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https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1

Compute $\sqrt{(31)(30)(29)(28)+1}$

Note that the four numbers to multiply are symmetric with the center at $29.5$ . Multiply the symmetric pairs to get $31\cdot 28=868$ and $30\cdot 29=870$ $\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1

Compute $\sqrt{(31)(30)(29)(28)+1}$

Similar to Solution 1 above, call the consecutive integers $\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)$ to make use of symmetry. Note that $n$ itself is not an integer - in this case, $n = 29.5$ . The expression becomes $\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}$ . Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives $\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}$ . The inside is a perfect square trinomial, since $b^2 = 4ac$ . It's equal to $\sqrt{\left(n^2 - \frac{5}{4}\right)^2}$ , which simplifies to $n^2 - \frac{5}{4}$ . You can plug in the value of $n$ from there, or further simplify to $\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1$ , which is easier to compute. Either way, plugging in $n=29.5$ gives $\boxed{869}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1

Compute $\sqrt{(31)(30)(29)(28)+1}$

We have $(31)(30)(29)(28)+1=755161.$ Since the alternating sum of the digits $7-5+5-1+6-1=11$ is divisible by $11,$ we conclude that $755161$ is divisible by $11.$ We evaluate the original expression by prime factorization: \begin{align*} \sqrt{(31)(30)(29)(28)+1}&=\sqrt{755161} \\ &=\sqrt{11\cdot68651} \\ &=\sqrt{11^2\cdot6241} \\ &=\sqrt{11^2\cdot79^2} \\ &=11\cdot79 \\ &=\boxed{869} ~Vrjmath (Fundamental Logic)

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1

Compute $\sqrt{(31)(30)(29)(28)+1}$

The last digit under the radical is $1$ , so the square root must either end in $1$ or $9$ , since $x^2 = 1\pmod {10}$ means $x = \pm 1$ . Additionally, the number must be near $29 \cdot 30 = 870$ , narrowing the reasonable choices to $869$ and $871$ Continuing the logic, the next-to-last digit under the radical is the same as the last digit of $28 \cdot 29 \cdot 3 \cdot 31$ , which is $6$ . Quick computation shows that $869^2$ ends in $61$ , while $871^2$ ends in $41$ . Thus, the answer is $\boxed{869}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_2

Ten points are marked on a circle . How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices

Any subset of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are $2^{10} = 1024$ total subsets of a ten-member set , but of these ${10 \choose 0} = 1$ have 0 members, ${10 \choose 1} = 10$ have 1 member and ${10 \choose 2} = 45$ have 2 members. Thus the answer is $1024 - 1 - 10 - 45 = \boxed{968}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3

Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if

Repeating decimals represent rational numbers . To figure out which rational number, we sum an infinite geometric series $0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$ . Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$ . Since 750 and 37 are relatively prime $4d + 1$ must be divisible by 37, and the only digit for which this is possible is $d = 9$ . Thus $4d + 1 = 37$ and $n = \boxed{750}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3

Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if

To get rid of repeating decimals, we multiply the equation by 1000. We get $\frac{1000n}{810} = d25.d25d25...$ We subtract the original equation from the second to get $\frac{999n}{810}=d25$ We simplify to $\frac{37n}{30} = d25$ Since $\frac{37n}{30}$ is an integer, $n=(30)(5)(2k+1)$ because $37$ is relatively prime to $30$ , and d25 is divisible by $5$ but not $10$ . The only odd number that yields a single digit $d$ and 25 at the end of the three digit number is $k=2$ , so the answer is $\boxed{750}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3

Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if

Similar to Solution 2, we start off by writing that $\frac{1000n}{810} = d25.d25d25 \dots$ .Then we subtract this from the original equation to get: \[\frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30\] Since n is an integer, we have that $37 \mid d25 \cdot 30$ Since $37$ is prime, we can apply Euclid's Lemma (which states that if $p$ is a prime and if $a$ and $b$ are integers and if $p \mid ab$ , then $p \mid a$ or $p \mid b$ ) to realize that $37 \mid d25$ , since $37 \nmid 30$ . Then we can expand $d25$ as $25 \cdot (4d +1)$ . Since $37 \nmid 25$ , by Euclid, we can arrive at $37 \mid 4d+1 \Longrightarrow d=9$ . From this we know that $n= 25 \cdot 30 = \boxed{750}$ . (This is true because $37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3

Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if

Write out these equations: $\frac{n}{810} = \frac{d25}{999}$ $\frac{n}{30} = \frac{d25}{37}$ $37n = 30(d25)$ Thus $n$ divides 25 and 30. The only solution for this under 1000 is $\boxed{750}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4

If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$

Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$ . Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ denoting an integers $c$ is minimized if it’s prime factorization contains only $3,5$ , and since there is a cubed term in $5^2 \cdot y^3$ $3^3$ must be a factor of $c$ $3^35^2 = \boxed{675}$ , which works as the solution.

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4

If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$

Let $b$ $c$ $d$ , and $e$ equal $a+1$ $a+2$ $a+3$ , and $a+4$ , respectively. Call the square and cube $k^2$ and $m^3$ , where both k and m are integers. Then: $5a + 10 = m^3$ Now we know $m^3$ is a multiple of 125 and $m$ is a multiple of 5. The lower $m$ is, the lower the value of $c$ will be. Start from 5 and add 5 each time. $m = 5$ gives no solution for k $m = 10$ gives no solution for k $m = 15$ gives a solution for k. $10 + 5a = 15^3$ $2 + a = 675$ $c = \boxed{675}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4

If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$

Let the numbers be $a,a+1,a+2,a+3,a+4.$ When then know $3a+6$ is a perfect cube and $5a+10$ is perfect cube. Since $5a+10$ is divisible by $5$ we know that $5a+10 = (5k)^3$ since otherwise we get a contradiction. This means $a = 25k^3 - 2$ in which plugging into the other expression we know $3(25k^3 - 2) + 6 = 75k^3$ is a perfect square. We know $75 = 5^2 \cdot 3$ so we let $k = 3$ to obtain the perfect square. This means that $c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4

If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$

(This is literally a combination of 1 and 3) Since $a$ $b$ $c$ $d$ , and $e$ are consecutive, $a = c-2$ $b = c-1$ $c=c$ $d = c+1$ , and $e = c+2$ Because $b+c+d = 3c$ is a perfect square, and $a+b+c+d+e = 5c$ is a perfect cube, we can express $c$ as $c = 3^{n} \cdot 5^{k}$ Now, by the problem's given information, $k \equiv 0 \text{(mod 2)}$ $n \equiv 0 \text{(mod 3)}$ and because ALL exponents have to be cubes/squares, $k \equiv 2 \text{(mod 3)}$ $n \equiv 1 \text{(mod 2)}$ Therefore, $k = 2$ $n = 3$ $c = 3^3 \cdot 5^2 = \boxed{675}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_5

When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$ , in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$

Denote the probability of getting a heads in one flip of the biased coin as $h$ . Based upon the problem, note that ${5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3$ . After canceling out terms, we get $1 - h = 2h$ , so $h = \frac{1}{3}$ . The answer we are looking for is ${5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}$ , so $i+j=40+243=\boxed{283}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_5

When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$ , in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$

Denote the probability of getting a heads in one flip of the biased coins as $h$ and the probability of getting a tails as $t$ . Based upon the problem, note that ${5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3$ . After cancelling out terms, we end up with $t = 2h$ . To find the probability getting $3$ heads, we need to find ${5\choose3}\dfrac{(h)^3(t)^2}{(h + t)^5} =10\cdot\dfrac{(h)^3(2h)^2}{(h + 2h)^5}$ (recall that $h$ cannot be $0$ ). The result after simplifying is $\frac{40}{243}$ , so $i + j = 40 + 243 = \boxed{283}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_6

Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

Label the point of intersection as $C$ . Since $d = rt$ $AC = 8t$ and $BC = 7t$ . According to the law of cosines \begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*} Since we are looking for the earliest possible intersection, $20$ seconds are needed. Thus, $8 \cdot 20 = \boxed{160}$ meters is the solution.

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_6

Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

Let $P$ be the point of intersection between the skaters, Allie and Billie. We can draw a line that goes through $P$ and is parallel to $\overline{AB}$ . Letting this line be the $x$ -axis, we can reflect $B$ over the $x$ -axis to get $B'$ . As reflections preserve length, $B'X = XB$ We then draw lines $BB'$ and $PB'$ . We can let the foot of the perpendicular from $P$ to $BB'$ be $X$ , and we can let the foot of the perpendicular from $P$ to $AB$ be $Y$ . In doing so, we have constructed rectangle $PXBY$ By $d=rt$ , we have $AP = 8t$ and $PB = PB' = 7t$ , where $t$ is the number of seconds it takes the skaters to meet. Furthermore, we have $30-60-90$ triangle $PAY$ , so $AY = 4t$ , and $PY = 4t\sqrt{3}$ . Since we have $PY = XB = B'X$ $B'X = 4t\sqrt{3}$ . By Pythagoras, $PX = t$ As $PXBY$ is a rectangle, $PX = YB$ . Thus $AY + YB = AB \Rightarrow AY + PX = AB$ , so we get $4t + t = 100$ . Solving for $t$ , we find $t = 20$ Our answer, $AP$ , is equivalent to $8t$ . Thus, $AP = 8 \cdot 20 = \boxed{160}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_6

Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$ . Then we can produce the following diagram: [asy] draw((0,0)--(100,0)--(80,139)--cycle); label("8x",(0,0)--(80,139),NW); label("7x",(100,0)--(80,139),NE); label("100",(0,0)--(100,0),S); dot((0,0)); label("A",(0,0),S); dot((100,0)); label("B",(100,0),S); dot((80,139)); label("M",(80,139),N); [/asy] Now, if we drop an altitude from point $M$ , we get : [asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label("8x",(0,0)--(80,139),NW); label("7x",(100,0)--(80,139),NE); dot((0,0)); label("A",(0,0),S); dot((100,0)); label("B",(100,0),SE); dot((80,139)); label("M",(80,139),N); draw((80,139)--(80,0),dashed); label("4x",(0,0)--(80,0),S); label("100-4x",(80,0)--(100,0),S); label("$4x \sqrt{3}$",(80,139)--(80,0),W); label("$60^{\circ}$", (3,1), NE); [/asy] We know this from the $30-60-90$ triangle that is formed. From this we get that: \[(7x)^2 = (4 \sqrt{3} x)^2 + (100-4x)^2\] \[\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2\] \[\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)\] Therefore, we get that $x = \frac{100}{3}$ or $x = 20$ . Since $20< \frac{100}{3}$ , we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \cdot x = 8 \cdot 20 = \boxed{160}$ meters.

