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5611776d841d723d0718847e18dbe2f6 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square).
Probability of rolling 4 duds: $\left(\frac{1}{3}\right)^4$
Probability of rolling 3 duds: $4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3}$
Probability of rolling 2 duds: $6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2$
Probability of rolling 1 dud: $4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3$
Probability of rolling 0 duds: $\left(\frac{2}{3}\right)^4$
Now we will find the probability of a square product given we have rolled each amount of duds
Probability of getting a square product given 4 duds: 1
Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)
Probability of getting a square product given 2 duds: $\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 dud: $\frac{3!}{4^3}$ $\frac{3}{32}$ (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: $\frac{40}{4^4}$ $\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).
We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values. \[\left(\frac{1}{3}\right)^4 * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}.\]
$25+162$ $\boxed{187}$ | null | 187 |
5611776d841d723d0718847e18dbe2f6 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:
If there are four 1/4's, then there are $2^4=16$ combinations.
If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square.
If there are two 1/4's, there are $2^2=4$ ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and $\frac{4!}{2!2!}=6$ ways to arrange, so there are $4\cdot 4\cdot 6=96$ combinations for this case.
If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\cdot 4!=48$ combinations for this case.
Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\binom{4}{2}$ to choose the numbers and $\frac{4!}{2!2!}=6$ ways to arrange them, so $6\cdot 6=36$ . If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case.
Hence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\frac{200}{1296}=\frac{25}{162}$ , yielding an answer of $25+162=\boxed{187}$ | null | 187 |
5611776d841d723d0718847e18dbe2f6 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Another way to solve this problem is to do casework on all the perfect squares from $1^2$ to $36^2$ , and how many ways they can be ordered $1^2$ $1,1,1,1$ $1$ way. $2^2$ $4,1,1,1$ or $2,2,1,1$ $\binom{4}{2}+4=10$ ways. $3^2$ $3,3,1,1$ $\binom{4}{2}=6$ ways. $4^2$ $4,4,1,1$ $2,2,2,2$ , or $2,2,4,1$ $\binom{4}{2}+1+12=19$ ways. $5^2$ $5,5,1,1$ $\binom{4}{2}=6$ ways. $6^2$ $6,6,1,1$ $1,2,3,6$ $2,3,2,3$ $3,3,4,1$ $2*\binom{4}{2}+4!+12=48$ ways. $7^2$ - Since there is a prime greater than 6 in its prime factorization there are $0$ ways. $8^2$ $4,4,4,1$ or $2,4,2,4$ $\binom{4}{2}+4=10$ ways. $9^2$ $3,3,3,3$ $1$ way. $10^2$ $2,2,5,5$ or $1,4,5,5$ $6+12=18$ ways. $11^2$ $0$ ways for the same reason as $7^2$ $12^2$ $6,6,2,2$ $4,4,3,3$ $2,3,4,6$ , or $1,4,6,6$ $2*\binom{4}{2}+4!+12=48$ ways. $13^2$ $0$ ways. $14^2$ $0$ ways. $15^2$ $3,3,5,5$ $\binom{4}{2}=6$ ways. $16^2$ $4,4,4,4$ $1$ way. $17^2$ $0$ ways. $18^2$ $3,3,6,6$ $\binom{4}{2}=6$ ways. $19^2$ $0$ ways. $20^2$ $4,4,5,5$ $\binom{4}{2}=6$ ways. $21^2$ $0$ ways. $22^2$ $0$ ways. $23^2$ $0$ ways. $24^2$ $4,4,6,6$ $\binom{4}{2}=6$ ways. $25^2$ $5,5,5,5$ $1$ way. $26^2$ $0$ ways. $27^2$ $0$ ways. $28^2$ $0$ ways. $29^2$ $0$ ways. $30^2$ $5,5,6,6$ $\binom{4}{2}$ ways. $31^2$ $0$ ways. $32^2$ $0$ ways. $33^2$ $0$ ways. $34^2$ $0$ ways. $35^2$ $0$ ways. $36^2$ $6,6,6,6$ $1$ way.
There are $6^4=1296$ ways that the dice can land. Summing up the ways, it is easy to see that there are $200$ ways.
This results in a probability of $\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}$ -superninja2000 | null | 187 |
5611776d841d723d0718847e18dbe2f6 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can do recursion on the number of rolls to find the number of ways we can get $4$ rolls to multiply to a square.
After $n$ rolls, let us say that the product is $p = 2^a3^b5^c$
Define the following:
$A_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is odd, and $b$ $c$ are even
$B_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is odd, and $a$ $c$ are even
$C_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is odd, and $a$ $b$ are even
$D_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is even, and $a$ $b$ are odd
$E_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is even, and $a$ $c$ are odd
$F_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is even, and $b$ $c$ are odd
$G_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all odd
$S_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all even (square!)
We have the following equations after considering the possible values of the nth roll:
\[A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}\]
\[B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}\]
\[C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}\]
\[D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}\]
\[E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}\]
\[F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}\]
\[G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}\]
\[S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}\]
We have the following values after considering the possible values of the 1st roll:
\[A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2\]
After applying recursion twice, we get:
\[A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8\]
\[A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38\]
Finally, we have $S_4 = 200$ $\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}$ meaning our answer is $\boxed{187}$ | null | 187 |
5611776d841d723d0718847e18dbe2f6 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Consider all the distinct "fundamental" groups of integers from $1$ to $6$ whose product is a perfect square. A "fundamental" group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, $\{2,2\}$ is a fundamental group, while $\{3,3,4\}$ is not, because it can be broken up into $\{3,3\}$ and $\{4\}$
$1$ and $4$ are already perfect squares, so they each form a "fundamental" group and cannot belong in any other group. Pairs of the other $4$ numbers ( $\{2,2\}$ $\{3,3\}$ , etc. ) form "fundamental" groups as well. The last "fundamental" group is $\{2,3,6\}$ . It can be easily seen that no more groups exist.
Thus, we have the "fundamental" groups $\{1\}$ $\{4\}$ $\{2,2\}$ $\{3,3\}$ $\{5,5\}$ $\{6,6\}$ , and $\{2,3,6\}$
We now consider the ways to use these groups to form a sequence of $4$ numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size $2$ or $3$ and fill in any remaining spots with $1$ s and $4$ s. We can do this in one of $5$ ways: Using only $1$ s and $4$ s, using one group of size $2$ , using one group of size $3$ , using two different groups of size $2$ , and using the same group of size $2$ twice.
If we only use $1$ s and $4$ s, each of the $4$ slots can be filled with one of the $2$ numbers, so there are $2^4=16$ possibilities.
If we use one group of size $2$ , there are $4$ options for the group to use, $\binom{4}{2}$ ways to place the two numbers (since they are identical), and $2^2$ ways to fill in the remaining slots with $1$ s and $4$ s, so there are $4\cdot\binom{4}{2}\cdot2^2=96$ possibilities.
If we use one group of size $3$ , there is only $1$ option for the group to use, $4\cdot3\cdot2$ ways to place the three numbers (since they are distinct), and $2$ ways to fill in the remaining slot, so there are $4\cdot3\cdot2\cdot2=48$ possibilities.
If we use two different groups of size $2$ , there are $\binom{4}{2}$ options for the groups to use and $\binom{4}{2}$ ways to place the four numbers (since there are $2$ groups of identical numbers, and one group's placement uniquely determines the other's), so there are $\binom{4}{2}\cdot\binom{4}{2}=36$ possibilities.
If we use the same group of size $2$ twice, there are $4$ options for the group to use and $1$ way to place the four numbers (since they are all identical), so there are $4=4$ possibilities.
This gives us a total of $16+96+48+36+4=200$ possibilities, and since there are $6^4=1296$ total sequences that can be rolled, the probability is equal to $\frac{200}{1296}=\frac{25}{162}$ , so the answer is $25+162=\boxed{187}$ . ~ emerald_block | null | 187 |
5611776d841d723d0718847e18dbe2f6 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | There are a total of $2^4=1296$ possible die rolls.
We use casework:
Case 1 : All 4 numbers are the same.
There are obviously $6$ ways.
Case 2 : Sets of 2 different numbers.
A set of two different numbers is basically $(x,x,y,y)$ . There are a total of $\frac{4!}{2!\cdot 2!}=6$ ways to arrange the numbers.
By listing these cases, we quickly see a pattern:
$(1,1,2,2)$
$(1,1,3,3)$
$...$
$(1,1,6,6)$
$(2,2,3,3)$
$...$
$(2,2,6,6)$
$...$
$(5,5,6,6)$
There are a total of $5+4+3+2+1=15$ cases. Multiplying this by $6$ yields $15\cdot 6=90$ ways.
Case 3 : Sets of numbers in the form of $(x,x,1,4)$
A special case must be made for the number $4$ because $4$ itself is a perfect square.
$(1,1,1,4)$ $4$
$(2,2,1,4)$ $12$
$(3,3,1,4)$ $12$
$(4,4,1,4)$ $4$
$(5,5,1,4)$ $12$
$(6,6,1,4)$ $12$
Summing these up yields a total of $4+12+12+4+12+12=56$ ways.
Case 4 : Sets with all 4 numbers different
Note that the sets
$(1,2,3,6)$
$(2,3,4,6)$
Multiply to perfect square. The total of these cases are $24+24=48$
Adding all these cases together yields $6+90+56+48=200$ ways that the product of the values of the die can be a perfect square.
Therefore the probability is
$\frac{200}{1296}=\frac{25}{162}$
$m+n = 25+162 = \boxed{187}$ | null | 187 |
547d609da123a0206a5b2ed3e1420d5c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_5 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table under these conditions. Find the remainder when $N$ is divided by $1000$ | Let us first consider the $4$ ambassadors and the $6$ even-numbered chairs. If we consider only their relative positions, they can sit in one of $3$ distinct ways: Such that the $2$ empty even-numbered chairs are adjacent, such that they are separated by one occupied even-numbered chair, and such that they are opposite each other. For each way, there are $4!=24$ ways to assign the $4$ ambassadors to the $4$ selected seats.
In the first way, there are $6$ distinct orientations. The $4$ advisors can be placed in any of the $5$ odd-numbered chairs adjacent to the ambassadors, and for each placement, there is exactly one way to assign them to the ambassadors. This means that there are $24\cdot6\cdot\binom{5}{4}=720$ total seating arrangements for this way.
In the second way, there are $6$ distinct orientations. $3$ advisors can be placed in any of the $4$ chairs adjacent to the "chunk" of $3$ ambassadors, and $1$ advisor can be placed in either of the $2$ chairs adjacent to the "lonely" ambassador. Once again, for each placement, there is exactly one way to assign the advisors to the ambassadors. This means that there are $24\cdot6\cdot\binom{4}{3}\cdot\binom{2}{1}=1152$ total seating arrangements for this way.
In the third way, there are $3$ distinct orientations. For both "chunks" of $2$ ambassadors, $2$ advisors can be placed in any of the $3$ chairs adjacent to them, and once again, there is exactly one way to assign them for each placement. This means that there are $24\cdot3\cdot\binom{3}{2}\cdot\binom{3}{2}=648$ total seating arrangements for this way.
Totalling up the arrangements, there are $720+1152+648=2520$ total ways to seat the people, and the remainder when $2520$ is divided by $1000$ is $\boxed{520}$ . ~ emerald_block | null | 520 |
547d609da123a0206a5b2ed3e1420d5c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_5 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table under these conditions. Find the remainder when $N$ is divided by $1000$ | In the diagram, the seats are numbered 1...12. Rather than picking seats for each person, however, each ambassador/assistant team picks a gap between the seats (A...L) and the ambassador sits in the even seat while the assistant sits in the odd seat. For example, if team 1 picks gap C then Ambassador 1 will sit in seat 2 while assistant 1 will sit in seat 3. No two teams can pick adjacent gaps. For example, if team 1 chooses gap C then team 2 cannot pick gaps B or D. In the diagram, the teams have picked gaps C, F, H and J. Note that the gap-gaps - distances between the chosen gaps - (in the diagram, 2, 1, 1, 4) must sum to 8. So, to get the number of seatings, we:
So, the total is $12\cdot35\cdot6=2520$
Therefore, the remainder is $\boxed{520}$ | null | 520 |
547d609da123a0206a5b2ed3e1420d5c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_5 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table under these conditions. Find the remainder when $N$ is divided by $1000$ | There are $6\cdot5\cdot4\cdot3\cdot2^4$ (placing the ambassadors and then choosing the locations of the advisors) total ways to let everyone sit. However, this may lead to advisors sitting in the same chair, leading to awkward situations. So we find how many ways this happens. There are $\dbinom{4}{2} = 6$ ways to choose which advisors end up sitting together, times $6\cdot2=12$ ways to find neighboring even seats (order matters), and for the other two ambassadors, there are $\dbinom{4}{2} \cdot 2 = 12$ ways to sit and 4 ways for their advisors to sit. This results in $6\cdot12\cdot12\cdot4=3456$ ways for this to happen. However, we overcounted the case when all four of the advisors run out of room to sit, where there are $\frac{6\cdot3}{2!}=9$ to choose the places the ambassadors sit and 4! ways to arrange the ambassadors. $9\cdot4!=216$ ways to happen. So the answer is $5760 - 3456 + 216 = 2520$ , which has a remainder of $\boxed{520}$ when divided by 1000. | null | 520 |
547d609da123a0206a5b2ed3e1420d5c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_5 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table under these conditions. Find the remainder when $N$ is divided by $1000$ | We see that for every 2 adjacent spots on the table, there is exactly one way for an ambassador and his or her partner to sit. There are 2 cases:
For the first case, there are 3 pairs and 4 empty spaces, which results in $\frac{7!}{3!4!} = 35$ ways to arrange the pairs.
For the second case, there are 4 pairs and 4 empty spaces, which results in $\frac{8!}{4!4!} = 70$ ways to arrange the pairs.
Finally, we have to assign ambassadors and their advisors to the pairs, which has $4!=24$ ways, so $N = (35+70)\cdot 24 = 2520$ , or $\boxed{520}$ ~PenguinMoosey | null | 520 |
5054d0dfdb85ea1df46d2f014b87c3cf | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | Using change of base on the second equation to base b, \[\frac{\log x}{\log \log_{b}{x}}=54\] \[\log x = 54 \cdot \log\log_{b}{x}\] \[x = (\log_{b}{x})^{54}\] Note by dolphin7 - you could also just rewrite the second equation in exponent form.
