id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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6,601 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4 | 1 | The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33... | Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$ , and let $l,w$ represent the dimensions of the rectangle.
The picture shows that \begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a... | 260 |
6,602 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4 | 2 | The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33... | We can just list the equations: \begin{align*} s_3 &= s_1 + s_2 \\ s_4 &= s_3 + s_1 \\ s_5 &= s_4 + s_3 \\ s_6 &= s_5 + s_4 \\ s_7 &= s_5 + s_3 + s_2 \\ s_8 &= s_7 + s_2 \\ s_9 &= s_8 + s_2 - s_1 \\ s_9 + s_8 &= s_7 + s_6 + s_5 \end{align*} We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \begin{alig... | 260 |
6,603 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4 | 3 | The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33... | We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some "side length chasing" and get $4a - 4 = 2a + 5$ . Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$ . Thus, the perimeter of the rectangle is $2(61 + 69) = \boxed{260}.$ | 260 |
6,604 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | 1 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | \begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}
Because $y > x$ , we only consider $+2$
For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.
The maximum that $\sqrt{y}$ can... | 997 |
6,605 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | 2 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | Let $a^2$ $x$ and $b^2$ $y$ , where $a$ and $b$ are positive.
Then \[\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2\] \[a^2 + b^2 = 2ab + 4\] \[(a-b)^2 = 4\] \[(a-b) = \pm 2\]
This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.
Because $\sqrt{10^6} = 10^3$ , then we can use a... | 997 |
6,606 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | 3 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$ . We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$ . Subtracting 4 and squaring gives \[((x+y)-4)^2 = 4xy\] \[((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy\] \[x^2 - 2xy + y^2 + 16 - 8x - 8y = 0\]
Notice that $((x-y)-4)^2 = x^... | 997 |
6,607 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6 | 4 | For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ | Rearranging our conditions to
\[x^2-2xy+y^2+16-8x-8y=0 \implies\] \[(y-x)^2=8(x+y-2).\]
Thus, $4|y-x.$
Now, let $y = 4k+x.$ Plugging this back into our expression, we get
\[(k-1)^2=x-1.\]
There, a unique value of $x, y$ is formed for every value of $k$ . However, we must have
\[y<10^6 \implies (k+1)^2< 10^6-1\]
and
\[x... | 997 |
6,608 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7 | 2 | Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symme... | Let $r = \frac{m}{n} = z + \frac {1}{y}$
\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \f... | 5 |
6,609 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_10 | 1 | sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$ | Let the sum of all of the terms in the sequence be $\mathbb{S}$ . Then for each integer $k$ $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \ldots, 100$
\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb... | 173 |
6,610 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_10 | 2 | sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$ | Consider $x_k$ and $x_{k+1}$ . Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\dfrac{1}{2}.$
In terms of $x_{50},$ the sequence is $x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{... | 173 |
6,611 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11 | 1 | Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$ | Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$ , it follows that $\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$ , where $-3 \le x,y \le 3$ . Thus every number in the form of $a/b$ will be expressed one time in the product
\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2... | 248 |
6,612 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11 | 2 | Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$ | Essentially, the problem asks us to compute \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}\] which is pretty easy: \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b} = \sum_{a=-3}^3 2^a \sum_{b=-3}^3 \frac{1}{5^b} = \sum_{a=-3}^3 2^a 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1... | 248 |
6,613 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11 | 4 | Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$ | We can organize the fractions and reduce them in quantities to reach our answer. First, separate the fractions with coprime parts into those that are combinations of powers of 2 and 5, and those that are a combination of a 1 and another divisor.
