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3eafd85dc85bb5b34d71f2af315b8122 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | $a_n=a_{n-1} + n \pmod{99}$ . Using the steps of the previous solution we get up to $n^2+n \equiv 90 \pmod{99}$ . This gives away the fact that $(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}$ so either $n$ or $n+1$ must be a multiple of 9.
Case 1 ( $9 \mid n$ ): Say $n=9x$ and after simplification $x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}$
Case 2: ( $9 \mid n+1$ ): Say $n=9a-1$ and after simplification $(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}$
As a result $a$ must be a divisor of $10$ and after doing some testing in both cases the smallest value that works is $x=5 \implies \boxed{045}$ | null | 045 |
3eafd85dc85bb5b34d71f2af315b8122 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | We just notice that $100 \equiv 1 \pmod{99}$ , so we are just trying to find $10 + 11 + 12 + \cdots + n$ modulo $99$ , or $\dfrac{n(n+1)}{2} - 45$ modulo $99$ . Also, the sum to $44$ is divisible by $99$ , and is the first one that is. Thus, if we sum to $45$ the $45$ is cut off and thus is just a sum to $44$
Without checking whether there are other sums congruent to $45 \pmod{99}$ , we can just write the answer to be $\boxed{045}$ | null | 045 |
3eafd85dc85bb5b34d71f2af315b8122 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | Let $b_n = 2a_{n+10}$ . We can find a formula for $b_n$
$b_n = (20+n)(n+1)$
Notice that both can't have a factor of 3. Thus we can limit our search range of n to $n \equiv 7,8 \pmod{9}$ . Testing values for n in our search range (like 7,8,16,17,25,26...), we get that 35 is the least n. But, don't write that down! Remember, $b_n = 2a_{n+10}$ , so, the 35th term in b corresponds to the 45th term in a. Thus our answer is $\boxed{045}$ | null | 045 |
3eafd85dc85bb5b34d71f2af315b8122 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | The first thing you should realize is that each term after the tenth is another two-digit number chained to the last number. $10, 1011, 101112, \dots$ . Now the fact that the sequence starts at $10$ can be completely discarded for this solution. Just consider $a(10)$ then same as $a(1)$ , and we can add nine to the answer at the end.
The second step is to split $99$ as $9$ and $11$ and solve for divisibility rules individually. Let's start with $11$ because it gives us the most information to continue.
In any number generated, if the numbers don't go beyond $20$ , then the highest number we can get is $10111213141516171819$ , with every odd digit being $1$ . This is a little risky because we are assuming that it doesn't exceed $20$ . If someone wanted to be absolutely sure they could continue, but this is unnecessary later and a big hassle. Anyways, now we write an equation to check for divisibility by $11$ . The expression being $\frac{((n-1) + 0)\cdot n)}{2}-n$
The concept here is to add $0$ to the $n-1^{th}$ term altogether, then subtract the number of ones in it, which is $n$ . Simplify to $\frac{n(n-3)}{2}$ congruent to $0 \pmod{11}$ . Now notice the divide by two can be discarded because one of $n$ or $(n-3)$ will be even. So if $n$ or $n-3$ is to be divisible by $11$ , we can make a simple list.
\[n = 0, 3, 11, 14, 22, 25, 33, 36, 44, 47, \dots\]
Now we test each $n$ for divisibility by $9$ .
This is done by making a list that ultimately calculates the sum of every digit in the large number. $n(1)$ to $n(10)$ has the first digit $1$ $n(11)$ to $n(20)$ has the first digit $2$ , and so on.
The necessary thing to realize is that the sum of all digits $0-9$ is divisible by $9$ , so we only have to solve for the sum of the first digits, and then the short list of second digits.
For example, let's test $n=25$
So we know that $25$ include both $1-10$ and $11-20$ , so that's $10 + 20$ right away. $21-25$ contains $5$ numbers that have the first digit $3$ , so $+15$ . Then we add $0-4$ together, which is $10$ $10+20+15+10=55$ , which is not divisible by $9$ , so it is not the answer.
Do this for just a minute you get that $36$ sums to $99$ , a multiple of nine! So $n(36)$ is the answer, right? Don't forget we have to add $9$ because we translated $n(10)$ to $n(1)$ at the very beginning!
Finally, after a short bash, we get $\boxed{045}$ | null | 045 |
3eafd85dc85bb5b34d71f2af315b8122 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | Taking the recurrence mod $99$ , we have \[a_n=a_{n-1}+n\] for $a_{10}=10$ . Then, we have \[a_n=10+11+12+\cdots+n \implies a_n=\frac{n(n+1)}2-45 \equiv 0 \pmod{99} \implies n(n+1)-90 \equiv 0 \pmod{99} \implies n(n+1)+9 \equiv 0 \pmod{99} \implies n \equiv 0 \pmod{9}.\] Then, we simply can test these values of $n$ to find that $n=\boxed{045}$ produces a value that is also divisible by $11$ | null | 045 |
12f7ea57fff6d9a682ee505ca5819b22 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_10 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$ | Now that we’ve proven this fact, we know that all four points lie on a circle (intuitively one can also observe this because $z=z_1$ and $z=z_2$ satisfy the property in the question, and $z=z_3$ techincally gives no imaginary part), so let’s draw that in;
[asy] size(1200,300); real xMin = 0; real xMax = 100; real yMin = 0; real yMax = 120; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$R$",(xMax,0),(2,0)); label("$Im$",(0,yMax),(0,2)); pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56,104.908997); dot(A); dot(B); dot(C); dot(D); draw(A--C); draw(A--B); draw(D--C); draw(C--B); draw(D--B); label("$Z_1$", A, W); label("$Z_2$", B, W); label("$Z_3$", C, N); label("$Z$", D, N); draw(circle((56,61), 43.9089968)); [/asy]
While $Z_1, Z_2, Z_3$ are fixed, $Z$ can be anywhere on the circle because those are the only values of $Z$ that satisfy the problem requirements. However, we want to find the real part of the $Z$ with the maximum imaginary part. This Z would lie directly above the center of the circle, and thus the real part would be the same as the x-value of the center of the circle. So all we have to do is find this value and we’re done.
Consider the perpendicular bisectors of $Z_1Z_2$ and $Z_2Z_3$ . Since any chord can be perpendicularly bisected by a radius of a circle, these two lines both intersect at the center. Since $Z_1Z_2$ is vertical, the perpendicular bisector will be horizontal and pass through the midpoint of this line, which is (18, 61). Therefore the equation for this line is $y=61$ $Z_2Z_3$ is nice because it turns out the differences in the x and y values are both equal (60) which means that the slope of the line is 1. The slope of the perpendicular bisector is therefore -1 and it passes through the midpoint, (48,69), so the equation of this line is $y=-x+117$ . Finally, equating the two yields \[61=-x+117 \implies x = \boxed{056}\] | null | 056 |
12f7ea57fff6d9a682ee505ca5819b22 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_10 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$ | We know that \[z_3-z_1 = (78+99i)-(16+83i) = 60 + 16i.\] \[z_2-z_1=(18+39i)-(18+83i) = -44i.\] Hence, \[\frac{z_3-z_1}{z_2-z_1} = \frac{60 + 16i}{-44i} = \frac{15i-4}{11} = \frac{c}{15i+4}\] where $c \in R$ .
Let $z = a+bi$ . Then, \[\frac{z-z_2}{z-z_3} = \frac{(a+bi)-(18+39i)}{(a+bi)-(78+99i)} =\] \[\frac{z-z_2}{z-z_3} = \frac{(a-18)+i(b-39)}{(a-78)+i(b-99)} =\] \[\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{((a-78)+i(b-99))((a-78)+i(99-b)} =\] \[\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{k}\] The numerator is: \[(a-18)(a-78)-(b-39)(99-b)+i((b-39)(a-78) + (99-b)(a-18))=\] \[a^2+b^2-96a-138b+18 \cdot 78 + 39 \cdot 99 + i(60a - 60b + 39 \cdot 78 + 18 \cdot 99)\] The ratio of the imaginary part to the real part must be $\frac{15}{4}$ because $\frac{z_3-z_1}{z_2-z_1} = \frac{c}{15i+4}.$ Hence, \[\frac{60a - 60b + 39 \cdot 78 - 18 \cdot 99}{a^2+b^2-96a-138b+18 \cdot 78 + 39 \cdot 99} = \frac{15}{4} \implies\] \[\frac{4a-4b+84}{a^2+b^2-96a-138b+9(2 \cdot 78 + 39 \cdot 11)} = \frac{1}{4} \implies\] \[16a - 16b + 336 = a^2 + b^2 -96a - 138b +5265 \implies\] \[0 = a^2 + b^2 -112a-122b+4929.\] Evidently, $b$ is maximized when $112a-a^2$ is maximized or when $a = \boxed{056}.$ | null | 056 |
12f7ea57fff6d9a682ee505ca5819b22 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_10 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$ | Algebra Bash
First we calculate $\frac{z_3 - z_1}{z_2 - z_1}$ , which becomes $\frac{15i-4}{11}$
Next, we define $z$ to be $a+bi$ for some real numbers $a$ and $b$ . Then, $\frac {z-z_2}{z-z_3}$ can be written as $\frac{(a-18)+(b-39)i}{(a-78)+(b-99)i}.$ Multiplying both the numerator and denominator by the conjugate of the denominator, we get:
$\frac {[(a-18)(a-78)+(b-39)(b-99)]+[(a-78)(b-39)-(a-18)(b-99)]i}{(a-78)^2+(b-99)^2}$
In order for the product to be a real number, since both denominators are real numbers, we must have the numerator of $\frac {z-z_2}{z-z_3}$ be a multiple of the conjugate of $15i-4$ , namely $-15i-4$ So, we have $(a-18)(a-78)+(b-39)(b-99) = -4k$ and $(a-78)(b-39)-(a-18)(b-99) = -15k$ for some real number $k$
Then, we get: $(a-18)(a-78)+(b-39)(b-99) = \frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)]$
Expanding both sides and combining like terms, we get:
$a^2 - 112a +b^2 - 122b + 4929 = 0$
which can be rewritten as:
$(a-56)^2 + (b-61)^2 = 1928$
Now, common sense tells us that to maximize $b$ , we would need to maximize $(b-61)^2$ . Therefore, we must set $(a-56)^2$ to its lowest value, namely 0. Therefore, $a$ must be $\boxed{056}.$ | null | 056 |
12f7ea57fff6d9a682ee505ca5819b22 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_10 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$ | We see that $\frac{z_3-z_1}{z_2-z_1}=\frac{60+16i}{-44i}=\frac{15i-4}{11}$
Now, let $z-z_3=a+bi$ , in which case $z=(a+78)+(b+99)i$ and $z-z_2=(a+60)+(b+60)i$
We now have that $\left(\frac{(a+60)+(b+60)i}{a+bi}\right)\left(\frac{15i-4}{11}\right)$ is real, meaning that: \[((a+60)+(b+60)i)\left(\frac{15i-4}{a+bi}\right)\] is also real.
We see that: \[\frac{15i-4}{a+bi}=\frac{(15i-4)(a-bi)}{a^2+b^2}=\frac{(15b-4a)+(15a+4b)i}{a^2+b^2},\] so therefore, $x=((a+60)+(b+60)i)((15b-4a)+(15a+4b)i)$ is real.
This means that $Im(x)=0$ , so we now have that: \[(a+60)(15a+4b)+(b+60)(15b-4a)=15a^2+15b^2+660a+1140b=0 \Rightarrow a^2+b^2+44a+76b=0.\]
This can be rewritten as: \[(a+22)^2+(b+38)^2=22^2+38^2.\] In order to maximize $Im(z)$ we want to maximize $b$ , and in order to maximize $b$ we want $a+22=0$ and $a=-22$ , so $Re(z)=a+78=-22+78=\boxed{056}$ .
(Note: $Im(\omega)$ is the imaginary part of $\omega$ , and $Re(\omega)$ is the real part of $\omega$ )
~Stormersyle | null | 056 |
12f7ea57fff6d9a682ee505ca5819b22 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_10 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$ | We will just bash. Let $z=a+bi$ where $a,b\in\mathbb{R}$ . We see that $\frac{z_3-z_1}{z_2-z_1}=\frac{-4+15i}{11}$ after doing some calculations. We also see that $\frac{[(a-18)+(b-39)i][(a-78)-(b-99)i]}{\text{some real stuff}}.$ We note that $[(a-18)+(b-39)i][(a-78)-(b-99)i]$ is a multiple of $-4-15i$ because the numerator has to be real. Thus, expanding it out, we see that $(a-18)(a-78)+(b-39)(b-99)=-4k \\ (a-78)(b-39)-(a-18)(b-99)=-15k.$ Hence, $(a-18)(a-78)+(b-39)(b-99)=\frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)] \implies a^2-96a+b^2-138b+5265=16a-16b+336 \\ (a-56)^2+(b-61)^2=1928.$ To maximize the imaginary part, $(a-56)^2$ must equal $0$ so hence, $a=\boxed{56}$ | null | 56 |
43a39e153ec59d470d8794021eb01299 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$ | Assume that $5 \in \{a_1, a_2, a_3\}$ $m \neq 5$ , and WLOG, $\max{(a_1, a_2, a_3)} = 5$ . Then we know that the other two medians in $\{a_1, a_2, a_3\}$ and the smallest number of rows 1, 2, and 3 are all less than 5. But there are only 4 numbers less than 5 in $1, 2, 3, \dots, 9$ , a Contradiction. Thus, if $5 \in \{a_1, a_2, a_3\}$ , then $m = 5$
WLOG, assume $5$ is in the upper left corner. One of the two other values in the top row needs to be below $5$ , and the other needs to be above $5$ . This can be done in $4\cdot4\cdot2=32$ ways.
The other $6$ can be arranged in $6!=720$ ways.
Finally, accounting for when $5$ is in every other space, our answer is $32\cdot720\cdot9$ , which is $207360$ . But we only need the last $3$ digits, so $\boxed{360}$ is our answer. | null | 360 |
43a39e153ec59d470d8794021eb01299 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$ | (Complementary Counting with probability)
Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric.
WLOG let $m=4$
1. There is a $\frac{15}{28}$ chance that exactly one of 1, 2, 3 is in the same row with 4.
There are 3 ways to select which of the smaller numbers will get in the row, and then 5
ways to select the number larger than 4.
$\frac{\dbinom{3}{1}\cdot\dbinom{5}{1}}{\dbinom{8}{2}} = \frac{15}{28}$
2. There is a $\frac{2}{5}$ chance that the other two smaller numbers end up in the same row.
There are 2 ways to select the row that the two smaller number are in, and then $\dbinom{3}{2}$ ways
to place the smaller numbers in the row.
$\frac{\dbinom{2}{1}\cdot\dbinom{3}{2}}{\dbinom{6}{2}} = \frac{2}{5}$
$9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}$ | null | 360 |
43a39e153ec59d470d8794021eb01299 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$ | We will make sure to multiply by $3!$ in the end to account for all the possible permutation of the rows.
WLOG, let $5$ be present in the Row # $1$
Notice that $5$ MUST be placed with a number lower than it and a number higher than it.
This happens in $4\cdot4$ ways. You can permutate Row # $1$ in $3!$ ways.
Now, take a look at Row $2$ and Row $3$
Because there are $6$ numbers to choose from now, you can assign #'s to Row's #2&3 in
$\frac{\binom{6}{3}\cdot\binom{3}{3}}{2}$ ways. There are $3!\cdot3!$ ways to permute the numbers in the individual Rows.