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_7

If the integer $k$ is added to each of the numbers $36$ $300$ , and $596$ , one obtains the squares of three consecutive terms of an arithmetic series. Find $k$

Call the terms of the arithmetic progression $a,\ a + d,\ a + 2d$ , making their squares $a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2$ We know that $a^2 = 36 + k$ and $(a + d)^2 = 300 + k$ , and subtracting these two we get $264 = 2ad + d^2$ (1). Similarly, using $(a + d)^2 = 300 + k$ and $(a + 2d)^2 = 596 + k$ , subtraction yields $296 = 2ad + 3d^2$ (2). Subtracting the first equation from the second, we get $2d^2 = 32$ , so $d = 4$ . Substituting backwards yields that $a = 31$ and $k = \boxed{925}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_7

If the integer $k$ is added to each of the numbers $36$ $300$ , and $596$ , one obtains the squares of three consecutive terms of an arithmetic series. Find $k$

Since terms in an arithmetic progression have constant differences, \[\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}\] \[\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}\] \[\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}\] \[\implies 2k+568=2\sqrt{(596+k)(36+k)}\] \[\implies k+284=\sqrt{(596+k)(36+k)}\] \[\implies k^2+568k+80656=k^2+632k+21456\] \[\implies 568k+80656=632k+21456\] \[\implies 64k = 59200\] \[\implies k = \boxed{925}\]

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

Note that each given equation is of the form \[f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7\] for some $k\in\{1,2,3\}.$ When we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ \[f(k)=ak^2+bk+c,\] where $a,b,$ and $c$ are linear combinations of $x_1,x_2,x_3,x_4,x_5,x_6,$ and $x_7.$ We are given that \begin{alignat*}{10} f(1)&=\phantom{42}a+b+c&&=1, \\ f(2)&=4a+2b+c&&=12, \\ f(3)&=9a+3b+c&&=123, \end{alignat*} and we wish to find $f(4).$ We eliminate $c$ by subtracting the first equation from the second, then subtracting the second equation from the third: \begin{align*} 3a+b&=11, \\ 5a+b&=111. \end{align*} By either substitution or elimination, we get $a=50$ and $b=-139.$ Substituting these back produces $c=90.$ Finally, the answer is \[f(4)=16a+4b+c=\boxed{334}.\]

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

For simplicity purposes, we number the given equations $(1),(2),$ and $(3),$ in that order. Let \[16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=S. \hspace{29.5mm}(4)\] Subtracting $(1)$ from $(2),$ subtracting $(2)$ from $(3),$ and subtracting $(3)$ from $(4),$ we obtain the following equations, respectively: \begin{align*} 3x_1 + 5x_2 + 7x_3 + 9x_4 + 11x_5 + 13x_6 + 15x_7 &=11, \hspace{20mm}&(5) \\ 5x_1 + 7x_2 + 9x_3 + 11x_4 + 13x_5 + 15x_6 + 17x_7 &=111, &(6) \\ 7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 + 19x_7 &=S-123. &(7) \\ \end{align*} Subtracting $(5)$ from $(6)$ and subtracting $(6)$ from $(7),$ we obtain the following equations, respectively: \begin{align*} 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \\ 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \hspace{20mm}&(9) \end{align*} Finally, applying the Transitive Property to $(8)$ and $(9)$ gives $S-234=100,$ from which $S=\boxed{334}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares: [asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=1; i<=10; ++i) { label("\boldmath{$"+string(i^2)+"$}",(i-1,0)); } for (real i=1; i<=9; ++i) { label("$"+string(1+2*i)+"$",(i-0.5,-0.75)); } for (real i=1; i<=8; ++i) { label("$2$",(i,-1.5)); } for (real i=1; i<=9; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=8; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=9; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=8; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label("\textbf{First Differences}",(-0.75,-0.75),align=W); label("\textbf{Second Differences}",(-0.75,-1.5),align=W); [/asy] Label equations $(1),(2),(3),$ and $(4)$ as Solution 2 does. Since the coefficients of $x_1,x_2,x_3,x_4,x_5,x_6,x_7,$ or $(1,4,9,16),(4,9,16,25),(9,16,25,36),(16,25,36,49),(25,36,49,64),(36,49,64,81),(49,64,81,100),$ respectively, all form quadratic sequences with second differences $2,$ we conclude that the second differences of equations $(1),(2),(3),(4)$ must be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label("\boldmath{$1$}",(0,0)); label("\boldmath{$12$}",(1,0)); label("\boldmath{$123$}",(2,0)); label("\boldmath{$S$}",(3,0)); label("$11$",(0.5,-0.75)); label("$111$",(1.5,-0.75)); label("$d_1$",(2.5,-0.75)); label("$100$",(1,-1.5)); label("$d_2$",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label("\textbf{First Differences}",(-0.75,-0.75),align=W); label("\textbf{Second Differences}",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \begin{align*} S&=123+d_1 \\ &=123+(111+d_2) \\ &=\boxed{334} ~MRENTHUSIASM

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

Notice that we may rewrite the equations in the more compact form as: \begin{align*} \sum_{i=1}^{7}i^2x_i&=c_1, \\ \sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\ \sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\ \sum_{i=1}^{7}(i+3)^2x_i&=c_4, \end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$ . Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\boxed{334}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

The idea is to multiply the first, second and third equations by $a,b,$ and $c,$ respectively. We can only consider the coefficients of $x_1,x_2,$ and $x_3:$ \begin{align} a+4b+9c&=16, \\ 4a+9b+16c&=25, \\ 9a+16b+25c&=36. \end{align} Subtracting $(1)$ from $(2),$ we get \[3a+5b+7c=9. \hspace{15mm}(4)\] Subtracting $3\cdot(4)$ from $(3),$ we get \[b+4c=9. \hspace{25.5mm}(5)\] Subtracting $(1)$ from $4\cdot(5),$ we get \[7c-a=20. \hspace{23mm}(6)\] From $(5)$ and $(6),$ we have $(a,b,c)=(7c-20,9-4c,c).$ Substituting this into $(2)$ gives $(a,b,c)=(1,-3,3).$ Therefore, the answer is $1\cdot1+12\cdot(-3) + 123\cdot3 = \boxed{334}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

We let $(x_4,x_5,x_6,x_7)=(0,0,0,0)$ . Thus, we have \begin{align*} x_1+4x_2+9x_3&=1,\\ 4x_1+9x_2+16x_3&=12,\\ 9x_1+16x_2+25x_3&=123.\\ \end{align*} Grinding this out, we have $(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)$ which gives $\boxed{334}$ as our final answer.

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

Let $s_n = n^2$ be the sequence of perfect squares. By either expanding or via finite differences, one can prove the miraculous recursion \[s_n = 3s_{n-1} - 3s_{n-2} + s_{n-3}.\] Hence, the answer is simply \[3 \cdot 123 - 3 \cdot 12 + 1 = \boxed{334}.\]

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$

Taking the given equation modulo $2,3,$ and $5,$ respectively, we have \begin{align*} n^5&\equiv0\pmod{2}, \\ n^5&\equiv0\pmod{3}, \\ n^5&\equiv4\pmod{5}. \end{align*} By either Fermat's Little Theorem (FLT) or inspection, we get \begin{align*} n&\equiv0\pmod{2}, \\ n&\equiv0\pmod{3}, \\ n&\equiv4\pmod{5}. \end{align*} By either the Chinese Remainder Theorem (CRT) or inspection, we get $n\equiv24\pmod{30}.$ It is clear that $n>133,$ so the possible values for $n$ are $144,174,204,\ldots.$ Note that \begin{align*} n^5&=133^5+110^5+84^5+27^5 \\ &<133^5+110^5+(84+27)^5 \\ &=133^5+110^5+111^5 \\ &<3\cdot133^5, \end{align*} from which $\left(\frac{n}{133}\right)^5<3.$ If $n\geq174,$ then \begin{align*} \left(\frac{n}{133}\right)^5&>1.3^5 \\ &=1.3^2\cdot1.3^2\cdot1.3 \\ &>1.6\cdot1.6\cdot1.3 \\ &=2.56\cdot1.3 \\ &>2.5\cdot1.2 \\ &=3, \end{align*} which arrives at a contradiction. Therefore, we conclude that $n=\boxed{144}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$

Note that $n$ is even, since the LHS consists of two odd and two even numbers. By Fermat's Little Theorem, we know $n^5\equiv n\pmod{5}.$ Hence, \[n\equiv3+0+4+2\equiv4\pmod{5}.\] Continuing, we examine the equation modulo $3,$ \[n\equiv1-1+0+0\equiv0\pmod{3}.\] Thus, $n$ is divisible by three and leaves a remainder of four when divided by $5.$ It is obvious that $n>133,$ so the only possibilities are $n = 144$ or $n \geq 174.$ It quickly becomes apparent that $174$ is much too large, so $n$ must be $\boxed{144}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$

We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, $n^5\equiv n\pmod{5},$ and it is easy to see that $n^5\equiv n\pmod 2.$ Therefore, $133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10},$ so the last digit of $n$ is $4.$ We notice that $133,110,84,$ and $27$ are all very close or equal to multiples of $27.$ We can rewrite $n^5$ as approximately equal to $27^5(5^5+4^5+3^5+1^5) = 27^5(4393).$ This means $\frac{n^5}{27^5}$ must be close to $4393.$ Note that $134$ will obviously be too small, so we try $144$ and get $\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5.$ Bashing through the division, we find that $\frac{1048576}{243}\approx 4315,$ which is very close to $4393.$ It is clear that $154$ will not give any closer of an answer, given the rate that fifth powers grow, so we can safely assume that $\boxed{144}$ is the answer.

