Datasets:

imta-ai

/

MathCode

like 0

Follow

imta-ai 3

493c7a5b42a716a3f254d62a2ef89c37

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_8

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 $p$ be 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. (We will call this event "a win", and the opposite event will be "a loss".) Let both players roll their first die. With probability $\frac 1{36}$ , both throw a six and we win. With probability $\frac{10}{36}$ exactly one of them throws a six. In this case, we win if the remaining player throws a six in their next throw, which happens with probability $\frac 16$ Finally, with probability $\frac{25}{36}$ none of them throws a six. Now comes the crucial observation: At this moment, we are in exactly the same situation as in the beginning. Hence in this case we will win with probability $p$ We just derived the following linear equation: \[p = \frac 1{36} + \frac{10}{36} \cdot \frac 16 + \frac{25}{36} \cdot p\] Solving for $p$ , we get $p=\frac 8{33}$ , hence the answer is $8+33 = \boxed{041}$

041

493c7a5b42a716a3f254d62a2ef89c37

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_8

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

79dee697857529c4ec89843f8f5133a9

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_9

Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$ , and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$ . Find the remainder when $m-n$ is divided by $1000$

We can avoid computing $m$ and $n$ , instead we will compute $m-n$ directly. Note that $4x+3y+2z=2009$ if and only if $4(x-1)+3(y-1)+2(z-1)=2000$ . Hence there is an almost 1-to-1 correspondence between the positive integer solutions of the two equations. The only exceptions are the solutions of the first equation in which at least one of the variables is equal to $1$ . The value $m-n$ is the number of such solutions. If $x=1$ , we get the equation $3y+2z=2005$ . The variable $y$ must be odd, and it must be between $1$ and $667$ , inclusive. For each such $y$ there is exactly one valid $z$ . Hence in this case there are $334$ valid solutions. If $y=1$ , we get the equation $4x+2z=2006$ , or equivalently $2x+z=1003$ . The variable $x$ must be between $1$ and $501$ , inclusive, and for each such $x$ there is exactly one valid $z$ . Hence in this case there are $501$ valid solutions. If $z=1$ , we get the equation $4x+3y=2007$ . The variable $y$ must be odd, thus let $y=2u-1$ . We get $4x+6u=2010$ , or equivalently, $2x+3u=1005$ . Again, we see that $u$ must be odd, thus let $u=2v-1$ . We get $2x+6v=1008$ , which simplifies to $x+3v=504$ . Now, we see that $v$ must be between $1$ and $167$ , inclusive, and for each such $v$ we have exactly one valid $x$ . Hence in this case there are $167$ valid solutions. Finally, we must note that there are two special solutions: one with $x=y=1$ , and one with $y=z=1$ . We counted each of them twice, hence we have to subtract two from the total. Therefore $m-n = 334 + 501 + 167 - 2 = 1000$ , and the answer is $1000\bmod 1000 = \boxed{000}$

000

ebea218ac08322c2303c900bff60e33a

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_11

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

81b32e6bc1e4d53632963b52b81fb310

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_12

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

254cff1701fd79be281fc85e346f9b30

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_13

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

254cff1701fd79be281fc85e346f9b30

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_13

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

a6d13adee56f16911770791ea3e03d59

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_14

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

a6d13adee56f16911770791ea3e03d59

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_14

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

611d0748ca67290bc832a75850d20ad2

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_15

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$ , where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$

Let $V = \overline{NM} \cap \overline{AC}$ and $W = \overline{NM} \cap \overline{BC}$ . Further more let $\angle NMC = \alpha$ and $\angle MNC = 90^\circ - \alpha$ . Angle chasing reveals $\angle NBC = \angle NAC = \alpha$ and $\angle MBC = \angle MAC = 90^\circ - \alpha$ . Additionally $NB = \frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem. By the Angle Bisector Formula, \[\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)\] \[\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)\] As $NV + MV =MW + NW = 1$ we compute $NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}$ and $MV = \frac{1}{1+\tan (\alpha)}$ , and finally $VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1$ . Taking the derivative of $VW$ with respect to $\alpha$ , we arrive at \[VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}\] Clearly the maximum occurs when $\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)$ . Plugging this back in, using the fact that $\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$ and $\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}$ , we get $VW = 7 - 4\sqrt{3}$ with $7 + 4 + 3 = \boxed{014}$

014

611d0748ca67290bc832a75850d20ad2

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_15

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$ , where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$

Since $MA = \frac{\sqrt{2}}{2} \approx 0.707 > \frac{3}{5}$ , point $B$ lies between $M$ and $A$ on the semicircular arc. We will first compute the length of $\overline{AB}$ . By the law of cosines, $\cos \angle MOB = \frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \frac{7}{25}$ , so $\cos \angle AOB = \sin \angle MOB = \frac{24}{25}$ . Then $AB^2 = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2 \cdot \frac{24}{25} = \frac{1}{50}$ , so $AB = \frac{1}{5\sqrt{2}}$ Let $P = AC \cap MN$ and $Q = BC \cap MN$ , and let $MQ = x$ $PQ = d$ $PN = y$ . Note that \[(M, P; Q, N) \stackrel{C}{=} (M, A; B, N),\] that is, \[\frac{QP}{QM} \div \frac{NP}{NM} = \frac{BA}{BM} \div \frac{NA}{NM}\] or \[\frac{d}{xy} = \frac{1/(5\sqrt{2})}{(3/5) \cdot (\sqrt{2}/2)} = \frac{1}{3}.\] Hence $d = \frac{1}{3}xy$ , and we also know $d+x+y=1$ . Now AM-GM gives \[\frac{x+y}{2} \ge \sqrt{xy} \implies \frac{1-d}{2} \ge \sqrt{3d}.\] This gives the quadratic inequality $d^2 - 14d + 1 \ge 0$ , which solves as \[d \in \left(-\infty, 7-4\sqrt3\right] \cup \left[7+4\sqrt3, \infty\right).\] But $d \le 1$ , so the greatest possible value of $d$ is $7-4\sqrt3$ . The answer is $7+4+3=\boxed{014}$

