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f57e23a3227324264a5d93cb24506175 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $BM = d, AM = x, CM = 15 - x$ . Observe that we have the equation by the incircle formula: \[\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.\] Now let $X$ be the point of tangency between the incircle of $\triangle ABC$ and $AC$ . Additionally, let $P$ and $Q$ be the points of tangency between the incircles of $\triangle ABM$ and $\triangle CBM$ with $AC$ respectively. Some easy calculation yields $AX = 7, CX = 8$ . By homothety we have \[\frac{AP}{7} = \frac{CQ}{8} \implies 8(AP) = 7(CQ) \implies 8(12 + x - d) = 7(13 + 15 - x - d) \implies d = 15x - 100.\] Substituting into the first equation derived earlier it is left to solve \[\frac{x}{15 - x} = \frac{15x - 88}{15x - 87} \implies 3x^2 - 40x + 132 \implies (x - 6)(3x - 22) = 0.\] Now $x = 6$ yields $d = -10$ which is invalid, hence $x = \frac{22}{3}$ so $\frac{AM}{CM} = \frac{\frac{22}{3}}{15 - \frac{22}{3}} = \frac{22}{23}.$ The requested sum is $22 + 23 = \boxed{45}$ . ~blueprimes | null | 45 |
c7811420543755fda69ab9667961fa5f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_2 | A point $P$ is chosen at random in the interior of a unit square $S$ . Let $d(P)$ denote the distance from $P$ to the closest side of $S$ . The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Any point outside the square with side length $\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\frac{3}{5}$ that has the same center and orientation as the unit square has $\frac{1}{5}\le d(P)\le\frac{1}{3}$
Since the area of the unit square is $1$ , the probability of a point $P$ with $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is the area of the shaded region, which is the difference of the area of two squares.
$\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}$
Thus, the answer is $56 + 225 = \boxed{281}.$ | null | 281 |
c7811420543755fda69ab9667961fa5f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_2 | A point $P$ is chosen at random in the interior of a unit square $S$ . Let $d(P)$ denote the distance from $P$ to the closest side of $S$ . The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | First, let's figure out $d(P) \geq \frac{1}{3}$ which is \[\left(\frac{3}{5}\right)^2=\frac{9}{25}.\] Then, $d(P) \geq \frac{1}{5}$ is a square inside $d(P) \geq \frac{1}{3}$ , so \[\left(\frac{1}{3}\right)^2=\frac{1}{9}.\] Therefore, the probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is \[\frac{9}{25}-\frac{1}{9}=\frac{56}{225}\] So, the answer is $56+225=\boxed{281}$ | null | 281 |
c7811420543755fda69ab9667961fa5f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_2 | A point $P$ is chosen at random in the interior of a unit square $S$ . Let $d(P)$ denote the distance from $P$ to the closest side of $S$ . The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | First, lets assume that point $P$ is closest to a side $S$ of the square. If it is $\frac{1}{5}$ far from $S$ , then it should be at least $\frac{1}{5}$ from both the adjacent sides of $S$ in the square. This leaves a segment of $1 - 2 \cdot \frac{1}{5} = \frac{3}{5}$ . If the distance from $P$ to $S$ is $\frac{1}{3}$ , then notice the length of the side-ways segment for $P$ is $1 - 2 \cdot \frac{1}{3} = \frac{1}{3}$ . Notice that as the distance from $P$ to $S$ increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides $\frac{3}{5}$ and $\frac{1}{3}$ with height $\frac{1}{3} - \frac{1}{5} = \frac{2}{15}$ . This trapezoid has area (or probability for one side) $\frac{1}{2} \cdot \left(\frac{1}{3}+\frac{3}{5}\right)\cdot \frac{2}{15} = \frac{14}{225}$ . Since the square has $4$ sides, we multiply by $4$ . Hence, the probability is $\frac{56}{225}$ . The answer is $\boxed{281}$ . ~Saucepan_man02 | null | 281 |
47c1b27e1a555699e42f58a0423f8394 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_3 | Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$ . Find the greatest positive integer $n$ such that $2^n$ divides $K$ | In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$ . Thus, the product is $(1^{19})(2^{18})\cdots(19^1)$ (or alternatively, $19! \cdot 18! \cdots 1!$ .)
When we count the number of factors of $2$ , we have 4 groups, factors that are divisible by $2$ at least once, twice, three times and four times.
Summing these give an answer of $\boxed{150}$ | null | 150 |
fdc14e3366370108644d0ff2d506b696 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_4 | Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $400$ feet or less to the new gate be a fraction $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | There are $12 \cdot 11 = 132$ possible situations ( $12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart.
If we number the gates $1$ through $12$ , then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$ $6$ $7$ $8$ have eight. Therefore, the number of valid gate assignments is \[2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76\] so the probability is $\frac{76}{132} = \frac{19}{33}$ . The answer is $19 + 33 = \boxed{052}$ | null | 052 |
fdc14e3366370108644d0ff2d506b696 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_4 | Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $400$ feet or less to the new gate be a fraction $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | As before, derive that there are $132$ possibilities for Dave's original and replacement gates.
Now suppose that Dave has to walk $100k$ feet to get to his new gate. This means that Dave's old and new gates must be $k$ gates apart. (For example, a $100$ foot walk would consist of the two gates being adjacent to each other.) There are $12-k$ ways to pick two gates which are $k$ gates apart, and $2$ possibilities for gate assignments, for a total of $2(12-k)$ possible assignments for each $k$
As a result, the total number of valid gate arrangements is \[2\cdot 11 + 2\cdot 10 + 2\cdot 9 + 2\cdot 8 = 76\] and so the requested probability is $\tfrac{19}{33}$ for a final answer of $\boxed{052}$ | null | 052 |
833ca79aa700a9e8f2c52b453b2da674 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_5 | Positive numbers $x$ $y$ , and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$ . Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$ | Using the properties of logarithms, $\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468$ by using the fact that $\log_{10}ab = \log_{10}a + \log_{10}b$
Through further simplification, we find that $\log_{10}x+\log_{10}y+\log_{10}z = 81$ . It can be seen that there is enough information to use the formula $\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$ , as we have both $\ a+b+c$ and $\ 2ab+2ac+2bc$ , and we want to find $\sqrt {a^2 + b^2 + c^2}$
After plugging in the values into the equation, we find that $\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2$ is equal to $\ 6561 - 936 = 5625$
However, we want to find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$ , so we take the square root of $\ 5625$ , or $\boxed{075}$ | null | 075 |
833ca79aa700a9e8f2c52b453b2da674 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_5 | Positive numbers $x$ $y$ , and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$ . Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$ | Let $a=\log_{10}x$ $b=\log_{10}y,$ and $c=\log_{10}z$
We have $a+b+c=81$ and $a(b+c)+bc=ab+ac+bc=468$ . Since these two equations look a lot like Vieta's for a cubic, create the polynomial $x^3-81x^2+468x=0$ (leave the constant term as $0$ to make things easy). Dividing by $x$ yields $x^2-81x+468=0$
Now we use the quadratic formula:
$x=\frac{81\pm\sqrt{81^2-4\cdot468}}{2}$
$x=\frac{81\pm\sqrt{4689}}{2}$
$x=\frac{81+3\sqrt{521}}{2}$ $x=\frac{81-3\sqrt{521}}{2}$
Since the question asks for $\sqrt{a^2+b^2+c^2}$ (remember one of the values was the solution $x=0$ that we divided out in the beginning), we find:
$\sqrt{\left(\frac{81+3\sqrt{521}}{2}\right)^2+\left(\frac{81-3\sqrt{521}}{2}\right)^2}$
$=\sqrt{2\cdot\frac{81^2+(3\sqrt{521})^2}{4}}$
$=\sqrt{\frac{11250}{2}}$
$=\boxed{075}$ | null | 075 |
acf08e3f64dbc1dc7ed0add40d766076 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_6 | Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients. | You can factor the polynomial into two quadratic factors or a linear and a cubic factor.
For two quadratic factors, let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that
\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\]
Therefore, again setting coefficients equal, $a + c = 0\Longrightarrow a=-c$ $b + d + ac = 0\Longrightarrow b+d=a^2$ $ad + bc = - n$ , and so $bd = 63$
Since $b+d=a^2$ , the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$ . From this we find that the possible values for $n$ are $\pm 8 \cdot 62$ and $\pm 4 \cdot 2$
For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest $n$ in that case to be $48$
Therefore, the answer is $4 \cdot 2 = \boxed{008}$ | null | 008 |
acf08e3f64dbc1dc7ed0add40d766076 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_6 | Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients. | Let $x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$ . From this, we get that $bd=63\implies d=\frac{63}{b}$ and $a+c=0\implies c=-a$ . Plugging this back into the equation, we get $x^4-nx+63=(x^2+ax+b)\left(x^2-ax+\frac{63}{b}\right)$ . Expanding gives us $x^4-nx+63=x^4+\left(-a^2+b+\frac{63}{b}\right)x^2+\left(\frac{63a}{b}-ab\right)x+63$ . Therefore $-a^2+b+\frac{63}{b}=0$ . Simplifying gets us $b(a^2-b)=63$ . Since $a$ and $b$ must be integers, we can use guess and check for values of $b$ because $b$ must be a factor of $63$ . Note that $b$ cannot be negative because $a$ would be imaginary. After guessing and checking, we find that the possible values of $(a,b)$ are $(\pm 8, 1), (\pm 4, 7), (\pm 4, 9),$ and $(\pm 8, 63)$ . We have that $n=ab-\frac{63a}{b}$ . Plugging in our values for $a$ and $b$ , we get that the smallest positive integer value $n$ can be is $\boxed{008}$ .
-Heavytoothpaste | null | 008 |
bbbe80f71be5dd3e8176e2adb420d8b9 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_7 | Let $P(z)=z^3+az^2+bz+c$ , where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ | Set $w=x+yi$ , so $x_1 = x+(y+3)i$ $x_2 = x+(y+9)i$ $x_3 = 2x-4+2yi$
Since $a,b,c\in{R}$ , the imaginary part of $a,b,c$ must be $0$
Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$
and therefore: $x_1 = x$ $x_2 = x+6i$ $x_3 = 2x-4-6i$
Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: $x(x+6i)(2x-4-6i)$ . The imaginary part is $6x^2-24x$ , which is 0, and therefore $x=4$ , since $x=0$ doesn't work.
So now, $x_1 = 4, x_2 = 4+6i, x_3 = 4-6i$
and therefore: $a=-12, b=84, c=-208$ . Finally, we have $|a+b+c|=|-12+84-208|=\boxed{136}$ | null | 136 |
bbbe80f71be5dd3e8176e2adb420d8b9 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_7 | Let $P(z)=z^3+az^2+bz+c$ , where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ | Note that at least one of $w+3i$ $w+9i$ , or $2w-4$ is real by complex conjugate roots. We now separate into casework based on which one.
Let $w=x+yi$ , where $x$ and $y$ are reals.
Case 1: $w+3i$ is real. This implies that $x+yi+3i$ is real, so by setting the imaginary part equal to zero we get $y=-3$ , so $w=x-3i$ . Now note that since $w+3i$ is real, $w+9i$ and $2w-4$ are complex conjugates. Thus $\overline{w+9i}=2w-4$ , so $\overline{x+6i}=2(x-3i)-4$ , implying that $x=4$ , so $w=4-3i$
Case 2: $w+9i$ is real. This means that $x+yi+9i$ is real, so again setting imaginary part to zero we get $y=-9$ , so $w=x-9i$ . Now by the same logic as above $w+3i$ and $2w-4$ are complex conjugates. Thus $\overline{w+3i}=2w-4$ , so $\overline{x-6i}=2(x-9i)-4$ , so $x+6i=2x-4-18i$ , which has no solution as $x$ is real.
Case 3: $2w-4$ is real. Going through the same steps, we get $y=0$ , so $w=x$ . Now $w+3i$ and $w+6i$ are complex conjugates, but $w=x$ , which means that $\overline{x+3i}=x+6i$ , so $x-3i=x+6i$ , which has no solutions.
