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9c27c0bb1060f65d64c2269b8b8a054b | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) | Again, let the total number of steps be $x$ , the speed of the escalator be $e$ and the speed of Bob be $b$ (all "per unit time").
Then this can be interpreted as a classic chasing problem: Bob is "behind" by $x$ steps, and since he moves at a pace of $b+e$ relative to the escalator, it will take $\frac{x}{b+e}=\frac{75}{e}$ time to get to the top.
Similarly, Al will take $\frac{x}{3b-e}=\frac{150}{e}$ time to get to the bottom.
From these two equations, we arrive at $150=\frac{ex}{3b-e}=2\cdot75=\frac{2ex}{b+e}=\frac{6ex}{3b+3e}=\frac{(ex)-(6ex)}{(3b-e)-(3b+3e)}=\frac{5x}{4}$ $\implies600=5x\implies x=\boxed{120}$ (the proportion manipulations are motivated by the desire to isolate $x$ , prompting the isolation of the $150$ on one side, and the fact that if we could cancel out the $b$ 's, then the $e$ 's in the numerator and denominator would cancel out, resulting in an equation with $x$ by itself). | null | 120 |
9c27c0bb1060f65d64c2269b8b8a054b | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) | Let $e$ and $b$ be the speeds of the escalator and Bob, respectively.
When Al was on his way down, he took $150$ steps with a speed of $3b-e$ per step. When Bob was on his way up, he took $75$ steps with a speed of $b+e$ per step. Since Al and Bob were walking the same distance, we have \[150(3b-e)=75(b+e)\] Solving gets the ratio $\frac{e}{b}=\frac{3}{5}$
Thus while Bob took $75$ steps to go up, the escalator contributed an extra $\frac{3}{5}\cdot75=45$ steps.
Finally, there is a total of $75+45=\boxed{120}$ steps in the length of the escalator. | null | 120 |
9c27c0bb1060f65d64c2269b8b8a054b | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) | Please understand the machinery of an escalator before proceeding to read this solution.
Let the number of steps that disappear at the top of the escalator equal $x.$ Assume that Al takes $3$ steps per second and that Bob takes $1$ step per second. Since Al counts $150$ steps, it takes him \[\frac{150}{3}=50\] seconds to traverse the distance of the escalator moving downwards. Since Bob counts $75$ steps, it takes him \[\frac{75}{1}=75\] seconds to traverse the distance of the escalator moving downwards.
For the sake of this solution, we activate the emergency stop button on the escalator.
Now, the escalator is not moving, or is simply a staircase. Imagine that Al is taking $3$ steps downwards every second, but we throw hands at him immediately after each second, such that he flinches and moves himself backwards $x$ steps. This is equivalent to Al taking $3-x$ steps downwards every second. Since we discovered that it takes him $50$ seconds to get from the top to the bottom of the escalator, and we are forcing Al to imitate the movement of the escalator, it also takes him $50$ seconds to move from the top to the bottom of the staircase. Thus, Al takes a total of \[(3-x) \cdot 50=150-50x \qquad (\heartsuit)\] steps.
The explanation for Bob is similar except now we pick him up and place him forward $x$ steps immediately after he takes his usual step per second, and since we discovered he does this for $75$ seconds, it takes him \[(1+x) \cdot 75=75+75x \qquad (\clubsuit)\] steps to get from the bottom to the top.
Note that because the escalator is broken and is now a staircase, Al and Bob must have had to take an equal amount of steps to get from the bottom to the top or from the top to the bottom. (Clearly, there are an equal amount of steps from the bottom to the top, and from the top to the bottom.) Therefore, we may equate $\heartsuit$ and $\clubsuit$ to get \[150-50x=75+75x\] \[x=\frac{3}{5}.\] Therefore, substituting $x$ in the expression we discovered in $\heartsuit,$ Al takes a total of \[150-50x=150-50\left(\frac{3}{5}\right)=\boxed{120}\] steps, and we are done. | null | 120 |
9c27c0bb1060f65d64c2269b8b8a054b | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) | WLOG, let Al's speed be $15$ steps per second, so Bob's speed is $5$ steps per second. Then, Al was on the escalator for $\frac{150}{15}\ = 10$ seconds and Bob was on the escalator for $\frac{75}{5}\ = 15$ seconds. Let $r$ be the rate of the escalator, in steps per second. Then, the total amount of steps is $150 - 10r = 75 + 15r$ . Al is getting $10$ seconds of resistance at rate $r$ from the escalator, while Bob is getting $15$ seconds of help at rate $r$ . Solving for $r$ , we have $r = 3$ steps per second. Then, we can plug $r$ into the previous equation or subtract/add it to Al/Bob's rate (respectively) then multiply by their respective time. Either way, we get $\boxed{120}$ and we are done. | null | 120 |
99bd25a10de003c1256de1acf926348a | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_11 | Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers | Let us write down one such sum, with $m$ terms and first term $n + 1$
$3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)$
Thus $m(2n + m + 1) = 2 \cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$ . However, because $n \geq 0$ we have $m^2 < m(m + 1) \leq 2\cdot 3^{11}$ so $m < \sqrt{2\cdot 3^{11}} < 3^6$ . Thus, we are looking for large factors of $2\cdot 3^{11}$ which are less than $3^6$ . The largest such factor is clearly $2\cdot 3^5 = 486$ ; for this value of $m$ we do indeed have the valid expression $3^{11} = 122 + 123 + \ldots + 607$ , for which $k=\boxed{486}$ | null | 486 |
99bd25a10de003c1256de1acf926348a | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_11 | Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers | Proceed as in Solution 1 until it is noted that $m$ is a divisor of $2\cdot 3^{11}$ . The divisors of $2\cdot 3^{11}$ are $3^{1} , 2\cdot 3^{1} , 3^{2} , 2\cdot 3^{2} , \ldots , 2\cdot 3^{10} , 3^{11}$ . Note that the factors of $m(2n + m + 1)$ are of opposite parity (if $m$ is odd, then $(2n + m + 1)$ is even and vice versa). Thus, one of the two factors will be a power of three, and the other will be twice a power of three. $(2n + m + 1)$ will represent the greater factor while $m$ will represent the lesser factor. Given this information, we need to find the factor pair that maximizes the lesser of the two factors, as this will maximize the value of $m$ . The factor pair which maximizes the lesser factor is $2\cdot 3^{5}$ and $3^{6}$ . It follows that $m$ $2\cdot 3^{5}$ $\boxed{486}$ | null | 486 |
a3e340db5e124d9bfe21b2089a245dfd | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_12 | Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$ | In order to keep $m$ as small as possible, we need to make $n$ as small as possible.
$m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3$ . Since $r < \frac{1}{1000}$ and $m - n^3 = r(3n^2 + 3nr + r^2)$ is an integer, we must have that $3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000$ . This means that the smallest possible $n$ should be less than 1000. In particular, $3nr + r^2$ should be less than 1, so $3n^2 > 999$ and $n > \sqrt{333}$ $18^2 = 324 < 333 < 361 = 19^2$ , so we must have $n \geq 19$ . Since we want to minimize $n$ , we take $n = 19$ . Then for any positive value of $r$ $3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000$ , so it is possible for $r$ to be less than $\frac{1}{1000}$ . However, we still have to make sure a sufficiently small $r$ exists.
In light of the equation $m - n^3 = r(3n^2 + 3nr + r^2)$ , we need to choose $m - n^3$ as small as possible to ensure a small enough $r$ . The smallest possible value for $m - n^3$ is 1, when $m = 19^3 + 1$ . Then for this value of $m$ $r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}$ , and we're set. The answer is $\boxed{019}$ | null | 019 |
a3e340db5e124d9bfe21b2089a245dfd | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_12 | Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$ | To minimize $m$ , we should minimize $n$ . We have that $(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}$ . For a given value of $n$ , if $(n + \frac{1}{1000})^3 - n^3 > 1$ , there exists an integer between $(n + \frac{1}{1000})^3$ and $n^3$ , and the cube root of this integer would be between $n$ and $n + \frac{1}{1000}$ as desired. We seek the smallest $n$ such that $(n + \frac{1}{1000})^3 - n^3 > 1$
\[(n + \frac{1}{1000})^3 - n^3 > 1\] \[\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1\] \[3n^2 + \frac{3}{10^3} n + \frac{1}{10^6} > 10^3\]
Trying values of $n$ , we see that the smallest value of $n$ that works is $\boxed{019}$ | null | 019 |
a3e340db5e124d9bfe21b2089a245dfd | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_12 | Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$ | Since $r$ is less than $1/1000$ , we have $\sqrt[3]{m} < n + \frac{1}{1000}$ . Notice that since we want $m$ minimized, $n$ should also be minimized. Also, $n^3$ should be as close as possible, but not exceeding $m$ . This means $m$ should be set to $n^3+1$ . Substituting and simplifying, we get \[\sqrt[3]{n^3+1} < n + \frac{1}{1000}\] \[n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}\] The last two terms in the right side can be ignored in the calculation because they are too small. This results in $1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}$ . The minimum positive integer $n$ that satisfies this is $\boxed{019}$ . ~ Hb10 | null | 019 |
61f63a1d82d44047fe05f49f78ca14ed | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_13 | A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$ , with its current predecessor and exchanging them if and only if the last term is smaller.
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.
Suppose that $n = 40$ , and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order. Let $p/q$ , in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\mbox{th}}$ place. Find $p + q$ | If any of $r_1, \ldots, r_{19}$ is larger than $r_{20}$ , one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position. Thus, $r_{20}$ must be the largest of the first 20 terms. In addition, $r_{20}$ must be larger than $r_{21}, r_{22}, \ldots, r_{30}$ but smaller than $r_{31}$ in order that it move right to the 30th position but then not continue moving right to the 31st.
Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position 31 and the second-largest is in position 20 (the other 29 numbers are irrelevant)?
This is much easier to solve: there are $31!$ ways to order the first thirty-one numbers and $29!$ ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of $\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}$ , so the answer is $\boxed{931}$ | null | 931 |
61f63a1d82d44047fe05f49f78ca14ed | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_13 | A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$ , with its current predecessor and exchanging them if and only if the last term is smaller.
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.
Suppose that $n = 40$ , and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order. Let $p/q$ , in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\mbox{th}}$ place. Find $p + q$ | Note that you can solve the restated problem differently:
We see that the numerator is $1$ because there is only $1$ way to make the $31^{\text{st}}$ term the largest and the $20^{\text{th}}$ the second-largest. To calculate the denominator, we note that there are $31\times30$ ways to pick the largest term and the second-largest term. Therefore, our answer is $\boxed{931}$ .
