https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Thisisasentence&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-26T11:35:07Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2011_UNCO_Math_Contest_II_Problems/Problem_6&diff=72633 2011 UNCO Math Contest II Problems/Problem 6 2015-10-28T04:42:22Z <p>Thisisasentence: /* Problem */</p> <hr /> <div>== Problem ==<br /> <br /> What is the remainder when &lt;math&gt;1! + 2! + 3! + \cdots + 2011!&lt;/math&gt; is divided by &lt;math&gt;18&lt;/math&gt;?<br /> <br /> == Solution ==<br /> Since all the terms past &lt;math&gt;5!&lt;/math&gt; are divisible by &lt;math&gt;18&lt;/math&gt;, it is only necessary to look at the remainder resulted from the first 5 terms.<br /> &lt;math&gt;1!+2!+3!+4!+5!=153\equiv 9\pmod{18}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2011|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2011_UNCO_Math_Contest_II_Problems/Problem_6&diff=72632 2011 UNCO Math Contest II Problems/Problem 6 2015-10-28T04:42:10Z <p>Thisisasentence: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> What is the remainder when &lt;math&gt;1! + 2! + 3! + ?+ 2011!&lt;/math&gt; is divided by &lt;math&gt;18&lt;/math&gt;?<br /> <br /> <br /> == Solution ==<br /> Since all the terms past &lt;math&gt;5!&lt;/math&gt; are divisible by &lt;math&gt;18&lt;/math&gt;, it is only necessary to look at the remainder resulted from the first 5 terms.<br /> &lt;math&gt;1!+2!+3!+4!+5!=153\equiv 9\pmod{18}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2011|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2005_Alabama_ARML_TST_Problems/Problem_8&diff=72631 2005 Alabama ARML TST Problems/Problem 8 2015-10-28T04:38:22Z <p>Thisisasentence: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the number of ordered pairs of integers &lt;math&gt;(x,y)&lt;/math&gt; which satisfy &lt;center&gt;&lt;math&gt;x^2+4xy+y^2=21&lt;/math&gt;.&lt;/center&gt;<br /> <br /> ==Solution==<br /> We look at &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y \pmod{3}&lt;/math&gt;, since &lt;math&gt;21&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> <br /> * Case 1: &lt;math&gt;x\equiv 0\pmod{3}&lt;/math&gt;<br /> ** Case 1a: &lt;math&gt;y\equiv 0\pmod{3}&lt;/math&gt;: Then &lt;math&gt;x^2+4xy+y^2&lt;/math&gt; is divisible by &lt;math&gt;3^2=9&lt;/math&gt;, but &lt;math&gt;21&lt;/math&gt; isn't.<br /> ** Case 1b: &lt;math&gt;y\equiv 1\pmod{3}&lt;/math&gt;: Then the LHS is &lt;math&gt;1\pmod{3}&lt;/math&gt;, while the RHS isn't.<br /> ** Case 1c: &lt;math&gt;y\equiv 2\pmod{3}&lt;/math&gt;: Then the LHS is &lt;math&gt;1\pmod{3}&lt;/math&gt;, while the RHS isn't.<br /> * Case 2: x=1mod3<br /> ** Case 2a: y=0mod3: This is equivalent to case 1b.<br /> ** Case 2b: y=1mod3: We let &lt;math&gt;x=3x_1+1&lt;/math&gt; and &lt;math&gt;y=3y_1+1&lt;/math&gt;:<br /> <br /> &lt;math&gt;x^2+4xy+y^2=21=(3x_1+1)^2+4(3x_1+1)(3y_1+1)+(3y_1+1)^2=9(x^2+y^2+4x_1y_1+2x_1+2y_1)+6&lt;/math&gt;<br /> <br /> But 21 isn't 6mod9, it's 3mod9.<br /> <br /> ** Case 2c: y=2mod3: Then the LHS is 1mod3 while the RHS isn't.<br /> * Case 3: x=2mod3<br /> ** Case 3a: y=0mod3: This is equivalent to case 1c.<br /> ** Case 3b: y=1mod3: This is equivalent to case 2c.<br /> ** Case 3c: y=2mod3: We let &lt;math&gt;x=3x_1+2&lt;/math&gt; and &lt;math&gt;y=3y_1+2&lt;/math&gt;:<br /> <br /> &lt;math&gt;x^2+4xy+y^2=21=(3x_1+2)^2+4(3x_1+2)(3y_1+2)+(3y_1+2)^2=9(x_1^2+y_1^2+4x_1y_1+4x_1+4y_1+2)+6&lt;/math&gt;<br /> <br /> But 21 isn't 6mod9, it's 3mod9.<br /> <br /> Therefore, there are absolutely no solutions to the above equation.