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d0cdd5adf9dc5527f01a6184eea8d22e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | 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 $\angle BCQ = \angle BQC = \angle BRS$ $BCRS$ is cyclic , which means $\angle RBS = \angle RCS = x$ .
Since $APRB$ is an isosceles trapezoid, $BP = AR$ , but since $AR$ bisects $\angle BAC$ $\angle ABR = \angle ACR = 2x$
Therefore we have that $\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y$ .
We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \frac {y}{2} = y$ to get $x = 10$ and $y = 80$ $\angle APQ = 180 - 4x = 140$ $\angle ACB = 80$ , so $r = \frac {80}{140} = \frac {4}{7}$ $\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$ | null | 571 |
d0cdd5adf9dc5527f01a6184eea8d22e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14 | 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 \[1=2\left(3\sin\theta-4\sin^3\theta\right)=2\sin3\theta\] by the triple angle formula! Thus $\theta=10^\circ$ and $\angle A = 2\theta=20^\circ$ .
It follows now that $\angle APQ=140^\circ$ $\angle ACB=80^\circ$ , giving $r=\tfrac{4}{7}$ , which implies that $1000r = 571 + \tfrac 37$ . So the answer is $\boxed{571}$ | null | 571 |
82ab388b90a21ce8d3d06bd9d8415b8e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15 | 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 the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$ | We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, $2(1024 - 1000) = 48$ , to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \times 2 = 928$ , meaning that there were $\boxed{927}$ cards are above the one labeled $1999$ | null | 927 |
82ab388b90a21ce8d3d06bd9d8415b8e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15 | 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 the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$ | 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 $1, 3, 5, \cdots, 95$ . Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$ . From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$ , so $\boxed{927}$ cards are above it. - Spacesam | null | 927 |
82ab388b90a21ce8d3d06bd9d8415b8e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15 | 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 the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$ | We work backwards. To reverse the process, we must move the bottom card to the top, and add a new number to the top. Let $d_n$ equal the number of cards below 1999 after $n$ process reversals. Reversing one process, our deck only has $2000$ , so we reverse again to obtain $1999, 2000$ . So, $d_2 = 1$ . When $d_{n-1} > 0$ , after a process reversal, the bottom card is moved to the top (it goes above $1999$ ), so we subtract by one to account for this (the addition of the new card doesn't matter since it goes above $1999$ ), giving us $d_n = d_{n-1} - 1$ . So, $d_3 = 0$ . Then, when $d_{n-1} = 0$ , after a process reversal, $1999$ moves to the top and one more card is added above $1999$ . Since at $d_{n}$ $n$ cards have been added, there must be $n - 2$ cards below $1999$ . So, $d_4 = 2$ . Then, consider all $d_{n-1} = 0$ . Then, $d_n = n-2$ as previously stated. After $n-2$ process reversals, we go back to $d_{2n - 2} = n - 2 - (n - 2) = 0$ . Then, $d_{2n-1} = 2n - 3$ . Next, we can simply calculate $2 \cdot 4 - 1 = 7$ $2 \cdot 7 - 1 = 13$ , and so on (which is a very simple bash, as $2000$ is a small number. If you don't want to do this, define sequence $a_n = 2a_{n-1} - 1$ , and solve for the closed form, which is very easy). Consequently, we derive $d_{1537} = 1537 - 2 = 1535$ $2000 - 1537 = 463$ steps remain. So, $d_{2000} = d_{1537} - 463 = 1072$ . Finally, there are $2000 - 1 - 1072 = \boxed{927}$ numbers above $1999$ | null | 927 |
82ab388b90a21ce8d3d06bd9d8415b8e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15 | 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 the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$ | 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 actions are ( $r$ for moving the card to the right, $b$ for moving the card to the bottom), the number of cards moved to the right, the number of cards left, and what the position of the cards moved to the right were in the original $2000$ card deck (as $n = a + ck$ where $n$ is the position, $c$ is the spacing of the cards moved, $a$ is an integer such that the correct first card is moved, and $k$ is an integer greater than or equal to $1$ which represents which card out of all the cards moved to the right you are finding the position of). Then,
Round 1: $r$ to $b$ $1000$ to right, $1000$ left in deck, $n = -1 + 2k$
Round 2: $r$ to $b$ $500$ to right, $500$ left in deck, $n = -2 + 4k$
Round 3: $r$ to $b$ $250$ to right, $250$ left in deck, $n = -4 + 8k$
Round 4: $r$ to $b$ $125$ to right, $125$ left in deck, $n = -8 + 16k$
Let us treat the remaining deck of $125$ cards as a totally independent deck, note that the positions of card in this deck are $n = 16k$ . Also note that the first action of a new round is never the same action as the last action of the previous round because actions alternate. Also note that for every new round, the spacing between cards moved doubles, and that the cards remaining in the beginning of a new round have position $n = a + c/2 + ck$ for the values $a, c$ of the previous round. Also note that if there are an odd number of cards in an initial deck, the first and last actions are the same. Then,
Round 5: $r$ to $r$ $63$ to right, $62$ left in deck, $n = -1 + 2k$
Round 6: $b$ to $r$ $31$ to right, $31$ left in deck, $n = 4k$ , because $n = 2k$ positioned cards are left at the beginning of this round and the first card is sent to the bottom, only every second card is sent to the right, and because spacing doubles every round,
Round 7: $b$ to $b$ $15$ to right, $16$ left in deck, $n = -2 + 8k$ , because $n = 2 + 4k$ positioned cards are left at the beginning and only every second card is sent to the right, and because spacing doubles every round,
Round 8: $r$ to $b$ $8$ to right, $8$ left in deck, $n = -14 + 16k$ , by similar reasoning, since the cards $n = 2 + 8k$ are left and spacing doubles every round, from here on things get real easy,
Round 9: $r$ to $b$ $4$ to right, $4$ left in deck, $n = -22 + 32k$
Round 10: $r$ to $b$ $2$ to right, $2$ left in deck, $n = -38 + 64k$
Round 11: $r$ to $b$ $1$ to right, $1$ left in deck, $n = 58$ , since $58 = -38 + 64/2 + 64$
This card must be labeled 1999 since it is second to last. Then, since it is the $58th$ in the deck of $125$ , it must be the $58 * 8 = 928th$ card in the original deck, and because we've been numbering from the top of the deck to the bottom (also because AIME answers are 000-999), there were $928 - 1 = \boxed{927}$ cards above it in the deck - Romulus and minimaxweii | null | 927 |
82ab388b90a21ce8d3d06bd9d8415b8e | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15 | 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 the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$ | Similar to Solution 5, we treat each run-through of the deck from the lowest-indexed card to the highest-indexed card as a separate round. Notice that after each round, approximately half the deck will remain, with the other half having been cast aside to the right in sorted order. Then, we can model each round as if we are running a "Sieve" through the deck with powers of 2s, filtering out the cards to put to the right and the cards to be pushed to the bottom.
At the beginning, when all $2000$ cards are still in the deck, notice that the first run-through of the deck consists of removing all the cards such that their index $I\equiv 1\pmod{2}$ , while we leave behind all the cards such that their index $I\equiv 2\pmod{2}$ . (Since the cards are not $0$ -indexed, using $2$ instead of $0\pmod{2}$ is simpler to keep track of the cards. This will appear again later.) The remaining $1000$ cards, then, will all share the attribute that their index numbers are even. To split this further for round 2, we look more specifically at each index, categorizing them further as $I\equiv 2\pmod 4$ or $I\equiv 4\pmod 4$ in that order (The card with the smaller remainder will always be the card on the top of the new deck). Then, in Round 2, we will remove all the cards with index values that leave a remainder of $2$ when divided by $4$ , while leaving the multiples of $4$ behind. Notice that each time, in each new deck, we are removing all the cards with an odd position in that specific deck, and leaving behind all the cards with an even position in that specific deck.
This process will continue until we reach Round 5, where the number of cards that remain ( $125$ ) is no longer a multiple of 2. Therefore, when we remove all the cards with indices $I\equiv 16\pmod{32}$ and leave behind all the cards with indices $I\equiv 32\pmod{32}$ , the last card in the deck will be removed instead of being placed to the bottom, with $125-63=62$ cards remaining. Because of this, when Round 6 begins, we cannot follow our normal procedure, since the top card of the Round 6 deck would have been moved to the bottom! Therefore, when we are using our sieve with $I\equiv 32\pmod{64}$ and $I\equiv 64\pmod{64}$ , the order at which the remainders are presented have flipped, so all the cards with indices that are multiples of 64 will be removed, while all the cards with indices that are 32 more than a multiple of 64 will remain. The $62$ cards from Round 6 will be reduced to $31$ , and the last card in the deck will be removed just like in Round 5, which means Round 7 will be affected the same way. Of the choices of $I\equiv 32\pmod{128}$ and $I\equiv 32+64=96\pmod{128}$ , we will remove the latter, leaving behind the cards with indices that are 32 more than a multiple of 128. However, since 31 is an odd number, the last card of the deck will NOT be removed, which means our sieving process will return to normal after Round 7, with $31-15=16$ cards remaining. Since $16$ is a perfect power of $2$ , we will not need to worry about this scenario again in the following rounds.
