id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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5,701 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | 4 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | In the problem, we are given that \[8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.\] We multiply by the LCM of the denominators, which is $576$ to get \[4608a+576b+72c+9d=6912b+576b+48b+4a.\] We then group like terms and factor to get \[4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.\]
Observe that the ... | 321 |
5,702 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | 1 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Let $c=6-(a+b)$ , and note that $\binom{6}{a + b}=\binom{6}{c}$ . The problem thus asks for the sum $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ . Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by cons... | 564 |
5,703 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | 3 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Treating $a+b$ as $n$ , this problem asks for \[\sum_{n=0}^{6} \left[\binom{6}{n}\sum_{m=0}^{n}\left[\binom{6}{m}\binom{6}{n-m}\right]\right].\] But \[\sum_{m=0}^{n} \left[\binom{6}{m} \binom{6}{n-m}\right]\] can be computed through the following combinatorial argument. Choosing $n$ elements from a set of size $12$ is... | 564 |
5,704 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | 4 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Case 1: $a<b$
Subcase 1: $a=0$ \[\binom{6}{0}\binom{6}{1}\binom{6}{1}=36\] \[\binom{6}{0}\binom{6}{2}\binom{6}{2}=225\] \[\binom{6}{0}\binom{6}{3}\binom{6}{3}=400\] \[\binom{6}{0}\binom{6}{4}\binom{6}{4}=225\] \[\binom{6}{0}\binom{6}{5}\binom{6}{5}=36\] \[\binom{6}{0}\binom{6}{6}\binom{6}{6}=1\] \[36+225+400+225+36+1=9... | 564 |
5,705 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | 5 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | We begin as in solution 1 to rewrite the sum as $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ .
Consider the polynomial $P(x)=\left(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6\right)^3$ .
We can see the sum we wish to compute is ... | 564 |
5,706 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | 6 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Let $c=6-(a+b)$ . Then $\binom{6}{a+b}=\binom{6}{c}$ , and $a+b+c=6$ . The problem thus asks for \[\sum_{a+b+c=6}\binom{6}{a}\binom{6}{b}\binom{6}{c} \pmod {1000}.\] Suppose we have $6$ red balls, $6$ green balls, and $6$ blue balls lined up in a row, and we want to choose $6$ balls from this set of $18$ balls by consi... | 564 |
5,707 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | 7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Since $\binom{6}{n}=\binom{6}{6-n}$ , we can rewrite $T(a,b)$ as $\binom{6}{a}\binom{6}{b}\binom{6}{6-(a+b)}$ . Consider the number of ways to choose a committee of 6 people from a group of 6 democrats, 6 republicans, and 6 independents. We can first pick $a$ democrats, then pick $b$ republicans, provided that $a+b \le... | 564 |
5,708 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | 2 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\ang... | Put $\triangle POQ$ and $\triangle POR$ with $O$ on the origin and the triangles on the $1^{st}$ quadrant.
The coordinates of $Q$ and $P$ is $(200 \cos^{2}a,200 \cos a\sin a )$ $(200\cos^{2}b,200\cos(b)\sin b)$ . So $PQ^{2}$ $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}$ , which we... | 41 |
5,709 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | 4 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\ang... | Impose a coordinate system as follows:
Let the midpoint of $\overline{OP}$ be the origin, and let $\overline{OP}$ be the x-axis. We construct a circle with center at the origin with radius 100. Since $\angle OQP$ and $\angle ORP$ are both right angles, points $Q$ and $R$ are on our circle. Place $Q$ and $R$ in the firs... | 41 |
5,710 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_10 | 5 | 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 nume... | 56 |
5,711 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | 1 | 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 $m = 5$ . ... | 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 \{... | 360 |
5,712 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | 2 | 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 $m = 5$ . ... | (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 ... | 360 |
5,713 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | 3 | 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 $m = 5$ . ... | 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 ... | 360 |
5,714 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | 5 | 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 $m = 5$ . ... | 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... | 360 |
5,715 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_11 | 6 | 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 $m = 5$ . ... | 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=2073... | 360 |
5,716 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | 1 | 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,... | 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 $S$ $\{3\}$ $\{... | 252 |
5,717 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | 2 | 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,... | 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 subse... | 252 |
5,718 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | 3 | 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,... | 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 1... | 252 |
5,719 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | 4 | 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,... | 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... | 252 |
5,720 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | 1 | 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 pri... | 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 $(x,y)$ ... | 145 |