<br /> <br /> ==See Also==<br /> {{ARML box|year=2005|state=Alabama|num-b=7|num-a=9}}<br /> <br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2010_UNCO_Math_Contest_II_Problems/Problem_8&diff=72456 2010 UNCO Math Contest II Problems/Problem 8 2015-10-13T01:12:24Z <p>Thisisasentence: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Simplify &lt;math&gt;(3^1+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\cdots (3^{1024}+1)&lt;/math&gt;, using exponential notation to<br /> express your answer. Generalize this result.<br /> <br /> <br /> == Solution ==<br /> Let the product = &lt;math&gt;P&lt;/math&gt;. Then <br /> &lt;math&gt;(3-1)P=3^{2048}-1\implies&lt;/math&gt;<br /> &lt;math&gt;P=\frac{3^{2048}-1}{2}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2010|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=1968_AHSME_Problems/Problem_12&diff=72455 1968 AHSME Problems/Problem 12 2015-10-13T01:02:58Z <p>Thisisasentence: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A circle passes through the vertices of a triangle with side-lengths &lt;math&gt;7\tfrac{1}{2},10,12\tfrac{1}{2}.&lt;/math&gt; The radius of the circle is:<br /> <br /> &lt;math&gt;\text{(A) } \frac{15}{4}\quad<br /> \text{(B) } 5\quad<br /> \text{(C) } \frac{25}{4}\quad<br /> \text{(D) } \frac{35}{4}\quad<br /> \text{(E) } \frac{15\sqrt{2}}{2}&lt;/math&gt;<br /> <br /> == Solution ==<br /> The triangle that goes through all the vertices of the triangle is the circumcircle of the triangle.<br /> &lt;math&gt;(7\frac{1}{2})^{2}+10^{2}=(12\frac{1}{2})^{2}&lt;/math&gt;, so the triangle is a right triangle.The radius of a circumcircle <br /> of a right triangle is half the hypotenuse. &lt;math&gt;\frac{1}{2}\cdot \frac{25}{2}=\frac{25}{4}\implies&lt;/math&gt; &lt;math&gt;\fbox{C}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1968|num-b=11|num-a=13}} <br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_5&diff=71310 2002 AIME II Problems/Problem 5 2015-07-25T06:15:15Z <p>Thisisasentence: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive integers &lt;math&gt;a=2^n3^m&lt;/math&gt; where &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; are non-negative integers, for which &lt;math&gt;a^6&lt;/math&gt; is not a divisor of &lt;math&gt;6^a&lt;/math&gt;<br /> <br /> == Solution ==<br /> Substitute &lt;math&gt;a=2^n3^m&lt;/math&gt; into &lt;math&gt;a^6&lt;/math&gt; and &lt;math&gt;6^a&lt;/math&gt;, and find all pairs of non-negative integers (n,m) for which &lt;math&gt;(2^n3^m)^{6}&lt;/math&gt; is not a divisor of &lt;math&gt;6^{2^n3^m}&lt;/math&gt;<br /> <br /> Simplifying both expressions:<br /> <br /> &lt;math&gt;2^{6n} \cdot 3^{6m}&lt;/math&gt; is not a divisor of &lt;math&gt;2^{2^n3^m} \cdot 3^{2^n3^m}&lt;/math&gt;<br /> <br /> Comparing both exponents (noting that there must be either extra powers of 2 or extra powers of 3 in the left expression):<br /> <br /> &lt;math&gt;6n &gt; 2^n3^m&lt;/math&gt; OR &lt;math&gt;6m &gt; 2^n3^m&lt;/math&gt;<br /> <br /> <br /> Using the first inequality &lt;math&gt;6n &gt; 2^n3^m&lt;/math&gt; and going case by case starting with n &lt;math&gt;\in&lt;/math&gt; {0, 1, 2, 3...}:<br /> <br /> n=0:<br /> &lt;math&gt;0&gt;1 \cdot 3^m&lt;/math&gt; which has no solution for non-negative integers m <br /> <br /> n=1:<br /> &lt;math&gt;6 &gt; 2 \cdot 3^m&lt;/math&gt; which is true for m=0 but fails for higher integers &lt;math&gt;\Rightarrow (1,0)&lt;/math&gt;<br /> <br /> n=2:<br /> &lt;math&gt;12 &gt; 4 \cdot 3^m&lt;/math&gt; which is true for m=0 but fails for higher integers &lt;math&gt;\Rightarrow (2,0)&lt;/math&gt;<br /> <br /> n=3:<br /> &lt;math&gt;18 &gt; 8 \cdot 3^m&lt;/math&gt; which is true for m=0 but fails for higher integers &lt;math&gt;\Rightarrow (3,0)&lt;/math&gt;<br /> <br /> n=4:<br /> &lt;math&gt;24 &gt; 16 \cdot 3^m&lt;/math&gt; which is true for m=0 but fails for higher integers &lt;math&gt;\Rightarrow (4,0)&lt;/math&gt;<br /> <br /> n=5:<br /> &lt;math&gt;30 &gt; 32 \cdot 3^m&lt;/math&gt; which has no solution for non-negative integers m <br /> <br /> There are no more solutions for higher &lt;math&gt;n&lt;/math&gt;, as polynomials like &lt;math&gt;6n&lt;/math&gt; grow slower than exponentials like &lt;math&gt;2^n&lt;/math&gt;.