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$

In this solution we take advantage of the large numbers and utilize parity properties to give us a very good guess at the answer. The units digits of $133^5, 110^5, 84^5, 27^5$ are $3, 0, 4, 7,$ respectively, so the units digit of $n^5$ is $4.$ This tells us $n$ is even. Since we are dealing with enormous numbers, $n$ should not be that far from $133.$ Note that $n$ 's units digit is $0, 2, 4, 6,$ or $8.$ When to the power of $5,$ they each give $0, 2, 4, 6,$ and $8$ as the units digits. This further clues us that $n$ ends in $4.$ Clearly, $n>133,$ so we start with $134.$ Now we need a way of distinguishing between numbers with units digit $4.$ We can do this by finding the last three digits for each of $133^5, 110^5, 84^5,$ and $27^5,$ which is not that difficult. For $133^5,$ we have \begin{align*} 133^5&=133^2\cdot133^2\cdot133 \\ &\equiv689\cdot689\cdot133 \\ &\equiv721\cdot133 \\ &\equiv893\pmod{1000}. \end{align*} By the same reasoning, we get \begin{align*} n^5&=133^5+110^5+84^5+27^5 \\ &\equiv893+0+424+907 \\ &\equiv224\pmod{1000}. \end{align*} Note that \begin{align*} 134^5&\equiv424\pmod{1000}, \\ 144^5&\equiv224\pmod{1000}, \\ 154^5&\equiv024\pmod{1000}, \\ 164^5&\equiv824\pmod{1000}, \\ 174^5&\equiv624\pmod{1000}, \\ 184^5&\equiv424\pmod{1000}, \\ 194^5&\equiv224\pmod{1000}. \end{align*} By observations, $n=194$ is obviously an overestimate. So, the answer is $n=\boxed{144}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$

First, we take mod $2$ on both sides to get $n^5\equiv 0\pmod{2}\implies n\equiv 0\pmod{2}$ . Mod $3$ gives $n^5\equiv 0\pmod{3}\implies n\equiv 0\pmod{3}$ . Also, mod $5$ gives $n^5\equiv -1\pmod{5}\implies n\equiv -1\pmod{5}$ (by FLT). Finally, note that mod $7$ gives $n^5\equiv 2\pmod{7}\implies n^{-1}\equiv 2\pmod{7}\implies n\equiv 4\pmod{7}$ . Thus, \begin{align*} n&\equiv 0\pmod{2}, \\ n&\equiv 0\pmod{3}, \\ n&\equiv -1\pmod{5}, \\ n&\equiv 4\pmod{7}. \end{align*} By CRT, $n\equiv 144\pmod{210}$ , so $n$ is one of $144, 354, ...$ . However, $133^5 + 110^5 + 84^5 + 27^5 < 4\cdot 133^5 < (2\cdot 133)^5 < 354^5$ , so $n < 354$ . Thus, $n = \boxed{144}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$

We have \[n^5 = 133^5 + 110^5 + 84^5 +27^5 = 61917364224,\] for which $n = \sqrt [5]{61917364224} = \boxed{144}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

We draw the altitude $h$ to $c$ , to get two right triangles Then $\cot{\alpha}+\cot{\beta}=\frac{c}{h}$ , from the definition of the cotangent Let $K$ be the area of $\triangle ABC.$ Then $h=\frac{2K}{c}$ , so $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2K}$ By identical logic, we can find similar expressions for the sums of the other two cotangents: \begin{align*} \cot \alpha + \cot \beta &= \frac{c^2}{2K} \\ \cot \beta + \cot \gamma &= \frac{a^2}{2K} \\ \cot \gamma + \cot \alpha &= \frac{b^2}{2K}. \end{align*} Adding the last two equations, subtracting the first, and dividing by 2, we get \[\cot \gamma = \frac{a^2 + b^2 - c^2}{4K}.\] Therefore \begin{align*} \frac{\cot \gamma}{\cot \alpha + \cot \beta} &= \frac{(a^2 + b^2 - c^2)/(4K)}{c^2/(2K)} \\ &= \frac{a^2 + b^2 - c^2}{2c^2} \\ &= \frac{1989 c^2 - c^2}{2c^2} \\ &= \frac{1988}{2} = \boxed{994}

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

We start as in solution 1, though we'll write $A$ instead of $K$ for the area. Now we evaluate the numerator: \[\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}\] From the Law of Cosines and the sine area formula, \begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\\ \sin{\gamma}&= \frac{2A}{ab}\\ \cot{\gamma}&= \frac{\cos \gamma}{\sin \gamma} = \frac{1988c^2}{4A} \end{align*} Then $\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

\begin{align*} \cot{\alpha} + \cot{\beta} &= \frac {\cos{\alpha}}{\sin{\alpha}} + \frac {\cos{\beta}}{\sin{\beta}} = \frac {\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}}\\ &= \frac {\sin{(\alpha + \beta)}}{\sin{\alpha}\sin{\beta}} = \frac {\sin{\gamma}}{\sin{\alpha}\sin{\beta}} \end{align*} By the Law of Cosines, \[a^2 + b^2 - 2ab\cos{\gamma} = c^2 = 1989c^2 - 2ab\cos{\gamma} \implies ab\cos{\gamma} = 994c^2\] Now \begin{align*}\frac {\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac {\cot{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\gamma}} = \frac {\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin^2{\gamma}} = \frac {ab}{c^2}\cos{\gamma} = \frac {ab}{c^2} \cdot \frac {994c^2}{ab}\\ &= \boxed{994}

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

Use Law of cosines to give us $c^2=a^2+b^2-2ab\cos(\gamma)$ or therefore $\cos(\gamma)=\frac{994c^2}{ab}$ . Next, we are going to put all the sin's in term of $\sin(a)$ . We get $\sin(\gamma)=\frac{c\sin(a)}{a}$ . Therefore, we get $\cot(\gamma)=\frac{994c}{b\sin a}$ Next, use Law of Cosines to give us $b^2=a^2+c^2-2ac\cos(\beta)$ . Therefore, $\cos(\beta)=\frac{a^2-994c^2}{ac}$ . Also, $\sin(\beta)=\frac{b\sin(a)}{a}$ . Hence, $\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}$ Lastly, $\cos(\alpha)=\frac{b^2-994c^2}{bc}$ . Therefore, we get $\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}$ Now, $\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sin a}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}$ . After using $a^2+b^2=1989c^2$ , we get $\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

Let $\gamma$ be $(180-\alpha-\beta)$ $\frac{\cot \gamma}{\cot \alpha+\cot \beta} = \frac{\frac{-\tan \alpha \tan \beta}{\tan(\alpha+\beta)}}{\tan \alpha + \tan \beta} = \frac{(\tan \alpha \tan \beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}$ WLOG, assume that $a$ and $c$ are legs of right triangle $abc$ with $\beta = 90^o$ and $c=1$ By the Pythagorean theorem, we have $b^2=a^2+1$ , and the given $a^2+b^2=1989$ . Solving the equations gives us $a=\sqrt{994}$ and $b=\sqrt{995}$ . We see that $\tan \beta = \infty$ , and $\tan \alpha = \sqrt{994}$ Our derived equation equals $\tan^2 \alpha$ as $\tan \beta$ approaches infinity. Evaluating $\tan^2 \alpha$ , we get $\boxed{994}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

As in Solution 1, drop an altitude $h$ to $c$ . Let $h$ meet $c$ at $P$ , and let $AP = x, BP = y$ Then, $\cot{\alpha} = \frac{1}{\tan{\alpha}} = \frac{x}{h}$ $\cot{\beta} = \frac{1}{\tan{\beta}} = \frac{y}{h}$ . We can calculate $\cot{\gamma}$ using the tangent addition formula , after noticing that $\cot{\gamma} = \frac{1}{\tan{\gamma}}$ . So, we find that \begin{align*} \cot{\gamma} &= \frac{1}{\tan{\gamma}} \\ &= \frac{1}{\frac{\frac{x}{h} + \frac{y}{h}}{1 - \frac{xy}{h^2}}} \\ &= \frac{1}{\frac{(x+y)h}{h^2 - xy}} \\ &= \frac{h^2 - xy}{(x+y)h}. \end{align*} So now we can simplify our original expression: \begin{align*} \frac{\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac{\frac{h^2 - xy}{(x+y)h}}{\frac{x + y}{h}} \\ &= \frac{h^2 - xy}{(x+y)^2}. \end{align*} But notice that $x+y = c$ , so this becomes \[\frac{h^2 - xy}{c^2}.\] Now note that we can use the Pythagorean theorem to calculate $h^2$ , we get that \[h^2 = \frac{a^2 - y^2 + b^2 - x^2}{2}.\] So our expression simplifies to \[\frac{1989c^2 - (x+y)^2}{2c^2}\] since $a^2 + b^2 = 1989c^2$ from the problem and that there is another $-\frac{2xy}{2}$ after the $h^2$ in our expression. Again note that $x+y = c$ , so it again simplifies to $\frac{1988c^2}{2c^2}$ , or $\boxed{994}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

Since no additional information is given, we can assume that triangle ABC is right with the right angle at B. We can use the Pythagorean theorem to say \[c^2+a^2=b^2\] We can now solve for $a$ in terms of $c$ \[c^2+a^2=1989c^2-a^2\] \[a^2=994c^2\] \[a=\sqrt{994}c\] Using the definition of cotangent \[cot(A)=\frac{c}{a}=\frac{1}{\sqrt{994}}\] \[cot(B)=cot(90)=0\] \[cot(C)=\frac{a}{c}=\sqrt{994}\] Plugging into our desired expression, we get $\boxed{994}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_11

A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$ ? (For real $x$ $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$ .)

