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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 of leg $CA$ , the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$ , and the circles are externally tangent to each other. The length of the radius either circle can be expressed as $p/q$ , where $p$ and $q$ are relatively prime positive integers . Find $p+q$
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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 between the 2 points equal to $2r$ . All points $r$ units away from $\overline{AB}$ are on the line with slope $-\frac{15}{8}$ , and y-intercept $30+ \frac{17}{8} r$
$O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us:
$O_1 = \left(r,\frac14 r+30\right)$ and $O_2 = \left(\frac35 r+16,r\right)$
By the distance formula and the fact that the circles and tangent, we have: $\left(16-\frac25 r\right)^2 + \left(30-\frac34 r\right)^2 = (2r)^2$
which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$
And by the quadratic equation, the solutions are: $\frac{-23120 \pm 54400}{2622}$ The solution including the " $-$ " is extraneous so we have the radius equal to $\frac{31280}{2622}$
Which simplifies to $\frac{680}{57}$ . The sum of the numerator and the denominator is $\boxed{737}$
| 737
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6,264
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9
| 2
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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 of leg $CA$ , the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$ , and the circles are externally tangent to each other. The length of the radius either circle can be expressed as $p/q$ , where $p$ and $q$ are relatively prime positive integers . Find $p+q$
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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 similarly for BG. Because tangents to a circle are equal, the hypotenuse can be expressed in terms of r: \[2r + \frac{47r}{17} - 30 + \frac{33r}{17} - 16 = 34\] Thus, r = $\frac{680}{57}$ , and the answer is $\boxed{737}.$
| 737
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6,265
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10
| 1
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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
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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 the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:
So there are $20(3+54+36) = 1860$ different shadings, and the solution is $\boxed{860}$
| 860
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6,266
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10
| 2
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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
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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 square in the first two columns shaded. That means that in total, the first two rows have $2+5=7$ squares shaded in- that is false since it should only be $6$ .
Thus, there exists another row that is complementary to the first. We remove those two and use a similar argument again to show that every group of $6$ rows can be grouped into $3$ complementary pairs.
Now we proceed with three cases.
Our answer is thus $720+1080+60=1860,$ leaving us with a final answer of $\boxed{860}.$
| 860
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6,267
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10
| 3
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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
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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 a "1-row") within the first 3 columns and which rows have 2 (we'll call these "2-rows") within the first 3 columns. Next, we do some casework:
In total, we have ${6\choose3}(1+18+54+20)=20*93=1860$ . Thus our answer is $\boxed{860}$
| 860
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6,268
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10
| 4
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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
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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 write as $P(a,b,c,d)=(ab+cd)+(a+b)(c+d)$ . Therefore, $P(a,b,c,d)^6$ represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of $a^3b^3c^3d^3$ in $P(a,b,c,d)^6$
By the Binomial Theorem, \[P(a,b,c,d)^6=\sum_{k=0}^6 \binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\tag{1}\] If we expand $(ab+cd)^k$ , then the powers of $a$ and $b$ are always equal. Therefore, to obtain terms of the form $a^3b^3c^3d^3$ , the powers of $a$ and $b$ in $(a+b)^{6-k}$ must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that $k$ must be even. We can use the same logic for $c$ and $d$ . Therefore, the coefficient of $a^3b^3c^3d^3$ in the following expression is the same as the coefficient of $a^3b^3c^3d^3$ in (1). \[\sum_{k=0}^3 \binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\binom{6-2k}{3-k}^2.\tag{2}\] Now we notice that the only way to obtain terms of the form $a^3b^3c^3d^3$ is if we take the central term in the binomial expansion of $(ab+cd)^{2k}$ . Therefore, the terms that contribute to the coefficient of $a^3b^3c^3d^3$ in (2) are \[\sum_{k=0}^3 \binom{6}{2k}\binom{2k}{k}\binom{6-2k}{3-k}^2(abcd)^3.\] This sum is $400+1080+360+20=1860$ so the answer is $\boxed{860}$
| 860
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6,269
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11
| 1
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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.