To begin with the first list, list powers of 2 and 5 from 0 to 3. In this ... | 248 |
6,614 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_12 | 1 | Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$ | \begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{align*}
Since $\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm
\[f(x) = f(352 + x)\]
So we need only to consider one period $f(0), f(1), ... f(351)$ , wh... | 177 |
6,615 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_13 | 1 | In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of... | Let the intersection of the highways be at the origin $O$ , and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\f... | 731 |
6,616 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | 1 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ | Let $\angle QPB=x^\circ$ . Because $\angle AQP$ is exterior to isosceles triangle $PQB$ its measure is $2x$ and $\angle PAQ$ has the same measure. Because $\angle BPC$ is exterior to $\triangle BPA$ its measure is $3x$ . Let $\angle PBC = y^\circ$ . It follows that $\angle ACB = x+y$ and that $4x+2y=180^\circ$ . Two of... | 571 |
6,617 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | 2 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ | Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$ . Then $PQBR$ is a rhombus , so $AB \parallel PR$ and $APRB$ is an isosceles trapezoid . Since $\overline{PB}$ bisects $\angle QBR$ , it follows by symmetry in trapezoid $APRB$ that $\overline{RA}$ bisects $\angle BAC$ . Thus $R$ lies on the perpendicular bis... | 571 |
6,618 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | 3 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ | Again, construct $R$ as above.
Let $\angle BAC = \angle QBR = \angle QPR = 2x$ and $\angle ABC = \angle ACB = y$ , which means $x + y = 90$ $\triangle QBC$ is isosceles with $QB = BC$ , so $\angle BCQ = 90 - \frac {y}{2}$ .
Let $S$ be the intersection of $QC$ and $BP$ . Since $\angle BCQ = \angle BQC = \angle BRS$ $BCR... | 571 |
6,619 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | 4 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ | Let $\angle BAC= 2\theta$ and $AP=PQ=QB=BC=x$ $\triangle APQ$ is isosceles, so $AQ=2x\cos 2\theta =2x(1-2\sin^2\theta)$ and $AB= AQ+x=x\left(3-4\sin^2\theta\right)$ $\triangle{ABC}$ is isosceles too, so $x=BC=2AB\sin\theta$ . Using the expression for $AB$ , we get \[1=2\left(3\sin\theta-4\sin^3\theta\right)=2\sin3\thet... | 571 |
6,620 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15 | 2 | A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from... | To simplify matters, we want a power of $2$ . Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.
Let the fake cards have positions $1, 3, 5, \cdots, 95$ .... | 927 |
6,621 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15 | 4 | A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from... | Let us treat each run through the deck as a separate "round". For example, in round one, you would go through all of the $2000$ cards initially in the deck once, in round two, you would go through all $1000$ cards initially in the deck once, so on and so forth. For each round, let us record what the initial and final a... | 927 |
6,622 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_3 | 1 | A deck of forty cards consists of four $1$ 's, four $2$ 's,..., and four $10$ 's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relati... | There are ${38 \choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in... | 758 |
6,623 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_4 | 1 | What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors? | We use the fact that the number of divisors of a number $n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}$ is $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$ . If a number has $18 = 2 \cdot 3 \cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization.
Dividing the greatest power of $2$ from $n$ , we have an odd i... | 180 |
6,624 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_4 | 2 | What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors? | Somewhat similar to the first solution, we see that the number $n$ has two even factors for every odd factor. Thus, if $x$ is an odd factor of $n$ , then $2x$ and $4x$ must be the two corresponding even factors. So, the prime factorization of $n$ is $2^2 3^a 5^b 7^c...$ for some set of integers $a, b, c, ...$
Since t... | 180 |
6,625 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_5 | 1 | Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $n$ | There are $\binom{8}{5}$ ways to choose the rings, and there are $5!$ distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. The number of ways to distribute the rings among the fingers is equivalent the number of wa... | 376 |
6,626 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_6 | 1 | One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$ . Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trape... | Let the shorter base have length $b$ (so the longer has length $b+100$ ), and let the height be $h$ . The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$ . The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h... | 181 |
6,627 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7 | 1 | Given that
find the greatest integer that is less than $\frac N{100}$ | Multiplying both sides by $19!$ yields:
\[\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.\]
\[\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\bi... | 137 |
6,628 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7 | 2 | Given that
find the greatest integer that is less than $\frac N{100}$ | Let $f(x) = (1+x)^{19}.$ Applying the binomial theorem gives us $f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.$ Since $\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},$ $N = \frac{2^{... | 137 |
6,629 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7 | 3 | Given that