Let the mode be $x$ , which we let appear $n > 1$ times. We let the arithmetic mean be $M$ , and the sum of the numbers $\neq x$ be $S$ . Then \begin{align*} D &= \left|M-x\right| = \left|\frac{S+xn}{121}-x\right| = \left|\frac{S}{121}-\left(\frac{121-n}{121}\right)x\right| \end{align*} As $S$ is essentially independent of $x$ , it follows that we wish to minimize or maximize $x$ (in other words, $x \in [1,1000]$ ). Indeed, $D(x)$ is symmetric about $x = 500.5$ ; consider replacing all of numbers $x_i$ in the sample with $1001-x_i$ , and the value of $D$ remains the same. So, without loss of generality , let $x=1$ . Now, we would like to maximize the quantity $S$ contains $121-n$ numbers that may appear at most $n-1$ times. Therefore, to maximize $S$ , we would have $1000$ appear $n-1$ times, $999$ appear $n-1$ times, and so forth. We can thereby represent $S$ as the sum of $n-1$ arithmetic series of $1000, 999, \ldots, 1001 - \left\lfloor \frac{121-n}{n-1} \right\rfloor$ . We let $k = \left\lfloor \frac{121-n}{n-1} \right\rfloor$ , so where $R(n)$ denotes the sum of the remaining $121-(n-1)k$ numbers, namely $R(n) = (121-(n-1)k)(1000-k)$ At this point, we introduce the crude estimate that $k=\frac{121-n}{n-1}$ , so $R(n) = 0$ and \begin{align*}2S+2n &= (121-n)\left(2001-\frac{121-n}{n-1}\right)+2n = (120-(n-1))\left(2002-\frac{120}{n-1}\right)\end{align*} Expanding (ignoring the constants, as these do not affect which $n$ yields a maximum) and scaling, we wish to minimize the expression $5(n-1) + \frac{36}{n-1}$ . By AM-GM , we have $5(n-1) + \frac{36}{n-1} \ge 2\sqrt{5(n-1) \cdot \frac{36}{n-1}}$ , with equality coming when $5(n-1) = \frac{36}{n-1}$ , so $n-1 \approx 3$ . Substituting this result and some arithmetic gives an answer of $\boxed{947}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_11

A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$ ? (For real $x$ $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$ .)

With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of numbers as $\underbrace{1,1,1...1,}_\text{n times}\underbrace{2,2,2...2,}_\text{n times}\underbrace{3,3,3...3,}_\text{n times}........\underbrace{1000,1000,1000....}_\text{n+1 times}$ And, we need to find the value of n that makes the sum as low as possible. And, we can create a formula to make it easier. It isn't hard to find the sum. The numbers which are not 1000, average to $\frac{120}{2n}$ or $\frac{60}{n}$ , and there are $120-n$ of them. So, they sum to $\frac{60}{n}(120-n)$ . And, the sum of the numbers that are 1000 is $1000(n+1)$ so, our total sum gets us $1000(n+1)+120/2n(120-n)$ We want to minimize it, since the mode will always be 1000. And, testing the values n = 1, n = 2, n = 3, n = 4, we get these results. $n = 1: 2000+60*119 = 9140$ $n = 2: 3000+30*118 = 6540$ $n = 3: 4000+20*117 = 6340$ $n = 4: 5000+15*116 = 6740$ And, as n grows larger and larger from 4, the values will start increasing. Thus, the lowest possible sum is 6340. Dividing by 121, the lowest possible mean is 52.396...., and thus, the highest possible value of $D$ is 947.604, and the floor of that is $\boxed{947}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_13

Let $S$ be a subset of $\{1,2,3,\ldots,1989\}$ such that no two members of $S$ differ by $4$ or $7$ . What is the largest number of elements $S$ can have?

We first show that we can choose at most 5 numbers from $\{1, 2, \ldots , 11\}$ such that no two numbers have a difference of $4$ or $7$ . We take the smallest number to be $1$ , which rules out $5,8$ . Now we can take at most one from each of the pairs: $[2,9]$ $[3,7]$ $[4,11]$ $[6,10]$ . Now, $1989 = 180\cdot 11 + 9$ . Because this isn't an exact multiple of $11$ , we need to consider some numbers separately. Notice that $1969 = 180\cdot11 - 11 = 179\cdot11$ (*). Therefore we can put the last $1969$ numbers into groups of 11. Now let's examine $\{1, 2, \ldots , 20\}$ . If we pick $1, 3, 4, 6, 9$ from the first $11$ numbers, then we're allowed to pick $11 + 1$ $11 + 3$ $11 + 4$ $11 + 6$ $11 + 9$ . This means we get 10 members from the 20 numbers. Our answer is thus $179\cdot 5 + 10 = \boxed{905}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Let $[RST]$ be the area of polygon $RST$ . We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$ , and line $XP$ intersects line $YZ$ at point $L$ , then $\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.$ This is true because triangles $XPY$ and $YPL$ have their areas in ratio $XP:PL$ (as they share a common height from $Y$ ), and the same is true of triangles $ZPY$ and $LPZ$ We'll also use the related fact that $\dfrac{[XPY]}{[ZPX]} = \dfrac{YL}{LZ}$ . This is slightly more well known, as it is used in the standard proof of Ceva's theorem Now we'll apply these results to the problem at hand. Since $AP = PD = 6$ , this means that $[APB] + [APC] = [BPC]$ ; thus $\triangle BPC$ has half the area of $\triangle ABC$ . And since $PE = 3 = \dfrac{1}{3}BP$ , we can conclude that $\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$ , and thus $\dfrac{1}{4}$ of the area of $\triangle ABC$ . This means that $\triangle APB$ is left with $\dfrac{1}{4}$ of the area of triangle $ABC$ \[[BPC]: [APC]: [APB] = 2:1:1.\] Since $[APC] = [APB]$ , and since $\dfrac{[APC]}{[APB]} = \dfrac{CD}{DB}$ , this means that $D$ is the midpoint of $BC$ Furthermore, we know that $\dfrac{CP}{PF} = \dfrac{[APC] + [BPC]}{[APB]} = 3$ , so $CP = \dfrac{3}{4} \cdot CF = 15$ We now apply Stewart's theorem to segment $PD$ in $\triangle BPC$ —or rather, the simplified version for a median. This tells us that \[2 BD^2 + 2 PD^2 = BP^2+ CP^2.\] Plugging in we know, we learn that \begin{align*} 2 BD^2 + 2 \cdot 36 &= 81 + 225 = 306, \\ BD^2 &= 117. \end{align*} Happily, $BP^2 + PD^2 = 81 + 36$ is also equal to 117. Therefore $\triangle BPD$ is a right triangle with a right angle at $B$ ; its area is thus $\dfrac{1}{2} \cdot 9 \cdot 6 = 27$ . As $PD$ is a median of $\triangle BPC$ , the area of $BPC$ is twice this, or 54. And we already know that $\triangle BPC$ has half the area of $\triangle ABC$ , which must therefore be $\boxed{108}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Because we're given three concurrent cevians and their lengths, it seems very tempting to apply Mass points . We immediately see that $w_E = 3$ $w_B = 1$ , and $w_A = w_D = 2$ . Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C = 1$ and $w_F = 3$ . Thus, $CP = 15$ and $PF = 5$ Recalling that $w_C = w_B = 1$ , we see that $DC = DB$ and $DP$ is a median to $BC$ in $\triangle BCP$ . Applying Stewart's Theorem $BC^2 + 12^2 = 2(15^2 + 9^2)$ , and $BC = 6\sqrt {13}$ . Now notice that $2[BCP] = [ABC]$ , because both triangles share the same base and the $h_{\triangle ABC} = 2h_{\triangle BCP}$ . Applying Heron's formula on triangle $BCP$ with sides $15$ $9$ , and $6\sqrt{13}$ $[BCP] = 54$ and $[ABC] = \boxed{108}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Using a different form of Ceva's Theorem , we have $\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}$ Solving $4y = x + y$ and $x + y = 20$ , we obtain $x = CP = 15$ and $y = FP = 5$ Let $Q$ be the point on $AB$ such that $FC \parallel QD$ . Since $AP = PD$ and $FP\parallel QD$ $QD = 2FP = 10$ . (Stewart's Theorem) Also, since $FC\parallel QD$ and $QD = \frac{FC}{2}$ , we see that $FQ = QB$ $BD = DC$ , etc. ( Stewart's Theorem ) Similarly, we have $PR = RB$ $= \frac12PB = 7.5$ ) and thus $RD = \frac12PC = 4.5$ $PDR$ is a $3-4-5$ right triangle , so $\angle PDR$ $\angle ADQ$ ) is $90^\circ$ . Therefore, the area of $\triangle ADQ = \frac12\cdot 12\cdot 6 = 36$ . Using area ratio, $\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = \boxed{108}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

First, let $[AEP]=a, [AFP]=b,$ and $[ECP]=c.$ Thus, we can easily find that $\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.$ Now, $\frac{[ABP]}{[BPD]}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.$ In the same manner, we find that $[CPD]=a+c.$ Now, we can find that $\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.$ We can now use this to find that $\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.$ Plugging this value in, we find that $\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.$ Now, since $\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},$ we can find that $2AE=EC.$ Setting $AC=b,$ we can apply Stewart's Theorem on triangle $APC$ to find that $(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).$ Solving, we find that $b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.$ But, $3^2+6^2=45,$ meaning that $\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.$ Since $[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,$ we conclude that the answer is $\boxed{108}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Firstly, since they all meet at one single point, denoting the mass of them separately. Assuming $M(A)=6;M(D)=6;M(B)=3;M(E)=9$ ; we can get that $M(P)=12;M(F)=9;M(C)=3$ ; which leads to the ratio between segments, \[\frac{CE}{AE}=2;\frac{BF}{AF}=2;\frac{BD}{CD}=1.\] Denoting that $CE=2x;AE=x; AF=y; BF=2y; CD=z; DB=z.$ Now we know three cevians' length, Applying Stewart theorem to them, getting three different equations: \begin{align} (3x)^2 \cdot 2y+(2z)^2 \cdot y&=(3y)(2y^2+400), \\ (3y)^2 \cdot z+(3x)^2 \cdot z&=(2z)(z^2+144), \\ (2z)^2 \cdot x+(3y)^2 \cdot x&=(3x)(2x^2+144). \end{align} After solving the system of equation, we get that $x=3\sqrt{5};y=\sqrt{13};z=3\sqrt{13}$ pulling $x,y,z$ back to get the length of $AC=9\sqrt{5};AB=3\sqrt{13};BC=6\sqrt{13}$ ; now we can apply Heron's formula here, which is \[\sqrt\frac{(9\sqrt{5}+9\sqrt{13})(9\sqrt{13}-9\sqrt{5})(9\sqrt{5}+3\sqrt{13})(9\sqrt{5}-3\sqrt{13})}{16}=108.\] Our answer is $\boxed{108}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

