https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ayushk&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-30T22:11:28Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems&diff=83279 2017 AMC 12A Problems 2017-02-09T14:26:11Z <p>Ayushk: /* Problem 8 */</p> <hr /> <div>NOTE: AS OF NOW A WORK IN PROGRESS (Problems are not accurate/might not be formatted correctly)<br /> <br /> {{AMC12 Problems|year=2017|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> Pablo buys popsicles for his friends. The store sells single popsicles for \$1 each, 3-popsicle boxes for \$2, and 5-popsicle boxes for \$3. What is the greatest number of popsicles that Pablo can buy with \$8?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which one of these statements necessarily follows logically?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{ If Lewis did not receive an A, then he got all of the multiple choice questions wrong.} \\ \qquad\textbf{(B)}\ \text{ If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.} \\ \qquad\textbf{(C)}\ \text{ If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A.} \\ \qquad\textbf{(D)}\ \text{ If Lewis received an A, then he got all of the multiple choice questions right.} \\ \qquad\textbf{(E)}\ \text{ If Lewis received an A, then he got at least one of the multiple choice questions right.} &lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> At a gathering of &lt;math&gt;30&lt;/math&gt; people, there are &lt;math&gt;20&lt;/math&gt; people who all know each other and &lt;math&gt;10&lt;/math&gt; people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joy has &lt;math&gt;30&lt;/math&gt; thin rods, one each of every integer length from &lt;math&gt;1 \text{ cm}&lt;/math&gt; through &lt;math&gt;30 \text{ cm}&lt;/math&gt;. She places the rods with lengths &lt;math&gt;3 \text{ cm}&lt;/math&gt;, &lt;math&gt;7 \text{ cm}&lt;/math&gt;, and &lt;math&gt;15 \text{cm}&lt;/math&gt; on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 19 \qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Define a function on the positive integers recursively by &lt;math&gt;f(1) = 2&lt;/math&gt;, &lt;math&gt;f(n) = f(n-1) + 2&lt;/math&gt; if &lt;math&gt;n&lt;/math&gt; is even, and &lt;math&gt;f(n) = f(n-2) + 2&lt;/math&gt; if &lt;math&gt;n&lt;/math&gt; is odd and greater than &lt;math&gt;1&lt;/math&gt;. What is &lt;math&gt;f(2017)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> The region consisting of all points in three-dimensional space within &lt;math&gt;3&lt;/math&gt; units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume &lt;math&gt;216 \pi&lt;/math&gt;. What is the length &lt;math&gt;AB&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x+2&lt;/math&gt;, and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} &lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Chloé chooses a real number uniformly at random from the interval &lt;math&gt; [ 0,2017 ]&lt;/math&gt;. Independently, Laurent chooses a real number uniformly at random from the interval &lt;math&gt;[ 0 , 4034 ]&lt;/math&gt;. What is the probability that Laurent's number is greater than Chloe's number? <br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8} &lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Each of the &lt;math&gt;100&lt;/math&gt; students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are &lt;math&gt;42&lt;/math&gt; students who cannot sing, &lt;math&gt;65&lt;/math&gt; students who cannot dance, and &lt;math&gt;29&lt;/math&gt; students who cannot act. How many students have two of these talents?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 6&lt;/math&gt;, &lt;math&gt;BC = 7&lt;/math&gt;, and &lt;math&gt;CA = 8&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt; bisects &lt;math&gt;\angle BAC&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BE}&lt;/math&gt; bisects &lt;math&gt;\angle ABC&lt;/math&gt;. The bisectors intersect at &lt;math&gt;F&lt;/math&gt;. What is the ratio &lt;math&gt;AF&lt;/math&gt; : &lt;math&gt;FD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7)), F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C);<br /> draw(A--B--C--A--D^^B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,1.5*N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Each vertex of a cube is to be labeled with an integer from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> The graphs of &lt;math&gt;y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,&lt;/math&gt; and &lt;math&gt;y=\log_x \dfrac{1}{3}&lt;/math&gt; are plotted on the same set of axes. How many points in the plane with positive &lt;math&gt;x&lt;/math&gt;-coordinates lie on two or more of the graphs? <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 /> [[2017 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> For some positive integer &lt;math&gt;n,&lt;/math&gt; the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3.