~Yiyj1 | null | 931 |
c20def49432511f9205c3592f13c0900 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | The Sophie Germain Identity states that $a^4 + 4b^4$ can be factored as $\left(a^2 + 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).$ Each of the terms is in the form of $x^4 + 324.$ Using Sophie Germain, we get that \begin{align*} x^4 + 324 &= x^4 + 4\cdot 3^4 \\ &= \left(x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x\right)\left(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x\right) \\ &= (x(x-6) + 18)(x(x+6)+18), \end{align*} so the original expression becomes \[\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]},\] which simplifies to \[\frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}.\] Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}.$ | null | 373 |
c20def49432511f9205c3592f13c0900 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$
We factor $N^4+18^2$ by completing the square, then applying the difference of squares: \begin{align*} N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\ &=\left(N^2+18\right)^2-(6N)^2 \\ &=\left(N^2-6N+18\right)\left(N^2+6N+18\right) \\ &=\left((N-3)^2+9\right)\left((N+3)^2+9\right). \end{align*} The original expression now becomes \[\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.\] ~MRENTHUSIASM | null | 373 |
c20def49432511f9205c3592f13c0900 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | We rewrite $N$ to the polar form \[N=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $N$ such that $r\geq0,$ and $\theta$ is the argument of $N$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem , we have \[N^4=r^4\operatorname{cis}(4\theta)=18^2(-1),\] from which
By the Factor Theorem , we get \begin{align*} N^4+18^2&=\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr) \\ &=\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr)\biggr]\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggr] \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$ | null | 373 |
c20def49432511f9205c3592f13c0900 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | We rewrite $N$ to the rectangular form \[N=a+bi\] for some real numbers $a$ and $b.$
Note that $N^2=\pm18i,$ so there are two cases:
By the Factor Theorem , we get \begin{align*} N^4+18^2&=(N-(3+3i))(N-(-3-3i))(N-(3-3i))(N-(-3+3i)) \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$ | null | 373 |
72666474f0409b8485a5d81423bc9553 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | 1987 AIME-15a.png
Because all the triangles in the figure are similar to triangle $ABC$ , it's a good idea to use area ratios . In the diagram above, $\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.$ Hence, $T_3 = \frac {440}{441}T_1$ and $T_4 = \frac {440}{441}T_2$ . Additionally, the area of triangle $ABC$ is equal to both $T_1 + T_2 + 441$ and $T_3 + T_4 + T_5 + 440.$
Setting the equations equal and solving for $T_5$ $T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}$ . Therefore, $441T_5 = 441 + T_1 + T_2$ . However, $441 + T_1 + T_2$ is equal to the area of triangle $ABC$ ! This means that the ratio between the areas $T_5$ and $ABC$ is $441$ , and the ratio between the sides is $\sqrt {441} = 21$ . As a result, $AB = 21\sqrt {440} = \sqrt {AC^2 + BC^2}$ . We now need $(AC)(BC)$ to find the value of $AC + BC$ , because $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2$
Let $h$ denote the height to the hypotenuse of triangle $ABC$ . Notice that $h - \frac {1}{21}h = \sqrt {440}$ . (The height of $ABC$ decreased by the corresponding height of $T_5$ ) Thus, $(AB)(h) = (AC)(BC) = 22\cdot 21^2$ . Because $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2$ $AC + BC = (21)(22) = \boxed{462}$ | null | 462 |
72666474f0409b8485a5d81423bc9553 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | Let $\tan\angle ABC = x$ . Now using the 1st square, $AC=21(1+x)$ and $CB=21(1+x^{-1})$ . Using the second square, $AB=\sqrt{440}(1+x+x^{-1})$ . We have $AC^2+CB^2=AB^2$ , or \[441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).\] Rearranging and letting $u=x+x^{-1} \Rightarrow u^2 - 2 = x^2 + x^{-2}$ gives us $u^2+2u-440=0.$ We take the positive root, so $u=20$ , which means $AC+CB=21(2+x+x^{-1})=21(2+u)=\boxed{462}$ | null | 462 |
72666474f0409b8485a5d81423bc9553 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | Let $\theta$ be the smaller angle in the triangle. Then the sum of shorter and longer leg is $\sqrt{441}(2+\tan{\theta}+\cot{\theta})$ . We observe that the short leg has length $\sqrt{441}(1+\tan{\theta}) = \sqrt{440}(\sec{\theta}+\sin{\theta})$ . Grouping and squaring, we get $\sqrt{\frac{440}{441}} = \frac{\sin{\theta}+\cos{\theta}}{1+\sin{\theta}\cos{\theta}}$ . Squaring and using the double angle identity for sine, we get, $110(\sin{2\theta})^2 + \sin{2\theta} - 1 = 0$ . Solving, we get $\sin{2\theta} = \frac{1}{10}$ . Now to find $\tan{\theta}$ , we find $\cos{2\theta}$ using the Pythagorean
Identity, and then use the tangent double angle identity. Thus, $\tan{\theta} = 10-3\sqrt{11}$ . Substituting into the original sum,
we get $\boxed{462}$ | null | 462 |
72666474f0409b8485a5d81423bc9553 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | [asy] size(200); pair A, B, C, D, E, F; A = (0, 5); B = (12, 0); C = (0, 0); D = (0, 60/17); E = (60/17, 60/17); F = (60/17, 0); draw(A--B--C--cycle); draw(D--E--F); label("$A$",A,N); label("$B$",B,dir(0)); label("$C$",C,SW); label("$D$",D,W); label("$E$",E,NE); label("$F$",F,S); label("$S_1$",(30/17,30/17)); [/asy]
[asy] size(200); pair A, B, C, W, X, Y, Z; A = (0, 5); B = (12, 0); C = (0, 0); real m = 1.31004366812; real n = 3.1441048035; W = (0,m); X = (m,m+n); Y = (m+n,n); Z = (n,0); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle); label("$A$",A,N); label("$B$",B,dir(0)); label("$C$",C,SW); label("$W$",W,dir(180)); label("$X$",X,NE); label("$Y$",Y,NE); label("$Z$",Z,S); label("$S_2$",(2.22707423581,2.22707423581)); [/asy]
Label points as above. Let $x=AC$ $y=BC$ $s_1 = 21$ be the side length of $S_1$ , and $s_2 = \sqrt{440}$ be the side length of $S_2$
Since $\triangle ABC\sim\triangle AED$ , we have $\frac{x}{y} = \frac{x-s_1}{s_1}$
$\implies xs_1=xy-ys_1$
$\implies xy=s_1(x+y)$
$\implies xy=21(x+y) \qquad \qquad (*)$
Since $\triangle ABC\sim\triangle AWX\sim\triangle ZBY$ , we have $s_2 + \frac{s_2x}{y} + \frac{s_2y}{x}=\sqrt{x^2+y^2}$
$\implies s_2(x^2+xy+y^2)=xy\sqrt{x^2+y^2}$
$\implies s_2^2(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$
$\implies 440(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$
Let $t=x+y$ . Repeatedly applying $(*)$ , we get \[440(t^2-21t)^2 = 441t^2(t^2 - 42t)\] \[440(t-21)^2 = 441(t^2-42t)\] \[440t^2 - 42\cdot 440t + 440\cdot 441 = 441t^2 - 441\cdot 42t\] \[t^2-42t-440\cdot 441=0\] \[(t-21)^2 = 441^2\] \[t-21=441\] \[t=\boxed{462}\] | null | 462 |
997095422bf9771cff682e186006d41d | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_1 | What is the sum of the solutions to the equation $\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}$ | Let $y = \sqrt[4]{x}$ . Then we have $y(7 - y) = 12$ , or, by simplifying, \[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.\]
This means that $\sqrt[4]{x} = y = 3$ or $4$
Thus the sum of the possible solutions for $x$ is $4^4 + 3^4 = \boxed{337}$ | null | 337 |
88b46587827612cc05f258d7ef80d5bf | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 | Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | More generally, let $(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right)$ so that $\left(x^2,y^2,z^2\right)=(5,6,7).$
We rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: \begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \left[((x+y)+z)((x+y)-z)\right]\left[((z+(x-y))(z-(x-y))\right] \\ &= \left[(x+y)^2-z^2\right]\left[z^2 - (x-y)^2\right] \\ &= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\ &= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= \left(2\cdot\sqrt5\cdot\sqrt6\right)^2 - \left(5+6-7\right)^2 \\ &= \boxed{104} Remark | null | 104 |
88b46587827612cc05f258d7ef80d5bf | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 | Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{42}\right)\right) \\ &= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\ &= \left(2\sqrt{42}\right)^2 - 8^2 \\ &= \boxed{104} ~Azjps (Solution) | null | 104 |
88b46587827612cc05f258d7ef80d5bf | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 | Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | Notice that in a triangle with side-lengths $2\sqrt5,2\sqrt6,$ and $2\sqrt7,$ by Heron's Formula, the area is the square root of the original expression.
Let $\theta$ be the measure of the angle opposite the $2\sqrt7$ side. By the Law of Cosines, \[\cos\theta=\frac{\left(2\sqrt5\right)^2+\left(2\sqrt{6}\right)^2-\left(2\sqrt7\right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}},\] so $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.$
The area of the triangle is then \[\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104},\] so our answer is $\left(\sqrt{104}\right)^2=\boxed{104}.$ | null | 104 |
8570e0ac7f66075261a9486a08e402df | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_3 | If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ | Since $\cot$ is the reciprocal function of $\tan$
$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$
Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$
Using the tangent addition formula:
$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}$ | null | 150 |
8570e0ac7f66075261a9486a08e402df | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_3 | If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ | Using the formula for tangent of a sum, $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}$ . We only need to find $\tan x \tan y$
We know that $25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}$ . Cross multiplying, we have $\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25$
Similarly, we have $30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}$
Dividing:
$\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}$ . Plugging in to the earlier formula, we have $\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}$ | null | 150 |
8570e0ac7f66075261a9486a08e402df | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_3 | If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ | Let $a=\tan x$ and $b=\tan y$ . This simplifies the equations to:
\[a + b = 25\]
\[\frac{1}{a} + \frac{1}{b} = 30\]
Taking the tangent of a sum formula from Solution 2, we get $\tan(x+y) = \frac{25}{1 - ab}$
We can use substitution to solve the system of equations from above: $b = -a + 25$ , so $\frac{1}{a} + \frac{1}{-a + 25} = 30$
Multiplying by $-a(a-25)$ , we get $a + (-a + 25) = -30a(a-25)$ , which is $-30a^2 + 750a = 25$ . Dividing everything by 5 and shifting everything to one side gives $6a^2 - 150a + 5 = 0$
Using the quadratic formula gives $a = \frac{150 \pm \sqrt {22380}}{12}$ . Since this looks too hard to simplify, we can solve for $b$ using $a + b = 25$ , which turns out to also be $b = \frac{150 \pm \sqrt {22380}}{12}$ , provided that the sign of the radical in $a$ is opposite the one in $b$
WLOG, assume $a = \frac{150 + \sqrt{22380}}{12}$ and $b = \frac{150 - \sqrt{22380}}{12}$ . Multiplying them gives $ab = \frac{22500 - 22380}{144}$ which simplifies to $\frac{5}{6}$
THe denominator of $\frac{25}{1 - ab}$ ends up being $\frac{1}{6}$ , so multiplying both numerator and denominator by 6 gives $\boxed{150}$ | null | 150 |
f77c062186b8908c402f8c9f4a119324 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_4 | Determine $3x_4+2x_5$ if $x_1$ $x_2$ $x_3$ $x_4$ , and $x_5$ satisfy the system of equations below. | Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ . Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$ , so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$ | null | 181 |
f77c062186b8908c402f8c9f4a119324 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_4 | Determine $3x_4+2x_5$ if $x_1$ $x_2$ $x_3$ $x_4$ , and $x_5$ satisfy the system of equations below. | Subtracting the first equation from every one of the other equations yields \begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*} Thus \begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*} Using the previous equations, \[3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}\] | null | 181 |
49c970caef8d32039a120f01174c1a17 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_6 | The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$ . When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$ . What was the number of the page that was added twice? | Denote the page number as $x$ , with $x < n$ . The sum formula shows that $\frac{n(n + 1)}{2} + x = 1986$ . Since $x$ cannot be very large, disregard it for now and solve $\frac{n(n+1)}{2} = 1986$ . The positive root for $n \approx \sqrt{3972} \approx 63$ . Quickly testing, we find that $63$ is too large, but if we plug in $62$ we find that our answer is $\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}$ | null | 033 |
49c970caef8d32039a120f01174c1a17 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_6 | The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$ . When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$ . What was the number of the page that was added twice? | Use the same method as above where you represent the sum of integers from $1$ to $n$ expressed as $\frac{n(n + 1)}{2}$ , plus the additional page number $k$ . We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.)