014

611d0748ca67290bc832a75850d20ad2

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_15

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$ , where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$

Let $O$ be the center of the circle. Define $\angle{MOC}=t$ $\angle{BOA}=2a$ , and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$ , respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \pi]$ Let $C'$ be the foot of the perpendicular from $C$ to $MN$ . We compute $XY$ as follows. (a) By the Extended Law of Sines in triangle $ABC$ , we have \[CA\] \[= \sin\angle{ABC}\] \[= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)\] \[= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)\] \[= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)\] \[= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\] (b) Note that $CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)$ and $AO = \frac{1}{2}$ . Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \sin(t)$ , and hence, \[CY/CA\] \[= \frac{CY}{CY + AY}\] \[= \frac{\sin(t)}{1 + \sin(t)}\] \[= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}\] \[= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}\] (c) We have $\angle{XCY} = \frac{\widehat{AB}}{2}=a$ and $\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}$ , and hence by the Law of Sines, \[XY/CY\] \[= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}\] \[= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}\] \[= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\] (d) Multiplying (a), (b), and (c), we have \[XY\] \[= CA * (CY/CA) * (XY/CY)\] \[= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\] \[= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}\] \[= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}\] which is a function of $t$ (and the constant $a$ ). Differentiating this with respect to $t$ yields \[\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}\] and the numerator of this is \[\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))\] \[= \sin(a) \times (\sin(a) + \cos(a)\cos(t))\] which vanishes when $\sin(a) + \cos(a)\cos(t) = 0$ . Therefore, the length of $XY$ is maximized when $t=t'$ , where $t'$ is the value in $[0, \pi]$ that satisfies $\cos(t') = -\tan(a)$ Note that \[\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}\] so $\tan(a) = \frac{1}{7}$ . We compute \[\sin(a) = \frac{\sqrt{2}}{10}\] \[\cos(a) = \frac{7\sqrt{2}}{10}\] \[\cos(t') = -\tan(a) = -\frac{1}{7}\] \[\sin(t') = \frac{4\sqrt{3}}{7}\] \[\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}\] so the maximum length of $XY$ is $\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}$ , and the answer is $7 + 4 + 3 = \boxed{014}$

014

611d0748ca67290bc832a75850d20ad2

https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_15

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$ , where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$

By Pythagoras in $\triangle BMN,$ we get $BN=\dfrac{4}{5}.$ Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \[\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\dfrac{(a+b)(b+c)}{b}.$ In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.$ Substituting, we get $\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}$ where we use $a+b+c=1.$ Again using $a+b+c=1,$ we have $a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.$ Then $b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \[(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.\] Substituting back yields $b=7-4\sqrt{3},$ so $7+4+3=\boxed{014}$ is the answer.

014

93bfc32f997fba96cc1a33c66e403814

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_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

93bfc32f997fba96cc1a33c66e403814

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_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?

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

93bfc32f997fba96cc1a33c66e403814

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_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?

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

6855fd50a9174e18193af5b607bbed90

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_2

Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$ , and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$

Note that if the altitude of the triangle is at most $10$ , then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$ . This implies that vertex G must be located outside of square $AIME$ Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$ . Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$ . Also, $\triangle GXY \sim \triangle GEM$ Let the height of $GXY$ be $h$ . By the similarity, $\dfrac{h}{6} = \dfrac{h + 10}{10}$ , we get $h = 15$ . Thus, the height of $GEM$ is $h + 10 = \boxed{025}$

025

c5db1fa834328147d298318500c6be44

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ 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.

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

c5db1fa834328147d298318500c6be44

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ 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.

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

c5db1fa834328147d298318500c6be44

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ 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.

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

c5db1fa834328147d298318500c6be44

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ 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.

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

e9e0969492e3958c43b926aeff979bb1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Completing the square $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$ . Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by difference of squares Since $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since $244 = 2^2 \cdot 61$ , the factors must be $2$ and $122$ . Since $x,y > 0$ , we have $y - x - 42 = 2$ and $y + x + 42 = 122$ ; the latter equation implies that $x + y = \boxed{080}$

080

e9e0969492e3958c43b926aeff979bb1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

We complete the square like in the first solution: $y^2 = (x+42)^2 + 244$ . Since consecutive squares differ by the consecutive odd numbers, we note that $y$ and $x+42$ must differ by an even number. We can use casework with the even numbers, starting with $y-(x+42)=2$ \begin{align*}2(x+42)+1+2(x+42)+3&=244\\ \Leftrightarrow x&=18\end{align*} Thus, $y=62$ and the answer is $\boxed{080}$

080

e9e0969492e3958c43b926aeff979bb1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

We see that $y^2 \equiv x^2 + 4 \pmod{6}$ . By quadratic residues , we find that either $x \equiv 0, 3 \pmod{6}$ . Also, $y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}$ , so $x \equiv 0, 2 \mod{4}$ . Combining, we see that $x \equiv 0 \mod{6}$ Testing $x = 6$ and other multiples of $6$ , we quickly find that $x = 18, y = 62$ is the solution. $18+62=\boxed{080}$

080

e9e0969492e3958c43b926aeff979bb1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