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$\left(k- \frac 12\right)^2=k^2-k+\frac 14$ 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 their full range. Summing this up with the formula for the sum of the first $n$ squares ( $\frac{n(n+1)(2n+1)}{6}$ ), we get $\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740$ . We need only consider the $740$ because we are working with modulo $1000$
Now consider the range of numbers such that $b(p)=45$ . These numbers are $\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981$ to $2007$ . There are $2007 - 1981 + 1 = 27$ (1 to be inclusive) of them. $27*45=1215$ , and $215+740= \boxed{955}$ , the answer.
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11
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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.
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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 $a+1$ terms in the first set of $p$ , and $a$ terms in the second set. Thus, the sum of $b(p)$ from $a^2 \le p <(a+1)^2$ is $2a(a+1)$ or $4\cdot\binom{a+1}{2}$ . For the time being, consider that $S = \sum_{p=1}^{44^2-1} b(p)$ . Then, the sum of the values of $b(p)$ is $4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)$ . We can collapse this to $4\binom{45}{3}=56760$ . Now, we have to consider $p$ from $44^2 \le p < 2007$ . Considering $p$ from just $44^2 \le p \le 1980$ , we see that all of these values have $b(p)=44$ . Because there are $45$ values of $p$ in that range, the sum of $b(p)$ in that range is $45\cdot44=1980$ . Adding this to $56760$ we get $58740$ or $740$ mod $1000$ . Now, take the range $1980 < p \le 2007$ . There are $27$ values of $p$ in this range, and each has $b(p)=45$ . Thus, that contributes $27*45=1215$ or $215$ to the sum. Finally, adding $740$ and $215$ we get $740+215=\boxed{955}$
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6,271
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12
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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 region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$
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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 $10 \tan 75 = 10 \tan (30 + 45)$ . The tangent addition rule yields $10(2 + \sqrt{3})$ (see above). Thus, $[\triangle ABC] = \frac{1}{2} 20 \cdot (20 + 10\sqrt{3}) = 200 + 100\sqrt{3}$
The area of $[\triangle ADC']$ (with a side on the y-axis) can be found by splitting it into two triangles, $30-60-90$ and $15-75-90$ right triangles $AC' = AC = \frac{10}{\sin 15}$ . The sine subtraction rule shows that $\frac{10}{\sin 15} = \frac{10}{\frac{\sqrt{6} - \sqrt{2}}4} = \frac{40}{\sqrt{6} - \sqrt{2}} = 10(\sqrt{6} + \sqrt{2})$ $AC'$ , in terms of the height of $\triangle ADC'$ , is equal to $h(\sqrt{3} + \tan 75) = h(\sqrt{3} + 2 + \sqrt{3})$ \begin{align*} [ADC'] &= \frac 12 AC' \cdot h \\ &= \frac 12 (10\sqrt{6} + 10\sqrt{2})\left(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}\right) \\ &= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}} \\ &= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50 \end{align*}
The area of $[\triangle EFB']$ was found in the previous solution to be $- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$
Therefore, $[ADEF]$ $= (200 + 100\sqrt{3}) - \left((50 + 50\sqrt{3}) + (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750)\right)$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$ , and our answer is $\boxed{875}$
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12
| 2
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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 region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$
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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 $AEDF$ (which we are trying to solve for) into two congruent triangles and an isosceles right triangle
Since $FE' = AE' = AE$ , we find that $[AE'F] = \frac 12 (5\sqrt{6} + 5\sqrt{2})^2 = 100 + 50\sqrt{3}$
Now, we need to find $[AED] = [AE'D]$ . The acute angles of the triangles are $\frac{15}{2}$ and $90 - \frac{15}{2}$ . By repeated application of the half-angle formula , we can find that $\tan \frac{15}{2} = \sqrt{2} - \sqrt{3} + \sqrt{6} - 2$
The area of $[AED] = \frac 12 \left(20 \cos 15\right)^2 \left(\tan \frac{15}{2}\right)$ . Thus, $[AED] + [AE'D] = 2\left(\frac 12((5\sqrt{6} + 5\sqrt{2})^2 \cdot (\sqrt{2} - \sqrt{3} + \sqrt{6} - 2))\right)$ , which eventually simplifies to $500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$
Adding them together, we find that the solution is $[AEDF] = [AE'F] + [AED] + [AE'D]$ $= 100 + 50\sqrt{3} + 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600=$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$ , and the answer is $\boxed{875}$
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12
| 3
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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 region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$
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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 shoelace, and the answer should come out to be $\boxed{875}$
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14
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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 $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$