Hence, our answer is $3!(4\cdot4\cdot3!\cdot{10}\cdot{3!}\cdot{3!})=3!(16\cdot60\cdot36)=3!(34560)\implies{207\boxed{360}$ | null | 360 |
43a39e153ec59d470d8794021eb01299 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$ | We see that if one of the medians is 5, then there are two remaining numbers greater than 5 and two less than 5, so it follows that $m=5$ . There are 3 ways to choose which row to have 5 in, $4\cdot 4=16$ ways to choose the other two numbers in that row, $3!=6$ ways to arrange the numbers in that row, and $6!=720$ ways for the remaining numbers, for our answer is $3\cdot 16\cdot 6\cdot 720=207\boxed{360}$ .
-Stormersyle | null | 360 |
43a39e153ec59d470d8794021eb01299 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$ | We take the grid, and we do a bunch of stuff with it. First, we sort each row, smallest on the left, largest on the right, then we arrange these 3 rows such that the middle #s are increasing, from top to bottom. Thus, we get that the cell in the very center of the grid must be 5. (We need to multiply by 1296 at the end)
Let S mean a number that is < 5, L mean a number > 5.
Bobby.jpg
In the 2 blank corners, one can be S, and the other one has to be L.
If the top right corner is S, there are 16 ways, otherwise, there are 144 ways. (This is left as an exercise to the reader).
Thus, there are 160 * 1296 total configurations, which gets us an answer of $207\boxed{360}$ | null | 360 |
43a39e153ec59d470d8794021eb01299 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$ | We note that if $5$ is a median of one of the rows, then $m=5$ . First, focus on the row with $5$ in it. There are $4^2$ ways to choose the other numbers in that row and then $3!$ ways to order it. Now, clearly, there are $6!$ ways to put the other $6$ numbers into the remaining slots so $Q=6!\cdot3!\cdot4^2\cdot3=207360$ . Hence, our answer is $\boxed{360}$ | null | 360 |
4c9fdf78c52a92e3331ac7599f40057f | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$ | We shall solve this problem by doing casework on the lowest element of the subset. Note that the number $1$ cannot be in the subset because $1*1=1$ . Let $S$ be a product-free set. If the lowest element of $S$ is $2$ , we consider the set $\{3, 6, 9\}$ . We see that 5 of these subsets can be a subset of $S$ $\{3\}$ $\{6\}$ $\{9\}$ $\{6, 9\}$ , and the empty set). Now consider the set $\{5, 10\}$ . We see that 3 of these subsets can be a subset of $S$ $\{5\}$ $\{10\}$ , and the empty set). Note that $4$ cannot be an element of $S$ , because $2$ is. Now consider the set $\{7, 8\}$ . All four of these subsets can be a subset of $S$ . So if the smallest element of $S$ is $2$ , there are $5*3*4=60$ possible such sets.
If the smallest element of $S$ is $3$ , the only restriction we have is that $9$ is not in $S$ . This leaves us $2^6=64$ such sets.
If the smallest element of $S$ is not $2$ or $3$ , then $S$ can be any subset of $\{4, 5, 6, 7, 8, 9, 10\}$ , including the empty set. This gives us $2^7=128$ such subsets.
So our answer is $60+64+128=\boxed{252}$ | null | 252 |
4c9fdf78c52a92e3331ac7599f40057f | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$ | We will consider the $2^9 = 512$ subsets that do not contain 1. A subset is product-free if and only if it does not contain one of the groups $\{2, 4\}, \{3, 9\}, \{2, 3, 6\},$ or $\{2, 5, 10\}$ . There are $2^7$ subsets that contain 2 and 4 since each of the seven elements other than 2 and 4 can either be in the subset or not. Similarly, there are $2^7$ subsets that contain 3 and 9, $2^6$ subsets that contain 2, 3, and 6, and $2^6$ subsets that contain 2, 5, and 10. The number of sets that contain one of the groups is: \[2^7 + 2^7 + 2^6 + 2^6 = 384\] For sets that contain two of the groups, we have: \[2^5 + 2^5 + 2^5 + 2^5 + 2^4 + 2^4 = 160\] For sets that contain three of the groups, we have: \[2^4 + 2^3 + 2^3 + 2^3 = 40\] For sets that contain all of the groups, we have: \[2^2 = 4\] By the principle of inclusion and exclusion, the number of product-free subsets is \[512 - 384 + 160 - 40 + 4 = \boxed{252}\] | null | 252 |
4c9fdf78c52a92e3331ac7599f40057f | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$ | Let $X$ be a product-free subset, and note that 1 is not in $x$ . We consider four cases:
1.) both 2 and 3 are not in $X$ . Then there are $2^7=128$ possible subsets for this case.
2.) 2 is in $X$ , but 3 is not. Then 4 in not in $X$ , so there are $2^6=64$ subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$ , so there are $64\cdot \frac{3}{4}=48$ subsets for this case.
3.) 2 is not in $X$ , but 3 is. Then, 9 is not in $X$ , so there are $2^6=64$ subsets for this case.
4.) 2 and 3 are both in $X$ . Then, 4, 6, and 9 are not in $X$ , so there are $2^4=16$ total subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$ , so there are $16\cdot \frac{3}{4}=12$ subsets for this case.
Hence our answer is $128+48+64+12=\boxed{252}$ . -Stormersyle | null | 252 |
4c9fdf78c52a92e3331ac7599f40057f | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$ | Note that if any element $s\geq6$ makes an invalid set, it must be $c$ of $ab=c$ . Otherwise, $ab\geq12>10$ so no $ab=c$ will suffice. Therefore, whether or not an element $s\geq6$ depends only on the previous elements $1\leq\omega\leq5$
First, if a set is product-free, it must never contain an element $\omega=1$ or $1\cdot1=1$
We do casework on the subset of $S\in{1,2,3,4,5}=s_1$ to determine $S\in{6,7,8,9,10}=s_2$ (There are only 12 to cover so we just list them out):
When it is empty, there are clearly no restrictions for $s_2$ so there are $2^5=32$ cases.
$s_1={2}, s_1={4}, s_1={5}$ has no restrictions $s_2$ so we get $2^5\cdot3=96$ total cases.
$s_1=3$ only cannot contain 9 so there are $2^4=16$ cases.
$s_1={2, 5}; s_1={3, 4}; s_1={3, 5}$ each have one restriction in $s_2$ so there are $2^4\cdot3=48$ ways.
$s_1={2, 3}$ has 2 restrictions (6 and 9) so there are $2^3=8$ ways.
$s_1={4, 5}$ has no restrictions so there are $2^5=32$ ways.
$s_1={2, 3, 5}$ has 3 restrictions (6, 9, and 10) so there are $2^2=4$ ways.
$s_1={3, 4, 5}$ has 1 restriction so there are $2^4=16$ cases.
Summing the numbers, we get $32+96+16+48+8+32+4+16=\boxed{252}$ | null | 252 |
f05f11e812263f689406dae473af8103 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_13 | For every $m \geq 2$ , let $Q(m)$ be the least positive integer with the following property: For every $n \geq Q(m)$ , there is always a perfect cube $k^3$ in the range $n < k^3 \leq m \cdot n$ . Find the remainder when \[\sum_{m = 2}^{2017} Q(m)\] is divided by 1000. | We claim that $Q(m) = 1$ when $m \ge 8$
When $m \ge 8$ , for every $n \ge Q(m) = 1$ , we need to prove there exists an integer $k$ , such that $n < k^3 \le m \cdot n$
Thats because $\sqrt[3]{m \cdot n} - \sqrt[3]{n} \ge 2\sqrt[3]{n} - \sqrt[3]{n} = \sqrt[3]{n} \ge 1$ , so k exists between $\sqrt[3]{m \cdot n}$ and $\sqrt[3]{n}$
$\sqrt[3]{n} < k \le \sqrt[3]{m \cdot n}$
We can then hand evaluate $Q(m)$ for $m = 2,3,4,5,6,7$ , and get $Q(2) = 32$ $Q(3) = 9$ , and all the others equal 2.
There are a total of 2010 integers from 8 to 2017.
\[\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{7} Q(m) + 2010 \equiv 32+9+2+2+2+2+10 = \boxed{059} \mod 1000\] | null | 059 |
48056892c0de1aa11eecf60c6ddef9ae | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_14 | Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$ | The first condition implies
\[a^{128} = \log_a\log_a 2 + \log_a 24 - 128\]
\[128+a^{128} = \log_a\log_a 2^{24}\]
\[a^{a^{128}a^{a^{128}}} = 2^{24}\]
\[\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8\]
So $a^{a^{128}} = 8$
Putting each side to the power of $128$
\[\left(a^{128}\right)^{\left(a^{128}\right)} = 8^{128} = 64^{64},\]
so $a^{128} = 64 \implies a = 2^{\frac{3}{64}}$ . Specifically,
\[\log_a(x) = \frac{\log_2(x)}{\log_2(a)} = \frac{64}{3}\log_2(x)\]
so we have that
\[256 = \log_a(\log_a(x)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[12 = \log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[2^{12} = \frac{64}{3}\log_2(x)\]
\[192 = \log_2(x)\]
\[x = 2^{192}\]
We only wish to find $x\bmod 1000$ . To do this, we note that $x\equiv 0\bmod 8$ and now, by the Chinese Remainder Theorem, wish only to find $x\bmod 125$ . By Euler's Totient Theorem
\[2^{\phi(125)} = 2^{100} \equiv 1\bmod 125\]
so
\[2^{192} \equiv \frac{1}{2^8} \equiv \frac{1}{256} \equiv \frac{1}{6} \bmod 125\]
so we only need to find the inverse of $6 \bmod 125$ . It is easy to realize that $6\cdot 21 = 126 \equiv 1\bmod 125$ , so
\[x\equiv 21\bmod 125, x\equiv 0\bmod 8.\]
Using Chinese Remainder Theorem , we get that $x\equiv \boxed{896}\bmod 1000$ , finishing the solution. | null | 896 |
48056892c0de1aa11eecf60c6ddef9ae | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_14 | Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$ | If you've found $x$ but you don't know that much number theory.
Note $192 = 3 * 2^6$ , so what we can do is take $2^3$ and keep squaring it (mod 1000).
\[2^3 = 8\] \[2^6 = 8*8 = 64\] \[2^{12} = 64*64 \equiv 96\bmod 1000\] \[2^{24} \equiv 96*96 \equiv 216\bmod 1000\] \[2^{48} \equiv 216*216 \equiv 656\bmod 1000\] \[2^{96} \equiv 656*656 \equiv 336\bmod 1000\] \[2^{192} \equiv 336*336 \equiv \boxed{896}\bmod 1000\] | null | 896 |
71512a86da3f43b1ab87764bcda82544 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy] | Lemma: If $x,y$ satisfy $px+qy=1$ , then the minimal value of $\sqrt{x^2+y^2}$ is $\frac{1}{\sqrt{p^2+q^2}}$
Proof: Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}$ . In particular, the distance between the origin and any point $(x,y)$ on the line $px+qy=1$ is at least $\frac{1}{\sqrt{p^2+q^2}}$
Let the vertices of the right triangle be $(0,0),(5,0),(0,2\sqrt{3}),$ and let $(a,0),(0,b)$ be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is $\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)$ . This point must lie on the hypotenuse $\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1$ , i.e. $a,b$ must satisfy \[\frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,\] which can be simplified to \[\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.\]
By the lemma, the minimal value of $\sqrt{a^2+b^2}$ is \[\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},\] so the minimal area of the equilateral triangle is \[\frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},\] and hence the answer is $75+3+67=\boxed{145}$ | null | 145 |
71512a86da3f43b1ab87764bcda82544 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy] | Let $AB=2\sqrt{3}, BC=5$ $D$ lies on $BC$ $F$ lies on $AB$ and $E$ lies on $AC$
Set $D$ as the origin, $BD=a,BF=b$ $F$ can be expressed as $-a+bi$ in argand plane, the distance of $CD$ is $5-a$
We know that $(-a+bi)\cdot\cos(-\frac{\pi}{3})=(-\frac{a+\sqrt{3}b}{2}+\frac{\sqrt{3}a+b}{2}i)$ . We know that the slope of $AC$ is $-\frac{2\sqrt{3}}{5}$ , we have that $\frac{5-a-(-\frac{a+\sqrt{3}b}{2})}{(\sqrt{3}a+b)/2}=\frac{5}{2\sqrt{3}}$ , after computation, we have $11b+7\sqrt{3}a=20\sqrt{3}$
Now the rest is easy with C-S inequality, $(a^2+b^2)(147+121)\geq (7\sqrt{3}a+11a)^2, a^2+b^2\geq \frac{300}{67}$ so the smallest area is $\frac{\sqrt{3}}{4}\cdot \frac{300}{67}=\frac{75\sqrt{3}}{67}$ , and the answer is $\boxed{145}$ | null | 145 |
71512a86da3f43b1ab87764bcda82544 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy] | Let $\triangle ABC$ be the right triangle with sides $AB = x$ $AC = y$ , and $BC = z$ and right angle at $A$
Let an equilateral triangle touch $AB$ $AC$ , and $BC$ at $D$ $E$ , and $F$ respectively, having side lengths of $c$
Now, call $AD$ as $a$ and $AE$ as $b$ . Thus, $DB = x-a$ and $EC = y-b$
By Law of Sines on triangles $\triangle DBF$ and $ECF$
$BF = \frac{z(a\sqrt{3}+b)} {2y}$ and $CF = \frac{z(a+b\sqrt{3})} {2x}$
Summing,
$BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z$
Now substituting $AB = x = 2\sqrt{3}$ $AC = y = 5$ , and $BC = \sqrt{37}$ and solving, $\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1$
We seek to minimize $[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}$
This is equivalent to minimizing $a^2+b^2$
Using the lemma from solution 1, we conclude that $\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}$
Thus, $[DEF] = \frac{75\sqrt{3}}{67}$ and our final answer is $\boxed{145}$ | null | 145 |
71512a86da3f43b1ab87764bcda82544 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy] | We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are $5$ and $2\sqrt{3}i$ , respectively. Now let the vertex of the equilateral triangle on the real axis be $a$ and let the vertex of the equilateral triangle on the imaginary axis be $bi$ . Then, the third vertex of the equilateral triangle is given by: \[(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{b}{2})i\]
For this to be on the hypotenuse of the right triangle, we also have the following: \[\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}\]
Note that the area of the equilateral triangle is given by $\frac{\sqrt{3}(a^2+b^2)}{4}$ , so we seek to minimize $a^2+b^2$ . This can be done by using the Cauchy Schwarz Inequality on the relation we derived above: \[1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}\]
Thus, the minimum we seek is simply $\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}$ , so the desired answer is $\boxed{145}$ | null | 145 |
71512a86da3f43b1ab87764bcda82544 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy] | We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be $a$ and the point on the imaginary axis be $bi$ . Then, we see that $(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).$ Now we switch back to Cartesian coordinates. The equation of the hypotenuse is $y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.$ This means that the point $\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)$ is on the line. Plugging the numbers in, we have $\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.$ Now, we note that the side length of the equilateral triangle is $a^2+b^2$ so it suffices to minimize that. By Cauchy-Schwarz, we have $(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.$ Thus, the area of the smallest triangle is $\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}$ so our desired answer is $\boxed{145}$ | null | 145 |
71512a86da3f43b1ab87764bcda82544 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy] | Let $S$ be the triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37}$
We will think about this problem backwards, by constructing a triangle as large as possible (We will call it $T$ , for convenience) which is similar to $S$ with vertices outside of a unit equilateral triangle $\triangle ABC$ , such that each vertex of the equilateral triangle lies on a side of $T$ . After we find the side lengths of $T$ , we will use ratios to trace back towards the original problem.
First of all, let $\theta = 90^{\circ}$ $\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)$ , and $\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)$ (These three angles are simply the angles of triangle $S$ ; out of these three angles, $\alpha$ is the smallest angle, and $\theta$ is the largest angle). Then let us consider a point $P$ inside $\triangle ABC$ such that $\angle APB = 180^{\circ} - \theta$ $\angle BPC = 180^{\circ} - \alpha$ , and $\angle APC = 180^{\circ} - \beta$ . Construct the circumcircles $\omega_{AB}, ~\omega_{BC},$ and $\omega_{AC}$ of triangles $APB, ~BPC,$ and $APC$ respectively.