find the greatest integer that is less than $\frac N{100}$ | Convert each denominator to $19!$ and get the numerators to be $9,51,204,612,1428,2652,3978,4862$ (refer to note). Adding these up we have $13796$ therefore $\boxed{137}$ is the desired answer. | 137 |
6,630 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 1 | In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ | Let $x = BC$ be the height of the trapezoid, and let $y = CD$ . Since $AC \perp BD$ , it follows that $\triangle BAC \sim \triangle CBD$ , so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$ . Then $AE = x$ , and $ADE$ is a right triang... | 110 |
6,631 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 2 | In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ | Let $BC=x$ . Dropping the altitude from $A$ and using the Pythagorean Theorem tells us that $CD=\sqrt{11}+\sqrt{1001-x^2}$ . Therefore, we know that vector $\vec{BD}=\langle \sqrt{11}+\sqrt{1001-x^2},-x\rangle$ and vector $\vec{AC}=\langle-\sqrt{11},-x\rangle$ . Now we know that these vectors are perpendicular, so thei... | 110 |
6,632 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 3 | In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ | Let $BC=x$ and $CD=y+\sqrt{11}$ . From Pythagoras with $AD$ , we obtain $x^2+y^2=1001$ . Since $AC$ and $BD$ are perpendicular diagonals of a quadrilateral, then $AB^2+CD^2=BC^2+AD^2$ , so we have \[\left(y+\sqrt{11}\right)^2+11=x^2+1001.\] Substituting $x^2=1001-y^2$ and simplifying yields \[y^2+\sqrt{11}y-990=0,\] an... | 110 |
6,633 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 4 | In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ | Let $E$ be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have \begin{align*} BC^2&=BE^2+CE^2 \\ &=(AB^2-AE^2)+(CD^2-DE^2) \\ &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\ &=CD^2+11-AD^2 \\ &=CD^2-990 \end{align*} Followed by dropping the perpendicular l... | 110 |
6,634 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10 | 1 | circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle. | Call the center of the circle $O$ . By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$ , or $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{... | 647 |
6,635 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10 | 2 | circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle. | Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ( $a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\] $r^2=\frac{A^2}{(a+b+c+d)^2} = \boxed{647}$ | 647 |
6,636 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10 | 3 | circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle. | Using the formulas established in solution 2, one notices: \[r^2=\frac{A^2}{(a+b+c+d)^2}\] \[r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\] \[r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}\] \[r^2=\boxed{647}\] | 647 |
6,637 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_11 | 1 | The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$ . The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$ ... | For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$ . Suppose $B$ has integer coordinates; then $\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\overline{CD}$ , and let $D' = (a,... | 131 |
6,638 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_11 | 2 | The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$ . The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$ ... | A very natural solution:
. Shift $A$ to the origin. Suppose point $B$ was $(x, kx)$ . Note $k$ is the slope we're looking for. Note that point $C$ must be of the form: $(x \pm 1, kx \pm 7)$ or $(x \pm 7, kx \pm 1)$ or $(x \pm 5, kx \pm 5)$ . Note that we want the slope of the line connecting $D$ and $C$ so also be $k$ ... | 131 |
6,639 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_12 | 1 | The points $A$ $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that $AB=13$ $BC=14$ $CA=15$ , and that the distance from $O$ to $\triangle ABC$ is $\frac{m\sqrt{n}}k$ , where $m$ $n$ , and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by t... | Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$ . By the Pythagorean Theorem on triangles $\triangle OAD$ $\triangle OBD$ and $\triangle OCD$ we get:
\[DA^2=DB^2=DC^2=20^2-OD^2\]
It follows that $DA=DB=DC$ , so $D$ is the circumcenter of $\triangle ABC$
By Heron's Formula the area of $\triangle ... | 118 |
6,640 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_12 | 2 | The points $A$ $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that $AB=13$ $BC=14$ $CA=15$ , and that the distance from $O$ to $\triangle ABC$ is $\frac{m\sqrt{n}}k$ , where $m$ $n$ , and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by t... | We know the radii to $A$ $B$ , and $C$ form a triangular pyramid $OABC$ . We know the lengths of the edges $OA = OB = OC = 20$ . First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$ . Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$ , and $z$ occupy the oth... | 118 |
6,641 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13 | 1 | The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$ | We may factor the equation as:
\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{align*}
Now $100x^4+10x^2+1\ge 1>0$ for real $x$ . Thus the real ... | 200 |
6,642 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13 | 2 | The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$ | It would be really nice if the coefficients were symmetrical. What if we make the substitution, $x = -\frac{i}{\sqrt{10}}y$ . The the polynomial becomes
$-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2$
It's symmetric! Dividing by $y^3$ and rearranging, we get
$-2(y^3 + \frac{1}... | 200 |
6,643 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13 | 3 | The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$ | Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$ .
