https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Interest&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-05T15:41:44Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems&diff=83247 2017 AMC 12A Problems 2017-02-09T05:20:48Z <p>Interest: /* Problem 3 */ spelling</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)}\ 14&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> Interest https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_22&diff=83245 2017 AMC 12A Problems/Problem 22 2017-02-09T05:10:13Z <p>Interest: /* Solution */ changed percentages to fractions</p> <hr /> <div>==Problem==<br /> <br /> A square is drawn in the Cartesian coordinate plane with vertices at &lt;math&gt;(2, 2)&lt;/math&gt;, &lt;math&gt;(-2, 2)&lt;/math&gt;, &lt;math&gt;(-2, -2)&lt;/math&gt;, &lt;math&gt;(2, -2)&lt;/math&gt;. A particle starts at &lt;math&gt;(0,0)&lt;/math&gt;. Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is &lt;math&gt;1/8&lt;/math&gt; that the particle will move from &lt;math&gt;(x, y)&lt;/math&gt; to each of &lt;math&gt;(x, y + 1)&lt;/math&gt;, &lt;math&gt;(x + 1, y + 1)&lt;/math&gt;, &lt;math&gt;(x + 1, y)&lt;/math&gt;, &lt;math&gt;(x + 1, y - 1)&lt;/math&gt;, &lt;math&gt;(x, y - 1)&lt;/math&gt;, &lt;math&gt;(x - 1, y - 1)&lt;/math&gt;, &lt;math&gt;(x - 1, y)&lt;/math&gt;, or &lt;math&gt;(x - 1, y + 1)&lt;/math&gt;. The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is &lt;math&gt;m/n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m + n&lt;/math&gt;?<br /> <br /> ==Solution==<br /> <br /> We let &lt;math&gt;c, e,&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; be the probability of reaching a corner before an edge when starting at an &quot;inside corner&quot; (e.g. &lt;math&gt;(1, 1)&lt;/math&gt;), an &quot;inside edge&quot; (e.g. &lt;math&gt;(1, 0)&lt;/math&gt;), and the middle respectively.<br /> <br /> Starting in the middle, there is a &lt;math&gt;\frac{4}{8}&lt;/math&gt; chance of moving to an inside edge and a &lt;math&gt;\frac{4}{8}&lt;/math&gt; chance of moving to an inside corner, so<br /> <br /> &lt;cmath&gt;m = \frac{1}{2}e + \frac{1}{2}c.&lt;/cmath&gt;<br /> <br /> Starting at an inside edge, there is a &lt;math&gt;\frac{2}{8}&lt;/math&gt; chance of moving to another inside edge, a &lt;math&gt;\frac{2}{8}&lt;/math&gt; chance of moving to an inside corner, a &lt;math&gt;\frac{1}{8}&lt;/math&gt; chance of moving into the middle, and a &lt;math&gt;\frac{3}{8}&lt;/math&gt; chance of reaching an outside edge and stopping. Therefore,<br /> <br /> &lt;cmath&gt;e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}*0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.&lt;/cmath&gt;<br /> <br /> Starting at an inside corner, there is a &lt;math&gt;\frac{2}{8}&lt;/math&gt; chance of moving to an inside edge, a &lt;math&gt;\frac{1}{8}&lt;/math&gt; chance of moving into the middle, a &lt;math&gt;\frac{4}{8}&lt;/math&gt; chance of moving to an outside edge and stopping, and finally a &lt;math&gt;\frac{1}{8}&lt;/math&gt; chance of reaching that elusive outside corner. This gives<br /> <br /> &lt;cmath&gt;c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}*1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.&lt;/cmath&gt;<br /> <br /> Solving this system of equations gives<br /> <br /> &lt;cmath&gt;m = \frac{4}{35},&lt;/cmath&gt;<br /> &lt;cmath&gt;e = \frac{1}{14},&lt;/cmath&gt;<br /> &lt;cmath&gt;c = \frac{11}{70}.&lt;/cmath&gt;<br /> <br /> Since the particle starts at &lt;math&gt;(0, 0),&lt;/math&gt; it is &lt;math&gt;m&lt;/math&gt; we are looking for, so the final answer is<br /> <br /> &lt;cmath&gt;4 + 35 = \textbf{(E) }39.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Interest https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_16&diff=69061 2015 AMC 10B Problems/Problem 16 2015-03-17T04:58:15Z <p>Interest: Created page with &quot;==Problem== Al, Bill, and Cal will each randomly be assigned a whole number from 1 to 10, inclusive, with no two of them getting the same number. What is the probability that ...