From here, we will prove the lemma that if we choose points $X$ $Y$ , and $Z$ on circumcircles $\omega_{AB}, ~\omega_{BC},$ and $\omega_{AC}$ respectively such that $X$ $B$ , and $Y$ are collinear and $Y$ $C$ , and $Z$ are collinear, then $Z$ $A$ , and $X$ must be collinear. First of all, if we let $\angle PAX = m$ , then $\angle PBX = 180^{\circ} - m$ (by the properties of cyclic quadrilaterals), $\angle PBY = m$ (by adjacent angles), $\angle PCY = 180^{\circ} - m$ (by cyclic quadrilaterals), $\angle PCZ = m$ (adjacent angles), and $\angle PAZ = 180^{\circ} - m$ (cyclic quadrilaterals). Since $\angle PAX$ and $\angle PAZ$ are supplementary, $Z$ $A$ , and $X$ are collinear as desired. Hence, $\triangle XYZ$ has an inscribed equilateral triangle $ABC$
In addition, now we know that all triangles $XYZ$ (as described above) must be similar to triangle $S$ , as $\angle AXB = \theta$ and $\angle BYC = \alpha$ , so we have developed $AA$ similarity between the two triangles. Thus, $\triangle XYZ$ is the triangle similar to $S$ which we were desiring. Our goal now is to maximize the length of $XY$ , in order to maximize the area of $XYZ$ , to achieve our original goal.
Note that, all triangles $PYX$ are similar to each other if $Y$ $B$ , and $X$ are collinear. This is because $\angle PYB$ is constant, and $\angle PXB$ is also a constant value. Then we have $AA$ similarity between this set of triangles. To maximize $XY$ , we can instead maximize $PY$ , which is simply the diameter of $\omega_{BC}$ . From there, we can determine that $\angle PBY = 90^{\circ}$ , and with similar logic, $PA$ $PB$ , and $PC$ are perpendicular to $ZX$ $XY$ , and $YZ$ respectively We have found our desired largest possible triangle $T$
All we have to do now is to calculate $YZ$ , and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within $S$ . First of all, we will prove that $\angle ZPY = \angle ACB + \angle AXB$ . By the properties of cyclic quadrilaterals, $\angle AXB = \angle PAB + \angle PBA$ , which means that $\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC$ . Now we will show that $\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC$ . Note that, by cyclic quadrilaterals, $\angle YZP = \angle PAC$ and $\angle ZYP = \angle PBC$ . Hence, $\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC$ (since $\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}$ ), proving the aforementioned claim. Then, since $\angle ACB = 60^{\circ}$ and $\angle AXB = \theta = 90^{\circ}$ $\angle ZPY = 150^{\circ}$
Now we calculate $PY$ and $PZ$ , which are simply the diameters of circumcircles $\omega_{BC}$ and $\omega_{AC}$ , respectively. By the extended law of sines, $PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}$ and $PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}$
We can now solve for $ZY$ with the law of cosines:
\[(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)\]
\[(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}\]
\[(ZY)^2 = \frac{37 \cdot 67}{300}\]
\[ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}\]
Now we will apply this discovery towards our original triangle $S$ . Since the ratio between $ZY$ and the hypotenuse of $S$ is $\frac{\sqrt{67}}{10\sqrt{3}}$ , the side length of the equilateral triangle inscribed within $S$ must be $\frac{10\sqrt{3}}{\sqrt{67}}$ (as $S$ is simply as scaled version of $XYZ$ , and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within $S$ is $\frac{75\sqrt{3}}{67}$ , implying that the answer is $\boxed{145}$ | null | 145 |
71512a86da3f43b1ab87764bcda82544 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy] | Let the right triangle's lower-left point be at $O(0,0)$ . Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the $x$ -axis ( $B$ ) and the $y$ -axis ( $A$ ) and label them $(b, 0)$ and $(0, a)$ respectively. The third point ( $C$ ) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of $a$ and $b$
1. Find the slope of $AB$ and take the negative reciprocal of it to find the slope of the line containing $C$ . Notice the line contains the midpoint of $AB$ so we can then have an equation of the line.
2. Let $AB=x.$ For $ABC$ to be an equilateral triangle, the altitude from $C$ to $AB$ must be $\frac{x\sqrt{3}}{2}.$
We then have two equations and two variables, so we can solve for $C$ 's coordinates.
We can find $C(\frac{a+b\sqrt{3}}{2}), (\frac{b+a\sqrt{3}}{2}).$ Also, note that $C$ must be on the hypotenuse of the triangle $\frac{x}{5}+\frac{y}{2\sqrt{3}}=1.$ We can plug in $x$ and $y$ as the coordinates of $C$ , which simplifies to
\[11b+7\sqrt{3}a=20\sqrt{3}.\]
We aim to minimize the side length of the triangle, which is $\sqrt{a^2+b^2}.$ Applying the Cauchy inequality gives us
\[(a^2+b^2)(7\sqrt{3}^2+11^2)\geq (11b+7\sqrt3a)^2 = 1200\]
From which we obtain $\sqrt{a^2+b^2} \geq \sqrt{\frac{300}{67}}.$ Thus, the area of the triangle = $\frac{75\sqrt{3}}{67}$ which leads to the answer $75+3+67=\boxed{145}.$ | null | 145 |
f5414e0f05138cc4876bb0d0d27340ae | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ | The number of subsets of a set with $n$ elements is $2^n$ . The total number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ is equal to $2^8$ . The number of sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ can be found using complementary counting. There are $2^5$ subsets of $\{1, 2, 3, 4, 5\}$ and $2^5$ subsets of $\{4, 5, 6, 7, 8\}$ . It is easy to make the mistake of assuming there are $2^5+2^5$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , but the $2^2$ subsets of $\{4, 5\}$ are overcounted. There are $2^5+2^5-2^2$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , so there are $2^8-(2^5+2^5-2^2)$ subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ $2^8-(2^5+2^5-2^2)=\boxed{196}$ | null | 196 |
f5414e0f05138cc4876bb0d0d27340ae | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ | Upon inspection, a viable set must contain at least one element from both of the sets $\{1, 2, 3, 4, 5\}$ and $\{4, 5, 6, 7, 8\}$ . Since 4 and 5 are included in both of these sets, then they basically don't matter, i.e. if set A is a subset of both of those two then adding a 4 or a 5 won't change that fact. Thus, we can count the number of ways to choose at least one number from 1 to 3 and at least one number from 6 to 8, and then multiply that by the number of ways to add in 4 and 5. The number of subsets of a 3 element set is $2^3=8$ , but we want to exclude the empty set, giving us 7 ways to choose from $\{1, 2, 3\}$ or $\{6, 7, 8\}$ . We can take each of these $7 \times 7=49$ sets and add in a 4 and/or a 5, which can be done in 4 different ways (by adding both, none, one, or the other one). Thus, the answer is $49 \times 4=\boxed{196}$ | null | 196 |
f5414e0f05138cc4876bb0d0d27340ae | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ | This solution is very similar to Solution $2$ . The set of all subsets of $\{1,2,3,4,5,6,7,8\}$ that are disjoint with respect to $\{4,5\}$ and are not disjoint with respect to the complements of sets (and therefore not a subset of) $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ will be named $S$ , which has $7\cdot7=49$ members. The union of each member in $S$ and the $2^2=4$ subsets of $\{4,5\}$ will be the members of set $Z$ , which has $49\cdot4=\boxed{196}$ members. $\blacksquare$ | null | 196 |
f5414e0f05138cc4876bb0d0d27340ae | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ | Consider that we are trying to figure out how many subsets are possible of $\{1,2,3,4,5,6,7,8\}$ that are not in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ . Assume that the number of numbers we pick from the subset $\{1,2,3,4,5,6,7,8\}$ is $n$ . Thus, we can compute this problem with simple combinatorics:
If $n=1$ $\binom{8}{1}$ $-$ $\binom{5}{1}$ $\binom{5}{1}$ $- 2$ ) [subtract $2$ to eliminate the overcounting of the subset $\{4\}$ or $\{5\}$ ] = $8 - 8$ $0$
If $n=2$ $\binom{8}{2}$ $-$ $\binom{5}{2}$ $\binom{5}{2}$ $- 1$ ) [subtract $1$ to eliminate the overcounting of the subset $\{4,5\}$ ] = $28 - 19$ $9$
If $n=3$ $\binom{8}{3}$ $-$ $\binom{5}{3}$ $\binom{5}{3}$ ) = $56 - 20$ $36$
If $n=4$ $\binom{8}{4}$ $-$ $\binom{5}{4}$ $\binom{5}{4}$ ) = $70 - 10$ $60$
If $n=5$ $\binom{8}{5}$ $-$ $\binom{5}{5}$ $\binom{5}{5}$ ) = $56 - 2$ $54$
If $n>5$ , then the set $\{1,2,3,4,5,6,7,8\}$ is never in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ . Thus,
If $n=6$ $\binom{8}{6}$ $28$
If $n=7$ $\binom{8}{7}$ $8$
If $n=8$ $\binom{8}{8}$ $1$
Adding these together, our solution becomes $0$ $9$ $36$ $60$ $54$ $28$ $8$ $1$ $=\boxed{196}$ | null | 196 |
7f008588289f7d842aacb11afdf55504 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_2 | The teams $T_1$ $T_2$ $T_3$ , and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$ , and $T_2$ plays $T_3$ . The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$ , the probability that $T_i$ wins is $\frac{i}{i+j}$ , and the outcomes of all the matches are independent. The probability that $T_4$ will be the champion is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | There are two scenarios in which $T_4$ wins. The first scenario is where $T_4$ beats $T_1$ $T_3$ beats $T_2$ , and $T_4$ beats $T_3$ , and the second scenario is where $T_4$ beats $T_1$ $T_2$ beats $T_3$ , and $T_4$ beats $T_2$ . Consider the first scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{4+1}$ , the probability $T_3$ beats $T_2$ is $\frac{3}{3+2}$ , and the probability $T_4$ beats $T_3$ is $\frac{4}{4+3}$ . Therefore the first scenario happens with probability $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}$ . Consider the second scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{1+4}$ , the probability $T_2$ beats $T_3$ is $\frac{2}{2+3}$ , and the probability $T_4$ beats $T_2$ is $\frac{4}{4+2}$ . Therefore the second scenario happens with probability $\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$ . By summing these two probabilities, the probability that $T_4$ wins is $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}+\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$ . Because this expression is equal to $\frac{256}{525}$ , the answer is $256+525=\boxed{781}$ | null | 781 |
a0e98681ac55311f8dd3d3cfb45ee7db | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | A triangle has vertices $A(0,0)$ $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | [asy] pair A,B,C,D,X,Z,P; A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4); fill(B--X--P--Z--cycle,lightgray); draw(A--B--C--cycle); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); draw(X--P,dashed); draw(Z--P,dashed); dot(X); label("$X$",X,NE); dot(Z); label("$Z$",Z,S); dot(P); label("$P$",P,NW); [/asy]
The set of all points closer to point $B$ than to point $A$ lie to the right of the perpendicular bisector of $AB$ (line $PZ$ in the diagram), and the set of all points closer to point $B$ than to point $C$ lie below the perpendicular bisector of $BC$ (line $PX$ in the diagram). Therefore, the set of points inside the triangle that are closer to $B$ than to either vertex $A$ or vertex $C$ is bounded by quadrilateral $BXPZ$ . Because $X$ is the midpoint of $BC$ and $Z$ is the midpoint of $AB$ $X=(10,5)$ and $Z=(6,0)$ . The coordinates of point $P$ is the solution to the system of equations defined by lines $PX$ and $PZ$ . Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope $m$ is $-\frac{1}{m}$ , the equation for line $PX$ is $y=\frac{2}{5}x+1$ and the equation for line $PZ$ is $x=6$ . The solution of this system is $P=\left(6,\frac{17}{5}\right)$ . Using the shoelace formula on quadrilateral $BXPZ$ and triangle $ABC$ , the area of quadrilateral $BXPZ$ is $\frac{109}{5}$ and the area of triangle $ABC$ is $60$ . Finally, the probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to vertex $A$ or vertex $C$ is the ratio of the area of quadrilateral $BXPZ$ to the area of $ABC$ , which is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$ . The answer is $109+300=\boxed{409}$ | null | 409 |
a0e98681ac55311f8dd3d3cfb45ee7db | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | A triangle has vertices $A(0,0)$ $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: $AB=12$ $AC=2\sqrt{41}$ , and $BC=2\sqrt{29}$ . We notice that the point where the three distances are the same is the circumcenter - so we use one of the triangle area formulas to find the circumradius, since we know what the area is. \[\frac{12 \cdot 2\sqrt{41} \cdot 2\sqrt{29}}{4 \cdot R}=\frac{12 \cdot 10}{2}.\] We rearrange to get \[R=\frac{\sqrt{41} \cdot \sqrt{29}}{5}.\] [asy] draw((0,0)--(12,0)--(8,10)--(0,0)); draw((6,0)--(6,3.4)--(10,5)); draw((6,3.4)--(4,5)); label("$A$", (0,0), SW); label("$B$", (12,0), SE); label("$C$", (8, 10), N); label("$P$", (6, 3.4), NNE); label("$R$", (10, 5), NE); label("$S$", (6, 0), S); label("$T$", (4, 5), NW); [/asy] We know that $AP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$ , and $AS=6$ , so using the Pythagorean Theorem gives $SP=\frac{17}{5}$ . This means $[ASP]=[BSP]=\frac{17}{5} \cdot 6 \cdot \frac{1}{2} = \frac{51}{5}$ .
Similarly, we know that $BP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$ , and $BR=\sqrt{29}$ , so we get that $PR=\frac{4\sqrt{29}}{5}$ , and so $[BRP]=[CRP]=\frac{4\sqrt{29}}{5} \cdot \sqrt{29} \cdot \frac{1}{2} = \frac{58}{5}$ .
Lastly, we know that $CP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$ , and $CT=\sqrt{41}$ , so we get that $PT=\frac{2\sqrt{41}}{5}$ , and $[ATP]=[CTP]=\frac{2\sqrt{41}}{5} \cdot \sqrt{41} \cdot \frac{1}{2} = \frac{41}{5}$ .
Therefore, our answer is $\frac{51+58}{2(51+58+41)}=\frac{109}{300} \rightarrow \boxed{409}$ | null | 409 |
a0e98681ac55311f8dd3d3cfb45ee7db | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | A triangle has vertices $A(0,0)$ $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | To start the problem, identify the two midpoints that connect $AB$ and $BC$ . This is because the midpoints of such lines is the mark at which the point will sway closer to vertex $A$ $C$ or vertex $B$ . The midpoint of $AB$ is $(6,0)$ , and the midpoint of $BC$ is $(10,5)$ . Then, determine the line at which the distance between vertex $B$ and vertex $C$ are the same. Assuming that $x$ is the real value of $x$ and $y$ is the real value of $y$ , we can create a simple equation:
$\sqrt{|x-8|^2 + (10-y)^2}$ $\sqrt{(12-x)^2 + y^2}$
where the left side of the equation is for the distance to vertex $C$ and the right side of the equation is the distance to vertex $B$
Squaring both sides and then distributing, we get
$x^2 - 16x + 64 + y^2 - 20y + 100 = x^2 - 24x + 144 + y^2$
Notice that $(x-8)^2 = (-x+8)^2$ , and thus there is no need to create another equation.