<br /> <br /> <br /> <br /> Using the second inequality &lt;math&gt;6m &gt; 2^n3^m&lt;/math&gt; and going case by case starting with m &lt;math&gt;\in&lt;/math&gt; {0, 1, 2, 3...}:<br /> <br /> m=0:<br /> &lt;math&gt;0&gt;2^n \cdot 1&lt;/math&gt; which has no solution for non-negative integers n <br /> <br /> m=1:<br /> &lt;math&gt;6&gt;2^n \cdot 3&lt;/math&gt; which is true for n=0 but fails for higher integers &lt;math&gt;\Rightarrow (0,1)&lt;/math&gt;<br /> <br /> m=2:<br /> &lt;math&gt;12&gt;2^n \cdot 9&lt;/math&gt; which is true for n=0 but fails for higher integers &lt;math&gt;\Rightarrow (0,2)&lt;/math&gt;<br /> <br /> m=3:<br /> &lt;math&gt;18&gt;2^n \cdot 27&lt;/math&gt; which has no solution for non-negative integers n <br /> <br /> There are no more solutions for higher &lt;math&gt;m&lt;/math&gt;, as polynomials like &lt;math&gt;6m&lt;/math&gt; grow slower than exponentials like &lt;math&gt;3^m&lt;/math&gt;.<br /> <br /> <br /> Thus there are six numbers corresponding to (1,0), (2,0), (3,0), (4,0), (0,1), and (0,2). <br /> Plugging them back into the original expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively. Their sum is &lt;math&gt;\framebox{042}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=<br /> 2002|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems&diff=70714 2012 AMC 10B Problems 2015-06-10T01:42:56Z <p>Thisisasentence: /* Problem 10 */</p> <hr /> <div>== Problem 1 ==<br /> <br /> Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); <br /> draw(circle((10,5),5));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> The point in the xy-plane with coordinates (1000, 2012) is reflected across the line y=2000. What are the coordinates of the reflected point?<br /> <br /> &lt;math&gt; \textbf{(A)}\ (998,2012)\qquad\textbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012) &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in dollars?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> In order to estimate the value of &lt;math&gt;x-y&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are real numbers with &lt;math&gt;x &gt; y &gt; 0&lt;/math&gt;, Xiaoli rounded &lt;math&gt;x&lt;/math&gt; up by a small amount, rounded &lt;math&gt;y&lt;/math&gt; down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct? <br /> <br /> &lt;math&gt; \textbf{(A)} &lt;/math&gt; Her estimate is larger than &lt;math&gt;x-y&lt;/math&gt; <br /> &lt;math&gt; \textbf{(B)} &lt;/math&gt; Her estimate is smaller than &lt;math&gt;x-y&lt;/math&gt;<br /> &lt;math&gt; \textbf{(C)} &lt;/math&gt; Her estimate equals &lt;math&gt;x-y&lt;/math&gt;<br /> &lt;math&gt; \textbf{(D)} &lt;/math&gt; Her estimate equals &lt;math&gt;y - x&lt;/math&gt;<br /> &lt;math&gt; \textbf{(E)} &lt;/math&gt; Her estimate is &lt;math&gt;0&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> <br /> == Problem 8 ==<br /> <br /> What is the sum of all integer solutions to &lt;math&gt;1&lt;(x-2)^2&lt;25&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> How many ordered pairs of positive integers &lt;math&gt;(M,N)&lt;/math&gt; satisfy the equation &lt;math&gt;\frac {M}{6}&lt;/math&gt; = &lt;math&gt;\frac{6}{N}?