Thus case 1 is the only one that works, so $w=4-3i$ and our polynomial is $(z-(4))(z-(4+6i))(z-(4-6i))$ . Note that instead of expanding this, we can save time by realizing that the answer format is $|a+b+c|$ , so we can plug in $z=1$ to our polynomial to get the sum of coefficients, which will give us $a+b+c+1$ . Plugging in $z=1$ into our polynomial, we get $(-3)(-3-6i)(-3+6i)$ which evaluates to $-135$ . Since this is $a+b+c+1$ , we subtract 1 from this to get $a+b+c=-136$ , so $|a+b+c|=\boxed{136}$ | null | 136 |
115f9929504b44d0823cf81a0d1d3eb7 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_8 | Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
Find $N$ | Let us partition the set $\{1,2,\cdots,12\}$ into $n$ numbers in $A$ and $12-n$ numbers in $B$
Since $n$ must be in $B$ and $12-n$ must be in $A$ $n\ne6$ , we cannot partition into two sets of 6 because $6$ needs to end up somewhere, $n\ne 0$ or $12$ either).
We have $\dbinom{10}{n-1}$ ways of picking the numbers to be in $A$
So the answer is $\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}$ | null | 772 |
115f9929504b44d0823cf81a0d1d3eb7 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_8 | Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
Find $N$ | Regardless of the size $n$ of $A$ (ignoring the case when $n = 6$ ), $n$ must not be in $A$ and $12 - n$ must be in $A$
There are $10$ remaining elements whose placements have yet to be determined. Note that the actual value of $n$ does not matter; there is always $1$ necessary element, $1$ forbidden element, and $10$ other elements that need to be distributed. There are $2$ places to put each of these elements, for $2^{10}$ possibilities.
However, there is the edge case of $n = 6; 6$ is forced not the be in either set, so we must subtract the $\dbinom{10}{5}$ cases where $A$ and $B$ have size $6$
Thus, our answer is $2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}$ | null | 772 |
115f9929504b44d0823cf81a0d1d3eb7 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_8 | Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
Find $N$ | The total number of possible subsets is $\sum_{i=1}^{11}\dbinom{12}{i}$ , which is $2^{12}-2$ . Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude $i=0$ and $i=12$ since they leave a set empty. We proceed with complementary counting and casework:
We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where $|A|$ is in $A$ , which violates the first condition. Then we find the ways where the elements $|B|$ and $12-|B|$ are in set $B$ , which violates only the second condition, and not the first. This ensures we do not overcount.
Case 1: $|A|$ is an element in $A$
There are $\sum_{i=1}^{11}\dbinom{11}{i-1}$ $2^{11}-1$ ways in this case.
Case 2: $|B|$ and $12-|B|$ are in $B$
We introduce a subcase where $|B|$ is not 6, since the other element would also be 6.
There are $B-2$ elements to choose from 10 elements. Therefore, $|B|$ can be from 2 to 11.
There are $\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211$ ways.
The other subcase is when $|B|$ is equal to 6. There are $\dbinom{11}{5}=462$ ways.
Adding the subcases gives us $2^{10}+251$
Adding this with case one gives us $2^{11}+2^{10}+250$ . Subtracting this from $2^{12}-2$ gives $1024-252=\boxed{772}$ | null | 772 |
e5b3c1eebd4b239188657fd05a97dae4 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_9 | Let $ABCDEF$ be a regular hexagon . Let $G$ $H$ $I$ $J$ $K$ , and $L$ be the midpoints of sides $AB$ $BC$ $CD$ $DE$ $EF$ , and $AF$ , respectively. The segments $\overline{AH}$ $\overline{BI}$ $\overline{CJ}$ $\overline{DK}$ $\overline{EL}$ , and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Without loss of generality, let $BC=2.$
Note that $\angle BMH$ is the vertical angle to an angle of the regular hexagon, so it has a measure of $120^\circ$
Because $\triangle ABH$ and $\triangle BCI$ are rotational images of one another, we get that $\angle{MBH}=\angle{HAB}$ and hence $\triangle ABH \sim \triangle BMH \sim \triangle BCI$
Using a similar argument, $NI=MH$ , and
\[MN=BI-NI-BM=BI-(BM+MH).\]
Applying the Law of cosines on $\triangle BCI$ $BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}$
\begin{align*}\frac{BC+CI}{BI}&=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH} \\ BM+MH&=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}} \\ MN&=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}} \\ \frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}&=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}
Thus, the answer is $4 + 7 = \boxed{011}.$ | null | 011 |
e5b3c1eebd4b239188657fd05a97dae4 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_9 | Let $ABCDEF$ be a regular hexagon . Let $G$ $H$ $I$ $J$ $K$ , and $L$ be the midpoints of sides $AB$ $BC$ $CD$ $DE$ $EF$ , and $AF$ , respectively. The segments $\overline{AH}$ $\overline{BI}$ $\overline{CJ}$ $\overline{DK}$ $\overline{EL}$ , and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We can use coordinates. Let $O$ be at $(0,0)$ with $A$ at $(1,0)$
then $B$ is at $(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$C$ is at $(\cos(120^\circ),\sin(120^\circ))=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$D$ is at $(\cos(180^\circ),\sin(180^\circ))=(-1,0)$
\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\ &I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}
Line $AH$ has the slope of $-\frac{\sqrt{3}}{2}$ and the equation of $y=-\frac{\sqrt{3}}{2}(x-1)$
Line $BI$ has the slope of $\frac{\sqrt{3}}{5}$ and the equation $y-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)$
Let's solve the system of equation to find $M$
\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\ -5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\ x&=\frac{1}{7} \\ y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}
Finally,
\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\ &\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}
Thus, the answer is $\boxed{011}$ | null | 011 |
10185faafbd5e2fa46eb23a059f3304f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_10 | Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ | Let $f(x) = a(x-r)(x-s)$ . Then $ars=2010=2\cdot3\cdot5\cdot67$ . First consider the case where $r$ and $s$ (and thus $a$ ) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$ $r$ , and $s$ . However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$ , we have $r=s$ . The other $80$ cases are double counting, so there are $40$
We must now consider the various cases of signs. For the $40$ cases where $|r|\neq |s|$ , there are a total of four possibilities, For the case $|r|=|s|=1$ , there are only three possibilities, $(r,s) = (1,1); (1,-1); (-1,-1)$ as $(-1,1)$ is not distinguishable from the second of those three.
You may ask: How can one of ${r, s}$ be positive and the other negative? $a$ will be negative as a result. That way, it's still $+2010$ that gets multiplied.
Thus the grand total is $4\cdot40 + 3 = \boxed{163}$ | null | 163 |
10185faafbd5e2fa46eb23a059f3304f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_10 | Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ | We use Burnside's Lemma . The set being acted upon is the set of integer triples $(a,r,s)$ such that $ars=2010$ . Because $r$ and $s$ are indistinguishable, the permutation group consists of the identity and the permutation that switches $r$ and $s$ . In cycle notation, the group consists of $(a)(r)(s)$ and $(a)(r \: s)$ . There are $4 \cdot 3^4$ fixed points of the first permutation (after distributing the primes among $a$ $r$ $s$ and then considering their signs. We have 4 ways since we can keep them all positive, first 2 negative, first and third negative, or last two negative) and $2$ fixed points of the second permutation ( $r=s=\pm 1$ ). By Burnside's Lemma, there are $\frac{1}{2} (4 \cdot 3^4+2)= \boxed{163}$ distinguishable triples $(a,r,s)$ .
Note: The permutation group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ | null | 163 |
10185faafbd5e2fa46eb23a059f3304f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_10 | Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ | The equation can be written in the form of $k(x-a)(x-b)$ , where $|k|$ is a divisor of $2010$ . Factor $2010=2\cdot 3\cdot 5\cdot 67$ . We divide into few cases to study.
Firstly, if $|k|=1$ , the equation can be $-(x-a)(x+b)$ or $(x-a)(x-b)$ or $(x+a)(x+b)$ , there are $2^4+2^4=32$ ways
If $|k|\in \{2,3,5,67\}$ . Take $|k|=2$ as an example, follow the procedure above, there are $2^3+2^3=16$ ways, and there are $\binom {4}{1}=4$ ways to choose $|k|$ , so there are $16\cdot 4=64$ ways
If $|k|$ is the product of two factors of $2010$ , there are $8$ ways for each. Thus there are $\binom {4}{2}\cdot 8=48$ ways
If $|k|$ is the product of three factors of $2010$ , there are $\binom {4}{3}\cdot 4=16$ ways
In the end, $|k|=2010$ , only $2010(x-1)(x-1); 2010(x+1)(x+1); -2010(x-1)(x+1)$ work, there are $3$ ways
Thus, $32+64+48+16+3=\boxed{163}$ | null | 163 |
2ba4d9f9648f0f4a2c363fdb83d3cbed | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_11 | Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Find the number of distinct T-grids | The T-grid can be considered as a tic-tac-toe board: five $1$ 's (or X's) and four $0$ 's (or O's).
There are only $\dbinom{9}{5} = 126$ ways to fill the board with five $1$ 's and four $0$ 's. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone has won twice.
Let three-in-a-row/column/diagonal be a "win" and let player $0$ be the one that fills in $0$ and player $1$ fills in $1$
Case $1$ Each player wins once.
If player X takes a diagonal, player Y cannot win. If either takes a row, all the columns are blocked, and visa versa. Therefore, they either both take a row or they both take a column.
Case $1$ $36$ cases
Case $2$ Player $1$ wins twice. (Player $0$ cannot win twice because he only has 4 moves.)
Case $2$ total: $22$
Thus, the answer is $126-22-36=\boxed{68}$ | null | 68 |
2ba4d9f9648f0f4a2c363fdb83d3cbed | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_11 | Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Find the number of distinct T-grids | We can use generating functions to compute the case that no row or column is completely filled with $1$ 's. Given a row, let $a$ $b$ $c$ be the events that the first, second, third respective squares are $1$ 's. Then the generating function representing the possible events that exclude a row of $1,1,1$ or $0,0,0$ from occuring is \[ab+bc+ca+a+b+c.\] Therefore, the generating function representing the possible grids where no row is filled with $0$ 's and $1$ 's is \[P(a,b,c)=((ab+bc+ca)+(a+b+c))^3.\] We expand this by the Binomial Theorem to find \[P(a,b,c)=(ab+bc+ca)^3+3(ab+bc+ca)^2(a+b+c)+3(ab+bc+ca)(a+b+c)^2+(a+b+c)^3.\] Recall that our grid has five $1$ 's, hence we only want terms where the sum of the exponents is $5$ . This is given by \[3(ab+bc+ca)^2(a+b+c).\] When we expand this, we find \[3(2abc(a+b+c)+a^2b^2+b^2c^2+c^2a^2)(a+b+c).\] We also want to make sure that each of $a$ $b$ $c$ appears at least once (so there is no column filled with $0$ 's) and the power of each of $a$ $b$ $c$ is not greater than or equal to $3$ (so there is no column filled with $1$ 's). The sum of the coefficients of the above polynomial is clearly $81$ (using $a,b,c=1$ ), and the sum of the coefficients of the terms $a^3bc$ $ab^3c$ , and $abc^3$ is $6+6+6+3+3+3+3+3+3=36$ , hence the sum of the coefficients of the desired terms is $81-36=45$ . This counts the number of grids where no column or row is filled with $0$ 's or $1$ 's. However, we could potentially have both diagonals filled with $1$ 's, but this is the only exception to our $45$ possibilities, hence the number of $T$ -grids with no row or column filled with the same digit is $44$
On the other hand, if a row (column) is filled with $0$ 's, then by the Pigeonhole Principle, another row (column) must be filled with $1$ 's. Hence this is impossible, so all other possible $T$ -grids have a row (column) filled with $1$ 's. If the top row is filled with $1$ 's, then we have two $1$ 's left to place. Clearly they cannot go in the same row, because then the other row is filled with $0$ 's. They also cannot appear in the same column. This leaves $3\cdot 2$ arrangements--3 choices for the location of the $1$ in the second row, and 2 choices for the location of the $1$ in the last row. However, two of these arrangements will fill a diagonal with $1$ 's. Hence there are only $4$ $T$ -grids where the top row is filled with $1$ 's. The same argument applies if any other row or column is filled with $1$ 's. Hence there are $4\cdot 6=24$ such $T$ -grids.