The expression in parentheses is a geometric series with common factor $10x^2$ . Using the geometric sum formula, we rewrite as $2000x^6-2+\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\neq0$ .
Factoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\... | 200 |
6,644 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14 | 1 | Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64... | Note that $1+\sum_{k=1}^{n-1} {k\cdot k!} = 1+\sum_{k=1}^{n-1} {((k+1)\cdot k!- k!)} = 1+\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!)) = n!$
Thus for all $m\in\mathbb{N}$
$(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\ri... | 495 |
6,645 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14 | 2 | Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64... | This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition.
Let us consider a base $n$ number system. It’s a well known fact that when we take the difference of two integral powers of $n$ , (such as $10000_{10} - 100_{10}$ ) the result will be an integer in base $n$ composed only ... | 495 |
6,646 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14 | 3 | Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64... | Let $S = 16!-32!+\cdots-1984!+2000!$ . Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \cdots$ is significantly smaller than $2000!$ ), it follows that $1999! < S < 2000!$ . Hence $f_{2000} = 0$ . Then $2000! = 2000 \cdot 1999! = 1999 \cdot 1999! + 1999!$ , and as $S - 2000! << 1999!$ , it follows tha... | 495 |
6,647 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2 | 1 | Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$ | Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$ . Let the second point on the line $x=28$ be $(28, 153-a)$ . For two given points, the line will pass the origin if the coordinates are proportional (such that $\frac{y_1}{x_1} = \frac{y_2}{x_2}$ ). Then, we can write that $\fr... | 118 |
6,648 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2 | 2 | Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$ | You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of $(10,45)$ , and $(28,153)$ gives $(19,99)$ , which is the center of the parallelogram. Thus the slope of the line must be $\frac{99}{19}$ , and the solution is $\bo... | 118 |
6,649 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2 | 3 | Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$ | Note that the area of the parallelogram is equivalent to $69 \cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$
The points where the line in question intersects the long side of the parallelo... | 118 |
6,650 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2 | 4 | Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$ | $(\Sigma x_i /4, \Sigma y_i /4)$ is the centroid, which generates $(19,99)$ , so the answer is $\boxed{118}$ . This is the fastest way because you do not need to find the opposite vertices by drawing. | 118 |
6,651 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_3 | 1 | Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square | If $n^2-19n+99=x^2$ for some positive integer $x$ , then rearranging we get $n^2-19n+99-x^2=0$ . Now from the quadratic formula,
Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$ . Rearranging gives $(2x+q)(2x-q)=35$ . Thus $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$ , giving $x=3$ or $9$ . Thi... | 38 |
6,652 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_3 | 2 | Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square | When $n \geq 12$ , we have \[(n-10)^2 < n^2 -19n + 99 < (n-8)^2.\]
So if $n \geq 12$ and $n^2 -19n + 99$ is a perfect square, then \[n^2 -19n + 99 = (n-9)^2\]
or $n = 18$
For $1 \leq n < 12$ , it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-... | 38 |
6,653 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4 | 1 | The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), ... | Triangles $AOB$ $BOC$ $COD$ , etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$ , etc.), and each area is $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}$ . Since the area of a triangle is... | 185 |
6,654 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4 | 2 | The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), ... | Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$ . The area of the octagon (by subtraction of areas) is $1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy$