&quot;</p> <hr /> <div>==Problem==<br /> Al, Bill, and Cal will each randomly be assigned a whole number from 1 to 10, inclusive, with no two of them getting the same number. What is the probability that Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's?<br /> <br /> ==Solution==<br /> We can solve this problem with a brute force approach.<br /> <br /> *If Cal's number is &lt;math&gt;1&lt;/math&gt;:<br /> **If Bill's number is &lt;math&gt;2&lt;/math&gt;, Al's can be any of &lt;math&gt;4, 6, 8, 10&lt;/math&gt;.<br /> **If Bill's number is &lt;math&gt;3&lt;/math&gt;, Al's can be any of &lt;math&gt;6, 9&lt;/math&gt;.<br /> **If Bill's number is &lt;math&gt;4&lt;/math&gt;, Al's can be &lt;math&gt;8&lt;/math&gt;.<br /> **If Bill's number is &lt;math&gt;5&lt;/math&gt;, Al's can be &lt;math&gt;10&lt;/math&gt;.<br /> **Otherwise, Al's number could not be a whole number multiple of Bill's.<br /> *If Cal's number is &lt;math&gt;2&lt;/math&gt;:<br /> **If Bill's number is &lt;math&gt;4&lt;/math&gt;, Al's can be &lt;math&gt;8&lt;/math&gt;.<br /> **Otherwise, Al's number could not be a whole number multiple of Bill's while Bill's number is still a whole number multiple of Cal's.<br /> *Otherwise, Bill's number must be greater than &lt;math&gt;5&lt;/math&gt;, i.e. Al's number could not be a whole number multiple of Bill's.<br /> <br /> Clearly, there are exactly &lt;math&gt;9&lt;/math&gt; cases where Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's. Since there are &lt;math&gt;10*9*8&lt;/math&gt; possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is &lt;math&gt;\frac9{10*9*8}=\text{(\textbf B) }\frac1{80}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Interest https://artofproblemsolving.com/wiki/index.php?title=User:Interest&diff=69052 User:Interest 2015-03-17T04:42:24Z <p>Interest: Created page with &quot;hi&quot;</p> <hr /> <div>hi</div> Interest https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_10&diff=69051 2015 AMC 10B Problems/Problem 10 2015-03-17T04:41:32Z <p>Interest: Created page with &quot;==Problem== What are the sign and units digit of the product of all the odd negative integers strictly greater than &lt;math&gt;-2015&lt;/math&gt;? ==Solution== Since &lt;math&gt;-5&gt;-2015&lt;/math...&quot;</p> <hr /> <div>==Problem==<br /> What are the sign and units digit of the product of all the odd negative integers strictly greater than &lt;math&gt;-2015&lt;/math&gt;?<br /> ==Solution==<br /> Since &lt;math&gt;-5&gt;-2015&lt;/math&gt;, the product must end with a &lt;math&gt;5&lt;/math&gt;.<br /> <br /> The multiplicands are the odd negative integers from &lt;math&gt;-1&lt;/math&gt; to &lt;math&gt;-2013&lt;/math&gt;. There are &lt;math&gt;\frac{2013+1}2=1007&lt;/math&gt; of these numbers. Since &lt;math&gt;(-1)^{1007}=-1&lt;/math&gt;, the product is negative.<br /> <br /> Therefore, the answer must be &lt;math&gt;\boxed{\text{(\textbf B) It is a negative number ending with a 5.}}&lt;/math&gt;<br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Interest https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_20&diff=69044 2015 AMC 10B Problems/Problem 20 2015-03-17T04:28:49Z <p>Interest: created page. no solution; included a cube in case someone wants it</p> <hr /> <div>==Problem==<br /> Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. how many paths are there meeting these conditions?<br /> ==Solution==<br /> &lt;asy&gt;<br /> import three;<br /> draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1));<br /> draw((0,0,1)--(1,0,1));<br /> draw((1,0,0)--(1,1,0));<br /> draw((0,0,0)--(0,1,0));<br /> &lt;/asy&gt;<br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Interest https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=69036 2015 AMC 10B Problems/Problem 22 2015-03-17T03:51:14Z <p>Interest: /* See Also */ fixed before and after</p> <hr /> <div>==Problem==<br /> In the figure shown below, &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon and &lt;math&gt;AG=1&lt;/math&gt;. What is &lt;math&gt;FG + JH + CD&lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br /> //(0,0) is a convenient point<br /> //E1 to prevent conflict with direction E(ast)<br /> pair F=intersectionpoints(D--A,E1--B), G=intersectionpoints(A--C,E1--B), H=intersectionpoints(B--D,A--C), I=intersectionpoints(C--E1,D--B), J=intersectionpoints(E1--C,D--A);<br /> draw(A--B--C--D--E1--A);<br /> draw(A--D--B--E1--C--A);<br /> draw(F--I--G--J--H--F);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,NW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,S);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> <br /> Triangle &lt;math&gt;AFG&lt;/math&gt; is isosceles, so &lt;math&gt;AG=AF=1&lt;/math&gt;. Using the symmetry of pentagon &lt;math&gt;FGHIJ&lt;/math&gt;, notice that &lt;math&gt;\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG&lt;/math&gt;. Therefore, &lt;math&gt;JH=AF=1&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle AJH \sim \triangle AFG&lt;/math&gt;, &lt;math&gt;\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1&lt;/math&gt;. From this, we get &lt;math&gt;FG=\frac{\sqrt{5} -1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle DIJ \cong \triangle AFG&lt;/math&gt;, &lt;math&gt;DJ=DI=AF=1&lt;/math&gt;. Since &lt;math&gt;\triangle AFG \sim ADC&lt;/math&gt;, &lt;math&gt; \frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac1{CD}&lt;/math&gt;. Solving for &lt;math&gt;CD&lt;/math&gt;, we get &lt;math&gt;CD = \frac{\sqrt{5} +1}{2}&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Interest https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=69035 2015 AMC 10B Problems/Problem 22 2015-03-17T03:48:55Z <p>Interest: Formatting, heavy phrasing edit, added &quot;problem&quot; and &quot;see also&quot; sections, removed &quot;solution by...&quot;</p> <hr /> <div>==Problem==<br /> In the figure shown below, &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon and &lt;math&gt;AG=1&lt;/math&gt;. What is &lt;math&gt;FG + JH + CD&lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br /> //(0,0) is a convenient point<br /> //E1 to prevent conflict with direction E(ast)<br /> pair F=intersectionpoints(D--A,E1--B), G=intersectionpoints(A--C,E1--B), H=intersectionpoints(B--D,A--C), I=intersectionpoints(C--E1,D--B), J=intersectionpoints(E1--C,D--A);<br /> draw(A--B--C--D--E1--A);<br /> draw(A--D--B--E1--C--A);<br /> draw(F--I--G--J--H--F);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,NW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,S);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> <br /> Triangle &lt;math&gt;AFG&lt;/math&gt; is isosceles, so &lt;math&gt;AG=AF=1&lt;/math&gt;. Using the symmetry of pentagon &lt;math&gt;FGHIJ&lt;/math&gt;, notice that &lt;math&gt;\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG&lt;/math&gt;. Therefore, &lt;math&gt;JH=AF=1&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle AJH \sim \triangle AFG&lt;/math&gt;, &lt;math&gt;\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1&lt;/math&gt;. From this, we get &lt;math&gt;FG=\frac{\sqrt{5} -1}{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle DIJ \cong \triangle AFG&lt;/math&gt;, &lt;math&gt;DJ=DI=AF=1&lt;/math&gt;. Since &lt;math&gt;\triangle AFG \sim ADC&lt;/math&gt;, &lt;math&gt; \frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac1{CD}&lt;/math&gt;. Solving for &lt;math&gt;CD&lt;/math&gt;, we get &lt;math&gt;CD = \frac{\sqrt{5} +1}{2}&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Interest https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_12&diff=51845 2013 AIME I Problems/Problem 12 2013-03-18T04:20:02Z <p>Interest: </p> <hr /> <div>== Problem 12 ==<br /> Let &lt;math&gt;\bigtriangleup PQR&lt;/math&gt; be a triangle with &lt;math&gt;\angle P = 75^\circ&lt;/math&gt; and &lt;math&gt;\angle Q = 60^\circ&lt;/math&gt;. A regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; with side length 1 is drawn inside &lt;math&gt;\triangle PQR&lt;/math&gt; so that side &lt;math&gt;\overline{AB}&lt;/math&gt; lies