We solve for x: $x^2 + 84x + 2008-y^2 = 0$ $x=\dfrac{-84+\sqrt{84^2-4\cdot 2008+4y^2}}{2}=-42+\sqrt{y^2-244}$ So $y^2-244$ is a perfect square. Since 244 is even, the difference $\sqrt{y^2-244} -y^2$ is even, so we try $y^2-244=(y-2)^2$ $-244=-4y+4$ $y=62$ Plugging into our equation, we find that $x=18$ , and $(x,y)=(18,62)$ indeed satisfies the original equation. $x+y=\boxed{080}$

080

e9e0969492e3958c43b926aeff979bb1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Let $y=x+d$ for some $d>0$ , substitute into the original equation to get $84x + 2008 = 2xd + d^2$ All terms except for the last one are even, hence $d^2$ must be even, hence let $d=2e$ . We obtain $21x + 502 = xe + e^2$ . Rearrange to $502-e^2 = x(e-21)$ Obviously for $0<e<21$ the right hand side is negative and the left hand side is positive. Hence $e\geq 21$ . Let $e=21+f$ , then $f\geq 0$ We have $502 - (21+f)^2 = xf$ . Left hand side simplifies to $61 - 42f + f^2$ . As $x$ must be an integer, $f$ must divide the left hand side. But $61$ is a prime, which only leaves two options: $f=1$ and $f=61$ Option $f=61$ gives us a negative $x$ . Option $f=1$ gives us $x=61/f - 42 + f = 18$ , and $y = x + d= x + 2e = x + 2(21+f) = 18 + 44 = 62$ , hence $x+y=\boxed{080}$

080

e9e0969492e3958c43b926aeff979bb1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

First complete the square to get $y^2 = (x+42)^2 + 244$ . Remember that squares are the sums of consecutive odd integers, so when the difference between the two squares is 244, the two squares must be an even number of odd integers apart. However, there is only one distinct solution, as the problem states, and very quickly you will realize that only two odd integers work. When there are four, then the numbers are not odd, and when it is any other even integer it does not divide. So we need two consecutive odd integers that sum to 244. Easily we find 121 and 123. 121 is the 61st odd integer and 123 is the 62nd odd integer, so $(x+42)^2$ is the sum of the first 60 odd integers, or $(60)^2$ , while $y^2$ is $62^2$ for the same reasons. That way we get $x=12$ $y=62$ , hence $x+y=\boxed{080}$

080

f1b8938f754da449f63d57acb414adb6

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_5

A right circular cone has base radius $r$ and height $h$ . The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making $17$ complete rotations. The value of $h/r$ can be written in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$

The path is a circle with radius equal to the slant height of the cone, which is $\sqrt {r^{2} + h^{2}}$ . Thus, the length of the path is $2\pi\sqrt {r^{2} + h^{2}}$ Also, the length of the path is 17 times the circumference of the base, which is $34r\pi$ . Setting these equal gives $\sqrt {r^{2} + h^{2}} = 17r$ , or $h^{2} = 288r^{2}$ . Thus, $\dfrac{h^{2}}{r^{2}} = 288$ , and $\dfrac{h}{r} = 12\sqrt {2}$ , giving an answer of $12 + 2 = \boxed{014}$

014

2f3976b4dd17593c661ee9087b21d9c7

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_6

triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of $67$

Let the $k$ th number in the $n$ th row be $a(n,k)$ . Writing out some numbers, we find that $a(n,k) = 2^{n-1}(n+2k-2)$ We wish to find all $(n,k)$ such that $67| a(n,k) = 2^{n-1} (n+2k-2)$ . Since $2^{n-1}$ and $67$ are relatively prime , it follows that $67|n+2k-2$ . Since every row has one less element than the previous row, $1 \le k \le 51-n$ (the first row has $50$ elements, the second $49$ , and so forth; so $k$ can range from $1$ to $50$ in the first row, and so forth). Hence it follows that $67| n - 2k + 2$ implies that $n-2k+2 = 67$ itself. Now, note that we need $n$ to be odd, and also that $n+2k-2 = 67 \le 100-n \Longrightarrow n \le 33$ We can check that all rows with odd $n$ satisfying $1 \le n \le 33$ indeed contains one entry that is a multiple of $67$ , and so the answer is $\frac{33+1}{2} = \boxed{017}$

017

2f3976b4dd17593c661ee9087b21d9c7

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_6

triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of $67$

The result above is fairly intuitive if we write out several rows and then divide all numbers in row $r$ by $2^{r-1}$ (we can do this because dividing by a power of 2 doesn't affect divisibility by $67$ ). The second row will be $2, 4, 6, \cdots , 98$ , the third row will be $3, 5, \cdots, 97$ , and so forth. Clearly, only the odd-numbered rows can have a term divisible by $67$ . However, with each row the row will have one less element, and the $99-67+1 = 33$ rd row is the last time $67$ will appear. Therefore the number of multiples of 67 in the entire array is $\frac{33+1}{2} = \boxed{017}$

017

f442ab68faf8808aa2f10a7dd4937b7a

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_7

Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$ . For example, $S_4$ is the set ${400,401,402,\ldots,499}$ . How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square?