By the Pythagorean theorem \[x^2 + y^2 = \left(\frac{43}{99}\right)^2\]
Also, \begin{... | 185 |
6,655 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_5 | 1 | For any positive integer $x_{}$ , let $S(x)$ be the sum of the digits of $x_{}$ , and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values of $T(x)$ do not exceed 1999? | For most values of $x$ $T(x)$ will equal $2$ . For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take $T(a999)$ as an example, \[|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|\] And in general, the values of $T(x)$ will then be in the form of $|2 - 9n| = 9n - 2$ . Fro... | 223 |
6,656 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_6 | 1 | A transformation of the first quadrant of the coordinate plane maps each point $(x,y)$ to the point $(\sqrt{x},\sqrt{y}).$ The vertices of quadrilateral $ABCD$ are $A=(900,300), B=(1800,600), C=(600,1800),$ and $D=(300,900).$ Let $k_{}$ be the area of the region enclosed by the image of quadrilateral $ABCD.$ Find the g... | \begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ B' = & (\sqrt {1800}, \sqrt {600})\\ C' = & (\sqrt {600}, \sqrt {1800})\\ D' = & (\sqrt {300}, \sqrt {900}) \end{eqnarray*}
First we see that lines passing through $AB$ and $CD$ have equations $y = \frac {1}{3}x$ and $y = 3x$ , respectively. Looking at the points abo... | 314 |
6,657 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_7 | 1 | There is a set of 1000 switches, each of which has four positions, called $A, B, C$ , and $D$ . When the position of any switch changes, it is only from $A$ to $B$ , from $B$ to $C$ , from $C$ to $D$ , or from $D$ to $A$ . Initially each switch is in position $A$ . The switches are labeled with the 1000 different in... | For each $i$ th switch (designated by $x_{i},y_{i},z_{i}$ ), it advances itself only one time at the $i$ th step; thereafter, only a switch with larger $x_{j},y_{j},z_{j}$ values will advance the $i$ th switch by one step provided $d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}$ divides $d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}$ . Let $... | 650 |
6,658 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 1 | A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relat... | Suppose we pick an arbitrary point on the complex plane , say $(1,1)$ . According to the definition of $f(z)$ \[f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,\] this image must be equidistant to $(1,1)$ and $(0,0)$ . Thus the image must lie on the line with slope $-1$ and which passes through $\left(\frac 12, \frac12\right)$ ,... | 259 |
6,659 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 2 | A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relat... | Plugging in $z=1$ yields $f(1) = a+bi$ . This implies that $a+bi$ must fall on the line $Re(z)=a=\frac{1}{2}$ , given the equidistant rule. By $|a+bi|=8$ , we get $a^2 + b^2 = 64$ , and plugging in $a=\frac{1}{2}$ yields $b^2=\frac{255}{4}$ . The answer is thus $\boxed{259}$ | 259 |
6,660 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 3 | A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relat... | We are given that $(a + bi)z$ is equidistant from the origin and $z.$ This translates to \begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\ |z(a - 1 + bi)| & = & |z(a + bi)| \\ |z|\cdot|(a - 1) + bi| & = & |z|\cdot|a + bi| \\ |(a - 1) + bi| & = & |a + bi| \\ (a - 1)^2 + b^2 & = & a^2 + b^2 \\ & \Rightarrow & a = \f... | 259 |
6,661 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 4 | A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relat... | Similarly to in Solution 3, we see that $|(a + bi)z - z| = |(a + bi)z|$ . Letting the point $z = c + di$ , we have $\sqrt{(ab+bc-d)^2+(ac-bd-c)^2} = \sqrt{(ac-bd)^2+(ad+bc)^2}$ . Expanding both sides of this equation (after squaring, of course) and canceling terms, we get $(d^2+c^2)(-2a+1) = 0$ . Of course, $(d^2+c^2)$... | 259 |
6,662 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 5 | A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relat... | This is a solution that minimizes the use of complex numbers, turning this into an introductory algebra analytic geometry problem.