Round 8: $I\equiv 32\pmod{256}$ removed; $I\equiv 32+128=160\pmod{256}$ remains; $16-8=8$ cards left.
Round 9: $I\equiv 160\pmod{512}$ removed; $I\equiv 160+256=416\pmod{512}$ remains; $8-4=4$ cards left.
Round 10: $I\equiv 416\pmod{1024}$ removed; $I\equiv 416+512=928\pmod{1024}$ remains; $4-2=2$ cards left.
Round 11: $I\equiv 928\pmod{2048}$ removed; $I\equiv 928+1024=1952\pmod{2048}$ remains; $2-1=1$ card left.
Round 12: The last card in position $1952$ is sorted into place (This is card # $2000$ !).
It is obvious that the single card put into place in Round 11 is the second-to-last card in the deck, which is our target # $1999$ . Then, its position in the original deck leaves a remainder of $928$ when divided by $2056$ . It is easy to see that $928$ is the only index that satisfies this, so there will be $928-1=\boxed{927}$ cards above card # $1999$ | null | 927 |
f1dd77b13cc816cd9140531c75cf26cb | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_1 | The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | $\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$
$=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}$
$=\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}$
$=\frac{\log{2000}}{\log{2000^6}}$
$=\frac{\log{2000}}{6\log{2000}}$
$=\frac{1}{6}$
Therefore, $m+n=1+6=\boxed{007}$ | null | 007 |
f1dd77b13cc816cd9140531c75cf26cb | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_1 | The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Alternatively, we could've noted that, because $\frac 1{\log_a{b}} = \log_b{a}$
\begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} &= 2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }\\ &=2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} \\ &={\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}}\\ &={\log_{2000^6}{4^2 \cdot 5^3}}\\ &={\log_{2000^6}{2000}}\\ &= {\frac{1}{6}}.\end{align*}
Therefore our answer is $1 + 6 = \boxed{007}$ | null | 007 |
f1dd77b13cc816cd9140531c75cf26cb | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_1 | The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We know that $2 = \log_4{16}$ and $3 = \log_5{125}$ , and by base of change formula, $\log_a{b} = \frac{\log_c{b}}{\log_c{a}}$ . Lastly, notice $\log a + \log b = \log ab$ for all bases. \begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} = \log_{2000^6}{16} + \log_{2000^6}{125} = \log_{2000^6}{2000} = \frac16 \implies \boxed{007} | null | 007 |
f1dd77b13cc816cd9140531c75cf26cb | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_1 | The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | \[\frac{2}{\log_4 2000^6} + \frac{3}{\log_5 2000^6}\] \[= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}\] \[= \frac{1}{3} \log_{2000} 4 + \frac{1}{2} \log_{2000} 5\] \[= \log_{2000} (\sqrt[3]{4} \cdot \sqrt{5}) = x\] \[\implies 2^{4x} \cdot 5^{3x} = 2^{\frac{2}{3}} \cdot 5^{\frac{1}{2}}\] \[\implies 4x + (3\log_2 5)x = \frac{2}{3}+\frac{1}{2} \log_2 5\] \[\implies x = \frac{\frac{2}{3} + \frac{1}{2} \log_2 5}{4 + 3\log_2 5}\] \[\implies 6x = \frac{4 + 3\log_2 5}{4 + 3\log_2 5}\] \[\implies x = \frac{1}{6}\] \[\implies m + n = \boxed{007}\] | null | 007 |
a7ecbc7a4c664c1708da5900d70ceac6 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_2 | A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$ | \[(x-y)(x+y)=2000^2=2^8 \cdot 5^6\]
Note that $(x-y)$ and $(x+y)$ have the same parities , so both must be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$ . We have $2^6 \cdot 5^6$ left. Since there are $7 \cdot 7=49$ factors of $2^6 \cdot 5^6$ , and since both $x$ and $y$ can be negative, this gives us $49\cdot2=\boxed{098}$ lattice points. | null | 098 |
687c41622a25d77fd4319e274428c487 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_3 | 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 relatively prime positive integers. Find $m + n.$ | 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 $1$ way. Thus, the answer is $\frac{54+1}{703} = \frac{55}{703}$ , and $m+n = \boxed{758}$ | null | 758 |
8c428d6f27c79c4ded4880e0e1bb1f08 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_4 | 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 integer with six positive divisors, which indicates that it either is ( $6 = 2 \cdot 3$ ) a prime raised to the $5$ th power, or two primes, one of which is squared. The smallest example of the former is $3^5 = 243$ , while the smallest example of the latter is $3^2 \cdot 5 = 45$
Suppose we now divide all of the odd factors from $n$ ; then we require a power of $2$ with $\frac{18}{6} = 3$ factors, namely $2^{3-1} = 4$ . Thus, our answer is $2^2 \cdot 3^2 \cdot 5 = \boxed{180}$ | null | 180 |
8c428d6f27c79c4ded4880e0e1bb1f08 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_4 | 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 there are $18$ factors of $n$ , we can write:
$(2+1)(a+1)(b+1)(c+1)... = 18$
$(a+1)(b+1)(c+1)... = 6$
Since $6$ only has factors from the set $1, 2, 3, 6$ , either $a=5$ and all other variables are $0$ , or $a=3$ and $b=2$ , with again all other variables equal to $0$ . This gives the two numbers $2^2 \cdot 3^5$ and $2^2 \cdot 3^2 \cdot 5$ . The latter number is smaller, and is equal to $\boxed{180}$ | null | 180 |
91ce2124bfa3c0b6e267d62f9ae4b4d7 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_5 | 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 ways we can drop five balls into 4 urns, or similarly dropping five balls into four compartments split by three dividers. The number of ways to arrange those dividers and balls is just $\binom {8}{3}$
Multiplying gives the answer: $\binom{8}{5}\binom{8}{3}5! = 376320$ , and the three leftmost digits are $\boxed{376}$ | null | 376 |
795aa37ba6af691dcc436d80fe55d4db | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_6 | 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 trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100$ | 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 $h/2$ . Then,
\[\frac{\frac 12 (h/2) (b + b+50)}{\frac 12 (h/2) (b + 50 + b + 100)} = \frac{2}{3} \Longrightarrow \frac{b + 75}{b + 25} = \frac 32 \Longrightarrow b = 75\]
Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$ . By similar triangles, we easily find that $\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}$
The area of the region including the shorter base must be half of the area of the entire trapezoid, so
\[2 \cdot \frac 12 h_1 (75 + x) = \frac 12 h (75 + 175) \Longrightarrow x = 125 \cdot \frac{h}{h_1} - 75\]
Substituting our expression for $\frac h{h_1}$ from above, we find that
\[x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125\]
The answer is $\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}$ | null | 181 |
b2bd3e88ea28e8e0c2819605a65ec43f | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7 | Given that
find the greatest integer that is less than $\frac N{100}$ | Multiplying both sides by $19!$ yields:
\[\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.\]
\[\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9} = 19N.\]
Recall the Combinatorial Identity $2^{19} = \sum_{n=0}^{19} {19 \choose n}$ . Since ${19 \choose n} = {19 \choose 19-n}$ , it follows that $\sum_{n=0}^{9} {19 \choose n} = \frac{2^{19}}{2} = 2^{18}$
Thus, $19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124$
So, $N=\frac{262124}{19}=13796$ and $\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}$ | null | 137 |
b2bd3e88ea28e8e0c2819605a65ec43f | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7 | 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 $\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},$ $N = \frac{2^{18}-20}{19}.$ After some fairly easy bashing, we get $\boxed{137}$ as the answer. | null | 137 |
b2bd3e88ea28e8e0c2819605a65ec43f | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7 | 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. | null | 137 |
83d3f11a974dc03c521069c2f1996455 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 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 triangle . By the Pythagorean Theorem
\[x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0\]
The positive solution to this quadratic equation is $x^2 = \boxed{110}$ | null | 110 |
83d3f11a974dc03c521069c2f1996455 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 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 their dot product is 0. \[\vec{BD}\cdot \vec{AC}=-11-\sqrt{11(1001-x^2)}+x^2=0\] \[(x^2-11)^2=11(1001-x^2)\] \[x^4-11x^2-11\cdot 990=0.\] As above, we can solve this quadratic to get the positve solution $BC^2=x^2=\boxed{110}$ | null | 110 |
83d3f11a974dc03c521069c2f1996455 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 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 \[y^2+\sqrt{11}y-990=0,\] and the quadratic formula gives $y=9\sqrt{11}$ . Then from $x^2+y^2=1001$ , we plug in $y$ to find $x^2=\boxed{110}$ | null | 110 |
83d3f11a974dc03c521069c2f1996455 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8 | 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 like in solution 1, we obtain system of equation \[BC^2=CD^2-990\] \[BC^2+CD^2-2\sqrt{11}CD=990\] Rearrange the first equation yields \[BC^2-CD^2=990\] Equating it with the second equation we have \[BC^2-CD^2=BC^2+CD^2-2\sqrt{11}CD\] Which gives $CD^2=\frac{BC^2}{11}$ .