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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 \[.\phantom{------------} \frac{a_{n+1}}{a_{n}} = \frac{a_{n}}{a_{n-1}} + \frac{2007}{a_{n}a_{n-1}} \phantom{------------} (1)\] and dividing through by $a_{n}a_{n+1}$ , we get \[.\phantom{------------} \frac{a_{n-1}}{a_{n}} = \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} \phantom{------------} (2)\] Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get \[\frac{a_{n+1}}{a_{n}} + \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} = \frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}} + \frac{2007}{a_{n}a_{n-1}}\] i.e. \[f(n)+ \frac{2007}{a_{n}a_{n+1}} = f(n-1) + \frac{2007}{a_{n}a_{n-1}}\] Summing over $n=1$ to $n=2006$ , we notice that most of the terms on each side cancel against the corresponding term on the other side. We are left with \[f(2006) + \frac{2007}{a_{2006}a_{2007}} = f(0) + \frac{2007}{a_{1}a_{0}}\] We have $f(0) = 2$ , and $2007/a_0a_1 = 2007/9 = 223$ . So \[f(2006) = 2 + 223 - \frac{2007}{a_{2006}a_{2007}} = 224 + \left( 1 - \frac{2007}{a_{2006}a_{2007}}\right)\] Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$ , this means that the sequence $(a_n)$ is increasing. Since $a_3=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$ . It follows that $\left\lfloor f(2006) \right\rfloor = \boxed{224}$
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| 2
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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 $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$
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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 $b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225$ . Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$ , notice that $b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}$ . This means that
$b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225$ . It is only a tiny bit less because all the $b_i$ are greater than $1$ , so we conclude that the floor of $\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}$ is $\boxed{224}$
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14
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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 $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$
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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 $b_n$ defined by $b_1=b_2=3$ and $b_{n+1}=\alpha b_n+\beta b_{n-1}$ for $n\ge 2$ . We wish to find $\alpha$ and $\beta$ such that $a_n=b_n$ for all $n\ge 1$ . To do this, we use the following matrix form of a linear recurrence relation
\[\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
When we take determinants, this equation becomes
\[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
We want \[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=2007\] for all $n$ . Therefore, we replace the two matrices by $2007$ to find that
\[2007=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\cdot 2007\] \[1=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)=-\beta.\] Therefore, $\beta=-1$ . Computing that $a_3=672$ , and using the fact that $b_3=\alpha b_2-b_1$ , we conclude that $\alpha=225$ . Clearly, $a_1=b_1$ $a_2=b_2$ , and $a_3=b_3$ . We claim that $a_n=b_n$ for all $n\ge 1$ . We proceed by induction . If $a_k=b_k$ for all $k\le n$ , then clearly, \[b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.\] We also know by the definition of $b_{n+1}$ that
\[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
We know that the RHS is $2007$ by previous work. Therefore, $b_{n+1}b_{n-1}-b_n^2=2007$ . After substuting in the values we know, this becomes $b_{n+1}a_{n-1}-a_n^2=2007$ . Thinking of this as a linear equation in the variable $b_{n+1}$ , we already know that this has the solution $b_{n+1}=a_{n+1}$ . Therefore, by induction, $a_n=b_n$ for all $n\ge 1$ . We conclude that $a_n$ satisfies the linear recurrence $a_{n+1}=225a_n-a_{n-1}$
It's easy to prove that $a_n$ is a strictly increasing sequence of integers for $n\ge 3$ . Now
\[\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.\]
\[=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.\]
\[=225-\frac{2007}{a_{2005}a_{2006}}.\]
The sequence certainly grows fast enough such that $\frac{2007}{a_{2005}a_{2006}}<1$ . Therefore, the largest integer less than or equal to this value is $\boxed{224}$
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| 4
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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 $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$
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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, $= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$ $= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}$
We can keep on using this method to get that $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}$
This telescopes to $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}$
or $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}$
Finding the first few values, we notice that they increase rapidly, so $\frac{2007}{a_{2006}a_{2007}} < 1$ . Calculating the other values, $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}$
The greatest number that does not exceed this is $\boxed{224}$
| 224
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| 5
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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 $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$
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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 expressions as I did momentarily.