In Solution 5, instead of finding all of $x, y, z$ , we only need $y, z$ . This is because after we solve for $y, z$ , we can notice that $\triangle BAD$ is isosceles with $AB = BD$ . Because $P$ is the midpoint of the base, $BP$ is an altitude of $\triangle BAD$ . Therefore, $[BAD] = \frac{(AD)(BP)}{2} = \frac{12 \cdot 9}{2} = 54$ . Using the same altitude property, we can find that $[ABC] = 2[BAD] = 2 \cdot 54 = \boxed{108}$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Set $AF=x,$ and use mass points to find that $PF=5$ and $BF=2x.$ Using Stewart's Theorem on $APB,$ we find that $AB=3\sqrt{13}.$ Then we notice that $APB$ is right, which means the area of $APB$ is $27.$ Because $CF=4\cdot PF,$ the area of $ABC$ is $4$ times the area of $APB,$ which means the area of $ABC=4\cdot 27=\boxed{108}.$

https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

We start with mass points as in Solution 2, and receive $BF:AF = 2$ $BD:CD = 1$ $CE:AE = 2$ Law of Cosines on triangles $ADB$ and $ADC$ with $\theta = \angle ADB$ and $BD=DC=x$ gives \[36+x^2-12x\cos \theta = 81\] \[36+x^2-12x\cos (180-\theta) = 36+x^2+12x\cos \theta = 225\] Adding them: $72+2x^2=306 \implies x=3\sqrt{13}$ , so $BC = 6\sqrt{13}$ . Similarly, $AB = 3\sqrt{13}$ and $AC = 9\sqrt{5}$ . Using Heron's, \[[ \triangle ABC ]= \sqrt{\left(\dfrac{9\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{9\sqrt{13}09\sqrt{5}}{2}\right)\left(\dfrac{3\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{-3\sqrt{13}+9\sqrt{5}}{2}\right)} = \boxed{108}.\]

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_1

One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has $\{1,2,3,6,9\}$ as its combination . Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow? 1988-1.png

Currently there are ${10 \choose 5}$ possible combinations. With any integer $x$ from $1$ to $9$ , the number of ways to choose a set of $x$ buttons is $\sum^{9}_{k=1}{10 \choose k}$ . Now we can use the identity $\sum^{n}_{k=0}{n \choose k}=2^{n}$ . So the number of additional combinations is just $2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_2

For any positive integer $k$ , let $f_1(k)$ denote the square of the sum of the digits of $k$ . For $n \ge 2$ , let $f_n(k) = f_1(f_{n - 1}(k))$ . Find $f_{1988}(11)$

We see that $f_{1}(11)=4$ $f_2(11) = f_1(4)=16$ $f_3(11) = f_1(16)=49$ $f_4(11) = f_1(49)=169$ $f_5(11) = f_1(169)=256$ $f_6(11) = f_1(256)=169$ Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_3

Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$

We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2. \begin{align*} {\log_2 (\frac{1}{3}\log_2 x)} &= \frac{1}{3}{\log_2 (\log_2 x)}\\ {\log_2 x = y}\\ {\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\ {3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\ {\log_2 (\frac{1}{3}y)^3} &= {\log_2 (y)}\\ \end{align*} Solving, we get $y^2 = 27$ , which is what we want. $\boxed{27}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_3

Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$

First we have \begin{align*} \log_2(\log_8x)&=\log_8(\log_2x)\\ \frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1 \end{align*} Changing the base in the numerator yields \begin{align*} \frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\ \frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\ \end{align*} Using the property $\frac{\log_ab}{\log_ac}=\log_cb$ yields \begin{align*} \log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ (\log_2x)^\frac{1}{3}&=\log_8x\\ \sqrt[3]{\log_2x}&=\frac{\log_2x}{3} \end{align*} Now setting $y=\log_2x$ , we have \[\sqrt[3]{y}=\frac{y}{3}\] Solving gets $y=\log_2x=3\sqrt{3}\Longrightarrow(\log_2x)^2=(3\sqrt{3})^2=\boxed{27}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_3

Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$

Say that $\log_{2^3}x=a$ and $\log_2x=b$ so we have $\log_2a=\log_{2^3}b$ . And we want $b^2$ $\\ \log_2a=\frac13 \log_{2}b \ \ \text{\tiny{(step 1)}}\\ \frac{\log_2a}{\log_{2}b}=\log_ba=\frac13\\ b^{1/3}=a.$ Because $3a=b$ (as $2^{3a}=x$ and $2^b=x$ from our setup), we have that $b^{1/3}=\frac{b}{3}\\ b^{-2/3}=\frac13\\ b=3^{3/2}\\ \\b^2=3^3=\boxed{27}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_5

Let $m/n$ , in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$ . Find $m + n$

$10^{99} = 2^{99}5^{99}$ , so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$ ; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$ . Our probability is $\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}$ , and $m + n = \boxed{634}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6

It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*). 1988 AIME-6.png

Let the coordinates of the square at the bottom left be $(0,0)$ , the square to the right $(1,0)$ , etc. Label the leftmost column (from bottom to top) $0, a, 2a, 3a, 4a$ and the bottom-most row (from left to right) $0, b, 2b, 3b, 4b$ . Our method will be to use the given numbers to set up equations to solve for $a$ and $b$ , and then calculate $(*)$ $\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & * & \\ \hline 3a & 74 & & & \\ \hline 2a & & & & 186 \\ \hline a & & 103 & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$ We can compute the squares at the intersections of two existing numbers in terms of $a$ and $b$ ; two such equations will give us the values of $a$ and $b$ . On the fourth row from the bottom, the common difference is $74 - 3a$ , so the square at $(2,3)$ has a value of $148 - 3a$ . On the third column from the left, the common difference is $103 - 2b$ , so that square also has a value of $2b + 3(103 - 2b) = 309 - 4b$ . Equating, we get $148 - 3a = 309 - 4b \Longrightarrow 4b - 3a = 161$ Now we compute the square $(2,2)$ . By rows, this value is simply the average of $2a$ and $186$ , so it is equal to $\frac{2a + 186}{2} = a + 93$ . By columns, the common difference is $103 - 2b$ , so our value is $206 - 2b$ . Equating, $a + 93 = 206 - 2b \Longrightarrow a + 2b = 113$ Solving \begin{align*}4b - 3a &= 161\\ a + 2b &= 113 \end{align*} gives $a = 13$ $b = 50$ . Now it is simple to calculate $(4,3)$ . One way to do it is to see that $(2,2)$ has $206 - 2b = 106$ and $(4,2)$ has $186$ , so $(3,2)$ has $\frac{106 + 186}{2} = 146$ . Now, $(3,0)$ has $3b = 150$ , so $(3,2) = \frac{(3,0) + (3,4)}{2} \Longrightarrow (3,4) = * = \boxed{142}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6

It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*). 1988 AIME-6.png

First, let $a =$ the number to be placed in the first column, fourth row. Let $b =$ the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of $a$ and $b$ $\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\ \hline 3a & & & & \\ \hline 2a & & & & \\ \hline a & & & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$ Next, let $a + b + c =$ the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of $a$ $b$ , and $c$ $\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & & & \\ \hline 3a & 3a + b + 3c & & & \\ \hline 2a & 2a + b + 2c & & & \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$ We have now determined at least two values in each row and column. We can finish the table without introducing any more variables: $\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & 4a + 2b + 8c & 4a + 3b + 12c & 4a + 4b + 16c \\ \hline 3a & 3a + b + 3c & 3a + 2b + 6c & 3a + 3b + 9c & 3a + 4b + 12c \\ \hline 2a & 2a + b + 2c & 2a + 2b + 4c & 2a + 3b + 6c & 2a + 4b + 8c \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$ We now have a system of equations. Solving, we find that $(a,b,c) = (13,50, - 5)$ . The number in the square marked by the asterisk is $4a + 3b + 12c = \boxed{142}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6

It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*). 1988 AIME-6.png

We begin with the table that was given to us and add in the following arithmetic progression on the bottom: $\begin{tabular}[b]{|c|c|c|c|c|}\hline & & & * & \\ \hline & 74 & & & \\ \hline & & & & 186 \\ \hline & & 103 & & \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$ Since all the rows and columns satisfy an arithmetic progression, we have the following: $\begin{tabular}[b]{|c|c|c|c|c|}\hline & & 412 - 6x & 392 - 5x & 372 - 4x \\ \hline & 74 & 309 - 4x & 294 - 3x & 279 - 2x \\ \hline & & 206 - 2x & 196 - x & 186 \\ \hline & & 103 & 98 + x & 2x + 93 \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$ We can solve for $x$ in the 2nd row, namely $324 - 5x = 74$ because the arithmetic progression from left to right has difference $x - 15$ . Therefore, we have $x = 50$ , and because the desired asterisk is $392 - 5x$ , the answer is $392 - 250$ $\boxed{142}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_7

In triangle $ABC$ $\tan \angle CAB = 22/7$ , and the altitude from $A$ divides $BC$ into segments of length 3 and 17. What is the area of triangle $ABC$

Call $\angle BAD$ $\alpha$ and $\angle CAD$ $\beta$ . So, $\tan \alpha = \frac {17}{h}$ and $\tan \beta = \frac {3}{h}$ . Using the tangent addition formula $\tan (\alpha + \beta) = \dfrac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}$ , we get $\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {22}{7}$ Simplifying, we get $\frac {20h}{h^2 - 51} = \frac {22}{7}$ . Cross-multiplying and simplifying, we get $11h^2-70h-561 = 0$ . Factoring, we get $(11h+51)(h-11) = 0$ , so we take the positive positive solution, which is $h = 11$ . Therefore, the answer is $\frac {20 \cdot 11}{2} = 110$ , so the answer is $\boxed{110}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_8