Substituting into equation 1 obtains the quadratic in terms of $BC^2$ \[(BC^2)^2-11BC^2-11\cdot990=0\] Solving the quadratic to obtain $BC^2=\boxed{110}$ | null | 110 |
0b53aea5ab39fd5966dec0bdd0a13ce5 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_9 | Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ | Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$ , we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$
There are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\theta} = \cos \theta + i\sin \theta$ and $\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta$ . We have $z+\frac 1z = 2\cos \theta = 2\cos 3^\circ$ and $\theta = 3^\circ$ . Alternatively, we could let $z = a + bi$ and solve to get $z=\cos 3^\circ + i\sin 3^\circ$
Using De Moivre's Theorem we have $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ$ $6000 = 16(360) + 240$ , so $z^{2000} = \cos 240^\circ + i\sin 240^\circ$
We want $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$
Finally, the least integer greater than $-1$ is $\boxed{000}$ | null | 000 |
0b53aea5ab39fd5966dec0bdd0a13ce5 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_9 | Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ | Let $z=re^{i\theta}$ . Notice that we have $2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.$
$r$ must be $1$ (or else if you take the magnitude would not be the same). Therefore, $z=e^{i\frac{\pi}{\theta}}$ and plugging into the desired expression, we get $e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1$ . Therefore, the least integer greater is $\boxed{000}.$ | null | 000 |
0b53aea5ab39fd5966dec0bdd0a13ce5 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_9 | Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ | For this solution, we assume that $z^{2000} + 1/z^{2000}$ and $z^{2048} + 1/z^{2048}$ have the same least integer greater than their solution. we have $z + 1/z = 2\cos 3$ . Since $\cos 3 < 1$ $2\cos 3 < 2$ . If we square the equation $z + 1/z = 2\cos 3$ , we get $z^2 + 2 + 1/(z^2) = 4\cos^2 3$ , or $z^2 + 1/(z^2) = 4\cos^2 3 - 2$ $4\cos^2 3 - 2$ is is less than $2$ , since $4\cos^2 3$ is less than $4$ . if we square the equation again, we get $z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2$ . since $4\cos^2 3 - 2$ is less than 2, $(4\cos^2 3 - 2)^2$ is less than 4, and $(4\cos^2 3 - 2)^2 -2$ is less than 2. However $(4\cos^2 3 - 2)^2 -2$ is also less than $4\cos^2 3 - 2$ . we can see that every time we square the equation, the right hand side gets smaller, and into the negatives. Since the smallest integer that is allowed as an answer is 0, thus smallest integer greater is $\boxed{000}.$ | null | 000 |
0b53aea5ab39fd5966dec0bdd0a13ce5 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_9 | Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ | First, let $z = a+bi$ where $a$ and $b$ are real numbers. We now have that \[a+bi + \frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}\] given the coniditons of the problem. Equating imaginary coefficients, we have that \[b \left( 1 - \frac{1}{a^2+b^2}\right) = 0\] giving us that either $b=0$ or $|z| = 1$ . Let's consider the latter case for now.
We now know that $a^2+b^2=1$ , so when we equate real coefficients we have that $2a = 2 \cos{3^{\circ}}$ , therefore $a = \cos{3^{\circ}}$ . So, $b = \cos{3^{\circ}}$ and then we can write $z = \text{cis}(3)^{\circ}$
By De Moivre's Theorem, \[z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}\] . The imaginary parts cancel, leaving us with $2 \cos{6000^{\circ}}$ which is $240 \pmod{360}$ . Therefore, we have it being $-1$ and our answer is $\boxed{000}$ | null | 000 |
de88cac363928e57e49493eb0a4b12c1 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10 | 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 $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$
Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get \[\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.\]
Use the identity for $\tan(A+B)$ again to get \[\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.\]
Solving gives $r^2=\boxed{647}$ | null | 647 |
de88cac363928e57e49493eb0a4b12c1 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10 | 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). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\] $r^2=\frac{A^2}{(a+b+c+d)^2} = \boxed{647}$ | null | 647 |
de88cac363928e57e49493eb0a4b12c1 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10 | 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: \[r^2=\frac{A^2}{(a+b+c+d)^2}\] \[r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\] \[r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}\] \[r^2=\boxed{647}\] | null | 647 |
184349616631b5df39c4009f39f1764d | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_11 | 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 $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | 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 $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram , and $\overrightarrow{AB} = \overrightarrow{D'C}$ . Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.
Let the slope of the perpendicular be $m$ . Then the midpoint of $\overline{DD'}$ lies on the line $y=mx$ , so $\frac{b+7}{2} = m \cdot \frac{a+1}{2}$ . Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$ . Combining these two equations yields
\[a^2 + \left(7 - (a+1)m\right)^2 = 50\]
Since $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $( \pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$ . We exclude the cases $(\pm 1, \pm 7)$ because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have
\[7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5\]
These yield $m = 1, \frac 34, -1, -\frac 43, 2, \frac 13, -3, - \frac 12$ . Therefore, the corresponding slopes of $\overline{AB}$ are $-1, -\frac 43, 1, \frac 34, -\frac 12, -3, \frac 13$ , and $2$ . The sum of their absolute values is $\frac{119}{12}$ . The answer is $m+n= \boxed{131}$ | null | 131 |
184349616631b5df39c4009f39f1764d | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_11 | 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 $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | 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 $k$ , since $AB$ and $CD$ are parallel.
Instead of dealing with the 12 cases, we consider
point $C$ of the form $(x \pm Y, kx \pm Z)$ where
we plug in the necessary values for $Y$ and $Z$ after simplifying.
Since the slopes of $AB$ and $CD$ must both be $k$ $\frac{7 - kx \pm Z}{1 - x \pm Y} = k \implies k = \frac{7 \pm Z}{1 \pm Y}$ . Plugging in the possible values of $\pm 7, \pm 1, \pm 5$ in their respective pairs and ruling out degenerate cases, we find the sum is $\frac{119}{12} \implies m + n = \boxed{131}$ - whatRthose | null | 131 |
8202e7860d90fc7d3aa42a3f5943e222 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_12 | 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 the square of any prime. Find $m+n+k$ | 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 $\triangle ABC$ is (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles ):
\[K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = 84\]
From $R = \frac{abc}{4K}$ , we know that the circumradius of $\triangle ABC$ is:
\[R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}\]
Thus by the Pythagorean Theorem again,
\[OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \frac{15\sqrt{95}}{8}.\]
So the final answer is $15+95+8=\boxed{118}$ | null | 118 |
8202e7860d90fc7d3aa42a3f5943e222 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_12 | 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 the square of any prime. Find $m+n+k$ | 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 other dimension, with the origin as $C$ . We look at vectors $OA$ and $OC$ . Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$ , hence it is $7$ units along the $x$ axis. Hence for vector $OC$ , in form $<x,y,z>$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$ , since $A$ is $9$ along $x$ , and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$ . We know both vector magnitudes are $20$ . Solving for $h$ yields $\frac{15\sqrt{95} }{8}$ , so Answer = $\boxed{118}$ | null | 118 |
cb6462dc48a56680c53ec4e27139224a | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13 | 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 roots must be the roots of the equation $20x^2+x-2=0$ . By the quadratic formula the roots of this are:
\[x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.\]
Thus $r=\frac{-1+\sqrt{161}}{40}$ , and so the final answer is $-1+161+40 = \boxed{200}$ | null | 200 |
cb6462dc48a56680c53ec4e27139224a | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13 | 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
$-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})$
Now, if we let $z = y + \frac{1}{y}$ , we can get the equations
$z = y + \frac{1}{y}$
$z^2 - 2 = y^2 + \frac{1}{y^2}$
and
$z^3 - 3z = y^3 + \frac{1}{y^3}$
(These come from squaring $z$ and subtracting $2$ , then multiplying that result by $z$ and subtracting $z$ )
Plugging this into our polynomial, expanding, and rearranging, we get
$-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})$
Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$ , so what squared equals 3? Plugging in $z = \sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\sqrt{3}$ , we see that it also works! Great, we use long division on the polynomial by
$(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)$ and we get
$2z -(\frac{i}{\sqrt{10}}) = 0$
We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant, then multiply by $-(\frac{i}{\sqrt{10}})$ . We get that $z = (\frac{i}{-2\sqrt{10}})$ . Solving for $y$ (using $y + \frac{1}{y} = z$ ) we get that $y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}$ , and multiplying this by $-(\frac{i}{\sqrt{10}})$ (because $x = -(\frac{i}{\sqrt{10}})y$ ) we get that $x = \frac{-1 \pm \sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \boxed{200}$ | null | 200 |
cb6462dc48a56680c53ec4e27139224a | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13 | 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 $2(1000x^6-1)+(1000x^6-1)\frac{x}{10x^2-1}=0, 10x^2-1\neq0 \implies$ $(1000x^6-1)(2+\frac{x}{10x^2-1})=0, 10x^2-1\neq0$ .
Note that setting $1000x^6-1=0$ gives $10x^2-1=0$ , which is clearly extraneous.