Simplifying, we get $8x + 20 = 20y$
Divide both sides by 20, then simplify, and the line that represents equivalent distance between vertex $B$ and vertex $C$ is $y = \frac{2x}{5} + 1$
This line starts at the midpoint of $BC$ , which is $(10,5)$ , and ends at the line $x=6$ , as $x=6$ represents equivalent distance between vertex $A$ and vertex $B$ . Plug in $x=6$ to the equation $y = \frac{2x}{5} + 1$ , and we get $y = \frac{17}{5}$ . Now that we have our four points that are $(6,0), (6,\frac{17}{5}), (10,5)$ , and $(12,0)$ , we can calculate the area of the quadrilateral in which a point is closer to vertex $B$ as opposed to either vertex $A$ or vertex $C$ . Simply draw a rectangle that has the points $(6,0), (6,5), (12,5)$ and $(12,0)$ , and then subtract the two triangles that appear in between.
Thus, the area of the quadrilateral is $6\cdot5 - (\frac{\frac{8}{5}\cdot4}{2} + \frac{5\cdot2}{2}) \rightarrow 30 - \frac{41}{5} = \frac{109}{5}$ .
Since the problem asks us for the probability that a point chosen inside the triangle is inside the quadrilateral, and because the area of $\triangle ABC$ is $\frac{12\cdot10}{2} = 60$ , the probability is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$ , which means the final answer is $109+300=\boxed{409}$ | null | 409 |
a0e98681ac55311f8dd3d3cfb45ee7db | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | A triangle has vertices $A(0,0)$ $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Calculate the area of the triangle using the Shoelace Theorem on $(0,0), (12,0), (8,10)$ \[\frac{1}{2}|(0+120+0)-(0+0+0)|=60\]
Get the four points $(6,0), (6,\frac{17}{5}), (10,5)$ , and $(12,0)$ by any method from the above solutions. Then use the Shoelace Theorem to find the area of the region we want: \[\frac{1}{2}|(0+60+34+0)-(0+0+30+\frac{102}{5})|=\frac{109}{5}\]
Therefore the probability is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$ . Thus giving the final answer of $109+300=\boxed{409}$ | null | 409 |
a0e98681ac55311f8dd3d3cfb45ee7db | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | A triangle has vertices $A(0,0)$ $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Draw the circumradii from the circumcenter to the three vertices. Drop perpendicular from the circumcenter to the sides. Note that since the triangle is isosceles, the perpendicular are in fact perpendicular bisectors. Therefore the region containing the points closer to B are in $\triangle{OBP_{1}} , \triangle{OBP_{2}}$ where $O$ is the circumcenter and $P_{1}, P_{2}$ points of contact of the perpendiculars and the sides. Therefore our fraction is $\frac{109}{300}$ . Our answer is then $\boxed{409}$ | null | 409 |
a0e98681ac55311f8dd3d3cfb45ee7db | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | A triangle has vertices $A(0,0)$ $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Using the same graph and methods as Solution 1, find the coordinates of $P$ $X$ $B$ , and $Z$ . Also, note that angles $PXB$ and $PZB$ are right angles, so $PXBZ$ is a cyclic quadrilateral. Then, use Brahmagupta's formula to determine the area of the quadrilateral, which is $\frac{109}{5}$ . Then find the area of triangle $ABC$ , which is $60$ . The probability is $\frac{109}{300}$ . Finally, we get our answer of $\boxed{409}$ | null | 409 |
118bdb1582b47cf4540faaeeea5b0aad | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_4 | Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$ | The base- $3$ representation of $2017_{10}$ is $2202201_3$ . Because any $7$ -digit base- $3$ number that starts with $22$ and has no digit equal to $0$ must be greater than $2017_{10}$ , all $7$ -digit numbers that have no digit equal to $0$ must start with $21$ or $1$ in base $3$ . Of the base- $3$ numbers that have no digit equal to $0$ , there are $2^5$ $7$ -digit numbers that start with $21$ $2^6$ $7$ -digit numbers that start with $1$ $2^6$ $6$ -digit numbers, $2^5$ $5$ -digit numbers, $2^4$ $4$ -digit numbers, $2^3$ $3$ -digit numbers, $2^2$ $2$ -digit numbers, and $2^1$ $1$ -digit numbers. Summing these up, we find that the answer is $2^5+2^6+2^6+2^5+2^4+2^3+2^2+2^1=\boxed{222}$ | null | 222 |
118bdb1582b47cf4540faaeeea5b0aad | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_4 | Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$ | Note that $2017=2202201_{3}$ , and $2187=3^7=10000000_{3}$ . There can be a $1,2,...,7$ digit number less than $2187$ , and each digit can either be $1$ or $2$ . So $2^1$ one digit numbers and so on up to $2^7$ $7$ digit.
Now we have to subtract out numbers from $2018$ to $2187$
Then either the number must begin $221...$ or $222...$ with four more digits at the end
Using $1$ s and $2$ s there are $2^4$ options for each so:
$2+4+8+16+32+64+128-2*16=256-2-32=\boxed{222}$ | null | 222 |
118bdb1582b47cf4540faaeeea5b0aad | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_4 | Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$ | Since the greatest power of $3$ that can be used is $3^6$ , we can do these cases.
Coefficient of $3^6=0$ : Then if the number has only $3^0$ , it has 2 choices (1 or 2). Likewise if the number has both a $3^1$ and $3^0$ term, there are 4 choices, and so on until $3^5$ , so the sum is $2+4+...+64=127-1=126$
Coefficient of $3^6=1$ : Any combination of $1$ or $2$ for the remaining coefficients works, so $2^6=64$ . Why is this less than $126$ ? Because the first time around, leading zeroes didn't count. But this time, all coefficients $3^n$ of $n<6$ need 1 and 2.
Coefficient of $3^6=2$ : Look at $3^5$ coefficient. If 1, all of them work because $3^7=2187-3^5=243<<2017$ . That's 32 cases. Now of this coefficient is 2, then at the coefficient of $3^4=81$ is at least 1. However, $3^6*2+3^5*2+3^4>>2017$ , so our answer is $126+64+32=\boxed{222}$ | null | 222 |
07c9f7b23681e2a0c1c585c701a025ad | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_5 | A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189$ $320$ $287$ $234$ $x$ , and $y$ . Find the greatest possible value of $x+y$ | Let these four numbers be $a$ $b$ $c$ , and $d$ , where $a>b>c>d$ $x+y$ needs to be maximized, so let $x=a+b$ and $y=a+c$ because these are the two largest pairwise sums. Now $x+y=2a+b+c$ needs to be maximized. Notice that $2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))$ . No matter how the numbers $189$ $320$ $287$ , and $234$ are assigned to the values $a+d$ $b+c$ $b+d$ , and $c+d$ , the sum $(a+d)+(b+c)+(b+d)+(c+d)$ will always be $189+320+287+234$ . Therefore we need to maximize $3((a+c)+(b+d))-(189+320+287+234)$ . The maximum value of $(a+c)+(b+d)$ is achieved when we let $a+c$ and $b+d$ be $320$ and $287$ because these are the two largest pairwise sums besides $x$ and $y$ . Therefore, the maximum possible value of $x+y=3(320+287)-(189+320+287+234)=\boxed{791}$ | null | 791 |
07c9f7b23681e2a0c1c585c701a025ad | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_5 | A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189$ $320$ $287$ $234$ $x$ , and $y$ . Find the greatest possible value of $x+y$ | Let the four numbers be $a$ $b$ $c$ , and $d$ , in no particular order. Adding the pairwise sums, we have $3a+3b+3c+3d=1030+x+y$ , so $x+y=3(a+b+c+d)-1030$ . Since we want to maximize $x+y$ , we must maximize $a+b+c+d$
Of the four sums whose values we know, there must be two sums that add to $a+b+c+d$ . To maximize this value, we choose the highest pairwise sums, $320$ and $287$ . Therefore, $a+b+c+d=320+287=607$
We can substitute this value into the earlier equation to find that $x+y=3(607)-1030=1821-1030=\boxed{791}$ | null | 791 |
07c9f7b23681e2a0c1c585c701a025ad | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_5 | A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189$ $320$ $287$ $234$ $x$ , and $y$ . Find the greatest possible value of $x+y$ | There are two cases we can consider. Let the elements of our set be denoted $a,b,c,d$ , and say that the largest sums $x$ and $y$ will be consisted of $b+d$ and $c+d$ . Thus, we want to maximize $b+c+2d$ , which means $d$ has to be as large as possible, and $a$ has to be as small as possible to maximize $b$ and $c$ . So, the two cases we look at are:
Case 1:
\[a+d = 287\] \[b+c = 320\] \[a+b = 234\] \[a+c = 189\]
Case 2:
\[a+d = 320\] \[b+c = 189\] \[a+b = 234\] \[a+c = 287\]
Note we have determined these cases by maximizing the value of $a+d$ determined by our previous conditions. So, the answers for each ( after some simple substitution ) will be:
Case 1:
\[(a,b,c,d) = (\frac{103}{2},\frac{365}{2},\frac{275}{2},\frac{471}{2})\]
Case 2:
\[(a,b,c,d) = (166,68,121,154)\]
See the first case has our largest $d$ , so our answer will be $471+\frac{640}{2} = \boxed{791}$ | null | 791 |
cfe0031d219a0e512030c666d5756781 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | Manipulating the given expression, $\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}$ . The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$ . Rearranging, $s^2-(2n+85)^2=843$ . By difference of squares, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$ . It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$ . The two values of $n$ that satisfy one of the equations are $168$ and $27$ . Summing these together gives us the answer ; $168+27=\boxed{195}$ | null | 195 |
cfe0031d219a0e512030c666d5756781 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | Clearly, the result when $n$ is plugged into the given expression is larger than $n$ itself. Let $x$ be the positive difference between that result and $n$ , so that $\sqrt{n^2+85n+2017}=n+x$ . Squaring both sides and canceling the $n^2$ terms gives $85n+2017=2xn+x^2$ . Combining like terms, $(85-2x)n=x^2-2017$ , so
\[n=\frac{x^2-2017}{85-2x}.\]
Since $n$ is positive, there are two cases, which are simple (luckily). Remembering that $x$ is a positive integer, then $x^2-2017$ and $85-2x$ are either both positive or both negative. The smallest value for which $x^2>2017$ is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that $x<45$ (from the numerator) and $85-2x<0$ , which means $x>42$ . This only gives two solutions, $x=43, 44$ . Plugging these into the expression for $n$ , we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is $168+27=\boxed{195}$ | null | 195 |
cfe0031d219a0e512030c666d5756781 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | Let the integer given by the square root be represented by $x$ . Then $0 = n^2 + 85n + 2017 - x^2$ . For this to have rational solutions for $n$ (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)
Thus, $b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2$ for some integer $y$ . Then $4x^2 - 843 = y^2$ . Rearranging this equation yields that $843 = (2x+y)(2x-y)$ . Noticing that there are 2 factor pairs of $843$ , namely, $1*843$ and $3*281$ , there are 2 systems to solve for $x$ and $y$ that create rational $n$ . These yield solutions $(x,y)$ of $(211, 421)$ and $(71, 139)$
The solution to the initial quadratic in $n$ must then be $\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}$ . Noticing that for each value of $x$ that has rational solutions for $n$ , the corresponding value of the square root of the discriminant is $y$ , the formula can be rewritten as $n = \frac{-85 \pm y}{2}$ . One solution is $\frac{421 - 85}{2} = 168$ and the other solution is $\frac{139 - 85}{2} = 27$ . Thus the answer is $168 + 27 = \boxed{195}$ as both rational solutions are integers. | null | 195 |
cfe0031d219a0e512030c666d5756781 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | Notice that $(n+42)^2= n^2+84n+1764$ . Also note that $(n+45)^2= n^2+90n+2025$ . Thus, \[(n+42)^2< n^2+85n+2017<(n+45)^2\] where $n^2+85n+2017$ is a perfect square. Hence, \[n^2+85n+2017= (n+43)^2\] or \[n^2+85n+2017= (n+44)^2.\] Solving the two equations yields the two solutions $n= 168, 27$ . Therefore, our answer is $\boxed{195}$ | null | 195 |
cfe0031d219a0e512030c666d5756781 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | Let the expression be equal to $a$ . This expression can be factored into $\sqrt{(n+44)^2-3n+81}$ . Then square both sides, and the expression becomes $(n+44)^2-3n+81=a^2$ . We have a difference of two squares. Rearranging, we have $(n+44+a)(n+44-a)=3(n-27)$ . By inspection, the only possible values for $(n+44-a)$ are 0 and 1. When $(n+44-a)=0$ , we must have $n-27=0$ . Therefore, $27$ is a solution. When we have $(n+44-a)=1$ , so $n=a-43$ . Plugging this back to $(n+44+a)=3(n-27)$ (since $(n+44-a)=1$ ), we find that $a=211 \implies n=168$ . Thus, the answer is $27+168= \boxed{195}$ | null | 195 |
cfe0031d219a0e512030c666d5756781 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | More intuitive, but a little bit slower considering the decimals.
Label the entire given expression as $k^2$
Instinctively we can do a crude completion of the square, resulting in $k^2$ $(n+42.5)^2+210.75$ Rearrange the equation to get a difference of squares.
$k^2-(n+42.5)^2 = 210.75$
$(k+n+42.5)(k-n-42.5) = 210.75$
Factor $21,075$ to get $3^1$ $5^2$ , and $281^1$
Now the two factors given are either divided by 10 each or one being divided by 100. Let's start with the former case.
If you try $281*3/10$ and $5*5/10$ , you quickly realize that $n$ becomes negative. Naturally, you will realize you want the number's difference to be larger. Try $281*5/10$ and $3*5/10$ . This gives an answer of $27$ for $x$ . The next largest possibility also works, giving an $n$ of $168$ . As you rise, some numbers don't work because it results in an n that is not an integer, as in the example of $\frac{281*5*5}{10}$ and $\frac{3}{10}$
Now if you continue on with the next case, where one factor is divided by $100$ , very swiftly will you realize most don't work simply because the difference is too small, or it doesn't give an integer. It helps a lot when you realize that the decimal does not end in a $5$ , the answer will not be an integer. After a few short tests, we get $168+27=\boxed{195}$ | null | 195 |
01c45b1c4b18eb2b052389d740631b3c | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_7 | Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution. | [asy] Label f; f.p=fontsize(5); xaxis(-3,3,Ticks(f,1.0)); yaxis(-3,26,Ticks(f,1.0)); real f(real x){return (x+2)^2;} real g(real x){return x*-1;} real h(real x){return x*-2;} real i(real x){return x*-3;} real j(real x){return x*8;} draw(graph(f,-2,3),green); draw(graph(g,-2,2),red); draw(graph(h,-2,1),red); draw(graph(i,-2,1/3),red); draw(graph(j,-0.25,3),red); [/asy] Note the equation $\log(kx)=2\log(x+2)$ is valid for $kx>0$ and $x>-2$ $\log(kx)=2\log(x+2)=\log((x+2)^2)$ . The equation $kx=(x+2)^2$ is derived by taking away the outside logs from the previous equation. Because $(x+2)^2$ is always non-negative, $kx$ must also be non-negative; therefore this takes care of the $kx>0$ condition as long as $k\neq0$ , i.e. $k$ cannot be $0$ . Now, we graph both $(x+2)^2$ (the green graph) and $kx$ (the red graph for $k=-1,k=-2,k=-3,k=8$ ) for $x>-2$ . It is easy to see that all negative values of $k$ make the equation $\log(kx)=2\log(x+2)$ have only one solution. However, there is also one positive value of $k$ that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation $(x+2)^2=kx$ to be $0$ and solving for $k$ . Therefore, there are $500$ negative solutions and $1$ positive solution, for a total of $\boxed{501}$ | null | 501 |
01c45b1c4b18eb2b052389d740631b3c | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_7 | Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution. | We use an algebraic approach. Since $\log(kx)=2\log(x+2)$ , then $kx = (x+2)^2$ (the converse isn't necessarily true!), or $x^2+(4-k)x+4=0$ . Our original equation has exactly one solution if and only if there is only one solution to the above equation, or one of the solutions is extraneous; it involves the computation of the log of a nonpositive number.