The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$

Let $z = x+y$ . By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\ f(x,z) &= \frac{z}{z-x} \cdot f(x,z-x). \end{align*} Using the properties of $f,$ we have \begin{align*} f(14,52) &= \frac{52}{38} \cdot f(14,38) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot f(14,24) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(14,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(10,14)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(10,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(4,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(4,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(2,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot f(2,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot 2\\ &=\boxed{364} ~MRENTHUSIASM (credit given to AoPS)

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_8

The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$

Since all of the function's properties contain a recursive definition except for the first one, we know that $f(x,x) = x$ in order to obtain an integer answer. So, we have to transform $f(14,52)$ to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one. Note that \[f(14,52) = f(14,14 + 38) = \frac{52}{38}\cdot f(14,38).\] Repeating the process several times, \begin{align*} f(14,52) & = f(14,14 + 38) \\ & = \frac{52}{38}\cdot f(14,38) \\ & = \frac{52}{38}\cdot \frac{38}{24}\cdot f(14,14 + 24) \\ & = \frac{52}{24}\cdot f(14,24) \\ & = \frac{52}{10}\cdot f(10,14) \\ & = \frac{52}{10}\cdot \frac{14}{4}\cdot f(10,4) \\ & = \frac{91}{5}\cdot f(4,10) \\ & = \frac{91}{3}\cdot f(4,6) \\ & = 91\cdot f(2,4) \\ & = 91\cdot 2 \cdot f(2,2) \\ & = \boxed{364}

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_8

The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$

Notice that $f(x,y) = \mathrm{lcm}(x,y)$ satisfies all three properties: For the first two properties, it is clear that $\mathrm{lcm}(x,x) = x$ and $\mathrm{lcm}(x,y) = \mathrm{lcm}(y,x)$ For the third property, using the identities $\gcd(x,y) \cdot \mathrm{lcm}(x,y) = x\cdot y$ and $\gcd(x,x+y) = \gcd(x,y)$ gives \begin{align*} y \cdot \mathrm{lcm}(x,x+y) &= \dfrac{y \cdot x(x+y)}{\gcd(x,x+y)} \\ &= \dfrac{(x+y) \cdot xy}{\gcd(x,y)} \\ &= (x+y) \cdot \mathrm{lcm}(x,y). \end{align*} Hence, $f(x,y) = \mathrm{lcm}(x,y)$ is a solution to the functional equation. Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value of $f(14,52)$ Therefore, we have \begin{align*} f(14,52) &= \mathrm{lcm}(14,52) \\ &= \mathrm{lcm}(2 \cdot 7,2^2 \cdot 13) \\ &= 2^2 \cdot 7 \cdot 13 \\ &= \boxed{364}

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9

Find the smallest positive integer whose cube ends in $888$

$n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8$ and $n^3 \equiv 13 \pmod{125}$ $n \equiv 2 \pmod 5$ due to the last digit of $n^3$ . Let $n = 5a + 2$ . By expanding, $125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}$ By looking at the last digit again, we see $a \equiv 3 \pmod5$ , so we let $a = 5a_1 + 3$ where $a_1 \in \mathbb{Z^+}$ . Plugging this in to $5a^2 + 12a \equiv 1 \pmod{25}$ gives $10a_1 + 6 \equiv 1 \pmod{25}$ . Obviously, $a_1 \equiv 2 \pmod 5$ , so we let $a_1 = 5a_2 + 2$ where $a_2$ can be any non-negative integer. Therefore, $n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67$ $n^3$ must also be a multiple of $8$ , so $n$ must be even. $125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2$ . Therefore, $a_2 = 2a_3 + 1$ , where $a_3$ is any non-negative integer. The number $n$ has form $125(2a_3+1)+67 = 250a_3+192$ . So the minimum $n = \boxed{192}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9

Find the smallest positive integer whose cube ends in $888$

Let $x^3 = 1000a + 888$ . We factor an $8$ out of the right hand side, and we note that $x$ must be of the form $x = 2y$ , where $y$ is a positive integer. Then, this becomes $y^3 = 125a + 111$ . Taking mod $5$ $25$ , and $125$ , we get $y^3 \equiv 1\pmod 5$ $y^3 \equiv 11\pmod{25}$ , and $y^3 \equiv 111\pmod{125}$ We can work our way up, and find that $y\equiv 1\pmod 5$ $y\equiv 21\pmod{25}$ , and finally $y\equiv 96\pmod{125}$ . This gives us our smallest value, $y = 96$ , so $x = \boxed{192}$ , as desired. - Spacesam

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9

Find the smallest positive integer whose cube ends in $888$

Let this integer be $x.$ Note that \[x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.\] We wish to find the residue of $x$ mod $125.$ Note that \[x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}\] using our congruence in mod $5.$ The residue that works must also satisfy $x^3 \equiv 13 \pmod{25}$ from our original congruence. Noting that $17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}$ (and bashing out the other residues perhaps but they're not that hard), we find that \[x \equiv 17 \pmod{25}.\] Thus, \[x \equiv 17,42,67,92,117 \pmod{125}.\] The residue that works must also satisfy $x^3 \equiv 13 \pmod{125}$ from our original congruence. It is easy to memorize that \[17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.\] Also, \[42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.\] Finally, \[67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},\] as desired. Thus, $x$ must satisfy \[x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.\] ~samrocksnature

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9

Find the smallest positive integer whose cube ends in $888$

This number is in the form of $10k+2$ , after binomial expansion, we only want $600k^2+120k\equiv 880 \pmod{1000}$ . We realize that $600,120$ are both multiples of $8$ , we only need that $600k^2+120k \equiv 5\pmod{125}$ , so we write $600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}$ Then, we write $5k^2+k=25m-1, 5k^2+k+1=25m$ so $k+1$ must be a multiple of $5$ at least, so $k\equiv {-1, -6, -11, -16, -21} \pmod {25}$ after checking, when $k=-6, 5k^2+k+1=175=25\cdot 7$ . So $k\equiv -6 \pmod{25}$ , smallest $k=19$ , the number is $\boxed{192}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10

convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face

The polyhedron described looks like this, a truncated cuboctahedron. The number of segments joining the vertices of the polyhedron is ${48\choose2} = 1128$ . We must now subtract out those segments that lie along an edge or a face. Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$ Each vertex is formed by the intersection of 3 edges. Since every edge is counted twice, once at each of its endpoints, the number of edges $E$ is $\frac{3}{2}V = 72$ Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes $\frac{n(n-3)}{2} = 2$ diagonals, each hexagon $9$ , and each octagon $20$ . The number of diagonals is thus $2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216$ Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = \boxed{840}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10

convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face

We first find the number of vertices on the polyhedron: There are 4 corners per square, 6 corners per hexagon, and 8 corners per octagon. Each vertex is where 3 corners coincide, so we count the corners and divide by 3. $\text{vertices} = \frac{12 \cdot 4 + 8 \cdot 6 + 6 \cdot 8}{3}=48$ We know that all vertices look the same (from the problem statement), so we should find the number of line segments originating from a vertex, and multiply that by the number of vertices, and divide by 2 (because each space diagonal is counted twice because it has two endpoints). Counting the vertices that are on the same face as an arbitrary vertex, we find that there are 13 vertices that aren't possible endpoints of a line originating from the vertex in the middle of the diagram. You can draw a diagram to count this better: 1988AIME10.png Since 13 aren't possible endpoints, that means that there are 35 possible endpoints per vertex. The total number of segments joining vertices that aren't on the same face is $48\cdot 35\cdot \frac 12 = 24 \cdot 35 = \boxed{840}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10

convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face

Since at each vertex one square, one hexagon, and one octagon meet, then there are a total of $12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$ vertices. This means that for each segment we have $48$ choices of vertices for the first endpoint of the segment. Since each vertex is the meeting point of a square, octagon, and hexagon, then there are $3$ other vertices of the square that are not the first one, and connecting the first point to any of these would result in a segment that lies on a face or edge. Similarly, there are $5$ points on the adjacent hexagon and $7$ points on adjoining octagon that, when connected to the first point, would result in a diagonal or edge. However, the square and hexagon share a vertex, as do the square and octagon, and the hexagon and octagon. Subtracting these from the $47$ vertices we have left to choose from, and adding the $3$ that we counted twice, we get \[48 \cdot (47 - 3 - 5 - 7 + 3) = 48 \cdot 35 = 1680\] We over-counted, however, as choosing vertex $A$ then $B$ is the same thing as choosing $B$ then $A$ , so we must divide $1680 / 2 = \boxed{840}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10

convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face

In the same ways as above, we find that there are 48 vertices. Now, notice that there are $\binom{48}{2}$ total possible ways to choose two vertices. However, we must remove the cases where the segments do not lie in the interior of the polyhedron. We get \[\binom{48}{2}-12\binom{4}{2}-8\binom{6}{2}-6\binom{8}{2}=768\] We remover all the possible edges of the squares, hexagons, and octagons. However, we have undercounted! We must add back the number of edges because when we subtracted the three binomials from $\binom{48}{2}$ we removed each edge twice (each edge is shared by two polygons). This means that we need to add back the number of edges, 72. Thus, we get $768+72=\boxed{840}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_11

Let $w_1, w_2, \dots, w_n$ be complex numbers . A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$ $w_2 = - 7 + 64i$ $w_3 = - 9 + 200i$ $w_4 = 1 + 27i$ , and $w_5 = - 14 + 43i$ , there is a unique mean line with $y$ -intercept 3. Find the slope of this mean line.