The difference between consecutive squares is $(x + 1)^2 - x^2 = 2x + 1$ , which means that all squares above $50^2 = 2500$ are more than $100$ apart. Then the first $26$ sets ( $S_0,\cdots S_{25}$ ) each have at least one perfect square because the differences between consecutive squares in them are all less than $100$ . 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

01a9e5767c644df91d24ecdb9d74912d

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_8

Find the positive integer $n$ such that \[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]

Since we are dealing with acute angles, $\tan(\arctan{a}) = a$ Note that $\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}$ , by tangent addition. Thus, $\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}$ Applying this to the first two terms, we get $\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}$ 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

01a9e5767c644df91d24ecdb9d74912d

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_8

Find the positive integer $n$ such that \[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]

Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\arctan\frac{1}{n}$ , is the argument of $n+i$ . The sum of these angles is then just the argument of the product \[(3+i)(4+i)(5+i)(n+i)\] and expansion give us $(48n-46)+(48+46n)i$ . Since the argument of 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

16cdd2bd27e9a33c787f70d99d8fc3f8

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9

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 $m$

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

16cdd2bd27e9a33c787f70d99d8fc3f8

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9

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 $m$

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

16cdd2bd27e9a33c787f70d99d8fc3f8

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9

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 $m$

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

16cdd2bd27e9a33c787f70d99d8fc3f8

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_9

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 $m$

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

273e52be74651f48c3b9455f066231dd

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_10

Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$ . The diagonals have length $10\sqrt {21}$ , and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$ , respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$ . The distance $EF$ can be expressed in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$

Key observation. $AD = 20\sqrt{7}$ Proof 1. By the triangle inequality , we can immediately see that $AD \geq 20\sqrt{7}$ . However, notice that $10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}$ , so by the law of sines, when $AD = 20\sqrt{7}$ $\angle ACD$ is right and the circle centered at $A$ with radius $10\sqrt{21}$ , which we will call $\omega$ , is tangent to $\overline{CD}$ . Thus, if $AD$ were increased, $\overline{CD}$ would have to be moved even farther outwards from $A$ to maintain the angle of $\frac{\pi}{3}$ and $\omega$ could not touch it, a contradiction. Proof 2. Again, use the triangle inequality to obtain $AD \geq 20\sqrt{7}$ . Let $x = AD$ and $y = CD$ . By the law of cosines on $\triangle ADC$ $2100 = x^2+y^2-xy \iff y^2-xy+(x^2-2100) = 0$ . Viewing this as a quadratic in $y$ , the discriminant $\Delta$ must satisfy $\Delta = x^2-4(x^2-2100) = 8400-3x^2 \geq 0 \iff x \leq 20\sqrt{7}$ . Combining these two inequalities yields the desired conclusion. This observation tells us that $E$ $A$ , and $D$ are collinear, in that order. Then, $\triangle ADC$ and $\triangle ACF$ are $30-60-90$ triangles. Hence $AF = 15\sqrt {7}$ , and Finally, the answer is $25+7=\boxed{032}$

032

273e52be74651f48c3b9455f066231dd

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_10

Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$ . The diagonals have length $10\sqrt {21}$ , and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$ , respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$ . The distance $EF$ can be expressed in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$

Extend $\overline {AB}$ through $B$ , to meet $\overline {DC}$ (extended through $C$ ) at $G$ $ADG$ is an equilateral triangle because of the angle conditions on the base. If $\overline {GC} = x$ then $\overline {CD} = 40\sqrt{7}-x$ , because $\overline{AD}$ and therefore $\overline{GD}$ $= 40\sqrt{7}$ By simple angle chasing, $CFD$ is a 30-60-90 triangle and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2}$ , and $\overline{CF} = \frac{40\sqrt{21} - \sqrt{3}x}{2}$ Similarly $CAF$ is a 30-60-90 triangle and thus $\overline{CF} = \frac{10\sqrt{21}}{2} = 5\sqrt{21}$ Equating and solving for $x$ $x = 30\sqrt{7}$ and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5\sqrt{7}$ $\overline{ED}-\overline{FD} = \overline{EF}$ $30\sqrt{7} - 5\sqrt{7} = 25\sqrt{7}$ and $25 + 7 = \boxed{032}$

032

13fb48e888c8704d11a66da1acf08d38

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_11

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?

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

13fb48e888c8704d11a66da1acf08d38

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_11

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?

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

13fb48e888c8704d11a66da1acf08d38

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_11

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?

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

0623894d856008c3d4440649a80ec7e8

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_12

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 $10$

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

0623894d856008c3d4440649a80ec7e8

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_12

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 $10$

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

0da37710d4b841bc2e7078868d1d9cca

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_13

Let \[p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.\] Suppose that \[p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.\] There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$ $b$ , and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$ . Find $a + b + c$

\begin{align*} p(0,0) &= a_0 \\ &= 0 \\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\ &= a_1 + a_3 + a_6 \\ &= 0 \\ p(-1,0) &= -a_1 + a_3 - a_6 \\ &= 0 \end{align*} Adding the above two equations gives $a_3 = 0$ , and so we can deduce that $a_6 = -a_1$ Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$ . Now, \begin{align*} p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\ &= a_4 + a_7 + a_8 \\ &= 0 \\ p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2 \\ &= -a_4 - a_7 + a_8 \\ &= 0 \end{align*} Therefore $a_8 = 0$ and $a_7 = -a_4$ . Finally, \begin{align*} p(2,2) &= 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 \\ &= -6 a_1 - 6 a_2 - 4 a_4 \\ &= 0 \end{align*} So, $3a_1 + 3a_2 + 2a_4 = 0$ , or equivalently $a_4 = -\frac{3(a_1 + a_2)}{2}$ Substituting these equations into the original polynomial $p$ , we find that at $\left(\frac{a}{c}, \frac{b}{c}\right)$ \[a_1x + a_2y + a_4xy - a_1x^3 - a_4x^2y - a_2y^3 = 0 \iff\] \[a_1x + a_2y - \frac{3(a_1 + a_2)}{2}xy - a_1x^3 + \frac{3(a_1 + a_2)}{2}x^2y - a_2y^3 = 0 \iff\] \[a_1x(x - 1)\left(x + 1 - \frac{3}{2}y\right) + a_2y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0\] . The remaining coefficients $a_1$ and $a_2$ are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible $p$ , we must have $x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0$ As the answer format implies that the $x$ -coordinate of the root is non-integral, $x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = 0 \iff x + 1 - \frac{3}{2}y = 0 \iff y = \frac{2}{3}(x + 1)\ (1)$ . The format also implies that $y$ is positive, so $y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0 \iff y^2 - 1 - \frac{3}{2}x(x - 1) = 0\ (2)$ . Substituting $(1)$ into $(2)$ and reducing to a quadratic yields $(19x - 5)(x - 2) = 0$ , in which the only non-integral root is $x = \frac{5}{19}$ , so $y = \frac{16}{19}$ The answer is $5 + 16 + 19 = \boxed{040}$