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> Point B is due east of point A. Point C is due north of point B. The distance between points A and C is &lt;math&gt;10\sqrt 2&lt;/math&gt;, and &lt;math&gt;\angle BAC = 45^\circ&lt;/math&gt;. Point D is 20 meters due north of point C. The distance AD is between which two integers?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Two equilateral triangles are contained in square whose side length is &lt;math&gt;2\sqrt 3&lt;/math&gt;. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?<br /> <br /> &lt;math&gt;<br /> \text{(A) } \frac{3}{2}<br /> \qquad<br /> \text{(B) } \sqrt 3<br /> \qquad<br /> \text{(C) } 2\sqrt 2 - 1 <br /> \qquad<br /> \text{(D) } 8\sqrt 3 - 12<br /> \qquad<br /> \text{(E)} \frac{4\sqrt 3}{3} <br /> &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> <br /> <br /> ==Problem 15==<br /> In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end on the tournament?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?<br /> <br /> &lt;asy&gt;<br /> filldraw((0,0)--(2,0)--(1,sqrt(3))--cycle,gray,gray);<br /> filldraw(circle((1,sqrt(3)),1),gray);<br /> filldraw(circle((0,0),1),gray);<br /> filldraw(circle((2,0),1),grey);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10\pi+4\sqrt{3}\qquad\textbf{(B)}\ 13\pi-\sqrt{3}\qquad\textbf{(C)}\ 12\pi+\sqrt{3}\qquad\textbf{(D)}\ 10\pi+9\qquad\textbf{(E)}\ 13\pi&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{\sqrt{10}}{10}\qquad\textbf{(D)}\ \frac{\sqrt{5}}{6}\qquad\textbf{(E)}\ \frac{\sqrt{5}}{5}&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Suppose that one of every 500 people in a certain population has a particular disease, which displays no symptoms. A blood test is available for screening for this disease. For a person who has this disease, the test always turns out positive. For a person who does not have the disease, however, there is a &lt;math&gt;2\%&lt;/math&gt; false positive rate--in other words, for such people, &lt;math&gt;98\%&lt;/math&gt; of the time the test will turn out negative, but &lt;math&gt;2\%&lt;/math&gt; of the time the test will turn out positive and will incorrectly indicate that the person has the disease. Let &lt;math&gt;p&lt;/math&gt; be the probability that a person who is chosen at random from this population and gets a positive test result actually has the disease. Which of the following is closest to &lt;math&gt;p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{98}\qquad\textbf{(B)}\ \frac{1}{9}\qquad\textbf{(C)}\ \frac{1}{11}\qquad\textbf{(D)}\ \frac{49}{99}\qquad\textbf{(E)}\ \frac{98}{99}&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=6&lt;/math&gt;, &lt;math&gt;AD=30&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;. Segment &lt;math&gt;AB&lt;/math&gt; is extended 2 units beyond &lt;math&gt;B&lt;/math&gt; to point &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is the intersection of &lt;math&gt;\overline{ED}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. What is the area of &lt;math&gt;BFDG&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}&lt;/math&gt;<br /> <br /> <br /> [[2012 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Bernardo and Silvia play the following game. An integer between 0 and 999, inclusive, is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let &lt;math&gt;N&lt;/math&gt; be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Four distinct points are arranged on a plane so that the segments connecting them have lengths &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;2a&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt;. What