For the first case, we note that this can only occur when it is a perfect square trinomal, or $k = 0, 8$ . However, $k = 0$ results in $\log(0)$ on the LHS, which is invalid. $k = 8$ yields $x = 2$ , so that is one solution.
For the second case, we can use the quadratic formula. We have \[x = \frac{k-4 \pm \sqrt{k^2-8k}}2,\] so in order for there to be at least one real solution, the discriminant must be nonnegative, or $k < 0$ or $k > 8$
Note that if $k > 8$ , then both solutions to $x$ will be positive, and therefore both valid, which means all such $k$ are unsatisfactory.
We now wish to show that if $k < 0$ , then there is exactly one solution that works. Note that whenever $k < 0$ , both "solutions" in $x$ are negative. One of the solutions to the equation is $x = \frac{k-4 + \sqrt{k^2-8k}}2$ . We wish to prove that $x + 2 > 0$ , or $x > -2$ (therefore the RHS in the original equation will be defined). Substituting, we have $\frac{k-4 + \sqrt{k^2-8k}}2 > -2$ , or $\sqrt{k^2 - 8k} > -k$ . Since both sides are positive, we can square both sides (if $k < 0$ , then $-k > 0$ ) to get $k^2-8k > k^2$ , or $8k < 0 \implies k < 0$ , which was our original assumption, so this solution satisfies the original equation. The other case is when $x = \frac{k-4 - \sqrt{k^2-8k}}2$ , which we wish to show is less that $-2$ , or $\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}$ . However, since the square root is defined to be positive, then this is always true, which implies that whenever $k < 0$ , there is exactly one real solution that satisfies the original equation. Combining this with $k \in [-500, 500]$ , we find that the answer is $500 + 1 = \boxed{501}$ | null | 501 |
01c45b1c4b18eb2b052389d740631b3c | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_7 | Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution. | Immediately we notice $k$ is non-zero, in fact we must have $kx, (x+2) > 0$ for our sole solution $x = x_0$ . Simplifying the logarithmic equation we get $kx = (x+2)^2 \rightarrow 0 = x^2 + (4-k)x + 4 \rightarrow x = \frac{k-4 \pm \sqrt{k^2 - 8k}}{2}$ . Then $k \leq 0$ or $k \geq 8$ . When $k = 8$ we have exactly one real solution (easily verifiable). Notice when $k > 8$ , both solutions to $x$ are positive, and so all such $k$ are not satisfactory. When $k < 0$ it can be shown that the greater solution to $x$ is in the interval $(0,-2)$ and the lesser solution is in the interval $(-2,-\infty)$ which is satisfactory. Then $k = 8,-1,-2,...,-500 \rightarrow \boxed{501}$ satisfactory integer values of $k$ | null | 501 |
b35fadbd7422993eb2821700404edc00 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer. | We start with the last two terms of the polynomial $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}$ , which are $\frac{n^5}{5!}+\frac{n^6}{6!}$ . This can simplify to $\frac{6n^5+n^6}{720}$ , which can further simplify to $\frac{n^5(6+n)}{720}$ . Notice that the prime factorization of $720$ is $5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2$ . In order for $\frac{n^5(6+n)}{720}$ to be an integer, one of the parts must divide $5, 3$ , and $2$ . Thus, one of the parts must be a multiple of $5, 3$ , and $2$ , and the LCM of these three numbers is $30$ . This means \[n^5 \equiv 0\pmod{30}\] or \[6+n \equiv 0\pmod{30}\] Thus, we can see that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$ . Note that as long as we satisfy $\frac{6n^5+n^6}{720}$ $2!, 3!$ , and $4!$ will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. $4! = 2\cdot2\cdot2\cdot2\cdot3$ , and this will be divisible by $2^4\cdot3^4\cdot5^4$ . Now, since we know that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$ in order for the polynomial to be an integer, $n\equiv0, 24\pmod{30}$ . To find how many integers fulfill the equation and are $<2017$ , we take $\left \lfloor\frac{2017}{30} \right \rfloor$ and multiply it by $2$ . Thus, we get $67\cdot2=\boxed{134}$ | null | 134 |
b35fadbd7422993eb2821700404edc00 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer. | Taking out the $1+n$ part of the expression and writing the remaining terms under a common denominator, we get $\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)$ . Therefore the expression $n^6+6n^5+30n^4+120n^3+360n^2$ must equal $720m$ for some positive integer $m$ .
Taking both sides mod $2$ , the result is $n^6 \equiv 0 \pmod{2}$ . Therefore $n$ must be even. If $n$ is even, that means $n$ can be written in the form $2a$ where $a$ is a positive integer. Replacing $n$ with $2a$ in the expression, $64a^6+192a^5+480a^4+960a^3+1440a^2$ is divisible by $16$ because each coefficient is divisible by $16$ . Therefore, if $n$ is even, $n^6+6n^5+30n^4+120n^3+360n^2$ is divisible by $16$
Taking the equation $n^6+6n^5+30n^4+120n^3+360n^2=720m$ mod $3$ , the result is $n^6 \equiv 0 \pmod{3}$ . Therefore $n$ must be a multiple of $3$ . If $n$ is a multiple of three, that means $n$ can be written in the form $3b$ where $b$ is a positive integer. Replacing $n$ with $3b$ in the expression, $729b^6+1458b^5+2430b^4+3240b^3+3240b^2$ is divisible by $9$ because each coefficient is divisible by $9$ . Therefore, if $n$ is a multiple of $3$ $n^6+6n^5+30n^4+120n^3+360n^2$ is divisibly by $9$
Taking the equation $n^6+6n^5+30n^4+120n^3+360n^2=720m$ mod $5$ , the result is $n^6+n^5 \equiv 0 \pmod{5}$ . The only values of $n (\text{mod }5)$ that satisfy the equation are $n\equiv0(\text{mod }5)$ and $n\equiv4(\text{mod }5)$ . Therefore if $n$ is $0$ or $4$ mod $5$ $n^6+6n^5+30n^4+120n^3+360n^2$ will be a multiple of $5$
The only way to get the expression $n^6+6n^5+30n^4+120n^3+360n^2$ to be divisible by $720=16 \cdot 9 \cdot 5$ is to have $n \equiv 0 \pmod{2}$ $n \equiv 0 \pmod{3}$ , and $n \equiv 0 \text{ or } 4 \pmod{5}$ . By the Chinese Remainder Theorem or simple guessing and checking, we see $n\equiv0,24 \pmod{30}$ . Because no numbers between $2011$ and $2017$ are equivalent to $0$ or $24$ mod $30$ , the answer is $\frac{2010}{30}\times2=\boxed{134}$ | null | 134 |
b35fadbd7422993eb2821700404edc00 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer. | Note that $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}$ will have a denominator that divides $5!$ . Therefore, for the expression to be an integer, $\frac{n^6}{6!}$ must have a denominator that divides $5!$ . Thus, $6\mid n^6$ , and $6\mid n$ . Let $n=6m$ . Substituting gives $1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}$ . Note that the first $5$ terms are integers, so it suffices for $\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}$ to be an integer. This simplifies to $\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)$ . It follows that $5\mid m^5(m+1)$ . Therefore, $m$ is either $0$ or $4$ modulo $5$ . However, we seek the number of $n$ , and $n=6m$ . By CRT, $n$ is either $0$ or $24$ modulo $30$ , and the answer is $67+67=\boxed{134}$ | null | 134 |
b35fadbd7422993eb2821700404edc00 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer. | Clearly $1+n$ is an integer. The part we need to verify as an integer is, upon common denominator, $\frac{360n^2+120n^3+30n^4+6n^5+n^6}{720}$ . Clearly, the numerator must be even for the fraction to be an integer. Therefore, $n^6$ is even and n is even, aka $n=2k$ for some integer $k$ . Then, we can substitute $n=2k$ and see that $\frac{n^2}{2}$ is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get $\frac{60k^3+30k^4+12k^5+4k^6}{45}$ . It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the $4k^6$ , and we see that $k=3b$ for some integer $b$ . From there we now know that $n=6b$ . If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that $n^5(6n+1) \equiv 0 \pmod5$ , so combining with divisibility by 6, $n$ is $24$ or $0\pmod{30}$ . There are $67$ cases for each, hence the answer $\boxed{134}$ | null | 134 |
a77f8cf8b95b33350b79fe482704d065 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_9 | A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$ $1$ $2$ $3$ $4$ $5$ $6$ , and $7$ , in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ because we cannot assign the two reds to the two $1$ 's.
In order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the $1$ s. The number of ways for this to happen is $2 \cdot 6 = 12$ (the first card she draws has to be a $1$ (2 choices), while the second card can be any card but the remaining card with a $1$ (6 choices)). Each of these assignments is equally likely, so desired probability is $\frac{12}{{8\choose2}-1}=\frac{4}{9} \implies 4 + 9 = 13 = \boxed{013}$ | null | 013 |
a77f8cf8b95b33350b79fe482704d065 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_9 | A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | First note that out of the $8$ selected cards, one pair of cards have to share the same number and another pair of cards have to share the same color. Now, these $2$ pairs of cards can't be the same or else there will be $2$ cards which are completely same. Then, WLOG let the numbers be $1,1,2,3,4,5,6,$ and $7$ and the colors be $a,a,b,c,d,e,f,$ and $g$ . We therefore obtain only $2$ cases:
Case One: $1a,1b,2a,3c,4d,5e,6f,$ and $7g$ In this case, we can discard $1a$ .
There are $2*6=12$ situations in this case.
Case Two: $1b,1c,2a,3a,4d,5e,6f,$ and $7g$ In this case, we can't discard.
There are $\dbinom{6}{2}=15$ situations in this case.
So the probability is $\frac{12}{12+15}=\frac{4}{9}$ , giving us the answer of $4+9=\boxed{013}$ | null | 013 |
a77f8cf8b95b33350b79fe482704d065 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_9 | A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | There are $7!$ ways to choose a set of 7 cards that have all the numbers from 1-7 and all 7 colors. There are then $42$ cards remaining. Thus, there are $7!(42)$ desired sets.
Now, the next thing to find is the number of ways to choose 8 cards where there is not a set of 7 such cards. In this case, one color must have 2 cards and one number must have 2 cards, and they can't be the same number/color card. The number of ways to pick this is equal to a multiplication of $\binom{7}{2}$ ways to pick 2 numbers, $7$ colors to assign them to, $\binom{6}{2}$ ways to pick 2 nonchosen colors, $5$ ways to pick a number to assign them to, and $4!$ ways to assign the rest.
Thus, the answer is $\frac{7!(42)}{7!(42) + 21(15)(7)(5!)}$ . Dividing out $5!$ yields $\frac{42(42)}{42(42) + 21(15)(7)}$ which is equal to $\frac{2(42)}{2(42) + 15(7)}$ which is equal to $\frac{12}{12 + 15}$ which is equal to $\frac{4}{9}$ giving a final answer of $\boxed{013}$ | null | 013 |
a77f8cf8b95b33350b79fe482704d065 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_9 | A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We can rewrite the problem as "What is the probability that, given 8 cards with numbers a, b, c, d, e, f, g, g and some assortment of seven colors A, B, C, D, E, F, G, G where G is repeated, one of the cards is Gg?" Note that this is a valid restatement because Sharon has to have two of one number and two of one color. She needs to be able to take away one of the cards with the duplicate number, but this also has to have the duplicate color. There are two cases.
Case I: One of the cards is Gg. This implies that the other card with color G can be placed in $6$ ways, and the rest of the colors can be paired with cards in $6!$ ways.
Case II: None of the cards are Gg. This implies that the cards with color G can be chosen in $\dbinom{6}{2}=15$ ways, and the rest of the colors can be paired with cards in $\frac{6!}{2}$ ways, with the divide by 2 because of the double-g.
Note that there is no case with Gg, Gg because all 49 cards are unique!
Therefore our answer is $\frac{6\times6!}{\frac{15}{2}\times6! + 6\times6!}=\frac{4}{9}$ , so $\boxed{013}$ | null | 013 |
a77f8cf8b95b33350b79fe482704d065 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_9 | A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We count the two entities: The number of sets of eight cards that contain all seven numbers and all seven colors; and The number of sets of eight cards that contain all seven numbers and colors, and one card can be removed and the property still holds.
For the first: It is equivalent to counting:
How many ways can (colors) $A, A, B, C, D, E, F, G$ be matched to (numbers) $1, 1, 2, 3, 4, 5, 6, 7$ such that the two $A's$ cannot both be matched to the two ones. (This would mean we've chosen two identical cards from the deck, both with color $A$ and number one.)
We consider two cases:
1) Neither $A$ is matched with a one. Then there are $\binom{6}{2}$ ways to choose which two numbers are matched with the $A's,$ and $6! / 2!$ ways to order the remaining numbers. (There are two one's, so we divide by two.)
2) One of the $A$ 's is matched with a one. There are $6 \cdot 6!$ ways to arrange the remaining color
Since there are seven ways to choose the extra color and likewise for the extra number,
we have in total $7 \cdot 7 \cdot \left( 6C2 \cdot \frac{6!}{2!} + 6 \cdot 6! \right)$ ways to choose eight cards satisfying the first condition.
For the second: View the cards as rooks on a seven by seven chess board. Seven cards that contain all colors and numbers is can be represented as placing seven rooks on the board so that no two rooks are attacking each other. There are $7!$ ways to do this. For each arrangement, we can choose any of the other 42 positions on the board (cards) to add an extra rook. Thus, there are $7! \cdot 42$ good sets of eight cards. Note that we are not overcounting.
Thus, the probability of getting a good set given the first condition is $\frac{ 7! \cdot 42 }{ 7 \cdot 7 \cdot \left( 6C2 \cdot \frac{6!}{2!} + 6 \cdot 6! \right) } = \frac{4}{9},$ and the answer is $\boxed{013}.$ | null | 013 |
a77f8cf8b95b33350b79fe482704d065 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_9 | A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We recast this problem to coloring 8 cells in a 7x7 grid, such that each row and column has at least one colored cell. If a cell in row a and column b is colored, that means we drew a card with color a and number b.
By pigeonhole principle, there exists a row that has 2 colored cells, and intuitively, that row is unique. There also exists a column with 2 colored cells. Call the unique row, "Row A", and Call the unique column, "Column B".
If we are able to remove a card so that we still have a card of each color and number, that means, we can erase a cell, so each row still has at least one colored cell, and so does each column. If we erase a cell not in Row A, then the erased cell's row will be empty. If we erase a cell not in Column B, the erased cell's column will be empty. Thus the cell that we erase must be in both Row A and Column B. There is only one cell in both Row A and Column B, and thus that cell must be colored.
We first count the number of successful outcomes. We must chose which row to be "Row A", and which column to be "Column B". There are 7 ways to chose "Row A", and 7 ways to chose "Column B". Because of the logic used above, the intersection of Row A and Column B must be colored. However, Row A needs to have 2 colored cells. There are 6 other uncolored cells to chose from. Similarly, there are 6 uncolored cells in Column B.
Note that we can ignore all the rows and columns that contain colored cells. This is because Row A already contains 2 colored cells, and thus we can't color any more cells in Row A, similarly for Column B. The other row that contains a colored cell is not Row A, so it can only contain 1 colored cell, which it has, thus we can't color anymore cells in it. Similarly for the other column.
We can actually remove these rows/columns. This leaves a 5x5 grid, full of uncolored cells. Each row can have 1 cell, and so can each column. We count the # of colorings from the perspective of the rows. We look at the first row, it has 5 choices. We can look at the 2nd row, it has 4 choices, as it's selected cell can't occupy the same column. We continue down the line, and see there are $5! = 120$ ways to color.
Thus, the # of successful outcomes is $5! \cdot 6 \cdot 6 \cdot 7 \cdot 7$
We can then count the # of possible outcomes.