We know that $\sum_{k=1}^5 w_k = 3 + 504i$ And because the sum of the 5 $z$ 's must cancel this out, $\sum_{k=1}^5 z_k = 3 + 504i$ We write the numbers in the form $a + bi$ and we know that $\sum_{k=1}^5 a_k = 3$ and $\sum_{k=1}^5 b_k = 504$ The line is of equation $y=mx+3$ . Substituting in the polar coordinates, we have $b_k = ma_k + 3$ Summing all 5 of the equations given for each $k$ , we get $504 = 3m + 15$ Solving for $m$ , the slope, we get $\boxed{163}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_11

Let $w_1, w_2, \dots, w_n$ be complex numbers . A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$ $w_2 = - 7 + 64i$ $w_3 = - 9 + 200i$ $w_4 = 1 + 27i$ , and $w_5 = - 14 + 43i$ , there is a unique mean line with $y$ -intercept 3. Find the slope of this mean line.

The mean line for $w_1, . . ., w_5$ must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is $(\frac{3}{5}, \frac{504i}{5})$ . Since we now have two points, namely that one and $(0, 3i)$ , we can simply find the slope between them, which is $\boxed{163}$ by the good ol' slope formula.

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12

Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$ 1988 AIME-12.png

Call the cevians AD, BE, and CF. Using area ratios ( $\triangle PBC$ and $\triangle ABC$ have the same base), we have: $\frac {d}{a + d} = \frac {[PBC]}{[ABC]}$ Similarily, $\frac {d}{b + d} = \frac {[PCA]}{[ABC]}$ and $\frac {d}{c + d} = \frac {[PAB]}{[ABC]}$ Then, $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} = \frac {[ABC]}{[ABC]} = 1$ The identity $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1$ is a form of Ceva's Theorem Plugging in $d = 3$ , we get \[\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1\] \[3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)\] \[3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27\] \[9(a + b + c) + 54 = abc=\boxed{441}\]

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12

Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$ 1988 AIME-12.png

Let $A,B,C$ be the weights of the respective vertices. We see that the weights of the feet of the cevians are $A+B,B+C,C+A$ . By mass points , we have that: \[\dfrac{a}{3}=\dfrac{B+C}{A}\] \[\dfrac{b}{3}=\dfrac{C+A}{B}\] \[\dfrac{c}{3}=\dfrac{A+B}{C}\] If we add the equations together, we get $\frac{a+b+c}{3}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\frac{43}{3}$ If we multiply them together, we get $\frac{abc}{27}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\frac{49}{3} \implies abc=\boxed{441}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12

Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$ 1988 AIME-12.png

You can use mass points to derive $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d}=1.$ Plugging it in yields $\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1.$ We proceed as we did in Solution 1 - however, to make the equation look less messy, we do the substitution $a'=a+3,b'=b+3,c'=c+3.$ Then we have $\frac{3}{a'}+\frac{3}{b'}+\frac{3}{c'}=1.$ Clearing fractions gives us $a'b'c'=3a'b'+3b'c'+3c'a'\to a'b'c'-3a'b'-3b'c'-3c'a'=0.$ Factoring yields $(a'-3)(b'-3)(c'-3)=9(a'+b'+c')-27,$ and the left hand side looks suspiciously like what we want to find. (It is.) Substituting yields our answer as $9\cdot 52-27=\boxed{441}.$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12

Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$ 1988 AIME-12.png

A cool identity derived from Ceva's Theorem is that: \[\frac{AP}{PA'}\frac{BP}{PB'}\frac{CP}{PC'} = 2 + \frac{AP}{PA'} + \frac{BP}{PB'} + \frac{CP}{PC'}\] To see this, we use another Ceva's Theorem identity (sometimes attributed to Gergonne): $\frac{AP}{PA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}$ , and similarly for cevians $BB'$ and $CC'$ . And then: $\frac{AP}{PA'}\frac{BP}{PB'}\frac{CP}{PC'} = \left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\right) \left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right) \left(\frac{CB'}{B'A}+\frac{CA'}{A'B}\right) = \\ \underbrace{\frac{AB'\cdot CA'\cdot BC'}{B'C\cdot A'B\cdot C'A}}_{Ceva} + \underbrace{\frac{AC'\cdot BA'\cdot CB'}{C'B\cdot A'C\cdot B'A}}_{Ceva} + \underbrace{\frac{AB'}{B'C} + \frac{AC'}{C'B}}_{Gergonne} + \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{Gergonne} + \underbrace{\frac{CA'}{A'B} + \frac{CB'}{B'A}}_{Gergonne} = \\ 2 + \frac{AP}{PA'} + \frac{BP}{PB'} + \frac{CP}{PC'}$ Inserting $a, b, c, d$ into our identity gives: \[\frac{a}{d}\frac{b}{d}\frac{c}{d}=2+\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\implies abc=d^3(2+\frac{a+b+c}{d})=3^3(2+\frac{43}{3})=\boxed{441}\]

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$

Let $F_n$ represent the $n$ th number in the Fibonacci sequence. Therefore, \begin{align*} x^2 - x - 1 = 0&\Longrightarrow x^n = F_n(x), \ n\in N \\ &\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N. \end{align*} The above uses the similarity between the Fibonacci recursion|recursive definition, $F_{n+2} - F_{n+1} - F_n = 0$ , and the polynomial $x^2 - x - 1 = 0$ \begin{align*} 0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1 &\Longrightarrow (aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q \\ &\Longrightarrow aF_{17} + bF_{16} = 0 \text{ and } aF_{16} + bF_{15} + 1 = 0 \\ &\Longrightarrow a = F_{16},\ b = - F_{17} \\ &\Longrightarrow a = \boxed{987}

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$

We can long divide and search for a pattern; then the remainder would be set to zero to solve for $a$ . Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is \[(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0.\] Since the coefficient of $x$ must be zero, this gives us two equations, $F_{16}b + F_{17}a = 0$ and $F_{15}b + F_{16}a + 1 = 0$ . Solving these two as above, we get that $a = \boxed{987}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$

Trying to divide $ax^{17} + bx^{16} + 1$ by $x^2-x-1$ would be very tough, so let's try to divide using smaller degrees of $x$ . Doing $\frac{ax^3+bx^2+1}{x^2-x-1}$ , we get the following systems of equations: \begin{align*} a+b &= -1, \\ 2a+b &= 0. \end{align*} Continuing with $\frac{ax^4+bx^3+1}{x^2-x-1}$ \begin{align*} 2a+b &= -1, \\ 3a+2b &= 0. \end{align*} There is somewhat of a pattern showing up, so let's try $\frac{ax^5+bx^4+1}{x^2-x-1}$ to verify. We get: \begin{align*} 3a+2b &= -1, \\ 5a+3b &= 0. \end{align*} Now we begin to see that our pattern is actually the Fibonacci Numbers! Using the previous equations, we can make a general statement about $\frac{ax^n+bx^{n-1}+1}{x^2-x-1}$ \begin{align*} af_{n-1}+bf_{n-2} &= -1, \\ af_n+bf_{n-1} &= 0. \end{align*} Also, noticing our solutions from the previous systems of equations, we can create the following statement: If $ax^n+bx^{n-1}+1$ has $x^2-x-1$ as a factor, then $a=f_{n-1}$ and $b = f_n.$ Thus, if $ax^{17}+bx^{16}+1$ has $x^2-x-1$ as a factor, we get that $a = 987$ and $b = -1597,$ so $a = \boxed{987}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$

Let's work backwards! Let $F(x) = ax^{17} + bx^{16} + 1$ and let $P(x)$ be the polynomial such that $P(x)(x^2 - x - 1) = F(x)$ Clearly, the constant term of $P(x)$ must be $- 1$ . Now, we have \[(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1),\] where $c_{i}$ is some coefficient. However, since $F(x)$ has no $x$ term, it must be true that $c_{15} = 1$ Let's find $c_{14}$ now. Notice that all we care about in finding $c_{14}$ is that $(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}$ . Therefore, $c_{14} = - 2$ . Undergoing a similar process, $c_{13} = 3$ $c_{12} = - 5$ $c_{11} = 8$ , and we see a nice pattern. The coefficients of $P(x)$ are just the Fibonacci sequence with alternating signs! Therefore, $a = c_1 = F_{16}$ , where $F_{16}$ denotes the 16th Fibonnaci number and $a = \boxed{987}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$

We are given that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1,$ so the roots of $x^2 - x - 1$ must also be roots of $ax^{17} + bx^{16} + 1.$ Let $x=r$ be a root of $x^2 - x - 1$ so that $r^2 - r - 1 = 0,$ or $r^2 = r + 1.$ It follows that \[ar^{17} + br^{16} + 1 = 0. \hspace{20mm} (\bigstar)\] Note that \begin{align*} r^4 &= (r+1)^2 \\ &= r^2 + 2r + 1 \\ &= (r+1) + 2r + 1 \\ &= 3r + 2, \\ r^8 &= (3r+2)^2 \\ &= 9r^2 + 12r + 4 \\ &= 9(r+1) + 12r + 4 \\ &= 21r + 13, \\ r^{16} &= (21r + 13)^2 \\ &= 441r^2 + 546r + 169 \\ &= 441(r+1) +546r + 169 \\ &= 987r + 610. \end{align*} We rewrite the left side of $(\bigstar)$ as a linear expression of $r:$ \begin{align*} (ar+b)r^{16} + 1 &= 0 \\ (ar+b)(987r + 610) + 1 &= 0 \\ 987ar^2 + (610a+987b)r + 610b + 1 &= 0 \\ 987a(r+1) + (610a+987b)r + 610b + 1 &= 0 \\ (1597a+987b)r + (987a + 610b + 1) &= 0. \end{align*} Since this linear equation has two solutions of $r,$ it must be an identity. Therefore, we have the following system of equations: \begin{align*} 1597a+987b &= 0, \\ 987a+610b &= -1. \end{align*} To eliminate $b,$ we multiply the first equation by $610$ and multiply the second equation by $987,$ then subtract the resulting equations: \begin{align*} 610(1597a)+610(987b) &= 0, \\ 987(987a)+987(610b) &= -987, \end{align*} from which \begin{align*} (610\cdot1597-987\cdot987)a&=987 \\ (974170-974169)a&=987 \\ a&=\boxed{987} ~MRENTHUSIASM