Thus the answer is $44+24=\boxed{68}$ | null | 68 |
2be89667986df734330d17548c96534d | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_12 | Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter. | Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$ . Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$ . By Heron's Formula, we have
Since $15$ and $338$ are coprime, to minimize, we must have $s=338$ and $x=15$ . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by $2$ , which gives us a final answer of $\boxed{676}$ | null | 676 |
2be89667986df734330d17548c96534d | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_12 | Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter. | Let the first triangle have sides $16n,a,a$ , so the second has sides $14n,a+n,a+n$ . The height of the first triangle is $\frac{7}{8}$ the height of the second triangle. Therefore, we have \[a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).\] Multiplying this, we get \[64a^2-4096n^2=49a^2+98an-2352n^2,\] which simplifies to \[15a^2-98an-1744n^2=0.\] Solving this for $a$ , we get $a=n\cdot\frac{218}{15}$ , so $n=15$ and $a=218$ and the perimeter is $15\cdot16+218+218=\boxed{676}$ | null | 676 |
9a5dfdc4bc2855d6b99ff70c03ffc19b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_13 | The $52$ cards in a deck are numbered $1, 2, \cdots, 52$ . Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards form a team, and the two persons with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$ , and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$ . where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Once the two cards are drawn, there are $\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$ , which occurs in $\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$ , which occurs in $\dbinom{43-a}{2}$ ways. Thus, \[p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}.\] Simplifying, we get $p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}$ , so we need $(43-a)(42-a)+(a-1)(a-2)\ge (1225)$ . If $a=22+b$ , then \begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225) \\ b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*} So $b> 13$ or $b< -13$ , and $a=22+b<9$ or $a>35$ , so $a=8$ or $a=36$ . Thus, $p(8) = \frac{616}{1225} = \frac{88}{175}$ , and the answer is $88+175 = \boxed{263}$ | null | 263 |
c0985c13fad91788ac0f64263aefff5e | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_14 | Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$ | Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$ . It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$ . Hence $ODP$ is isosceles and $OD = DP = 2$
Denote $E$ the projection of $O$ onto $CD$ . Now $CD = CP + DP = 3$ . By the Pythagorean Theorem $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$ . Now note that $EP = \frac {1}{2}$ . By the Pythagorean Theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$ . Hence it now follows that,
\[\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}\]
This gives that the answer is $\boxed{007}$ | null | 007 |
c0985c13fad91788ac0f64263aefff5e | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_14 | Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$ | Let $\angle{ACP}$ be equal to $x$ . Then by Law of Sines, $PB = -\frac{\cos{x}}{\cos{3x}}$ and $AP = \frac{\sin{x}}{\sin{3x}}$ . We then obtain $\cos{3x} = 4\cos^3{x} - 3\cos{x}$ and $\sin{3x} = 3\sin{x} - 4\sin^3{x}$ . Solving, we determine that $\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}$ . Plugging this in gives that $\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}$ . The answer is $\boxed{007}$ | null | 007 |
c0985c13fad91788ac0f64263aefff5e | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_14 | Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$ | Let $\alpha=\angle{ACP}$ $\beta=\angle{ABC}$ , and $x=BP$ . By Law of Sines,
$\frac{1}{\sin(\beta)}=\frac{x}{\sin(90-\alpha)}\implies \sin(\beta)=\frac{\cos(\alpha)}{x}$ (1), and
$\frac{4-x}{\sin(\alpha)}=\frac{4\sin(\beta)}{\sin(2\alpha)} \implies 4-x=\frac{2\sin(\beta)}{\cos(\alpha)}$ . (2)
Then, substituting (1) into (2), we get
$4-x=\frac{2}{x} \implies x^2-4x+2=0 \implies x=2-\sqrt{2} \implies \frac{4-x}{x}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$
The answer is $\boxed{007}$ .
~Rowechen | null | 007 |
c0985c13fad91788ac0f64263aefff5e | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_14 | Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$ | Let $\angle{ACP}=x$ . Then, $\angle{APC}=2x$ and $\angle{A}=180-3x$ . Let the foot of the angle bisector of $\angle{APC}$ on side $AC$ be $D$ . Then,
$CD=DP$ and $\triangle{DAP}\sim{\triangle{APC}}$ due to the angles of these triangles.
Let $CD=a$ . By the Angle Bisector Theorem, $\frac{1}{a}=\frac{AP}{AD}$ , so $AD=a\cdot{AP}$ . Moreover, since $CD=DP=a$ , by similar triangle ratios, $\frac{AP}{a+a\cdot{AP}}=a$ . Therefore, $AP = \frac{a^2}{1-a^2}$
Construct the perpendicular from $D$ to $AP$ and denote it as $F$ . Denote the midpoint of $CP$ as $M$ . Since $PD$ is an angle bisector, $PF$ is congruent to $PM$ , so $PF=\frac{1}{2}$
Also, $\triangle{DFA}\sim{\triangle{BCA}}$ . Thus, $\frac{FA}{AC}=\frac{AD}{AB}\Longrightarrow\frac{\frac{a^2}{1-a^2}-\frac{1}{2}}{a+\frac{a^3}{1-a^2}}=\frac{\frac{a^3}{1-a^3}}{4}$ . After some major cancellation, we have $7a^4-8a^2+2=0$ , which is a quadratic in $a^2$ . Thus, $a^2 = \frac{4\pm\sqrt{2}}{7}$
Taking the negative root implies $AP<BP$ , contradiction. Thus, we take the positive root to find that $AP=2+\sqrt{2}$ . Thus, $BP=2-\sqrt{2}$ , and our desired ratio is $\frac{2+\sqrt{2}}{2-\sqrt{2}}\implies{3+2\sqrt{2}}$
The answer is $\boxed{007}$ | null | 007 |
c0985c13fad91788ac0f64263aefff5e | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_14 | Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$ | Let $O$ be the circumcenter of $\triangle ABC$ . Since $\triangle ABC$ is a right triangle, $O$ will be on $\overline{AB}$ and $\overline{AO} \cong \overline{OB} \cong \overline{OC} = 2$ . Let $\overline{OP} = x$
Next, let's do some angle chasing. Label $\angle ACP = \theta^{\circ}$ , and $\angle APC = 2\theta^{\circ}$ . Thus, $\angle PAC = (180-3\theta)^{\circ}$ , and by isosceles triangles, $\angle ACO = (180-3\theta)^{\circ}$ . Then, by angle subtraction, $\angle OCP = (\theta - (180-3\theta))^{\circ} = (4\theta - 180)^{\circ}$
Using the Law of Sines: \[\frac{x}{\sin (4\theta-180)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\] Using trigonometric identies, $\sin (4\theta-180)^{\circ}=-\sin (4\theta)^{\circ}=-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}$ . Plugging this back into the Law of Sines formula gives us: \[\frac{x}{-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\]
\[-4\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}=x\sin (2\theta)^{\circ}\] \[-4\cos (2\theta)^{\circ}=x\] \[\cos(2\theta)^{\circ}=\frac{-x}4\]
Next, using the Law of Cosines: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \cos (2\theta)^{\circ}\] Substituting $\cos(2\theta)^{\circ}=\frac{-x}4$ gives us: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \frac{-x}4\] \[4=1+x^2+\frac{x^2}{2}\]
Solving for x gives $x=\sqrt 2$
Finally: $\frac{\overline{AP}}{\overline{BP}}=\frac{\overline{AO}+\overline{OP}}{\overline{BO}-\overline{OP}}=\frac{2+\sqrt 2}{2-\sqrt 2}=3+2\sqrt2$ , which gives us an answer of $3+2+2=\boxed{007}$ . ~adyj | null | 007 |
78cdddc82a03232350653c6f817635db | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_15 | In triangle $ABC$ $AC = 13$ $BC = 14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$ | Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{(AMN)}(X)-\text{Pow}_{(ADE)}(X)\] for points $X$ in the plane. Then $f$ is linear, so $\frac{BQ}{CQ}=\frac{f(B)-f(Q)}{f(Q)-f(C)}$ . But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$ $(ADE)$ thus \[\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}=-\frac{BN\cdot BA-BE\cdot BA}{CM\cdot CA-CD\cdot CA}\] Let $AC=b$ $BC=a$ and $AB=c$ . Note that $BN=\tfrac{c}{2}$ and $CM=\tfrac{b}{2}$ because they are midpoints, while $BE=\frac{ac}{a+b}$ and $CD=\frac{ab}{a+c}$ by Angle Bisector Theorem. Thus we can rewrite this expression as \begin{align*}&-\frac{\tfrac{c^{2}}{2}-\tfrac{ac^{2}}{a+b}}{\tfrac{b^{2}}{2}-\tfrac{ab^{2}}{a+c}} \\ =&-\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{\tfrac{1}{2}-\tfrac{a}{a+b}}{\tfrac{1}{2}-\tfrac{a}{a+c}}\right) \\ =&\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{a+c}{a+b}\right) \\ =&\left(\frac{225}{169}\right)\left(\frac{29}{27}\right) \\ =&~\frac{725}{507}\end{align*} so $m-n=\boxed{218}$ | null | 218 |
78cdddc82a03232350653c6f817635db | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_15 | In triangle $ABC$ $AC = 13$ $BC = 14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$ | Let $Y = MN \cap AQ$ $\frac {BQ}{QC} = \frac {NY}{MY}$ since $\triangle AMN \sim \triangle ACB$ . Since quadrilateral $AMPN$ is cyclic, $\triangle MYA \sim \triangle PYN$ and $\triangle MYP \sim \triangle AYN$ , yielding $\frac {YM}{YA} = \frac {MP}{AN}$ and $\frac {YA}{YN} = \frac {AM}{PN}$ . Multiplying these together yields * $\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)$
$\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}$
Now we claim that $\triangle PMD \sim \triangle PNE$ . To prove this, we can use cyclic quadrilaterals.
From $AMPN$ $\angle PNY \cong \angle PAM$ and $\angle ANM \cong \angle APM$ . So, $m\angle PNA = m\angle PNY + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA$ and $\angle PNA \cong \angle PMD$
From $ADPE$ $\angle PDE \cong \angle PAE$ and $\angle EDA \cong \angle EPA$ . Thus, $m\angle MDP = m\angle PDE + m\angle EDA = m\angle PAE + m\angle EPA = 180-m\angle PEA$ and $\angle PDM \cong \angle PEN$
Thus, from AA similarity, $\triangle PMD \sim \triangle PNE$
Therefore, $\frac {PN}{PM} = \frac {NE}{MD}$ , which can easily be computed by the angle bisector theorem to be $\frac {145}{117}$ . It follows that * $\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}$ , giving us an answer of $725 - 507 = \boxed{218}$ | null | 218 |
78cdddc82a03232350653c6f817635db | https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_15 | In triangle $ABC$ $AC = 13$ $BC = 14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$ | This problem can be solved with barycentric coordinates. Let triangle $ABC$ be the reference triangle with $A=(1,0,0)$ $B=(0,1,0)$ , and $C=(0,0,1)$ . Thus, $N=(1:1:0)$ and $M=(1:0:1)$ . Using the Angle Bisector Theorem, we can deduce that $D=(14:0:15)$ and $E = (14:13:0)$ . Plugging the coordinates for triangles $ANM$ and $AED$ into the circle formula, we deduce that the equation for triangle $ANM$ is $-a^2yz-b^2zx-c^2xy+(\frac{c^2}{2}y+\frac{b^2}{2}z)(x+y+z)=0$ and the equation for triangle $AED$ is $-a^2yz-b^2zx-c^2xy+(\frac{14c^2}{27}y+\frac{14b^2}{29}z)(x+y+z)=0$ . Solving the system of equations, we get that $\frac{c^2y}{54}=\frac{b^2z}{58}$ . This equation determines the radical axis of circles $ANM$ and $AED$ , on which points $P$ and $Q$ lie. Thus, solving for $\frac{z}{y}$ gets the desired ratio of lengths, and $\frac{z}{y}=\frac{58c^2}{54b^2}$ and plugging in the lengths $b=13$ and $c=15$ gets $\frac{725}{507}$ . From this we get the desired answer of $725-507=\boxed{218}$ .
-wertguk | null | 218 |
16a522e5dbe9ec22b6820cdfd2341c27 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_1 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | Assume that the largest geometric number starts with a $9$ . We know that the common ratio must be a rational of the form $k/3$ for some integer $k$ , because a whole number should be attained for the 3rd term as well. When $k = 1$ , the number is $931$ . When $k = 2$ , the number is $964$ . When $k = 3$ , we get $999$ , but the integers must be distinct. By the same logic, the smallest geometric number is $124$ . The largest geometric number is $964$ and the smallest is $124$ . Thus the difference is $964 - 124 = \boxed{840}$ | null | 840 |
16a522e5dbe9ec22b6820cdfd2341c27 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_1 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | Consider the three-digit number $abc$ . If its digits form a geometric progression, we must have that ${a \over b} = {b \over c}$ , that is, $b^2 = ac$
The minimum and maximum geometric numbers occur when $a$ is minimized and maximized, respectively. The minimum occurs when $a = 1$ ; letting $b = 2$ and $c = 4$ achieves this, so the smallest possible geometric number is 124.