on &lt;math&gt;\overline{PQ}&lt;/math&gt;, side &lt;math&gt;\overline{CD}&lt;/math&gt; lies on &lt;math&gt;\overline{QR}&lt;/math&gt;, and one of the remaining vertices lies on &lt;math&gt;\overline{RP}&lt;/math&gt;. There are positive integers &lt;math&gt;a, b, c, &lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; such that the area of &lt;math&gt;\triangle PQR&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{a+b\sqrt{c}}{d}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are relatively prime, and c is not divisible by the square of any prime. Find &lt;math&gt;a+b+c+d&lt;/math&gt;.<br /> <br /> == Solution ==<br /> First, find that &lt;math&gt;\angle R = 45^\circ&lt;/math&gt;.<br /> Draw &lt;math&gt;ABCDEF&lt;/math&gt;. Now draw &lt;math&gt;\bigtriangleup PQR&lt;/math&gt; around &lt;math&gt;ABCDEF&lt;/math&gt; such that &lt;math&gt;Q&lt;/math&gt; is adjacent to &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The height of &lt;math&gt;ABCDEF&lt;/math&gt; is &lt;math&gt;\sqrt{3}&lt;/math&gt;, so the length of base &lt;math&gt;QR&lt;/math&gt; is &lt;math&gt;2+\sqrt{3}&lt;/math&gt;. Let the equation of &lt;math&gt;\overline{RP}&lt;/math&gt; be &lt;math&gt;y = x&lt;/math&gt;. Then, the equation of &lt;math&gt;\overline{PQ}&lt;/math&gt; is &lt;math&gt;y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3&lt;/math&gt;. Solving the two equations gives &lt;math&gt;y = x = \frac{\sqrt{3} + 3}{2}&lt;/math&gt;. The area of &lt;math&gt;\bigtriangleup PQR&lt;/math&gt; is &lt;math&gt;(2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{2}&lt;/math&gt;. &lt;math&gt;a + b + c + d = 9 + 5 + 3 + 2 = 19&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2013|n=I|num-b=11|num-a=13}}</div> Interest https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_12&diff=51844 2013 AIME I Problems/Problem 12 2013-03-18T04:19:45Z <p>Interest: added solution. unsure of correctness</p> <hr /> <div>== Problem 12 ==<br /> Let &lt;math&gt;\bigtriangleup PQR&lt;/math&gt; be a triangle with &lt;math&gt;\angle P = 75^\circ&lt;/math&gt; and &lt;math&gt;\angle Q = 60^\circ&lt;/math&gt;. A regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; with side length 1 is drawn inside &lt;math&gt;\triangle PQR&lt;/math&gt; so that side &lt;math&gt;\overline{AB}&lt;/math&gt; lies on &lt;math&gt;\overline{PQ}&lt;/math&gt;, side &lt;math&gt;\overline{CD}&lt;/math&gt; lies on &lt;math&gt;\overline{QR}&lt;/math&gt;, and one of the remaining vertices lies on &lt;math&gt;\overline{RP}&lt;/math&gt;. There are positive integers &lt;math&gt;a, b, c, &lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; such that the area of &lt;math&gt;\triangle PQR&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{a+b\sqrt{c}}{d}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are relatively prime, and c is not divisible by the square of any prime. Find &lt;math&gt;a+b+c+d&lt;/math&gt;.<br /> <br /> == Solution ==<br /> First, find that &lt;math&gt;\angle R = 45^\circ&lt;/math&gt;.<br /> Draw &lt;math&gt;ABCDEF&lt;/math&gt;. Now draw &lt;math&gt;\bigtriangleup PQR&lt;/math&gt; around &lt;math&gt;ABCDEF&lt;/math&gt; such that &lt;math&gt;Q&lt;/math&gt; is adjacent to &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The height of &lt;math&gt;ABCDEF&lt;/math&gt; is &lt;math&gt;\sqrt{3}&lt;/math&gt;, so the length of base &lt;math&gt;QR&lt;/math&gt; is &lt;math&gt;2+\sqrt{3}&lt;/math&gt;. Let the equation of &lt;math&gt;\overline{RP}&lt;/math&gt; be &lt;math&gt;y = x&lt;/math&gt;. Then, the equation of &lt;math&gt;\overline{PQ}&lt;/math&gt; is &lt;math&gt;y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3&lt;/math&gt;. Solving the two equations gives &lt;math&gt;y = x = \frac{\sqrt{3} + 3}{2}&lt;/math&gt;. The area of &lt;math&gt;\bigtriangleup PQR is &lt;/math&gt;(2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{2}&lt;math&gt;. &lt;/math&gt;a + b + c + d = 9 + 5 + 3 + 2 = 19\$<br /> <br /> == See also ==<br /> {{AIME box|year=2013|n=I|num-b=11|num-a=13}}</div> Interest