&lt;/math&gt; How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 110\qquad\textbf{(B)}\ 191\qquad\textbf{(C)}\ 261\qquad\textbf{(D)}\ 325\qquad\textbf{(E)}\ 425&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Jerry starts at &lt;math&gt;0&lt;/math&gt; on the real number line. He tosses a fair coin &lt;math&gt;8&lt;/math&gt; times. When he gets heads, he moves &lt;math&gt;1&lt;/math&gt; unit in the positive direction; when he gets tails, he moves &lt;math&gt;1&lt;/math&gt; unit in the negative direction. The probability that he reaches &lt;math&gt;4&lt;/math&gt; at some time during this process is &lt;math&gt;\frac{a}{b},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b?&lt;/math&gt; (For example, he succeeds if his sequence of tosses is &lt;math&gt;HTHHHHHH.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> A binary operation &lt;math&gt;\diamondsuit &lt;/math&gt; has the properties that &lt;math&gt;a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\ \diamondsuit\ a = 1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c.&lt;/math&gt; (Here the dot &lt;math&gt;\cdot&lt;/math&gt; represents the usual multiplication operation.) The solution to the equation &lt;math&gt;2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100&lt;/math&gt; can be written as &lt;math&gt;\frac{p}{q},&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p + q?&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}.&lt;/math&gt; Three of the sides of this quadrilateral have length &lt;math&gt;200.&lt;/math&gt; What is the length of its fourth side? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Three numbers in the interval &lt;math&gt;\left[0,1\right]&lt;/math&gt; are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1}{6}\qquad\textbf{(B)}\ \dfrac{1}{3}\qquad\textbf{(C)}\ \dfrac{1}{2}\qquad\textbf{(D)}\ \dfrac{2}{3}\qquad\textbf{(E)}\ \dfrac{5}{6}&lt;/math&gt;<br /> <br /> [[2017 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=AMC_12_Problems_and_Solutions&diff=82781 AMC 12 Problems and Solutions 2017-02-08T19:09:04Z <p>Ayushk: </p> <hr /> <div>[[AMC 12]] problems and solutions.<br /> <br /> {| class=&quot;wikitable&quot;<br /> |-<br /> ! Year || Test A || Test B<br /> |-<br /> | 2017 || [[2017 AMC 12A | AMC 12A]] || [[2017 AMC 12B | AMC 12B]]<br /> |-<br /> | 2016 || [[2016 AMC 12A | AMC 12A]] || [[2016 AMC 12B | AMC 12B]]<br /> |-<br /> | 2015 || [[2015 AMC 12A | AMC 12A]] || [[2015 AMC 12B | AMC 12B]]<br /> |-<br /> | 2014 || [[2014 AMC 12A | AMC 12A]] || [[2014 AMC 12B | AMC 12B]]<br /> |-<br /> | 2013 || [[2013 AMC 12A | AMC 12A]] || [[2013 AMC 12B | AMC 12B]]<br /> |-<br /> | 2012 || [[2012 AMC 12A | AMC 12A]] || [[2012 AMC 12B | AMC 12B]]<br /> |-<br /> | 2011 || [[2011 AMC 12A | AMC 12A]] || [[2011 AMC 12B | AMC 12B]]<br /> |-<br /> | 2010 || [[2010 AMC 12A | AMC 12A]] || [[2010 AMC 12B | AMC 12B]]<br /> |-<br /> | 2009 || [[2009 AMC 12A | AMC 12A]] || [[2009 AMC 12B | AMC 12B]]<br /> |-<br /> | 2008 || [[2008 AMC 12A | AMC 12A]] || [[2008 AMC 12B | AMC 12B]]<br /> |-<br /> | 2007 || [[2007 AMC 12A | AMC 12A]] || [[2007 AMC 12B | AMC 12B]]<br /> |-<br /> | 2006 || [[2006 AMC 12A | AMC 12A]] || [[2006 AMC 12B | AMC 12B]]<br /> |-<br /> | 2005 || [[2005 AMC 12A | AMC 12A]] || [[2005 AMC 12B | AMC 12B]]<br /> |-<br /> | 2004 || [[2004 AMC 12A | AMC 12A]] || [[2004 AMC 12B | AMC 12B]]<br /> |-<br /> | 2003 || [[2003 AMC 12A | AMC 12A]] || [[2003 AMC 12B | AMC 12B]]<br /> |-<br /> | 2002 || [[2002 AMC 12A | AMC 12A]] || [[2002 AMC 12B | AMC 12B]]<br /> |-<br /> | 2001 || [[2001 AMC 12 | AMC 12]] ||<br /> |-<br /> | 2000 || [[2000 AMC 12 | AMC 12]] ||<br /> |}<br /> <br /> == Resources ==<br /> * [[American Mathematics Competitions]]<br /> * [[AMC Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Math Contest Problems]]</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems&diff=82074 2006 AIME I Problems 2016-12-30T19:40:36Z <p>Ayushk: /* Problem 14 */</p> <hr /> <div>{{AIME Problems|year=2006|n=I}}<br /> <br /> == Problem 1 ==<br /> In quadrilateral &lt;math&gt; ABCD , \angle B &lt;/math&gt; is a right angle, diagonal &lt;math&gt; \overline{AC} &lt;/math&gt; is perpendicular to &lt;math&gt; \overline{CD}, AB=18, BC=21, &lt;/math&gt; and &lt;math&gt; CD=14. &lt;/math&gt; Find the perimeter of &lt;math&gt; ABCD. &lt;/math&gt;<br /> <br /> [[2006 AIME I Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let set &lt;math&gt; \mathcal{A} &lt;/math&gt; be a 90-element subset of &lt;math&gt; \{1,2,3,\ldots,100\}, &lt;/math&gt; and let &lt;math&gt; S &lt;/math&gt; be the sum of the elements of &lt;math&gt; \mathcal{A}. &lt;/math&gt; Find the number of possible values of &lt;math&gt; S. &lt;/math&gt;<br /> <br /> [[2006 AIME I Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is &lt;math&gt;1/29&lt;/math&gt; of the original integer.<br /> <br /> [[2006 AIME I Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> Let &lt;math&gt; N &lt;/math&gt; be the number of consecutive 0's at the right end of the decimal representation of the product &lt;math&gt; 1!2!3!4!\cdots99!100!. &lt;/math&gt; Find the remainder when &lt;math&gt; N &lt;/math&gt; is divided by 1000.