Hence, we set $2+\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\frac{-1+\sqrt{161}}{40} \implies -1+161+40=\boxed{200}$ | null | 200 |
23c9a6eff71426ebf9b7abdd02743654 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14 | 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 $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$ | 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 $m\in\mathbb{N}$
$(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.$
So now,
\begin{align*} 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ &=16! +\sum_{m=1}^{62}(32m+16)!-(32m)!\\ &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! \end{align*}
Therefore we have $f_{16} = 1$ $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$
Therefore we have:
\begin{align*} f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ &= -1 + \sum_{m=1}^{62}\left[\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\right]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}1\\ &= -1 + \sum_{m=1}^{62}8\\ &= -1 + 8\cdot 62\\ &= \boxed{495} | null | 495 |
23c9a6eff71426ebf9b7abdd02743654 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14 | 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 $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$ | 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 of the digits $n - 1$ and $0$ (in this example, $9900$ ). More specifically, the difference $(n^k)_n - (n^j)_n$ $j<k$ , is an integer $k$ digits long (note that $(n^k)_n$ has $k + 1$ digits). This integer is made up of $(k-j)$ $(n - 1)$ ’s followed by $j$ $0$ ’s.
It should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base $n$ , the largest digit value is $n - 1$ , in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, $321_!$ is a valid factorial base number, as is $3210_!$ . However, $31_!$ is not, as $3$ is greater than the argument of the second place factorial, $2$ $31_!$ should be represented as $101_!$ , and is $7_{10}$ .) Therefore, for example, $1000000_! - 10000_!$ is not $990000_!$ , but rather is $650000_!$ . Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a $1$ in the argument of the factorial’s place and $0$ ’s everywhere else, and then using a standard carry/borrow system accounting for the place value.
With general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: $(2000! - 1984!) + (1968! - 1952!) + \cdots + (48! - 32!) + 16!$ we now apply our intuition from paragraph 2. $2000!_{10}$ is equivalent to $1$ followed by $1999$ $0$ ’s in the factorial base, and $1984!$ is $1$ followed by $1983$ $0$ ’s, and so on. Therefore, $2000! - 1984! = (1999)(1998)(1997)\cdots(1984)$ followed by $1983$ $0$ ’s in the factorial base. $1968! - 1952! = (1967)(1966)\cdots(1952)$ followed by $1951$ $0$ ’s, and so on for the rest of the terms, except $16!$ , which will merely have a $1$ in the $16!$ place followed by $0$ ’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value $f_k$ is $k$ for all $32m \leq k \leq 32m+15$ if $1\leq m \in\mathbb{Z} \leq 62$ $f_{16} = 1$ , and $f_k = 0$ for all other $k$
Therefore, to answer, we notice that $1999 - 1998 = 1997 - 1996 = 1$ , and this will continue. Therefore, $f_{1999} - f_{1998} + \cdots - f_{1984} = 8$ . We have 62 sets that sum like this, and each contains $8$ pairs of elements that sum to $1$ , so our answer is almost $8 \cdot 62$ . However, we must subtract the $1$ in the $f_{16}$ place, and our answer is $8 \cdot 62 - 1 = \boxed{495}$ | null | 495 |
23c9a6eff71426ebf9b7abdd02743654 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14 | 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 $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$ | 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 that $1999 \cdot 1999! < S < 2000 \cdot 1999!$ . Hence $f_{1999} = 1999$ , and we now need to find the factorial base expansion of
\[S_2 = S - 1999 \cdot 1999! = 1999! - 1984! + 1962! - 1946! + \cdots + 16!\]
Since $|S_2 - 1999!| << 1999!$ , we can repeat the above argument recursively to yield $f_{1998} = 1998$ , and so forth down to $f_{1985} = 1985$ . Now $S_{16} = 1985! - 1984! + 1962! + \cdots = 1984 \cdot 1984! + 1962! + \cdots$ , so $f_{1984} = 1984$
The remaining sum is now just $1962! - 1946! + \cdots + 16!$ . We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$ , and $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$
Now for each $m$ , we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \cdots + 1 + 1$ $= 8$ . Thus, our answer is $-f_{16} + 8 \cdot 62 = \boxed{495}$ | null | 495 |
10b50d55917dd542d2b311c91eb88e2d | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_15 | Find the least positive integer $n$ such that | We apply the identity
\begin{align*} \frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}
The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping
Thus our summation becomes
\[\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).\]
Since $\cot (180 - x) = - \cot x$ , the summation simply reduces to $\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}$ . Therefore, the answer is $\boxed{001}$ | null | 001 |
10b50d55917dd542d2b311c91eb88e2d | https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_15 | Find the least positive integer $n$ such that | Let S be the sum of the sequence. We begin the same as in Solution 1 to get $S\sin(1)=\cot(45)-\cot(46)+\cot(47)-\cot(48)+...+\cot(133)-\cot(134)$ . Observe that this "almost telescopes," if only we had some extra terms. Consider adding the sequence $\frac{1}{\sin(46)\sin(47)}+\frac{1}{\sin(48)\sin(49)}+...+\frac{1}{\sin(134)\sin(135)}$ . By the identity $\sin(x)=\sin(180-x)$ , this sequence is equal to the original one, simply written backwards. By the same logic as before, we may rewrite this sequence as $S\sin(1)=\cot(46)-\cot(47)+\cot(48)-\cot(49)+...+\cot(134)-\cot(135)$ ,
and when we add the two sequences, they telescope to give $2S\sin(1)=\cot(45)-\cot(135)=2$ .
Hence, $S=\sin(1)$ , and our angle is $\boxed{001}$ | null | 001 |
948bdd7f30ff00f07464c6bf40addbef | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_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 $m+n$ | 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 $\frac{45 + a}{10} = \frac{153 - a}{28}$ . Solving for $a$ yields that $1530 - 10a = 1260 + 28a$ , so $a=\frac{270}{38}=\frac{135}{19}$ . The slope of the line (since it passes through the origin) is $\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}$ , and the solution is $m + n = \boxed{118}$ | null | 118 |
948bdd7f30ff00f07464c6bf40addbef | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_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 $\boxed{118}$ | null | 118 |
948bdd7f30ff00f07464c6bf40addbef | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_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 $m+n$ | 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 parallelogram can be denoted as $(10, \frac{10m}{n})$ and $(28, \frac{28m}{n}),$ respectively. We see that $\frac{10m}{n} - 45 + \frac{28m}{n} - 84 = 69,$ so $\frac{38m}{n} = 198 \implies \frac{m}{n} = \frac{99}{19} \implies \boxed{118}.$ | null | 118 |
948bdd7f30ff00f07464c6bf40addbef | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_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 $m+n$ | $(\Sigma x_i /4, \Sigma y_i /4)$ 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. | null | 118 |
626b53c735b5103e44b0fd5ca96eaa2c | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_3 | 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 $9$ . This gives $n=1, 9, 10,$ or $18$ , and the sum is $1+9+10+18=\boxed{38}$ | null | 38 |
626b53c735b5103e44b0fd5ca96eaa2c | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_3 | 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 $n^2 -19n + 99 = (n-10)^2 + n - 1.)$
We conclude that the answer is $1 + 9 + 10 + 18 = \boxed{38}.$ | null | 38 |
66092325b121487d3e02472407f979a4 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4 | 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), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy] | 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 $bh/2$ , the area of all $8$ of them is $\frac{86}{99}$ and the answer is $\boxed{185}$ | null | 185 |
66092325b121487d3e02472407f979a4 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4 | 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), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy] | 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{align*}x + y + \frac{43}{99} &= 1\\ x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}
Substituting, \begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\ 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}
Thus, the area of the octagon is $1 - \frac{13}{99} = \frac{86}{99}$ , so $m + n = \boxed{185}$ | null | 185 |
4829c4112b4ed3289c400eabdc934c09 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_5 | 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 $|2 - 9n| = 9n - 2$ . From $7$ to $1999$ , there are $\left\lceil \frac{1999 - 7}{9}\right\rceil = 222$ solutions; including $2$ and there are a total of $\boxed{223}$ solutions. | null | 223 |
d1c47a2a5c1b9d2669ababb0399e4853 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_6 | 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 greatest integer that does not exceed $k_{}.$ | \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 above, we see the equations for $A'B'$ and $C'D'$ are $y^2 = \frac {1}{3}x^2$ and $y^2 = 3x^2$ , or, after manipulation $y = \frac {x}{\sqrt {3}}$ and $y = \sqrt {3}x$ , respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines.
Now take a look at $BC$ and $AD$ , which have the equations $y = - x + 2400$ and $y = - x + 1200$ . The image equations hence are $x^2 + y^2 = 2400$ and $x^2 + y^2 = 1200$ , respectively, which are the equations for circles
1999 AIME-6.png
To find the area between the circles (actually, parts of the circles), we need to figure out the angle of the arc . This could be done by $\arctan \sqrt {3} - \arctan \frac {1}{\sqrt {3}} = 60^\circ - 30^\circ = 30^\circ$ . So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = $\frac {30^\circ}{360^\circ}(R^2\pi - r^2\pi) = \frac {1}{12}(2400\pi - 1200\pi) = 100\pi$ . Hence the answer is $\boxed{314}$ | null | 314 |
ae057c22da59169a5551da4a4ff008b8 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_7 | 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 integers $(2^{x})(3^{y})(5^{z})$ , where $x, y$ , and $z$ take on the values $0, 1, \ldots, 9$ . At step i of a 1000-step process, the $i$ -th switch is advanced one step, and so are all the other switches whose labels divide the label on the $i$ -th switch. After step 1000 has been completed, how many switches will be in position $A$ | 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 $N = 2^{9}3^{9}5^{9}$ be the max switch label. To find the divisor multiples in the range of $d_{i}$ to $N$ , we consider the exponents of the number $\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}$ . In general, the divisor-count of $\frac{N}{d}$ must be a multiple of 4 to ensure that a switch is in position A:
We consider the cases where the 3 factors above do not contribute multiples of 4.