040

0da37710d4b841bc2e7078868d1d9cca

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_13

Let \[p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.\] Suppose that \[p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.\] There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$ $b$ , and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$ . Find $a + b + c$

Consider the cross section of $z = p(x, y)$ on the plane $z = 0$ . We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of $p(x, y)$ and they go over the eight given points. A simple way to do this would be to use the equations $x = 0$ $x = 1$ , and $y = \frac{2}{3}x + \frac{2}{3}$ , giving us $p_1(x, y) = x\left(x - 1\right)\left( \frac{2}{3}x - y + \frac{2}{3}\right) = \frac{2}{3}x + xy + \frac{2}{3}x^3-x^2y$ Another way to do this would to use the line $y = x$ and the ellipse, $x^2 + xy + y^2 = 1$ . This would give $p_2(x, y) = \left(x - y\right)\left(x^2 + xy + y^2 - 1\right) = -x + y + x^3 - y^3$ At this point, we see that $p_1$ and $p_2$ both must have $\left(\frac{a}{c}, \frac{b}{c}\right)$ as a zero. A quick graph of the 4 lines and the ellipse used to create $p_1$ and $p_2$ gives nine intersection points. Eight of them are the given ones, and the ninth is $\left(\frac{5}{9}, \frac{16}{9}\right)$ . The last intersection point can be found by finding the intersection points of $y = \frac{2}{3}x + \frac{2}{3}$ and $x^2 + xy + y^2 = 1$ . Finally, just add the values of $a$ $b$ , and $c$ to get $5 + 16 + 19 = \boxed{040}$

040

a44fd58256c930f8aa0ad314f84c74c1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14

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 $x = OC$ . Since $OT, AP \perp TC$ , it follows easily that $\triangle APC \sim \triangle OTC$ . Thus $\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}$ . By the Law of Cosines on $\triangle BAP$ \begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*} where $\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}$ , so: \begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*} Let $k = \frac{2x-27}{x^2} \Longrightarrow kx^2 - 2x + 27 = 0$ ; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(k)(27) \ge 0 \Longleftrightarrow k \le \frac{1}{27}$ . Thus, \[BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}\] Equality holds when $x = 27$

432

a44fd58256c930f8aa0ad314f84c74c1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14

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 $m^{2}$

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

a44fd58256c930f8aa0ad314f84c74c1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14

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 $m^{2}$

(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

a44fd58256c930f8aa0ad314f84c74c1

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_14

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

0c7345a9eaef04cefea1badb99284f4b

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_15

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

0c7345a9eaef04cefea1badb99284f4b

https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_15

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

c68ff8b1c4546b74ac17e6e314997075

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_1

Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$

Rewriting this sequence with more terms, we have Factoring this expression yields Next, we get Then, Dividing $10100$ by $1000$ yields a remainder of $\boxed{100}$

100

c68ff8b1c4546b74ac17e6e314997075

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_1

Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$

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

f26f4cbb26c729f31a74943123c30647

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_2

Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$ -mile mark at exactly the same time. How many minutes has it taken them?

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

f26f4cbb26c729f31a74943123c30647

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_2

Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$ -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

797dc9b803898b42b00aea8c1017276e

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_3

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?

Let the lengths of the three sides of the rectangular solid after the cutting be $a,b,c$ , so that the desired volume is $abc$ . Note that each cut reduces one of the dimensions by one, so that after ten cuts, $a+b+c = 10 + 13 + 14 - 10 = 27$ . By AM-GM $\frac{a+b+c}{3} = 9 \ge \sqrt[3]{abc} \Longrightarrow abc \le \boxed{729}$ . Equality is achieved when $a=b=c=9$ , which is possible if we make one slice perpendicular to the $10$ cm edge, four slices perpendicular to the $13$ cm edge, and five slices perpendicular to the $14$ cm edge.

729

797dc9b803898b42b00aea8c1017276e

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_3

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

70c5a4f1d230582ededa0409bcaf6a73

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_4

There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ $1\le k\le r$ ) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$

In base $3$ , we find that $\overline{2008}_{10} = \overline{2202101}_{3}$ . In other words, In order to rewrite as a sum of perfect powers of $3$ , we can use the fact that $2 \cdot 3^k = 3^{k+1} - 3^k$ The answer is $7+5+4+3+2+0 = \boxed{021}$

021

d9bac1cbb13d45287b80ee389f31efc2

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

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

d9bac1cbb13d45287b80ee389f31efc2

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Let $F,G,H$ 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

d9bac1cbb13d45287b80ee389f31efc2

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

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

d9bac1cbb13d45287b80ee389f31efc2

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $\boxed{504}$ . (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)

504

d9bac1cbb13d45287b80ee389f31efc2

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Let the height be h. Note that if $\overline{NH} = x$ then if we draw perpendiculars like in solution 1, $\overline{FN} = 500 - x, \overline{AF} = 504 + x, \overline{HG} = 500, \overline{GD} = 504 - x.$ Note that we wish to find $\overline{MN} = \sqrt{x^2 + h^2}.$ Let's find $\tan(53)$ in two ways. Finding $\tan(53)$ 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

d9bac1cbb13d45287b80ee389f31efc2

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_5

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

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

95bf978c5f5f2bb922892441add707ba

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_6

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

a7dab31a0b2abaa0561b710ebdcecc03

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_8

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

a7dab31a0b2abaa0561b710ebdcecc03

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_8

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

d535caed24674624f49b2aa32a4b931c

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_9

A particle is located on the coordinate plane at $(5,0)$ . Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle's position after $150$ moves is $(p,q)$ , find the greatest integer less than or equal to $|p| + |q|$