Substituting this into the $\sqrt x$ of the first equation, \[3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56\] \[3\log_{b}{(\log_{b}{x})^{28}} = 56\] \[\log_{b}{(\log_{b}{x})^{84}} = 56\]
We can manipulate this equation to be able to substitute $x = (\log_{b}{x})^{54}$ a couple more times: \[\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}\] \[\log_{b}{x} = 36\] \[(\log_{b}{x})^{54} = 36^{54}\] \[x = 6^{108}\]
However, since we found that $\log_{b}{x} = 36$ $x$ is also equal to $b^{36}$ . Equating these, \[b^{36} = 6^{108}\] \[b = 6^3 = \boxed{216}\] | null | 216 |
5054d0dfdb85ea1df46d2f014b87c3cf | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | We start by simplifying the first equation to \[3\log_{b}{(\sqrt{x}\log x)}=\log_{b}{(x^{\frac{3}{2}}\log^3x)}=56\] \[x^\frac{3}{2}\cdot \log_b^3x=b^{56}\] Next, we simplify the second equation to \[\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] \[\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))\] \[x=\log_b^{54}x\] Substituting this into the first equation gives \[\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}\] \[x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}\] Plugging this into $x=\log_b^{54}x$ gives \[b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}\] \[b^{\frac{2}{3}}=36\] \[b=36^{\frac{3}{2}}=6^3=\boxed{216}\] -ktong | null | 216 |
5054d0dfdb85ea1df46d2f014b87c3cf | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | Apply change of base to \[\log_{\log x}(x)=54\] to yield: \[\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] which can be rearranged as: \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] Apply log properties to \[3\log(\sqrt{x}\log x)=56\] to yield: \[3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}\] Substituting \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] into the equation $\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$ yields: \[\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}\] So \[\log_b(x)=36.\] Substituting this back in to \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] yields \[\frac{36}{54}=\log_b(36).\] So, \[b^{\frac{2}{3}}=36\Rightarrow \boxed{216}\] | null | 216 |
5054d0dfdb85ea1df46d2f014b87c3cf | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | 1st equation: \[\log (\sqrt{x}\log x)=\frac{56}{3}\] \[\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}\] 2nd equation: \[x=(\log x)^{54}\] So now substitute $\log x=a$ and $x=b^a$ \[b^a=a^{54}\] \[b=a^{\frac{54}{a}}\] We also have that \[\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}\] \[\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}\] This means that $\frac{14}{27}a=\frac{56}{3}$ , so \[a=36\] \[b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}\] | null | 216 |
5054d0dfdb85ea1df46d2f014b87c3cf | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | This system of equations looks complicated to work with, so we let $a=\log_bx$ to make it easier for us to read.
Now, the first equation becomes $3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3$
The second equation, $\log_{\log(x)}(x)=54$ gives us $\underline{a^{54} = x}$
Let's plug this back into the first equation to see what we get: $\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3$ , and simplifying, $\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}$ , so $b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}$
Combining this new finding with what we had above $a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}$
Now that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get $\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies$ $\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36$
Finally, that gives us that $\frac{b^{\frac{56}3}}{b^{18}}=36\implies b^{\frac{56}{3}-18}=b^{\frac{56}{3}-\frac{54}{3}}=b^{\frac 23}=36\implies b=36^{\frac 32}=6^3$ . Thus, $b=\boxed{216}$ | null | 216 |
5054d0dfdb85ea1df46d2f014b87c3cf | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | Using change of base on the second equation, we have
\[\frac{\log_{b} x}{\log_{b} \log_{b} x} = 54\]
Using log rules on the first equation, we have
\[\frac{3}{2} \log_{b} x + 3 \log_{b} \log_{b} x = 56\]
We notice that $\log_{b} x$ and $\log_{b} \log_{b} x$ are in both equations. Thus, we set $m = \log_{b} x$ and $n = \log_{b} \log_{b} x$ and we have
\[\frac{3}{2} m + 3n = 56\] \[\frac{m}{n} = 54\]
Solving this yields $m = 36$ $n = \frac{2}{3}$
Now, $n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36$ , so we have $\log_{b} 36 = \frac{2}{3}$ . Solving this yields $b = \boxed{216}$ | null | 216 |
5054d0dfdb85ea1df46d2f014b87c3cf | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | The second equation implies that \[\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}\] The first equation implies that \[3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x\] Substituting the first result into the second gives us \[b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.\] Because $b^{36}=x,$ $\log_b x=36$ by the definition of a logarithm.
Substituting this into the second equation, \[\log_{36} x=54 \implies x=36^{54}.\] Finally, \[b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.\] | null | 216 |
2e1405c82fd93aeb9c59617da6c20bab | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$ , and $15$ , respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$ , and $\ell_C$ | Let the points of intersection of $\ell_A, \ell_B,\ell_C$ with $\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$ . Furthermore, let the desired triangle be $\triangle XYZ$ , with $X$ closest to side $BC$ $Y$ closest to side $AC$ , and $Z$ closest to side $AB$ . Hence, the desired perimeter is $XE+EF+FY+YG+GH+HZ+ZI+ID+DX=(DX+XE)+(FY+YG)+(HZ+ZI)+115$ since $HG=55$ $EF=15$ , and $ID=45$
Note that $\triangle AHG\sim \triangle BID\sim \triangle EFC\sim \triangle ABC$ , so using similar triangle ratios, we find that $BI=HA=30$ $BD=HG=55$ $FC=\frac{45}{2}$ , and $EC=\frac{55}{2}$
We also notice that $\triangle EFC\sim \triangle YFG\sim \triangle EXD$ and $\triangle BID\sim \triangle HIZ$ . Using similar triangles, we get that \[FY+YG=\frac{GF}{FC}\cdot \left(EF+EC\right)=\frac{225}{45}\cdot \left(15+\frac{55}{2}\right)=\frac{425}{2}\] \[DX+XE=\frac{DE}{EC}\cdot \left(EF+FC\right)=\frac{275}{55}\cdot \left(15+\frac{45}{2}\right)=\frac{375}{2}\] \[HZ+ZI=\frac{IH}{BI}\cdot \left(ID+BD\right)=2\cdot \left(45+55\right)=200\] Hence, the desired perimeter is $200+\frac{425+375}{2}+115=600+115=\boxed{715}$ -ktong | null | 715 |
2e1405c82fd93aeb9c59617da6c20bab | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$ , and $15$ , respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$ , and $\ell_C$ | Let the diagram be set up like that in Solution 1.
By similar triangles we have \[\frac{AH}{AB}=\frac{GH}{BC}\Longrightarrow AH=30\] \[\frac{IB}{AB}=\frac{DI}{AC}\Longrightarrow IB=30\] Thus \[HI=AB-AH-IB=60\]
Since $\bigtriangleup IHZ\sim\bigtriangleup ABC$ and $\frac{HI}{AB}=\frac{1}{2}$ , the altitude of $\bigtriangleup IHZ$ from $Z$ is half the altitude of $\bigtriangleup ABC$ from $C$ , say $\frac{h}{2}$ . Also since $\frac{EF}{AB}=\frac{1}{8}$ , the distance from $\ell_C$ to $AB$ is $\frac{7}{8}h$ . Therefore the altitude of $\bigtriangleup XYZ$ from $Z$ is \[\frac{1}{2}h+\frac{7}{8}h=\frac{11}{8}h\]
By triangle scaling, the perimeter of $\bigtriangleup XYZ$ is $\frac{11}{8}$ of that of $\bigtriangleup ABC$ , or \[\frac{11}{8}(220+180+120)=\boxed{715}\] | null | 715 |
2e1405c82fd93aeb9c59617da6c20bab | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$ , and $15$ , respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$ , and $\ell_C$ | Notation shown on diagram. By similar triangles we have \[k_1 = \frac{EF}{BC} = \frac{AE}{AB} = \frac {AF}{AC} = \frac {1}{4},\] \[k_2 = \frac{F''E''}{AC} = \frac {BF''}{AB} = \frac{1}{4},\] \[k_3 = \frac{E'F'}{AB} = \frac{E'C }{AC} = \frac{1}{8}.\] So, \[\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 - k_1 - k_2,\] \[\frac{FY}{BC} = \frac{FE'}{AC} = \frac{AC - AF - CE'}{AC} = 1 - k_1 - k_3.\] \[k = \frac{ZY}{BC} = \frac{ZE + EF + FY}{BC} = (1 - k_1 - k_2) + k_1 + (1 - k_1 - k_3)\] \[k = 2 - k_1 - k_2 - k_3 = 2 - \frac{1}{4} - \frac{1}{4} - \frac{1}{8} = \frac{11}{8}.\] \[\frac{ZY+YX +XZ}{BC +AB + AC} = k \implies ZY + YX + XZ =\frac{11}{8} (220 + 120 + 180) = \boxed{715}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 715 |
773005f5adda1466ae7a25e01a467aff | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_8 | The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$ , and $f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i$ . Find the remainder when $f(1)$ is divided by $1000$ | We have $\frac{1+\sqrt{3}i}{2} = \omega$ where $\omega = e^{\frac{i\pi}{3}}$ is a primitive 6th root of unity. Then we have
\begin{align*} f(\omega) &= a\omega^{2018} + b\omega^{2017} + c\omega^{2016}\\ &= a\omega^2 + b\omega + c \end{align*}
We wish to find $f(1) = a+b+c$ . We first look at the real parts. As $\text{Re}(\omega^2) = -\frac{1}{2}$ and $\text{Re}(\omega) = \frac{1}{2}$ , we have $-\frac{1}{2}a + \frac{1}{2}b + c = 2015 \implies -a+b + 2c = 4030$ . Looking at imaginary parts, we have $\text{Im}(\omega^2) = \text{Im}(\omega) = \frac{\sqrt{3}}{2}$ , so $\frac{\sqrt{3}}{2}(a+b) = 2019\sqrt{3} \implies a+b = 4038$ . As $a$ and $b$ do not exceed 2019, we must have $a = 2019$ and $b = 2019$ . Then $c = \frac{4030}{2} = 2015$ , so $f(1) = 4038 + 2015 = 6053 \implies f(1) \pmod{1000} = \boxed{053}$ | null | 053 |
773005f5adda1466ae7a25e01a467aff | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_8 | The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$ , and $f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i$ . Find the remainder when $f(1)$ is divided by $1000$ | Denote $\frac{1+\sqrt{3}i}{2}$ with $\omega$
By using the quadratic formula ( $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ) in reverse, we can find that $\omega$ is the solution to a quadratic equation of the form $ax^2+bx+c=0$ such that $2a=2$ $-b=1$ , and $b^2-4ac=-3$ . This clearly solves to $a=1$ $b=-1$ , and $c=1$ , so $\omega$ solves $x^2-x+1=0$
Multiplying $x^2-x+1=0$ by $x+1$ on both sides yields $x^3+1=0$ . Muliplying this by $x^3-1$ on both sides yields $x^6-1=0$ , or $x^6=1$ . This means that $\omega^6=1$
We can use this to simplify the equation $a\omega^{2018}+b\omega^{2017}+c\omega^{2016}=f(\omega)=2015+2019\sqrt{3}i$ to $a\omega^2+b\omega+c=2015+2019\sqrt{3}i.$
As in Solution 1, we use the values $\omega=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\omega^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ to find that $-\frac{1}{2}a+\frac{1}{2}b+c=2015$ and $\frac{\sqrt{3}}{2}a+\frac{\sqrt{3}}{2}b=2019\sqrt{3} \implies a+b=4038.$ Since neither $a$ nor $b$ can exceed $2019$ , they must both be equal to $2019$ . Since $a$ and $b$ are equal, they cancel out in the first equation, resulting in $c=2015$
Therefore, $f(1)=a+b+c=2019+2019+2015=6053$ , and $6053\bmod1000=\boxed{053}$ . ~ emerald_block | null | 053 |
773005f5adda1466ae7a25e01a467aff | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_8 | The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$ , and $f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i$ . Find the remainder when $f(1)$ is divided by $1000$ | Calculate the first few powers of $\frac{1+\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^1=\frac{1+\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^2=\frac{-1+\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^3=-1$
$(\frac{1+\sqrt{3}i}{2})^4=\frac{-1-\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^5=\frac{1-\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^6=1$
We figure that the power of $\frac{1+\sqrt{3}i}{2}$ repeats in a cycle 6.
$f(\frac{1+\sqrt{3}i}{2})=(a(\frac{1+\sqrt{3}i}{2})^2+b(\frac{1+\sqrt{3}i}{2})+c)(\frac{1+\sqrt{3}i}{2})^{2016}$
Since 2016 is a multiple of 6, $(\frac{1+\sqrt{3}i}{2})^{2016}=1$
$f(\frac{1+\sqrt{3}i}{2})=a(\frac{-1+\sqrt{3}i}{2})+b(\frac{1+\sqrt{3}i}{2})+c$
$f(\frac{1+\sqrt{3}i}{2})=(-\frac{1}{2}a+\frac{1}{2}b+c)+(\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b)=2015+2019\sqrt{3}i$
Therefore, $-\frac{1}{2}a+\frac{1}{2}b+c=2015$ and $\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b=2019\sqrt{3}i$
Using the first equation, we can get that $-a+b+2c=4030$ , and using the second equation, we can get that $a+b=4038$
Since all coefficients are less than or equal to $2019$ $a=b=2019$
Therefore, $2c=4030$ and $c=2015$
$f(1)=a+b+c=2019+2019+2015=6053$ , and the remainder when it divides $1000$ is $\boxed{053}.$ | null | 053 |
5e4a211823b3a5d468a10a4fd42114de | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_9 | Call a positive integer $n$ $k$ pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ | Every 20-pretty integer can be written in form $n = 2^a 5^b k$ , where $a \ge 2$ $b \ge 1$ $\gcd(k,10) = 1$ , and $d(n) = 20$ , where $d(n)$ is the number of divisors of $n$ . Thus, we have $20 = (a+1)(b+1)d(k)$ , using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there are not many cases to check.