The number of switches in position A is $1000-125-225 = \boxed{650}$ | null | 650 |
f14eb190938dbee0905ee9a60df14c66 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_8 | Let $\mathcal{T}$ be the set of ordered triples $(x,y,z)$ of nonnegative real numbers that lie in the plane $x+y+z=1.$ Let us say that $(x,y,z)$ supports $(a,b,c)$ when exactly two of the following are true: $x\ge a, y\ge b, z\ge c.$ Let $\mathcal{S}$ consist of those triples in $\mathcal{T}$ that support $\left(\frac 12,\frac 13,\frac 16\right).$ The area of $\mathcal{S}$ divided by the area of $\mathcal{T}$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ | This problem just requires a good diagram and strong 3D visualization.
1999 AIME-8.png
The region in $(x,y,z)$ where $x \ge \frac{1}{2}, y \ge \frac{1}{3}$ is that of a little triangle on the bottom of the above diagram, of $y \ge \frac{1}{3}, z \ge \frac{1}{6}$ is the triangle at the right, and $x \ge \frac 12, z \ge \frac 16$ the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by $x+y+z=1,\ x,y,z \ge 0$ . We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third.
1999 AIME-8a.png
The side length of the large equilateral triangle is $\sqrt{2}$ , which we can find using 45-45-90 $\triangle$ with the axes. Using the formula $A = \frac{s^2\sqrt{3}}{4}$ for equilateral triangles , the area of the large triangle is $\frac{(\sqrt{2})^2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$ . Since the lines of the smaller triangles are parallel to those of the large triangle, by corresponding angles we see that all of the triangles are similar , so they are all equilateral triangles. We can solve for their side lengths easily by subtraction, and we get $\frac{\sqrt{2}}{6}, \frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{2}$ . Calculating their areas, we get $\frac{\sqrt{3}}{8}, \frac{\sqrt{3}}{18}, \frac{\sqrt{3}}{72}$ . The ratio $\frac{\mathcal{S}}{\mathcal{T}} = \frac{\frac{9\sqrt{3} + 4\sqrt{3} + \sqrt{3}}{72}}{\frac{\sqrt{3}}{2}} = \frac{14}{36} = \frac{7}{18}$ , and the answer is $m + n = \boxed{025}$ | null | 025 |
a75a31d954875014e53a991ca42c92da | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 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 relatively prime positive integers, find $m+n.$ | 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 $\left(\frac 12, \frac12\right)$ , so its graph is $x + y = 1$ . Substituting $x = (a-b)$ and $y = (a+b)$ , we get $2a = 1 \Rightarrow a = \frac 12$
By the Pythagorean Theorem , we have $\left(\frac{1}{2}\right)^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}$ , and the answer is $\boxed{259}$ | null | 259 |
a75a31d954875014e53a991ca42c92da | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 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 relatively prime positive integers, find $m+n.$ | 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 $\boxed{259}$ | null | 259 |
a75a31d954875014e53a991ca42c92da | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 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 relatively prime positive integers, find $m+n.$ | 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 = \frac 12 \end{eqnarray*} Since $|a + bi| = 8,$ $a^2 + b^2 = 64.$ Because $a = \frac 12,$ thus $b^2 = \frac {255}4.$ So the answer is $\boxed{259}$ | null | 259 |
a75a31d954875014e53a991ca42c92da | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 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 relatively prime positive integers, find $m+n.$ | 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, $(d^2+c^2)$ can't be zero because this property of the function holds for all complex $z$ . Therefore, $a = \frac{1}{2}$ and we proceed as above to get $\boxed{259}$ | null | 259 |
a75a31d954875014e53a991ca42c92da | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | 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 relatively prime positive integers, find $m+n.$ | 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 of point $P$ be $Q$ , after the point undergoes the function. Since each image is equidistant from the preimage and the origin, $Q$ must be on the perpendicular bisector of $OP$ .Given $z=c+di$ $f(z)=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ . Then $Q=(ac-bd,ad+bc)$ . The midpoint of $OP$ is $(0.5c, 0.5d)$ . Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of $-1$ , using the point-slope form, the equation of the perpendicular line to $OP$ is $y-0.5d=-\frac{c}{d}(x-0.5c)$ . Rearranging, we have $y=-\frac{cx}{d}+\frac{c^2}{2d}+\frac{d}{2}$
Since we know that $Q=(ac-bd,ad+bc)$ , thus we plug in $Q$ into the line: $ad+bc=-\frac{ac^2-bcd}{d}+\frac{c^2}{2d}+\frac{d}{2}$
Let's start canceling. $2ad^2+2bcd=-2ac^2+2bcd+c^2+d^2$ . Subtracting, $c^2+d^2-2ac^2=2ad^2$ . Thus $c^2+d^2=2ac^2+2ad^2$ . Since this is an identity for any $(c,d)$ , thus $2a=1$ $a=\frac{1}{2}$ . Since $|a+bi|=8$ , thus $a^2+b^2=64$ (or simply think of $a+bi$ as the point $(a,b)$ , and $|a+bi|$ being the distance of $(a,b)$ to the origin). Thus plug in $a=\frac{1}{2}, b^2=\frac{255}{4}$ . Since $255$ and $4$ are relatively prime, the final result is $255+4=\boxed{259}$ | null | 259 |
3fe5565580f13450519c0d9cb6897917 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_10 | 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 $n_{}$ are relatively prime positive integers . Find $m+n.$ | 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 that the fourth segment doesn't matter in this case.
Note that there are $(9 - 1) \times 2 = 16$ segments that share an endpoint with the first segment. The answer is then $4 \times \frac{16}{44} \times \frac{1}{43} = \frac{16}{11} \times \frac{1}{43} = \frac{16}{473} \implies m + n = \boxed{489}$ -whatRthose | null | 489 |
3fe5565580f13450519c0d9cb6897917 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_10 | 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 $n_{}$ are relatively prime positive integers . Find $m+n.$ | 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 will be "left out".
However, if we have $5$ endpoints, we can use three segments form a triangle with three of the points and connect the remaining two points with the last segment. There are ${10\choose5}$ ways to select these $5$ points from the original $10,$ and ${5\choose3}$ ways to decide which three points are in the triangle.
Finally, if we have $4$ endpoints, we can also form a triangle with three of the points, then use the remaining segment to connect the last point to either of the previous three. We have ${10\choose4}$ ways to select the $4$ points and ${4\choose3}$ ways to choose three points for the triangle. Finally, we must connect the last point to one vertex of the triangle; we can do this in $3$ ways.
As in Solution 1, there are ${45\choose4}$ total ways to select four segments. So, our desired probability is
\[\dfrac{{10\choose5}{5\choose3}+{10\choose4}{4\choose3}\cdot 3}{{45\choose4}}=\dfrac{16}{473} \implies m + n = \boxed{489}.\] | null | 489 |
d1fad871cd9c6861841854a2697c1bfb | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11 | 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 identity $\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))$ , we can rewrite $s$ as
\begin{align*} s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 &= \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\\ &= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180) \end{align*}
This telescopes to \[s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.\] Manipulating this to use the identity $\tan x = \frac{1 - \cos 2x}{\sin 2x}$ , we get \[s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},\] and our answer is $\boxed{177}$ | null | 177 |
d1fad871cd9c6861841854a2697c1bfb | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11 | 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}\\ &= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\ &= \frac{2 \sin 5}{2 - 2 \cos 5}\\ &= \frac{\sin 5}{1 - \cos 5}\\ &= \frac{\sin 175}{1 + \cos 175} \\ &= \tan \frac{175}{2}.\\ \end{align*} The desired answer is thus $175 + 2 = \boxed{177}$ | null | 177 |
d1fad871cd9c6861841854a2697c1bfb | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11 | 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 $\frac{1}{2i}\left(\frac{-\frac{1}{x^{35}}(x^{35}-1)}{x-1}+\frac{x(x^{35}-1)}{x-1}\right)$
However, we note that $x^{36}=-1$ , so $-\frac{1}{x^{35}}=x$ , so our two geometric series are actually the same. We combine the terms and simplify to get $\frac{1}{i}\left(\frac{x^{36}-x}{x-1}\right)$
Apply Euler's identity and simplify again to get $\frac{1}{i}\left(\frac{-x-1}{x-1}\right)$
Now, we need to figure out how to express this as the tangent of something. We note that $\tan(5k^\circ)=\frac{\sin(5k^\circ)}{\cos(5k^\circ)}=\frac{\frac{x^k-\frac{1}{x^k}}{2i}}{\frac{x^k+\frac{1}{x^k}}{2}}=\frac{1}{i}\frac{x^{2k}-1}{x^{2k}+1}$
So, we set the two equal to each other to solve for $k$ . Cross multiplying gets $(-x-1)(x^{2k}+1)=(x-1)(x^{2k}-1)$ . Expanding yields $-x^{2k+1}-x-x^{2k}-1=x^{2k+1}-x-x^{2k}+1$ . Simplifying yields $x^{2k+1}=-1$ . Since $2k+1=36$ is the smallest solution, we have $k=\frac{35}{2}$ , and the argument of tangent is $5k=\frac{175}{2}$ . The requested sum is $175+2=\boxed{177}$ | null | 177 |
7dc9bf3c8c5fe9dc382e006b7ce0d2ab | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_12 | 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 $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}$ . Equating and squaring both sides,
\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2} \end{eqnarray*}
We want the perimeter, which is $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$ | null | 345 |
7dc9bf3c8c5fe9dc382e006b7ce0d2ab | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_12 | 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 \beta & = & \frac {21}{23} \\ \tan \gamma & = & \frac {21}x. \end{eqnarray*} So naturally we look at $\tan \gamma.$ But since $\gamma = \frac \pi2 - (\beta + \alpha)$ we have \begin{eqnarray*} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\ & = & \frac 1{\tan(\alpha + \beta)} \\ \Rightarrow \frac {21}x & = & \frac {1 - \frac {21\cdot 21}{23\cdot 27}}{\frac {21}{27} + \frac {21}{23}} \end{eqnarray*} Doing the algebra, we get $x = \frac {245}2.$
The perimeter is therefore $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$ | null | 345 |
bf5b21db0e8e402482239e8defc20349 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_13 | 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 $\log_2 n.$ | 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, each team corresponds uniquely with some $k$ , with $0 \leq k \leq 39$ , where $k$ represents the number of games the team won. With this in mind, we see that there are a total of $40!$ outcomes in which no two teams win the same number of games. Further, note that these are all the valid combinations, as the team with 1 win must beat the team with 0 wins, the team with 2 wins must beat the teams with 1 and 0 wins, and so on; thus, this uniquely defines a combination.