Next, let's see what the expression $\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$ . For $n = 1$ , we get $\frac{1 + 224^2}{224}$ , the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}$ . It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$ , while the second-largest term in the denominator is smaller than $224^2$ . Thus, the floor of this expression will come out to be $224$ as well.
Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$ , the degrees of both the numerator and denominator increase by $2$ , because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2$ ). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\approx224:1$ ratio between the two.
For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$ , they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\geq2$ , whose $\lim_{k\to\infty}=$ $\boxed{224}$
| 224
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_15
| 1
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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 $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime . Find $r$
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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, $\sin 60 = \frac{\sqrt{3}}{2}$ ): $\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5 \cdot y + (x - 2)(x - 5) + 2(x - y)) + 14\sqrt{3}$ . This simplifies to $\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56)$ . Some terms will cancel out, leaving $y = \frac{5}{3}x - 22$
$\angle FEC$ is an exterior angle to $\triangle AEF$ , from which we find that $60 + \angle CED = 60 + \angle AFE$ , so $\angle CED = \angle AFE$ . Similarly, we find that $\angle EDC = \angle AEF$ . Thus, $\triangle AEF \sim \triangle CDE$ . Setting up a ratio of sides, we get that $\frac{5}{x-y} = \frac{y}{2}$ . Using the previous relationship between $x$ and $y$ , we can solve for $x$
Use the quadratic formula , though we only need the root of the discriminant . This is $\sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}$ $= \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}$ . The answer is $\boxed{989}$
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6,280
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_15
| 2
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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 $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime . Find $r$
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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 $\triangle{CED}, \triangle{AEF}$ separately, getting $4+x^2-2x=a^2; 25+(2+m-x)^2-5(2+m-x)=b^2.$ Add those terms together and use the equality $a^2+b^2=m^2-3m+65$ , we can find: $2x^2-(2m+1)x+2m-42=0$
According to basic angle chasing, $\angle{A}=\angle{C}; \angle{AFE}=\angle{CED}$ , so $\triangle{AFE}\sim \triangle{CED}$ , the ratio makes $\frac{5}{x}=\frac{2+m-x}{2}$ , getting that $x^2-(2+m)x+10=0$ Now we have two equations with $m$ , and $x$ values for both equations must be the same, so we can solve for $x$ in two equations. $x=\frac{2m+1 \pm \sqrt{4m^2+4m+1-16m+336}}{4}; x=\frac{4+2m \pm \sqrt{4m^2+16m-144}}{4}$ , then we can just use positive sign to solve, simplifies to $3+\sqrt{4m^2+16m-144}=\sqrt{4m^2-12m+337}$ , getting $m=\frac{211-3\sqrt{989}}{10}$ , since the triangle is equilateral, $AB=BC=2+m=\frac{231-3\sqrt{989}}{10}$ , and the desired answer is $\boxed{989}$
| 989
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6,281
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_1
| 1
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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. A set of plates in which each possible sequence appears exactly once contains N license plates. Find $\frac{N}{10}$
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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
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6,282
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_2
| 1
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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 $a+b+c=100$
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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 $x + y = n$ . Thus, there are $0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}$ solutions of $(a,b,c)$
| 200
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6,283
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_3
| 1
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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(A--E--B); draw(C--F--D); dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S);[/asy]
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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 $EF=EP+FP=2\cdot EP$ , the answer is $(17\sqrt{2})^2=\boxed{578}.$
| 578
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6,284
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_3
| 2
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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(A--E--B); draw(C--F--D); dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S);[/asy]
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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 $\angle EAB = \arccos{\left(\frac{12}{13}\right)}$ or $\arcsin{\left(\frac{5}{13}\right)}.$ Finally, we know that $AO =\frac{13\sqrt{2}}{2}.$ Now we apply laws of cosines on $\bigtriangleup AEO.$
We have $EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.$ We know that $\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.$ Thus we have $\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)$ which applying the cosine sum identity yields $\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} =\frac{12\sqrt{2}}{26} -\frac{5\sqrt{2}}{26} =\frac{7\sqrt{2}}{26}.$