For the maximum, we have that $b^2 = 9c$ $b$ is maximized when $9c$ is the greatest possible perfect square; this happens when $c = 4$ , yielding $b = 6$ . Thus, the largest possible geometric number is 964.
Our answer is thus $964 - 124 = \boxed{840}$ | null | 840 |
16a522e5dbe9ec22b6820cdfd2341c27 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_1 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | The smallest geometric number is $124$ because $123$ and any number containing a zero does not work. $964$ is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives $\boxed{840}.$ | null | 840 |
720933afdd40b794ae9c7b405266415d | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | Let $z = a + 164i$
Then \[\frac {a + 164i}{a + 164i + n} = 4i\] and \[a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.\]
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
we conclude that \[a = -656.\]
By equating the imaginary terms on each side of the equation,
we conclude that \[164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).\]
We now have an equation for $n$ \[4i \left (-656 + n \right ) = 164i,\]
and this equation shows that $n = \boxed{697}.$ | null | 697 |
720933afdd40b794ae9c7b405266415d | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | \[\frac {z}{z+n}=4i\]
\[1-\frac {n}{z+n}=4i\]
\[1-4i=\frac {n}{z+n}\]
\[\frac {1}{1-4i}=\frac {z+n}{n}\]
\[\frac {1+4i}{17}=\frac {z}{n}+1\]
Since their imaginary part has to be equal,
\[\frac {4i}{17}=\frac {164i}{n}\]
\[n=\frac {(164)(17)}{4}=697\]
\[n = \boxed{697}.\] | null | 697 |
720933afdd40b794ae9c7b405266415d | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$ [asy] unitsize(1cm); xaxis("Re",Arrows); yaxis("Im",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label("Im$(z)=164$",F(-330),N); pair z = (-656,164); dot(Label("$z$",align=N),z); dot(Label("$z+n$",align=N),z+(697,0)); draw(Label("$4x$"),z--(0,0)); draw(Label("$x$"),(0,0)--z+(697,0)); markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); [/asy] $z$ must lie on the line $\operatorname{Im}(z)=164$ $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$
Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)$ and $\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)$ , where $r$ is the magnitude and $\theta$ is the phase, and $z_n=r_n\angle\theta_n$
Since $4i$ has magnitude $4$ and phase $90^\circ$ (since the positive imaginary axis points in a direction $90^\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$ . We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$ . Additionally, $z$ , the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$
This means that $z$ , the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$ , since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\frac{x \cdot 4x}{2}$ , or using the hypotenuse and its corresponding altitude, as $\frac{164n}{2}$ , so $\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n$ . By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \implies 17x^2 = n^2$ . Substituting out $x^2$ using the earlier equation, we get $17\cdot41n = n^2 \implies n = \boxed{697}$ . ~ emerald_block | null | 697 |
720933afdd40b794ae9c7b405266415d | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | Taking the reciprocal of our equation gives us $1 + \frac{n}{z} = \frac{1}{4i}.$ Therefore, \[\frac{n}{z} = \frac{1-4i}{4i} = \frac{17}{-16+4i}.\] Since $z$ has an imaginary part of $164$ , we must multiply both sides of our RHS fraction by $\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: \[\frac{n}{z} = \frac{697}{-656 + 164i}.\] Therefore, we can conclude the the real part of $z$ is $-656$ and $n = \boxed{697}.$ (it wasn't necessary to find the real part) | null | 697 |
026b69379e4219c1a0d92b071de8e975 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_3 | A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$
\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25&=p^2 \\ 24p^2-50p+25&=0 \\ p&=\frac {5}{6}\end{align*}
Therefore, the answer is $5+6=\boxed{011}$ | null | 011 |
026b69379e4219c1a0d92b071de8e975 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_3 | A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We start as shown above. However, when we get to $25(1-p)^2=p^2$ , we square root both sides to get $5(1-p)=p$ . We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$ , so they are both positive. Now, we have:
\begin{align*} 5(1-p)&=p \\ 5-5p&=p \\ 5&=6p \\ p&=\frac {5}{6}\end{align*}
Now, we get $5+6=\boxed{011}$ | null | 011 |
026b69379e4219c1a0d92b071de8e975 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_3 | A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Rewrite it as : $(P)^3$ $(1-P)^5=\frac {1}{25}$ $(P)^5$ $(1-P)^3$
This can be simplified as $24P^2 -50P + 25 = 0$
This can be factored into $(4P-5)(6P-5)$
This yields two solutions: $\frac54$ (ignored because it would result in $1-p<0$ ) or $\frac56$
Therefore, the answer is $5+6$ $\boxed{011}$ | null | 011 |
b2832d5c34ed4064416901e6bd06b1cc | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | One of the ways to solve this problem is to make this parallelogram a straight line.
So the whole length of the line is $APC$ $AMC$ or $ANC$ ), and $ABC$ is $1000x+2009x=3009x.$
$AP$ $AM$ or $AN$ ) is $17x.$
So the answer is $3009x/17x = \boxed{177}$ | null | 177 |
b2832d5c34ed4064416901e6bd06b1cc | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | Draw a diagram with all the given points and lines involved. Construct parallel lines $\overline{DF_2F_1}$ and $\overline{BB_1B_2}$ to $\overline{MN}$ , where for the lines the endpoints are on $\overline{AM}$ and $\overline{AN}$ , respectively, and each point refers to an intersection. Also, draw the median of quadrilateral $BB_2DF_1$ $\overline{E_1E_2E_3}$ where the points are in order from top to bottom. Clearly, by similar triangles, $BB_2 = \frac {1000}{17}MN$ and $DF_1 = \frac {2009}{17}MN$ . It is not difficult to see that $E_2$ is the center of quadrilateral $ABCD$ and thus the midpoint of $\overline{AC}$ as well as the midpoint of $\overline{B_1}{F_2}$ (all of this is easily proven with symmetry). From more triangle similarity, $E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP$ $= \boxed{177}AP$ | null | 177 |
b2832d5c34ed4064416901e6bd06b1cc | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | Assume, for the ease of computation, that $AM=AN=17$ $AB=1000$ , and $AD=2009$ . Now, let line $MN$ intersect line $CD$ at point $X$ and let $Y$ be a point such that $XY\parallel AD$ and $AY\parallel DX$ . As a result, $ADXY$ is a parallelogram. By construction, $\triangle MAN\sim \triangle MYX$ so \[\frac{MY}{MA}=\frac{YX}{AN}=\frac{AD}{AN}=\frac{2009}{17}\implies MY=2009\] and $AY=DX=2009-17$ . Also, because $AM\parallel XC$ , we have $\triangle PAM\sim \triangle PCX$ so \[\frac{PC}{PA}=\frac{CX}{AM}=\frac{DX+CD}{AM}=\frac{2009-17+1000}{17}=176.\] Hence, $\frac{AC}{AP}=\frac{PC}{PA}+1=\boxed{177}.$ | null | 177 |
b2832d5c34ed4064416901e6bd06b1cc | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | Assign $A = (0,0)$ . Since there are no constraints in the problem against this, assume $ABCD$ to be a rectangle with dimensions $1000 \times 2009.$ Now, we can assign \[A=(0,0)\] \[B=(1000, 0)\] \[C=(1000,-2009)\] \[D=(0,-2009).\]
Then, since $\frac{AM}{AB} = \frac{17}{1000}$ and $AB = 1000$ , we can place $M$ at $(17, 0).$ Similarly, place $AN$ at $(0, 17).$ Then, the equation of line $MN$ is $y=x-17,$ and the equation of $AC$ is $y=\frac{-2009}{1000}x.$ Solve to find point $P$ at $\left( \frac{1000}{177}, \frac{-2009}{177} \right)$
We can calculate vectors to represent the distances: \[\overrightarrow{AC}= <1000, -2009>\] \[\overrightarrow{AP}= \frac{1}{177}<1000, -2009>.\] In this way, we can see that \[AC:AP = 177:1,\] and our answer is $\boxed{177}.$ | null | 177 |
dfe507f63763b1281b02a1b61ac9a183 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_5 | Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$ | Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$ , quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$
Thus,
\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]
Now let's apply the angle bisector theorem.
\[\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}\]
\[\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}\]
\[\frac {180}{LP}=\frac {5}{2}\]
\[LP=\boxed{072}\] | null | 072 |
dfe507f63763b1281b02a1b61ac9a183 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_5 | Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$ | Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: \[\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL\] So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$ . Since $K$ is the midpoint of $A$ and $C$ , the weight of $A$ is equal to the weight of $C$ , which equals $2$ .
Also, since the weight of $L$ is $5$ and $C$ is $2$ , we can weight $P$ as $7$
By the definition of mass points, \[\frac{LP}{CP}=\frac{2}{5}\implies LP=\frac{2}{5}CP\] By vertical angles, angle $MKA =$ angle $PKC$ .
Also, it is given that $AK=CK$ and $PK=MK$
By the SAS congruence, $\triangle MKA$ $\triangle PKC$ . So, $MA$ $CP$ $180$ .
Since $LP=\frac{2}{5}CP$ $LP = \frac{2}{5}(180) = \boxed{072}$ | null | 072 |
dfe507f63763b1281b02a1b61ac9a183 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_5 | Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$ | Using the diagram from solution $1$ , we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$
Applying the angle bisector theorem to $\triangle CKB$ , we get that $\frac{KP}{PB} = \frac{225}{300} = \frac{3}{4}.$ So, we can let $MK = KP = 3x$ and $BP = 4x$
Now, apply law of cosines on $\triangle CKP$ and $\triangle CPB.$
If we let $\angle KCP = \angle PCB = \alpha$ , then the law of cosines gives the following system of equations:
\[9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha\] \[16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos \alpha.\]
Bashing those out, we get that $x = 15 \sqrt{13}$ and $\cos \alpha = \frac{7}{10}.$
Since $\cos \alpha = \frac{7}{10}$ , we can use the double angle formula to calculate that $\cos \left(2 \alpha \right) = -\frac{1}{50}.$
Now, apply Law of Cosines on $\triangle ABC$ to find $AB$
We get: \[AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).\]
Bashing gives $AB = 30 \sqrt{331}.$
From the angle bisector theorem on $\triangle ABC$ , we know that $\frac{AL}{BL} = \frac{450}{300} = \frac{3}{2}.$ So, $AL = 18 \sqrt{331}$ and $BL = 12 \sqrt{331}.$
Now, we apply Law of Cosines on $\triangle ALC$ and $\triangle BLC$ in order to solve for the length of $LC$
We get the following system:
\[(18 \sqrt{331})^2 = 450^2 + LC^2 - 2 \cdot 450 \cdot LC \cdot \frac{7}{10}\] \[(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}\]
The first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$
The only value that satisfies both equations is $LC = 252$ , and since $LP = LC - PC$ , we have \[LC = 252 - 180 = \boxed{072}.\] | null | 072 |
c06962ddf3f4a19e7df9731bd5fe304b | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_6 | How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ | First, $x$ must be less than $5$ , since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$
Because ${\lfloor x\rfloor}$ must be an integer, let’s do case work based on ${\lfloor x\rfloor}$
For ${\lfloor x\rfloor}=0$ $N=1$ as long as $x \neq 0$ . This gives us $1$ value of $N$
For ${\lfloor x\rfloor}=1$ $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$
Therefore, $N=1$ . However, we got $N=1$ in case 1 so it got counted twice.