The desired probability is thus $\frac{40!}{2^{780}}$ . We wish to simplify this into the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime. The only necessary step is to factor out all the powers of 2 from $40!$ ; the remaining number is clearly relatively prime to all powers of 2.
The number of powers of 2 in $40!$ is $\left \lfloor \frac{40}{2} \right \rfloor + \left \lfloor \frac{40}{4} \right \rfloor + \left \lfloor \frac{40}{8} \right \rfloor + \left \lfloor \frac{40}{16} \right \rfloor + \left \lfloor \frac{40}{32} \right \rfloor = 20 + 10 + 5 + 2 + 1 = 38.$
$780-38 = \boxed{742}$ | null | 742 |
0538434dec1f9deb86e5f8158afb08ab | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 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 tool of calculating area in two ways \begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\ &=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\ &=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*} On the other hand, \begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21\cdot6\cdot7\cdot8}\\ &=84\end{align*} We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras . So we give it a shot: \begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\ z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\ y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align} Adding $(1) + (2) + (3)$ gives \begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\ \Rightarrow13z+14x+15y&=295\end{align*} Recall that we found that $[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84$ . Plugging in $13z+14x+15y=295$ , we get $\tan\theta=\frac{168}{295}$ , giving us $\boxed{463}$ for an answer. | null | 463 |
0538434dec1f9deb86e5f8158afb08ab | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 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: \[a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1)\] Now $[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84$ , by Heron's formula
Now the area of a triangle, $[A] = \frac {mn\sin{\beta}}{2}$ , where $m$ and $n$ are sides on either side of an angle, $\beta$ . So, \begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\ [ABP] &= \frac {cx\sin{\theta}}{2}\\ [BCP] &= \frac {ay\sin{\theta}}{2}\end{align*} Adding these equations yields: \begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\ \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*} Dividing $(2)$ by $(1)$ , we have: \begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\ \Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{align*} Thus, $m + n = 168 + 295 = \boxed{463}$ | null | 463 |
0538434dec1f9deb86e5f8158afb08ab | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 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 \[(13a + 14b + 15c) \cos x = 295.\] Also, using $[ABC] = \frac{1}{2}ab \sin \angle C$ we have \[[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.\] Divide the two equations to obtain $\tan x = \frac{168}{295} \iff \boxed{463}.~\square$ | null | 463 |
0538434dec1f9deb86e5f8158afb08ab | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | 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 computed, so we have $\cos C=\frac{3}{5}$
Using Law of Sines, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)} \end{align*} hence, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin B} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin C} \end{align*} Now, computation carries the rest. \begin{align*} \frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C} \\ \frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \\ 169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \\ 169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \\ 169 \sin x &= 168 \cos x - 126 \sin x \\ 295 \sin x &= 168 \cos x \\ \tan x &= \frac{168}{295} \end{align*} Extracting yields $168 + 295 = \boxed{463}$ | null | 463 |
3099c641c14c88457c8760eaa5ed4430 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 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 point $O$ is the orthocenter of $\triangle ABC$
Proof. Observe that \[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$ . Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$ . Analogously, $O$ lies on the $B$ -altitude and $C$ -altitude of $\triangle ABC$ , and so $O$ is, indeed, the orthocenter of $\triangle ABC$
To find the coordinates of $O$ , we need to find the intersection point of altitudes $BE$ and $AD$ . The equation of $BE$ is simply $x=16$ $AD$ is perpendicular to line $BC$ , so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$ $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$ , therefore $y=\frac{3}{4} x$ . These two lines intersect at $(16,12)$ , so that's the base of the height of the tetrahedron.
Let $S$ be the foot of altitude $BS$ in $\triangle BPQ$ . From the Pythagorean Theorem $h=\sqrt{BS^2-SO^2}$ . However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$
The area of the base is $102$ , so the volume is $\frac{102*12}{3}=\boxed{408}$ | null | 408 |
3099c641c14c88457c8760eaa5ed4430 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 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, the height of the pyramid is $z$ . The desired volume is thus $\frac{102z}{3} = 34z$
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$ $VP = PB$ , and $VQ = QC$ . We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$ . The desired volume is thus $34 \times 12 = \boxed{408}$ | null | 408 |
3099c641c14c88457c8760eaa5ed4430 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 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 Pythagoras)
$p^2+q^2=4^2\cdot{13}$
$q^2+r^2=15^2$
$r^2+p^2=17^2$
to find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\cdot{17},2^3\cdot{17},2^3\cdot{3^2}).$
Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\tfrac{1}{3}$ and then the volume is
$\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}$ | null | 408 |
3099c641c14c88457c8760eaa5ed4430 | https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15 | 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 $\triangle DEP \cong \triangle EDF$ . Create point $F'$ on the plane of $DEF$ such that $\triangle DEP \cong \triangle DEF'$ (i.e by reflecting $F$ over the perpendicular bisector of $DE$ ). Project $F, P$ onto $DE$ as $X, Y$ . Note by the definition of $F'$ then $\angle PYF'$ is the dihedral angle between planes $DEP, DEF$
Now see that by Heron's, \[[DEP] = [DEF] = \sqrt{(16 + 2 \sqrt{13})(16 - 2 \sqrt{13})(1 + 2 \sqrt{13})(-1 + 2 \sqrt{13})} = 102.\] So $PY$ , the hypotenuse $DEP$ has length $\frac{102 \cdot 2}{4 \sqrt{13}} = \frac{51}{\sqrt{13}}$ . Similarly $F'Y = \frac{51}{\sqrt{13}}.$ Further from Pythagoras $DY = \sqrt{DP^2 - PY^2} = \frac{18}{\sqrt{13}}.$ Symmetrically $EX = \frac{18}{\sqrt{13}}.$ Therefore $XY = DE - DY - EX = \frac{16}{\sqrt{13}}.$
By Law of Cosines on $\triangle PYF'$ \begin{align*} PF'^2 &= PY^2 + F'Y^2 - 2 \cdot PY \cdot F'Y \cos{\angle PYF'} \\ PF^2 - XY^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ (4\sqrt{13})^2 - (\frac{16}{\sqrt{13}})^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ \cos{\angle PYF'} &= \frac{9}{17} \\ \sin{\angle PYF'} &= \frac{4 \sqrt{13}}{17}. \end{align*}.
Therefore the altitude of the tetrahedron from vertex $P$ to base $DEF$ is $PY \sin{\angle PYF'} = \frac{51}{\sqrt{13}} \frac{4 \sqrt{13}}{17} = 12.$ So the area is $\frac{1}{3}bh = \frac{1}{3} 12 \cdot 102 = \boxed{408}.$ | null | 408 |
10abbe485a7e30114a7909d9aa0062ba | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_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 $12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$ , and $b = 12$ . Since $0 \le a \le 24$ , there are $\boxed{25}$ values of $k$ | null | 25 |
f550d28232ab1258e2ac75f6b98e2b49 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_3 | 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 $y = 20$ or $y = -2x - 20$ . If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is $\boxed{800}$ | null | 800 |
0db69c362d8ac2f9c088a74a8630c71b | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_4 | Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m+n.$ | In order for a player to have an odd sum, he must have an odd number of odd tiles: that is, he can either have three odd tiles, or two even tiles and an odd tile. Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ even tiles and $1$ odd tile. We count the number of ways this can happen. (We will count assuming that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)
$\dbinom{5}{3} = 10$ choices for the tiles that he gets. The other two odd tiles can be distributed to the other two players in $2$ ways, and the even tiles can be distributed between them in $\dbinom{4}{2} \cdot \dbinom{2}{2} = 6$ ways. This gives us a total of $10 \cdot 2 \cdot 6 = 120$ possibilities in which all three people get odd sums.