If $a+1 = 4$ , then $b+1 = 5$ . But this leads to no solutions, as $(a,b) = (3,4)$ gives $2^3 5^4 > 2019$
If $a+1 = 5$ , then $b+1 = 2$ or $4$ . The first case gives $n = 2^4 \cdot 5^1 \cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23$ . The sum of all such $n$ is $80(3+7+11+13+17+19+23) = 7440$ . In the second case $b+1 = 4$ and $d(k) = 1$ , and there is one solution $n = 2^4 \cdot 5^3 = 2000$
If $a+1 = 10$ , then $b+1 = 2$ , but this gives $2^9 \cdot 5^1 > 2019$ . No other values for $a+1$ work.
Then we have $\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}$ | null | 472 |
5e4a211823b3a5d468a10a4fd42114de | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_9 | Call a positive integer $n$ $k$ pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ | For $n$ to have exactly $20$ positive divisors, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$ where $p,q,r$ are primes. No number that is a multiple of $20$ can be expressed in the first form because 20 has two primes in its prime factorization, while the first form has only one , and the only integer divisible by $20$ that has the second form is $2^{9}5$ , which is 2560, greater than $2019$
For the third form, the only $20$ -pretty numbers are $2^45^3=2000$ and $2^35^4=5000$ , and only $2000$ is small enough.
For the fourth form, any number of the form $2^45r$ where $r$ is a prime other than $2$ or $5$ will satisfy the $20$ -pretty requirement. Since $n=80r<2019$ $r\le 25$ . Therefore, $r$ can take on $3, 7, 11, 13, 17, 19,$ or $23$
Thus, $\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}$ | null | 472 |
e6df1391ec3831b1be81816f7f07adfa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integers. Find $p+q$ | Note that if $\tan \theta$ is positive, then $\theta$ is in the first or third quadrant, so $0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}$
Furthermore, the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ that are multiples of $3$ is when: \[2^0\theta \equiv 2^3\theta \equiv 2^6\theta \equiv ... \pmod{180^{\circ}}.\] (This is because if it isn't the same value, the terminal angle will gradually shift from the first quadrant into different quadrants, making the condition for positive tan untrue. This must also be true in order for $\theta$ to be unique.)
This is the case if $2^3\theta \equiv 2^0\theta \pmod{180^{\circ}}$ , so $7\theta \equiv 0^{\circ} \pmod{180^{\circ}}$ . Therefore, recalling that $0^{\circ}<\theta<90^{\circ},$ the possible $\theta$ are: \[\frac{180}{7}^{\circ}, \frac{360}{7}^{\circ}, \frac{540}{7}^{\circ}.\]
$\frac{180}{7}^{\circ}$ does not work since $\tan\left(2 \cdot \frac{180}{7}^{\circ}\right)$ is positive.
$\frac{360}{7}^{\circ}$ does not work because $\tan\left(4 \cdot \frac{360}{7}^{\circ}\right)$ is positive.
Thus, $\theta = \frac{540}{7}^{\circ}$ , and a quick check verifies that it does work. Our desired answer is $540 + 7 = \boxed{547}$ | null | 547 |
e6df1391ec3831b1be81816f7f07adfa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integers. Find $p+q$ | As in the previous solution, we note that $\tan \theta$ is positive when $\theta$ is in the first or third quadrant. In order for $\tan\left(2^n\theta\right)$ to be positive for all $n$ divisible by $3$ , we must have $\theta$ $2^3\theta$ $2^6\theta$ , etc to lie in the first or third quadrants. We already know that $\theta\in(0,90)$ . We can keep track of the range of $2^n\theta$ for each $n$ by considering the portion in the desired quadrants, which gives \[n=1 \implies (90,180)\] \[n=2\implies (270,360)\] \[n=3 \implies (180,270)\] \[n=4 \implies (90,180)\] \[n=5\implies(270,360)\] \[n=6 \implies (180,270)\] \[\cdots\] at which point we realize a pattern emerging. Specifically, the intervals repeat every $3$ after $n=1$ . We can use these repeating intervals to determine the desired value of $\theta$ since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.
Initially, the lower bound is $0$ (at $n=0$ ), then increases to $\frac{90}{2}=45$ at $n=1$ . This then becomes $45+\frac{45}{2}$ at $n=2$ $45+\frac{45}{2}$ at $n=3$ $45+\frac{45}{2}+\frac{45}{2^3}$ at $n=4$ $45+\frac{45}{2}+\frac{45}{2^3}+\frac{45}{2^4}$ at $n=5$ , etc. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as $n$ approaches infinity, the lower bound converges to \[\sum_{k=0}^{\infty}\left(45+\frac{45}{2}\right)\cdot \left(\frac{1}{8}\right)^k=\frac{45+\frac{45}{2}}{1-\frac{1}{8}}=\frac{\frac{135}{2}}{\frac{7}{8}}=\frac{540}{7}\implies p+q=540+7=\boxed{547}\] -ktong | null | 547 |
e6df1391ec3831b1be81816f7f07adfa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integers. Find $p+q$ | Since $\tan\left(\theta\right) > 0$ $0 < \theta < 90$ . Since $\tan\left(2\theta\right) < 0$ $\theta$ has to be in the second half of the interval (0, 90) ie (45, 90). Since $\tan\left(4\theta\right) < 0$ $\theta$ has to be in the second half of that interval ie (67.5, 90). And since $\tan\left(8\theta\right) > 0$ $\theta$ has to be in the first half of (67.5, 90). Inductively, the pattern repeats: $\theta$ is in the second half of the second half of the first half of the second half of the second half... of the interval (0, 90). Consider the binary representation of numbers in the interval (0, 1). Numbers in the first half of the interval start with 0.0... and numbers in the second half start with 0.1... . Similarly, numbers in the second half of the second half start with 0.11... etc. So if we want a number in the first half of the second half of the second half... of the interval, we want its binary representation to be $0.11011011011..._2 = \frac{6}{7}_{10}$ . So we want the number which is 6/7 of the way through the interval (0, 90) so $\theta = \frac{6}{7}\cdot 90 = \frac{540}{7}$ and $p+q = 540 + 7 = \boxed{547}$ | null | 547 |
e6df1391ec3831b1be81816f7f07adfa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integers. Find $p+q$ | With some simple arithmetic and guess and check, we can set the lower bound and upper bounds for the "first round of $3$ powers of two", which are $\frac{540}{8} = 67.5$ and $\frac{630}{8} = 78.75$ . Going on to the "second round of $3$ powers of two, we set the new lower and upper bounds as $\frac{360 \times 12.5}{64} = 70.3125$ and $\frac{360 \times 13.75}{64} = 77.34375$ using some guess and check and bashing. Now, it is obvious that the bounds for the "zeroth round of $3$ powers of two" are $0$ and $90$ , and notice that $90 - 78.75 = 11.25$ and $78.75 - 77.34375 = 1.40625$ and $\frac{11.25}{1.40625} = 8$ .
This is obviously a geometric series, so setting $11.25$ as $u$ , we obtain $90 - (u + \frac{u}{8} + \frac{u}{64} + ...)$ $90 - \frac{u}{1-\frac{1}{8}}$ $\frac{u}{\frac{7}{8}}$ $\frac{45}{4} \times \frac{8}{7}$ which simplifies to $\frac{90}{7}$ . We can now finally subtract $\frac{90}{7}$ from $\frac{630}{7}$ and then we get $\frac{540}{7}$ as the unique angle, so $\boxed{547}$ is our answer.
-fidgetboss_4000 | null | 547 |
e6df1391ec3831b1be81816f7f07adfa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_10 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integers. Find $p+q$ | Since $7$ is the only number n such that f(x) = $2^{\lfloor x\rfloor}$ $\text{(mod 7)}$ has a period of 3, we find that $\theta$ is a multiple of ${\frac{180}{7}}^\circ$ . Note that the tangents of ${\frac{180}{7}}^\circ$ ${\frac{360}{7}}^\circ$ ${\frac{540}{7}}^\circ$ are positive while those of ${\frac{720}{7}}^\circ$ ${\frac{900}{7}}^\circ$ , and ${\frac{1080}{7}}^\circ$ are negative. With a bit of trial and error, we find $\theta$ is ${\frac{540}{7}}^\circ$ and $540+7$ is $\boxed{547}$ | null | 547 |
9e6dde14ff8440a32ab504bf529457ee | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | [asy] unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label("$A$",A,dir(105)); label("$B$",B,dir(-135)); label("$C$",C,dir(-75)); dot((2.68,2.25)); label("$K$",(2.68,2.25),2*down); label("$\omega_1$",(-4.5,1)); label("$\omega_2$",(12.75,6)); label("$7$",(A+B)/2,dir(140)); label("$8$",(B+C)/2,dir(-90)); label("$9$",(A+C)/2,dir(60)); [/asy] -Diagram by Brendanb4321
Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$ , respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$ $^{(*)}$ , so $\triangle AKB\sim \triangle CKA$ . Using similarity ratios, we can easily find \[AK^2=BK*KC\] However, since $AB=7$ and $CA=9$ , we can use similarity ratios to get \[BK=\frac{7}{9}AK, CK=\frac{9}{7}AK\]
Giving us \[AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49\] \[\implies \frac{196}{81}AK^2=49\] \[AK=\frac{9}{2}\] so our answer is $9+2=\boxed{011}$ | null | 011 |
9e6dde14ff8440a32ab504bf529457ee | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Consider an inversion with center $A$ and radius $r=AK$ . Then, we have $AB\cdot AB^*=AK^2$ , or $AB^*=\frac{AK^2}{7}$ . Similarly, $AC^*=\frac{AK^2}{9}$ . Notice that $AB^*KC^*$ is a parallelogram, since $\omega_1$ and $\omega_2$ are tangent to $AC$ and $AB$ , respectively. Thus, $AC^*=B^*K$ . Now, we get that \[\cos(\angle AB^*K)=\cos(180-\angle BAC)=-\frac{11}{21}\] so by Law of Cosines on $\triangle AB^*K$ we have \[(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)\] \[\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}\] \[\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{22AK^2}{63\cdot21}\] \[\Rightarrow AK=\frac{9}{2}\] Then, our answer is $9+2=\boxed{11}$ .
-brianzjk | null | 11 |
9e6dde14ff8440a32ab504bf529457ee | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Let the centers of the circles be $O_{1}$ and $O_{2}$ where the $O_{1}$ has the side length $7$ contained in the circle. Now let $\angle BAC =x.$ This implies \[\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x\] by the angle by by tangent. Then we also know that \[\angle AO_{1}B = \angle AO_{2}C = 2x\] Now we first find $\cos x.$ We use law of cosines on $\bigtriangleup ABC$ to obtain \[64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}\] \[\implies \cos{x} =\frac{11}{21}\] \[\implies \sin{x} =\frac{8\sqrt{5}}{21}\] Then applying law of sines on $\bigtriangleup AO_{1}B$ we obtain \[\frac{7}{\sin{2x}} =\frac{O_{1}B}{\sin{90^{\circ}-x}}\] \[\implies\frac{7}{2\sin{x}\cos{x}} =\frac{O_{1}B}{\cos{x}}\] \[\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}\] Using similar logic we obtain $O_{2}A =\frac{189}{16\sqrt{5}}.$
Now we know that $\angle O_{1}AO_{2}=180^{\circ}-x.$ Thus using law of cosines on $\bigtriangleup O_{1}AO_{2}$ yields \[O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}\] While this does look daunting we can write the above expression as \[\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}\] Then factoring yields \[\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}\] The area \[[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] Now $AK$ is twice the length of the altitude of $\bigtriangleup O_{1}AO_{2}$ so we let the altitude be $h$ and we have \[\frac{1}{2} \cdot h \cdot\frac{147}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] \[\implies h =\frac{9}{4}\] Thus our desired length is $\frac{9}{2} \implies m+n = \boxed{11}.$ | null | 11 |
9e6dde14ff8440a32ab504bf529457ee | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | By the definition of $K$ , it is the spiral center mapping $BA\to AC$ , which means that it is the midpoint of the $A$ -symmedian chord. In particular, if $M$ is the midpoint of $BC$ and $M'$ is the reflection of $A$ across $K$ , we have $\triangle ABM'\sim\triangle AMC$ . By Stewart's Theorem, it then follows that \[AK = \frac{AM'}{2} = \frac{AC\cdot AB}{2AM} = \frac{7\cdot 9}{2\sqrt{\frac{9^2\cdot 4 + 7^2\cdot 4 - 4^2\cdot 8}{8}}} = \frac{7\cdot 9}{2\sqrt{49}} = \frac{9}{2}\implies m + n = \boxed{11}.\] | null | 11 |
9e6dde14ff8440a32ab504bf529457ee | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We start by assigning coordinates to point $A$ , labeling it $(0,0)$ and point $B$ at $(7,0)$ , and letting point $C$ be above the $x$ -axis. Through an application of the Pythagorean Theorem and dropping an altitude to side $AB$ , it is easy to see that $C$ has coordinates $(33/7, 24\sqrt{5}/7)$
Let $O1$ be the center of circle $\omega_1$ and $O2$ be the center of circle $\omega_2$ . Since circle $\omega_1$ contains both points $A$ and $B$ $O1$ must lie on the perpendicular bisector of line $AB$ , and similarly $O2$ must lie on the perpendicular bisector of line $AC$ . Through some calculations, we find that the perpendicular bisector of $AB$ has equation $x = 3.5$ , and the perpendicular bisector of $AC$ has equation $y = {-11\sqrt{5}/40 \cdot x} + 189\sqrt{5}/80$
Since circle $\omega_1$ is tangent to line $AC$ at $A$ , its radius must be perpendicular to $AC$ at $A$ .
