Difference between revisions of "Mock AIME II 2012 Problems"
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==Problem 1== | ==Problem 1== | ||
Given that <cmath>\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\right)=\dfrac{m}{n},</cmath> where <math>m</math> and <math>n</math> are positive relatively prime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>. | Given that <cmath>\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\right)=\dfrac{m}{n},</cmath> where <math>m</math> and <math>n</math> are positive relatively prime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>. | ||
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==Problem 7== | ==Problem 7== | ||
+ | Given <math> x, y </math> are positive real numbers that satisfy <math> 3x+4y+1=3\sqrt{x}+2\sqrt{y} </math>, then the value <math> xy </math> can be expressed as <math> \frac{m}{n} </math>, where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n </math>. | ||
[[Mock AIME II 2012 Problems/Problem 7| Solution]] | [[Mock AIME II 2012 Problems/Problem 7| Solution]] | ||
+ | ==Problem 8== | ||
+ | Let <math>A</math> be a point outside circle <math>\Omega</math> with center <math>O</math> and radius <math>9</math> such that the tangents from <math>A</math> to <math>\Omega</math>, <math>AB</math> and <math>AC</math>, form <math>\angle BAO=15^{\circ}</math>. Let <math>AO</math> first intersect the circle at <math>D</math>, and extend the parallel to <math>AB</math> from <math>D</math> to meet the circle at <math>E</math>. The length <math>EC^2=m+k\sqrt{n}</math>, where <math>m</math>,<math>n</math>, and <math>k</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+k</math>. | ||
[[Mock AIME II 2012 Problems/Problem 8| Solution]] | [[Mock AIME II 2012 Problems/Problem 8| Solution]] | ||
+ | ==Problem 9== | ||
+ | In <math>\triangle ABC</math>, <math>AB=12</math>, <math>AC=20</math>, and <math>\angle ABC=120^\circ</math>. <math>D, E,</math> and <math>F</math> lie on <math>\overline{AC}, \overline{AB}</math>, and <math>\overline{BC}</math>, respectively. If <math>AE=\frac{1}{4}AB, BF=\frac{1}{4}BC</math>, and <math>AD=\frac{1}{4}AC</math>, the area of <math>\triangle DEF</math> can be expressed in the form <math>\frac{a\sqrt{b}-c\sqrt{d}}{e}</math> where <math>a, b, c, d, e</math> are all positive integers, and <math>b</math> and <math>d</math> do not have any perfect squares greater than <math>1</math> as divisors. Find <math>a+b+c+d+e</math>. | ||
[[Mock AIME II 2012 Problems/Problem 9| Solution]] | [[Mock AIME II 2012 Problems/Problem 9| Solution]] | ||
+ | ==Problem 10== | ||
+ | Call a set of positive integers <math>\mathcal{S}</math> <math>\textit{lucky}</math> if it can be split into two nonempty disjoint subsets <math>\mathcal{A}</math> and <math>\mathcal{B}</math> with <math>A\cap B=S</math> such that the product of the elements in <math>\mathcal{A}</math> and the product of the elements in <math>\mathcal{B}</math> sum up to the cardinality of <math>\mathcal{S}</math>. Find the number of <math>\textit{lucky}</math> sets such that the largest element is less than <math>15</math>. (Disjoint subsets have no elements in common, and the cardinality of a set is the number of elements in the set.) | ||
[[Mock AIME II 2012 Problems/Problem 10| Solution]] | [[Mock AIME II 2012 Problems/Problem 10| Solution]] | ||
+ | ==Problem 11== | ||
[[Mock AIME II 2012 Problems/Problem 11| Solution]] | [[Mock AIME II 2012 Problems/Problem 11| Solution]] | ||
Revision as of 02:14, 5 April 2012
Contents
Problem 1
Given that where and are positive relatively prime integers, find the remainder when is divided by .
Problem 2
Let be a recursion defined such that , and where , and is an integer. If for being a positive integer greater than and being a positive integer greater than 2, find the smallest possible value of .
Problem 3
The of a number is defined as the result obtained by repeatedly adding the digits of the number until a single digit remains. For example, the of is (). Find the of .
Problem 4
Let be a triangle, and let , , and be the points where the angle bisectors of , , and , respectfully, intersect the sides opposite them. Given that , , and , then the ratio can be written in the form where and are positive relatively prime integers. Find .
Problem 5
A fair die with sides numbered through inclusive is rolled times. The probability that the sum of the rolls is is nonzero and is equivalent to the probability that a sum of is rolled. Find the minimum value of k.
Problem 6
A circle with radius and center in the first quadrant is placed so that it is tangent to the -axis. If the line passing through the origin that is tangent to the circle has slope , then the -coordinate of the center of the circle can be written in the form where , , and are positive integers, and . Find .
Problem 7
Given are positive real numbers that satisfy , then the value can be expressed as , where and are relatively prime positive integers. Find . Solution
Problem 8
Let be a point outside circle with center and radius such that the tangents from to , and , form . Let first intersect the circle at , and extend the parallel to from to meet the circle at . The length , where ,, and are positive integers and is not divisible by the square of any prime. Find . Solution
Problem 9
In , , , and . and lie on , and , respectively. If , and , the area of can be expressed in the form where are all positive integers, and and do not have any perfect squares greater than as divisors. Find . Solution
Problem 10
Call a set of positive integers if it can be split into two nonempty disjoint subsets and with such that the product of the elements in and the product of the elements in sum up to the cardinality of . Find the number of sets such that the largest element is less than . (Disjoint subsets have no elements in common, and the cardinality of a set is the number of elements in the set.) Solution