$\frac{n(n + 1)}{2} = 1986$ is the quadratic you must solve to obtain the upper bound of $n$ and $\frac{n(n + 1)}{2} + n = 1986$ is the quadratic you must solve to obtain the lower bound of $n$
Solving the two equations gives values that are respectively around $62.5$ and $61.5$ with the quadratic formula, and the only integer between the two is $62$
This implies that we can plug in $62$ and come to the same conclusion as the above solution where $x = \boxed{033}$ | null | 033 |
06eff45200e3aeb927bbd46e913e56ba | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal to $64 + 32 + 4$ , so in binary form we get $1100100$ . However, we must change it back to base 10 for the answer, which is $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed{981}$ | null | 981 |
06eff45200e3aeb927bbd46e913e56ba | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | Notice that the first term of the sequence is $1$ , the second is $3$ , the fourth is $9$ , and so on. Thus the $64th$ term of the sequence is $729$ . Now out of $64$ terms which are of the form $729$ $'''S'''$ $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$ , i.e. $972$ , is greater than the largest term which does not, or $854$ . So the $96$ th term will be $972$ , then $973$ , then $975$ , then $976$ , and finally $\boxed{981}$ | null | 981 |
06eff45200e3aeb927bbd46e913e56ba | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | After the $n$ th power of 3 in the sequence, the number of terms after that power but before the $(n+1)$ th power of 3 is equal to the number of terms before the $n$ th power, because those terms after the $n$ th power are just the $n$ th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of $3$ and the terms that come after them, we see that the $100$ th term is after $729$ , which is the $64$ th term. Also, note that the $k$ th term after the $n$ th power of 3 is equal to the power plus the $k$ th term in the entire sequence. Thus, the $100$ th term is $729$ plus the $36$ th term. Using the same logic, the $36$ th term is $243$ plus the $4$ th term, $9$ . We now have $729+243+9=\boxed{981}$ | null | 981 |
06eff45200e3aeb927bbd46e913e56ba | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the $2^{nth}$ term is equal to $3^n$ . From here, we can ballpark the range of the 100th term. The 64th term is $3^6$ $729$ and the 128th term is $3^7$ $2187$ . Writing out more terms of the sequence until the next power of 3 again (81) we can see that the ( $2^n$ $2^{n+1}$ )/2 term is equal to $3^n$ $3^{n-1}$ . From here, we know that the 96th term is $3^6$ $3^5$ $972$ . From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is $972 + 1 = 973$ , the 98th term is $972 + 3 = 975$ , the 99th term is $972 + 3 + 1 = 976$ , and finally the 100th term is $972 + 9 = \boxed{981}$ | null | 981 |
06eff45200e3aeb927bbd46e913e56ba | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | The number of terms $3^n$ produces includes each power of 3 ( $1, 3^1, ..., 3^n$ ), the sums of two power of 3s(ex. $3^1 + 1$ ), three power of 3s (ex. $3^1 + 1 + 3^n$ ), all the way to the sum of them all. Since there are $n+1$ powers of 3, the one number sum gives us ${n+1\choose 1}$ terms, the two number ${n+1\choose 2}$ terms, all the way to the sum of all the powers which gives us ${n+1\choose n+1}$ terms. Summing all these up gives us $2^{n+1} - 1 ^ {*}$ according to the theorem
Since $2^6$ is the greatest power $<100$ , then $n=5$ and the sequence would look like { $3^0, ..., 3^5$ }, where $3^5$ or $243$ would be the $2^5 - 1 = 63$ rd number. The next largest power $729$ would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces.
Noting that $729$ plus any combination of lower powers ${1, 3^1 . . .3^4}$ is < $729 + 243$ , so we can add all those terms( $2^5 - 1 = 31$ ) into our sequence:
Our sequence now has $63 + 1 + 31 = 95$ terms. The remaining $5$ would just be the smallest sums starting with $729 + 243$ or $972$
Hence the 100th term would be $972 + 9 = \boxed{981}$ . ~SoilMilk | null | 981 |
bf69d697afa9ff4b480df2cc02f2982f | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | The prime factorization of $1000000 = 2^65^6$ , so there are $(6 + 1)(6 + 1) = 49$ divisors, of which $48$ are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers.
Writing out the first few terms, we see that the answer is equal to \[\log 1 + \log 2 + \log 4 + \log 5 +\ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots (2^65^6).\] Each power of $2$ appears $7$ times; and the same goes for $5$ . So the overall power of $2$ and $5$ is $7(1+2+3+4+5+6) = 7 \cdot 21 = 147$ . However, since the question asks for proper divisors, we exclude $2^65^6$ , so each power is actually $141$ times. The answer is thus $S = \log 2^{141}5^{141} = \log 10^{141} = \boxed{141}$ | null | 141 |
bf69d697afa9ff4b480df2cc02f2982f | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | The formula for the product of the divisors of $n$ is $n^{(d(n))/2}$ , where $d(n)$ is the number of divisor of $n$ .
We know that $\log_{10} a + \log_{10} b + \log_{10} c + \log_{10} d...$ and so on equals $\log_{10} (abcd...)$ by sum-product rule of logs, so the problem is reduced to finding the logarithm base 10 of the product of the proper divisors of $10^6$ . The product of the divisors, by the earlier formula, is ${(10^6)}^{49/2} = 10^{49*3}$ , and since we need the product of only the proper divisors, which means the divisors NOT including the number, $10^6$ , itself, we divide $10^{(49*3)}$ by $10^6$ to get $10^{(49*3-6)} = 10^{(141)}$ . The base-10 logarithm of this value, in base 10, is clearly $\boxed{141}$ | null | 141 |
bf69d697afa9ff4b480df2cc02f2982f | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | Since the prime factorization of $10^6$ is $2^6 \cdot 5^6$ , the number of factors in $10^6$ is $7 \cdot 7=49$ . You can pair them up into groups of two so each group multiplies to $10^6$ . Note that $\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6$ . Thus, the sum of the logs of the divisors is half the number of divisors of $10^6 \cdot 6 -6$ (since they are asking only for proper divisors), and the answer is $(49/2)\cdot 6-6=\boxed{141}$ | null | 141 |
bf69d697afa9ff4b480df2cc02f2982f | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | Note that we can just pair terms up such that the product is $10^{6}.$ Now, however, note that $10^{3}$ is not included. Therefore we first exclude. We have $\displaystyle\frac{49-1}{2} = 24$ pairs that all multiply to $10^{6}.$ Now we include $10^{3}$ so our current product is $24 \cdot 6 - 3.$ However we dont want to include $10^6$ since we are considering proper factors only so the final answer is $144 + 3 - 6 = \boxed{141}.$ | null | 141 |
73ae0f7d7212ea3239750f5f64e62ce1 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar $\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$ ). The remaining three sections are parallelograms
By similar triangles, $BE'=\frac{d}{510}\cdot450=\frac{15}{17}d$ and $EC=\frac{d}{425}\cdot450=\frac{18}{17}d$ . Since $FD'=BC-EE'$ , we have $900-\frac{33}{17}d=d$ , so $d=\boxed{306}$ | null | 306 |
73ae0f7d7212ea3239750f5f64e62ce1 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar $\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$ ). The remaining three sections are parallelograms
Since $PDAF'$ is a parallelogram, we find $PF' = AD$ , and similarly $PE = BD'$ . So $d = PF' + PE = AD + BD' = 425 - DD'$ . Thus $DD' = 425 - d$ . By the same logic, $EE' = 450 - d$
Since $\triangle DPD' \sim \triangle ABC$ , we have the proportion
Doing the same with $\triangle PEE'$ , we find that $PE' =510 - \frac{17}{15}d$ . Now, $d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}$ | null | 306 |
73ae0f7d7212ea3239750f5f64e62ce1 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Refer to the diagram in solution 2; let $a^2=[E'EP]$ $b^2=[D'DP]$ , and $c^2=[F'FP]$ . Now, note that $[E'BD]$ $[D'DP]$ , and $[E'EP]$ are similar, so through some similarities we find that $\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2$ . Similarly, we find that $[D'AF]=(b+c)^2$ and $[F'CE]=(c+a)^2$ , so $[ABC]=(a+b+c)^2$ . Now, again from similarity, it follows that $\frac{d}{510}=\frac{a+b}{a+b+c}$ $\frac{d}{450}=\frac{b+c}{a+b+c}$ , and $\frac{d}{425}=\frac{c+a}{a+b+c}$ , so adding these together, simplifying, and solving gives $d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}$ $=\frac{10}{\frac{10}{306}}=\boxed{306}$ | null | 306 |
73ae0f7d7212ea3239750f5f64e62ce1 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Refer to the diagram above. Notice that because $CE'PF$ $AF'PD$ , and $BD'PE$ are parallelograms, $\overline{DD'} = 425-d$ $\overline{EE'} = 450-d$ , and $\overline{FF'} = 510-d$
Let $F'P = x$ . Then, because $\triangle ABC \sim \triangle F'PF$ $\frac{AB}{AC}=\frac{F'P}{F'F}$ , so $\frac{425}{510}=\frac{x}{510-d}$ . Simplifying the LHS and cross-multiplying, we have $6x=2550-5d$ . From the same triangles, we can find that $FP=\frac{18}{17}x$
$\triangle PEE'$ is also similar to $\triangle F'PF$ . Since $EF'=d$ $EP=d-x$ . We now have $\frac{PE}{EE'}=\frac{F'P}{FP}$ , and $\frac{d-x}{450-d}=\frac{17}{18}$ . Cross multiplying, we have $18d-18x=450 \cdot 17-17d$ . Using the previous equation to substitute for $x$ , we have: \[18d-3\cdot2550+15d=450\cdot17-17d\] This is a linear equation in one variable, and we can solve to get $d=\boxed{306}$ | null | 306 |
fda83f07379c6dfa25917fe6d19ff2bf | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$ | Let $m$ be the number $100a+10b+c$ . Observe that $3194+m=222(a+b+c)$ so
\[m\equiv -3194\equiv -86\equiv 136\pmod{222}\]
This reduces $m$ to one of $136, 358, 580, 802$ . But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$ .
Recall that $a, b, c$ refer to the digits the three digit number $(abc)$ , so of the four options, only $m = \boxed{358}$ satisfies this inequality. | null | 358 |
fda83f07379c6dfa25917fe6d19ff2bf | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$ | As in Solution 1, $3194 + m \equiv 222(a+b+c) \pmod{222}$ , and so as above we get $m \equiv 136 \pmod{222}$ .
We can also take this equation modulo $9$ ; note that $m \equiv a+b+c \pmod{9}$ , so
\[3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.\]
Therefore $m$ is $7$ mod $9$ and $136$ mod $222$ . There is a shared factor in $3$ in both, but the Chinese Remainder Theorem still tells us the value of $m$ mod $666$ , namely $m \equiv 358$ mod $666$ . We see that there are no other 3-digit integers that are $358$ mod $666$ , so $m = \boxed{358}$ | null | 358 |
fda83f07379c6dfa25917fe6d19ff2bf | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$ | Let $n=abc$ then \[N=222(a+b+c)-n\] \[N=222(a+b+c)-100a-10b-c=3194\] Since $0<100a+10b+c<1000$ , we get the inequality \[N<222(a+b+c)<N+1000\] \[3194<222(a+b+c)<4194\] \[14<a+b+c<19\] Checking each of the multiples of $222$ from $15\cdot222$ to $18\cdot222$ by subtracting $N$ from each $222(a+b+c)$ , we quickly find $n=\boxed{358}$ | null | 358 |
fda83f07379c6dfa25917fe6d19ff2bf | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$ | The sum of the five numbers is $222(a+b+c)-100a-10b-c=122a+212b+221c=122(a+b+c)+9(10b+11c)=3194$ We can see that $3194 \equiv 8$ （mod $9$ ） and $122 \equiv 5$ （mod $9$ ） so we need to make sure that $a+b+c \equiv 7$ （mod $9$ ） by some testing. So we let $a+b+c=9k+7$
Then, we know that $1\leq a+b+c \leq 27$ so only $7,16,25$ lie in the interval
When we test $a+b+c=25, 10b+11c=16$ , impossible
When we test $a+b+c=16, 10b+11c=138, b=5,c=8,a=3$
When we test $a+b+c=7, 10b+11c=260$ , well, it's impossible
The answer is $\boxed{358}$ then | null | 358 |
c0965ebdc2ff3d68aeac56ad17eb5967 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the numerator by $1$ ). By the Binomial Theorem , this is $(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}$ | null | 816 |
c0965ebdc2ff3d68aeac56ad17eb5967 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Again, notice $x = y - 1$ . So
\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.