Therefore, the radius has equation $y = {{-11\cdot\sqrt{5}/40} \cdot x}$ . Substituting the $x$ -coordinate of $O1$ into this, we find the y-coordinate of $O1 = {{-11 \cdot \sqrt{5}/40} \cdot 7/2} = {-77 \cdot \sqrt{5}/80}$
Similarly, since circle $\omega_2$ is tangent to line $AB$ at $A$ , its radius must be perpendicular to $AB$ at $A$ . Therefore, the radius has equation $x = 0$ and combining with the previous result for $O2$ we get that the coordinates of $O2$ are $(0, 189\sqrt{5}/80)$
We now find the slope of $O1O2$ , the line joining the centers of circles $\omega_1$ and $\omega_2$ , which turns out to be ${({266 \cdot \sqrt{5} / 80}) \cdot -2/7} = {-19 \cdot \sqrt{5}/20}$ . Since the $y$ -intercept of that line is at $O2(0,189\sqrt{5}/80)$ , the equation is $y = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}$ . Since circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $K$ , line $AK$ is the radical axis of those circles, and since the radical axis is always perpendicular to the line joining the centers of the circles, $AK$ has slope ${4 \cdot \sqrt{5}/19}$ . Since point $A$ is $(0,0)$ , this line has a $y$ -intercept of $0$ , so it has equation $y$ ${{4 \cdot \sqrt{5}/19} \cdot x}$
We set ${{4 \cdot \sqrt{5}/19} \cdot x} = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}$ in order to find the intersection $I$ of the radical axis $AK$ and $O1O2$ . Through some moderate bashing, we find that the intersection point is $I(57/28, {3 \cdot \sqrt{5}/7})$ . We know that either intersection point of two circles is the same distance from the intersection of radical axis and line joining the centers of those circles, so reflecting $A$ over $I$ yields $K$ and $AK$ $2AI$ = (This is the most tedious part of the bash) ${2 \cdot \sqrt{(57/28)^2 + ({3 \cdot \sqrt{5}/7)^2)}}} = {2 \cdot \sqrt{3969/784}} = {2 \cdot 63/28} = {2 \cdot 9/4} = 9/2$ . Therefore the answer is $9 + 2 = \boxed{011}.$ | null | 011 |
ae766be0d6154f6207013632b06eb153 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_12 | For $n \ge 1$ call a finite sequence $(a_1, a_2 \ldots a_n)$ of positive integers progressive if $a_i < a_{i+1}$ and $a_i$ divides $a_{i+1}$ for all $1 \le i \le n-1$ . Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360$ | If the first term is $x$ , then dividing through by $x$ , we see that we can find the number of progressive sequences whose sum is $\frac{360}{x} - 1$ , and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the answer $N$ as follows:
\begin{align*}N &= a(359) + a(179) + a(119) + a(89) + a(71) + a(59) + a(44) + a(39) \\ &+ a(35) + a(29) + a(23) + a(19) + a(17) + a(14) + a(11) + a(9) \\ &+ a(8) + a(7) + a(5) + a(4) + a(3) + a(2) + a(1) + 1 \end{align*}
The $+1$ at the end accounts for the sequence whose only term is 360. Fortunately, many of these numbers are prime; we have $a(p) = 1$ for primes $p$ as the only such sequence is " $p$ " itself. Also, $a(1) = 0$ . So we have
\[N = 15 + a(119) + a(44) + a(39) + a(35) + a(14) + a(9) + a(8) + a(4)\]
For small $k$ $a(k)$ is easy to compute: $a(4) = 1$ $a(8) = 2$ $a(9) = 2$ . For intermediate $k$ (e.g. $k=21$ below), $a(k)$ can be computed recursively using previously-computed values of $a(\cdot)$ , similar to dynamic programming. Then we have \begin{align*} a(14) &= 1+a(6) = 1+2 = 3\\ a(35) &= 1+a(6)+a(4) = 1 + 2 + 1 = 4 \\ a(39) &= 2 + a(12) = 2+4 = 6 \\ a(44) &= 2 + a(21) + a(10) = 2+4+2=8 \\ a(119) &= 1 + a(16) + a(6) = 1 + 3 + 2 = 6 \end{align*} Thus the answer is $N = 15+6+8+6+4+3+2+2+1 = \boxed{47}$ | null | 47 |
8539b2c26fe760b150088f8816bc5b82 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_13 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac{1}{9}.$ There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6},\overline{PA_7},$ and the minor arc $\widehat{A_6A_7}$ of the circle is equal to $\tfrac{1}{8}-\tfrac{\sqrt2}{n}.$ Find $n.$ | The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, \(1\), and assume the side length of the octagon is \(2\).
Let \(r\) denote the radius of the circle, \(O\) be the center of the circle. Then: \[r^2= 1^2 + \left(\sqrt{2}+1\right)^2= 4+2\sqrt{2}.\]
Now, we need to find the "D" shape, the small area enclosed by one side of the octagon and \(\frac{1}{8}\) of the circumference of the circle: \[D= \frac{1}{8} \pi r^2 - [A_1 A_2 O]=\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\]
Let \(PU\) be the height of \(\triangle A_1 A_2 P\), \(PV\) be the height of \(\triangle A_3 A_4 P\), \(PW\) be the height of \(\triangle A_6 A_7 P\). From the \(\frac{1}{7}\) and \(\frac{1}{9}\) condition we have \[\triangle P A_1 A_2= \frac{\pi r^2}{7} - D= \frac{1}{7} \pi \left(4+2\sqrt{2}\right)-\left(\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\right)\] \[\triangle P A_3 A_4= \frac{\pi r^2}{9} - D= \frac{1}{9} \pi \left(4+2\sqrt{2}\right)-\left(\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\right)\] which gives \(PU= \left(\frac{1}{7}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\) and \(PV= \left(\frac{1}{9}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\).
Now, let \(A_1 A_2\) intersects \(A_3 A_4\) at \(X\), \(A_1 A_2\) intersects \(A_6 A_7\) at \(Y\),\(A_6 A_7\) intersects \(A_3 A_4\) at \(Z\).
Clearly, \(\triangle XYZ\) is an isosceles right triangle, with right angle at \(X\) and the height with regard to which shall be \(3+2\sqrt2\). Now \(\frac{PU}{\sqrt{2}} + \frac{PV}{\sqrt{2}} + PW = 3+2\sqrt2\) which gives:
\begin{align*}
PW&= 3+2\sqrt2-\frac{PU}{\sqrt{2}} - \frac{PV}{\sqrt{2}} \\
&=3+2\sqrt{2}-\frac{1}{\sqrt{2}}\left(\left(\frac{1}{7}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1+\left(\frac{1}{9}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\right)\\
&=1+\sqrt{2}- \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\pi\left(4+2\sqrt{2}\right)
\end{align*}
Now, we have the area for \(D\) and the area for \(\triangle P A_6 A_7\), so we add them together:
\begin{align*}
\text{Target Area} &= \frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right) + \left(1+\sqrt{2}\right)- \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\pi\left(4+2\sqrt{2}\right)\\
&=\left(\frac{1}{8} - \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\right)\text{Total Area}
\end{align*}
The answer should therefore be \(\frac{1}{8}- \frac{\sqrt{2}}{2}\left(\frac{16}{63}-\frac{16}{64}\right)=\frac{1}{8}- \frac{\sqrt{2}}{504}\). The answer is \(\boxed{504}\). | null | 504 |
8539b2c26fe760b150088f8816bc5b82 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_13 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac{1}{9}.$ There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6},\overline{PA_7},$ and the minor arc $\widehat{A_6A_7}$ of the circle is equal to $\tfrac{1}{8}-\tfrac{\sqrt2}{n}.$ Find $n.$ | Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point $P$ from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram, where the octagon is oriented so as $A_1A_2$ is horizontal (and therefore $A_3A_4$ is vertical). Note that the area bounded by $\overline{A_iA_j}$ and the arc $\widehat{A_iA_j}$ is fixed, so we only need to consider the relevant triangles.
[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label("$P$", P, dir(-75)); label("$A_1$", dir(112.5), dir(112.5)); label("$A_2$", dir(112.5-45), dir(112.5-45)); label("$A_3$", dir(112.5-90), dir(112.5-90)); label("$A_4$", dir(112.5-135), dir(112.5-135)); label("$A_5$", dir(112.5-180), dir(112.5-180)); label("$A_6$", dir(112.5-225), dir(112.5-225)); label("$A_7$", dir(112.5-270), dir(112.5-270)); label("$A_8$", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]
Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\triangle PA_1A_2$ by $1$ . We can see that $P$ was moved down by $\tfrac{1}{7}-\tfrac{1}{8}=\tfrac{1}{56}$ units to make the area defined by $P$ $A_1$ , and $A_2$ $\tfrac{1}{7}$ . Similarly, $P$ was moved right by $\tfrac{1}{8}-\tfrac{1}{9}=\tfrac{1}{72}$ to make the area defined by $P$ $A_3$ , and $A_4$ $\tfrac{1}{9}$ . This means that $P$ has coordinates $(\tfrac{1}{72},-\tfrac{1}{56})$
Now, we need to consider how this displacement in $P$ affected the area defined by $P$ $A_6$ , and $A_7$ . This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\tfrac{\sqrt{2}}{2}(\tfrac{1}{56}-\tfrac{1}{72})=\tfrac{\sqrt{2}}{504}$
Remembering the definition of our unit, this yields a final area of \[\frac{1}{8}-\frac{\sqrt{2}}{\boxed{504}.\] | null | 504 |
0f4c0c19b5aba246156a9010b1018ffa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_14 | Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed. | By the Chicken McNugget theorem, the least possible value of $n$ such that $91$ cents cannot be formed satisfies $5n - (5 + n) = 91 \implies n = 24$ , so $n$ must be at least $24$
For a value of $n$ to work, we must not only be unable to form the value $91$ , but we must also be able to form the values $92$ through $96$ , as with these five values, we can form any value greater than $96$ by using additional $5$ cent stamps.
Notice that we must form the value $96$ without forming the value $91$ . If we use any $5$ cent stamps when forming $96$ , we could simply remove one to get $91$ . This means that we must obtain the value $96$ using only stamps of denominations $n$ and $n+1$
Recalling that $n \geq 24$ , we can easily figure out the working $(n,n+1)$ pairs that can used to obtain $96$ , as we can use at most $\frac{96}{24}=4$ stamps without going over. The potential sets are $(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)$ , and $(96, 97)$
The last two obviously do not work, since they are too large to form the values $92$ through $95$ . By a little testing, only $(24, 25)$ and $(47, 48)$ can form the necessary values, so $n \in \{24, 47\}$ $24 + 47 = \boxed{071}$ | null | 071 |
0f4c0c19b5aba246156a9010b1018ffa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_14 | Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed. | Notice that once we hit all residues $\bmod 5$ , we'd be able to get any number greater (since we can continually add $5$ to each residue). Furthermore, $n\not\equiv 0,1\pmod{5}$ since otherwise $91$ is obtainable (by repeatedly adding $5$ to either $n$ or $n+1$ ) Since the given numbers are $5$ $n$ , and $n+1$ , we consider two cases: when $n\equiv 4\pmod{5}$ and when $n$ is not that.
When $n\equiv 4 \pmod{5}$ , we can only hit all residues $\bmod 5$ once we get to $4n$ (since $n$ and $n+1$ only contribute $1$ more residue $\bmod 5$ ). Looking at multiples of $4$ greater than $91$ with $n\equiv 4\pmod{5}$ , we get $n=24$ . It's easy to check that this works. Furthermore, any $n$ greater than this does not work since $91$ isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem).
Now, if $n\equiv 2,3\pmod{5}$ , then we'd need to go up to $2(n+1)=2n+2$ until we can hit all residues $\bmod 5$ since $n$ and $n+1$ create $2$ distinct residues $\bmod{5}$ . Checking for such $n$ gives $n=47$ and $n=48$ . It's easy to check that $n=47$ works, but $n=48$ does not (since $92$ is unobtainable). Furthermore, any $n$ greater than this does not work since $91$ isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the $3 \pmod{5}$ case, the residue $2 \pmod{5}$ has will not be produced until $3(n+1)$ while the $1\pmod5$ case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to $1 \pmod5$
Since we've checked all residues $\bmod 5$ , we can be sure that these are all the possible values of $n$ . Hence, the answer is $24+47=\boxed{071}$ . - ktong | null | 071 |
0f4c0c19b5aba246156a9010b1018ffa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_14 | Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed. | Obviously $n\le 90$ . We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If $n\equiv 0\pmod{5}$ , then 91 can be formed by using $n+1$ and some 5's, so there are no solutions for this case. If $n\equiv 1\pmod{5}$ , then 91 can be formed by using $n$ and some 5's, so there are no solutions for this case either.
For $n\equiv 2\pmod{5}$ $2n+2$ is the smallest value that can be formed which is 1 mod 5, so $2n+2=96$ and $n=47$ . We see that $92=45+47$ $93=48+45$ , and $94=47+47$ , so $n=47$ does work. If $n\equiv 3\pmod{5}$ , then the smallest value that can be formed which is 1 mod 5 is $2n$ , so $2n=96$ and $n=48$ . We see that $94=49+45$ and $93=48+45$ , but 92 cannot be formed, so there are no solutions for this case. If $n\equiv 4\pmod{5}$ , then we can just ignore $n+1$ since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that $5n-n-5=91$ meaning $4n=96$ and $n=24$ .
Hence, the only two $n$ that work are $n=24$ and $n=47$ , so our answer is $24+47=\boxed{071}$ .
-Stormersyle | null | 071 |
8cbb77862f37ace65f2c3ed8bfbe84d5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | First we have $a\cos A=PQ=25$ , and $(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400$ by PoP. Similarly, $(a\cos A)(b\cos B)=15(10+25)=525,$ and dividing these each by $a\cos A$ gives $b\cos B=21,c\cos C=16$
It is known that the sides of the orthic triangle are $a\cos A,b\cos B,c\cos C$ , and its angles are $\pi-2A$ $\pi-2B$ , and $\pi-2C$ . We thus have the three sides of the orthic triangle now.
Letting $D$ be the foot of the altitude from $A$ , we have, in $\triangle DPQ$ \[\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.\] \[\Rightarrow \cos B=\cos\left(\tfrac 12 (\pi-P)\right)=\sin\tfrac 12 P =\sqrt{\frac{4}{35}},\] similarly, we get \[\cos C=\cos\left(\tfrac 12 (\pi-Q)\right)=\sin\tfrac 12 Q=\sqrt{\frac{9}{40}}.\] To finish, \[bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.\] The requested sum is $\boxed{574}$ .