Note that we are looking for $4EO^2$ so we multiply $EO^2 = 12^2 + \left(\frac{13\sqrt{2}}{2}\right)^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}$ by $4$ obtaining $4EO^2 = 576 + 338 - 8 \cdot\left(\frac{13\sqrt{2}}{2}\right) \cdot 12 \cdot\frac{7\sqrt{2}}{26} = 576 + 338 - 4 \cdot 12 \cdot 7 = \boxed{578}.$
| 578
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6,285
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_4
| 1
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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 $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m$
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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 $(x,y) = \left(\frac{1}{7}, \frac{2}{7}\right)$ . Substitute this into the third equation to find that $1050 = 150 + 2m$ , so $m = \boxed{450}$
| 450
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6,286
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_6
| 1
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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?
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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 if you want) so the recursion is $S_n=4S_{n-1}$ where $S_n$ stands for the number of such numbers with $n$ digits. Since $S_1=10$ the answer is $\boxed{640}$
| 640
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6,287
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_8
| 1
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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 rectangles determined. Find the remainder when $N$ is divided by 1000.
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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, $-\frac 54x^2 + 502x - \frac{2003}4$ . Use $\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201$ . However, this gives a non-integral answer for $y$ . The closest two values that work are $(199,253)$ and $(203,248)$
We see that $252 \cdot 198 = 49896 > 202 \cdot 247 = 49894$ . The solution is $\boxed{896}$
| 896
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6,288
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_9
| 1
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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 $PQ.$
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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 are congruent; therefore, $EP = FQ = \frac{287 - PQ}{2}$ . By the Two Tangent theorem, note that $EP = EX = \frac{287 - PQ}{2}$ , making $BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$ . Also, $BX = BY$ $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$
Finally, $FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}$ . Also, $FP = FQ + PQ = \frac{287 - PQ}{2} + PQ$ . Equating, we see that $\frac{805 - PQ}{2} = \frac{287 + PQ}{2}$ , so $PQ = \boxed{259}$
| 259
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6,289
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_10
| 1
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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 positive integers $n$ is prime , and $m$ and $n$ are relatively prime . Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$
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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 order for $B$ to be in neither $A$ nor $S-A$ $B$ must have at least one element from both $A$ and $S-A$ . In other words, $B$ must contain any subset of $A$ and $S-A$ except for the empty set $\{\}$ . This can be done in $\binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1)$ ways. As $|A|$ ranges from $0$ to $6$ , we can calculate the total number of unsuccessful outcomes to be \[\sum_{|A| = 0}^{6} \binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1) = 2702.\] So our desired answer is \[1 - \dfrac{2702}{2^{12}} = \dfrac{697}{2^{11}} \Rightarrow \boxed{710}.\]
| 710
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6,290
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_11
| 1
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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 rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi+b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime . Find $a+b+c.$
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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 triangle in the diagram has a length of $72 + 24 = 96$ . The horizontal line divides the radius of the larger circle into $72 - 24 = 48$ on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a $30-60-90 \triangle$
Find the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree central angle indicates to take $\frac{60}{360} = \frac 16$ of the circumference of the larger circle (twice), while the $180 - 2(30) = 120^{\circ}$ central angle in the smaller circle indicates to take $\frac{120}{360} = \frac 13$ of the circumference. This adds up to $2 \cdot \frac 16 144\pi + \frac 13 48\pi = 64\pi$
The actual horizontal distance it takes can be found by using the $30-60-90 \triangle$ s. The missing leg is equal in length to $48\sqrt{3}$ . Thus, the total horizontal distance covered is $96\sqrt{3}$
Thus, we get $144\pi - 64\pi + 96\sqrt{3} = 80\pi + 96\sqrt{3}$ , and our answer is $\boxed{179}$
| 179
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6,291
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_13
| 1
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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, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$ 's and $1$ 's in the bottom row is the number in the top square a multiple of $3$
[asy] for (int i=0; i<12; ++i){ for (int j=0; j<i; ++j){ //dot((-j+i/2,-i)); draw((-j+i/2,-i)--(-j+i/2+1,-i)--(-j+i/2+1,-i+1)--(-j+i/2,-i+1)--cycle); } } [/asy]
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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 top position going only up is equal to the number of times it will be counted in the final sum.)