is the ratio of &lt;math&gt;b&lt;/math&gt; to &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Let &lt;math&gt;(a_1,a_2, \dots ,a_{10})&lt;/math&gt; be a list of the first 10 positive integers such that for each &lt;math&gt;2 \le i \le 10&lt;/math&gt; either &lt;math&gt;a_i+1&lt;/math&gt; or &lt;math&gt;a_i-1&lt;/math&gt; or both appear somewhere before &lt;math&gt;a_i&lt;/math&gt; in the list. How many such lists are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> <br /> == Problem 23 ==<br /> <br /> A solid tetrahedron is sliced off a wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2} &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> A bug travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400&lt;/math&gt;<br /> <br /> [[2012 AMC 10B Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2012|ab=B|before=[[2012 AMC 10A Problems]]|after=[[2013 AMC 10A Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2012 AMC 10B]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_14&diff=70698 2015 AIME II Problems/Problem 14 2015-06-08T20:55:38Z <p>Thisisasentence: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers satisfying &lt;math&gt;x^4y^5+y^4x^5=810&lt;/math&gt; and &lt;math&gt;x^3y^6+y^3x^6=945&lt;/math&gt;. Evaluate &lt;math&gt;2x^3+(xy)^3+2y^3&lt;/math&gt;.<br /> <br /> ==Solution==<br /> The expression we want to find is &lt;math&gt;2(x^3+y^3) + x^3y^3&lt;/math&gt;.<br /> <br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x^3+y^3)=945&lt;/math&gt;, respectively. Dividing the latter by the former equation yields &lt;math&gt;\frac{x^2-xy+y^2}{xy} = \frac{945}{810}&lt;/math&gt;. Adding 3 to both sides and simplifying yields &lt;math&gt;\frac{(x+y)^2}{xy} = \frac{25}{6}&lt;/math&gt;. Solving for &lt;math&gt;x+y&lt;/math&gt; and substituting this expression into the first equation yields &lt;math&gt;\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810&lt;/math&gt;. Solving for &lt;math&gt;xy&lt;/math&gt;, we find that &lt;math&gt;xy = 3\sqrt{2}&lt;/math&gt;, so &lt;math&gt;x^3y^3 = 54&lt;/math&gt;. Substituting this into the second equation and solving for &lt;math&gt;x^3+y^3&lt;/math&gt; yields &lt;math&gt;x^3+y^3=\frac{35}{2}&lt;/math&gt;. So, the expression to evaluate is equal to &lt;math&gt;2 \times \frac{35}{2} + 54 = \boxed{089}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x+y)(x^2-xy+y^2)=945&lt;/math&gt;, respectively. By the first equation, &lt;math&gt;x+y=\frac{810}{x^4y^4}&lt;/math&gt;. Plugging this in to the second equation and simplifying yields &lt;math&gt;(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}&lt;/math&gt;. Now substitute &lt;math&gt;\frac{x}{y}=a&lt;/math&gt;. Solving the quadratic in &lt;math&gt;a&lt;/math&gt;, we get &lt;math&gt;a=\frac{x}{y}=\frac{2}{3}&lt;/math&gt; or &lt;math&gt;\frac{3}{2}&lt;/math&gt; As both of the original equations were symmetric in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, WLOG, let &lt;math&gt;\frac{x}{y}=\frac{2}{3}&lt;/math&gt;, so &lt;math&gt;x=\frac{2}{3}y&lt;/math&gt;. Now plugging this in to either one of the equations, we get the solutions &lt;math&gt;y=\frac{3(2^{\frac{2}{3}})}{2}&lt;/math&gt;, &lt;math&gt;x=2^{\frac{2}{3}}&lt;/math&gt;. Now plugging into what we want, we get &lt;math&gt;8+54+27=\boxed{089}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Add three times the first equation to the second equation and factor to get &lt;math&gt;(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375&lt;/math&gt;. Taking the cube root yields &lt;math&gt;xy(x+y)=15&lt;/math&gt;. Noting that the first equation is &lt;math&gt;(xy)^3\cdot(xy(x+y))=810&lt;/math&gt;, we