<br /> <br /> [[2006 AIME I Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> The number &lt;math&gt; \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}&lt;/math&gt; can be written as &lt;math&gt; a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers. Find &lt;math&gt; a\cdot b\cdot c. &lt;/math&gt;<br /> <br /> [[2006 AIME I Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt; \mathcal{S} &lt;/math&gt; be the set of real numbers that can be represented as repeating decimals of the form &lt;math&gt; 0.\overline{abc} &lt;/math&gt; where &lt;math&gt; a, b, c &lt;/math&gt; are distinct digits. Find the sum of the elements of &lt;math&gt; \mathcal{S}. &lt;/math&gt;<br /> <br /> [[2006 AIME I Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region &lt;math&gt; \mathcal{C} &lt;/math&gt; to the area of shaded region &lt;math&gt; \mathcal{B} &lt;/math&gt; is 11/5. Find the ratio of shaded region &lt;math&gt; \mathcal{D} &lt;/math&gt; to the area of shaded region &lt;math&gt; \mathcal{A}. &lt;/math&gt;<br /> <br /> [[Image:2006AimeA7.PNG]]<br /> <br /> [[2006 AIME I Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Hexagon &lt;math&gt; ABCDEF &lt;/math&gt; is divided into five rhombuses, &lt;math&gt; \mathcal{P, Q, R, S,} &lt;/math&gt; and &lt;math&gt; \mathcal{T,} &lt;/math&gt; as shown. Rhombuses &lt;math&gt; \mathcal{P, Q, R,} &lt;/math&gt; and &lt;math&gt; \mathcal{S} &lt;/math&gt; are congruent, and each has area &lt;math&gt; \sqrt{2006}. &lt;/math&gt; Let &lt;math&gt; K &lt;/math&gt; be the area of rhombus &lt;math&gt; \mathcal{T}. &lt;/math&gt; Given that &lt;math&gt; K &lt;/math&gt; is a positive integer, find the number of possible values for &lt;math&gt; K. &lt;/math&gt; <br /> <br /> [[Image:2006AimeA8.PNG]]<br /> <br /> [[2006 AIME I Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> The sequence &lt;math&gt; a_1, a_2, \ldots &lt;/math&gt; is geometric with &lt;math&gt; a_1=a &lt;/math&gt; and common ratio &lt;math&gt; r, &lt;/math&gt; where &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; r &lt;/math&gt; are positive integers. Given that &lt;math&gt; \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, &lt;/math&gt; find the number of possible ordered pairs &lt;math&gt; (a,r). &lt;/math&gt;<br /> <br /> [[2006 AIME I Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose greatest common divisor is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> <br /> &lt;asy&gt;<br /> unitsize(0.50cm);<br /> draw((0,-1)--(0,6));<br /> draw((-1,0)--(6,0));<br /> draw(shift(1,1)*unitcircle);<br /> draw(shift(1,3)*unitcircle);<br /> draw(shift(1,5)*unitcircle);<br /> draw(shift(3,1)*unitcircle);<br /> draw(shift(3,3)*unitcircle);<br /> draw(shift(3,5)*unitcircle);<br /> draw(shift(5,1)*unitcircle);<br /> draw(shift(5,3)*unitcircle);<br /> &lt;/asy&gt;<br /> <br /> [[2006 AIME I Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> A collection of 8 cubes consists of one cube with edge-length &lt;math&gt; k &lt;/math&gt; for each integer &lt;math&gt; k, 1 \le k \le 8. &lt;/math&gt; A tower is to be built using all 8 cubes according to the rules: <br /> <br /> * Any cube may be the bottom cube in the tower.<br /> * The cube immediately on top of a cube with edge-length &lt;math&gt; k &lt;/math&gt; must have edge-length at most &lt;math&gt; k+2. &lt;/math&gt; <br /> <br /> Let &lt;math&gt; T &lt;/math&gt; be the number of different towers than can be constructed. What is the remainder when &lt;math&gt; T &lt;/math&gt; is divided by 1000?<br /> <br /> [[2006 AIME I Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Find the sum of the values of &lt;math&gt; x &lt;/math&gt; such that &lt;math&gt; \cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x, &lt;/math&gt; where &lt;math&gt; x &lt;/math&gt; is measured in degrees and &lt;math&gt; 100&lt; x&lt; 200. &lt;/math&gt;<br /> <br /> [[2006 AIME I Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> For each even positive integer &lt;math&gt; x, &lt;/math&gt; let &lt;math&gt; g(x) &lt;/math&gt; denote the greatest power of 2 that divides &lt;math&gt; x. &lt;/math&gt; For example, &lt;math&gt; g(20)=4 &lt;/math&gt; and &lt;math&gt; g(16)=16. &lt;/math&gt; For each positive integer &lt;math&gt; n, &lt;/math&gt; let &lt;math&gt; S_n=\sum_{k=1}^{2^{n-1}}g(2k). &lt;/math&gt; Find the greatest integer &lt;math&gt; n &lt;/math&gt; less than 1000 such that &lt;math&gt; S_n &lt;/math&gt; is a perfect square.