- crazyeyemoody907 | null | 574 |
8cbb77862f37ace65f2c3ed8bfbe84d5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | Let $BC=a$ $AC=b$ , and $AB=c$ . Let $\cos\angle A=k$ . Then $AP=bk$ and $AQ=ck$
By Power of a Point theorem, \begin{align} AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\ AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 \end{align} Thus $bck = (bk)^2+400=(ck)^2+525 = u$ . Then $bk=\sqrt{u-400}$ $ck=\sqrt{u-525}$ , and \[k=\sqrt{\frac{(u-400)(u-525)}{u^2}}\] Use the Law of Cosines in $\triangle APQ$ to get $25^2=b^2k^2+c^2k^2-2bck^3 = 2bck-925-2bck^3$ , which rearranges to \[775=bck - k^2\cdot bck = u-\frac{(u-400)(u-525)}{u}\] Upon simplification, this reduces to a linear equation in $u$ , with solution $u=1400$ . Then \[AB\cdot AC = bc = \frac 1{k}\cdot bck = \frac{u^2}{\sqrt{(u-400)(u-525)}}=560 \sqrt{14}\] So the final answer is $560 + 14 = \boxed{574}$ | null | 574 |
8cbb77862f37ace65f2c3ed8bfbe84d5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | Let $AP=p$ $PB=q$ $AQ=r$ , and $QC=s$ . By Power of a Point, \begin{align} AP\cdot PB=XP\cdot YP \quad &\Longrightarrow \quad pq=400\\ AQ\cdot QC=YQ\cdot XQ \quad &\Longrightarrow \quad rs=525 \end{align} Points $P$ and $Q$ lie on the circle, $\omega$ , with diameter $BC$ , and pow $(A,\omega) = AP\cdot AB = AQ\cdot AC$ , so \[p(p+q)=r(r+s)\quad \Longrightarrow \quad p^2-r^2=125\] Use Law of Cosines in $\triangle APQ$ to get $25^2=p^2+r^2-2pr\cos A$ ; since $\cos A = \frac r{p+q}$ , this simplifies as \[500 \ =\ 2r^2-\frac{2pr^2}{p+q} \ =\ 2r^2-\frac{2p^2r^2}{p^2+400} \ =\ \frac{800r^2}{r^2+525}\] We get $r=5\sqrt{35}$ and thus \[r=5\sqrt{35}, \quad p = \sqrt{r^2+125} = 10\sqrt{10}, \quad q = \frac{400}{p} =4\sqrt{10}, \quad s= \frac{525}{r} = 3\sqrt{35}.\] Therefore $AB\cdot AC = (p+q)\cdot(r+s) = 560\sqrt{14}$ . So the answer is $560 + 14 = \boxed{574}$ | null | 574 |
8cbb77862f37ace65f2c3ed8bfbe84d5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | This solution is directly based of @CantonMathGuy's solution.
We start off with a key claim.
Claim. $XB \parallel AC$ and $YC \parallel AB$
Proof.
Let $E$ and $F$ denote the reflections of the orthocenter over points $P$ and $Q$ , respectively. Since $EF \parallel XY$ and \[EF = 2 PQ = XP + PQ + QY = XY,\] we have that $E X Y F$ is a rectangle. Then, since $\angle XYF = 90^\circ$ we obtain $\angle XBF = 90^\circ$ (which directly follows from $XBYF$ being cyclic); hence $\angle XBQ = \angle AQB$ , or $XB \parallel AQ \Rightarrow XB \parallel AC$
Similarly, we can obtain $YC \parallel AB$ $\ \blacksquare$
A direct result of this claim is that $\triangle BPX \sim \triangle APQ \sim \triangle CYQ$
Thus, we can set $AP = 5k$ and $BP = 2k$ , then applying Power of a Point on $P$ we get $10 \cdot 40 = 10k^2 \implies k = 2\sqrt{10} \implies AB = 14 \sqrt{10}$ . Also, we can set $AQ = 5l$ and $CQ = 3l$ and once again applying Power of a Point (but this time to $Q$ ) we get
$\phantom{...................}15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}$
Hence,
$\phantom{...................}AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}$
and the answer is $560 + 14 = \boxed{574}$ . ~rocketsri | null | 574 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Let the linear factors be $(x+c)(x+d)$
Then, $a=c+d$ and $b=cd$
We know that $1\le a\le 100$ and $b\ge 0$ , so $c$ and $d$ both have to be non-negative
However, $a$ cannot be $0$ , so at least one of $c$ and $d$ must be greater than $0$ , ie positive.
Also, $a$ cannot be greater than $100$ , so $c+d$ must be less than or equal to $100$
Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices $(0,0), (0, 100),$ and $(100,0)$ . Remember that $(0,0)$ does not work, so there is a square with the top right corner $(1,1)$
Note that $c$ and $d$ are interchangeable since they end up as $a$ and $b$ in the end anyways. Thus, we simply draw a line from $(1,1)$ to $(50,50)$ , designating one of the halves as our solution (since the other side is simply the coordinates flipped).
We note that the pattern from $(1,1)$ to $(50,50)$ is $2+3+4+\dots+51$ solutions and from $(51, 49)$ to $(100,0)$ is $50+49+48+\dots+1$ solutions, since we can decrease the $y$ -value by $1$ until $0$ for each coordinate.
Adding up gives \[\dfrac{2+51}{2}\cdot 50+\dfrac{50+1}{2}\cdot 50.\] This gives us $2600$ , and $2600\equiv 600 \bmod{1000}.$
Thus, the answer is: \[\boxed{600}.\] | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Similar to the previous solution, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is $101+51+51-3$ , or just $200$ . Using Pick's theorem, we know that the area of the half-triangle, which is $2500$ , is just $I+100-1$ . That means that there are $2401$ interior points, plus $200$ boundary points, which is $2601$ . However, $(0,0)$ does not work, so the answer is \[\boxed{600}.\] | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Notice that for $x^2+ax+b$ to be true, for every $a$ $b$ will always be the product of the possibilities of how to add two integers to $a$ . For example, if $a=3$ $b$ will be the product of $(3,0)$ and $(2,1)$ , as those two sets are the only possibilities of adding two integers to $a$ . Note that order does not matter. If we just do some simple casework, we find out that:
if $a$ is odd, there will always be $\left\lceil\frac{a}{2}\right\rceil$ $\left(\text{which is also }\frac{a+1}{2}\right)$ possibilities of adding two integers to $a$
if $a$ is even, there will always be $\frac{a}{2}+1$ possibilities of adding two integers to $a$
Using the casework, we have $1+2+2+3+3+...50+50+51$ possibilities. This will mean that the answer is \[\frac{(1+51)\cdot100}{2}\Rightarrow52\cdot50=2600\] possibilities.
Thus, our solution is $2600\bmod {1000}\equiv\boxed{600}$ | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | We see the pattern $1, 2; 2, 3; 3, 4; ...$ . There are 50 pairs of $i, i+1$ in this pattern, and each pair sums to $2i+1$ . So the pattern condenses to $3, 5, 7, ...$ for 50 terms. This is just $1+3+5+...$ for 51 terms, minus $1$ , or $51^2-1=2601-1=2600\implies\boxed{600}$ | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | The following link is the URL to the graph I drew showing the relationship between a-values and b-values http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file
The pattern continues until $a=100$ , and in total, there are $49$ pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the ( $a=1$ , amount of b-values=1) in the beginning, and ( $a=100$ , amount of b-values=51) in the end.
Then, we add numbers from the opposite ends of the spectrum, and quickly notice that there are $50$ pairs each with a sum of $52$ $52\cdot50$ gives $2600$ ordered pairs:
$1+51, 2+50, 2+50, 3+49, 3+49, 4+48, 4+48…$
When divided by $1000$ , it gives the remainder $\boxed{600}$ , the answer. | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Let's say that the quadratic $x^2 + ax + b$ can be factored into $(x+c)(x+d)$ where $c$ and $d$ are non-negative numbers. We can't have both of them zero because $a$ would not be within bounds. Also, $c+d \leq 100$ . Assume that $c < d$ $d$ can be written as $c + x$ where $x \geq 0$ . Therefore, $c + d = 2c + x \leq 100$ . To find the amount of ordered pairs, we must consider how many values of $x$ are possible for each value of $c$ . The amount of possible values of $x$ is given by $100 - 2c + 1$ . The $+1$ is the case where $c = d$ . We don't include the case where $c = d = 0$ , so we must subtract a case from our total. The amount of ordered pairs of $(a,b)$ is: \[\left(\sum_{c=0}^{50} (100 - 2c + 1)\right) - 1\] This is an arithmetic progression. \[\frac{(101 + 1)(51)}{2} - 1 = 2601 - 1 = 2600\] When divided by $1000$ , it gives the remainder $\boxed{600}$ | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | By Vietas, the sum of the roots is $-a$ and the product is $b$ . Therefore, both roots are nonpositive. For each value of $a$ from $1$ to $100$ , the number of $b$ values is the number of ways to sum two numbers between $0$ and $a-1$ inclusive to $a$ . This is just $1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600$ . Thus, the answer is $\boxed{600}$ | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Similar to solution 1 we plot the triangle and half it. From dividing the triangle in half we are removing the other half of answers that are just flipped coordinates. We notice that we can measure the length of the longest side of the half triangle which is just from $0$ to $100$ , so the number of points on that line is is $101$ . The next row has length $99$ , the one after that has length $97$ , and so on. We simply add this arithmetic series of odd integers $101+99+97+...+1$ .
This is \[\frac{(101+1)(51)}{2} = 2601\] Or you can notice that this is the sum of the first $51$ odd terms, which is just $51^2 = 2601$ .
However, $(0,0)$ is the singular coordinate that does not work, so the answer is $(2601-1)\bmod {1000}\equiv\boxed{600}$ | null | 600 |
e487fc456e366e3cd30ee60ddd08d03d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ | Letting $-r$ and $-q$ be the roots, we must find the number of unordered pairs $(r,q)$ such that $1 \le r+q \le 100.$ To do this we define three boxes: $r, q,$ and $t$ for "trash." For example, if we had $t=77,$ we would have any arrangement of $r$ and $q$ summing to $23.$ Note that we cannot have $t=100,$ subtracting one from our total. By stars and bars, we have that the number of ordered pairs $(r,q)$ is ${102 \choose 2} - 1 = 5150.$ However, we are not done: we have double-counted cases such as $(1,2)$ and $(2,1)$ but not cases such as $(4,4).$ Thus our answer will be $\frac{5150 - 50}{2} + 50 \pmod {100} = \boxed{600}.$ ~ab2024 | null | 600 |
cfaf298e37191c75f3b931c457e0cc76 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_2 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , where $a > 0$ . Find the base- $10$ representation of $n$ | We have these equations: $196a+14b+c=225a+15c+b=222a+37c$ .
Taking the last two we get $3a+b=22c$ . Because $c \neq 0$ otherwise $a \ngtr 0$ , and $a \leq 5$ $c=1$
Then we know $3a+b=22$ .
Taking the first two equations we see that $29a+14c=13b$ . Combining the two gives $a=4, b=10, c=1$ . Then we see that $222 \times 4+37 \times1=\boxed{925}$ | null | 925 |
cfaf298e37191c75f3b931c457e0cc76 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_2 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , where $a > 0$ . Find the base- $10$ representation of $n$ | We know that $196a+14b+c=225a+15c+b=222a+37c$ . Combining the first and third equations give that $196a+14b+c=222a+37c$ , or \[7b=13a+18c\] The second and third gives $222a+37c=225a+15c+b$ , or \[22c-3a=b\] \[154c-21a=7b=13a+18c\] \[4c=a\] We can have $a=4,8,12$ , but only $a=4$ falls within the possible digits of base $6$ . Thus $a=4$ $c=1$ , and thus you can find $b$ which equals $10$ . Thus, our answer is $4\cdot225+1\cdot15+10=\boxed{925}$ | null | 925 |
cfaf298e37191c75f3b931c457e0cc76 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_2 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , where $a > 0$ . Find the base- $10$ representation of $n$ | We're given that $196a+14b+c=225a+15c+b=222a+37c.$ By taking the difference of the first $2$ equalities, we receive $29a+14c=13b.$ Taking $\pmod{13}$ , we receive $3a+c \equiv 0 \pmod{13}.$ We receive the following cases: $(a,c)=(4,1)$ or $(3,4).$ (Note that $(2,7)$ doesn't work since $a,c<6$ by third condition). We can just check these two, and find that $(a,b,c)=(4,10,1),$ and just plugging in $(a,c)$ into the third expression we receive $888+37=\boxed{925}.$ | null | 925 |
da9d0ba58966d7d984933f456162258c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We have $2+4\cdot 2$ cases total.
The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order, so times two.
Obviously the denominator is $10\cdot 9\cdot 8\cdot 7\cdot 6$ , since we are choosing a card without replacement.
Then, we have for the numerator for the two of all red and green: \[5\cdot 4\cdot 3\cdot 2\cdot 1.\]
For the 4 and 1, we have: \[5\cdot 4\cdot 3\cdot 2\cdot 5.\]
For the 3 and 2, we have: \[5\cdot 4\cdot 3\cdot 5\cdot 4.\]
For the 2 and 3, we have: \[5\cdot 4\cdot 5\cdot 4\cdot 3.\]
For the 1 and 4, we have: \[5\cdot 5\cdot 4\cdot 3\cdot 2.\]
Adding up and remembering to double all of them, since they can be reversed and the 5's can be red or green, we get, after simplifying: $\dfrac{31}{126}$
Thus the answer is $31 + 126 = \boxed{157}$ .
-gorefeebuddie | null | 157 |
da9d0ba58966d7d984933f456162258c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Our probability will be $\dfrac{\text{number of "happy" configurations of cards}}{\text{total number of configurations of cards}}.$
First of all, we have $10$ choices for the first card, $9$ choices for the second card, $8$ choices for the third card, $7$ choices for the fourth card, and $6$ choices for the last card. This gives a total of $10\cdot 9\cdot 8\cdot 7\cdot 6$ possible ways for five cards to be chosen.
Finding the number of configurations that make Kathy happy is a more difficult task, however, and we will resort to casework to do it.
First, let's look at the appearances of the "happy configurations" that Kathy likes. Based on the premise of the problem, we can realize that there are ten cases for the appearance of the configurations: \[\text{RRRRR},\] \[\text{GGGGG},\] \[\text{RRRRG},\] \[\text{GGGGR},\] \[\text{RRRGG},\] \[\text{GGGRR},\] \[\text{RRGGG},\] \[\text{GGRRR},\] \[\text{RGGGG},\] \[\text{GRRRR}.\] But this doesn't mean there are 10 "happy configurations" in total-- remember that we've been treating these cards as distinguishable, so we must continue to do so.