We want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem is ${2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}$ . The Hockey Stick Identity tells us that this quantity is equal to ${18\choose 3} = \boxed{816}$ | null | 816 |
c0965ebdc2ff3d68aeac56ad17eb5967 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Again, notice $x=y-1$ . Substituting $y-1$ for $x$ in $f(x)$ gives: \begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}. From binomial theorem, the coefficient of the $y^2$ term is ${2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}$ . This is actually the sum of the first 16 triangular numbers, which evaluates to $\frac{(16)(17)(18)}{6} = \boxed{816}$ | null | 816 |
c0965ebdc2ff3d68aeac56ad17eb5967 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Let $f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ and $g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$
Then, since $f(x)=g(y)$ \[\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}\] $\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}$ by the power rule.
Similarly, $\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\cdots+a_{17}(17\cdot 16y^{15})$
Now, notice that if $x = -1$ , then $y = 0$ , so $f^{''}(-1) = g^{''}(0)$
$g^{''}(0)= 2a_2$ , and $f^{''}(-1) = 2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17$
Now, we can use the hockey stick theorem to see that $2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}$
Thus, $2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}$ | null | 816 |
c0965ebdc2ff3d68aeac56ad17eb5967 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Let $V$ be the vector space of polynomials of degree $\leq 17,$ and let $B = \{1, x, x^2, ..., x^{17} \}$ and $C = \{1, (x+1), (x+1)^2, ..., (x+1)^{17} \}$ be two bases for $V$ .
Let $\vec{v} \in V$ be the polynomial given in the problem, and it is easy to see that $[ \vec{v} ]_B = \langle 1, -1, 1, -1, ... , 1, -1 \rangle.$
Note that the transformation matrix from $C$ to $B$ can be easily found to be $P_{C \to B} = [ [\vec{c_1}]_B [\vec{c_2}]_B ... [\vec{c_3}]_B ] = \begin{bmatrix} \tbinom{0}{0} & \tbinom{1}{0} & \tbinom{2}{0} & \cdots & \tbinom{17}{0} \\ 0 & \tbinom{1}{1} & \tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & \tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$
I claim that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} ,$ where the term $\dbinom{n}{k}$ is negated if $n+k$ is odd.
One can prove that the $i$ th row of $P_{C \to B}$ dotted with the $j$ th column of $P_{B \to C}$ is $\delta_{i, j}$ by using combinatorial identities, which is left as an exercise for the reader. Thus, since the two matrices multiply to form $\mathbb{I}_{18},$ we have proved that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$
To find the coordinates of $\vec{v}$ under basis $C,$ we compute the product $[ \vec{v} ]_C = P_{B \to C} [\vec{v} ]_B = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 1 \\ \vdots \\ -1 \end{bmatrix} = \begin{bmatrix} \sum_{n=0}^{17} \tbinom{n}{0} \\ -\sum_{n=1}^{17} \tbinom{n}{1} \\ \sum_{n=2}^{17} \tbinom{n}{2} \\ \vdots \\ -\sum_{n=17}^{17} \tbinom{n}{17} \end{bmatrix} = \begin{bmatrix} \tbinom{18}{1} \\ - \tbinom{18}{2} \\ \tbinom{18}{3} \\ \vdots \\ -\tbinom{18}{18} \end{bmatrix},$ where the last equality was obtained via Hockey Stick Identity.
Thus, our answer is $a_2 = [ [ \vec{v} ]_C ]_3 = \dbinom{18}{3} = \boxed{816}.$ | null | 816 |
3020bf5a55c9b91bd41325ea84d70a51 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_13 | In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH HH , and etc. For example, in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH , three HT , four TH , and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH , three HT , four TH , and five TT subsequences? | Let's consider each of the sequences of two coin tosses as an operation instead; this operation takes a string and adds the next coin toss on (eg, THHTH HT THHTHT ). We examine what happens to the last coin toss. Adding HH or TT is simply an identity for the last coin toss, so we will ignore them for now. However, adding HT or TH switches the last coin. switches to three times, but switches to four times; hence it follows that our string will have a structure of THTHTHTH
Now we have to count all of the different ways we can add the identities back in. There are 5 TT subsequences, which means that we have to add 5 into the strings, as long as the new s are adjacent to existing s. There are already 4 s in the sequence, and since order doesn’t matter between different tail flips this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives ${{5+3}\choose3} = 56$ combinations. We do the same with 2 s to get ${{2+3}\choose3} = 10$ combinations; thus there are $56 \cdot 10 = \boxed{560}$ possible sequences. | null | 560 |
09be60a5324c8de43f19c8d0701fb0d2 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_14 | The shortest distances between an interior diagonal of a rectangular parallelepiped $P$ , and the edges it does not meet are $2\sqrt{5}$ $\frac{30}{\sqrt{13}}$ , and $\frac{15}{\sqrt{10}}$ . Determine the volume of $P$ | In the above diagram, we focus on the line that appears closest and is parallel to $BC$ . All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$ , the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and reach out to $AC$ , and have the same lengths as their corresponding blue lines. So we want to find the shortest distance between $AC$ and that corner, which is $\frac {wl}{\sqrt {w^2 + l^2}}$
So we have: \[\frac {lw}{\sqrt {l^2 + w^2}} = \frac {10}{\sqrt {5}}\] \[\frac {hw}{\sqrt {h^2 + w^2}} = \frac {30}{\sqrt {13}}\] \[\frac {hl}{\sqrt {h^2 + l^2}} = \frac {15}{\sqrt {10}}\]
Notice the familiar roots: $\sqrt {5}$ $\sqrt {13}$ $\sqrt {10}$ , which are $\sqrt {1^2 + 2^2}$ $\sqrt {2^2 + 3^2}$ $\sqrt {1^2 + 3^2}$ , respectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.)
\[\frac {l^2w^2}{l^2 + w^2} = \frac {1}{\frac {1}{l^2} + \frac {1}{w^2}} = 20\] \[\frac {h^2w^2}{h^2 + w^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{w^2}} = \frac {900}{13}\] \[\frac {h^2l^2}{h^2 + l^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{l^2}} = \frac {45}{2}\]
We invert the above equations to get a system of linear equations in $\frac {1}{h^2}$ $\frac {1}{l^2}$ , and $\frac {1}{w^2}$ \[\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}\] \[\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}\] \[\frac {1}{h^2} + \frac {1}{l^2} = \frac {40}{900}\]
We see that $h = 15$ $l = 5$ $w = 10$ . Therefore $V = 5 \cdot 10 \cdot 15 = \boxed{750}$ | null | 750 |
2411fcbb4d7054fcf6b06dd2d53f4b96 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_15 | Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$ . Given that the length of the hypotenuse $AB$ is $60$ , and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$ | We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$ [asy] import graph; size(150); Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 2.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return (x + 3); } real g( real x){ return (2x + 4); } draw(graph(f,-8,5),red+linewidth(1)); draw(graph(g,-6,2),blue+linewidth(1)); [/asy] Let the acute angle the red line makes with the $x-$ axis be $\alpha$ and the acute angle the blue line makes with the $x-$ axis be $\beta$ . Then, we know that $\tan \alpha = 1$ and $\tan \beta = 2$ . Note that the acute angle between the red and blue lines is clearly $\beta - \alpha$ . Therefore, we have that: \[\tan (\beta - \alpha) = \frac{2 - 1}{1 + 2} = \frac{1}{3}\] It follows that $\cos (\beta - \alpha) = \frac{1}{\sqrt{10}}$ and $\sin (\beta - \alpha) = \frac{3}{\sqrt{10}}$ . From now on, refer to $\theta = \beta - \alpha$ [asy] import graph; size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.96, xmax = 24.22, ymin = -10.06, ymax = 14.62; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw(arc((2,2.6666666666666665),0.6,290.55604521958344,326.3099324740202)--(2,2.6666666666666665)--cycle, linewidth(1)); draw((0,8)--(0,0), linewidth(1) + wrwrwr); draw((0,0)--(6,0), linewidth(1) + wrwrwr); draw((6,0)--(0,8), linewidth(1) + wrwrwr); draw((0,8)--(3,0), linewidth(1) + wrwrwr); draw((0,4)--(6,0), linewidth(1) + wrwrwr); dot((0,0),dotstyle); label("$A$", (0.08,0.2), NE * labelscalefactor); dot((6,0),dotstyle); label("$B$", (6.08,0.2), NE * labelscalefactor); dot((0,8),dotstyle); label("$C$", (0.08,8.2), NE * labelscalefactor); dot((3,0),linewidth(4pt) + dotstyle); label("$D$", (3.08,0.16), NE * labelscalefactor); dot((0,4),linewidth(4pt) + dotstyle); label("$E$", (0.08,4.16), NE * labelscalefactor); label("theta ", (2.14,2.3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] Suppose $ABC$ is our desired triangle. Let $E$ be the midpoint of $CA$ and $D$ be the midpoint of $AB$ such that $CE = EA = m$ and $AD = DB = n$ . Let $G$ be the centroid of the triangle (in other words, the intersection of $EB$ and $CD$ ). It follows that if $CG = 2x$ $GD = x$ and if $GB = 2y$ $GE = y$ . By the Law of Cosines on $\triangle GEB$ , we get that: \[GE^2 + GC^2 - 2GE \cdot GC \cdot \cos \theta = CE^2 \Longleftrightarrow\] \[(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2\] By the Law of Cosines on $\triangle GDB$ , we get that: \[DG^2 + GB^2 -2 DG \cdot GB \cdot \cos \theta = DB^2 \Longleftrightarrow\] \[(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2\] Adding yields that: \[5x^2+5y^2 - 8xy\cos \theta = m^2 + n^2\] However, note that $(2m)^2 + (2n)^2 = 60^2 \Longleftrightarrow m^2+n^2 = 30^2$ . Therefore, \[5x^2+5y^2 - 8xy\cos \theta = 30^2\] We also know that by the Law of Cosines on $\triangle CGB$ \[CG^2 + GB^2 - 2CG \cdot GB \cdot \cos (180 - \theta) = CB^2 \Longleftrightarrow\] \[(2x)^2 + (2y)^2 + 2(2x)(2y) \cos \theta = 60^2 \Longleftrightarrow\] \[x^2+y^2+2xy \cos \theta = 30^2 \Longleftrightarrow\] \[5x^2+5y^2 + 10xy \cos \theta = 30^2 \cdot 5\] Subtracting this from the $5x^2+5y^2 - 8xy\cos \theta = 30^2$ we got earlier yields that: \[xy \cos \theta = 200\] but recall that $\cos \theta = \frac{3}{\sqrt{10}}$ to get that: \[xy = \frac{200 \sqrt{10}}{3}\] Plugging this into the $x^2 + y^2 + 2xy \cos \theta = 30^2$ , we get that: \[x^2 + y^2 = 500\] Aha! How convenient. Recall that $(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2$ and $(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2$ . Then, we clearly have that: \[m^2n^2 = ((2x)^2 + y^2 - (2x)(y)(2)\cos \theta)((2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2) \Longleftrightarrow\] \[m^2n^2 = 17x^2y^2 - 4xy \cos \theta (5x^2 + 5y^2) + 16x^2y^2 \cos ^2 \theta + 4x^4 + 4y^4 \Longleftrightarrow\] \[m^2n^2 = 17 \cdot \frac{200^2 \cdot 10}{9} - 800(5 \cdot 500) + 16 \cdot 200^2 + 4 \left( (x^2+y^2)^2-2x^2y^2\right) \Longleftrightarrow\] \[\frac{m^2n^2}{100^2} = \frac{680}{9} - 8 \cdot 5 \cdot 5 + 16 \cdot 2^2 + 4 \left( 5^2 - 2 \cdot \frac{2^2 \cdot 10}{9}\right) \Longleftrightarrow\] \[\frac{m^2n^2}{100^2} = 4\] But note that the area of our triangle is $2m \cdot 2n \cdot \frac {1}{2} = 2mn$ . As $mn = 200$ , we get a final answer of $\boxed{400}$ | null | 400 |
4791c3bb3ce545b6ef04cd88b1e05fbc | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_1 | Let $x_1=97$ , and for $n>1$ , let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$ | Since $x_n=\frac{n}{x_{n-1}}$ $x_n \cdot x_{n - 1} = n$ . Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$ $x_3x_4 = 4$ $x_5x_6 = 6$ and $x_7x_8 = 8$ so \[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.\] Notice that the value of $x_1$ was completely unneeded! | null | 384 |
4791c3bb3ce545b6ef04cd88b1e05fbc | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_1 | Let $x_1=97$ , and for $n>1$ , let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$ | Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:
\[x_1x_2x_3x_4x_5x_6x_7x_8=x_1\cdot\dfrac{2}{x_1}\cdot\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}\cdot\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}\cdot\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}}\] \[=\left (x_1\cdot\dfrac{2}{x_1} \right )\cdot \left (\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}} \right )\cdot \left (\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}} \right )\cdot \left (\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}} \right )\] \[=(2)\cdot (4)\cdot (6)\cdot (8)=\boxed{384}\] | null | 384 |
0587a4b394c9d2fa3ff4acc8b4b2d06c | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_2 | When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle? | Let one leg of the triangle have length $a$ and let the other leg have length $b$ . When we rotate around the leg of length $a$ , the result is a cone of height $a$ and radius $b$ , and so of volume $\frac 13 \pi ab^2 = 800\pi$ . Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and radius $a$ and so of volume $\frac13 \pi b a^2 = 1920 \pi$ . If we divide this equation by the previous one, we get $\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}$ , so $a = \frac{12}{5}b$ . Then $\frac{1}{3} \pi \left(\frac{12}{5}b\right)b^2 = 800\pi$ so $b^3 = 1000$ and $b = 10$ so $a = 24$ . Then by the Pythagorean Theorem , the hypotenuse has length $\sqrt{a^2 + b^2} = \boxed{026}$ | null | 026 |
0587a4b394c9d2fa3ff4acc8b4b2d06c | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_2 | When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle? | Let $a$ $b$ be the $2$ legs, we have the $2$ equations \[\frac{a^2b\pi}{3}=800\pi,\frac{ab^2\pi}{3}=1920\pi\] Thus $a^2b=2400, ab^2=5760$ . Multiplying gets \begin{align*} (a^2b)(ab^2)&=2400\cdot5760\\ (ab)^3&=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5)\\ ab&=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \end{align*} Adding gets \begin{align*} a^2b+ab^2=ab(a+b)&=2400+5760\\ 240(a+b)&=240\cdot(10+24)\\ a+b&=34 \end{align*} Let $h$ be the hypotenuse then \begin{align*} h&=\sqrt{a^2+b^2}\\ &=\sqrt{(a+b)^2-2ab}\\ &=\sqrt{34^2-2\cdot240}\\ &=\sqrt{676}\\ &=\boxed{26} | null | 26 |
0587a4b394c9d2fa3ff4acc8b4b2d06c | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_2 | When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle? | Let $a$ and $b$ be the two legs of the equation. We can find $\frac{a}{b}$ by doing $\frac{1920\pi}{800\pi}$ . This simplified is $\frac{12}{5}$ . We can represent the two legs as $12x$ and $5x$ for $a$ and $b$ respectively.
Since the volume of the first cone is $800\pi$ , we use the formula for the volume of a cone and get $100\pi x^3=800 \pi$ . Solving for $x$ , we get $x=2$
Plugging in the side lengths to the Pythagorean Theorem, we get an answer of $\boxed{026}$ | null | 026 |
2146f8d5bb0df05fbf9c8b25b5f495e8 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_3 | Find $c$ if $a$ $b$ , and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$ , where $i^2 = -1$ | Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$ . Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$ . Since $a, b$ are integers , this means $b$ is a divisor of 107, which is a prime number . Thus either $b = 1$ or $b = 107$ . If $b = 107$ $3a^2 - 107^2 = 1$ so $3a^2 = 107^2 + 1$ , but $107^2 + 1$ is not divisible by 3, a contradiction. Thus we must have $b = 1$ $3a^2 = 108$ so $a^2 = 36$ and $a = 6$ (since we know $a$ is positive). Thus $c = 6^3 - 3\cdot 6 = \boxed{198}$ | null | 198 |
c16fb0e4349ade90096f16df1c945f10 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_4 | A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png | The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram . Using the smaller side of the parallelogram, $1/n$ , as the base, where the height is 1, we find that the area of the parallelogram is $A = \frac{1}{n}$ . By the Pythagorean Theorem , the longer base of the parallelogram has length $l = \sqrt{1^2 + \left(\frac{n - 1}{n}\right)^2} = \frac{1}{n}\sqrt{2n^2 - 2n + 1}$ , so the parallelogram has height $h = \frac{A}{l} = \frac{1}{\sqrt{2n^2 - 2n + 1}}$ . But the height of the parallelogram is the side of the little square, so $2n^2 - 2n + 1 = 1985$ . Solving this quadratic equation gives $n = \boxed{32}$ | null | 32 |
c16fb0e4349ade90096f16df1c945f10 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_4 | A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png | Aime.png
Surrounding the square with area $\frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + \frac{1}{1985} = 1$ , where $X$ is the area of the of the 4 triangles.
We can thus use proportions to solve this problem. \begin{eqnarray*} \frac{GF}{BE}=\frac{CG}{CB}\implies \frac{\frac{1}{\sqrt{1985}}}{BE}=\frac{\frac{1}{n}}{1}\implies BE=\frac{n\sqrt{1985}}{1985} \end{eqnarray*} Also, \begin{eqnarray*} \frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies EC=\frac{\sqrt{1985}}{1985}(n-1) \end{eqnarray*} Thus, \begin{eqnarray*} 2(BE)(EC)+\frac{1}{1985}=1\\ 2n^{2}-2n+1=1985\\ n(n-1)=992 \end{eqnarray*} Simple factorization and guess and check gives us $\boxed{32}$ | null | 32 |
c16fb0e4349ade90096f16df1c945f10 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_4 | A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png | AIME 1985 Problem 4 Solution 3 Diagram.png
Line Segment $DE = \frac{1}{n}$ , so $EC = 1 - \frac{1}{n} = \frac{n-1}{n}$ . Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\frac{1}{\sqrt{1985}}$ , as it is the same length as the sides of the square. Notice that $\triangle CEL$ is similar to $\triangle HDE$ by $AA$ similarity. Thus, $\frac{LC}{HE} = \frac{EC}{DE} = n-1$ , so $LC = \frac{n-1}{\sqrt{1985}}$ . Notice that $\triangle CEL$ is also similar to $\triangle CDF$ by $AA$ similarity. Thus, $\frac{FC}{EC} = \frac{DC}{LC}$ , and the expression simplifies into a quadratic equation $n^2 - n - 992 = 0$ . Solving this quadratic equation yields $n =\boxed{32}$ | null | 32 |
7364d57ce803ce5bf3f00443d1720a75 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_5 | sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$ . What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492? | The problem gives us a sequence defined by a recursion , so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$ . Then $a_3 = b - a$ $a_4 = (b - a) - b = -a$ $a_5 = -a - (b - a) = -b$ $a_6 = -b - (-a) = a - b$ $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$ . Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular $a_{j + 6} = a_j$ for all $j$ , and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$
Because of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 \cdot 248$ is the largest multiple of 6 less than 1492, so \[\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}\] \[=(a + b + (b - a) + (-a)) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_j = 2b - a,\] where we can make this last step because $\sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero
Similarly, since $1980 = 6 \cdot 330$ $\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$
Finally, $\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$
Then by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2\cdot 493 = \boxed{986}$ .