5,721 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | 2 | 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 pri... | 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 $A... | 145 |
5,722 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | 3 | 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 pri... | 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 tri... | 145 |
5,723 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | 4 | 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 pri... | 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,... | 145 |
5,724 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | 5 | 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 pri... | 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(\... | 145 |
5,725 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | 6 | 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 pri... | 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 vert... | 145 |
5,726 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | 7 | 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 pri... | 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 pr... | 145 |
5,727 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | 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, ... | 196 |
5,728 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | 2 | 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 c... | 196 |
5,729 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | 3 | 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.... | 196 |
5,730 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_1 | 4 | 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 combi... | 196 |
5,731 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_2 | 1 | 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 out... | 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 pr... | 781 |
5,732 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | 1 | 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);... | 409 |
5,733 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | 2 | 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 wha... | 409 |
5,734 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | 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 dist... | 409 |
5,735 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | 4 | 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}|... | 409 |
5,736 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | 5 | 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}... | 409 |
5,737 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_3 | 6 | 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 trian... | 409 |
5,738 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_4 | 1 | 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 ... | 222 |
5,739 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_4 | 2 | 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..... | 222 |
5,740 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_4 | 3 | 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... | 222 |
5,741 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_5 | 1 | 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 num... | 791 |
5,742 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_5 | 2 | 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 thi... | 791 |
5,743 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_5 | 3 | 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 $c$ . S... | 791 |
5,744 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | 1 | 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+... | 195 |
5,745 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | 2 | 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
\[... | 195 |
5,746 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | 3 | 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, ... | 195 |
5,747 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | 4 | 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... | 195 |
5,748 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | 5 | 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 ... | 195 |
5,749 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_6 | 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.
$k^2-(n+42.5)^2 = 210.75$
$(k+n+42.5)(k-n-42.5) = 210.75$
F... | 195 |
5,750 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_7 | 1 | 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(grap... | 501 |
5,751 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_7 | 2 | 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... | 501 |
5,752 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_7 | 3 | 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 r... | 501 |
5,753 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | 1 | 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 $7... | 134 |
5,754 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | 2 | 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... | 134 |
5,755 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | 3 | 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!}... | 134 |
5,756 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_8 | 4 | 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 $n=2k$ ... | 134 |
5,757 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_9 | 6 | 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 c... | 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 uniqu... | 13 |
5,758 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_10 | 1 | 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 ... | [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); labe... | 546 |
5,759 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | 1 | 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 $3$... | 544 |
5,760 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | 2 | 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 $A$ ,... | 544 |