For ${\lfloor x\rfloor}=2$ $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$
This gives us $3^2-2^2=5$ $N$ 's
For ${\lfloor x\rfloor}=3$ $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$
This gives us $4^3-3^3=37$ $N$ 's
For ${\lfloor x\rfloor}=4$ $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$
This gives us $5^4-4^4=369$ $N$ 's
Since $x$ must be less than $5$ , we can stop here and the answer is $1+5+37+369= \boxed{412}$ possible values for $N$ | null | 412 |
c06962ddf3f4a19e7df9731bd5fe304b | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_6 | How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ | For a positive integer $k$ , we find the number of positive integers $N$ such that $x^{\lfloor x\rfloor}=N$ has a solution with ${\lfloor x\rfloor}=k$ . Then $x=\sqrt[k]{N}$ , and because $k \le x < k+1$ , we have $k^k \le x^k < (k+1)^k$ , and because $(k+1)^k$ is an integer, we get $k^k \le x^k \le (k+1)^k-1$ . The number of possible values of $x^k$ is equal to the number of integers between $k^k$ and $(k+1)^k-1$ inclusive, which is equal to the larger number minus the smaller number plus one or $((k+1)^k-1)-(k^k)+1$ , and this is equal to $(k+1)^k-k^k$ . If $k>4$ , the value of $x^k$ exceeds $1000$ , so we only need to consider $k \le 4$ . The requested number of values of $N$ is the same as the number of values of $x^k$ , which is $\sum^{4}_{k=1} [(k+1)^k-k^k]=2-1+9-4+64-27+625-256=\boxed{412}$ | null | 412 |
e359ac5a68e46a4e8738e97294d0fe40 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_7 | The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$ | The best way to solve this problem is to get the iterated part out of the exponent: \[5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1\] \[5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}\] \[5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}\] \[a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}\] \[a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}\] Plug in $n = 1, 2, 3, 4$ to see the first few terms of the sequence: \[\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.\] We notice that the terms $5, 8, 11, 14$ are in arithmetic progression. Since $a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}$ , we can easily use induction to show that $a_n = \log_5{(3n + 2)}$ . So now we only need to find the next value of $n$ that makes $\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$ . We test $25$ \[3n + 2 = 25\] \[3n = 23\] This has no integral solutions, so we try $125$ \[3n + 2 = 125\] \[3n = 123\] \[n = \boxed{041}\] | null | 041 |
e359ac5a68e46a4e8738e97294d0fe40 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_7 | The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$ | We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$ , we get: \[5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}\] This simplifies to \[5^{a_n}=3n+2.\] We can now test powers of $5$
$5$ - that gives us $n=1$ , which is useless.
$25$ - that gives a non-integer $n$
$125$ - that gives $n=\boxed{41}$ | null | 41 |
13bf6111ee5b617a97c8291f7a0865ab | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_8 | Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ | Find the positive differences in all $55$ pairs and you will get $\boxed{398}$ .
(This is not recommended unless you can't find any other solutions to this problem) | null | 398 |
13bf6111ee5b617a97c8291f7a0865ab | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_8 | Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ | When computing $N$ , the number $2^x$ will be added $x$ times (for terms $2^x-2^0$ $2^x-2^1$ , ..., $2^x - 2^{x-1}$ ), and subtracted $10-x$ times. Hence $N$ can be computed as $N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0$ . Evaluating $N \bmod {1000}$ yields:
\begin{align*} N & = 10(2^{10}-1) + 8(2^9 - 2^1) + 6(2^8-2^2) + 4(2^7-2^3) + 2(2^6-2^4) \\ & = 10(1023) + 8(510) + 6(252) + 4(120) + 2(48) \\ & = 10(1000+23) + 8(500+10) + 6(250+2) + 480 + 96 \\ & \equiv (0 + 230) + (0 + 80) + (500 + 12) + 480 + 96 \\ & \equiv \boxed{398} | null | 398 |
13bf6111ee5b617a97c8291f7a0865ab | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_8 | Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ | This solution can be generalized to apply when $10$ is replaced by other positive integers.
Extending from Solution 2, we get that the sum $N$ of all possible differences of pairs of elements in $S$ when $S = \{2^0,2^1,2^2,\ldots,2^{n}\}$ is equal to $\sum_{x=0}^{n} (2x-n) 2^x$ . Let $A = \sum_{x=0}^{n} x2^x$ $B=\sum_{x=0}^{n} 2^x$ . Then $N=2A - nB$
For $n = 10$ $B = 2^{11}-1 = 2047 \equiv 47 \pmod{1000}$ by the geometric sequence formula.
$2A = \sum_{x=1}^{n+1} (x-1)2^x$ , so $2A - A = A = n2^{n+1} - \sum_{x=1}^{n} 2^x$ . Hence, for $n = 10$ $A = 10 \cdot 2^{11} - 2^{11} + 2 = 9 \cdot 2^{11} + 2 \equiv 48 \cdot 9 + 2 =$
$434 \pmod{1000}$ , by the geometric sequence formula and the fact that $2^{10} = 1024 \equiv 24 \pmod{1000}$
Thus, for $n = 10$ $N = 2A - 10B \equiv 2\cdot 434 - 10\cdot 47 = 868 - 470 = \boxed{398}$ | null | 398 |
2aec2b70742d0fb21a1d13c4160b0d7e | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_9 | A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $$ 1$ to $$ 9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$ . Find the total number of possible guesses for all three prizes consistent with the hint. | [Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$ , that could have been on that day.]
Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$ , then the string is
\[1131331.\]
Since the strings have seven digits and three threes, there are $\binom{7}{3}=35$ arrangements of all such strings.
In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.
Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to
\[x+y+z=7, x,y,z>0.\]
This gives us
\[\binom{6}{2}=15\]
ways by balls and urns. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$
Thus, each arrangement has \[\binom{6}{2}-3=12\] ways per arrangement, and there are $12\times35=\boxed{420}$ ways. | null | 420 |
0fd50a589087e10c29d9f6b2b44998ce | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_10 | The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $1$ to $15$ in clockwise order. Committee rules state that a Martian must occupy chair $1$ and an Earthling must occupy chair $15$ , Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is $N \cdot (5!)^3$ . Find $N$ | Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form $N*(5!)^3$ because for each $M, V, E$ sequence we have $5!$ arrangements within the Ms, Vs, and Es.
Pretend the table only seats $3$ "people", with $1$ "person" from each planet. Counting clockwise, only the arrangement M, V, E satisfies the given constraints. Therefore, in the actual problem, the members must sit in cycles of M, V, E, but not necessarily with one M, one V, and one E in each cycle(for example, MMVVVE, MVVVEEE, MMMVVVEE all count as cycles). These cycles of MVE must start at seat $1$ , since an M is at seat $1$ . We simply count the number of arrangements through casework.
1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE
2. Two cycles - There are 3 Ms, Vs and Es left to distribute among the existing MVEMVE. Using stars and bars, we get $\binom{4}{1}=4$ ways for the members of each planet. Therefore, there are $4^3=64$ ways in total.
3. Three cycles - 2 Ms, Vs, Es left, so $\binom{4}{2}=6$ , making there $6^3=216$ ways total.
4. Four cycles - 1 M, V, E left, each M can go to any of the four MVE cycles and likewise for V and E, $4^3=64$ ways total
5. Five cycles - MVEMVEMVEMVEMVE is the only possibility, so there is just $1$ way.
Combining all these cases, we get $1+1+64+64+216= \boxed{346}$ | null | 346 |
0fd50a589087e10c29d9f6b2b44998ce | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_10 | The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $1$ to $15$ in clockwise order. Committee rules state that a Martian must occupy chair $1$ and an Earthling must occupy chair $15$ , Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is $N \cdot (5!)^3$ . Find $N$ | The arrangements must follow the pattern MVEMVE....... where each MVE consists of some Martians followed by some Venusians followed by some Earthlings, for $1, 2, 3, 4,$ or $5$ MVE's. If there are $k$ MVE's, then by stars and bars, there are ${4 \choose k-1}$ choices for the Martians in each block, and the same goes for the Venusians and the Earthlings. Thus, we have $N = 1^3+4^3+6^3+4^3+1^3 = \boxed{346}$ - aops5234 | null | 346 |
78b2801f0c4a5a7e8b41f51bf1e192c5 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_11 | Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$ . Find the number of such distinct triangles whose area is a positive integer. | Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1\\x_2&y_2\end{array}\right).$
Since the triangle has half the area of the parallelogram, we just need the determinant to be even.
The determinant is
\[(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))\]
Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$ 's must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$ . There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles. | null | 600 |
78b2801f0c4a5a7e8b41f51bf1e192c5 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_11 | Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$ . Find the number of such distinct triangles whose area is a positive integer. | As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either
We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$
Using the point-to-line distance formula, we can calculate the height of $\triangle OPQ$ from vertex $O$ (the origin) to be:
$\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}$
Let $b$ be the base of the triangle that is part of the line $41x+y=2009$
The area is calculated as: $\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b$
Let the numerical area of the triangle be $k$
So, $k = \dfrac{2009}{58\sqrt2}\times b$
We know that $k$ is an integer. So, $b = 58\sqrt2 \times z$ , where $z$ is also an integer.
We defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
Changing the y-coordinates to be in terms of x, we get:
$P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$
The distance between them equals $b$
Using the distance formula, we get
$PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z$ $(*)$
WLOG, we can assume that $x_2 > x_1$
Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$ , we get
$29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z$
Dividing both sides by $29\sqrt2$ , we get
$x_2-x_1 = 2z$
As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well.
...
Summing them up, we get that there are $2+4+\dots + 48 = \boxed{600}$ triangles. | null | 600 |
1aa753c405e15ce7d12d376d4a5d3cf9 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | First, note that $AB=37$ ; let the tangents from $I$ to $\omega$ have length $x$ . Then the perimeter of $\triangle ABI$ is equal to \[2(x+AD+DB)=2(x+37).\] It remains to compute $\dfrac{2(x+37)}{37}=2+\dfrac{2}{37}x$
Observe $CD=\dfrac{12\cdot 35}{37}=\dfrac{420}{37}$ , so the radius of $\omega$ is $\dfrac{210}{37}$ . We may also compute $AD=\dfrac{12^{2}}{37}$ and $DB=\dfrac{35^{2}}{37}$ by similar triangles. Let $O$ be the center of $\omega$ ; notice that \[\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}\] so it follows \[\sin(\angle DAO)=\dfrac{35}{\sqrt{35^{2}+24^{2}}}=\dfrac{35}{\sqrt{1801}}\] while $\cos(\angle DAO)=\dfrac{24}{\sqrt{1801}}$ . By the double-angle formula $\sin(2\theta)=2\sin\theta\cos\theta$ , it turns out that \[\sin(\angle BAI)=\sin(2\angle DAO)=\dfrac{2\cdot 35\cdot 24}{1801}=\dfrac{1680}{1801}\]
Using the area formula $\dfrac{1}{2}ab\sin(C)$ in $\triangle ABI$ \[[ABI]=\left(\dfrac{1}{2}\right)\left(\dfrac{144}{37}+x\right)(37)\left(\dfrac{1680}{1801}\right)=\left(\dfrac{840}{1801}\right)(144+37x).\] But also, using $rs$ \[[ABI]=\left(\dfrac{210}{37}\right)(37+x).\] Now we can get \[\dfrac{[ABI]}{210}=\dfrac{4(144+37x)}{1801}=\dfrac{37+x}{37}\] so multiplying everything by $37\cdot 1801=66637$ lets us solve for $x$ \[21312+5476x=66637+1801x.\] We have $x=\dfrac{66637-21312}{5476-1801}=\dfrac{45325}{3675}=\dfrac{37}{3}$ , and now \[2+\dfrac{2}{37}x=2+\dfrac{2}{3}=\dfrac{8}{3}\] giving the answer, $\boxed{011}$ | null | 011 |
1aa753c405e15ce7d12d376d4a5d3cf9 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | As in Solution $1$ , let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively.
First, by pythagorean theorem, $AB = \sqrt{12^2+35^2} = 37$ . Now the area of $ABC$ is $1/2*12*35 = 1/2*37*CD$ , so $CD=\frac{420}{37}$ and the inradius of $\triangle ABI$ is $r=\frac{210}{37}$
Now from $\triangle CDB \sim \triangle ACB$ we find that $\frac{BC}{BD} = \frac{AB}{BC}$ so $BD = BC^2/AB = 35^2/37$ and similarly, $AD = 12^2/37$
Note $IP=IQ=x$ $BP=BD$ , and $AQ=AD$ . So we have $AI = 144/37+x$ $BI = 1225/37+x$ . Now we can compute the area of $\triangle ABI$ in two ways: by heron's formula and by inradius times semiperimeter, which yields
$rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/37)x}$
$210/37(37+x) = 12*35/37 \sqrt{x(37+x)}$
$37+x = 2 \sqrt{x(x+37)}$
$x^2+74x+1369 = 4x^2 + 148x$
$3x^2 + 74x - 1369 = 0$
The quadratic formula now yields $x=37/3$ . Plugging this back in, the perimeter of $ABI$ is $2s=2(37+x)=2(37+37/3) = 37(8/3)$ so the ratio of the perimeter to $AB$ is $8/3$ and our answer is $8+3=\boxed{011}$ | null | 011 |
1aa753c405e15ce7d12d376d4a5d3cf9 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | As in Solution $2$ , let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively.