Consider any complex number $z=c+di$ . Let $z$ denote point $P$ on the complex plane. Then $P=(c,d)$ on the complex plane. The equation for the line $OP$ is $y=\frac{d}{c}x$
Let the image o... | 259 |
6,663 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_10 | 1 | Ten points in the plane are given, with no three collinear . Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m_{}$ and $n_{}... | Note that 4 points can NEVER form 2 triangles. Therefore, we just need to multiply the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that share an endpoint with the first one, then the one segment that completes the triangle. Note... | 489 |
6,664 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_10 | 2 | Ten points in the plane are given, with no three collinear . Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m_{}$ and $n_{}... | Instead of working with the four segments, let's focus on their endpoints. When we select these segments, we are working with $4, 5, 6, 7,$ or $8$ endpoints in total.
If we have $6, 7,$ or $8$ endpoints, it is easy to see that we cannot form a triangle by drawing four segments between them, because at least one point w... | 489 |
6,665 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11 | 1 | Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$ | Let $s = \sum_{k=1}^{35}\sin 5k = \sin 5 + \sin 10 + \ldots + \sin 175$ . We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identit... | 177 |
6,666 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11 | 2 | Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$ | We note that $\sin x = \mbox{Im } e^{ix}\text{*}$ . We thus have that \begin{align*} \sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\\ &= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\\ &= \mbox{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}}\\ &= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\ &=... | 177 |
6,667 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11 | 3 | Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$ | Let $x=e^{\frac{i\pi}{36}}$ . By Euler's Formula, $\sin{5k^\circ}=\frac{x^k-\frac{1}{x^{k}}}{2i}$
The sum we want is thus $\frac{x-\frac{1}{x}}{2i}+\frac{x^2-\frac{1}{x^{2}}}{2i}+\cdots+\frac{x^{35}-\frac{1}{x^{35}}}{2i}$
We factor the $\frac{1}{2i}$ and split into two geometric series to get $\frac{1}{2i}\left(\frac{-... | 177 |
6,668 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_12 | 1 | The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$ . Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle. | Let $Q$ be the tangency point on $\overline{AC}$ , and $R$ on $\overline{BC}$ . By the Two Tangent Theorem $AP = AQ = 23$ $BP = BR = 27$ , and $CQ = CR = x$ . Using $rs = A$ , where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$ , we get $(21)(50 + x) = A$ . By Heron's formula $A = \sqrt{s(s-a)(s-b)(s-c)}... | 345 |
6,669 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_12 | 2 | The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$ . Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle. | Let the incenter be denoted $I$ . It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let $\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$
We have that \begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan... | 345 |
6,670 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_13 | 1 | Forty teams play a tournament in which every team plays every other team exactly once. No ties occur, and each team has a $50 \%$ chance of winning any game it plays. The probability that no two teams win the same number of games is $\frac mn,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $\l... | There are ${40 \choose 2} = 780$ total pairings of teams, and thus $2^{780}$ possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games. Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total... | 742 |
6,671 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 1 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ | Drop perpendiculars from $P$ to the three sides of $\triangle ABC$ and let them meet $\overline{AB}, \overline{BC},$ and $\overline{CA}$ at $D, E,$ and $F$ respectively.
Let $BE = x, CF = y,$ and $AD = z$ . We have that \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*} We can then use the t... | 463 |
6,672 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 2 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ | Let $AB=c$ $BC=a$ $AC=b$ $PA=x$ $PB=y$ , and $PC=z$
So by the Law of Cosines , we have: \begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\ y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\ z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{align*} Adding these equations and rearranging, we have: \[a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta... | 463 |
6,673 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 3 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ | Let $\angle{PAB} = \angle{PBC} = \angle{PCA} = x.$ Then, using Law of Cosines on the three triangles containing vertex $P,$ we have \begin{align*} b^2 &= a^2 + 169 - 26a \cos x \\ c^2 &= b^2 + 196 - 28b \cos x \\ a^2 &= c^2 + 225 - 30c \cos x. \end{align*} Add the three equations up and rearrange to obtain \[(13a + 14b... | 463 |
6,674 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 4 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ | Firstly, denote angles $ABC$ $BCA$ , and $CAB$ as $B$ $A$ , and $C$ respectively. Let $\angle{PAB}=x$ .
Notice that by angle chasing that $\angle{BPC}=180-C$ and $\angle{BPA}=180-B$ .