find that &lt;math&gt;(xy)^3=\frac{810}{15}=54&lt;/math&gt;. Plugging this into the second equation and dividing yields &lt;math&gt;x^3+y^3 = \frac{945}{54} = \frac{35}{2}&lt;/math&gt;. Thus the sum required, as noted in Solution 1, is &lt;math&gt;54+\frac{35}{2}\cdot2 = \boxed{089}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_14&diff=70697 2015 AIME II Problems/Problem 14 2015-06-08T20:55:12Z <p>Thisisasentence: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers satisfying &lt;math&gt;x^4y^5+y^4x^5=810&lt;/math&gt; and &lt;math&gt;x^3y^6+y^3x^6=945&lt;/math&gt;. Evaluate &lt;math&gt;2x^3+(xy)^3+2y^3&lt;/math&gt;.<br /> <br /> ==Solution==<br /> The expression we want to find is &lt;math&gt;2(x^3+y^3) + x^3y^3&lt;/math&gt;.<br /> <br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x^3+y^3)=945&lt;/math&gt;, respectively. Dividing the latter by the former equation yields &lt;math&gt;\frac{x^2-xy+y^2}{xy} = \frac{945}{810}&lt;/math&gt;. Adding 3 to both sides and simplifying yields &lt;math&gt;\frac{(x+y)^2}{xy} = \frac{25}{6}&lt;/math&gt;. Solving for &lt;math&gt;x+y&lt;/math&gt; and substituting this expression into the first equation yields &lt;math&gt;\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810&lt;/math&gt;. Solving for &lt;math&gt;xy&lt;/math&gt;, we find that &lt;math&gt;xy = 3\sqrt{2}&lt;/math&gt;, so &lt;math&gt;x^3y^3 = 54&lt;/math&gt;. Substituting this into the second equation and solving for &lt;math&gt;x^3+y^3&lt;/math&gt; yields &lt;math&gt;x^3+y^3=\frac{35}{2}&lt;/math&gt;. So, the expression to evaluate is equal to &lt;math&gt;2 \times \frac{35}{2} + 54 = \boxed{089}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x+y)(x^2-xy+y^2)=945&lt;/math&gt;, respectively. By the first equation, &lt;math&gt;x+y=\frac{810}{x^4y^4}&lt;/math&gt;. Plugging this in to the second equation and simplifying yields &lt;math&gt;(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}&lt;/math&gt;. Now substitute &lt;math&gt;\frac{x}{y}=a&lt;/math&gt;. Solving the quadratic in &lt;math&gt;a&lt;/math&gt;, we get &lt;math&gt;a=\frac{x}{y}=\frac{2}{3}&lt;/math&gt; or &lt;math&gt;\frac{3}{2}&lt;/math&gt; As both of the original equations were symmetric in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, WLOG, let &lt;math&gt;\frac{x}{y}=\frac{2}{3}&lt;/math&gt;, so &lt;math&gt;x=\frac{2}{3}y&lt;/math&gt;. Now plugging this in to either one of the equations, we get the solutions &lt;math&gt;y=\frac{3(2^{\frac{2}{3}})}{2}&lt;/math&gt;, &lt;math&gt;x=2^{\frac{2}{3}}&lt;/math&gt;. Now plugging into what we want, we get &lt;math&gt;8+54+27=\boxed{089}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Add three times the first equation to the second equation and factor to get &lt;math&gt;(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375&lt;/math&gt;. Taking the cube root yields &lt;math&gt;xy(x+y)=15&lt;/math&gt;. Noting that the first equation is &lt;math&gt;(xy)^3\cdot(xy(x+y))=810&lt;/math&gt;, we find that &lt;math&gt;(xy)^3=\frac{810}{15}=54&lt;/math&gt;. Plugging this into the second equation and dividing yields &lt;math&gt;x^3+y^3 = \frac{945}{54} = \frac{35}{2}&lt;/math&gt;. Thus the sum required, as noted in Solution 1, is &lt;math&gt;54+\frac{35}{2} * 2 = \boxed{089}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Thisisasentence https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems/Problem_2&diff=69952 2001 IMO Problems/Problem 2 2015-04-17T02:07:14Z <p>Thisisasentence: /* Alternate Solution using Jensen's */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers. Prove that &lt;math&gt;\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution using Holder's ===<br /> By Holder's inequality,<br /> &lt;math&gt;\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum a(a^{2}+8bc)\right)\ge (a+b+c)^{3}&lt;/math&gt;<br /> Thus we need only show that<br /> &lt;math&gt;(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc&lt;/math&gt;<br /> Which is obviously true since &lt;math&gt;(a+b)(b+c)(c+a)\ge 8abc&lt;/math&gt;.