Examine the equation $\mod 3$ . All of the coefficients from $x_2 \ldots x_8$ will be multiples of $3$ (since the numerator will have a $9$ ). Thus, the expression boils down to $x_0 + 10x_1 + 10x_9 + x_{10} \equiv 0 \mod 3$ . Reduce to find that $x_0 + x_1 + x_9 + x_{10} \equiv 0 \mod 3$ . Out of $x_0,\ x_1,\ x_9,\ x_{10}$ , either all are equal to $0$ , or three of them are equal to $1$ . This gives ${4\choose0} + {4\choose3} = 1 + 4 = 5$ possible combinations of numbers that work.
The seven terms from $x_2 \ldots x_8$ can assume either $0$ or $1$ , giving us $2^7$ possibilities. The answer is therefore $5 \cdot 2^7 = \boxed{640}$
| 640
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6,292
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14
| 1
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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 $f(5).$
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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 $f(2\lambda^3 + \lambda) = 0$ . Assume that there exists a root $\lambda$ with $|\lambda| \neq 1$ . Then there must be such a root $\lambda_1$ with $|\lambda_1| > 1$ . Then \[|2\lambda^3 + \lambda| \ge 2|\lambda|^3 - |\lambda| > 2|\lambda| - |\lambda| = |\lambda|.\] But then $f(x)$ would have infinitely many roots, given by $\lambda_{k+1} = 2\lambda_k^3 + \lambda_k$ , for $k \ge 1$ .
Therefore $|\lambda| = 1$ for all of the roots of the polynomial. Thus $\lambda\overline{\lambda}=1$ , and $(2\lambda^3 + \lambda)\overline{(2\lambda^3 + \lambda)} = 1$ .
Solving these equations simultaneously for $\lambda = a + bi$ yields $a = 0$ $b^2 = 1$ , and so $\lambda^2 = -1$ . Because the polynomial has real coefficients, the polynomial must have the form $f(x) = (1 + x^2)^n$ for some integer $n \ge 1$ . The condition $f(2) + f(3) = 125$ implies $n = 2$ , giving $f(5) = \boxed{676}$
| 676
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6,293
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14
| 2
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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 $f(5).$
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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 real roots.
Note that $f(x)$ is not constant. We then find two complex roots: $r = \pm i$ . We find that $f(i)f(-2) = f(-i)$ , and that $f(-i)f(-2) = f(i)$ . This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0$ . Thus, $\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial. (Note: This requires the assumption that $f(-2)\neq1$ . Clearly, $f(-2)\neq-1$ , because that would imply the existence of a real root.)