<br /> <br /> [[2006 AIME I Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let &lt;math&gt; h &lt;/math&gt; be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then &lt;math&gt; h &lt;/math&gt; can be written in the form &lt;math&gt; \frac m{\sqrt{n}}, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are positive integers and &lt;math&gt; n &lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt; \lfloor m+\sqrt{n}\rfloor. &lt;/math&gt; (The notation &lt;math&gt; \lfloor x\rfloor &lt;/math&gt; denotes the greatest integer that is less than or equal to &lt;math&gt; x. &lt;/math&gt;)<br /> <br /> [[2006 AIME I Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Given that a sequence satisfies &lt;math&gt; x_0=0 &lt;/math&gt; and &lt;math&gt; |x_k|=|x_{k-1}+3| &lt;/math&gt; for all integers &lt;math&gt; k\ge 1, &lt;/math&gt; find the minimum possible value of &lt;math&gt; |x_1+x_2+\cdots+x_{2006}|. &lt;/math&gt;<br /> <br /> [[2006 AIME I Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=144 2006 AIME I Math Jam Transcript]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems&diff=82070 2005 AIME II Problems 2016-12-30T04:25:19Z <p>Ayushk: /* Problem 11 */</p> <hr /> <div>{{AIME Problems|year=2005|n=II}}<br /> <br /> == Problem 1 ==<br /> A game uses a deck of &lt;math&gt; n &lt;/math&gt; different cards, where &lt;math&gt; n &lt;/math&gt; is an integer and &lt;math&gt; n \geq 6. &lt;/math&gt; The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find &lt;math&gt; n. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is &lt;math&gt; \frac mn, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime integers, find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is &lt;math&gt; \frac mn &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime integers. Find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> Find the number of positive integers that are divisors of at least one of &lt;math&gt; 10^{10},15^7,18^{11}. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Determine the number of ordered pairs &lt;math&gt; (a,b) &lt;/math&gt; of integers such that &lt;math&gt; \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, &lt;/math&gt; and &lt;math&gt; 2 \leq b \leq 2005. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> The cards in a stack of &lt;math&gt; 2n &lt;/math&gt; cards are numbered consecutively from 1 through &lt;math&gt; 2n &lt;/math&gt; from top to bottom. The top &lt;math&gt; n &lt;/math&gt; cards are removed, kept in order, and form pile &lt;math&gt; A. &lt;/math&gt; The remaining cards form pile &lt;math&gt; B. &lt;/math&gt; The cards are then restacked by taking cards alternately from the tops of pile &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; A, &lt;/math&gt; respectively. In this process, card number &lt;math&gt; (n+1) &lt;/math&gt; becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles &lt;math&gt; A &lt;/math&gt; and &lt;math&gt; B &lt;/math&gt; are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. Find the number of cards in the magical stack in which card number 131 retains its original position.<br /> <br /> [[2005 AIME II Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Let &lt;math&gt; x=\frac{4}{(\sqrt{5}+1)(\sqrt{5}+1)(\sqrt{5}+1)(\sqrt{5}+1)}. &lt;/math&gt; Find &lt;math&gt; (x+1)^{48}. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Circles &lt;math&gt; C_1 &lt;/math&gt; and &lt;math&gt; C_2 &lt;/math&gt; are externally tangent, and they are both internally tangent to circle &lt;math&gt; C_3. &lt;/math&gt; The radii of &lt;math&gt; C_1 &lt;/math&gt; and &lt;math&gt; C_2 &lt;/math&gt; are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of &lt;math&gt; C_3 &lt;/math&gt; is also a common external tangent of &lt;math&gt; C_1 &lt;/math&gt; and &lt;math&gt; C_2. &lt;/math&gt; Given that the length of the chord is &lt;math&gt; \frac{m\sqrt{n}}p &lt;/math&gt; where &lt;math&gt; m,n, &lt;/math&gt; and &lt;math&gt; p &lt;/math&gt; are positive integers, &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; p &lt;/math&gt; are relatively prime, and &lt;math&gt; n &lt;/math&gt; is not divisible by the square of any prime, find &lt;math&gt; m+n+p. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> For how many positive integers &lt;math&gt; n &lt;/math&gt; less than or equal to 1000 is &lt;math&gt; (\sin t + i \cos t)^n = \sin nt + i \cos nt &lt;/math&gt; true for all real &lt;math&gt; t &lt;/math&gt;?<br /> <br /> [[2005 AIME II Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Given that &lt;math&gt; O &lt;/math&gt; is a regular octahedron, that &lt;math&gt; C &lt;/math&gt; is the cube whose vertices are the centers of the faces of &lt;math&gt; O, &lt;/math&gt; and that the ratio of the volume of &lt;math&gt; O &lt;/math&gt; to that of &lt;math&gt; C &lt;/math&gt; is &lt;math&gt; \frac mn, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime integers, find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Let &lt;math&gt; m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt; a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt; m. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has center &lt;math&gt; O, AB=900, E &lt;/math&gt; and &lt;math&gt; F &lt;/math&gt; are on &lt;math&gt; AB &lt;/math&gt; with &lt;math&gt; AE&lt;BF &lt;/math&gt; and &lt;math&gt; E &lt;/math&gt; between &lt;math&gt; A &lt;/math&gt; and &lt;math&gt; F, m\angle EOF =45^\circ, &lt;/math&gt; and &lt;math&gt; EF=400. &lt;/math&gt; Given that &lt;math&gt; BF=p+q\sqrt{r}, &lt;/math&gt; where &lt;math&gt; p,q, &lt;/math&gt; and &lt;math&gt; r &lt;/math&gt; are positive integers and &lt;math&gt; r &lt;/math&gt; is not divisible by the square of any prime, find &lt;math&gt; p+q+r. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Let &lt;math&gt; P(x) &lt;/math&gt; be a polynomial with integer coefficients that satisfies &lt;math&gt; P(17)=10 &lt;/math&gt; and &lt;math&gt; P(24)=17. &lt;/math&gt; Given that &lt;math&gt; P(n)=n+3 &lt;/math&gt; has two distinct integer solutions &lt;math&gt; n_1 &lt;/math&gt; and &lt;math&gt; n_2, &lt;/math&gt; find the product &lt;math&gt; n_1\cdot n_2. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> In triangle &lt;math&gt; ABC, AB=13, BC=15, &lt;/math&gt; and &lt;math&gt;CA = 14. &lt;/math&gt; Point &lt;math&gt; D &lt;/math&gt; is on &lt;math&gt; \overline{BC} &lt;/math&gt; with &lt;math&gt; CD=6. &lt;/math&gt; Point &lt;math&gt; E &lt;/math&gt; is on &lt;math&gt; \overline{BC} &lt;/math&gt; such that &lt;math&gt; \angle BAE\cong \angle CAD. &lt;/math&gt; Given that &lt;math&gt; BE=\frac pq &lt;/math&gt; where &lt;math&gt; p &lt;/math&gt; and &lt;math&gt; q &lt;/math&gt; are relatively prime positive integers, find &lt;math&gt; q. &lt;/math&gt;<br /> <br /> [[2005 AIME II Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Let &lt;math&gt; w_1 &lt;/math&gt; and &lt;math&gt; w_2 &lt;/math&gt; denote the circles &lt;math&gt; x^2+y^2+10x-24y-87=0 &lt;/math&gt; and &lt;math&gt; x^2 +y^2-10x-24y+153=0, &lt;/math&gt; respectively. Let &lt;math&gt; m &lt;/math&gt; be the smallest positive value of &lt;math&gt; a &lt;/math&gt; for which the line &lt;math&gt; y=ax &lt;/math&gt; contains the center of a circle that is externally tangent to &lt;math&gt; w_2 &lt;/math&gt; and internally tangent to &lt;math&gt; w_1. &lt;/math&gt; Given that &lt;math&gt; m^2=\frac pq, &lt;/math&gt; where &lt;math&gt; p &lt;/math&gt; and &lt;math&gt; q &lt;/math&gt; are relatively prime integers, find &lt;math&gt; p+q. &lt;/math&gt; <br /> <br /> [[2005 AIME II Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=51 2005 AIME II Math Jam Transcript]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_11&diff=81763 2000 AIME I Problems/Problem 11 2016-12-04T01:46:26Z <p>Ayushk: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the sum of all numbers of the form &lt;math&gt;a/b,&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are [[relatively prime]] positive [[divisor]]s of &lt;math&gt;1000.&lt;/math&gt; What is the [[floor function|greatest integer]] that does not exceed &lt;math&gt;S/10&lt;/math&gt;?<br /> <br /> == Solution 1 ==<br /> Since all divisors of &lt;math&gt;1000 = 2^35^3&lt;/math&gt; can be written in the form of &lt;math&gt;2^{m}5^{n}&lt;/math&gt;, it follows that &lt;math&gt;\frac{a}{b}&lt;/math&gt; can also be expressed in the form of &lt;math&gt;2^{x}5^{y}&lt;/math&gt;, where &lt;math&gt;-3 \le x,y \le 3&lt;/math&gt;. Thus every number in the form of &lt;math&gt;a/b&lt;/math&gt; will be expressed one time in the product<br /> <br /> &lt;cmath&gt;(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)&lt;/cmath&gt;<br /> <br /> Using the formula for a [[geometric series]], this reduces to &lt;math&gt;S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}&lt;/math&gt;, and &lt;math&gt;\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Essentially, the problem asks us to compute &lt;cmath&gt;\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}&lt;/cmath&gt; which is pretty easy: &lt;cmath&gt;\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b} = \sum_{a=-3}^3 2^a \sum_{b=-3}^3 \frac{1}{5^b} = \sum_{a=-3}^3 2^a 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) \sum_{a=-3}^3 2^a = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg)2^{-3} \bigg( \frac{1-2^7}{1-2} \bigg) = 2480 + \frac{437}{1000}&lt;/cmath&gt; so our answer is &lt;math&gt;\left\lfloor \frac{2480 + \frac{437}{1000}}{10} \right\rfloor = \boxed{248}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=80231 1988 AIME Problems/Problem 9 2016-09-09T05:42:34Z <p>Ayushk: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the smallest positive integer whose [[perfect cube|cube]] ends in &lt;math&gt;888&lt;/math&gt;.<br /> <br /> == Solution ==<br /> A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of &lt;math&gt;(10k + 2)^3&lt;/math&gt;; using the [[binomial theorem]] gives us &lt;math&gt;1000k^3 + 600k^2 + 120k + 8&lt;/math&gt;. Since we are looking for the tens digit, &lt;math&gt;\mod{100}&lt;/math&gt; we get &lt;math&gt;20k + 8 \equiv 88 \pmod{100}&lt;/math&gt;. This is true if the tens digit is either &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. Casework:<br /> *&lt;math&gt;4&lt;/math&gt;: Then our cube must be in the form of &lt;math&gt;(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}&lt;/math&gt;. Hence the lowest possible value for the hundreds digit is &lt;math&gt;4&lt;/math&gt;, and so &lt;math&gt;442&lt;/math&gt; is a valid solution. <br /> *&lt;math&gt;9&lt;/math&gt;: Then our cube is &lt;math&gt;(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}&lt;/math&gt;. The lowest possible value for the hundreds digit is &lt;math&gt;1&lt;/math&gt;, and we get &lt;math&gt;192&lt;/math&gt;. Hence, since &lt;math&gt;192 &lt; 442&lt;/math&gt;, the answer is &lt;math&gt;\fbox{192}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_15&diff=80096 2005 AIME I Problems/Problem 15 2016-08-28T18:58:47Z <p>Ayushk: /* Solution */</p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt; ABC &lt;/math&gt; has &lt;math&gt; BC=20. &lt;/math&gt; The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] &lt;math&gt; AD. &lt;/math&gt; If the area of the triangle is &lt;math&gt; m \sqrt{n} &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are integers and &lt;math&gt; n &lt;/math&gt; is not [[divisor | divisible]] by the [[perfect square | square]] of a prime, find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10);<br /> pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C);<br /> path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir);<br /> D(MP(&quot;A&quot;,A,s)--MP(&quot;B&quot;,B,s)--MP(&quot;C&quot;,C,N,s)--cycle); D(cir); <br /> D(A--MP(&quot;D&quot;,D,NE,s)); D(MP(&quot;E&quot;,E1,NE,s)); D(MP(&quot;F&quot;,F,NW,s)); D(MP(&quot;G&quot;,G,s)); D(MP(&quot;P&quot;,P,SW,s)); D(MP(&quot;Q&quot;,Q,SE,s));<br /> MP(&quot;10&quot;,(B+D)/2,NE); MP(&quot;10&quot;,(C+D)/2,NE);<br /> &lt;/asy&gt;&lt;/center&gt;&lt;!-- Asymptote replacement for Image:2005_I_AIME-15.png by azjps --&gt;<br /> <br /> Let &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; be the points of tangency of the incircle with &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Without loss of generality, let &lt;math&gt;AC &lt; AB&lt;/math&gt;, so that &lt;math&gt;E&lt;/math&gt; is between &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. Let the length of the median be &lt;math&gt;3m&lt;/math&gt;. Then by two applications of the [[Power of a Point Theorem]], &lt;math&gt;DE^2 = 2m \cdot m = AF^2&lt;/math&gt;, so &lt;math&gt;DE = AF&lt;/math&gt;. Now, &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt; are two tangents to a circle from the same point, so &lt;math&gt;CE = CF = c&lt;/math&gt; and thus &lt;math&gt;AC = AF + CF = DE + CE = CD = 10&lt;/math&gt;. Then &lt;math&gt;DE = AF = AG = 10 - c&lt;/math&gt; so &lt;math&gt;BG = BE = BD + DE = 20 - c&lt;/math&gt; and thus &lt;math&gt;AB = AG + BG = 30 - 2c&lt;/math&gt;.<br /> <br /> Now, by [[Stewart's Theorem]] in triangle &lt;math&gt;\triangle ABC&lt;/math&gt; with [[cevian]] &lt;math&gt;\overline{AD}&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;(3m)^2\cdot 20 + 20\cdot10\cdot10 = 1^2\cdot0 + (30 - 2c)^2\cdot 10.&lt;/cmath&gt;<br /> <br /> Our earlier result from Power of a Point was that &lt;math&gt;2m^2 = (10 - c)^2&lt;/math&gt;, so we combine these two results to solve for &lt;math&gt;c&lt;/math&gt; and we get<br /> <br /> &lt;cmath&gt;(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;c = 2&lt;/math&gt; or &lt;math&gt; = 10&lt;/math&gt;. We discard the value &lt;math&gt;c = 10&lt;/math&gt; as extraneous (it gives us an [[equilateral triangle]]) and are left with &lt;math&gt;c = 2&lt;/math&gt;, so our triangle has area &lt;math&gt;\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}&lt;/math&gt; and so the answer is &lt;math&gt;24 + 14 = \boxed{038}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_10&diff=80095 2005 AIME I Problems/Problem 10 2016-08-28T18:52:26Z <p>Ayushk: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> [[Triangle]] &lt;math&gt; ABC &lt;/math&gt; lies in the [[cartesian plane]] and has an [[area]] of &lt;math&gt;70&lt;/math&gt;. The coordinates of &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; C &lt;/math&gt; are &lt;math&gt; (12,19) &lt;/math&gt; and &lt;math&gt; (23,20), &lt;/math&gt; respectively, and the coordinates of &lt;math&gt; A &lt;/math&gt; are &lt;math&gt; (p,q). &lt;/math&gt; The [[line]] containing the [[median of a triangle | median]] to side &lt;math&gt; BC &lt;/math&gt; has [[slope]] &lt;math&gt; -5. &lt;/math&gt; Find the largest possible value of &lt;math&gt; p+q. &lt;/math&gt;<br /> <br /> __TOC__<br /> &lt;center&gt;&lt;asy&gt;defaultpen(fontsize(8));<br /> size(170);<br /> pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22);<br /> draw(A--B--C--A);draw(A--M);draw(B--P--C);<br /> label(&quot;A (p,q)&quot;,A,(1,1));label(&quot;B (12,19)&quot;,B,(-1,-1));label(&quot;C (23,20)&quot;,C,(1,-1));label(&quot;M&quot;,M,(0.2,-1));<br /> label(&quot;(17,22)&quot;,P,(1,1));<br /> dot(A^^B^^C^^M^^P);&lt;/asy&gt;&lt;/center&gt;<br /> <br /> === Solution 1 ===<br /> The [[midpoint]] &lt;math&gt;M&lt;/math&gt; of [[line segment]] &lt;math&gt;\overline{BC}&lt;/math&gt; is &lt;math&gt;\left(\frac{35}{2}, \frac{39}{2}\right)&lt;/math&gt;. The equation of the median can be found by &lt;math&gt;-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}&lt;/math&gt;. Cross multiply and simplify to yield that &lt;math&gt;-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}&lt;/math&gt;, so &lt;math&gt;q = -5p + 107&lt;/math&gt;.