We can count the # of unsuccessful outcomes. We select "Row A" and "Column B" as usual, with $7 \cdot 7$ ways. We don't color the intersection, so we have to color 2 cells in each. There are 6 other cells that can be colored in each, so we have ${6 \choose 2} = 15$ ways to color each. Then, we can get rid of rows and columns using the method above and we are left with a 4x4 grid. There are $4! = 24$ ways to color.
Thus, the # of unsuccessful outcomes is $4! \cdot 15 \cdot 15 \cdot 7 \cdot 7$
After some calculation, we get $\frac{successful}{possible} = \frac{successful}{successful+unsuccessful} = \frac{5! \cdot 6^2 \cdot 7^2}{7^2 \cdot(5! \cdot 6^2 + 4! \cdot 15^2} = \frac{5! \cdot 6^2}{5! \cdot 6^2 +4! \cdot 15^2} = \frac{5 \cdot 6^2}{5 \cdot 6^2 + 15^2} = \frac{180}{180+225} = \frac{4}{9} => 4+9 = \boxed{13}$ | null | 13 |
91f773a22faafc3d16c2b30962b92782 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_10 | Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$ . Point $M$ is the midpoint of $\overline{AD}$ , point $N$ is the trisection point of $\overline{AB}$ closer to $A$ , and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$ . Point $P$ lies on the quadrilateral $BCON$ , and $\overline{BP}$ bisects the area of $BCON$ . Find the area of $\triangle CDP$ | [asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,blue); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,W); label("$N$",n,N); label("$O$",O,(-0.5,1)); label("$P$",P,N); dot(A); dot(B); dot(C); dot(D); dot(M); dot(n); dot(O); dot(P); label("28",(0,42)--(28,42),N); label("56",(28,42)--(84,42),N); [/asy] Impose a coordinate system on the diagram where point $D$ is the origin. Therefore $A=(0,42)$ $B=(84,42)$ $C=(84,0)$ , and $D=(0,0)$ . Because $M$ is a midpoint and $N$ is a trisection point, $M=(0,21)$ and $N=(28,42)$ . The equation for line $DN$ is $y=\frac{3}{2}x$ and the equation for line $CM$ is $\frac{1}{84}x+\frac{1}{21}y=1$ , so their intersection, point $O$ , is $(12,18)$ . Using the shoelace formula on quadrilateral $BCON$ , or drawing diagonal $\overline{BO}$ and using $\frac12bh$ , we find that its area is $2184$ . Therefore the area of triangle $BCP$ is $\frac{2184}{2} = 1092$ . Using $A = \frac 12 bh$ , we get $2184 = 42h$ . Simplifying, we get $h = 52$ . This means that the x-coordinate of $P = 84 - 52 = 32$ . Since P lies on $\frac{1}{84}x+\frac{1}{21}y=1$ , you can solve and get that the y-coordinate of $P$ is $13$ . Therefore the area of $CDP$ is $\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}$ | null | 546 |
91f773a22faafc3d16c2b30962b92782 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_10 | Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$ . Point $M$ is the midpoint of $\overline{AD}$ , point $N$ is the trisection point of $\overline{AB}$ closer to $A$ , and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$ . Point $P$ lies on the quadrilateral $BCON$ , and $\overline{BP}$ bisects the area of $BCON$ . Find the area of $\triangle CDP$ | Since the problem tells us that segment $\overline{BP}$ bisects the area of quadrilateral $BCON$ , let us compute the area of $BCON$ by subtracting the areas of $\triangle{AND}$ and $\triangle{DOC}$ from rectangle $ABCD$
To do this, drop altitude $\overline{OE}$ onto side $\overline{DC}$ and draw a segment $\overline{MQ}$ parallel to $\overline{AN}$ from side $\overline{AD}$ to $\overline{ND}$ . Since $M$ is the midpoint of side $\overline{AD}$ \[\overline{MQ}=14\] Denote $\overline{OE}$ as $a$ . Noting that $\triangle{MOQ}~\triangle{COD}$ , we can write the statement \[\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}\] \[\implies \frac{84}{a}=\frac{14}{21-a}\] \[\implies a=18\] Using this information, the area of $\triangle{DOC}$ and $\triangle{AND}$ are \[\frac{18\cdot 84}{2}=756\] and \[\frac{28\cdot 42}{2}=588\] respectively. Thus, the area of quadrilateral $BCON$ is \[84\cdot 42-588-756=2184\] Now, it is clear that point $P$ lies on side $\overline{MC}$ , so the area of $\triangle{BPC}$ is \[\frac{2184}{2}=1092\] Given this, drop altitude $\overline{PF}$ (let's call it $b$ ) onto $\overline{BC}$ . Therefore, \[\frac{42b}{2}=1092\implies b=52\] From here, drop an altitude $\overline{PG}$ onto $\overline{DC}$ . Recognizing that $\overline{PF}=\overline{GC}$ and that $\triangle{MDC}$ and $\triangle{PGC}$ are similar, we write \[\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}\] \[\implies \frac{\overline{PG}}{52}=\frac{21}{84}\] \[\implies \overline{PG}=13\] The area of $\triangle{CDP}$ is given by \[\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}\] ~blitzkrieg21 and jdong2006 | null | 546 |
850aff50aa732cfd4360ce0a2c438cf9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | It is obvious that any configuration of one-way roads which contains a town whose roads all lead into it or lead out of it cannot satisfy the given. We claim that any configuration which does not have a town whose roads all lead into it or lead out of it does satisfy the given conditions. Now we show that a loop of $3$ or more towns exist. Pick a town, then choose a neighboring town to travel to $5$ times. Of these $6$ towns visited, at least two of them must be the same; therefore there must exist a loop of $3$ or more towns because a loop of $2$ towns cannot exist. We want to show that the loop can be reached from any town, and any town can be reached from the loop.
$\textbf{Case 1}$ . The loop has $5$ towns.
Clearly every town can be reached by going around the loop.
$\textbf{Case 2}$ . The loop has $4$ towns.
The town not on the loop must have a road leading to it. This road comes from a town on the loop. Therefore this town can be reached from the loop. This town not on the loop must also have a road leading out of it. This road leads to a town on the loop. Therefore the loop can be reached from the town.
$\textbf{Case 3}$ . The loop has $3$ towns.
There are two towns not on the loop; call them Town $A$ and Town $B$ . Without loss of generality assume $A$ leads to $B$ . Because a road must lead to $A$ , the town where this road comes from must be on the loop. Therefore $A$ and therefore $B$ can be reached from the loop. Because a road must lead out of $B$ , the town it leads to must be on the loop. Therefore the loop can be reached from $B$ and also $A$
The number of good configurations is the total number of configurations minus the number of bad configurations. There are $2^{{5\choose2}}$ total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are $5$ ways to choose this town and $2^6$ ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are $5 \cdot 2^6+5 \cdot 2^6$ bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are $5$ ways to choose the town for which all roads lead to it, $4$ ways to choose the town for which all roads lead out of it, and $2^3$ ways to assign the remaining $3$ roads not connected to either of these towns. Therefore, the answer is $2^{{5\choose2}}-(5 \cdot 2^6+5 \cdot 2^6-5\cdot 4 \cdot 2^3)=\boxed{544}$ | null | 544 |
850aff50aa732cfd4360ce0a2c438cf9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | The only way a town does not meet the conditions in the question is if the town has either all roads leading towards it or all roads leading away from it. For example, if all roads lead away from Town $A$ , there is no way to reach the town starting from Towns $B$ $C$ $D$ , or $E$ . If all roads lead towards Town $A$ , there is no way to reach any other town starting from Town $A$ . Thus, we will count the ways this occurs. (To show this, first note that there must always be a directed path that visits every town once, say $A \rightarrow C \rightarrow B \rightarrow E \rightarrow D.$ (Redei's Theorem). Taking cases, it can be seen that if $A$ does not have all its roads leading, then $E$ must have its roads leading into it.)
$\textbf{Case 1}$ . WLOG, Town $A$ has all roads leading towards it.
In this case, the four roads leading to Town $A$ have already been determined. There are $6$ roads that have not been given directions. Each of these roads have $2$ options: it can lead towards one town or the other town. Thus, there are $2^6$ arrangements of roads.
However, we must consider the $5$ towns in which this scenario can occur: there are $5 \cdot 2^6=320$ arrangements.
$\textbf{Case 2}$ . WLOG, Town $A$ has all roads leading away.
Notice that this case is symmetrical to the first case. Thus, there are $320$ arrangements.
$\textbf{Case 3}$ . WLOG, Town $A$ has all roads leading towards it and Town $B$ has all roads leading away.
We must check for double-counted cases. Drawing a flow diagram, we see that this case determines $7$ roads. For the undetermined $3$ roads, there are $2^3=8$ arrangements. However, we must again consider the different ways that this case can occur. There are $5$ choices for towns with roads leading towards it $4$ choices for roads leading away. Thus, the total double-counted cases are $5 \cdot 4 \cdot 8=160$ arrangements.
We must subtract the cases we counted from the total. There are a total of $10$ roads with $2$ possible arrangements each.
Therefore the total number of cases is $2^{10} =1024$ , and the number of cases that meet the conditions is $1024 - (320+320-160) = \boxed{544}$ arrangements. | null | 544 |
850aff50aa732cfd4360ce0a2c438cf9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | Assume the five towns are named A, B, C, D, and E. Draw roads that connect each of the five towns. First, figure out how many total roads there are with no restrictions. This would be $2^{10}$ , or $1024$ . Next, add together the number of restrictions (or incorrect cases) there are. One incorrect case happens if we can't get to one of the five towns. When that happens, any roads that connect that town with any of the other four towns must be one way. Four roads are invalidated (because they are one way) as, assuming we can't get to town A, roads AB, AC, AD, and AE would all be one way. Since there are ${{5\choose1}}$ ways of choosing which town to restrict, there are $5\cdot2^{10-4}$ , or $320$ restrictions in this case. The next case is when we can't get to two of the towns. Let's assume we don't want to get to towns A and B. Then, roads AC, AD, AE, BC, BD, and BE are all invalidated. Since there are ${{5\choose2}}$ of choosing which two towns to restrict, there are $10\cdot2^{10-6}$ , or $160$ restrictions in this case. Notice that when we restrict three towns, it is the same thing as restricting two towns. For example, if we restricted entry to towns C, D, and E from towns A and B, it's the same thing as if we restricted entry to towns A and B from towns C, D, and E, as the exact same roads are invalidated in both cases. Thus, our final answer is $1024-(320+160) = \boxed{544}$ arrangements. | null | 544 |
850aff50aa732cfd4360ce0a2c438cf9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | As noted before, you can see that there are 2 ways that the condition cannot be met; it either has all roads leading into one city, or all roads leading out of one city. We then use complementary counting to count these cases, with PIE. Obviously there are $2^5*2^5$ ways to make roads (just draw a pentagon with all of the diagonals and assume that they are roads, and count the sides and then the diagonals). Then, make all roads leading in to another city. There are 4 roads leading into that city, so there are $2^6$ ways to designate that. Given that there are 5 cities, and it can also be the ways to go out of the city, you get $10*2^6$ ways NOT to get into the city. However, this disregards the case that 2 cities are like this, which is the PIE case. Quick inspection shows that there are no cases where 3 are like that, and inspection with le pentagon can also show you that there are $\binom{5}{2}*2$ (because they can all lead in and all lead out) $*2^3$ (because there are 3 roads left to choose). In the end, this comes out to $2^5*2^5-(10*2^6-2*\binom{5}{2}*2^3)=1024-(640-160)=\boxed{544}$ | null | 544 |
850aff50aa732cfd4360ce0a2c438cf9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | We claim the following conditions:
Call the vertices $\omega_1, \omega_2, \omega_3, \omega_4, \omega_5$ . WLOG, let $\omega_1 \to \omega_2$ . Then, WLOG let $\omega_2 \to \omega_3$ . We have two ways to go from here:
If $\omega_3 \to \omega_1$ , we call it a "cycle".
We will prove (3):
Notice that once $\omega_4, \omega_5$ can reach one of the vertices on the "cycle", it can reach any of the other vertices on the same cycle simply by travelling along the arcs of the cycle. Additionally, one of $\omega_4, \omega_5$ must be able to travel to the cycle. Consider that it cannot, then $\omega_3 \to \omega_4$ and $\omega_5 \to \omega_4$ are forced. Now, $\omega_4 \to \omega_1, \omega_2, \omega_3$ which is on the cycle. Now, if $\omega_5 \to \omega_4$ , it will also be able to travel to the cycle. Otherwise, the directed arc will directly travel to the cycle. Also, each of the vertices $\omega_1, \omega_2, \omega_3$ will travel to $\omega_4, \omega_5$
If our graph does not satisfy (1) and (2), $\omega_4, \omega_5$ will connect to the cycle on both directions. This can be shown below:
Clearly $\omega_4, \omega_5$ are connected. Assuming the graph does not satisfy (1) and (2), $\omega_4, \omega_5$ are connected to the cycle consisting of $\omega_1, \omega_2, \omega_3$ . WLOG, let $\omega_4 \to \omega_5$
Again, to avoid (2), $\omega_5 \to \omega_a$ where $a\in{1,2,3}$ . Now to avoid (1), one of $\omega_a \to \omega_4, a\in{1,2,3}$
Now it is clear that each vertex can go to any other vertex. This concludes our proof.
Say we try to avoid (3) so that WLOG $\omega_5 \to \omega_1$ . If we keep trying to avoid a "cycle", $\omega_3 \to \omega_4$ and $\omega_4 \to \omega_5$ which is a cycle.
Therefore, if there is any path $\omega_a \to \omega_b \to \omega_c$ , it will lead to a cycle and ultimately a valid path. The only way to avoid this is to have (1) or (2).
Consider (1):
WLOG, let $\omega_1$ have no arcs leading to it. Then all other vertices will have an arc directed toward them away from $\omega_1$ . Therefore, one path can have only 1 unique vertex that has no arcs leading to it. There are 5 ways to choose the vertex and $2^{\binom{5}{2}-4}=2^6$ ways for the remaining 6 arcs since they can be directed in any way. There are $2^6\cdot5=320$ paths for this condition.
Consider (2):
WLOG, let $\omega_1$ have all arcs leading to it. Similarly, all other vertices will have an arc that is directed away from them and points to $\omega_1$ . Therefore, only one unique vertex can have all arcs leading to it. There are 5 ways to choose the vertex and $2^{\binom{5}{2}-4}=2^6$ ways for the remaining arcs. There are also $2^6\cdot5=320$ paths for this condition.
However, we have overcounted paths that satisfy both (1) and (2). These paths will have one vertex that has all arcs leading to it and another that has all arcs directed away from it. There are $5\cdot4=20$ ways to choose these 2 vertices, and the remaining $\binom{3}{2}=3$ arcs each have 2 ways to be directed toward each other, so there are a total of $2^3\cdot 20=160$ overlapping paths.