In order to calculate the probability, we need to know the total number of possible distributions for the tiles. The first player needs three tiles which we can give him in $\dbinom{9}{3} = 84$ ways, and the second player needs three of the remaining six, which we can give him in $\dbinom{6}{3} = 20$ ways. Finally, the third player will simply take the remaining tiles in $1$ way. So, there are $\dbinom{9}{3} \cdot \dbinom{6}{3} \cdot 1 = 84 \cdot 20 = 1680$ ways total to distribute the tiles.
We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is $\frac{360}{1680} = \frac{3}{14},$ so the answer is $3 + 14 = \boxed{017}$ | null | 017 |
0db69c362d8ac2f9c088a74a8630c71b | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_4 | Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m+n.$ | Let $O$ stand for an odd number and $E$ an even.
Therefore, one person must pick $OOO$ , the other person must pick $OEE$ and the last person must pick $OEE$ . Since any permutation of the order of who is picking or change in the order of the even numbers (e.g. $EOE$ instead of $OEE$ ) doesn't change the probability, we just need to multiply the probability of one case by $\binom{3}{2}^3 = 27$ as there are 27 such cases (by cases I mean ordered triples of ordered multisets $A, B, C$ such that one of them has 3 $O$ 's and the other two have two $E$ 's and an $O$ in them, respectively.) . Let's do the case $OOO$ $OEE$ $OEE$ $\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} = \frac{1}{126}$ . We now multiply by 27 to get $\frac{3}{14} \implies m + n = \boxed{017}$ - whatRthose | null | 017 |
0db69c362d8ac2f9c088a74a8630c71b | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_4 | Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m+n.$ | For this problem, let's think about parity. There are $5$ odd numbers from $1-9$ and there are four even numbers from $1-9$ . Since this problem is asking for the probability that the each player gets an odd sum, we also have to calculate the total numeber of ways.
In this case, there are only two ways to get an odd sum. Either have the sequence $OOO$ or $OEE$ where the letters $O$ and $E$ stand for odd and even respectively.
Since we constrained to only $5$ odds, the only way to do the pairing is \[OOO-OEE-OEE\] There are of couse three ways to choose who gets the three odds.
Once we have chosen who has gotten the three odds, we can actually reorder the sequence like this: \[OEE-OEE-OOO\] Now since we are choosing in groups, we can ignor order of these terms.
For the first $OEE$ , there are $5$ ways to choose which odd to use, and $\dbinom{4}{2}$ or $6$ ways to choose the evens.
Great, let's move on to the second $OEE$ . There are $4$ odds left, so there are $4$ ways choose the odds. Since there are only two even numbers, they have to go here, so there is only $1$ way to choose the evens.
The three odds will now fall in place. We can now multiply all the numbers since they are independent, and we have $5*6*4*3$ or $360$
Since this is a probability question, we have to ask ourselves how many ways are there to distribute the 9 tiles equally among 3 players.
Fortunately for us, this is not hard as the first player has $\dbinom{9}{3}$ options and the second player has $\dbinom{6}{3}$ . When we multiply these, we get $1680$ . This is our denominator.
When we make the fraction, we have $\frac{360}{1680}$ . When we simplify it, we have: \[\frac{3}{14}\] We are asked to find the sum of the numerator and the denominator, so summing these, we have: \[\boxed{017}\] -Pi_is_3.14 | null | 017 |
3cd53a0450dc4b2ee2854bffcbe7bc57 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_5 | Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$ | Though the problem may appear to be quite daunting, it is actually not that difficult. $\frac {k(k-1)}2$ always evaluates to an integer ( triangular number ), and the cosine of $n\pi$ where $n \in \mathbb{Z}$ is 1 if $n$ is even and -1 if $n$ is odd. $\frac {k(k-1)}2$ will be even if $4|k$ or $4|k-1$ , and odd otherwise.
So our sum looks something like:
$\left|\sum_{i=19}^{98} A_i\right| =- \frac{19(18)}{2} + \frac{20(19)}{2} + \frac{21(20)}{2} - \frac{22(21)}{2} - \frac{23(22)}{2} + \frac{24(23)}{2} \cdots - \frac{98(97)}{2}$
If we group the terms in pairs, we see that we need a formula for $-\frac{(n)(n-1)}{2} + \frac{(n+1)(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n$ . So the first two fractions add up to $19$ , the next two to $-21$ , and so forth.
If we pair the terms again now, each pair adds up to $-2$ . There are $\frac{98-19+1}{2 \cdot 2} = 20$ such pairs, so our answer is $|-2 \cdot 20| = \boxed{040}$ | null | 040 |
f942660dd79f957e1ff5177c6bff33cc | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_6 | 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 $RC=\boxed{308}$ | null | 308 |
9cd682f98dfcaaeaee661e0ac6a3fcef | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 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 want $x_1 +x_2+x_3+x_4 =98$ . This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$ , as for all integers $y_i$ $2y_i + 1$ will be odd. Substituting we get \[2y_1+2y_2+2y_3+2y_4 +4 = 98 \implies y_1+y_2+y_3+y_4 =47\] Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us $n= {50\choose3}$ . Computing this and dividing by 100 gives us an answer of $\boxed{196}$ .
~smartguy888 | null | 196 |
9cd682f98dfcaaeaee661e0ac6a3fcef | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 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; thus the answer is $n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600$ , and $\frac n{100} = \boxed{196}$ | null | 196 |
9cd682f98dfcaaeaee661e0ac6a3fcef | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 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. Because we want an odd number in each box, we pair the stones, creating $47$ sets of $2$ . Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).
Our problem can now be restated: how many ways are there to partition a line of $47$ stones? We can easily solve this by using $3$ sticks to separate the stones into $4$ groups, and this is the same as arranging a line of $3$ sticks and $47$ stones. \[\frac{50!}{47! \cdot 3!} = 19600\] \[\frac{50 * 49 * 48}{3 * 2} = 19600\] Our answer is therefore $\frac{19600}{100} = \boxed{196}$ | null | 196 |
9cd682f98dfcaaeaee661e0ac6a3fcef | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 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) + (24.5 - 22.5)(24.5 + 22.5) + ... + (24.5 + 23.5)(24.5 - 23.5)\\ &= 48(24.5)^2 - 2(0.5^2 + 1.5^2 + ... + 23.5^2)\\ &= 28812 - \frac{1^2 + 3^2 + ... + 47^2}{2}\\ &= 28812 - \frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2}\\ &= 28812 - \frac{\frac{47(47 + 1)(2(47) + 1)}{6} - \frac{4(23)(23 + 1)(2(23) + 1)}{6}}{2}\\ &= 19600 \end{align*}
Therefore, $\frac{n}{100} = \frac{19600}{100} = \boxed{196}$ | null | 196 |
9cd682f98dfcaaeaee661e0ac6a3fcef | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | 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 $\left(\frac{1}{1-x^2}\right)^4=\binom{3}{3} +\binom{4}{3}x^2+\binom{5}{3}x^4 \cdots$ and in general we want that the coefficient of $x^{2k}$ is $\binom{k+3}{3}.$ We see the $x^{94}$ coefficient so we let $k=47$ and so the coefficient is $\binom{50}{3}=19600$ in which $\frac{n}{100}=\boxed{196}.$ | null | 196 |
887f4c9e41840e8858de62161578ec2e | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_8 | 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):
$x < \frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}$
$x > \frac{F_{8}}{F_{9}} \cdot 1000 = \frac{21}{34} \cdot 1000 \approx 617.977$
Thus the sequence is maximized when $x = \boxed{618}.$ | null | 618 |
887f4c9e41840e8858de62161578ec2e | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_8 | 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 $x = \boxed{618}.$ | null | 618 |
2c3890690c8dfe3523cbc8ba2e9c6be8 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_10 | 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 $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime . Find $a + b + c$ | 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 examine the relation between one of the outside 8 spheres and the center one (with radius $r$ ):
1998 AIME-10b.png
If we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid ; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$ . Then by the Pythagorean Theorem
\[x^2 + (r-100)^2 = (r+100)^2 \Longrightarrow x = 20\sqrt{r}\]
1998 AIME-10c.png