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
Let $x = \overline{AD} = \overline{AQ}$ . Let $y = \overline{BD} = \overline{BP}$ . Let $z = \overline{PI} = \overline{QI}$ . The semi-perimeter of $ABI$ is $x + y + z$ .
Since the lengths of the sides of $ABI$ are $x + y$ $y + z$ and $x + z$ , the square of its area by Heron's formula is $(x+y+z)xyz$
The radius $r$ of $\omega$ is $\overline{CD}/2$ . Therefore $r^2 = xy/4$ . As $\omega$ is the in-circle of $ABI$ , the area of $ABI$ is also $r(x+y+z)$ , and so the square area is $r^2(x+y+z)^2$
Therefore \[(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}\] Dividing both sides by $xy(x+y+z)/4$ we get: \[4z = (x+y+z),\] and so $z = (x+y)/3$ . The semi-perimeter of $ABI$ is therefore $\frac{4}{3}(x+y)$ and the whole perimeter is $\frac{8}{3}(x+y)$ . Now $x + y = \overline{AB}$ , so the ratio of the perimeter of $ABI$ to the hypotenuse $\overline{AB}$ is $8/3$ and our answer is $8+3=\boxed{011}$ | null | 011 |
1aa753c405e15ce7d12d376d4a5d3cf9 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,O,X; C=origin; A=(0,12); B=(18,0); D=foot(C,A,B); O = (C+D)/2; real r = length(D-C)/2; path c = CR(O, r); pair OA = (O+A)/2; real rA = length(A-O)/2; pair Ap = OP(CR(OA,rA), c); pair OB = (O+B)/2; real rB = length(B-O)/2; pair Bp = OP(CR(OB,rB), c); X=extension(A,Ap,B,Bp); draw(A--B--C--A, s); draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); draw(c^^A--X--B); dot("$A$", A, N); dot("$B$", B, SE); dot("$C$", C, SW); dot("$D$", D, 0.2*(D-C)); dot("$I$", X, 0.5*(X-C)); dot("$P$", Ap, 0.3*(Ap-O)); dot("$Q$", Bp, 0.3*(Bp-O)); dot("$O$", O, W); label("$\beta$",B,10*dir(157)); label("$\alpha$",A,5*dir(-55)); label("$\theta$",X,5*dir(55)); [/asy] Let $AP=AD=x$ , let $BQ=BD=y$ , and let $IP=IQ=z$ . Let $OD=r$ . We find $AB=37$ . Let $\alpha$ $\beta$ , and $\theta$ be the angles $OAD$ $OBD$ , and $OPI$ respectively. Then $\alpha + \beta + \theta = 90^\circ$ , so \[\theta = 90^\circ - (\alpha+\beta).\] The perimeter of $\triangle ABI$ is $2(x+y+z)=2(37+z)$ . The desired ratio is then \[\rho = 2\left(1+\frac z{37}\right)\] We need to find $z$ . In $\triangle OPI$ $z=r\cot\theta = r\tan (\alpha+\beta)$ . We get \[\tan\alpha = \frac{OD}{AD} = \frac 12 \frac{CD}{AD} = \frac 12 \tan A = \frac 12 \frac{BC}{AC} = \frac{35}{24}.\] Similarly, $\tan\beta = \tfrac 6{35}$ . Then \[z = r\cdot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}\] Computing $[ABC]$ in two ways we get $CD = \tfrac{12\cdot 35}{37}$ , so $r=\tfrac{6\cdot 35}{37}$ . Using this value of $r$ we get $z=\tfrac {37}3$ . Thus \[\rho = 2\left(1+\frac 1{3}\right) = \frac 8{3},\] and $8+3=\boxed{011}$ | null | 011 |
1aa753c405e15ce7d12d376d4a5d3cf9 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | This solution is not a real solution and is solving the problem with a ruler and compass.
Draw $AC = 4.8, BC = 14, AB = 14.8$ . Then, drawing the tangents and intersecting them, we get that $IA$ is around $6.55$ and $IB$ is around $18.1$ . We then find the ratio to be around $\frac{39.45}{14.8}$ . Using long division, we find that this ratio is approximately 2.666, which you should recognize as $\frac{8}{3}$ . Since this seems reasonable, we find that the answer is $\boxed{11}$ ~ilp | null | 11 |
1aa753c405e15ce7d12d376d4a5d3cf9 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_12 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ $AC = 12$ $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Denoting three tangents has length $h_1,h_2,h_3$ while $h_1,h_3$ lies on $AB$ with $h_1>h_3$ .The area of $ABC$ is $1/2*12*35 = 1/2*37*CD$ , so $CD=\frac{420}{37}$ and the inradius of $\triangle ABI$ is $r=\frac{210}{37}$ .As we know that the diameter of the circle is the height of $\triangle ACB$ from $C$ to $AB$ . Assume that $\tan\alpha=\frac{h_1}{r}$ and $\tan\beta=\frac{h_3}{r}$ and $\tan\omega=\frac{h_2}{r}$ . But we know that $\tan(\alpha+\beta)=-\tan(180-\alpha-\beta)=-\tan\omega$ According to the basic computation, we can get that $\tan(\alpha)=\frac{35}{6}$ $\tan(\beta)=\frac{24}{35}$ So we know that $\tan(\omega)=\frac{1369}{630}$ according to the tangent addition formula. Hence, it is not hard to find that the length of $h_2$ is $\frac{37}{3}$ . According to basic addition and division, we get the answer is $\frac{8}{3}$ which leads to $8+3=\boxed{11}$ ~bluesoul | null | 11 |
1bde6215cb9817e3f7f9a9b6455eddd4 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_13 | The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ | This question is guessable but let's prove our answer
\[a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}\]
\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\]
\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\]
lets put $n+1$ into $n$ now
\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\]
and set them equal now
\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n\]
\[a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n\]
let's rewrite it
\[(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n\]
Let's make it look nice and let $b_n=a_{n + 2}-a_n$
\[(b_{n+1})(a_{n + 2}+1)= b_n\]
Since $b_n$ and $b_{n+1}$ are integers, we can see $b_n$ is divisible by $b_{n+1}$
But we can't have an infinite sequence of proper factors, unless $b_n=0$
Thus, $a_{n + 2}-a_n=0$
\[a_{n + 2}=a_n\]
So now, we know $a_3=a_1$
\[a_{3} = \frac {a_1 + 2009} {1 + a_{2}}\]
\[a_{1} = \frac {a_1 + 2009} {1 + a_{2}}\]
\[a_{1}+a_{1}a_{2} = a_1 + 2009\]
\[a_{1}a_{2} = 2009\]
To minimize $a_{1}+a_{2}$ , we need $41$ and $49$
Thus, our answer $= 41+49=\boxed{090}$ | null | 090 |
1bde6215cb9817e3f7f9a9b6455eddd4 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_13 | The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ | If $a_{n} \ne \frac {2009}{a_{n+1}}$ , then either \[a_{n} = \frac {a_{n}}{1} < \frac {a_{n} + 2009}{1 + a_{n+1}} < \frac {2009}{a_{n+1}}\]
or
\[\frac {2009}{a_{n+1}} < \frac {2009 + a_{n}}{a_{n+1} + 1} < \frac {a_{n}}{1} = a_{n}\]
All the integers between $a_{n}$ and $\frac {2009}{a_{n+1}}$ would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.
So $a_{n} = \frac {2009}{a_{n+1}}$ , which $a_{n} \cdot a_{n+1} = 2009$ . When $n = 1$ $a_{1} \cdot a_{2} = 2009$ . The smallest sum of two factors which have a product of $2009$ is $41 + 49=\boxed{090}$ | null | 090 |
1bde6215cb9817e3f7f9a9b6455eddd4 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_13 | The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ | Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms. \begin{align*} a_{1} &= a \\ a_{2} &= b \\ a_{3} &=\frac{a+2009}{1+b} \\ a_{4} &=\frac{(b+1)(b+2009)}{a+b+2010} \\ \end{align*} The terms get more and more wacky, so we just solve for $a,b$ such that $a_{1}=a_{3}$ and $a_{2}=a_{4}.$
Solving we find both equations end up to the equation $ab=2009$ in which we see to minimize we see that $a = 49$ and $b=41$ or vice versa for an answer of $\boxed{90}.$ This solution is VERY non rigorous and not recommended. | null | 90 |
4c0f5e1d00eb799e1d88c40451b8113f | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_14 | For $t = 1, 2, 3, 4$ , define $S_t = \sum_{i = 1}^{350}a_i^t$ , where $a_i \in \{1,2,3,4\}$ . If $S_1 = 513$ and $S_4 = 4745$ , find the minimum possible value for $S_2$ | Because the order of the $a$ 's doesn't matter, we simply need to find the number of $1$ $2$ $3$ s and $4$ s that minimize $S_2$ . So let $w, x, y,$ and $z$ represent the number of $1$ s, $2$ s, $3$ s, and $4$ s respectively. Then we can write three equations based on these variables. Since there are a total of $350$ $a$ s, we know that $w + x + y + z = 350$ . We also know that $w + 2x + 3y + 4z = 513$ and $w + 16x + 81y + 256z = 4745$ . We can now solve these down to two variables: \[w = 350 - x - y - z\] Substituting this into the second and third equations, we get \[x + 2y + 3z = 163\] and \[15x + 80y + 255z = 4395.\] The second of these can be reduced to \[3x + 16y + 51z = 879.\] Now we substitute $x$ from the first new equation into the other new equation. \[x = 163 - 2y - 3z\] \[3(163 - 2y - 3z) + 16y + 51z = 879\] \[489 + 10y + 42z = 879\] \[5y + 21z = 195\] Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$ .