5,761 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | 3 | 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 ca... | 544 |
5,762 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | 4 | 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 ... | 544 |
5,763 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | 5 | 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, \omeg... | 544 |
5,764 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | 1 | 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 tang... | 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 ... | 110 |
5,765 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | 2 | 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 tang... | 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 $49+61=\boxed{110}$ | 110 |
5,766 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | 3 | 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 tang... | 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 rotati... | 110 |
5,767 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | 4 | 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 tang... | 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((-0.1448535... | 110 |
5,768 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | 5 | 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 tang... | 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... | 110 |
5,769 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_13 | 1 | 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 r... | 245 |
5,770 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_13 | 2 | 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 th... | 245 |
5,771 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_13 | 3 | 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 ... | 245 |
5,772 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | 1 | $10\times10\times10$ 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 $(10,10,10)$ .... | 168 |
5,773 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | 2 | $10\times10\times10$ 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 th... | 168 |
5,774 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | 3 | $10\times10\times10$ 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 ... | 168 |
5,775 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | 4 | $10\times10\times10$ 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), (... | 168 |
5,776 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_15 | 1 | 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{... | 682 |
5,777 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_1 | 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 $\frac{2016}{6}=\boxed{336}$ | 336 |
5,778 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 | 1 | 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, ... | 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 fro... | 810 |
5,779 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 | 2 | 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, ... | 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 ... | 810 |
5,780 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 | 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, ... | 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 $\boxed{810}$ | 810 |
5,781 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_4 | 1 | 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 t... | 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... | 108 |
5,782 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_4 | 2 | 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 t... | 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 $AM=6$ $30-6... | 108 |
5,783 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | 9 | 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 relativ... | Without loss of generality, let $\triangle ABC$ be isosceles. Note that by the incenter-excenter lemma, $DI = DA = DB.$ Hence, $DA=DB=5.$ Let the point of tangency of the incircle and $\overline{BC}$ be $F$ and the point of tangency of the incircle and $\overline{AC}$ be $E.$ We note that $\angle ALC = \angle BLC = 90^... | 13 |
5,784 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_7 | 1 | For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i\]
Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number. | We consider two cases:
Case 1: $ab \ge -2016$
In this case, if \[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\frac{\sqrt{|a+b|}}{ab+100}\] then $ab \ne -100$ and $|a + b| = 0 = a + b$ . Thus $ab = -a^2$ so $a^2 < 2016$ . Thus $a = -44,-43, ... , -1, 0, 1, .... | 103 |
5,785 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | 1 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | We start by drawing a diagram;
[asy] size(400); import olympiad; import geometry; pair A = (0, 20) ,B=(30,10) ,C=(15,0), Q=(30,20) ,R=(30,0), S=(0,0); draw(A--B--C--cycle); draw(A--Q); draw(Q--R); draw(R--S); draw(S--A); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$Q$", Q, E); label("$R$", R... | 744 |
5,786 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | 2 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\text{cis} A$ , and let $z$ be a complex number with $|z|=1$ $\text{Arg}(z)\ge 0^\circ$ and $\text{Arg}(z... | 744 |
5,787 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | 3 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | Let $\theta$ be the angle $\angle BAQ$ . The height of the rectangle then can be expressed as $h = 31 \sin (A+\theta)$ , and the length of the rectangle can be expressed as $l = 40\cos \theta$ . The area of the rectangle can then be written as a function of $\theta$ $[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos ... | 744 |
5,788 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | 4 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | Let $\alpha$ be the angle $\angle CAS$ and $\beta$ be the angle $\angle BAQ$ . Then \[\alpha + \beta + \angle A = 90^\circ \Rightarrow \alpha + \beta = 90^\circ - \angle A\] \[\cos(\alpha + \beta) = \cos(90^\circ - \angle A)\] \[\cos(\alpha + \beta) = \sin(\angle A) = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sin\alpha\si... | 744 |