Let the particle's position be represented by a complex number. Recall that multiplying a number by cis $\left( \theta \right)$ rotates the object in the complex plane by $\theta$ counterclockwise. In this case, we use $a = cis(\frac{\pi}{4})$ . Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to where a is cis $\left( \theta \right)$ . By De-Moivre's theorem, $\left(cis( \theta \right)^n )$ =cis $\left(n \theta \right)$ . Therefore, Furthermore, $5a^{150} = - 5i$ . Thus, the final answer is As before, consider $z$ as a complex number. Consider the transformation $z \to (z-\omega)e^{i\theta} + \omega$ . This is a clockwise rotation of $z$ by $\theta$ radians about the points $\omega$ . Let $f(z)$ denote one move of $z$ . Then Therefore, $z$ rotates along a circle with center $\omega = 5+(5+5\sqrt2)i$ . Since $8 \cdot \frac{\pi}{4} = 2\pi$ $f^9(z) = f(z) \implies f^{150}(z) = f^6(z) \implies p+q = \boxed{019}$ , as desired (the final algebra bash isn't bad).

019

d535caed24674624f49b2aa32a4b931c

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_9

A particle is located on the coordinate plane at $(5,0)$ . Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle's position after $150$ moves is $(p,q)$ , find the greatest integer less than or equal to $|p| + |q|$

Let $T:\begin{pmatrix}x\\y\end{pmatrix}\rightarrow R(\frac{\pi}{4})\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix}$ . We assume that the rotation matrix $R(\frac{\pi}{4}) = R$ here. Then we have This simplifies to Since $R+R^{7}=O, R^2+R^6=O, R^3+R^5=O, I+R^4=O$ , so we have $R^6\begin{pmatrix}5\\0\end{pmatrix}+(-R^6-R^7)\begin{pmatrix}10\\0\end{pmatrix}$ , giving $p=-5\sqrt{2}, q=5\sqrt{2}+5$ . The answer is yet $\lfloor10\sqrt{2}+5\rfloor=\boxed{019}$

019

0bae60ab2e56841bc406422b9b90bf07

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_10

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

ae19aa2f17063c4d4d563e077cb3a952

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_11

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 $m + nk$

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

ae19aa2f17063c4d4d563e077cb3a952

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_11

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 $m + nk$

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

504936132945277c2dc0b8ad868860df

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_13

regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$ . Then the area of $S$ has the form $a\pi + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

If a point $z = r\text{cis}\,\theta$ is in $R$ , then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$ ). Since $R$ is symmetric every $60^{\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \le \theta \le 30$ , and then to multiply by $6$ to account for the entire area. We note that if the region $S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace$ , where $R_2$ is the region (in green below) outside the circle of radius $1/\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R-R_2$ (in blue below). The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$ , which is transformed to $\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta =$ 2 . Let $r\text{cis}\,\theta = a+bi$ where $a,b \in \mathbb{R}$ ; then $r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}$ , so the equation becomes $a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1$ . Hence the side is sent to an arc of the unit circle centered at $(1,0)$ , after considering the restriction that the side of the hexagon is a segment of length $1/\sqrt{3}$ Including $S_2$ , we find that $S$ is the union of six unit circles centered at $\text{cis}\, \frac{k\pi}{6}$ $k = 0,1,2,3,4,5$ , as shown below. The area of the regular hexagon is $6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}$ . The total area of the six $120^{\circ}$ sectors is $6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}$ . Their sum is $2\pi + \sqrt{27}$ , and $a+b = \boxed{029}$ . - Th3Numb3rThr33

029

504936132945277c2dc0b8ad868860df

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_13

regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$ . Then the area of $S$ has the form $a\pi + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

We can describe the line parallel to the imaginary axis $x=\frac{1}{2}$ using polar coordinates as $r(\theta)=\dfrac{1}{2\cos{\theta}},$ which rearranges to $z=\left(\dfrac{1}{2\cos{\theta}}\right)(cis{\theta})\implies \frac{1}{z}=2\cos{\theta}cis(-\theta).$ Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral: $[S] = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta.$ Thankfully, this is a routine computation: $[S] = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta$ $[S] = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12\left[\frac{1}{2}\sin{2\theta}+\theta\right]_0^{\frac{\pi}{6}}=12\left(\frac{\sqrt{3}}{4}+\frac{\pi}{6}\right)=2\pi+3\sqrt{3}=2\pi + \sqrt{27}$ $a+b = \boxed{029}$

029

2b611c40a02c1185ddfb3a15df89c058

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_14

Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Notice that the given equation implies We have $2by \ge y^2$ , so $2ax \le a^2 \implies x \le \frac {a}{2}$ Then, notice $b^2 + x^2 = a^2 + y^2 \ge a^2$ , so $b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \le \frac {4}{3}$ The solution $(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \frac {1}{2}, 0\right)$ satisfies the equation, so $\rho^2 = \frac {4}{3}$ , and the answer is $3 + 4 = \boxed{007}$