If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$ $x = 112$ $y = 18$ , and $z = 5$ , so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \boxed{905}$ | null | 905 |
195c4e9f543be99d4db251ab484f31cf | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_15 | In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area of $\triangle BPC$ can be expressed in the form $a - b\sqrt {c}$ , where $a$ $b$ , and $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a + b + c$ | First, by the Law of Cosines , we have \[\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},\] so $\angle BAC = 60^\circ$
Let $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$ , respectively. We first compute \[\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.\] Because $\angle BDI_B$ and $\angle I_BBD$ are half of $\angle BDA$ and $\angle ABD$ , respectively, the above expression can be simplified to \[\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA.\] Similarly, $\angle CO_2D = \angle ACD + \angle CDA$ . As a result
\begin{align*}\angle CPB &= \angle CPD + \angle BPD \\&= \frac {1}{2} \cdot \angle CO_2D + \frac {1}{2} \cdot \angle BO_1D \\&= \frac {1}{2}(\angle ABD + \angle BDA + \angle ACD + \angle CDA) \\&= \frac {1}{2} (2 \cdot 180^\circ - \angle BAC) \\&= \frac {1}{2} \cdot 300^\circ = 150^\circ.\end{align*}
Therefore $\angle CPB$ is constant ( $150^\circ$ ). Also, $P$ is $B$ or $C$ when $D$ is $B$ or $C$ . Let point $L$ be on the same side of $\overline{BC}$ as $A$ with $LC = LB = BC = 14$ $P$ is on the circle with $L$ as the center and $\overline{LC}$ as the radius, which is $14$ . The shortest (and only) distance from $L$ to $\overline{BC}$ is $7\sqrt {3}$
When the area of $\triangle BPC$ is the maximum, the distance from $P$ to $\overline{BC}$ has to be the greatest. In this case, it's $14 - 7\sqrt {3}$ . The maximum area of $\triangle BPC$ is \[\frac {1}{2} \cdot 14 \cdot (14 - 7\sqrt {3}) = 98 - 49 \sqrt {3}\] and the requested answer is $98 + 49 + 3 = \boxed{150}$ | null | 150 |
195c4e9f543be99d4db251ab484f31cf | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_15 | In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area of $\triangle BPC$ can be expressed in the form $a - b\sqrt {c}$ , where $a$ $b$ , and $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a + b + c$ | From Law of Cosines on $\triangle{ABC}$ \[\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.\] Now, \[\angle{CI_CD}+\angle{BI_BD}=180^\circ+\frac{\angle{A}}{2}=210^\circ.\] Since $CI_CDP$ and $BI_BDP$ are cyclic quadrilaterals, it follows that \[\angle{BPC}=\angle{CPD}+\angle{DPB}=(180^\circ-\angle{CI_CD})+(180^\circ-\angle{BI_BD})=360^\circ-210^\circ=150^\circ.\] Next, applying Law of Cosines on $\triangle{CPB}$ \begin{align*} & BC^2=14^2=PC^2+PB^2+2\cdot PB\cdot PC\cdot\frac{\sqrt{3}}{2} \\ & \implies \frac{PC^2+PB^2-196}{PC\cdot PB}=-\sqrt{3} \\ & \implies \frac{PC}{PB}+\frac{PB}{PC}-\frac{196}{PC\cdot PB}=-\sqrt{3} \\ & \implies PC\cdot PB = 196\left(\frac{1}{\frac{PC}{PB}+\frac{PB}{PC}+\sqrt{3}}\right). \end{align*} By AM-GM, $\frac{PC}{PB}+\frac{PB}{PC}\geq{2}$ , so \[PB\cdot PC\leq 196\left(\frac{1}{2+\sqrt{3}}\right)=196(2-\sqrt{3}).\] Finally, \[[\triangle{BPC}]=\frac12 \cdot PB\cdot PC\cdot\sin{150^\circ}=\frac14 \cdot PB\cdot PC,\] and the maximum area would be $49(2-\sqrt{3})=98-49\sqrt{3},$ so the answer is $\boxed{150}$ | null | 150 |
195c4e9f543be99d4db251ab484f31cf | https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_15 | In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area of $\triangle BPC$ can be expressed in the form $a - b\sqrt {c}$ , where $a$ $b$ , and $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a + b + c$ | Proceed as in Solution 2 until you find $\angle CPB = 150$ . The locus of points $P$ that give $\angle CPB = 150$ is a fixed arc from $B$ to $C$ $P$ will move along this arc as $D$ moves along $BC$ ) and we want to maximise the area of [ $\triangle BPC$ ]. This means we want $P$ to be farthest distance away from $BC$ as possible, so we put $P$ in the middle of the arc (making $\triangle BPC$ isosceles). We know that $BC=14$ and $\angle CPB = 150$ , so $\angle PBC = \angle PCB = 15$ . Let $O$ be the foot of the perpendicular from $P$ to line $BC$ . Then the area of [ $\triangle BPC$ ] is the same as $7OP$ because base $BC$ has length $14$ . We can split $\triangle BPC$ into two $15-75-90$ triangles $BOP$ and $COP$ , with $BO=CO=7$ and $OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3$ . Then, the area of [ $\triangle BPC$ ] is equal to $7 \cdot OP=98-49\sqrt{3}$ , and so the answer is $98+49+3=\boxed{150}$ | null | 150 |
d78f3bab136a5edf4836f847917edd11 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | Let $x$ be the amount of paint in a strip. The total amount of paint is $130+164+188=482$ . After painting the wall, the total amount of paint is $482-4x$ . Because the total amount in each color is the same after finishing, we have
\[\frac{482-4x}{3} = 130-x\] \[482-4x=390-3x\] \[x=92\]
Our answer is $482-4\cdot 92 = 482 - 368 = \boxed{114}$ | null | 114 |
d78f3bab136a5edf4836f847917edd11 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | Let the stripes be $b, r, w,$ and $p$ , respectively. Let the red part of the pink be $\frac{r_p}{p}$ and the white part be $\frac{w_p}{p}$ for $\frac{r_p+w_p}{p}=p$
Since the stripes are of equal size, we have $b=r=w=p$ . Since the amounts of paint end equal, we have $130-b=164-r-\frac{r_p}{p}=188-w-\frac{w_p}{p}$ . Thus, we know that \[130-p=164-p-\frac{r_p}{p}=188-p-\frac{w_p}{p}\] \[130=164-\frac{r_p}{p}=188-\frac{w_p}{p}\] \[r_p=34p, w_p=58p\] \[\frac{r_p+w_p}{p}=92=p=b.\] Each paint must end with $130-92=38$ oz left, for a total of $3 \cdot 38 = \boxed{114}$ oz. | null | 114 |
d78f3bab136a5edf4836f847917edd11 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)x = 188 - 130 = 58$ . The second equation becomes $x - ax = 58$ and substituting the first equation into this one we get $x - 34 = 58$ so $x = 92$ . The amount of each color left over at the end is thus $130 - 92 = 38$ and $38 * 3 = \boxed{114}$ | null | 114 |
d78f3bab136a5edf4836f847917edd11 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | We know that all the stripes are of equal size. We can then say that $r$ is the amount of paint per stripe. Then $130 - r$ will be the amount of blue paint left. Now for the other two stripes. The amount of white paint left after the white stripe and the amount of red paint left after the blue stripe are $188 - r$ and $164 - r$ respectively. The pink stripe is also r ounces of paint, but let there be $k$ ounces of red paint in the mixture and $r - k$ ounces of white paint. We now have two equations: $164 - r - k = 188 - r - (r-k)$ and $164 - r - k = 130 - r$ . Solving yields k = 34 and r = 92. We now see that there will be $130 - 92 = 38$ ounces of paint left in each can. $38 * 3 = \boxed{114}$ | null | 114 |
d78f3bab136a5edf4836f847917edd11 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | Let the amount of paint each stripe painted used be $x$ . Also, let the amount of paint of each color left be $y$ . 1 stripe is drawn with the blue paint, and 3 stripes are drawn with the red and white paints. Add together the amount of red and white paint, $164 + 188 = 352$ and obtain the following equations : $352 - 3x = 2y$ and $130 - x = y$ . Solve to obtain $x = 92$ . Therefore $y$ is $130 - 92 = 38$ , with three cans of equal amount of paint, the answer is $38 * 3 = \boxed{114}$ | null | 114 |
d78f3bab136a5edf4836f847917edd11 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | Let $x$ be the number of ounces of paint needed for a single stripe. We know that in the end, the total amount of red and white paint will equal double the amount of blue paint.
After painting, the amount of red and white paint remaining is equal to $164+188-2x$ , and then minus another $x$ for the pink stripe. The amount of blue paint remaining is equal to $130-x$ .
So, we get the equation $2\cdot(130-x)=164+188-3x$ . Simplifying, we get $x=38$ and $3x=\boxed{114}$ | null | 114 |
d78f3bab136a5edf4836f847917edd11 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | Just like in solution 1, we note that all colors will be used equally, except for the pink stripe. This must bring red and white down to $130$ each, so $34$ red and $58$ white are used, making for a total of $92$ for the pink stripe. Thus, the other stripes also use $92$ . The answer is $130+164+188-4(92)=\boxed{114}$ | null | 114 |
f35f634f2750d3a83590f0dd70a70cd0 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_2 | Suppose that $a$ $b$ , and $c$ are positive real numbers such that $a^{\log_3 7} = 27$ $b^{\log_7 11} = 49$ , and $c^{\log_{11}25} = \sqrt{11}$ . Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\] | First, we have: \[x^{(\log_y z)^2} = x^{\left( (\log_y z)^2 \right) } = x^{(\log_y z) \cdot (\log_y z) } = \left( x^{\log_y z} \right)^{\log_y z}\]
Now, let $x=y^w$ , then we have: \[x^{\log_y z} = \left( y^w \right)^{\log_y z} = y^{w\log_y z} = y^{\log_y (z^w)} = z^w\]
This is all we need to evaluate the given formula. Note that in our case we have $27=3^3$ $49=7^2$ , and $\sqrt{11}=11^{1/2}$ . We can now compute:
\[a^{(\log_3 7)^2} = \left( a^{\log_3 7} \right)^{\log_3 7} = 27^{\log_3 7} = (3^3)^{\log_3 7} = 7^3 = 343\]
Similarly, we get \[b^{(\log_7 11)^2} = (7^2)^{\log_7 11} = 11^2 = 121\]
and \[c^{(\log_{11} 25)^2} = (11^{1/2})^{\log_{11} 25} = 25^{1/2} = 5\]
and therefore the answer is $343+121+5 = \boxed{469}$ | null | 469 |
f35f634f2750d3a83590f0dd70a70cd0 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_2 | Suppose that $a$ $b$ , and $c$ are positive real numbers such that $a^{\log_3 7} = 27$ $b^{\log_7 11} = 49$ , and $c^{\log_{11}25} = \sqrt{11}$ . Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\] | We know from the first three equations that $\log_a27 = \log_37$ $\log_b49 = \log_711$ , and $\log_c\sqrt{11} = \log_{11}25$ . Substituting, we find
\[a^{(\log_a27)(\log_37)} + b^{(\log_b49)(\log_711)} + c^{(\log_c\sqrt {11})(\log_{11}25)}.\]
We know that $x^{\log_xy} =y$ , so we find
\[27^{\log_37} + 49^{\log_711} + \sqrt {11}^{\log_{11}25}\]
\[(3^{\log_37})^3 + (7^{\log_711})^2 + ({11^{\log_{11}25}})^{1/2}.\]
The $3$ and the $\log_37$ cancel to make $7$ , and we can do this for the other two terms. Thus, our answer is
\[7^3 + 11^2 + 25^{1/2}\] \[= 343 + 121 + 5\] \[= \boxed{469}.\] | null | 469 |
f42bea685006b8c359e00b0e3139bdb2 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_3 | In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ | From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$ , and $ABE$ are also right triangles. By $AA$ $\triangle FBA \sim \triangle BCA$ , and $\triangle FBA \sim \triangle ABE$ , so $\triangle ABE \sim \triangle BCA$ . This gives $\frac {AE}{AB}= \frac {AB}{BC}$ $AE=\frac{AD}{2}$ and $BC=AD$ , so $\frac {AD}{2AB}= \frac {AB}{AD}$ , or $(AD)^2=2(AB)^2$ , so $AD=AB \sqrt{2}$ , or $100 \sqrt{2}$ , so the answer is $\boxed{141}$ | null | 141 |
f42bea685006b8c359e00b0e3139bdb2 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_3 | In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ | Let $x$ be the ratio of $BC$ to $AB$ . On the coordinate plane, plot $A=(0,0)$ $B=(100,0)$ $C=(100,100x)$ , and $D=(0,100x)$ . Then $E=(0,50x)$ . Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$ . They are perpendicular, so they multiply to $-1$ , that is, \[x\cdot-\frac{x}{2}=-1,\] which implies that $-x^2=-2$ or $x=\sqrt 2$ . Therefore $AD=100\sqrt 2\approx 141.42$ so $\lfloor AD\rfloor=\boxed{141}$ | null | 141 |
f42bea685006b8c359e00b0e3139bdb2 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_3 | In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ | Draw $CX$ and $EX$ to form a parallelogram $AEXC$ . Since $EX \parallel AC$ $\angle BEX=90^\circ$ by the problem statement, so $\triangle BEX$ is right.
Letting $AE=y$ , we have $BE=\sqrt{100^2+y^2}$ and $AC=EX=\sqrt{100^2+(2y)^2}$ . Since $CX=EA$ $\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2$ . Solving this, we have \[100^2+ 100^2 + y^2 + 4y^2 = 9y^2\] \[2\cdot 100^2 = 4y^2\] \[\frac{100^2}{2}=y^2\] \[\frac{100}{\sqrt{2}}=y\] \[\frac{100\sqrt{2}}{2}=y\] \[100\sqrt{2}=2y=AD\] , so the answer is $\boxed{141}$ | null | 141 |
6381ebaab6ebb311f1df953c4148ef05 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_4 | A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ -th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$ . Find the smallest possible value of $n$ | The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term $n$ (the number of grapes eaten by the child in $1$ -st place), difference $d=-2$ , and number of terms $c$ . We can easily compute that this sum is equal to $c(n-c+1)$
Hence we have the equation $2009=c(n-c+1)$ , and we are looking for a solution $(n,c)$ , where both $n$ and $c$ are positive integers, $n\geq 2(c-1)$ , and $n$ is minimized. (The condition $n\geq 2(c-1)$ states that even the last child had to eat a non-negative number of grapes.)
The prime factorization of $2009$ is $2009=7^2 \cdot 41$ . Hence there are $6$ ways how to factor $2009$ into two positive terms $c$ and $n-c+1$
The smallest valid solution is therefore $c=41$ $n=\boxed{089}$ | null | 089 |
6381ebaab6ebb311f1df953c4148ef05 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_4 | A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ -th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$ . Find the smallest possible value of $n$ | If the first child ate $n=2m$ grapes, then the maximum number of grapes eaten by all the children together is $2m + (2m-2) + (2m-4) + \cdots + 4 + 2 = m(m+1)$ . Similarly, if the first child ate $2m-1$ grapes, the maximum total number of grapes eaten is $(2m-1)+(2m-3)+\cdots+3+1 = m^2$
For $m=44$ the value $m(m+1)=44\cdot 45 =1980$ is less than $2009$ . Hence $n$ must be at least $2\cdot 44+1=89$ . For $n=89$ , the maximum possible sum is $45^2=2025$ . And we can easily see that $2009 = 2025 - 16 = 2025 - (1+3+5+7)$ , hence $2009$ grapes can indeed be achieved for $n=89$ by dropping the last four children.