Using the nice properties of the 13-14-15 triangle, we have $\sin B = \frac{12}{13}$ and $\sin C = \frac{4}{5}$ $\cos C$ is easily compu... | 463 |
6,675 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 1 | Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? | As shown in the image above, let $D$ $E$ , and $F$ be the midpoints of $\overline{BC}$ $\overline{CA}$ , and $\overline{AB}$ , respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$ . The crux of this problem is the following lemma.
Lemma: The poi... | 408 |
6,676 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 2 | Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? | Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$ $(8, 12, 0)$ , and $(25, 12, 0)$ . We can compute the area of this triangle as $102$ . Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$ . Clearly... | 408 |
6,677 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 3 | Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? | The formed tetrahedron has pairwise parallel planar and oppositely equal length ( $4\sqrt{13},15,17$ ) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces. Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pyth... | 408 |
6,678 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 4 | Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? | Let $A = (0,0), B = (16, 24), C = (34,0).$ Then define $D,E,F$ as the midpoints of $BC, AC, AB$ . By Pythagorean theorem, $EF = \frac{1}{2} BC = 15, DE = \frac{1}{2}AB = 4 \sqrt{13}, DF = \frac{1}{2} AC = 17.$ Then let $P$ be the point in space which is the vertex of the tetrahedron with base $DEF$
Note that $\triangle... | 408 |
6,679 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_1 | 1 | For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ $8^8$ , and $k$ | It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$
The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$ . Therefore $12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$... | 25 |
6,680 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_3 | 1 | The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions. What is the area of the bounded region? | The equation given can be rewritten as:
We can split the equation into a piecewise equation by breaking up the absolute value
Factoring the first one: (alternatively, it is also possible to complete the square
Hence, either $y = -20$ , or $2x = 20 - y \Longrightarrow y = -2x + 20$
Similarily, for the second one, we get... | 800 |
6,681 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_6 | 1 | Let $ABCD$ be a parallelogram . Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$ | We have $\triangle BRQ\sim \triangle DRC$ so $\frac{112}{RC} = \frac{BR}{DR}$ . We also have $\triangle BRC \sim \triangle DRP$ so $\frac{ RC}{847} = \frac {BR}{DR}$ . Equating the two results gives $\frac{112}{RC} = \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$ | 308 |
6,682 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 2 | Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$ | Define $x_i = 2y_i - 1$ . Then $2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98$ , so $\sum_{i = 1}^4 y_i = 51$
So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50;... | 196 |
6,683 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 3 | Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$ | Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into $4$ boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have $94$ stones left. Becau... | 196 |
6,684 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 4 | Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$ | Let $x = a + b$ and $y = c + d$ . Then $x + y = 98$ , where $x, y$ are positive even integers ranging from $2-98$
We quickly see that the total number of acceptable ordered pairs $(a, b, c, d)$ is:
\begin{align*} &\mathrel{\phantom{=}} 1 \cdot 48 + 2 \cdot 47 + 3 \cdot 46 + ... + 48 \cdot 1\\ &= (24.5 - 23.5)(24.5) + (... | 196 |
6,685 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 5 | Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$ | We write the generating functions for each of the terms, and obtain $(x+x^3+x^5\cdots)^4$ as the generating function for the sum of the $4$ numbers. We seek the $x^{98}$ coefficient, or the $x^{94}$ coefficient in $(1+x^2+x^4...)^4.$ Now we simplify this as $\left(\frac{1}{1-x^2}\right)^4=\binom{3}{3} +\binom{4}{3}x^2+... | 196 |
6,686 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_8 | 1 | Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encounted. What positive integer $x$ produces a sequence of maximum length? | We can start to write out some of the inequalities now:
And in general,
It is apparent that the bounds are slowly closing in on $x$ , so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11):
$x < \frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}$
$x > \frac{F_{8}}{F_{9}} ... | 618 |
6,687 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_8 | 2 | Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encounted. What positive integer $x$ produces a sequence of maximum length? | It is well known that $\lim_{n\rightarrow\infty} \frac{F_{n-1}}{F_n} = \phi - 1 =\frac{1 + \sqrt{5}}{2} - 1 \approx .61803$ , so $1000 \cdot \frac{F_{n-1}}{F_n}$ approaches $x = \boxed{618}.$ | 618 |
6,688 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_10 | 1 | Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon . A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ wher... | The key is to realize the significance that the figures are spheres, not circles . The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.