Minor edit by: PlainOldNumberTheory | null | 986 |
c960c1f3877946a8ca5a4027b40eb185 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | Let the interior point be $P$ , let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively. Let $x$ be the area of $\triangle APE$ and $y$ be the area of $\triangle CPD$ . Note that $\triangle APF$ and $\triangle BPF$ share the same altitude from $P$ , so the ratio of their areas is the same as the ratio of their bases. Similarly, $\triangle ACF$ and $\triangle BCF$ share the same altitude from $C$ , so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have: $\frac{40}{30} = \frac{124 + x}{65 + y}$ or equivalently $372 + 3x = 260 + 4y$ and so $4y = 3x+ 112$
Applying identical reasoning to the triangles with bases $\overline{CD}$ and $\overline{BD}$ , we get $\frac{y}{35} = \frac{x+y+84}{105}$ so that $3y = x + y + 84$ and $2y = x + 84$ . Substituting from this equation into the previous one gives $x = 56$ , from which we get $y = 70$ and so the area of $\triangle ABC$ is $56 + 40 + 30 + 35 + 70 + 84 = \Rightarrow \boxed{315}$ | null | 315 |
c960c1f3877946a8ca5a4027b40eb185 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight $\times$ side) = (Other weight) $\times$ (The other side), the problem yields the answer $\boxed{315}$ | null | 315 |
c960c1f3877946a8ca5a4027b40eb185 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | Let the interior point be $P$ and let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively. Also, let $[APE]=x,[CPD]=y.$ Then notice that by Ceva's, $\frac{FB\cdot DC\cdot EA}{DB\cdot CE\cdot AF}=1.$ However, we can deduce $\frac{FB}{AF}=\frac{3}{4}$ from the fact that $[AFP]$ and $[BPF]$ share the same height. Similarly, $x=\frac{84CE}{EA}$ and $y=\frac{35DC}{BD}.$ Plug and chug and you get $xy=84\cdot 35\cdot \frac{3}{4}=2205.$ Then notice by the same height reasoning, $\frac{84}{x}=\frac{119+y}{x+70}.$ Clear the fractions and combine like terms to get $35x=5880-xy.$ We know $xy=2205$ so subtraction yields $35x=3675,$ or $x=105.$ Plugging this in to our previous ratio statement yields $\frac{84}{105}=\frac{4}{5}=\frac{119+y}{175},$ so $y=21.$ Basic addition gives us $105+84+21+35+30+40=\boxed{315}.$ | null | 315 |
c960c1f3877946a8ca5a4027b40eb185 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | Let the interior point be $P$ and let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively. Then the cevians $AD,BF,CE$ are concurrent, so we can use Ceva's Theorem, letting $\frac{BD}{DC}=\frac{a}{b}$ and $\frac{CF}{FA}=\frac{c}{d}$ . Notice that $\frac{AE}{EB}=\frac{[\Delta APE]}{[\Delta EPB]}=\frac43.$ \[\frac{4}{3}\cdot \frac{a}{b}\cdot \frac{c}{d}=1\implies \frac{d}{c}=\frac{a}{b}\cdot \frac{4}{3}.\]
We know that $[\Delta CPD]=35\cdot \frac ba$ and $[\Delta APF]=84\cdot \frac dc,$ so \[[\Delta ABC] = 84+84\cdot \frac dc + 35\cdot \frac ba + 40+30+35 = \left(1+\frac ba\right)(40+30+35).\] We will now solve for $\frac ba$
\[84+84\cdot \frac ab\cdot \frac 43 + 35\cdot \frac ba = \frac ba\cdot 105.\] \[12+16\cdot \frac ab + 5\cdot \frac ba = \frac ba\cdot 15\] \[10\left(\frac ba\right)^2-12\cdot \frac ba-16=0\] Factoring this gives $\left(\frac ba-2\right)\left(10\cdot \frac ba - 8\right)=0,$ so the area of $\triangle ABC$ is \[\left(1+\frac ba\right)(40+30+35)=3\cdot 105=\boxed{315.}\] | null | 315. |
c13b397ad1ff8ac49e101e705f31cef8 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_7 | Assume that $a$ $b$ $c$ , and $d$ are positive integers such that $a^5 = b^4$ $c^3 = d^2$ , and $c - a = 19$ . Determine $d - b$ | It follows from the givens that $a$ is a perfect fourth power $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube . Thus, there exist integers $s$ and $t$ such that $a = t^4$ $b = t^5$ $c = s^2$ and $d = s^3$ . So $s^2 - t^4 = 19$ . We can factor the left-hand side of this equation as a difference of two squares, $(s - t^2)(s + t^2) = 19$ . 19 is a prime number and $s + t^2 > s - t^2$ so we must have $s + t^2 = 19$ and $s - t^2 = 1$ . Then $s = 10, t = 3$ and so $d = s^3 = 1000$ $b = t^5 = 243$ and $d-b=\boxed{757}$ | null | 757 |
74d88254a73367e9aa806606c6bd766d | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_8 | The sum of the following seven numbers is exactly 19: $a_1 = 2.56$ $a_2 = 2.61$ $a_3 = 2.65$ $a_4 = 2.71$ $a_5 = 2.79$ $a_6 = 2.82$ $a_7 = 2.86$ . It is desired to replace each $a_i$ by an integer approximation $A_i$ $1\le i \le 7$ , so that the sum of the $A_i$ 's is also 19 and so that $M$ , the maximum of the "errors" $\| A_i-a_i\|$ , the maximum absolute value of the difference, is as small as possible. For this minimum $M$ , what is $100M$ | If any of the approximations $A_i$ is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the $A_i$ are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3. Then there must be five 3s and two 2s. It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the $a_i$ , so our approximations are $A_1 = A_2 = 2$ and $A_3 = A_4 = A_5 = A_6 = A_7 = 3$ and the largest error is $|A_2 - a_2| = 0.61$ , so the answer is $\boxed{061}$ | null | 061 |
dd4f351d85e9b712a07426905b750a2c | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_9 | In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator? | All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle
This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}$ . The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = \frac{abc}{4R}$ , so $\frac6R = \frac{3}{4}\sqrt{15}$ and $R = \frac8{\sqrt{15}}$
Now, consider the triangle formed by two radii and the chord of length 2. This isosceles triangle has vertex angle $\alpha$ , so by the Law of Cosines
\[2^2 = R^2 + R^2 - 2R^2\cos \alpha \Longrightarrow \cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}\] and the answer is $17 + 32 = \boxed{049}$ | null | 049 |
dd4f351d85e9b712a07426905b750a2c | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_9 | In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator? | It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is $\frac{\alpha}{2}$ , and using the Law of Cosines , we get: \[2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}\] Which, rearranges to: \[21 = 24\cos\frac{\alpha}{2}\] And, that gets us: \[\cos\frac{\alpha}{2} = 7/8\] Using $\cos 2\theta = 2\cos^2 \theta - 1$ , we get that: \[\cos\alpha = 17/32\] Which gives an answer of $\boxed{049}$ | null | 049 |
dd4f351d85e9b712a07426905b750a2c | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_9 | In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator? | Using the first diagram above, \[\sin \frac{\alpha}{2} = \frac{1}{r}\] \[\sin \frac{\beta}{2} = \frac{1.5}{r}\] \[\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}\] by the Pythagorean trig identities, \[\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}\] \[\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}\] so by the composite sine identity \[\frac{2}{r}=\frac{1}{r}\sqrt{1-\frac{2.25}{r^2}}+\frac{1.5}{r}\sqrt{1-\frac{1}{r^2}}\] multiply both sides by $2r$ , then subtract $\sqrt{4-\frac{9}{r^2}}$ from both sides
squaring both sides, we get \[16 - 8\sqrt{4-\frac{9}{r^2}} + 4 - \frac{9}{r^2}=9 - \frac{9}{r^2}\] \[\Longrightarrow 16+4=9+8\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{11}{8}=\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{121}{64}=4-\frac{9}{r^2}\] \[\Longrightarrow\frac{(256-121)r^2}{64}=9\Longrightarrow r^2= \frac{64}{15}\] plugging this back in, \[\cos^2(\frac{\alpha}{2})=1-\frac{15}{64}=\frac{49}{64}\] so \[\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}\] and the answer is $17+32=\boxed{049}$ | null | 049 |
4d936f930f55b511c5e7468c02d9e49f | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Noting that all of the numbers are even, we can reduce this to any real number $x$ between $0$ to $\frac 12$ , as this will be equivalent to $\frac n2$ to $\frac {n+1}2$ for any integer $n$ (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10.
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):
We can match up the greatest integer functions with one of the partitions of the integer. If we let $x = \frac 12$ then we get the solution $10$ ; now consider when $x < \frac 12$ $\lfloor 2x \rfloor = 0$ $\lfloor 4x \rfloor \le 1$ $\lfloor 6x \rfloor \le 2$ $\lfloor 8x \rfloor \le 3$ . But according to this the maximum we can get is $1+2+3 = 6$ , so we only need to try the first 6 numbers.
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers $1,2,4,5,6,10$ ; hence our solution is $6 \cdot 100 = \boxed{600}$ | null | 600 |
4d936f930f55b511c5e7468c02d9e49f | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | As we change the value of $x$ , the value of our expression changes only when $x$ crosses rational number of the form $\frac{m}{n}$ , where $n$ is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form $\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}$ . This gives us 24 calculations to make; we summarize the results here:
$\frac{1}{24}, \frac{2}{24} \to 0$
$\frac{3}{24} \to 1$
$\frac{4}{24}, \frac{5}{24} \to 2$
$\frac{6}{24}, \frac{7}{24} \to 4$
$\frac{8}{24} \to 5$
$\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6$
$\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10$
$\frac{15}{24} \to 11$
$\frac{16}{24},\frac{17}{24} \to 12$
$\frac{18}{24}, \frac{19}{24} \to 14$
$\frac{20}{24}\to 15$
$\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16$
$\frac{24}{24} \to 20$
Thus, we hit 12 of the first 20 integers and so we hit $50 \cdot 12 = \boxed{600}$ of the first $1000$ | null | 600 |
4d936f930f55b511c5e7468c02d9e49f | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Because $2,4,6,8$ are all multiples of $2$ , we can speed things up. We only need to check up to $\frac{12}{24}$ , and the rest should repeat. As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{12}{24}$ . Similarly, this should repeat 100 times, for $\boxed{600}$ | null | 600 |
4d936f930f55b511c5e7468c02d9e49f | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | We only need to check the numbers where it increments, namely $\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}$ . As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{1}{2}$ . Similarly, this should repeat 100 times, for $\boxed{600}$ | null | 600 |
4d936f930f55b511c5e7468c02d9e49f | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Recall from Hermite's Identity that $\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor$ . Then we can rewrite $\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor$ $+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor$ . There are $12$ terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from $20$ ). Starting from every integer $x$ , we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by $20$ while only achieving $12$ of those $20$ values. We can conveniently shift the $1000$ (since it can be achieved) to the position of the $0$ so that there are only complete cycles of $20$ , and the answer is $\frac {12}{20}\cdot1000 = \boxed{600}$ | null | 600 |
4d936f930f55b511c5e7468c02d9e49f | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Let $x=\lfloor x\rfloor+\{x\}$ then \begin{align*} \lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor&=\lfloor 2(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 4(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 6(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 8(\lfloor x\rfloor+\{x\})\rfloor\\ &=2\lfloor x\rfloor+4\lfloor x\rfloor+6\lfloor x\rfloor+8\lfloor x\rfloor+\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor\\ &=20\lfloor x\rfloor+(\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor) \end{align*} Similar to the previous solutions, the value of $\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor$ changes when $\{x\}=\frac{m}{n}$ , where $m\in\{1,2,3,...,n-1\}$ $n\in\{2,4,6,8\}$ . Using Euler's Totient Function \[\sum\limits_{k=0}^4 \phi(2k)\] to obtain $12$ different values for $\{x\}=\frac{m}{n}$ . (note that here Euler's Totient Function counts the number of $\{x\}=\frac{m}{n}$ where $m$ $n$ are relatively prime so that the values of $\{x\}$ won't overlap.).
Thus if $k$ can be expressed as $\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor$ , then $k=20a+b$ for some non-negative integers $a$ $b$ , where there are $12$ values for $b$
Exclusively, there are $49$ values for $a$ in the range $0<k<1000$ , or $49\cdot12=588$ ordered pairs $(a,b)$
If $a=0$ $b\neq0$ , which includes $11$ ordered pairs.
If $a=50$ $b=0$ , which includes $1$ ordered pair.