5,789 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_10 | 2 | A strictly increasing sequence of positive integers $a_1$ $a_2$ $a_3$ $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ $a_{2k}$ $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ $a_{2k+1}$ $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ | The thirteenth term of the sequence is $2016$ , which makes that fourteenth term of the sequence $2016+r$ and the $15^{\text{th}}$ term $\frac{(2016+r)^2}{2016}$ . We note that $r$ is an integer so that means $\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$ , which is $168$ . We bash all the ... | 504 |
5,790 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_10 | 3 | A strictly increasing sequence of positive integers $a_1$ $a_2$ $a_3$ $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ $a_{2k}$ $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ $a_{2k+1}$ $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ | Let $a_{2k-1}=s$ where $k=1$ . Then, $a_{2k}=sr \Longrightarrow a_{2k+1}=sr^2$ . Continuing on, we get $a_{2k+2}=sr(r-1)+sr^2=sr(2r-1) \Longrightarrow a_{2k+3}=sr^2(\frac{2r-1}{r})^2=s(2r-1)^2$ . Moreover, $a_{2k+4}=s(2r-1)(r-1)+s(2r-1)^2=s(2r-1)(3r-2) \Longrightarrow a_{2k+5}=s(2r-1)^2(\frac{3r-2}{2r-1})^2=s(3r-2)^2$
... | 504 |
5,791 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | 1 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Plug in $x=1$ to get $(1-1)P(1+1) = 0 = (1+2)P(1) \Rightarrow P(1) = 0$ . Plug in $x=0$ to get $(0-1)P(0+1) = (0+2)P(0)\Rightarrow P(0) = -\frac{1}{2}P(1) = 0$ . Plug in $x=-1$ to get $(-1-1)P(-1+1) = (-1+2)P(-1)\Rightarrow (-2)P(0)=P(-1)\Rightarrow P(-1) = 0$ . So $P(x) = x(x-1)(x+1)Q(x)$ for some polynomial $Q(x)$ . ... | 109 |
5,792 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | 2 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | From the equation we see that $x-1$ divides $P(x)$ and $(x+2)$ divides $P(x+1)$ so we can conclude that $x-1$ and $x+1$ divide $P(x)$ (if we shift the function right by 1, we get $(x-2)P(x) = (x+1)P(x-1)$ , and from here we can see that $x+1$ divides $P(x)$ ). This means that $1$ and $-1$ are roots of $P(x)$ . Plug in ... | 109 |
5,793 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | 3 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Although this may not be the most mathematically rigorous answer, we see that $\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}$ . Using a bit of logic, we can make a guess that $P(x+1)$ has a factor of $x+2$ , telling us $P(x)$ has a factor of $x+1$ . Similarly, we guess that $P(x)$ has a factor of $x-1$ , which means $P(x+1)$ has... | 109 |
5,794 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | 4 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Substituting $x=2$ into the given equation, we find that $P(3)=4P(2)=P(2)^2$ . Therefore, either $P(2)=0$ or $P(2)=4$ . Now for integers $n\ge 2$ , we know that \[P(n+1)=\frac{n+2}{n-1}P(n).\] Applying this repeatedly, we find that \[P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).\] If $P(2)=0$ , this shows that $P(x)$ has infini... | 109 |
5,795 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | 5 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We can find zeroes of the polynomial by making the first given equation $0 = 0$ . Plugging in $x = 1$ and $x = -2$ gives us the zeroes $1$ and $-1$ , respectively. Now we can plug in these zeros to get more zeroes. $x = -1$ gives us the zero $0$ (no pun intended). $x = 1$ makes the equation $0 \cdot P(2) = 0$ , which m... | 109 |
5,796 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | 6 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Plug in $x=2$ yields $P(3)=4P(2)$ . Since also $(P(2))^2=P(3)$ , we have $P(2)=4$ and $P(3)=16$ . Plug in $x=3$ yields $2P(4)=5P(3)$ so $P(4)=40$
Repeat the action gives $P(2)=4$ $P(3)=16$ $P(4)=40$ $P(5)=80$ , and $P(6)=140$
Since $P(x)$ is a polynomial, the $k$ th difference is constant, where $k=\deg(P(x))$ . Thus w... | 109 |
5,797 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_12 | 1 | Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes. | Suppose $p=11$ ; then $m^2-m+11=11qrs$ . Reducing modulo 11, we get $m\equiv 1,0 \pmod{11}$ so $k(11k\pm 1)+1 = qrs$
Suppose $q=11$ . Then we must have $11k^2\pm k + 1 = 11rs$ , which leads to $k\equiv \mp 1 \pmod{11}$ , i.e., $k\in \{1,10,12,21,23,\ldots\}$
$k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs... | 132 |
5,798 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_12 | 2 | Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes. | Let $m^2 - m + 11 = pqrs$ for primes $p, q, r, s\ge11$ . If $p, q, r, s = 11$ , then $m^2-m+11=11^4$ . We can multiply this by $4$ and complete the square to find $(2m-1)^2=4\cdot 11^4-43$ . But \[(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,\] hence we have pinned a perfect square $(2m-1... | 132 |
5,799 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_13 | 1 | Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$ . A fence is located at the horizontal line $y = 0$ . On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a poi... | Clearly Freddy's $x$ -coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$ -coordinate. Observe that $E(24)=0$ , and \[E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}\] for all $y$ such that $1\le y\le 23$ . Also note that $E(0)=1+\frac{2E(0)+... | 273 |
5,800 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14 | 1 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ | First note that $1001 = 143 \cdot 7$ and $429 = 143 \cdot 3$ so every point of the form $(7k, 3k)$ is on the line. Then consider the line $l$ from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$ . Translate the line $l$ so that $(7k, 3k)$ is now the origin. There is one square and one circle that intersect the line around $(... | 574 |
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