To lighten the load of 10 cases on the human brain, we can note that in the eyes of what we will soon do, $RRRRR$ and $GGGGG$ are effectively equivalent, and therefore may be treated in the same case. We will have to multiply by 2 at the end, though.
Similarly, we can equate $RRRRG,$ $GGGGR,$ $RGGGG,$ and $GRRRR,$ as well as $RRRGG,$ $GGGRR,$ $RRGGG,$ and $GGRRR,$ so that we just have three cases. We can approach each of these cases with constructive counting.
Case 1: $RRRRR$ -type.
For this case, there are $5$ available choices for the first card, $4$ available choices for the second card, $3$ for the third card, $2$ for the fourth card, and $1$ for the last card. This leads to a total of $5\cdot 4\cdot 3\cdot 2\cdot 1=120$ configurations for this case. There are $2$ cases of this type.
Case 2: $RRRRG$ -type.
For this case, there are $5$ available choices for the first card, $4$ available choices for the second card, $3$ for the third card, $2$ for the fourth card, and $5$ choices for the last card (not $1$ , because we're doing a new color). This leads to a total of $5\cdot 4\cdot 3\cdot 2\cdot 5=600$ configurations for this case. There are $4$ cases of this type.
Case 3: $RRRGG$ -type.
For this case, there are $5$ available choices for the first card, $4$ available choices for the second card, $3$ for the third card, $5$ for the fourth card, and $4$ choices for the last card. This leads to a total of $5\cdot 4\cdot 3\cdot 5\cdot 4=1200$ configurations for this case. There are $4$ cases of this type.
Adding the cases up gives $2\cdot 120+4\cdot 600+4\cdot 1200=7440$ "happy" configurations in total.
This means that the probability that Kathy is happy will be $\dfrac{7440}{10\cdot 9\cdot 8\cdot 7\cdot 6},$ which simplifies to $\dfrac{31}{126}.$
So the answer is $31+126=\boxed{157}$ | null | 157 |
da9d0ba58966d7d984933f456162258c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | As the problem states, some examples of valid are $RRGGG$ $GGGGR$ , and $RRRRR$ . Let's use each of these as more general cases.
Let $RRGGG$ be the case when there are 2 adjacents of one color, and 3 adjacents of the other color. This yields $4$ combinations ( $RRGGG$ $GGRRR$ $RRRGG$ , and $GGGRR$ ). The probability of each of these is equal, equating to $\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}=\frac{5}{126}$ , and since there are $4$ combinations, the probability of this case is $4\cdot \frac{5}{126}=\frac{10}{63}$
Next case is $GGGGR$ . Let this be when there are 4 adjacents of one color, and 1 individual color. Once again, this yields $4$ combinations ( $GGGGR$ $RRRRG$ $RGGGG$ , and $GRRRR$ ). The probability of each is the same, equating to $\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\cdot \frac{2}{7}\cdot \frac{5}{6}=\frac{5}{252}$ , and since there are $4$ combinations, the probability of this case is $4\cdot \frac{5}{252}=\frac{5}{63}$
The final case is $RRRRR$ , in which there is just an adjacent block of $5$ colors. There are only $2$ combinations this time, each equating to the probability $\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\cdot \frac{2}{7}\cdot \frac{1}{6}=\frac{1}{252}$ , and since there are $2$ combinations, the probability of this case is $2\cdot \frac{1}{252}=\frac{1}{126}$
Thus, the total probability is $\frac{10}{63}+\frac{5}{63}+\frac{1}{126}=\frac{31}{126} \implies m+n=\boxed{157}$ | null | 157 |
da9d0ba58966d7d984933f456162258c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Kathy will draw the 10 cards in some order, but the restriction of having all greens in a row and all reds in a row only applies to the first 5 cards drawn. The total number of ways the 10 cards can be drawn is simply 10 choose 5 which is 252. Now we just count the number of possible successful configurations of the ten cards. The first 5 cards can start either be $GRRRR$ $GGRRR$ $GGGRR$ $GGGGR$ $GGGGG$ or the same thing except starting with a red. The number of ways to order $GRRRR$ is the number of ways to order the last 5 cards, which is 5C1. Doing all of the other cases, the total is $(\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5})*2 = 62$ $\frac{62}{252} = \frac{31}{126},$ so the solution is $31 + 126 = \boxed{157}$ | null | 157 |
da9d0ba58966d7d984933f456162258c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_3 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Assume without loss of generality that the first card laid out is red. Then the arrangements that satisfy Kathy’s requirements are RRRRR, RRRRG, RRRGG, RRGGG, and RGGGG. The probability that Kathy will lay out one of these
arrangements is \[\frac49\cdot\frac38\cdot\frac27\cdot\frac16\] \[\frac49\cdot\frac38\cdot\frac27\cdot\frac56\] \[\frac49\cdot\frac38\cdot\frac57\cdot\frac46\] \[\frac49\cdot\frac58\cdot\frac47\cdot\frac36\] \[+\frac59\cdot\frac48\cdot\frac37\cdot\frac26\] \[\overline{..........................}\] \[\frac{31}{126}\] The requested sum is $31+126=\boxed{157}.$ | null | 157 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | By the Law of Cosines on $\triangle ABC$ , we have: \[\cos(A) = \frac{10^2+10^2-12^2}{2*10*10} = \frac{7}{25}\] By the Law of Cosines on $\triangle ADE$ , then \[\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}\] So, our answer is $250+39=\boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We draw the altitude from $B$ to $\overline{AC}$ to get point $F$ . We notice that the triangle's height from $A$ to $\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\overline{BF}$ , we let $h$ represent $\overline{BF}$ and set up an equation by finding two ways to express the area. The equation is $(8)(12)=(10)(h)$ , which leaves us with $h=9.6$ . We then solve for the length $\overline{AF}$ , which is done through pythagorean theorm and get $\overline{AF}$ $2.8$ . We can now see that $\triangle AFB$ is a $7-24-25$ Right Triangle. Thus, we set $\overline{AG}$ as $5-$ $\tfrac{x}{2}$ , and yield that $\overline{AD}$ $=$ $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$ . Now, we can see $x$ $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$ . Solving this equation, we yield $39x=250$ , or $x=$ $\tfrac{250}{39}$ . Thus, our final answer is $250+39=\boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We start by adding a few points to the diagram. Call $F$ the midpoint of $AE$ , and $G$ the midpoint of $BC$ . (Note that $DF$ and $AG$ are altitudes of their respective triangles). We also call $\angle BAC = \theta$ . Since triangle $ADE$ is isosceles, $\angle AED = \theta$ , and $\angle ADF = \angle EDF = 90 - \theta$ . Since $\angle DEA = \theta$ $\angle DEC = 180 - \theta$ and $\angle EDC = \angle ECD = \frac{\theta}{2}$ . Since $FDC$ is a right triangle, $\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}$
Since $\angle BAG = \frac{\theta}{2}$ and $\angle ABG = \frac{180-m}{2}$ , triangles $ABG$ and $CDF$ are similar by Angle-Angle similarity. Using similar triangle ratios, we have $\frac{AG}{BG} = \frac{CF}{DF}$ $AG = 8$ and $BG = 6$ because there are $2$ $6-8-10$ triangles in the problem. Call $AD = x$ . Then $CE = x$ $AE = 10-x$ , and $EF = \frac{10-x}{2}$ . Thus $CF = x + \frac{10-x}{2}$ . Our ratio now becomes $\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}$ . Solving for $DF$ gives us $DF = \frac{30+3x}{8}$ . Since $DF$ is a height of the triangle $ADE$ $FE^2 + DF^2 = x^2$ , or $DF = \sqrt{x^2 - (\frac{10-x}{2})^2}$ . Solving the equation $\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}$ gives us $x = \frac{250}{39}$ , so our answer is $250+39 = \boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | As in the diagram, let $DE = x$ . Consider point $G$ on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on $DG, GC$ , and $DC$ . Let $GE = \frac{10-x}{2}$ . Therefore, it is trivial to see that $GC^2 = \Big(x + \frac{10-x}{2}\Big)^2$ (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle $DGE$ , we know that $DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2$ . Finally, we apply Law of Cosines on Triangle $DBC$ . We know that $\cos(\angle DBC) = \frac{3}{5}$ . Therefore, we get that $DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}$ . We can now do our final calculation: \[DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}\] After some quick cleaning up, we get \[30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}\] Therefore, our answer is $250+39=\boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let $B = (0, 0)$ $C = (12, 0)$ , and $A = (6, 8)$ . Then, let $x$ be in the interval $0<x<2$ and parametrically define $D$ and $E$ as $(6-3x, 8-4x)$ and $(12-3x, 4x)$ respectively. Note that $AD = 5x$ , so $DE = 5x$ . This means that \begin{align*} \sqrt{36+(8x-8)^2} &= 5x\\ 36+(8x-8)^2 &= 25x^2\\ 64x^2-128x+100 &= 25x^2\\ 39x^2-128x+100 &= 0\\ x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\ x &= \dfrac{100}{78}, 2\\ \end{align*} However, since $2$ is extraneous by definition, $x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}$ ~ mathwiz0803 | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | As shown in the diagram, let $x$ denote $\overline{AD}$ . Let us denote the foot of the altitude of $A$ to $\overline{BC}$ as $F$ . Note that $\overline{AE}$ can be expressed as $10-x$ and $\triangle{ABF}$ is a $6-8-10$ triangle . Therefore, $\sin(\angle{BAF})=\frac{3}{5}$ and $\cos(\angle{BAF})=\frac{4}{5}$ . Before we can proceed with the Law of Cosines, we must determine $\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}$ . Using LOC, we can write the following statement: \[(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies\] \[x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies\] \[0=(10-x)^2-\frac{14x}{25}(10-x)\implies\] \[0=10-x-\frac{14x}{25}\implies\] \[10=\frac{39x}{25}\implies x=\frac{250}{39}\] Thus, the desired answer is $\boxed{289}$ ~ blitzkrieg21 | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | In isosceles triangle, draw the altitude from $D$ onto $\overline{AD}$ . Let the point of intersection be $X$ . Clearly, $AE=10-AD$ , and hence $AX=\frac{10-AD}{2}$
Now, we recognise that the perpendicular from $A$ onto $\overline{AD}$ gives us two $6$ $8$ $10$ triangles. So, we calculate $\sin \angle ABC=\frac{4}{5}$ and $\cos \angle ABC=\frac{3}{5}$
$\angle BAC = 180-2\cdot\angle ABC$ . And hence,
$\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC) = -\cos (2\cdot\angle ABC) = \sin^2 \angle ABC - \cos^2 \angle ABC = 2\cos^2 \angle ABC - 1 = \frac{32}{25}-\frac{25}{25}=\frac{7}{25}$
Inspecting $\triangle ADX$ gives us $\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}$ Solving the equation $\frac{10-x}{2x}=\frac{7}{25}$ gives $x= \frac{250}{39} \implies\boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We can have 2 Law of Cosines applied on $\angle A$ (one from $\triangle ADE$ and one from $\triangle ABC$ ),
$x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}$ and $12^2=10^2+10^2-2(10)(10)\cdot \cos{A}$
Solving for $\cos{A}$ in both equations, we get
$\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}$ and $cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}$ , so the answer is $\boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let $B=(0, 0)$ , and $C=(12, 0)$ . From there, we know that $A=(6, 8)$ , so line $AB$ is $y=\frac{4}{3}x$ . Hence, $D=(a, \frac{4}{3}a)$ for some $a$ , and $BD=\frac{5}{3}a$ so $AD=10-\frac{5}{3}a$ . Now, notice that by symmetry, $E=(6+a, 8-\frac{4}{3}a)$ , so $ED^2=6^2+(8-\frac{8}{3}a)^2$ . Because $AD=ED$ , we now have $(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2$ , which simplifies to $\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100$ , so $\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}$ , and $a=\frac{28}{13}$ .
It follows that $AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}$ , and our answer is $250+39=\boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let's label $\angle A = \theta$ and $\angle ECD = \alpha$ . Using isosceles triangle properties and the triangle angle sum equation, we get \[180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.\] Solving, we find $\theta = 2 \alpha$
Relabelling our triangle, we get $\angle ABC = 90 - \alpha$ . Dropping an altitude from $A$ to $BC$ and using the Pythagorean theorem, we find $[ABC] = 48$ . Using the sine area formula, we see $\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48$ . Plugging in our sine angle cofunction identity, $\sin(90-\alpha) = \cos(\alpha)$ , we get $\alpha = \cos{^{-1}}{\frac45}$
Now, using the Law of Sines on $\triangle ADE$ , we get \[\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.\] After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as $\sin{(180-4\alpha)}=\sin{4\alpha}$ and $\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35$ , we find $\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}$
Therefore, our answer is $250 + 39 = \boxed{289}$ | null | 289 |
8eb217a34843c22d64eabf54a617fea9 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We start by labelling a few angles (all of them in degrees). Let $\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha$ . Also let $AD=a$ . By sine rule in $\triangle{ADE},$ we get $\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}$ Using sine rule in $\triangle{ABC}$ , we get $\sin{\alpha}=\frac{3}{5}$ . Hence we get $\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}$ . Hence $\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}$ . Therefore, our answer is $\boxed{289}$ | null | 289 |
29d409980ecb643bd33ac91d47491462 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_5 | For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$ | Using the logarithmic property $\log_{a^n}b^n = \log_{a}b$ , we note that \[(2x+y)^2 = x^2+xy+7y^2\] That gives \[x^2+xy-2y^2=0\] upon simplification and division by $3$ . Factoring $x^2+xy-2y^2=0$ gives \[(x+2y)(x-y)=0\] Then, \[x=y \text{ or }x=-2y\] From the second equation, \[9x^2+6xy+y^2=3x^2+4xy+Ky^2\] If we take $x=y$ , we see that $K=9$ . If we take $x=-2y$ , we see that $K=21$ . The product is $\boxed{189}$ | null | 189 |
29d409980ecb643bd33ac91d47491462 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_5 | For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$ | Do as done in Solution 1 to get \[x^2+xy-2y^2=0\] \[\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\] \[\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2\] Do as done in Solution 1 to get \[9x^2+6xy+y^2=3x^2+4xy+Ky^2\] \[\implies 6x^2+2xy+(1-K)y^2=0\] \[\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\] \[\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}\] \[\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}\] If $\frac{x}{y}=1$ then \[1=\frac{-1\pm \sqrt{6K-5}}{6}\] \[\implies 6=-1\pm \sqrt{6K-5}\] \[\implies 7=\pm \sqrt{6K-5}\] \[\implies 49=6K-5\] \[\implies K=9\] If $\frac{x}{y}=-2$ then \[-2=\frac{-1\pm \sqrt{6K-5}}{6}\] \[\implies -12=-1\pm \sqrt{6K-5}\] \[\implies -11=\sqrt{6K-5}\] \[\implies 121=6K-5\] \[\implies 126=6K\] \[\implies K=21\] Hence our final answer is $21\cdot 9=\boxed{189}$ -vsamc $\newline$ -minor edit:einsteinstudent | null | 189 |
34337042d3f4be5e252157a0b8902615 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | Let $a=z^{120}$ . This simplifies the problem constraint to $a^6-a \in \mathbb{R}$ . This is true if $\text{Im}(a^6)=\text{Im}(a)$ . Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$ . We are given $\sin\theta = \sin{6\theta}$ . Note that $\sin \theta = \sin (\pi - \theta)$ and $\sin \theta = \sin (\theta + 2\pi)$ . We can use these facts to create two types of solutions:
\[\sin \theta = \sin ((2m + 1)\pi - \theta)\]
which implies that $(2m+1)\pi-\theta = 6\theta$ and reduces to $\frac{(2m + 1)\pi}{7} = \theta$ . There are 7 solutions for this.