By PIE, the number of invalid cases is $320+320-160=480$
The total number of cases is $2^{\binom{5}{2}}=2^{10}=1024$ since there are 10 edges and each can go 2 ways. By complementary counting, the number of valid paths is just $1024-480=\boxed{544}$ | null | 544 |
a9194da7a2e9adce2c69d4b00c480832 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | Circle $C_0$ has radius $1$ , and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$ . Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$ . Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$ . In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$ , and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$ , as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$ , the distance from the center $C_0$ to $B$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("$A_0$",(125,0),E); dot((25,100)); label("$A_1$",(25,100),SE); dot((-55,20)); label("$A_2$",(-55,20),E); [/asy] | Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$ . Therefore $A_1=(1-r,r)$ $A_2=(1-r-r^2,r-r^2)$ $A_3=(1-r-r^2+r^3,r-r^2-r^3)$ $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$ , and so on, where the signs alternate in groups of $2$ . The limit of all these points is point $B$ . Using the sum of infinite geometric series formula on $B$ and reducing the expression: $(1-r)-(r^2-r^3)+(r^4-r^5)-...=\frac{1-r}{1+r^2}$ $(r-r^2)-(r^3-r^4)+(r^5-r^6)-...=\frac{r-r^2}{1+r^2}$ . Thus, we get $B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)$ . The distance from $B$ to the origin is $\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.$ Let $r=\frac{11}{60}$ , and the distance from the origin is $\frac{49}{61}$ $49+61=\boxed{110}$ | null | 110 |
a9194da7a2e9adce2c69d4b00c480832 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | Circle $C_0$ has radius $1$ , and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$ . Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$ . Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$ . In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$ , and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$ , as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$ , the distance from the center $C_0$ to $B$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("$A_0$",(125,0),E); dot((25,100)); label("$A_1$",(25,100),SE); dot((-55,20)); label("$A_2$",(-55,20),E); [/asy] | Let the center of circle $C_i$ be $O_i$ . Note that $O_0BO_1$ is a right triangle, with right angle at $B$ . Also, $O_1B=\frac{11}{60}O_0B$ , or $O_0B = \frac{60}{61}O_0O_1$ . It is clear that $O_0O_1=1-r=\frac{49}{60}$ , so $O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}$ . Our answer is $49+61=\boxed{110}$ | null | 110 |
a9194da7a2e9adce2c69d4b00c480832 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | Circle $C_0$ has radius $1$ , and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$ . Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$ . Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$ . In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$ , and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$ , as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$ , the distance from the center $C_0$ to $B$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("$A_0$",(125,0),E); dot((25,100)); label("$A_1$",(25,100),SE); dot((-55,20)); label("$A_2$",(-55,20),E); [/asy] | Note that there is an invariance, Consider the entire figure $\mathcal{F}$ . Perform a $90^\circ$ counterclockwise rotation, then scale by $r$ with respect to $(1, 0)$ . It is easy to see that the new figure $\mathcal{F}' \cup S^1 = \mathcal{F}$ , so $B$ is invariant.
Using the invariance, Let $B = (x,y)$ . Then rotating and scaling, $B = (1-r(1+y), rx)$ . Equating, we find $x = \frac{1-r}{r^2+1}, y = \frac{r-r^2}{r^2+1}$ . The distance is thus $\frac{49}{61}$ . Our answer is $49+61=\boxed{110}$ | null | 110 |
a9194da7a2e9adce2c69d4b00c480832 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | Circle $C_0$ has radius $1$ , and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$ . Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$ . Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$ . In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$ , and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$ , as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$ , the distance from the center $C_0$ to $B$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("$A_0$",(125,0),E); dot((25,100)); label("$A_1$",(25,100),SE); dot((-55,20)); label("$A_2$",(-55,20),E); [/asy] | Using the invariance again as in Solution 3, assume $B$ is $d$ away from the origin. The locus of possible points is a circle with radius $d$ . Consider the following diagram. [asy] size(7cm); draw(circle((0,0), 49/61)); draw((0,0)--(0.790110185, 0.144853534)); draw((0,0)--(-0.144853534, 0.790110185)); draw((-0.144853534, 0.790110185)--(1,0)); draw((0,0)--(1,0)); draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3)); label("$O$", (0,0), SW); label("$(1,0)$", (1,0), E); label("$B$", (0.790110185, 0.144853534), NE); label("$B'$", (-0.144853534, 0.790110185), N); label("$d$", (0.5 * 49/61, 0), S); [/asy]
Let the distance from $B$ to $(1,0)$ be $x$ . As $B$ is invariant, $x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}$ . Then by Power of a Point, $x(BB' + x) = (1-d)(1+d) \implies xr(BB' + x) = r(1-d)(1+d) \implies x^2 = r(1-d^2) \implies d^2 = \left(1 + \frac{2r}{(1-r)^2}\right)$ . Solving, $d = \frac{49}{61}$ . Our answer is $49+61=\boxed{110}$ | null | 110 |
a9194da7a2e9adce2c69d4b00c480832 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | Circle $C_0$ has radius $1$ , and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$ . Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$ . Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$ . In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$ , and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$ , as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$ , the distance from the center $C_0$ to $B$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("$A_0$",(125,0),E); dot((25,100)); label("$A_1$",(25,100),SE); dot((-55,20)); label("$A_2$",(-55,20),E); [/asy] | Let $A_0$ be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to $r$ . Now accounting for rotation by $\frac{\pi}{2}$ radians, we see that the common ratio is $ri$ . Thus since our first term is $A_1=-r+ri$ , the total sum (by geometric series formula) is $\frac{-r+ri}{1-ri}=\frac{-781+538i}{3721}$ . We need the distance from $C_0=-1$ so our distance is $|B-C_0|=\sqrt{\left(\frac{-781}{3721}-(-1)\right)^2+\left(\frac{538i}{3721}\right)^2}=\sqrt{\frac{2401}{3721}}=\frac{49}{61}$ . Our answer is $49+61=\boxed{110}$ | null | 110 |
e5fff45be1ba37f3f483125784b3c172 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_13 | For each integer $n\geq3$ , let $f(n)$ be the number of $3$ -element subsets of the vertices of the regular $n$ -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$ | Considering $n \pmod{6}$ , we have the following formulas:
$n\equiv 0$ $\frac{n(n-4)}{2} + \frac{n}{3}$
$n\equiv 2, 4$ $\frac{n(n-2)}{2}$
$n\equiv 3$ $\frac{n(n-3)}{2} + \frac{n}{3}$
$n\equiv 1, 5$ $\frac{n(n-1)}{2}$
To derive these formulas, we note the following:
Any isosceles triangle formed by the vertices of our regular $n$ -sided polygon $P$ has its sides from the set of edges and diagonals of $P$ . Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of $n$ , unless $n$ is even when one set of diagonals (those which bisect the polygon) comes in a group of $\frac{n}{2}$ . Three properties hold true of $f(n)$
When $n$ is odd there are $\frac{n(n-1)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-1}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, and we must divide by 2 for over-count).*
When $n$ is even there are $\frac{n(n-2)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-2}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).
When $n$ is a multiple of three we additionally over-count equilateral triangles, of which there are $\frac{n}{3}$ . As we count them three times, we are two times over, so we subtract $\frac{2n}{3}$
Considering the six possibilities $n \equiv 0,1,2,3,4,5 \pmod{6}$ and solving, we find that the only valid solutions are $n = 36, 52, 157$ , from which the answer is $36 + 52 + 157 = \boxed{245}$ | null | 245 |
e5fff45be1ba37f3f483125784b3c172 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_13 | For each integer $n\geq3$ , let $f(n)$ be the number of $3$ -element subsets of the vertices of the regular $n$ -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$ | In the case that $n\equiv 0\pmod 3$ , there are $\frac{n}{3}$ equilateral triangles. We will now count the number of non-equilateral isosceles triangles in this case.
Select a vertex $P$ of a regular $n$ -gon. We will count the number of isosceles triangles with their vertex at $P$ . (In other words, we are counting the number of isosceles triangles $\triangle APB$ with $A, B, P$ among the vertices of the $n$ -gon, and $AP=BP$ .)
If the side $AP$ spans $k$ sides of the $n$ -gon (where $k<\frac{n}{2}$ ), the side $BP$ must span $k$ sides of the $n$ -gon, and, thus, the side $AB$ must span $n-2k$ sides of the $n$ -gon. As $\triangle ABP$ has three distinct vertices, the side $AB$ must span at least one side, so $n-2k \ge 1$ . Combining this inequality with the fact that $1\le k<\frac{n}{2}$ and $k\not = \frac{n}{3}$ (as $\triangle ABP$ cannot be equilateral), we find that there are $\lceil\frac{n}{2}\rceil-2$ possible $k$
As each of the $n$ vertices can be the vertex of a given triangle $\triangle ABP$ , there are $\left(\lceil \frac{n}{2} \rceil -2 \right)\cdot n$ non-equilateral isosceles triangles.
Adding in the $\frac{n}{3}$ equilateral triangles, we find that for $n\equiv 0\pmod 3$ $f(n) = \frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n$
On the other hand, if $n\equiv 1, 2\pmod 3$ , there are no equilateral triangles, and we may follow the logic of the paragraph above to find that $f(n)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n$
We may now rewrite the given equation, based on the remainder $n$ leaves when divided by 3.
Case 1: $n\equiv 0\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n+78$
In this case, $n$ is of the form $6k$ or $6k+3$ , for some integer $k$
Subcase 1: $n=6k$ Plugging into the equation above yields $k=6\rightarrow n=36$
Subcase 2: $n=6k+3$ Plugging into the equation above yields $7k=75$ , which has no integer solutions.
Case 2: $n\equiv 1\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$
In this case, $n$ is of the form $6k+1$ or $6k+4$ , for some integer $k$
Subcase 1: $n=6k+1$ In this case, the equation above yields $k=8\rightarrow n=52$
Subcase 2: $n=6k+4$ In this case, the equation above yields $k=26\rightarrow n=157$
Case 3: $n\equiv 2\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\frac{n+1}{3}+\left(\lceil \frac{n+1}{2} \rceil -2\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$
In this case, $n$ is of the form $6k+2$ or $6k+5$ , for some integer $k$
Subcase 1: $n=6k+2$ The equation above reduces to $5k=77$ , which has no integer solutions.
Subcase 2: $n=6k+5$ The equation above reduces to $k=-80$ , which does not yield a positive integer solution for $n$
In summary, the possible $n$ are $36, 52, 157$ , which add to $\boxed{245}$ | null | 245 |
e5fff45be1ba37f3f483125784b3c172 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_13 | For each integer $n\geq3$ , let $f(n)$ be the number of $3$ -element subsets of the vertices of the regular $n$ -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$ | We first notice that when a polygon has $s$ sides where $s\not\equiv 0\pmod{3}$ , there cannot exist any three vertices that form an equilateral triangle. Also, the parity of $s$ and $s+1$ also matters, since they influence how many isosceles triangles including equilateral triangles exist in the polygon. We can model an equation $2x+y=s$ , where the lines that are congruent connect the vertices that are $x$ vertices apart and the other line has vertices that are $y$ vertices apart. If $s$ is even, there are $\frac{s-2}{2}$ solutions for $(x,y)$ which would create a unique type of isosceles triangle. We subtract two since $y$ cannot be zero. If $s$ is odd, there are $\frac{s-1}{2}$ solutions for $(x,y)$ .
Next, we do casework on the congruence of $s\pmod{3}$ and the parody of $s$ using the information above:
Case 1:
$s\not\equiv 0\pmod{3}$ $s+1\not\equiv 0\pmod{3}$
$s$ is even, $s+1$ is odd
There are $\frac{s-2}{2}$ unique types of isosceles triangles in the polygon with $s$ sides. Each isosceles triangle has a unique point which connects the two congruent sides. Therefore, for each type of isosceles triangle, there exists $s$ of those triangles since the unique point can be any of the $s$ vertices. There are $\frac{s}{2}$ types of isosceles triangles in the polygon with $s+1$ sides and $s+1$ unique points for each type of triangle. Therefore, $\frac{s}{2}\cdot{(s+1)}-\frac{s-2}{2}\cdot{(s)}=78$ . Solving, we get $s=52$
Case 2:
$s\not\equiv 0\pmod{3}$ $s+1\not\equiv 0\pmod{3}$
$s$ is odd, $s+1$ is even
Using similar logic, there are $\frac{s-1}{2}\cdot{(s)}$ possible isosceles triangles in the $s$ sided polygon. There are $\frac{s-1}{2}\cdot{(s+1)}$ possible isosceles triangles in the $s+1$ sided polygon. The difference should be $78$ , so $\frac{s-1}{2}\cdot{(s+1)}-\frac{s-1}{2}\cdot{(s)}=78$ . Solving gives $s=157$
For both of these cases, we don't have to worry about equilateral triangles since in both cases $s,s+1\nmid3$
Case 3:
$s\not\equiv 0\pmod{3}$ $s+1\equiv 0\pmod{3}$
$s$ is even, $s+1$ is odd
There are $\frac{s-2}{2}\cdot{(s)}$ possible isosceles triangles in the $s$ sided polygon. The $s+1$ case is a bit more complicated, as we have to consider equilateral triangles as well. In this case, there is one solution where $x=y$ , which would produce an equilateral triangle. Therefore, we subtract that case to calculate only isosceles triangles with 2 congruent sides. Only isosceles triangles with exactly 2 congruent sides have a unique point, while there exist only $\frac{s+1}{3}$ distinct equilateral triangles in a polygon with $s+1$ sides, the rest are equivalent and symmetrical. Therefore, there are $(\frac{s}{2}-1)\cdot{(s+1)}+\frac{s+1}{3}$ isosceles triangles with at least 2 congruent sides in a polygon with $s+1$ sides. Putting it together, $\frac{s}{2}-1\cdot{(s+1)}+(\frac{s+1}{3})-\frac{s-2}{2}\cdot{(s)}=78$ . Solving yields $5s=772$ which is impossible since $s$ has to be an integer. Therefore, this case is not valid.
Case 4:
$s\not\equiv 0\pmod{3}$ $s+1\equiv 0\pmod{3}$
$s$ is odd, $s+1$ is even
There are $\frac{s-1}{2}\cdot{(s)}$ isosceles triangles in the $s$ sided polygon. Using the idea above, there are $(\frac{s-1}{2}-1)\cdot{(s+1)}$ isosceles triangles with 2 congruent sides in an $s+1$ sided polygon. There are $\frac{s+1}{3}$ equilateral triangles. Therefore, $(\frac{s-1}{2}-1)\cdot{(s+1)}+(\frac{s+1}{3})-\frac{s-1}{2}\cdot{(s)}=78$ . Looking at the left hand side, it is clear that $s$ has to be negative, which is not valid. Therefore, this case is not valid.
Case 5:
$s\equiv 0\pmod{3}$ $s+1\not\equiv 0\pmod{3}$
$s$ is even, $s+1$ is odd
There are a total of $(\frac{s-2}{2}-1)\cdot{(s)}+(\frac{s}{3})$ isosceles triangles in a polygon with $s$ sides. There are a total of $\frac{s}{2}\cdot{(s+1)}$ isosceles triangles in a polygon with $s+1$ sides. Therefore, $\frac{s}{2}\cdot{(s+1)}-(\frac{s-2}{2}-1)\cdot{(s)}-(\frac{s}{3})=78$ . Simplifying, we get $s=36$
Case 6:
$s\equiv 0\pmod{3}$ $s+1\not\equiv 0\pmod{3}$
$s$ is odd, $s+1$ is even
There are a total of $(\frac{s-1}{2}-1)\cdot{(s)}+(\frac{s}{3})$ isosceles triangles in the polygon with $s$ sides. There are $\frac{s-1}{2}\cdot{(s+1)}$ isosceles triangles in the polygon with $s+1$ sides. Therefore, $\frac{s-1}{2}\cdot{(s+1)}-(\frac{s-1}{2}-1)\cdot{(s)}-(\frac{s}{3})=78$ . Simplifying, we get $7s=471$ , which means $s$ is not an integer. Thus, this case is invalid.
Adding all the valid cases, we obtain $52+157+36=\boxed{245}$ | null | 245 |
16bc3e16d55e7089e51ff8d477edc3ab | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points. | $Case \textrm{ } 1:$ The lines are not parallel to the faces
A line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.
We look at the one from $(1,1,1)$ to $(10,10,10)$ . The lower endpoint of the desired lines must contain both a 1 and a 3, so it can be $(1,1,3), (1,2,3), (1,3,3)$ . If $\textrm{min} > 0$ then the point $(a-1,b-1,c-1)$ will also be on the line for example, 3 applies to the other end.
Accounting for permutations, there are $12$ ways, so there are $12 \cdot 4 = 48$ different lines for this case.
$Case \textrm{ } 2:$ The lines where the $x$ $y$ , or $z$ is the same for all the points on the line.