In total, there are $588+11+1=\boxed{600}$ values for $k$ | null | 600 |
4d936f930f55b511c5e7468c02d9e49f | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | To simplify the question, let $y = 2x$ . Then, the expression in the question becomes $\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$
Let $\{x\}$ represent the non-integer part of $x$ (For example, $\{2.8\} = 0.8$ ). Then,
\begin{align*} \lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor &= y - \{y\} + 2y - \{2y\} + 3y - \{3y\} + 4y - \{4y\} \\ &= 10y - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10(\lfloor y \rfloor + \{y\}) - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 10\{y\} - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) \\ \end{align*}
Since $\lfloor y \rfloor$ is always an integer, $10\lfloor y \rfloor$ will be a multiple of 10. Thus, we look for the range of the other part of the expression. We will be able to reach the same numbers when $y$ ranges from $0$ to $1$ , because the curly brackets ( $\{\}$ ) gets rid of any integer part. Let the combined integer part of $2y$ $3y$ , and $4y$ be $k$ (In other words, $k = \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$ ). Then,
\begin{align*} 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) &= 9\{y\} - (2\{y\} + 3\{y\} + 4\{y\} - k) \\ &= 9\{y\} - (9\{y\} - k) \\ &= k \end{align*}
The maximum value of $k$ will be when $y$ is slightly less than $1$ , which means $k = 1 + 2 + 3 = 6$ . As $y$ increases from $0$ to $1$ $k$ will increase whenever $2y$ $3y$ , or $4y$ is an integer, which happens when $y$ hits one of the numbers in the set $\left\{\dfrac14, \dfrac13, \dfrac12, \dfrac23, \dfrac34 \right\}$ . When $y$ reaches $\dfrac12$ , both $2y$ and $4y$ will be an integer, so $k$ will increase by $2$ . For all the other numbers in the set, $k$ increases by $1$ since only $1$ number in the set will be an integer. Thus, the possible values of $k$ are $\{0, 1, 2, 4, 5, 6\}$
Finally, looking back at the original expression, we plug in $k$ to get a multiple of $10$ plus any number in $\{0, 1, 2, 4, 5, 6\}$ . Thus, we hit all numbers ending in $\{0, 1, 2, 4, 5, 6\}$ , of which there are $\boxed{600}$ | null | 600 |
023ae51992b528990407683227644ac0 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_11 | An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$ -plane and is tangent to the $x$ -axis. What is the length of its major axis | An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant. Let $F_1 = (9, 20)$ $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$ -axis. Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is minimal. (The last claim begs justification: Let $F'_2$ be the reflection of $F_2$ across the $x$ -axis. Let $Y$ be where the line through $F_1$ and $F’_2$ intersects the ellipse. We will show that $X=Y$ . Note that $X F_2 = X F’_2$ since $X$ is on the $x$ -axis. Also, since the entire ellipse is on or above the $x$ -axis and the line through $F_2$ and $F’_2$ is perpendicular to the $x$ -axis, we must have $F_2 Y \leq F’_2 Y$ with equality if and only if $Y$ is on the $x$ -axis. Now, we have \[F_1 X + F'_2 X = F_1 X + F_2 X = F_1 Y + F_2 Y \leq F_1 Y + F’_2 Y\] But the right most sum is the straight-line distance from $F_1$ to $F’_2$ and the left is the distance of some path from $F_1$ to $F_2$ ., so this is only possible if we have equality and thus $X = Y$ ). Finding the optimal location for $X$ is a classic problem: for any path from $F_1$ to $X$ and then back to $F_2$ , we can reflect (as above) the second leg of this path (from $X$ to $F_2$ ) across the $x$ -axis. Then our path connects $F_1$ to the reflection $F_2'$ of $F_2$ via some point on the $x$ -axis, and this path will have shortest length exactly when our original path has shortest length. This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the $x-$ axis.
The sum of the two distances $F_1 X$ and $F_2X$ is therefore equal to the length of the segment $F_1F_2'$ , which by the distance formula is just $d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85$
Finally, let $A$ and $B$ be the two endpoints of the major axis of the ellipse. Then by symmetry $AF_1 = F_2B$ so $AB = AF_1 + F_1B = F_2B + F_1B = d$ (because $B$ is on the ellipse), so the answer is $\boxed{085}$ | null | 085 |
023ae51992b528990407683227644ac0 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_11 | An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$ -plane and is tangent to the $x$ -axis. What is the length of its major axis | An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The length of major axis is equal to the sum of these distances $(2a)$ . Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:
$k = \sqrt{(x - 9)^2 + 20^2} + \sqrt{(x - 49)^2 + 55^2}$
This is the equation of the ellipse expressed in terms of $x$ . The line tangent to the ellipse at the given point $P(x, 0)$ will thus have slope $0$ . Taking the derivative gives us the slope of this line. To simplify, let $f(x) = (x - 9)^2 + 20^2$ and $g(x) = (x - 49)^2 + 55^2$ . Then we get:
$0 = \frac{f^\prime(x)}{2\sqrt{f(x)}} + \frac{g^\prime(x)}{2\sqrt{g(x)}}$
Next, we multiply by the conjugate to remove square roots. We next move the resulting $a^2 - b^2$ form expression into form $a^2 = b^2$
$\frac{(f^\prime(x))^2}{4\cdot f(x)} = \frac{(g^\prime(x))^2}{4\cdot g(x)}$
We know $f^\prime(x) = 2x - 18$ and $g^\prime(x) = 2x - 98$ . Simplifying yields:
$\frac{(x - 9)^2}{(x - 9)^2 + 20^2} = \frac{(x - 49)^2}{(x - 49)^2 + 55^2}$
To further simplify, let $a = (x - 9)^2$ and $b = (x - 49)^2$ . This means $\frac{a}{a + 400} = \frac{b}{b + 3025}$ . Solving yields that $16b = 121a$ .
Substituting back $a$ and $b$ yields:
$16 \cdot (x - 49)^2 = 121 \cdot (x - 9)^2$
Solving for $x$ yields $x = \frac{59}{3}$ . Substituting back into our original distance formula, solving for $k$ yields $\boxed{085}$ | null | 085 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | We evaluate $P(7)$ recursively: \begin{alignat*}{6} P(0)&=1, \\ P(1)&=\frac13(1-P(0))&&=0, \\ P(2)&=\frac13(1-P(1))&&=\frac13, \\ P(3)&=\frac13(1-P(2))&&=\frac29, \\ P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\ P(5)&=\frac13(1-P(4))&&=\frac{20}{81}, \\ P(6)&=\frac13(1-P(5))&&=\frac{61}{243},\\ P(7)&=\frac13(1-P(6))&&=\frac{182}{729}. \end{alignat*} Therefore, the answer is $n=\boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | Let $P(k)=Q(k)+c$ for some function $Q(k)$ and constant $c.$ For all $k\geq1,$ the recursive formula for $P(k)$ becomes \[Q(k)+c=\frac13(1-(Q(k-1)+c))=\frac13-\frac13Q(k-1)-\frac13c.\] Solving for $Q(k),$ we get \[Q(k)=\frac13-\frac13Q(k-1)-\frac43c.\] For simplicity purposes, we set $c=\frac14,$ which gives \[Q(k)=-\frac13Q(k-1).\] Clearly, $Q(0),Q(1),Q(2),\ldots$ is a geometric sequence with the common ratio $\frac{Q(k)}{Q(k-1)}=-\frac13.$ Substituting $P(0)=1$ and $c=\frac14$ into $P(0)=Q(0)+c$ produces $Q(0)=\frac34,$ the first term of the geometric sequence.
For all nonnegative integers $k,$ the explicit formula for $Q(k)$ is \[Q(k)=\frac34\left(-\frac13\right)^k,\] and the explicit formula for $P(k)$ is \[P(k)=\frac34\left(-\frac13\right)^k+\frac14.\] Finally, the requested probability is $p=P(7)=\frac{182}{729},$ from which $n=\boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | Denominator
There are $3^7$ ways for the bug to make $7$ independent crawls without restrictions.
Numerator
Let $V_k$ denote the number of ways for the bug to crawl exactly $k$ meters starting from vertex $V$ and ending at vertex $A,$ where $V\in\{A,B,C,D\}$ and $k$ is a positive integer. We wish to find $A_7.$
Since the bug must crawl to vertex $B,C,$ or $D$ on the first move, we have \begin{align*} A_7&=B_6+C_6+D_6 \\ &=(A_5+C_5+D_5)+(A_5+B_5+D_5)+(A_5+B_5+C_5) \\ &=A_5+2(A_5+B_5+C_5+D_5) \\ &=A_5+2S_5, \end{align*} where $S_k=A_k+B_k+C_k+D_k.$
More generally, we get \[A_{k+2}=A_k+2S_k. \qquad\qquad (\spadesuit)\] For $S_k,$ note that
Clearly, $S_1,S_2,S_3,\ldots$ is a geometric sequence with the first term $S_1=3$ and the common ratio $\frac{S_{k+1}}{S_k}=3.$ Therefore, its explicit formula is \[S_k=3^k. \qquad\qquad (\clubsuit)\] Recall that $A_1=0.$ By $(\spadesuit)$ and $(\clubsuit),$ we rewrite $A_7$ recursively: \begin{align*} A_7 &= A_5+2S_5 \\ &=\left(A_3+2S_3\right)+2S_5 \\ &=\left(\left(A_1+2S_1\right)+2S_3\right)+2S_5 \\ &=2S_1+2S_3+2S_5 \\ &=2(3)+2\left(3^3\right)+2\left(3^5\right). \end{align*} Answer
The requested probability is \[p=\frac{A_7}{3^7}=\frac{2(3)+2\left(3^3\right)+2\left(3^5\right)}{3^7}=\frac{2(1)+2\left(3^2\right)+2\left(3^4\right)}{3^6}=\frac{182}{729},\] from which $n=\boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | Define notation $V_k$ as Solution 2 does.
In fact, we can generalize the following relationships for all nonnegative integers $k:$ \begin{align*} A_0&=1, \\ B_0&=0, \\ C_0&=0, \\ D_0&=0, \\ A_{k+1}&=B_k+C_k+D_k, \\ B_{k+1}&=A_k+C_k+D_k, \\ C_{k+1}&=A_k+B_k+D_k, \\ D_{k+1}&=A_k+B_k+C_k. \\ \end{align*} Using these equations, we recursively fill out the table below: \[\begin{array}{c||c|c|c|c|c|c|c|c} \hspace{7mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&& \\ [-2.5ex] \boldsymbol{k} & \boldsymbol{0} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} \\ \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{A_k} &1&0&3&6&21&60&183&546 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{B_k} &0&1&2&7&20&61&182&547 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{C_k} &0&1&2&7&20&61&182&547 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{D_k} &0&1&2&7&20&61&182&547 \\ \end{array}\] Note that the paths from $V$ to $A$ and the paths from $A$ to $V$ have one-to-one correspondence. So, we must get \[A_k+B_k+C_k+D_k=3^k\] for all values of $k.$
The requested probability is \[p=\frac{A_7}{3^7}=\frac{546}{2187}=\frac{182}{729},\] from which $n=\boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | Let $a_n$ denotes the number of ways that the bug arrives at $A$ after crawling $n$ meters, then we have $a_1=0$
Notice that there is respectively $1$ way to arrive at $A$ for each of the different routes after the previous $n-1$ crawls, excluding the possibility that the bug ends up at $A$ after the $(n-1)$ th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation \[a_n=3^{n-1}-a_{n-1}.\] Quick calculations yield \begin{align*} a_7&=3^6-a_6\\ &=3^6-\left(3^5-3^4+3^3-3^2+3-a_1\right)\\ &=546. \end{align*} Thus, $p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | Let $A(n)$ be the probability the bug lands on vertex $A$ after crawling $n$ meters, $B(n)$ be the probability the bug lands on vertex $B$ after crawling $n$ meters, and etc.