\[\sin \theta = \sin (2n\pi + \theta)\]
which implies that $2n\pi+\theta=6\theta$ and reduces to $\frac{2n\pi}{5} = \theta$ . There are 5 solutions for this, totaling 12 values of $a$
For each of these solutions for $a$ , there are necessarily $120$ solutions for $z$ . Thus, there are $12\cdot 120=1440$ solutions for $z$ , yielding an answer of $\boxed{440}$ | null | 440 |
34337042d3f4be5e252157a0b8902615 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to $0$ . Since $|z|=1$ , let $z=\cos \theta + i\sin \theta$ , then we can write the imaginary part of $\Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0$ . Using the sum-to-product formula, we get $\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0$ or $\sin\left(\frac{600\theta}{2}\right)=0$ . The former yields $840$ solutions, and the latter yields $600$ solutions, giving a total of $840+600=1440$ solution, so our answer is $\boxed{440}$ | null | 440 |
34337042d3f4be5e252157a0b8902615 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let $z = e^{i \theta}$ . We have two cases to consider. Either $z^{6!} = z^{5!}$ , or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis.
If $z^{6!} = z^{5!}$ , then $e^{6! \theta i} = e^{5! \theta i}$ . Thus, $720 \theta = 120 \theta$ or $600\theta = 0$ , giving us 600 solutions. (Equalities are taken modulo $2 \pi$ )
For the second case, $e^{6! \theta i} = e^{(\pi - 5!\theta)i}$ . This means $840 \theta = \pi$ , giving us 840 solutions.
Our total count is thus $1440$ , yielding a final answer of $\boxed{440}$ | null | 440 |
34337042d3f4be5e252157a0b8902615 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_6 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ | Because $|z| = 1,$ we know that $z\overline{z} = 1^2 = 1.$ Hence $\overline{z} = \frac 1 {z}.$ Because $z^{6!}-z^{5!}$ is real, it is equal to its complex conjugate. Hence $z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.$ Substituting the expression we that we derived earlier, we get $z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.$ This leaves us with a polynomial whose leading term is $z^{1440}.$ Hence our answer is $\boxed{440}$ | null | 440 |
43f56aacfd2cca043f22f0aded4a8762 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_7 | A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). | We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases.
Case 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \sqrt{3}$ with 6 cases or $\sqrt{3}, \sqrt{3}, \sqrt{3}$ with 2 cases. This can be repeated on the other base for $16$ cases.
Case 2: The vertices span two bases. WLOG call the only vertex on one of the bases $X$ . Call the closest vertex on the other base $B$ , and label clockwise $C, D, E, F, G$ . We will multiply the following scenarios by $12$ , because the top vertex can have $6$ positions and the top vertex can be on the other base. We can have $XCG, XDF$ , but we are not done! Don't forget that the problem statement implies that the longest diagonal in a base is $2$ and the height is $2$ , so $XBE$ is also correct! Those are the only three cases, so there are $12*3=36$ cases for this case.
In total there's $\boxed{052}$ cases. | null | 052 |
43f56aacfd2cca043f22f0aded4a8762 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_7 | A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). | If there are two edges on a single diameter, there would be six diameters. There are four ways to put the third number, and four equilateral triangles. There are $4+ 2 \cdot 2\cdot 6 = 28$ ways. Then, if one length was $\sqrt{3}$ but no side on the diameter, there would be twelve was to put the $\sqrt{3}$ side, and two ways to put the other point. $2 \cdot 12 = 24$ for four ways to put the third point. Adding the number up, the final answer is $24+28 = \boxed{052}.$ | null | 052 |
43f56aacfd2cca043f22f0aded4a8762 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_7 | A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). | To start, let's find the distances from any vertex (call it A, doesn't matter since the prism is symmetrical) to all the other 11 vertices. Using Pythagorean relations, we find that the distances are $1$ $\sqrt{3}$ $2$ $\sqrt{3}$ $1$ $\sqrt{7}$ $\sqrt{5}$ $2$ $\sqrt{5}$ $\sqrt{7}$ , and $\sqrt{8}$
We can clearly form an isosceles triangle using any vertex and any two neighboring edges that are equal. There are 12 total vertices, all of which are symmetrical to the vertex A. Thus, for each vertex, we can form 5 isosceles triangles (with neighboring edges $1$ $1$ $2$ $2$ $\sqrt{3}$ $\sqrt{3}$ $\sqrt{5}$ $\sqrt{5}$ , and $\sqrt{7}$ $\sqrt{7}$ ) , so have so far $12 \times 5 = 60$ isosceles triangles.
To check for overcounting, note that isosceles triangles can be split into two major categories: equilateral isosceles, and non-equilateral isosceles.
We only counted non-equilateral isosceles triangles once since there is only one vertex whose two neighboring edges are equal.
But equilateral triangles were counted three times (namely once for each vertex).
By observation, there are only 4 equilateral triangles (1-1-1 side lengths), two on each hexagonal face. Since we counted each two more times than we should of (three instead of once), we will subtract ( $4$ eq. triangles) $\times$ $2$ overcounts per eq. triangle) = $8$ overcounts.
Finally, we have $60 - 8$ overcounts = $\boxed{052}$ isoceles triangles. | null | 052 |
43f56aacfd2cca043f22f0aded4a8762 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_7 | A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). | Start by drawing the hexagonal prism. Start from one of the points, let's call it $A$ . From simple inspection, we can see that each point we choose creates $4$ isosceles triangles; $A$ and the two adjacent points, $A$ and the two points adjacent to those, $A$ and the adjacent point on the other face, and $A$ and the two points adjacent to those. Here, we have $12\cdot4\Rightarrow48$ isosceles triangles.
Next, notice that you can create an isosceles triangle with the bases of the hexagonal prism. Let's say that our base, starting from the very left-most point of the hexagon, has points $A, B, C, D, E, F$ , rotating counter-clockwise. We can create a triangle with either points $\bigtriangleup{BCE}$ or $\bigtriangleup{BCF}$ . WLOG, assuming we use $\bigtriangleup{BCE}$ , we can use the Law of Cosines and find that $\overline{CE}$ $\sqrt{3}$ . Thus, this would mean $\overline{CF}$ is $2$ , which is the exactly the same length as the height of the hexagonal prism. There are $4$ possibilities of this case, so our final solution is $48+4=\boxed{052}$ | null | 052 |
d53ee033f8a48feb9025a7d785324358 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_9 | Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets. | This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set $\{a, b, c, d\}$
Note that there are only two cases: 1 where $a + b = 16$ and $c + d = 24$ or 2 where $a + b = 16$ and $a + c = 24$ . Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you $a=d$ , which cannot be true.
Case 1.
This is probably the simplest: just make a list of possible combinations for $\{a, b\}$ and $\{c, d\}$ . We get $\{1, 15\}\dots\{7, 9\}$ for the first and $\{4, 20\}\dots\{11, 13\}$ for the second. That appears to give us $7*8=56$ solutions, right? NO. Because elements can't repeat, take out the supposed sets \[\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 4, 20\}, \{5, 11, 5, 19\},\] \[\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}\] That's ten cases gone. So $46$ for Case 1.
Case 2.
We can look for solutions by listing possible $a$ values and filling in the blanks. Start with $a=4$ , as that is the minimum. We find $\{4, 12, 20, ?\}$ , and likewise up to $a=15$ . But we can't have $a=8$ or $a=12$ because $a=b$ or $a=c$ , respectively! Now, it would seem like there are $10$ values for $a$ and $17$ unique values for each $?$ , giving a total of $170$ , but that is once again not true because there are some repeated values!
There are two cases of overcounting:
case 1) (5,11,13,19) & (5.11.19.13)
The same is for (6,10,14,18) and (7,9,15,17)
case 2) those that have the same b and c values
this case includes:
(1,15,9,7) and (7,9,15,1)
(2,14,10,6) and (6,10,14,2)
(3,13,11,5) and (5,11,13,3)
So we need to subtract 6 overcounts.
So, that's $164$ for Case 2.
Total gives $\boxed{210}$ | null | 210 |
d53ee033f8a48feb9025a7d785324358 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_9 | Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets. | Let's say our four elements in our subset are $a,b,c,d$ . We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
$\textrm{Case } 1 \textrm{:}$ $a+b = 16$ and $c+d = 24$
List out possibilities for $a+b$ $(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})$ but don't list $8+8$ because those are the same elements and that is restricted.
Then list out the possibilities for $c+d \text{ }(\text{i.e. } 4+20, 5+19, 6+18, \text{ etc.})$ but don't list $12+12$ because they are the same elements.
This will give you $7 \cdot 8$ elements, which is $56$ . However, as stated above, we have overlap. Just count starting from $a+b$ $15,14,13,11,10,9,7,6,5,4$ all overlap once, which is $10$ , thus $56 - 10 = 46$ cases in this case. Note that $12$ wasn't included because again, if $c+d = 24$ $c$ and $d$ cannot be $12$
$\textrm{Case } 2 \textrm{:}$ $a+b = 16$ and $b+c = 24$
Here, $b$ is included in both equations. We can easily see that $a, b, c$ will never equal each other.
Furthermore, there are 17 choices for $d$ $20 - 3$ included elements) for each $b$ . Listing out the possible $b$ s, we go from $15,14,13,11,10,9,7,6,5,4$ . Do not include $8$ or $12$ because if they are included, then $a/c$ will be the same as $b$ , which is restricted.
There are $10$ options there, and thus $10 \cdot 17 = 170$ . But, note that if $d = b+8$ $a+d = a+b+8 = 24$ , and so we have a double-counted set. Starting with $b=15$ , we have $15, 14, 13, 11, 10, 9$ (where $d$ is $7, 6, 5, 3, 2, 1)$ . That means there are $6$ double-counted cases. Thus $170 - 6 = 164$ cases in this case.
Adding these up, we get $46+164 = \boxed{210}.$ | null | 210 |
a0e1a885ddd7f38fc8b82a9710e24f6a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_10 | The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$ , which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A$ . Find the remainder when $n$ is divided by $1000.$ | We divide this up into casework. The "directions" the bug can go are $\text{Clockwise}$ $\text{Counter-Clockwise}$ , and $\text{Switching}$ . Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circles. An example string of length fifteen that gets the bug back to $A$ would be $ISSIIISOOSISSII$ .