<br /> <br /> === Alternate Solution using Jensen's ===<br /> This inequality is homogeneous so we can assume without loss of generality &lt;math&gt;a+b+c=1&lt;/math&gt; and apply Jensen's inequality for &lt;math&gt;f(x)=\frac{1}{\sqrt{x}}&lt;/math&gt;, so we get:<br /> &lt;cmath&gt;\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}&lt;/cmath&gt;<br /> but<br /> &lt;cmath&gt;1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc&lt;/cmath&gt; by AM-GM, and thus the inequality is proven.<br /> <br /> === Alternate Solution using Isolated Fudging ===<br /> We claim that <br /> &lt;cmath&gt;\frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}&lt;/cmath&gt;<br /> Cross-multiplying, squaring both sides and expanding, we have<br /> &lt;cmath&gt;a^{\frac{14}{3}}+a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq a^{\frac{14}{3}}+8a^{\frac{8}{3}}bc&lt;/cmath&gt;<br /> After cancelling the &lt;math&gt; a^{\frac{14}{3}}&lt;/math&gt; term, we apply AM-GM to RHS and obtain<br /> &lt;cmath&gt;a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq 8(a^{\frac{64}{3}}b^8c^8)^{\frac{1}{8}}=8a^{\frac{8}{3}}bc&lt;/cmath&gt;<br /> as desired, completing the proof of the claim.<br /> <br /> Similarly &lt;math&gt;\frac{b}{\sqrt{b^2+8ca}} \geq \frac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}&lt;/math&gt; and &lt;math&gt;\frac{c}{\sqrt{c^2+8ab}} \geq \frac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}&lt;/math&gt;.<br /> Summing the three inequalities, we obtain the original inequality.<br /> <br /> === Alternate Solution using Cauchy ===<br /> <br /> We want to prove &lt;cmath&gt;\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\ge 1&lt;/cmath&gt;<br /> <br /> Note that since this inequality is homogenous, assume &lt;math&gt;a+b+c=3&lt;/math&gt;.<br /> <br /> By Cauchy, &lt;cmath&gt;\left(\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\right)\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)\ge (a+b+c)^2=9&lt;/cmath&gt;<br /> <br /> Dividing both sides by &lt;math&gt;\sum_{cyc}a\sqrt{a^2+8bc}&lt;/math&gt;, we see that we want to prove &lt;cmath&gt;\dfrac{9}{\sum\limits_{cyc}a\sqrt{a^2+8bc}}\ge 1&lt;/cmath&gt; or equivalently &lt;cmath&gt;\sum\limits_{cyc}a\sqrt{a^2+8bc}\le 9&lt;/cmath&gt;<br /> <br /> Squaring both sides, we have &lt;cmath&gt;\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le 81&lt;/cmath&gt;<br /> <br /> Now use Cauchy again to obtain &lt;cmath&gt;\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le (a+b+c)\left(\sum_{cyc}a(a^2+8bc)\right)\le 81&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a+b+c=3&lt;/math&gt;, the inequality becomes &lt;cmath&gt;\sum_{cyc}a^3+8abc\le 27&lt;/cmath&gt; after some simplifying.<br /> <br /> But this equals &lt;cmath&gt;(a+b+c)^3-3\left(\sum_{sym}a^2b\right)+18abc\le 27&lt;/cmath&gt; and since &lt;math&gt;a+b+c=3&lt;/math&gt; we just want to prove &lt;cmath&gt;\left(\sum_{sym}a^2b\right)\ge 6abc&lt;/cmath&gt; after some simplifying.<br /> <br /> But that is true by AM-GM. Thus, proved. &lt;math&gt;\Box&lt;/math&gt;<br /> <br /> == See also ==<br /> {{IMO box|year=2001|num-b=1|num-a=3}}<br /> <br /> [[Category:Olympiad Algebra Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Thisisasentence