Note that $A(1)=0$ and $B(1)=C(1)=D(1)=\frac13.$ For $n\geq2,$ the probability that the bug land on each vertex after $n$ meters is $\frac13$ the sum of the probability the bug land on other vertices after crawling $n-1$ meters. So, we have \begin{align*} A(n) &= \frac13 \cdot [B(n-1) + C(n-1) + D(n-1)], \\ B(n) &= \frac13 \cdot [A(n-1) + C(n-1) + D(n-1)], \\ C(n) &= \frac13 \cdot [A(n-1) + B(n-1) + D(n-1)], \\ D(n) &= \frac13 \cdot [A(n-1) + B(n-1) + C(n-1)]. \end{align*} It follows that $A(n) = B(n-1) = C(n-1) = D(n-1).$
We construct the following table: \[\begin{array}{c|cccc} & & & & \\ [-2ex] n & A(n) & B(n) & C(n) & D(n) \\ [1ex] \hline & & & & \\ [-1ex] 1 & 0 & \frac13 & \frac13 & \frac13 \\ & & & & \\ 2 & \frac13 & \frac29 & \frac29 & \frac29 \\ & & & & \\ 3 & \frac29 & \frac{7}{27} & \frac{7}{27} & \frac{7}{27} \\ & & & & \\ 4 & \frac{7}{27} & \frac{20}{81} & \frac{20}{81} & \frac{20}{81}\\ & & & & \\ 5 & \frac{20}{81} & \frac{61}{243} & \frac{61}{243} & \frac{61}{243} \\ & & & & \\ 6 & \frac{61}{243} & \frac{182}{729} & \frac{182}{729} & \frac{182}{729} \\ & & & & \\ 7 & \frac{182}{729} & \frac{547}{2187} & \frac{547}{2187} & \frac{547}{2187} \\ [1ex] \end{array}\] Therefore, the answer is $n = \boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | Let $\omega = e^{i\pi / 2}$ . We have that if $G(x) = (x+x^2+x^3)^7$ , then \[\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)}{4} = \frac{2187-1-1-1}{4} = 546.\] From here, the desired probability is $\frac{546}{2187} = \frac{182}{729}$ . Therefore, the answer is $n=\boxed{182}$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | We can factor $(x+x^2+x^3)^7$ as $x^7(1+x+x^2)^7.$ The $x^{4n}$ coefficients of $(x+x^2+x^3)^7$ will be the same as the $x^{4n+1}$ coefficients of $(1+x+x^2)^7.$ The possible values for $4n+1$ would then be $1,$ $5,$ $9,$ and $13.$ Because $1+13=5+9=14,$ the coefficients of $x^1$ and $x^{13}$ are equal and so are the coefficients of $x^5$ and $x^9.$ To make an $x$ term, we need $6$ $1$ " terms and one " $x$ " term multiplied together. There are $7$ ways to arrange these numbers. The coefficient of the $x^5$ term will be the sum of the number of the possible arrangements for these numbers: \begin{align*} 0000122&=105\text{ arrangements}, \\ 0001112&=140\text{ arrangements}, \\ 0011111&=21\text{ arrangements}. \end{align*} Thus, the coefficient of the $x^5$ term is $105+140+21=266.$ From here, the desired probability is $\frac{2(7+266)}{2187}=\frac{546}{2187}=\frac{182}{729}.$ Thus, our answer is $\boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | We can find the number of different times the bug reaches vertex $A$ before the $7$ th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A.$
Define $f(x)$ to be the number of paths of length $x$ which start and end at $A$ but do not pass through $A$ otherwise. Obviously $f(1) = 0.$ In general for $f(x),$ the bug has three initial edges to pick from. From there, since the bug cannot return to $A$ by definition, the bug has exactly two choices. This continues from the $2$ nd move up to the $(x-1)$ th move. The last move must be a return to $A,$ so this move is determined. So $f(x) = 2^{x-2}3.$
Now we need to find the number of cycles by which the bug can reach $A$ at the end. Since $f(1) = 0,$ we know that $f(6)$ cannot be used, as on the $7$ th move the bug cannot move from $A$ to $A.$ So we need to find the number of partitions of $7$ using only $2,3,4,5,$ and $7.$ These are $f(2)f(2)f(3),f(2)f(5),f(3)f(4),$ and $f(7).$ We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition: \[{3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7) = 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53) = 546.\] Finally, this is a probability question, so we divide by $3^7,$ or $\frac{546}{3^7} = \frac{182}{3^6}.$ The answer is $n=\boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | Note that this problem is basically equivalent to the following:
How many distinct sequences of $8$ integers $a_1, a_2, a_3, \ldots, a_8$ are there such that $a_1 = a_8 = 1,$ $a_i \in \{1, 2, 3, 4\}$ for all $2 \leq i\leq 8,$ and $a_i \neq a_{i+1}$ for all $1 \leq i \leq 7$
Now consider the $8$ integers modulo $4.$ Let $b_1, b_2, b_3, \ldots, b_7$ be a new sequence of integers such that $b_i = a_{i+1} - a_i \mod 4$ for all $1 \leq i \leq 7.$
Note that the condition is equivalent to that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7,$ and since $a_1 \mod 4 = a_8 \mod 4,$ $b_1 + b_2 + \cdots + b_7$ must be a multiple of $4.$
Thus, our desired number of paths is equivalent to the number of ordered septuples of positive integers $(b_1, b_2, \ldots, b_7)$ such that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7$ and $b_1 + b_2 + \cdots + b_7$ is congruent to $0 \mod 4.$
We will now proceed with casework on the number of $2$ s in the septuple.
One $2$ : There are $\dbinom{7}{1} = 7$ ways to arrange the $2$ , and $\dbinom{6}{0} + \dbinom{6}{2} + \dbinom{6}{4} + \dbinom{6}{6} = 2^5$ valid ways (the proof of this combinatorial identity is left as an exercise to the reader) to arrange the $1$ s and $3$ s.
Three $2$ s: There are $\dbinom{7}{3} = 35$ ways to arrange the $2$ s, and $\dbinom{4}{1} + \dbinom{4}{3} = 2^3$ valid ways to arrange the $1$ s and $3$ s.
Five $2$ s: There are $\dbinom{7}{5} = 21$ ways to arrange the $2$ s, and $\dbinom{2}{0} + \dbinom{2}{2} = 2$ valid ways to arrange the $1$ s and $3$ s.
Adding up our cases, we obtain $7 \cdot 32 + 35 \cdot 8 + 21 \cdot 2 = 546$ valid sequences.
Dividing by the total number of paths without restrictions, $3^7 = 2187,$ our desired probability is $\frac{546}{2187} = \frac{182}{729},$ giving an answer of $\boxed{182}.$ | null | 182 |
b03ff4731ebcf390ad43660ebae16d91 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$ | We instead find the probability that the bug is NOT at vertex $A$ after crawling $n$ meters (equivalent to moving $n$ times). Call $A_n$ the probability that the bug IS at vertex $A$ after $n$ moves; call $O_n$ the probability that the bug is on some other vertex. We have the following recurrence relations. \[A_n = \frac{1}{3}O_{n-1}\] \[O_n = A_{n-1} + \frac{2}{3}O_{n-1}\] However, we can calculate $A_{n-1}$ in terms of $O_n$ ; take $n = k-1$ , and we have \[A_{k-1} = \frac{1}{3}O_{k-2}\] . Substituting this into our equation for $O$ , we have that \[O_n = \frac{1}{3}O_{n-2} + \frac{2}{3}O_{n-1}\] . We also have the conditions that $O_0 = 0$ (the bug will not be at vertex other than $A$ on the "0th" move) and $O_1 = 1$ (the bug will be at a vertex other than $A$ after the $1st$ move). Iteratively calculating $O_7$ , we find that it is equal to $\frac{547}{729}$ (I will not do this calculation here; you can do it manually if you wish to check). However, this is the probability that the ant is NOT at vertex $A$ after $7$ moves; then its complement, $\frac{182}{729}$ is the probability that the ant IS at vertex $A$ after $7$ moves, so our answer is $\boxed{182}$ | null | 182 |
b2a8d5207a0fe01157eb3669d8b3bbb5 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_13 | The numbers in the sequence $101$ $104$ $109$ $116$ $\ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,\ldots$ For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers | If $(x,y)$ denotes the greatest common divisor of $x$ and $y$ , then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$ . Now assuming that $d_n$ divides $100+n^2$ , it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$
Thus the equation turns into $d_n=(100+n^2,2n+1)$ . Now note that since $2n+1$ is odd for integral $n$ , we can multiply the left integer, $100+n^2$ , by a power of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multiply by $4$ so that we can get a $(2n+1)^2$ in there.
So $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$ . It simplified the way we wanted it to!
Now using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$ . Thus $d_n$ must divide $\boxed{401}$ for every single $n$ . This means the largest possible value for $d_n$ is $401$ , and we see that it can be achieved when $n = 200$ | null | 401 |
b2a8d5207a0fe01157eb3669d8b3bbb5 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_13 | The numbers in the sequence $101$ $104$ $109$ $116$ $\ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,\ldots$ For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers | We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$ . Since we want to find the GCD of $a_n$ and $a_{n+1}$ , we can use the Euclidean algorithm
$a_{n+1}-a_n = 2n+1$
Now, the question is to find the GCD of $2n+1$ and $100+n^2$ .
We subtract $2n+1$ $100$ times from $100+n^2$ \[(100+n^2)-100(2n+1)\] \[=n^2+100-200n-100\] This leaves us with \[n^2-200n.\] Factoring, we get \[n(n-200)\] Because $n$ and $2n+1$ will be coprime, the only thing stopping the GCD from being $1$ is $n-200.$ We want this to equal 0, because that will maximize our GCD.
Solving for $n$ gives us $n=200$ . The last remainder is 0, thus $200*2+1 = \boxed{401}$ is our GCD. | null | 401 |
5ce1eaf329eda088363619910acce041 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_14 | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament? | Let us suppose for convenience that there were $n + 10$ players overall. Among the $n$ players not in the weakest 10 there were $n \choose 2$ games played and thus $n \choose 2$ points earned. By the givens, this means that these $n$ players also earned $n \choose 2$ points against our weakest 10. Now, the 10 weakest players playing amongst themselves played ${10 \choose 2} = 45$ games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger $n$ players. Since every point earned falls into one of these categories, It follows that the total number of points earned was $2{n \choose 2} + 90 = n^2 - n + 90$ . However, there was one point earned per game, and there were a total of ${n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}$ games played and thus $\frac{(n + 10)(n + 9)}{2}$ points earned. So we have $n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}$ so $2n^2 - 2n + 180 = n^2 + 19n + 90$ and $n^2 -21n + 90 = 0$ and $n = 6$ or $n = 15$ . Now, note that the top $n$ players got $n(n - 1)$ points in total (by our previous calculation) for an average of $n - 1$ , while the bottom 10 got 90 points total, for an average of 9. Thus we must have $n > 10$ , so $n = 15$ and the answer is $15 + 10 = \boxed{25}$ | null | 25 |
5ce1eaf329eda088363619910acce041 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_14 | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament? | Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people. Thus, the $n-10$ people earned a total of $10(n-10)-45 = 10n-145$ points playing vs. this group of 10 people, and also earned a total of $10n-145$ playing against themselves. Since each match gives a total of one point, we must have that $\dbinom{n-10}{2}=10n-145$ . Expanding and simplifying gives us $n^2-41n+400=0$ . Thus, $n=16$ or $n=25$ . Note however that if $n=16$ , then the strongest $16$ people get a total of $16*10-145=15$ playing against the weakest $10$ who gained $45$ points vs them, which is a contradiction since it must be larger. Thus, $n=\boxed{25}$ | null | 25 |
5ce1eaf329eda088363619910acce041 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_14 | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament? | Note that the total number of points accumulated must sum to ${p \choose 2} = \frac{p(p-1)}{2}$ . Say the number of people is $n$ . Consider the number of points gained when the 10 lowest scoring people play each other. The problem tells us that each of these 10 people must earn exactly half of the total number of points they will earn during the whole game. This implies that this group of 10 people must accumulate half their total combined points after they (the 10 people) all play each other, meaning they must earn the other half of their points by playing the $n-10$ stronger players. The problem also tells us that the $n-10$ people who aren't part of the losers group will earn half of their points by playing the $10$ losers. Since the $n-10$ group and $10$ losers will earn half their points by playing each other, the sum of the number of points that they gain playing each other must then be half of the total amount of points earned by everyone in the game. Therefore, $\frac{p(p-1)}{4} = 10(p-10)$ . This equation is the same as above, and by the same logic, the answer is $n=\boxed{25}$ | null | 25 |
92b528f5ff3db244505bcf70f9fa573c | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_15 | Three 12 cm $\times$ 12 cm squares are each cut into two pieces $A$ and $B$ , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon , as shown in the second figure, so as to fold into a polyhedron . What is the volume (in $\mathrm{cm}^3$ ) of this polyhedron?
AIME 1985 Problem 15.png | Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\circ$ from each other) yields a cube , so the volume is $\frac12 \cdot 12^3 = 864$ , so our answer is $\boxed{864}$ | null | 864 |