Hence we found a solution with $n=89$ and $45-4=41$ kids, and we also showed that no smaller solution exists. Therefore the answer is $\boxed{089}$ | null | 089 |
6381ebaab6ebb311f1df953c4148ef05 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_4 | A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ -th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$ . Find the smallest possible value of $n$ | If the winner ate n grapes, then 2nd place ate $n+2-4=n-2$ grapes, 3rd place ate $n+2-6=n-4$ grapes, 4th place ate $n-6$ grapes, and so on. Our sum can be written as $n+(n-2)+(n-4)+(n-6)\dots$ . If there are x places, we can express this sum as $(x+1)n-x(x+1)$ , as there are $(x+1)$ occurrences of n, and $(2+4+6+\dots)$ is equal to $x(x+1)$ . This can be factored as $(x+1)(n-x)=2009$ . Our factor pairs are (1,2009), (7,287), and (41,49). To minimize n we take (41,49). If $x+1=41$ , then $x=40$ and $n=40+49=\boxed{089}$ . (Note we would have come upon the same result had we used $x+1=49$ .)
~MC413551 | null | 089 |
578d6a933ad0867bc64c3ad77eda5c02 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_5 | Equilateral triangle $T$ is inscribed in circle $A$ , which has radius $10$ . Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$ . Circles $C$ and $D$ , both with radius $2$ , are internally tangent to circle $A$ at the other two vertices of $T$ . Circles $B$ $C$ , and $D$ are all externally tangent to circle $E$ , which has radius $\dfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90); pair Ep=(0,4-27/5); pair[] dotted={A,B,C,D,Ep}; draw(Circle(A,10)); draw(Circle(B,3)); draw(Circle(C,2)); draw(Circle(D,2)); draw(Circle(Ep,27/5)); dot(dotted); label("$E$",Ep,E); label("$A$",A,W); label("$B$",B,W); label("$C$",C,W); label("$D$",D,E); [/asy] | Let $X$ be the intersection of the circles with centers $B$ and $E$ , and $Y$ be the intersection of the circles with centers $C$ and $E$ . Since the radius of $B$ is $3$ $AX =4$ . Assume $AE$ $p$ . Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$ $AC = 8$ , and angle $CAE = 60$ degrees because we are given that triangle $T$ is equilateral. Using the Law of Cosines on triangle $CAE$ , we obtain
$(6+p)^2 =p^2 + 64 - 2(8)(p) \cos 60$
The $2$ and the $\cos 60$ terms cancel out:
$p^2 + 12p +36 = p^2 + 64 - 8p$
$12p+ 36 = 64 - 8p$
$p =\frac {28}{20} = \frac {7}{5}$ . The radius of circle $E$ is $4 + \frac {7}{5} = \frac {27}{5}$ , so the answer is $27 + 5 = \boxed{032}$ | null | 032 |
ec2dc00af3775214b1329a646b126c80 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_6 | Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ | We can use complementary counting. We can choose a five-element subset in ${14\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$ s and 5 $B$ s, thereby showing that there are ${10\choose 5}$ such sets.
Given a five-element subset $S$ of $\{1,2,\dots,14\}$ in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the $i$ -th character is $A$ if $i\in S$ and $B$ otherwise. Now we got a string with 5 $A$ s and 9 $B$ s. As no two numbers were consecutive, we know that in our string no two $A$ s are consecutive. We can now remove exactly one $B$ from between each pair of $A$ s to get a string with 5 $A$ s and 5 $B$ s. And clearly this is a bijection, as from each string with 5 $A$ s and 5 $B$ s we can reconstruct one original set by reversing the construction.
Hence we have $m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750$ , and the answer is $1750 \bmod 1000 = \boxed{750}$ | null | 750 |
ec2dc00af3775214b1329a646b126c80 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_6 | Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ | We will approach this problem using complementary counting. First it is obvious, there are $\binom{14}{5}$ total sets. To find the number of sets with NO consecutive elements, we do a little experimentation. Consider the following:
Start of with any of the $14$ numbers, say WLOG, $1$ . Then we cannot have $2$ . So again WLOG, we may pick $3$ . Then we cannot have $4$ , so again WLOG, we may pick $5$ . Then not $6$ , but $7$ , then not $8$ , but $9$ . Now we have the set ${1,3,5,7,9}$ , and we had to remove $4$ digits from the $14$ total amount, so there wasn't any consecutive numbers. So we have that the number of non-consecutive cardinality $5$ sets, is $\binom{14-4}{5}=\binom{10}{5}$ . Now you already read the solutions above, and if you are here, you are either confused, or looking for more. So now the answer is simply $1750$ , which is $\boxed{750}$ (mod $1000$ ). | null | 750 |
ec2dc00af3775214b1329a646b126c80 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_6 | Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ | We will proceed by complementary counting. We can easily see that there are $\binom{14}{5}$ ways to select $5$ element subsets from the original set. Now, we must find the number of subsets part of that group that do NOT have any consecutive integers.
We can think of this situation as stars and bars. We have $9$ stars and $5$ bars, where the stars represent the integers NOT in the subset while the bars ARE the integers in the chosen subset. We want the amount of combinations where there is at least one star between each bar (meaning that none of the integers in the subset are consecutive). To do this, we remove $4$ stars from the total $9$ stars ( $4$ because that is the amount that is between each bar). Now, we have $5$ stars and $5$ bars with no restrictions, so we calculate this as $\binom{10}{5}$ , which is $252$ . Hence, $\binom{14}{15} - \binom{10}{5} = 2002 - 252 = 1750$ , which becomes $\boxed{750}$ | null | 750 |
7cee9d4bb93b10837de0bff19c266aa3 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ | First, note that $(2n)!! = 2^n \cdot n!$ , and that $(2n)!! \cdot (2n-1)!! = (2n)!$
We can now take the fraction $\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$ . We get that this fraction is equal to $\dfrac{(2i)!}{(2i)!!^2} = \dfrac{(2i)!}{2^{2i}(i!)^2}$
Now we can recognize that $\dfrac{(2i)!}{(i!)^2}$ is simply ${2i \choose i}$ , hence this fraction is $\dfrac{{2i\choose i}}{2^{2i}}$ , and our sum turns into $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$
Let $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ .
Obviously $c$ is an integer, and $S$ can be written as $\dfrac{c}{2^{2\cdot 2009}}$ .
Hence if $S$ is expressed as a fraction in lowest terms, its denominator will be of the form $2^a$ for some $a\leq 2\cdot 2009$
In other words, we just showed that $b=1$ .
To determine $a$ , we need to determine the largest power of $2$ that divides $c$
Let $p(i)$ be the largest $x$ such that $2^x$ that divides $i$
We can now return to the observation that $(2i)! = (2i)!! \cdot (2i-1)!! = 2^i \cdot i! \cdot (2i-1)!!$ . Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$
It immediately follows that $p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)$ ,
and hence $p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)$
Obviously, for $i\in\{1,2,\dots,2009\}$ the function $f(i)=2\cdot 2009 - i - p(i!)$ is is a strictly decreasing function.
Therefore $p(c) = p\left( {2\cdot 2009\choose 2009} \right) = 2009 - p(2009!)$
We can now compute $p(2009!) = \sum_{k=1}^{\infty} \left\lfloor \dfrac{2009}{2^k} \right\rfloor = 1004 + 502 + \cdots + 3 + 1 = 2001$ .
Hence $p(c)=2009-2001=8$
And thus we have $a=2\cdot 2009 - p(c) = 4010$ , and the answer is $\dfrac{ab}{10} = \dfrac{4010\cdot 1}{10} = \boxed{401}$ | null | 401 |
7cee9d4bb93b10837de0bff19c266aa3 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ | Using the steps of the previous solution we get $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ and if you do the small cases(like $1, 2, 3, 4, 5, 6$ ) you realize that you can "thin-slice" the problem and simply look at the cases where $i=2009, 2008$ (they're nearly identical in nature but one has $4$ with it) since $\dbinom{2i}{I}$ hardly contains any powers of $2$ or in other words it's very inefficient and the inefficient cases hold all the power so you can just look at the highest powers of $2$ in $\dbinom{4018}{2009}$ and $\dbinom{4016}{2008}$ and you get the minimum power of $2$ in either expression is $8$ so the answer is $\frac{4010}{10} \implies \boxed{401}$ since it would violate the rules of the AIME and the small cases if $b>1$ | null | 401 |
7cee9d4bb93b10837de0bff19c266aa3 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ | We can logically deduce that the $b$ value will be 1. Listing out the first few values of odd and even integers, we have: $1, 3, 5, 7, 9, 11, 13...$ and $2, 4, 6, 8, 10, 12, 14, 16...$ . Obviously, none of the factors of $2$ in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of $3$ occurs every $3$ terms, and starting out on the third term of the denominator, a factor of $3$ appears also every $3$ terms. Thus, the factors of $3$ on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form $1 \cdot 2^a$ . Since there will be no odd factors in the denominators, all the denominators will be in the form $2^a$ where $a$ is the number of factors of $2$ in $(2009 \cdot 2)! = 4018!$ . This is simply $\sum_{n=1}^{11} \left \lfloor{\frac{4018}{2^n}}\right \rfloor = 4010$ . Therefore, our answer is $\boxed{401}$ | null | 401 |
7cee9d4bb93b10837de0bff19c266aa3 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_7 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ | Using the initial steps from Solution 1, $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$ . Clearly $b = 1$ as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find $a$ . Note that $2^a$ would be the largest denominator in all the summands, so when they are summed it is the common denominator.
Taking the p-adic valuations of each term, the powers of 2 in the denominator for $i$ is $(2i) - v_2(\binom{2i}{i}).$ We can use Kummers theorem to see that $v_2(\binom{2i}{i})$ is the number of digits carried over when $i$ is added to $i$ in base $2$ . This is simply the number of $1$ 's in the binary representation of $i$
Looking at the binary representations of some of the larger $i$ we see $2009 = 11111011001_2$ having eight $1$ 's. So the power of two is $2 \cdot 2009 - 8 = 4010$ . Experimenting with $2008, 2007, 2006$ we see that the power of two are all $< 4010$ , and under $2005$ the power of two $2i - v_2(\binom{2i}{i}) < 2i < 4010$ . Therefore $a = 4010$ and $\frac{ab}{10} = \boxed{401}.$ | null | 401 |
493c7a5b42a716a3f254d62a2ef89c37 | https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_8 | Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\dfrac mn$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $m+n$ | There are many almost equivalent approaches that lead to summing a geometric series. For example, we can compute the probability of the opposite event. Let $p$ be the probability that Dave will make at least two more throws than Linda. Obviously, $p$ is then also the probability that Linda will make at least two more throws than Dave, and our answer will therefore be $1-2p$
How to compute $p$
Suppose that Linda made exactly $t$ throws. The probability that this happens is $(5/6)^{t-1}\cdot (1/6)$ , as she must make $t-1$ unsuccessful throws followed by a successful one. In this case, we need Dave to make at least $t+2$ throws. This happens if his first $t+1$ throws are unsuccessful, hence the probability is $(5/6)^{t+1}$
Thus for a fixed $t$ the probability that Linda makes $t$ throws and Dave at least $t+2$ throws is $(5/6)^{2t} \cdot (1/6)$
Then, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$ . Hence:
\begin{align*} p & = \sum_{t=1}^\infty \left(\frac 56\right)^{2t} \cdot \frac 16 \\ & = \frac 16 \cdot \left(\frac 56\right)^2 \cdot \sum_{x=0}^\infty \left(\frac{25}{36}\right)^x \\ & = \frac 16 \cdot \frac{25}{36} \cdot \frac 1{1 - \dfrac{25}{36}} \\ & = \frac 16 \cdot \frac{25}{36} \cdot \frac{36}{11} \\ & = \frac {25}{66} \end{align*}
Hence the probability we were supposed to compute is $1 - 2p = 1 - 2\cdot \frac{25}{66} = 1 - \frac{25}{33} = \frac 8{33}$ , and the answer is $8+33 = \boxed{041}$ | null | 041 |