<br /> <br /> Use [[determinant]]s to find that the [[area]] of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{1}{2} \begin{vmatrix}p &amp; 12 &amp; 23 \\ q &amp; 19 &amp; 20 \\ 1 &amp; 1 &amp; 1\end{vmatrix} = 70&lt;/math&gt; (note that there is a missing [[absolute value]]; we will assume that the other solution for the triangle will give a smaller value of &lt;math&gt;p+q&lt;/math&gt;, which is provable by following these steps over again). We can calculate this determinant to become &lt;math&gt;140 = \begin{vmatrix} 12 &amp; 23 \\ 19 &amp; 20 \end{vmatrix} - \begin{vmatrix} p &amp; q \\ 23 &amp; 20 \end{vmatrix} + \begin{vmatrix} p &amp; q \\ 12 &amp; 19 \end{vmatrix}&lt;/math&gt; &lt;math&gt;\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q&lt;/math&gt; &lt;math&gt;= -197 - p + 11q&lt;/math&gt;. Thus, &lt;math&gt;q = \frac{1}{11}p - \frac{337}{11}&lt;/math&gt;.<br /> <br /> Setting this equation equal to the equation of the median, we get that &lt;math&gt;\frac{1}{11}p - \frac{337}{11} = -5p + 107&lt;/math&gt;, so &lt;math&gt;\frac{56}{11}p = \frac{107 \cdot 11 + 337}{11}&lt;/math&gt;. Solving produces that &lt;math&gt;p = 15&lt;/math&gt;. [[Substitution|Substituting]] backwards yields that &lt;math&gt;q = 32&lt;/math&gt;; the solution is &lt;math&gt;p + q = \boxed{047}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Using the equation of the median from above, we can write the [[coordinate]]s of &lt;math&gt;A&lt;/math&gt; as &lt;math&gt;(p,\ -5p + 107)&lt;/math&gt;. The equation of &lt;math&gt;\overline{BC}&lt;/math&gt; is &lt;math&gt;\frac{20 - 19}{23 - 12} = \frac{y - 19}{x - 12}&lt;/math&gt;, so &lt;math&gt;x - 12 = 11y - 209&lt;/math&gt;. In [[general form]], the line is &lt;math&gt;x - 11y + 197 = 0&lt;/math&gt;. Use the equation for the distance between a line and point to find the distance between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; (which is the height of &lt;math&gt;\triangle ABC&lt;/math&gt;): &lt;math&gt;\frac{|1(p) - 11(-5p + 107) + 197|}{1^2 + 11^2} = \frac{|56p - 980|}{\sqrt{122}}&lt;/math&gt;. Now we need the length of &lt;math&gt;BC&lt;/math&gt;, which is &lt;math&gt;\sqrt{(23 - 12)^2 + (20 - 19)^2} = \sqrt{122}&lt;/math&gt;. The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;70 = \frac{1}{2}bh = \frac{1}{2}\left(\frac{|56p - 980|}{\sqrt{122}}\right) \cdot \sqrt{122}&lt;/math&gt;. Thus, &lt;math&gt;|28p - 490| = 70&lt;/math&gt;, and &lt;math&gt;p = 15,\ 20&lt;/math&gt;. We are looking for &lt;math&gt;p + q = -4p + 107 = 47,\ 27&lt;/math&gt;. The maximum possible value of &lt;math&gt;p + q = \fbox{047}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Again, the [[midpoint]] &lt;math&gt;M&lt;/math&gt; of [[line segment]] &lt;math&gt;\overline{BC}&lt;/math&gt; is at &lt;math&gt;\left(\frac{35}{2}, \frac{39}{2}\right)&lt;/math&gt;. Let &lt;math&gt;A'&lt;/math&gt; be the point &lt;math&gt;(17, 22)&lt;/math&gt;, which lies along the line through &lt;math&gt;M&lt;/math&gt; of slope &lt;math&gt;-5&lt;/math&gt;. The area of triangle &lt;math&gt;A'BC&lt;/math&gt; can be computed in a number of ways (one possibility: extend &lt;math&gt;A'B&lt;/math&gt; until it hits the line &lt;math&gt;y = 19&lt;/math&gt;, and subtract one triangle from another), and each such calculation gives an area of 14. This is &lt;math&gt;\frac{1}{5}&lt;/math&gt; of our needed area, so we simply need the point &lt;math&gt;A&lt;/math&gt; to be 5 times as far from &lt;math&gt;M&lt;/math&gt; as &lt;math&gt;A'&lt;/math&gt; is. Thus &lt;math&gt;A = \left(\frac{35}{2}, \frac{39}{2}\right) \pm 5\left(-\frac{1}{2}, \frac{5}{2}\right)&lt;/math&gt;, and the sum of coordinates will be larger if we take the positive value, so &lt;math&gt;A = \left(\frac{35}{2} - \frac{5}2, \frac{39}{2} + \frac{25}{2}\right)&lt;/math&gt; and the answer is &lt;math&gt;\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = \fbox{047}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_10&diff=80094 2005 AIME I Problems/Problem 10 2016-08-28T18:52:04Z <p>Ayushk: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> [[Triangle]] &lt;math&gt; ABC &lt;/math&gt; lies in the [[cartesian plane]] and has an [[area]] of &lt;math&gt;70&lt;/math&gt;. The coordinates of &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; C &lt;/math&gt; are &lt;math&gt; (12,19) &lt;/math&gt; and &lt;math&gt; (23,20), &lt;/math&gt; respectively, and the coordinates of &lt;math&gt; A &lt;/math&gt; are &lt;math&gt; (p,q). &lt;/math&gt; The [[line]] containing the [[median of a triangle | median]] to side &lt;math&gt; BC &lt;/math&gt; has [[slope]] &lt;math&gt; -5. &lt;/math&gt; Find the largest possible value of &lt;math&gt; p+q. &lt;/math&gt;<br /> <br /> __TOC__<br /> &lt;center&gt;&lt;asy&gt;defaultpen(fontsize(8));<br /> size(170);<br /> pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22);<br /> draw(A--B--C--A);draw(A--M);draw(B--P--C);<br /> label(&quot;A (p,q)&quot;,A,(1,1));label(&quot;B (12,19)&quot;,B,(-1,-1));label(&quot;C (23,20)&quot;,C,(1,-1));label(&quot;M&quot;,M,(0.2,-1));<br /> label(&quot;(17,22)&quot;,P,(1,1));<br /> dot(A^^B^^C^^M^^P);&lt;/asy&gt;&lt;/center&gt;<br /> <br /> === Solution 1 ===<br /> The [[midpoint]] &lt;math&gt;M&lt;/math&gt; of [[line segment]] &lt;math&gt;\overline{BC}&lt;/math&gt; is &lt;math&gt;\left(\frac{35}{2}, \frac{39}{2}\right)&lt;/math&gt;. The equation of the median can be found by &lt;math&gt;-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}&lt;/math&gt;. Cross multiply and simplify to yield that &lt;math&gt;-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}&lt;/math&gt;, so &lt;math&gt;q = -5p + 107&lt;/math&gt;.