007

2b611c40a02c1185ddfb3a15df89c058

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_14

Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Consider the points $(a,y)$ and $(x,b)$ . They form an equilateral triangle with the origin. We let the side length be $1$ , so $a = \cos{\theta}$ and $b = \sin{\left(\theta + \frac {\pi}{3}\right)}$ Thus $f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}$ and we need to maximize this for $0 \le \theta \le \frac {\pi}{6}$ Taking the derivative shows that $-f'(\theta) = \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta + \frac {\pi}{3}\right)}} \ge 0$ , so the maximum is at the endpoint $\theta = 0$ . We then get Then, $\rho^2 = \frac {4}{3}$ , and the answer is $3+4=\boxed{007}$

007

2b611c40a02c1185ddfb3a15df89c058

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_14

Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Consider a cyclic quadrilateral $ABCD$ with $\angle B = \angle D = 90^{\circ}$ , and $AB = y, BC = a, CD = b, AD = x$ . Then \[AC^2 = a^2 + y^2 = b^2 + x^2\] From Ptolemy's Theorem $ax + by = AC(BD)$ , so \[AC^2 = (a - x)^2 + (b - y)^2 = a^2 + y^2 + b^2 + x^2 - 2(ax + by) = 2AC^2 - 2AC*BD\] Simplifying, we have $BD = AC/2$ Note the circumcircle of $ABCD$ has radius $r = AC/2$ , so $BD = r$ and has an arc of $60^{\circ}$ , so $\angle C = 30^{\circ}$ . Let $\angle BDC = \theta$ $\frac ab = \frac{BC}{CD} = \frac{\sin \theta}{\sin(150^{\circ} - \theta)}$ , where both $\theta$ and $150^{\circ} - \theta$ are $\leq 90^{\circ}$ since triangle $BCD$ must be acute . Since $\sin$ is an increasing function over $(0, 90^{\circ})$ $\frac{\sin \theta}{\sin(150^{\circ} - \theta)}$ is also increasing function over $(60^{\circ}, 90^{\circ})$ $\frac ab$ maximizes at $\theta = 90^{\circ} \Longrightarrow \frac ab$ maximizes at $\frac 2{\sqrt {3}}$ . This squared is $(\frac 2{\sqrt {3}})^2 = \frac4{3}$ , and $4 + 3 = \boxed{007}$

007

2b611c40a02c1185ddfb3a15df89c058

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_14

Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

The problem is looking for an intersection in the said range between parabola $P$ $y = \tfrac{(x-a)^2 + b^2-a^2}{2b}$ and the hyperbola $H$ $y^2 = x^2 + b^2 - a^2$ . The vertex of $P$ is below the x-axis and it's x-coordinate is a, which is to the right of the vertex of the $H$ , which is $\sqrt{a^2 - b^2}$ . So for the intersection to exist with $x<a$ and $y \geq 0$ $P$ needs to cross x-axis between $\sqrt{a^2 - b^2}$ , and $a$ , meaning, \[(\sqrt{a^2 - b^2}-a)^2 + b^2-a^2 \geq 0\] Divide both side by $b^2$ \[(\sqrt{\rho^2 - 1}-\rho)^2 + 1-\rho^2 \geq 0\] which can be easily solved by moving $1-\rho^2$ to RHS and taking square roots. Final answer $\rho^2 \leq \frac{4}{3}$ $\boxed{007}$

007

2b611c40a02c1185ddfb3a15df89c058

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_14

Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

The given system is equivalent to the points $(a,y)$ and $(x,b)$ forming an equilateral triangle with the origin. WLOG let this triangle have side length $1$ , so $x=\sqrt{1-a^2}$ . The condition $x<a$ implies $(x,b)$ lies to the left of $(a,y)$ , so $(x,b)$ is the top vertex. Now we can compute (by complex numbers, or the sine angle addition identity) that $b = \frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}$ , so $\frac{a}{b} = \frac{a}{\frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}} = \frac{1}{\frac{\sqrt{3}}{2} + \frac{1}{2a}\sqrt{1-a^2}}$ . Minimizing this is equivalent to minimizing the denominator, which happens when $\sqrt{1-a^2} = 0$ and thus $a=1$ , resulting in $\rho = \frac{2}{\sqrt{3}}$ , so $\rho^2 = \frac{4}{3}$ and the answer is $\boxed{007}$

007

e9eb94fa0b3d66056cb5d98a7227448d

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15

Find the largest integer $n$ satisfying the following conditions:

Write $n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1$ , or equivalently, $(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2$ Since $2n + 1$ and $2n - 1$ are both odd and their difference is $2$ , they are relatively prime . But since their product is three times a square, one of them must be a square and the other 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

e9eb94fa0b3d66056cb5d98a7227448d

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15

Find the largest integer $n$ satisfying the following conditions:

Suppose that the consecutive cubes are $m$ and $m + 1$ . We can use completing the square and the first condition to get: \[(2n)^2 - 3(2m + 1)^2 = 1\equiv a^2 - 3b^2\] where $a$ and $b$ are non-negative integers. Now this is a Pell equation , with solutions in the form $(2 + \sqrt {3})^k = a_k + \sqrt {3}b_k,$ $k = 0,1,2,3,...$ . 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

e9eb94fa0b3d66056cb5d98a7227448d

https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15

Find the largest integer $n$ satisfying the following conditions:

Let us generate numbers $1$ to $1000$ for the second condition, for squares. We know for $N$ to be integer, the squares must be odd. So we generate $N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801$ $N$ cannot exceed $1000$ since it is AIME problem. Now take the first criterion, let $a$ be the smaller consecutive 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

9a061727dff4c78e490b8508744804ef

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_1

How many positive perfect squares less than $10^6$ are multiples of $24$

The prime factorization of $24$ is $2^3\cdot3$ . Thus, each square must have at least $3$ factors of $2$ and $1$ factor of $3$ and its square root must have $2$ factors of $2$ and $1$ factor of $3$ . This means that each square is in the form $(12c)^2$ , where $12 c$ is a positive integer less than $\sqrt{10^6}$ . There are $\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}$ solutions.