1998 AIME-10a.png
Let us e... | 152 |
6,689 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_10 | 2 | Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon . A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ wher... | Isolate a triangle, with base length $200$ (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as $x$ . Since the interior angle is $45$ degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get: \begin{eqnarray*} 200^2 &=& 2x^2 - 2x^2*... | 152 |
6,690 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_14 | 1 | An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$ | \[2mnp = (m+2)(n+2)(p+2)\]
Let’s solve for $p$
\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\] \[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\] \[p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\]
Clearly, we want to minimize the denominator, so we test $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$ . The... | 130 |
6,691 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_14 | 2 | An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$ | Similarly as above, we solve for $p,$ but we express the denominator differently:
\[p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.\] Hence, it suffices to maximize $\dfrac{m+n+2}{(m+2)(n+2)},$ under the conditions that $p$ is a positive integer.
Then since $\... | 130 |
6,692 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_15 | 1 | Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $... | We can draw a comparison between the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes.
You need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the p... | 761 |
6,693 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_1 | 1 | How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers? | Notice that all odd numbers can be obtained by using $(a+1)^2-a^2=2a+1,$ where $a$ is a nonnegative integer. All multiples of $4$ can be obtained by using $(b+1)^2-(b-1)^2 = 4b$ , where $b$ is a positive integer. Numbers congruent to $2 \pmod 4$ cannot be obtained because squares are $0, 1 \pmod 4.$ Thus, the answer is... | 750 |
6,694 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_2 | 1 | The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles , of which $s$ are squares . The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\choose 2} = 36$ . Similarly, there are ${9\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles.
For $s$ , there are $8^2$ unit squares $7^2$ of the $2\times2$ squares, an... | 125 |
6,695 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3 | 1 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum ... | Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \Longrightarrow 9xy-1000x-y=0$ . Using SFFT , this factorizes to $(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}$ , and $(9x-1)(9y-1000)=1000$
Since $89 < 9x-1 < 890$ , we can use trial and error on fa... | 126 |
6,696 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3 | 2 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum ... | As shown above, we have $1000x+y=9xy$ , so $1000/y=9-1/x$ $1000/y$ must be just a little bit smaller than 9, so we find $y=112$ $x=14$ , and the solution is $\boxed{126}$ | 126 |
6,697 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3 | 3 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum ... | To begin, we rewrite $(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z$
as
$(90a+9b-1)(100x+10y+z) = 10000a + 1000b$
and
$(90a+9b-1)(100x+10y+z) = 1000(10a + b)$
This is the most important part: Notice $(90a+9b-1)$ is $-1 \pmod{10a+b}$ and $1000(10a + b)$ is $0\pmod{10a+b}$ . That means $(100x+10y+z)$ is also $... | 126 |
6,698 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_4 | 1 | Circles of radii $5, 5, 8,$ and $\frac mn$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 1997 AIME-4.png
If (in the diagram above) we draw the line going through the centers of the circles with radii $8$ and $\frac mn = r$ , that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii $5$ . Then we form two right triangles , of lengths $5, x, 5+r$ and $5, 8+r+... | 17 |
6,699 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_5 | 1 | The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits , any of which could be zero. It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$ What is the number of p... | The nearest fractions to $\frac 27$ with numerator $1$ are $\frac 13, \frac 14$ ; and with numerator $2$ are $\frac 26, \frac 28 = \frac 13, \frac 14$ anyway. For $\frac 27$ to be the best approximation for $r$ , the decimal must be closer to $\frac 27 \approx .28571$ than to $\frac 13 \approx .33333$ or $\frac 14 \app... | 417 |
6,700 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_7 | 1 | A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the s... | We set up a coordinate system, with the starting point of the car at the origin . At time $t$ , the car is at $\left(\frac 23t,0\right)$ and the center of the storm is at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right)$ . Using the distance formula,
\begin{eqnarray*} \sqrt{\left(\frac{2}{3}t - \frac 12t\right)^2 + \left(1... | 198 |
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