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$

For simplicity, let $f(x) =ax^{17} + bx^{16} + 1$ and $g(x) = x^2-x-1$ . Notice that the roots of $g(x)$ are also roots of $f(x)$ . Let these roots be $u,v$ . We get the system \begin{align*} au^{17} + bu^{16} + 1 &= 0, \\ av^{17} + bv^{16} + 1 &= 0. \end{align*} If we multiply the first equation by $v^{16}$ and the second by $u^{16}$ we get \begin{align*} u^{17} v^{16} a + u^{16} v^{16} b + v^{16} &= 0, \\ u^{16} v^{17} a + u^{16} v^{16} b + u^{16} &= 0. \end{align*} Now subtracting, we get \[a(u^{17}v^{16} -u^{16} v^{17}) = u^{16}-v^{16} \implies a = \frac{u^{16} - v^{16}}{u^{17}v^{16} -u^{16} v^{17}}.\] By Vieta's, $uv=-1$ so the denominator becomes $u-v$ . By difference of squares and dividing out $u-v$ we get \[a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).\] A simple exercise of Vieta's gets us $a= \boxed{987}.$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_15

In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order $1, 2, 3, 4, 5, 6, 7, 8, 9$ While leaving for lunch, the secretary tells a colleague that letter $8$ has already been typed but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based on the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)

Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$ , or the last element of $S$ can be removed. The question asks for the number of different orders in which all of the remaining elements of $S$ can be removed, given that $8$ had been removed already. Since $8$ had already been added to the pile, the numbers $1 \ldots 7$ had already been added at some time to the pile; $9$ might or might not have been added yet. So currently $S$ is a subset of $\{1, 2, \ldots 7\}$ , possibly with $9$ at the end. Given that $S$ has $k$ elements, there are $k+1$ intervals for $9$ to be inserted, or $9$ might have already been placed, giving $k+2$ different possibilities. Thus, the answer is $\sum_{k=0}^{7} {7 \choose k}(k+2)$ $= 1 \cdot 2 + 7 \cdot 3 + 21 \cdot 4 + 35 \cdot 5 + 35 \cdot 6 + 21 \cdot 7 + 7 \cdot 8 + 1 \cdot 9$ $= \boxed{704}$

https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_15

In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order $1, 2, 3, 4, 5, 6, 7, 8, 9$ While leaving for lunch, the secretary tells a colleague that letter $8$ has already been typed but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based on the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)

$1 \cdot 2 + 7 \cdot 3 + 21 \cdot 4 + 35 \cdot 5 + 35 \cdot 6 + 21 \cdot 7 + 7 \cdot 8 + 1 \cdot 9$ $=1 \cdot (2+9) + 7 \cdot (3+8) + 21 \cdot (4+7) + 35 \cdot (5+6) = 64 \cdot 11 = \boxed{704}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_1

An ordered pair $(m,n)$ of non-negative integers is called "simple" if the addition $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$

Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective digit in $1492$ (the values of $n$ are then fixed). Thus, the number of ordered pairs will be $(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_2

What is the largest possible distance between two points , one on the sphere of radius 19 with center $(-2,-10,5)$ and the other on the sphere of radius 87 with center $(12,8,-16)$

The distance between the two centers of the spheres can be determined via the distance formula in three dimensions: $\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \sqrt{14^2 + 18^2 + 21^2} = 31$ . The largest possible distance would be the sum of the two radii and the distance between the two centers, making it $19 + 87 + 31 = \boxed{137}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_3

By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called nice if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?

Let $p(n)$ denote the product of the distinct proper divisors of $n$ . A number $n$ is nice in one of two instances: We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form $n = pqr$ (with $p,q$ prime and $r > 1$ ) or $n = p^e$ (with $e \neq 3$ ). In the former case, it suffices to note that $p(n) \ge (pr) \cdot (qr) = pqr^2 > pqr = n$ In the latter case, then $p(n) = p \cdot p^2 \cdots p^{(e-1)} = p^{(e-1)e/2}$ For $p(n) = n$ , we need $p^{(e-1)e/2} = p^e \Longrightarrow e^2 - e = 2e \Longrightarrow$ $e = 0$ or $e = 3$ Since $e \neq 3$ , in the case $e = 0 \Longrightarrow n = 1$ does not work. Thus, listing out the first ten numbers to fit this form, $2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,$ $\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,$ $\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,$ $\ 3^3 = 27,\ 3 \cdot 11 = 33$ Summing these yields $\boxed{182}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_4

Find the area of the region enclosed by the graph of $|x-60|+|y|=\left|\frac{x}{4}\right|.$

1987 AIME-4.png Since $|y|$ is nonnegative $\left|\frac{x}{4}\right| \ge |x - 60|$ . Solving this gives us two equations: $\frac{x}{4} \ge x - 60\ \mathrm{and} \ -\frac{x}{4} \le x - 60$ . Thus, $48 \le x \le 80$ . The maximum and minimum y value is when $|x - 60| = 0$ , which is when $x = 60$ and $y = \pm 15$ . Since the graph is symmetric about the y-axis, we just need casework upon $x$ $\frac{x}{4} > 0$ , so we break up the condition $|x-60|$ The area of the region enclosed by the graph is that of the quadrilateral defined by the points $(48,0),\ (60,15),\ (80,0), \ (60,-15)$ . Breaking it up into triangles and solving or using the Shoelace Theorem , we get $2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_5

Find $3x^2 y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2 y^2 = 30x^2 + 517$

If we move the $x^2$ term to the left side, it is factorable with Simon's Favorite Factoring Trick \[(3x^2 + 1)(y^2 - 10) = 517 - 10\] $507$ is equal to $3 \cdot 13^2$ . Since $x$ and $y$ are integers, $3x^2 + 1$ cannot equal a multiple of three. $169$ doesn't work either, so $3x^2 + 1 = 13$ , and $x^2 = 4$ . This leaves $y^2 - 10 = 39$ , so $y^2 = 49$ . Thus, $3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_6

Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm. AIME 1987 Problem 6.png

Since $XY = WZ$ $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area $\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)$ . This number is also equal to one quarter the area of the entire rectangle, which is $\frac{19\cdot AB}{4}$ , so we have $AB = XY + 87$ In addition, we see that the perimeter of the rectangle is $2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY$ , so $AB + 19 = 2XY$ Solving these two equations gives $AB = \boxed{193}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_6

Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm. AIME 1987 Problem 6.png

Let $YB=a$ $CZ=b$ $AX=c$ , and $WD=d$ . First we drop a perpendicular from $Q$ to a point $R$ on $BC$ so $QR=h$ . Since $XY = WZ$ and $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, the heights of the trapezoids are both $\frac{19}{2}$ .From here, we have that $[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}$ . We are told that this area is equal to $[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}$ . Setting these equal to each other and solving gives $h=53$ . In the same way, we find that the perpendicular from $P$ to $AD$ is $53$ . So $AB=53*2+87=\boxed{193}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_6

Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm. AIME 1987 Problem 6.png

Since $XY = YB + BC + CZ = ZW = WD + DA + AX$ . Let $a = AX + DW = BY + CZ$ . Since $2AB - 2a = XY = WZ$ , then $XY = AB - a$ .Let $S$ be the midpoint of $DA$ , and $T$ be the midpoint of $CB$ . Since the area of $PQWZ$ and $PQYX$ are the same, then their heights are the same, and so $PQ$ is equidistant from $AB$ and $CD$ . This means that $PS$ is perpendicular to $DA$ , and $QT$ is perpendicular to $BC$ . Therefore, $PSCW$ $PSAX$ $QZCT$ , and $QYTB$ are all trapezoids, and $QT = (AB - 87)/$ 2. This implies that \[((a + 2((AB - 87)/2)/2) \cdot 19 = (((AB - a) + 87)/2) \cdot 19\] \[(a + AB - 87) = AB - a + 87\] \[2a = 174\] \[a = 87\] Since $a + CB = XY$ $XY = 19 + 87 = 106$ , and $AB = 106 + 87 = \boxed{193}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_8

What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$

Flip the fractions and subtract one from all sides to yield \[\frac{7}{8}>\frac{k}{n}>\frac{6}{7}.\] Multiply both sides by $56n$ to get \[49n>56k>48n.\] This is equivalent to find the largest value of $n$ such that there is only one multiple of 56 within the open interval between $48n$ and $49n$ . If $n=112,$ then $98>k>96$ and $k=97$ is the unique value. For $n\geq 113,$ there is at least $(49\cdot 113-48\cdot 113)-1=112$ possible numbers for $k$ and there is one $k$ every 56 numbers. Hence, there must be at least two values of $k$ that work. So, the largest value of $n$ is $\boxed{112}$

https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_8

What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$

Notice that in order for $k$ to be unique, $\frac{n}{n + k+ 1} \le \frac{8}{15}$ and $\frac{n}{n+ k-1} \ge \frac{7}{13}$ must be true. Solving these inequalities for $k$ yields $\frac{7}{6}(k-1) \le n \le \frac{8}{7}(k+1)$ Thus, we want to find $k$ such that $\frac{7}{6}(k-1)\le \frac{8}{7}(k+1)$ . Solving this inequality yields $k \le 97$ , and plugging this into $\frac{n}{n+k} < \frac{7}{13}$ in the original equation yields $n \le 112$ so the answer is $\boxed{112}$

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 the total number of steps be $x$ , the speed of the escalator be $e$ and the speed of Bob be $b$ In the time it took Bob to climb up the escalator he saw 75 steps and also climbed the entire escalator. Thus the contribution of the escalator must have been an additional $x - 75$ steps. Since Bob and the escalator were both moving at a constant speed over the time it took Bob to climb, the ratio of their distances covered is the same as the ratio of their speeds, so $\frac{e}{b} = \frac{x - 75}{75}$ Similarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved $150 - x$ steps in that time. Thus $\frac{e}{3b} = \frac{150 - x}{150}$ or $\frac{e}{b} = \frac{150 - x}{50}$ Equating the two values of $\frac{e}{b}$ we have $\frac{x - 75}{75} = \frac{150 - x}{50}$ and so $2x - 150 = 450 - 3x$ and $5x = 600$ and $x = \boxed{120}$ , the answer.

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.

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.

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.