<br /> <br /> Use [[determinant]]s to find that the [[area]] of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{1}{2} \begin{vmatrix}p &amp; 12 &amp; 23 \\ q &amp; 19 &amp; 20 \\ 1 &amp; 1 &amp; 1\end{vmatrix} = 70&lt;/math&gt; (note that there is a missing [[absolute value]]; we will assume that the other solution for the triangle will give a smaller value of &lt;math&gt;p+q&lt;/math&gt;, which is provable by following these steps over again). We can calculate this determinant to become &lt;math&gt;140 = \begin{vmatrix} 12 &amp; 23 \\ 19 &amp; 20 \end{vmatrix} - \begin{vmatrix} p &amp; q \\ 23 &amp; 20 \end{vmatrix} + \begin{vmatrix} p &amp; q \\ 12 &amp; 19 \end{vmatrix}&lt;/math&gt; &lt;math&gt;\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q&lt;/math&gt; &lt;math&gt;= -197 - p + 11q&lt;/math&gt;. Thus, &lt;math&gt;q = \frac{1}{11}p - \frac{337}{11}&lt;/math&gt;.<br /> <br /> Setting this equation equal to the equation of the median, we get that &lt;math&gt;\frac{1}{11}p - \frac{337}{11} = -5p + 107&lt;/math&gt;, so &lt;math&gt;\frac{56}{11}p = \frac{107 \cdot 11 + 337}{11}&lt;/math&gt;. Solving produces that &lt;math&gt;p = 15&lt;/math&gt;. [[Substitution|Substituting]] backwards yields that &lt;math&gt;q = 32&lt;/math&gt;; the solution is &lt;math&gt;p + q = \boxed{047}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Using the equation of the median from above, we can write the [[coordinate]]s of &lt;math&gt;A&lt;/math&gt; as &lt;math&gt;(p,\ -5p + 107)&lt;/math&gt;. The equation of &lt;math&gt;\overline{BC}&lt;/math&gt; is &lt;math&gt;\frac{20 - 19}{23 - 12} = \frac{y - 19}{x - 12}&lt;/math&gt;, so &lt;math&gt;x - 12 = 11y - 209&lt;/math&gt;. In [[general form]], the line is &lt;math&gt;x - 11y + 197 = 0&lt;/math&gt;. Use the equation for the distance between a line and point to find the distance between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; (which is the height of &lt;math&gt;\triangle ABC&lt;/math&gt;): &lt;math&gt;\frac{|1(p) - 11(-5p + 107) + 197|}{1^2 + 11^2} = \frac{|56p - 980|}{\sqrt{122}}&lt;/math&gt;. Now we need the length of &lt;math&gt;BC&lt;/math&gt;, which is &lt;math&gt;\sqrt{(23 - 12)^2 + (20 - 19)^2} = \sqrt{122}&lt;/math&gt;. The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;70 = \frac{1}{2}bh = \frac{1}{2}\left(\frac{|56p - 980|}{\sqrt{122}}\right) \cdot \sqrt{122}&lt;/math&gt;. Thus, &lt;math&gt;|28p - 490| = 70&lt;/math&gt;, and &lt;math&gt;p = 15,\ 20&lt;/math&gt;. We are looking for &lt;math&gt;p + q = -4p + 107 = 47,\ 27&lt;/math&gt;. The maximum possible value of &lt;math&gt;p + q = 47&lt;/math&gt;. <br /> <br /> === Solution 3 ===<br /> Again, the [[midpoint]] &lt;math&gt;M&lt;/math&gt; of [[line segment]] &lt;math&gt;\overline{BC}&lt;/math&gt; is at &lt;math&gt;\left(\frac{35}{2}, \frac{39}{2}\right)&lt;/math&gt;. Let &lt;math&gt;A'&lt;/math&gt; be the point &lt;math&gt;(17, 22)&lt;/math&gt;, which lies along the line through &lt;math&gt;M&lt;/math&gt; of slope &lt;math&gt;-5&lt;/math&gt;. The area of triangle &lt;math&gt;A'BC&lt;/math&gt; can be computed in a number of ways (one possibility: extend &lt;math&gt;A'B&lt;/math&gt; until it hits the line &lt;math&gt;y = 19&lt;/math&gt;, and subtract one triangle from another), and each such calculation gives an area of 14. This is &lt;math&gt;\frac{1}{5}&lt;/math&gt; of our needed area, so we simply need the point &lt;math&gt;A&lt;/math&gt; to be 5 times as far from &lt;math&gt;M&lt;/math&gt; as &lt;math&gt;A'&lt;/math&gt; is. Thus &lt;math&gt;A = \left(\frac{35}{2}, \frac{39}{2}\right) \pm 5\left(-\frac{1}{2}, \frac{5}{2}\right)&lt;/math&gt;, and the sum of coordinates will be larger if we take the positive value, so &lt;math&gt;A = \left(\frac{35}{2} - \frac{5}2, \frac{39}{2} + \frac{25}{2}\right)&lt;/math&gt; and the answer is &lt;math&gt;\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = \fbox{047}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ayushk https://artofproblemsolving.com/wiki/index.php?title=Rational_function&diff=79873 Rational function 2016-08-06T05:46:43Z <p>Ayushk: Created page with &quot;A rational function &lt;math&gt;R(x)&lt;/math&gt; is defined as any function that can be written as &lt;math&gt;\frac{P(x)}{Q(x)}&lt;/math&gt; for polynomials &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;.&quot;</p> <hr /> <div>A rational function &lt;math&gt;R(x)&lt;/math&gt; is defined as any function that can be written as &lt;math&gt;\frac{P(x)}{Q(x)}&lt;/math&gt; for polynomials &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;.</div> Ayushk