$x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40\sqrt{r}$ . We can draw another right triangle as shown above. One leg has a length of $200$ . The other can be found by partitioning the leg into three sections and using $45-45-90 \triangle$ s to see that the leg is $100\sqrt{2} + 200 + 100\sqrt{2} = 200(\sqrt{2} + 1)$ . Pythagorean Theorem:
\begin{eqnarray*} (40\sqrt{r})^2 &=& 200^2 + [200(\sqrt{2}+1)]^2\\ 1600r &=& 200^2[(1 + \sqrt{2})^2 + 1] \\ r &=& 100 + 50\sqrt{2} \end{eqnarray*}
Thus $a + b + c = 100 + 50 + 2 = \boxed{152}$ | null | 152 |
2c3890690c8dfe3523cbc8ba2e9c6be8 | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_10 | 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 $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime . Find $a + b + c$ | 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*cos(45^\circ) \\ &=& 2x^2 - 2x^2*\frac{\sqrt{2}}{2} \\ &=& (2-\sqrt{2})x^2 \end{eqnarray*}
And thus \[x = \frac{200}{\sqrt{2-\sqrt{2}}}\]
From the above, $x = 20\sqrt{r}$ , so we get
\begin{eqnarray*} r &=& (\frac{200}{20(\sqrt{2-\sqrt{2}})})^2 \\ &=& (\frac{10}{\sqrt{2-\sqrt{2}})})^2 \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} \\ &=& \frac{200 + 100\sqrt{2}}{2} \\ &=& 100 + 50\sqrt{2} \end{eqnarray*}
And hence the answer is $100 + 50 + 2 \Rightarrow \boxed{152}$ | null | 152 |
d5bb96fb478fd5954dd03e8a0b98adfc | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_12 | Let $ABC$ be equilateral , and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + b\sqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime . What is $a^{2} + b^{2} + c^{2}$ | 1998 AIME-12.png
We let $x = EP = FQ$ $y = EQ$ $k = PQ$ . Since $AE = \frac {1}{2}AB$ and $AD = \frac {1}{2}AC$ $\triangle AED \sim \triangle ABC$ and $ED \parallel BC$
By alternate interior angles, we have $\angle PEQ = \angle BFQ$ and $\angle EPQ = \angle FBQ$ . By vertical angles, $\angle EQP = \angle FQB$
Thus $\triangle EQP \sim \triangle FQB$ , so $\frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y$
Since $\triangle EDF$ is equilateral, $EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1$ . Solving for $x$ and $y$ using $x^{2} = y$ and $x + y = 1$ gives $x = \frac {\sqrt {5} - 1}{2}$ and $y = \frac {3 - \sqrt {5}}{2}$
Using the Law of Cosines , we get
We want the ratio of the squares of the sides, so $\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}$ so $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}$ | null | 083 |
d5bb96fb478fd5954dd03e8a0b98adfc | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_12 | Let $ABC$ be equilateral , and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + b\sqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime . What is $a^{2} + b^{2} + c^{2}$ | WLOG, let $\Delta ABC$ have side length $2.$ Then, $DE = EF = FD = 1.$ We also notice that $\angle CEP = \angle DEF = 60^{\circ},$ meaning $\angle CEF = \angle CEP + \angle DEF = 120^{\circ}.$
Let $EP = x.$ Since $FQ = x$ by congruent triangles $\Delta EPC$ and $\Delta FQA,$ $EQ = EF - FQ = 1-x.$ We can now apply Law of Cosines to $\Delta CEP, \Delta PEQ,$ and $\Delta CEQ.$
By LoC on $\Delta CEP,$ we get \[CP^2 = 1^2 + x^2 - 2\cdot 1\cdot x\cdot \left(\frac{1}{2}\right) = x^2 - x + 1.\]
In a similar vein, using LoC on $\Delta PEQ$ and $\Delta CEQ,$ respectively, earns \[PQ^2 = x^2 + (1-x)^2 - 2\cdot x\cdot (1-x)\cdot \left(\frac{1}{2}\right) = 3x^2 - 3x + 1\] \[CQ^2 = 1^2 + (1-x)^2 - 2\cdot 1\cdot (1-x)\cdot \left(-\frac{1}{2}\right) = x^2 - 3x + 3\]
We have $CP^2, PQ^2,$ and $CQ^2.$ Additionally, by segment addition, $CP + PQ = CQ.$ Solving for $CP, PQ,$ and $CQ$ from the Law of Cosines expressions and plugging them into the segment addition gets the (admittedly-ugly) equation \[\sqrt{x^2-3x+3} = \sqrt{x^2-x+1} + \sqrt{3x^2-3x+1}.\]
Since the equation is ugly, we look at what the problem is asking for us to solve. We want $\frac{[ABC]}{[PQR]}.$ We see that $[ABC] = \sqrt{3}$ and $[PQR] = [DEF] - [PDR] - [RFQ] - [QEP] = \frac{\sqrt{3}}{4} - \frac{3}{2}\left(\frac{\sqrt{3}}{2}x(1-x)\right),$ since $[PDR] = [RFQ] = [QEP] = \frac{1}{2}x(1-x)\frac{\sqrt{3}}{2}$ from the sine area formula. Simplifying $\frac{[ABC]}{[PQR]}$ gets us wanting to find $\frac{4}{3x^2-3x+1}.$
We see $3x^2-3x+1$ in both the denominator of what we want and under a radicand in our algebraic expression, which leads us to think the calculations may not be that bad. Isolate $\sqrt{3x^2-3x+1}$ and square to get \[3x^2-3x+1 = 2x^2-4x+4-2\sqrt{(x^2-3x+3)(x^2-x+1)}\]
Isolate the radicand and square and expand to get $x^4+2x^3-5x^2-6x+9=4x^4-16x^3+28x^2-24x+12,$ and moving terms to one side and dividing by $3,$ we get \[x^4-6x^3+11x^2-6x+1=0.\]
This can be factored into $(x^2-3x+1)^2 = 0 \rightarrow x^2-3x+1 = 0 \rightarrow x = \frac{3 \pm \sqrt{5}}{2}.$ From the equation $x^2-3x+1=0,$ we have $x^2=3x+1,$ so plugging that value into the expression we want to find, we get $\frac{4}{3(3x+1)-3x+1} = \frac{4}{6x+2}.$
Substituting $x = \frac{3-\sqrt{5}}{2}$ into $\frac{4}{6x+2}$ gets an expression of $7+3\sqrt{5},$ so $a^2+b^2+c^2 = \boxed{083}$ | null | 083 |
799a603bf6aa179eb8620f592ce1a2cd | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_14 | 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 possible pairs of factors of $9$ are $(1,9)(3,3)$ . These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$ , while the second pair gives $98$ . We now check that $130$ is optimal, setting $a=m-2$ $b=n-2$ in order to simplify calculations. Since \[0 \le (a-1)(b-1) \implies a+b \le ab+1\] We have \[p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130\] Where we see $(m,n)=(3,11)$ gives us our maximum value of $\boxed{130}$ | null | 130 |
799a603bf6aa179eb8620f592ce1a2cd | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_14 | 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 $\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}$ for $m=1,2,$ we fix $m=3.$ \[\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+5)}{5(n+2)}=\dfrac{n-10}{10(n+2)},\] where we simply let $n=11$ to achieve $p=\boxed{130}.$ | null | 130 |
425bebea4a8b35482b2f0f140f28130d | https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_15 | 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 $j$ . Let $D_{40}$ be the set of all dominos whose coordinates are no larger than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}.$ | 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 point. (Reasoning for this: Everytime you go to a vertex, you have to leave the vertex, so every vertex reached is equivalent to adding 2 more segments. So the degree of each vertex must be even, with the exception of endpoints) Since there are 39 segments coming from each point it is impossible to touch every segment.
But you can get up to 38 on each segment because you go in to the point then out on a different segment. Counting going out from the starting and ending at the ending point we have:
$\frac{38\cdot 38 + 2\cdot 39}2 = \boxed{761}$ | null | 761 |
64dd886006d0059e85c058add28b9c41 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_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 $500+250 = \boxed{750}.$ | null | 750 |
60b81b484a69c92f39611e5a78522e64 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_2 | 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, and so on until $1^2$ of the $8\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204$
Thus $\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}$ , and $m+n=\boxed{125}$ | null | 125 |
60b81b484a69c92f39611e5a78522e64 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_2 | 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 $m + n.$ | First, to find the number of squares, we can look case by case by the side length of the possible squares on the checkerboard. We see that there are $8^2$ ways to place a $1$ $1$ square and $7^2$ for a $2$ $2$ square. This pattern can be easily generalized and we see that the number of squares is just $\sum^8_{i=1}{i^2}$ . This can be simplified by using the well-known formula for the sum of consecutive squares $\frac{n(n+1)(2n+1)}{6}$ to get $204$
Then, to find the number of rectangles, first note that a square falls under the definition of a rectangle. We can break up the rectangles into cases for the length x width. As we note down the cases for $1$ $1, 1$ $2 , 2$ $1, 2$ $2,...,$ we see they are respectively $8$ $8, 8$ $7, 7$ $8, 7$ $7, ...$ . We can quickly generalize this pattern to basically just ${\sum^8_{i=1}{i}}\cdot{\sum^8_{i=1}{i}}$ . This gets us ${(\frac{9\cdot8}{2})}^2,$ which is just $1296.$
Now, to calculate the ratio of $s/r,$ we divide $204$ by $1296$ to get a simplified fraction of $\frac{17}{108}.$
Thus, our answer is just $s + r = 17+108 = \boxed{125}$ ~MathWhiz35 | null | 125 |