083

9a061727dff4c78e490b8508744804ef

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_1

How many positive perfect squares less than $10^6$ are multiples of $24$

The perfect squares divisible by $24$ are all multiples of $12$ $12^2$ $24^2$ $36^2$ $48^2$ , etc... Since they all have to be less than $10^6$ , or $1000^2$ , the closest multiple of $12$ to $1000$ is $996$ $12*83$ ), so we know that this is the last term in the sequence. Therefore, we know that there are $\boxed{083}$ perfect squares divisible by $24$ that are less than $10^6$

083

34eee30465c2d0a7e7b803d9c4f70fe9

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_2

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.

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

5153e2ab29a7f38310181b0e0de2d212

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_3

The complex number $z$ is equal to $9+bi$ , where $b$ is a positive real number and $i^{2}=-1$ . Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to?

Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$ . Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i$ Setting these two equal, we get that $18bi = 243bi - b^3i$ , so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$ . Since $b > 0$ , the solution is $\boxed{015}$

015

7d4062496b7ae40fb419364c5e94767f

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_4

Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are 60, 84, and 140 years. The three planets and the star are currently collinear . What is the fewest number of years from now that they will all be collinear again?

Denote the planets $A, B, C$ respectively. Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years. These are given by $a(t) = \frac{t \pi}{30}$ $b(t) = \frac{t \pi}{42}$ $c(t) = \frac{t \pi}{70}$ In order for the planets and the central star to be 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

5bf3f45ec7fc867594e2467770b0b060

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5

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?

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

5bf3f45ec7fc867594e2467770b0b060

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5

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?

Notice that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$ holds if $k=\left[ \frac{9}{5}x\right]$ for some integer $x$ . Thus, after translating from $F\to F-32$ we want count how many values of $x$ there are such that $k=\left[ \frac{9}{5}x\right]$ is an integer from $0$ to $968$ . This value is computed as $\left[968*\frac{5}{9}\right]+1 = \boxed{539}$ , adding in the extra solution corresponding to $0$

539

5bf3f45ec7fc867594e2467770b0b060

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5

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?

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

6d8b76872afa9dd5a8b4904ee031d89f

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6

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?

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

6d8b76872afa9dd5a8b4904ee031d89f

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6

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?

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

6d8b76872afa9dd5a8b4904ee031d89f

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6

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?

Another way you can visualize the problem is by thinking of points $13$ $26$ , and $39$ as planets and all multiples of 3 as points at which your spaceship can jump to hyperspace. Given that you wish to visit planet $39$ , you can choose to visit planets $13$ or $26$ along the way. Case 1: Neither There are $4$ ways to jump to $39$ from hyperspace. Case 2: Only planet $13$ There are $5$ ways to jump to planet $13$ and $4$ ways to jump to planet $39$ $20$ ways total. Case 3: Only planet $26$ There are $4$ ways to jump to planet $26$ and $5$ ways to jump to planet $39$ $20$ ways total. Case 4: Both There are $5$ ways to jump to $13$ $5$ ways to jump to $26$ , and $5$ ways to jump to $39$ $125$ ways total. Therefore, there are $4+20+20+125=\boxed{169}$ ways to jump to 39 through various journeys in hyperspace.

169

36e1000816f90efeb209e18121a8c210

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_7

Let $N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )$ Find the remainder when $N$ is divided by 1000. ( $\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$ , and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$ .)

The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer ); otherwise, it is equal to 1. Thus, we need to find when or not $\log_{\sqrt{2}} k$ is an integer. The change of base formula shows that $\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}$ . For the $\log 2$ 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

94dc8d468511cc7eefedd8bc329e9a83

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_8

The polynomial $P(x)$ is cubic . What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$

We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic. Let this root be $a$ We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \frac{-k}{5}$ We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get: $\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$ or $k^{2}=30k$ , so $k=30$ is the highest. We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$

030

94dc8d468511cc7eefedd8bc329e9a83

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_8

The polynomial $P(x)$ is cubic . What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$

Again, let the common root be $a$ ; let the other two roots be $m$ and $n$ . We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)$ Therefore, we can write four equations (and we have four variables ), $a + m = 29 - k$ $a + n = \frac{43}{2} - k$ $am = -k$ , and $an = \frac{k}{2}$ The first two equations show that $m - n = 29 - \frac{43}{2} = \frac{15}{2}$ . The last two equations show that $\frac{m}{n} = -2$ . Solving these show that $m = 5$ and that $n = -\frac{5}{2}$ . Substituting back into the equations, we eventually find that $k = \boxed{030}$

030

94dc8d468511cc7eefedd8bc329e9a83

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_8

The polynomial $P(x)$ is cubic . What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$

Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$ , which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$ , where $R(x) = ax + b$ and $S(x) = cx + d$ . Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$ . Equating coefficients, we get the following system of equations: \begin{align} a = 2c \\ b = -d \\ 2c(k - 29) - d = c(2k - 43) + 2d \\ -d(k - 29) - 2ck = d(2k - 43) + ck \end{align} Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$ , we see that the $k$ 's drop out and we're left with $d = -5c$ . Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\boxed{030}$

030

94dc8d468511cc7eefedd8bc329e9a83

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_8

The polynomial $P(x)$ is cubic . What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$

Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$ Thus, we see that we have \[r^2 + (k - 29)r - k = 0,\] \[2r^2 + (2k - 43)r + k = 0.\] Adding the two equations gives us \[3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.\] Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that \[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.\] This can easily be checked to see that it does indeed work, and we're done!

030

02841d381c965e9b346c41df7fd4cd82

https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_9

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 $p+q$

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