The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$ . Substituting into the given expression, we have
\[(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)\] \[(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)\]
Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$ , or $g(x)$ satisfies the same constraints as $f(x)$ . Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$
Since $f(2)+f(3)=125=5^n+10^n$ for some $n$ , we have $n=2$ ; so $f(5) = \boxed{676}$
| 676
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6,294
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_14
| 3
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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 $f(5).$
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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 \[|2r^3+r|\geq|2r^3|-|r|=|r|(2|r|^2-1)>|r|(2\cdot1^2-1)=|r|.\] It follows that $|2r^3+r|>|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 2: $|r|<1$
This means that $|2r^3+r|\geq2|r^3|-|r|=|r|(|2r^2|-1)<|r|.$ It follows that $|2r^3+r|<|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 3: $|r|=1.$
Since $|r|$ is not greater than or less than 1, $|r|=1.$ This means that all the roots of the polynomial have a magnitude of $1.$ More specifically, $|2r^3+r|$ has a magnitude of one. Since this would mean an equality condition from the triangle inequality, $2r^3$ and $r$ are collinear with the origin in the complex plane. In other words, $\frac{2r^3}{r}=\pm c\Leftrightarrow cr=2r^3\Leftrightarrow 2r^2=c\Leftrightarrow r=\pm\sqrt{\pm\frac{c}{2}},$ for some real constant $c.$ Now, from $|r|=1,$ we find that $\left|\pm\sqrt{\pm\frac{c}{2}}\right|=1\Leftrightarrow \sqrt{\pm\frac{c}{2}}=1\Leftrightarrow \pm\frac{c}{2}=1\Leftrightarrow c=\pm2.$ Putting this back into the equation, we find that $r=1,-1,i,-i.$ Now, this means that $2r^3+r=3,-3,i,-i.$ $3$ and $-3$ obviously doesn't have a magnitude of $1.$ Thus, $i,-i$ are the only possible roots of the polynomial. Since roots come in conjugate pairs, $f(x)=[(x-i)(x+i)]^n=(x^2+1)^n,$ works for all constants $n\neq0.$
Case 2: $\deg f=0.$
This means that $f(x)=c,$ for some constant $c.$ In other words, $c^2=c.$ We can easily find that this means that $c=0,1.$ Combining all the cases, we conclude that $f(x)=(x^2+1)^n,0,1$ are the only polynomials that satisfy this equation.
Now, we can test! $f(x)=0,1$ obviously don't satisfy $f(2)+f(3)=125.$ Thus, $f(x)=(x^2+1)^n.$ Substituting, we find that $5^n+10^n=125\Leftrightarrow n=2.$ We conclude that $f(5)=(5^2+1)^2=26^2=\boxed{676}.$
| 676
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6,295
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15
| 1
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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 $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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[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,Y,Z); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6)); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,down,p); dot("$O$",Zp,dir(-45),p+red); dot("$I$",P,left+up,p); [/asy]
First, apply Heron's formula to find that $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$ . The semiperimeter is $21$ , so the inradius is $\frac{A}{s} = \frac{84}{21} = 4$
Now consider the incenter $I$ of $\triangle ABC$ . Let the radius of one of the small circles be $r$ . Let the centers of the three little circles tangent to the sides of $\triangle ABC$ be $O_A$ $O_B$ , and $O_C$ . Let the center of the circle tangent to those three circles be $O$ . The homothety $\mathcal{H}\left(I, \frac{4-r}{4}\right)$ maps $\triangle ABC$ to $\triangle XYZ$ ; since $OO_A = OO_B = OO_C = 2r$ $O$ is the circumcenter of $\triangle XYZ$ and $\mathcal{H}$ therefore maps the circumcenter of $\triangle ABC$ to $O$ . Thus, $2r = R \cdot \frac{4 - r}{4}$ , where $R$ is the circumradius of $\triangle ABC$ . Substituting $R = \frac{abc}{4[ABC]} = \frac{65}{8}$ $r = \frac{260}{129}$ and the answer is $\boxed{389}$
| 389
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6,296
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15
| 2
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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 $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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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 $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, $v$
Cut and combine the triangles, as shown. Then solve for $4u$
The solution is $260+129=\boxed{389}$
| 389
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6,297
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15
| 3