WLOG, let the $x$ value stay the same throughout. Let the line be parallel to the diagonal from $(1,1,1)$ to $(1,10,10)$ . For the line to have 8 points, the $y$ and $z$ must be 1 and 3 in either order, and the $x$ value can be any value from 1 to 10. In addition, this line can be parallel to 6 face diagonals. So we get $2 \cdot 10 \cdot 6 = 120$ possible lines for this case.
The answer is, therefore, $120 + 48 = \boxed{168}$ | null | 168 |
16bc3e16d55e7089e51ff8d477edc3ab | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points. | Look at one pair of opposite faces of the cube. There are $4$ lines say $l_1, l_2, l_3, l_4$ with exactly $8$ collinear points on the top face. For each of these lines, draw a rectangular plane that consists of one of the $l_i$ for $1 \leq i \leq 4$ and perpendicular to the top face.
There are $16$ lines in total on this plane. $10$ of which are parallel to one of the edges of the rectangular plane and $6$ of which are diagonals. There are $3$ pairs of opposite faces. So $3 \cdot 4 \cdot 16=192$ lines.
But we are overcounting the lines of the diagonals of those rectangular planes twice. There are $4$ rectangular planes perpendicular to one pair of opposite faces. Thus $4 \cdot 6=24$ lines are overcounted.
So the answer is $192-24=\boxed{168}$ | null | 168 |
16bc3e16d55e7089e51ff8d477edc3ab | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points. | Considered the cases $(1, 2, ..., 8), (2, 3, ...,9), (3, 4, ..., 10)$ and reverse. Also, consider the constant subsequences of length 8 $(1, 1, ..., 1), (2, 2, ..., 2), ..., (10, 10, ..., 10)$ . Of all the triplets that work they cannot be extended to form another point on the line in the $10 \times 10 \times 10$ grid but we need to divide by 2 because reversing all the subsequences gives the same line. Thus the answer is \[\frac{16^3 - 14^3 - 14^3 + 12^3}{2} = \boxed{168}\] | null | 168 |
16bc3e16d55e7089e51ff8d477edc3ab | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points. | The lines can be defined as starting from $(a, b, c)$ with "slope" (vector) $(d, e, f)$ . We impose the condition that at least one of $a - d, b - e, \textrm{ or } c - f$ is outside the range of $[1, 10]$ in order to ensure that $(a, b, c)$ is the first valid point on this line. Then, the line ranges from $(a, b, c), (a + d, b + e, c + f), \ldots, (a + 7d, b + 7e, c + 7f)$ , where $1 \le a + 7d, b + 7e, c + 7f \le 10$ , in which case at least one of $a + 8d, b + 8e, \textrm{ or } c + 8f$ is outside of the range $[1, 10]$ to ensure the line does not contain more than 8 points. For $1 \le a + 7d, b + 7e, c + 7f \le 10$ to be satisfied, the pairs $(a, d), (b, e), \textrm{ and } (c, f)$ can only be $(1, 0), (2, 0), \ldots (10, 0), (1, 1), (2, 1), (3, 1), (10, -1), (9, -1), \textrm{ and } (8, -1)$ . Notice that there are only two pairs such that $a + 8d, b + 8e, \textrm{ or } c + 8f \not \in [1, 10]$ , namely $(3, 1)$ and $(8, -1)$ . Thus, our line must contain at least one of these two pairs. In addition, only the two pairs $(1, 1)$ and $(10, -1)$ satisfy $a - d, b - e, \textrm{ or } c - f \not \in [1, 10]$ . Thus, we must also include at least one of these two pairs as well. Let us call these 4 pairs "important" pairs. Finally, we can include any valid pair for our third pair.
Case 1: We repeat one of the important pairs
There are 2 ways to choose from $(3, 1)$ and $(8, -1)$ , and 2 ways to choose from $(1, 1)$ and $(10, -1)$ . Then, there are 2 ways to choose the repeated pair. Next, we can arrange these pairs $3!/2! = 3$ ways. So, we have $2 \cdot 2 \cdot 2 \cdot 3 = 24.$
Case 2: We use 3 distinct important pairs
There are ${4 \choose 3}$ ways to choose the pairs (note that by pigeonhole principle, this guarantees we get at least one of each required pair). Then, there are $3! = 6$ ways to arrange it. We obtain $4 \cdot 6 = 24$
Case 3: We use 3 unique pairs, one that is not important.
Again, there are $2 \cdot 2 = 4$ ways to choose the important pairs. Then, there are 12 ways to choose the non-important pair. Again, there are $3! = 6$ ways to arrange it. So, we get $4 \cdot 12 \cdot 6 = 288$
Summing all of it up and dividing by two (since we over counted each line and its reversal), we get $\frac{24 + 24 + 288}{2} = \boxed{168}.$ | null | 168 |
99b60e3b7e667381614ac5b7d4d6a6ba | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_15 | Tetrahedron $ABCD$ has $AD=BC=28$ $AC=BD=44$ , and $AB=CD=52$ . For any point $X$ in space, suppose $f(X)=AX+BX+CX+DX$ . The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | Let $M$ and $N$ be midpoints of $\overline{AB}$ and $\overline{CD}$ . The given conditions imply that $\triangle ABD\cong\triangle BAC$ and $\triangle CDA\cong\triangle DCB$ , and therefore $MC=MD$ and $NA=NB$ . It follows that $M$ and $N$ both lie on the common perpendicular bisector of $\overline{AB}$ and $\overline{CD}$ , and thus line $MN$ is that common perpendicular bisector. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to line $MN$ . If $X$ is a point in space and $X'$ is the point symmetric to $X$ with respect to line $MN$ , then $BX=AX'$ and $CX=DX'$ , so $f(X) = AX+AX'+DX+DX'$
Let $Q$ be the intersection of $\overline{XX'}$ and $\overline{MN}$ . Then $AX+AX'\geq 2AQ$ , from which it follows that $f(X) \geq 2(AQ+DQ) = f(Q)$ . It remains to minimize $f(Q)$ as $Q$ moves along $\overline{MN}$
Allow $D$ to rotate about $\overline{MN}$ to point $D'$ in the plane $AMN$ on the side of $\overline{MN}$ opposite $A$ . Because $\angle DNM$ is a right angle, $D'N=DN$ . It then follows that $f(Q) = 2(AQ+D'Q)\geq 2AD'$ , and equality occurs when $Q$ is the intersection of $\overline{AD'}$ and $\overline{MN}$ . Thus $\min f(Q) = 2AD'$ . Because $\overline{MD}$ is the median of $\triangle ADB$ , the Length of Median Formula shows that $4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2$ and $MD^2 = 684$ . By the Pythagorean Theorem $MN^2 = MD^2 - ND^2 = 8$
Because $\angle AMN$ and $\angle D'NM$ are right angles, \[(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.\] It follows that $\min f(Q) = 2AD' = 4\sqrt{678}$ . The requested sum is $4+678=\boxed{682}$ | null | 682 |
d28f945e3141b62c36c690b201e72203 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_1 | For $-1<r<1$ , let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$ . Find $S(a)+S(-a)$ | The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$ . The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$ $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$ , so the answer is $\frac{2016}{6}=\boxed{336}$ | null | 336 |
bf6fa5b06c3a9fdc12d7cce12728dff1 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_2 | Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc. Then there are $21^2=441$ possible rolls. There are $2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\frac{56}{441}=\frac{8}{63}$ . The answer is $8+63=\boxed{071}$ | null | 071 |
bf6fa5b06c3a9fdc12d7cce12728dff1 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_2 | Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling $n$ is $\frac{n}{21}$ because $21=\frac{6 \cdot 7}{2}$ Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice.
The probability that 7 is rolled is now $2(\frac{1}{21} \cdot \frac{6}{21}+\frac{2}{21} \cdot \frac{5}{21} + \frac{3}{21} \cdot \frac{4}{21})$ which is equal to $\frac{56}{441}=\frac{8}{63}$ . Therefore the answer is $8+63=\boxed{071}$ ~PEKKA | null | 071 |
bf6fa5b06c3a9fdc12d7cce12728dff1 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_2 | Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Since the probability of rolling a $1$ is $\frac{1}{21}$ , the probability of rolling a $2$ is $\frac{2}{21}$ the probability of rolling a $3$ is $\frac{3}{21}$ and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of $1$ and $6$ $2$ and $5$ , and $3$ and $4$ , and the rest is symmetry and the others are irrelevant.
We have: $2 \cdot (\frac{1}{21} \cdot \frac{2}{7}$ $+$ $\frac{2}{21} \cdot \frac{5}{21}$ $+$ $\frac{1}{7} \cdot \frac{4}{21})$ $=$ $2 \cdot (\frac{2}{147} + \frac{10}{441} + \frac{4}{147})$ $2 \cdot \frac{4}{63} = \frac{8}{63}$ . Therefore, the answer is $8 + 63$ $\boxed{071}$ | null | 071 |
96c434e660faa817a9c01f2e70ef135d | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 | regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy] | Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to go to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 paths from that point down to the third plane. Within the third plane, there are 9 paths as well (consider the logic from the second plane). Finally, there is only one way down to the bottom vertex. Therefore there are $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1=\boxed{810}$ paths. | null | 810 |
96c434e660faa817a9c01f2e70ef135d | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 | regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy] | Assume an ant is on the top of this icosahedron. Note that the icosahedron has two pentagon planes and two points where the ant starts and ends. Also note that when the ant hits a vertex of the pentagon, there is only two ways to go down. When the ant ends up at the last vertex and is about to head down, there is only one way to go down.
Case $1$ : The ant move down to the first pentagon, then the next pentagon, and finally to the last point in a total of four moves; there are a total of $5 \cdot 2 \cdot 1 = 10$ ways to achieve this.
Case $2$ : The ant goes to the first pentagon and moves to another vertex in the same pentagon. The ant has a total of $8$ paths doing this since it may go through $1, 2, 3,$ or $4$ edges from either two directions when going across the edges of the pentagon (recall the ant may not repeat a move to another vertex) . Now the ant moves to the second pentagon below. The ant again has a total of $8$ moves if he wanders around the pentagon. Finally, the ant moves down after wandering to the last vertex. There is a total of $5 \cdot 8 \cdot 2 \cdot 8 = 640$ ways.
Case $3$ : Assume the ant goes to the first pentagon and wanders around. Then the ant goes the next pentagon and then heads directly to the last vertex. There are $5 \cdot 8 \cdot 2 = 80$ ways.
Case $4$ : Now let the ant go to the first pentagon and then go directly down to the next pentagon. The ant wanders around the second pentagon before heading to the last vertex. Like case $3$ above, there are $5 \cdot 2 \cdot 8 = 80$ ways.
Adding up all four cases, we get $10 + 640 + 80 + 80 =\boxed{810}$ total paths, as desired. ~skyscraper | null | 810 |
96c434e660faa817a9c01f2e70ef135d | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 | regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy] | Go to 2020 AMC 10A #19 , and connect all of the centers of the faces on the dodecahedron to get the icosahedron. The answer is $\boxed{810}$ | null | 810 |
f5e22abc369eb0d8da053c89ef008993 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_4 | A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60$ degrees. Find $h^2$ | Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid. Then $\angle CAB = 120^\circ$ . Let $X$ be the foot of the altitude from $A$ to $\overline{BC}$ . Then since $\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$ . Since the dihedral angle between $\triangle ABC$ and $\triangle BCD$ is $60^\circ$ $\triangle AXD$ is a $30-60-90$ triangle and $AD = 6\sqrt{3} = h$ . Thus $h^2 = \boxed{108}$ | null | 108 |
f5e22abc369eb0d8da053c89ef008993 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_4 | A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60$ degrees. Find $h^2$ | Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$ . Then, we have $AM=6$ $30-60-90$ ) triangles and $DM=\sqrt{36+h^2}$ by the Pythagorean Theorem. Applying the Law of Cosines, since $\angle AMD=60^{\circ}$ , we get \[h^2=36+h^2+36-\frac12 \cdot 12 \sqrt{36+h^2} \implies h^2=\boxed{108},\] as desired. | null | 108 |
ef60e92824e71892ae6e79a32ac51687 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_5 | Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$ | We could see that both $374$ and $319$ are divisible by $11$ in the outset, and that $34$ and $29$ , the quotients, are relatively prime. Both are the $average$ number of minutes across the $11$ days, so we need to subtract $\left \lfloor{\frac{11}{2}}\right \rfloor=5$ from each to get $(n,t)=(29,24)$ and $29+24=\boxed{053}$ | null | 053 |
ef60e92824e71892ae6e79a32ac51687 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_5 | Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$ | If we let $k$ be equal to the number of days it took to read the book, the sum of $n$ through $n+k$ is equal to $(2n+k)(k+1)=748$ Similarly, $(2t+k)(k+1)=638$ We know that both factors must be integers and we see that the only common multiple of $748$ and $638$ not equal to $1$ that will get us positive integer solutions for $n$ and $t$ is $11$ . We set $k+1=11$ so $k=10$ . We then solve for $n$ and $t$ in their respective equations, getting $2n+10=68$ $n=29$ We also get $2t+10=58$ $t=24$ . Our final answer is $29+24=\boxed{053}$ | null | 053 |
ef60e92824e71892ae6e79a32ac51687 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_5 | Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$ | Notice $374=34\cdot 11$ and $319=29\cdot 11$ . Also, note the sum of an arithmetic series is $\frac{2n+k}{2} \cdot b$ , where $n$ is our first term, $n+k$ is our final term, and $b$ is the number of terms. Since we know both sequences of $n$ and $t$ have the same length, and since $11$ is prime and shared by both $319$ and $374$ , we deduce that $b=11$ . Thus from here we know $2n+k=68$ and $2t+k=58$ by using our other factors $34$ and $29$ . Finally, we add the two systems up and we get $2t+2k+2n=126$ . But, notice that $k=b-1$ , since the first term has $k=0$ , and our last term has $k=b-1$ . Plugging this back into our equation we get $2n+2t=106 \implies n+t=\boxed{053}$ | null | 053 |
ef60e92824e71892ae6e79a32ac51687 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_5 | Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$ | We list two equations:
\begin{align*}
n+(n+1)+...+(n+k)&=374\\
t+(t+1)+...+(t+k)&=319.
\end{align*}
Subtracting the two, we get: \[(n-t)(k+1)=374-319=55.\] Manipulating the first and second equation, we get:
\begin{align*}
n(k+1)+\frac{k(k+1)}{2}&=374 \\
t(k+1)+\frac{k(k+1)}{2}&=319.
\end{align*}
We factor out the common factor $k+1$ :
\begin{align*}
(k+1)\left(n+\frac{k}{2}\right)&=374 \\
(k+1)\left(t+\frac{k}{2}\right)&=319.
\end{align*}
Note that $374$ and $319$ have a GCD of $11,$ now combining this with our equation that $(n-t)(k+1)=55,$ we see that $k+1$ has to equal $11.$ Thus, we get: \[(n,t)=(29,24) \Rightarrow n+t=29+24=\boxed{053}.\] Note: We see that because n-t=5, it becomes impossible for n+k/2 and t+k/2 to both be multiples of 11. Thus, this satisfies our condition. Thus k+1 must be 11 to satisfy the common factor 11 constraint. mathboy282 | null | 053 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Suppose we label the angles as shown below. [asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$ . Similarly, $\angle ABD=\gamma$ . Also, using $\triangle ICA$ , we find $\angle CIA=180-\alpha-\gamma$ . Therefore, $\angle AID=\alpha+\gamma$ . Therefore, $\angle DAI=\angle AID=\alpha+\gamma$ , so $\triangle AID$ must be isosceles with $AD=ID=5$ . Similarly, $BD=ID=5$ . Then $\triangle DLB \sim \triangle ALC$ , hence $\frac{AL}{AC} = \frac{3}{5}$ . Also, $AI$ bisects $\angle LAC$ , so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$ . Thus $CI = \frac{10}{3}$ , and the answer is $\boxed{013}$ | null | 013 |