For the bug to end up back at $A$ , the difference between the number of $I$ 's and $O$ 's must be a multiple of $5$
So, the total number of ways is $1+2002+1001=3004$ which gives $\boxed{004}$ as the answer. | null | 004 |
a0e1a885ddd7f38fc8b82a9710e24f6a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_10 | The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$ , which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A$ . Find the remainder when $n$ is divided by $1000.$ | Define $A_n$ to be the number of sequences of length $n$ that ends at $A$ and similarly for the other spokes. Also let \[S_n=A_n+B_n+C_n+D_n+E_n+F_n+G_n+H_n+I_n+J_n\] Apparently everytime the bug has $2$ choices for its next move, thus we have $S_n=2^n$ . Now we attempt to find a recursive formula for $A_n$ \begin{align*} A_n&=J_{n-1}+E_{n-1} \\ &=(I_{n-2}+A_{n-2})+(D_{n-2}+F_{n-2}) \\ &=(H_{n-3}+B_{n-3})+(J_{n-3}+E_{n-3})+(G_{n-3}+C_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(B_{n-3}+C_{n-3}+E_{n-3}+G_{n-3}+H_{n-3}+J_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(S_{n-3}-(I_{n-3}+A_{n-3}+D_{n-3}+F_{n-3}))+A_{n-2} \\ &=2^{n-3}-(J_{n-2}+E_{n-2})+A_{n-2} \\ &=2^{n-3}-A_{n-1}+A_{n-2} \\ \end{align*} Computing a few easy terms we have $A_0=0$ $A_1=0$ $A_2=1$ $A_3=0$ $A_4=3$ . Continuing the process yields $A_{15}=3\boxed{004}$ | null | 004 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Note that the given condition is equivalent to $3^n \equiv 1 \pmod{143^2}$ and $143=11\cdot 13$ . Because $\gcd(11^2, 13^2) = 1$ , the desired condition is equivalent to $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$
If $3^n \equiv 1 \pmod{121}$ , one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9...$ so $5|n$
Now if $3^n \equiv 1 \pmod{169}$ , it is harder. But we do observe that $3^3 \equiv 1 \pmod{13}$ , therefore $3^3 = 13a + 1$ for some integer $a$ . So our goal is to find the first number $p_1$ such that $(13a+1)^ {p_1} \equiv 1 \pmod{169}$ . Then, $p_1 \equiv 0 \pmod{13},$ which follows from the binomial theorem. It is not difficult to see that the smallest $p_1=13$ , so ultimately $3^{39} \equiv 1 \pmod{169}$ . Therefore, $39|n$
The first $n$ satisfying both criteria is thus $5\cdot 39=\boxed{195}$ | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Note that Euler's Totient Theorem would not necessarily lead to the smallest $n$ and that in this case that $n$ is greater than $1000$
We wish to find the least $n$ such that $3^n \equiv 1 \pmod{143^2}$ . This factors as $143^2=11^{2}*13^{2}$ . Because $\gcd(121, 169) = 1$ , we can simply find the least $n$ such that $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$
Quick inspection yields $3^5 \equiv 1 \pmod{121}$ and $3^3 \equiv 1 \pmod{13}$ . Now we must find the smallest $k$ such that $3^{3k} \equiv 1 \pmod{169}$ . Euler's gives $3^{156} \equiv 1 \pmod{169}$ . So $3k$ is a factor of $156$ . This gives $k=1,2, 4, 13, 26, 52$ . Some more inspection yields $k=13$ is the smallest valid $k$ . So $3^5 \equiv 1 \pmod{121}$ and $3^{39} \equiv 1 \pmod{169}$ . The least $n$ satisfying both is $lcm(5, 39)=\boxed{195}$ . (RegularHexagon) | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Listing out the powers of $3$ , modulo $169$ and modulo $121$ , we have: \[\begin{array}{c|c|c} n & 3^n\mod{169} & 3^n\mod{121}\\ \hline 0 & 1 & 1\\ 1 & 3 & 3\\ 2 & 9 & 9\\ 3 & 27 & 27\\ 4 & 81 & 81\\ 5 & 74 & 1\\ 6 & 53\\ 7 & 159\\ 8 & 139\\ 9 & 79\\ 10 & 68\\ 11 & 35\\ 12 & 105\\ 13 & 146\\ 14 & 100\\ 15 & 131\\ 16 & 55\\ 17 & 165\\ 18 & 157\\ 19 & 133\\ 20 & 61\\ 21 & 14\\ 22 & 42\\ 23 & 126\\ 24 & 40\\ 25 & 120\\ 26 & 22\\ 27 & 66\\ 28 & 29\\ 29 & 87\\ 30 & 92\\ 31 & 107\\ 32 & 152\\ 33 & 118\\ 34 & 16\\ 35 & 48\\ 36 & 144\\ 37 & 94\\ 38 & 113\\ 39 & 1\\ \end{array}\]
The powers of $3$ repeat in cycles of $5$ and $39$ in modulo $121$ and modulo $169$ , respectively. The answer is $\text{lcm}(5, 39) = \boxed{195}$ | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | We have that \[3^n \equiv 1 \pmod{143^2}.\] Now, $3^{110} \equiv 1 \pmod{11^2}$ so by the Fundamental Theorem of Orders, $\text{ord}_{11^2}(3)|110$ and with some bashing, we get that it is $5$ . Similarly, we get that $\text{ord}_{13^2}(3)=39$ . Now, $\text{lcm}(39,5)=\boxed{195}$ which is our desired solution. | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | We wish to find the smallest $n$ such that $3^n\equiv 1\pmod{143^2}$ , so we want $n\equiv 1\pmod{121}$ and $n\equiv 1\pmod{169}$ . Note that $3^5\equiv 1\pmod{121}$ , so $3^n$ repeats $121$ with a period of $5$ , so $5|n$ . Now, in order for $n\equiv 1\pmod{169}$ , then $n\equiv 1\pmod{13}$ . Because $3^3\equiv 1\pmod{13}$ $3^n$ repeats with a period of $3$ , so $3|n$ .
Hence, we have that for some positive integer $p$ $3^n\equiv (3^3)^p\equiv (26+1)^p\equiv \binom{p}{0}26^p+\binom{p}{1}26^{p-1}....+\binom{p}{p-2}26^2+\binom{p}{p-1}26+\binom{p}{p}\equiv 26p+1\equiv 1\pmod{169}$ , so $26p\equiv 0\pmod{169}$ and $p\equiv 0\pmod{13}$ . Thus, we have that $5|n$ $3|n$ , and $13|n$ , so the smallest possible value of $n$ is $3\times5\times13=\boxed{195}$ .
-Stormersyle | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | We can see that $3^n-1 = 143^2*x$ , which means that $v_{11}(3^n-1) \geq 2$ $v_{13}(3^n-1) \geq 2$ $v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})$ $v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})$ by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as $v_{13}(3^n-1) \geq 2$ . Therefore the minimum possible value of n is $3\times5\times13=\boxed{195}$ | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | Note that the problem is basically asking for the least positive integer $n$ such that $11^2 \cdot 13^2 | 3^n - 1.$ It is easy to see that $n = \text{lcm } (a, b),$ where $a$ is the least positive integer satisfying $11^2 | 3^a - 1$ and $b$ the least positive integer satisfying $13^2 | 3^b - 1$ . Luckily, finding $a$ is a relatively trivial task, as one can simply notice that $3^5 = 243 \equiv 1 \mod 121$ . However, finding $b$ is slightly more nontrivial. The order of $3^k$ modulo $13$ (which is $3$ ) is trivial to find, as one can either bash out a pattern of remainders upon dividing powers of $3$ by $13$ , or one can notice that $3^3 = 27 \equiv 1 \mod 13$ (the latter which is the definition of period/orders by FLT). We can thus rewrite $3^3$ as $(2 \cdot 13 + 1) \mod 13^2$ . Now suppose that \[3^{3k} \equiv (13n + 1) \mod 13^2.\] I claim that $3^{3(k+1)} \equiv (13(n+2) + 1) \mod 13^2.$
Proof:
To find $3^{3(k+1)},$ we can simply multiply $3^{3k}$ by $3^3,$ which is congruent to $2 \cdot 13 + 1$ modulo $13^2$ .
By expanding the product out, we obtain \[3^{3(k+1)} \equiv (13n + 1)(2 \cdot 13 + 1) = 13^2 \cdot 2n + 13n + 2 \cdot 13 + 1 \mod 13^2,\] and since the $13^2$ on the LHS cancels out, we're left with \[13n + 2 \cdot 13 + 1 \mod 13^2 \implies 13(n+2) + 1 \mod 13^2\] . Thus, our claim is proven.
Let $f(n)$ be the second to last digit when $3^{3k}$ is written in base $13^2$ .
Using our proof, it is easy to see that $f(n)$ satisfies the recurrence $f(1) = 2$ and $f(n+1) = f(n) + 2$ . Since this implies $f(n) = 2n,$ we just have to find the least positive integer $n$ such that $2n$ is a multiple of $13$ , which is trivially obtained as $13$ .
The least integer $n$ such that $3^n - 1$ is divisible by $13^2$ is $3 \cdot 13 = 39,$ so our final answer is $\text{lcm } (5, 39) = \boxed{195}.$ | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | The requested positive integer is the least value of $n>0$ such that $3^n\equiv 1\pmod{143^2}.$ Note that $143=11\cdot 13.$ The least power of $3$ that is congruent to $1$ modulo $11^2$ is $3^5=243=2\cdot 11^2+1.$ It follows that. $3^n\equiv 1\pmod {11^2}$ if and only if $n=5j$ for some positive integer $j$
The least power of $3$ that is congruent to $1$ modulo $13$ is $3^3=27=2\cdot 13+1.$ It follows that $3^n\equiv 1\pmod{13}$ if and only if $n=3k$ for some positive integer $k$ . Additionally, for some positive integer $k$ , the Binomial Theorem shows that $3^{3k}=(26+1)^k=26\cdot k+1 \pmod{13^2}$ . In particular, $3^n=3^{3k}\equiv 1\pmod {13^2}$ if and only if $k=13m$ for some positive integer $m$ , that is, if and only if $n=39m.$
Because $11^2$ and $13^2$ are relatively prime, $3^n\equiv 1\pmod {143^2}$ if and only if $3^n\equiv 1\pmod{11^2}$ and $3^n \equiv 1\pmod {13^2}$ . This occurs if and only if $n$ is a multiple of both of the relatively prime integers $5$ and $39$ , so the least possible value of $n$ is $5\cdot 39=\boxed{195}.$ | null | 195 |
e35159ec0dad238ff3af9656468c6e81 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ | We first note that we wish to find $3^n \equiv 1 \pmod{11^2}$ and $3^n \equiv 1 \pmod{13^2}.$ Not thinking of anything else, we try a few numbers for the first condition to get that $5 \mid n.$ For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Exponent from our toolkit to get that \[v_{13}(27^m-1^m)=v_{13}(26)+v(m)=1+v(m), 3m = n\] which clearly implies $m=13$ and $39 | n.$ Our answer is then obviously $39 \cdot 5 = \boxed{195}.$ | null | 195 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | The question asks us for the probability that a randomly chosen subset of the set of the first 18 positive integers has the property that the sum of its elements is divisible by 3. Note that the total number of subsets is $2^{18}$ because each element can either be in or not in the subset. To find the probability, we will find the total numbers of ways the problem can occur and divide by $2^{18}$
To simplify the problem, let’s convert the set to mod 3:
\[U' = \{1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0\}\]
Note that there are six elements congruent to 0 mod 3, 6 congruent to 1 mod 3, and 6 congruent to 2 mod 3. After conversion to mod three, the problem is the same but we’re dealing with much simpler numbers. Let’s apply casework on this converted set based on $S = s(T')$ , the sum of the elements of a subset $T'$ of $U'$
$\textbf{Case 1: }S=0$
In this case, we can restrict the subsets to subsets that only contain 0. There are six 0’s and each one can be in or out of the subset, for a total of $2^{6} = 64$ subsets. In fact, for every case we will have, we can always add a sequence of 0’s and the total sum will not change, so we will have to multiply by 64 for each case. Therefore, let’s just calculate the total number of ways we can have each case and remember to multiply it in after summing up the cases. This is equivalent to finding the number of ways you can choose such subsets without including the 0's. Therefore this case gives us $\boxed{1}$ way. | null | 1 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | Consider the elements of $U$ modulo $3.$
Ignore the $0$ 's because we're gonna multiply $\binom{6}{0}+..+\binom{6}{6}=2^6$ at the end. Let $a$ be the $1's$ and $b$ be the $2's$ . The key here is that $2 \equiv -1 \pmod{3}$ so the difference between the number of $a$ and $b$ is a multiple of $3$
1. Counted twice because $a$ and $b$ can be switched:
$6a$
$6a,3b$
$5a,2b$
$4a,b$
$3a$
2. Counted once:
$6a,6b$ ... $0a,0b$
By Vandermonde's Identity on the second case, this is $2^6 \left( 2\left(1+20+90+90+20\right) + \binom{12}{6} \right)\implies \boxed{683}$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | Elements $0 \pmod{3}$ can either be included or excluded, for a total of $2^6$ . Then, like the previous solution, let $a$ be the number of elements $1 \pmod{3}$ and $b$ be the number of elements $2 \pmod{3}$ . Then, $a + 2b \equiv 0 \pmod{3} \implies a - b \equiv 0 \pmod{3}$ . Since $0 \le a, b \le 6$ , there are 3 cases:
$\>\>\>\> 1. \> a - b = 0$
Then, we have, \[\binom{6}{0}\binom{6}{0} + \binom{6}{1}\binom{6}{1} + \cdots + \binom{6}{6} \binom{6}{6}\]
Using the substitution $\binom{6}{k} = \binom{6}{6-k}$ on every second binomial coefficient, it is clear that the bottom numbers sum to $6$ , and the ones above sum to $12$ . Apply Vandermonde's, we obtain $\binom{12}{6}$
$\>\>\>\> 2. \> a - b = \pm 3$
For $a - b = 3$ , using the same substitution and applying Vandermonde's, we get: \[\binom{6}{3} \binom{6}{0} + \binom{6}{4} \binom{6}{1} + \cdots + \binom{6}{6} \binom{6}{3} = \binom{12}{3}\]
Then, for $a - b = -3$ , it is completely symmetric with $a - b = 3$ . We get $\binom{12}{3}$
$\>\>\>\> 3. \> a - b = \pm 6$
We have $\binom{6}{6} \binom{6}{0} + \binom{6}{0} \binom{6}{6} = 2$
We get $\frac{2^6 (\binom{12}{6} + 2 \binom{12}{3} + 2)}{2^{18}} = \frac{683}{2^{11}} \implies m = \boxed{683}$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | Note that in general, the answer will be around $1/3$ of $2^{18}$ . We will use this to our advantage. Partition $U$ into disjoint subsets $\{1,2,3\}, \{4,5,6\}, \cdots, \{16,17,18\}$ . Within each of these subsets, there are 3 possible remainders $\mod 3$ depending on what elements we choose to include into $T$ : (Using set $\{1,2,3\}$ for demonstration purposes)
Remainder of zero: Include $\{3\}, \{1,2\}, \{\}, \{1,2,3\}$
Remainder of one: Include $\{1\}, \{1,3\}$
Remainder of two: Include $\{2\}, \{2,3\}$
Suppose all subsets are of the form $\{\}, \{1,2,3\}$ . Then, since both choices are $0 \pmod{3}$ , we can choose either one for all six subsets, giving us $2^6$ combinations.
On the other hand, suppose there exists a subset $V$ that isn't of the form $\{\}, \{1,2,3\}$ . Then, for $V$ , it is equally likely that a remainder of zero, one, or two is chosen, since each remainder has $2$ ways to be achieved, i.e. there is a $1/3$ chance for each. Consider the sum of the other subsets $T \setminus V$ . The sum is either $0, 1, 2 \pmod{3}$ . Whatever that remainder $r$ might be, we can always choose $3-r$ as the remainder for our set $V$ , giving us a total of $r + 3 - r \equiv 0 \pmod{3}$ . The probability we choose remainder $3-r$ is $1/3$ as previously mentioned. So, we get $\frac{1}{3}(2^{18} - 2^6)$ total combinations.
Therefore, we get $\frac{2^6 + \frac{1}{3}(2^{18} - 2^6)}{2^{18}} = \frac{683}{2^{11}} \implies m = \boxed{683}$ | null | 683 |