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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 $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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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 $OA'=OB'=OC'=2x$ $O$ is the circumcenter of triangle $A'B'C'$ and its circumradius is $2x$ . Let $I$ denote the incenter of triangle $ABC$ and $r$ the inradius of $ABC$ . Then the inradius of $A'B'C'=r-x$ , so now we compute r. Computing the inradius by $A=rs$ , we find that the inradius of $ABC$ is $4$ . Additionally, using the circumradius formula $R=\frac{abc}{4K}$ where $K$ is the area of $ABC$ and $R$ is the circumradius, we find $R=\frac{65}{8}$ . Now we can equate the ratio of circumradius to inradius in triangles $ABC$ and $A'B'C'$
\[\frac{\frac{65}{8}}{4}=\frac{2x}{4-x}\]
Solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$
| 389
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6,298
|
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15
| 4
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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 $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
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 to $BC$ . It is easy to know that $tanB=\frac{12}{5}$ , so $tan\angle OBJ=\frac{2}{3}$ according to half angle formula. Similarly, we can find $tan\angle NCK=\frac{1}{2}$ . So we can see that $JK=ON=14-\frac{7x}{2}$ . Obviously, $\frac{2x}{14-\frac{7x}{2}}=\frac{65}{112}$ . After solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$ . ~bluesoul
| 389
|
6,299
|
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_15
| 5
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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 $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
[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); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p); dot("$H$",(7,0),down,p); [/asy]
Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$ , respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$ . Since $AB||O_AO_B$ $BC||O_BO_C$ , and $CA||O_CO_A$ , we know that
\[\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\angle CAI = \angle O_CO_AI, \angle ABI = \angle O_AO_BI, \angle BCI = \angle O_BO_CI.\]
That means $\angle ABC = \angle O_AO_BO_C$ $\angle BAC = \angle O_BO_AO_C$ , and $\angle ACB = \angle O_AO_CO_B$ . Thus, $\triangle ABC \sim \triangle O_AO_BO_C$ . We also know that we are scaling each side of $\triangle ABC$ (from $AB$ to $O_AO_B$ for instance), about $I$ (since A,O_A,I are collinear ; same apply with $B$ and $C$ ).
Now, let the homothety $\mathcal{H} (I, x)$ map $\triangle ABC$ to $\triangle O_AO_BO_C$ . To start off, we know the circumradius of $O_AO_BO_C$ is $O$ , since $OO_A = OO_B = OO_C = 2r$ . Since $O_AO_B = 13x$ $O_BO_C = 14x$ $O_CO_A = 15x$ , we can get an relationship involving $x$ and $r$ via another way to find the circumradius:
\[[\triangle O_AO_BO_C ] =\frac{abc}{4R} \Longrightarrow 84x^2 =\frac{13x\cdot 14x\cdot 15x}{4\cdot 2r} \Longrightarrow r=\frac{65x}{16}\]
Take notice of the inradius of $ABC$ . We get the inradius to be $[\triangle ABC ] = sr_0 \Longrightarrow r_0=4$ . Let the tangency point of the incircle and side $BC$ be $H$ . We know $IH = 4$ . We also know that we can cut off the part of $IH$ that is outside of $\triangle O_AO_BO_C$ to get the inradius of $\triangle O_AO_BO_C$ . To part that is outside $\triangle O_AO_BO_C$ turns out just to be the radius of circle $w_B$ (as seen in the picture). That means the inradius of $\triangle O_AO_BO_C$ is just $4-r$ . We can calculate that incradius in another way, though. We know that the inradius of $\triangle ABC$ is $4$ , which means the inradius of $\triangle O_AO_BO_C$ is just $4x$ (by our homethety ratio).
Thus, we have $4x = 4-r = 4-\dfrac{65x}{16} \Longrightarrow x = \dfrac{64}{129} \Longrightarrow r = \dfrac{260}{129}$ . That gives $\boxed{389}$ as our final answer.
| 389
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6,300
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_2
| 1
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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 $S.$
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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 subset of $\{1,2,3,\ldots,100\}$ , and let $T$ be the sum of the elements of $\mathcal{B}$ . Note that the number of possible $S$ is the number of possible $T=5050-S$ . The smallest possible $T$ is $1+2+ \ldots +10 = 55$ and the largest is $91+92+ \ldots + 100 = 955$ , so the number of possible values of T, and therefore S, is $955-55+1=\boxed{901}$
| 901
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