https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bgav&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-23T00:36:49Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_1&diff=103096 2019 AMC 10B Problems/Problem 1 2019-02-17T00:39:20Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #1]] and [[2019 AMC 12B Problems|2019 AMC 12B #1]]}}<br /> <br /> ==Problem==<br /> <br /> Alicia had two containers. The first was &lt;math&gt;\tfrac{5}{6}&lt;/math&gt; full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; full of water. What is the ratio of the volume of the first container to the volume of the second container?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the first jar's volume be &lt;math&gt;A&lt;/math&gt; and the second's be &lt;math&gt;B&lt;/math&gt;. It is given that &lt;math&gt;\frac{3}{4}A=\frac{5}{6}B&lt;/math&gt;. We find that &lt;math&gt;\frac{B}{A}=\frac{3/4}{5/6}=\boxed{\frac{9}{10}}.&lt;/math&gt; <br /> <br /> We already know that this is the ratio of smaller to larger volume because it is less than &lt;math&gt;1.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can set up a ratio to solve this problem. If x is the volume of the first container, and y is the volume of the second container, then:<br /> &lt;cmath&gt;\frac{5}{6}x = \frac{3}{4}y&lt;/cmath&gt;<br /> <br /> Cross Multiplying allows us to get &lt;math&gt;\frac{x}{y} = \frac{3}{4} \cdot \frac{6}{5} \Rightarrow \frac{18}{20} \Rightarrow \frac{9}{10}&lt;/math&gt;. Thus the ratio of the volume of the first container to the second container is &lt;math&gt;\boxed{(\text{D})\frac{9}{10}}&lt;/math&gt;<br /> <br /> An alternate solution is to plug in some maximum volume for the first container - let's say &lt;math&gt;72&lt;/math&gt;, so there was a volume of 60 in the first container, and then the second container also has a volume of &lt;math&gt;60&lt;/math&gt;, so you get &lt;math&gt;60 \cdot \frac{4}{3} \Rightarrow 80&lt;/math&gt;. Thus, &lt;math&gt;\frac{72}{80} \Rightarrow \frac{9}{10}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|before=First Problem|num-a=2}}<br /> {{AMC12 box|year=2019|ab=B|before=First Problem|num-a=2}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_3&diff=103095 2019 AMC 10B Problems/Problem 3 2019-02-17T00:38:58Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> <br /> In a high school with &lt;math&gt;500&lt;/math&gt; students, &lt;math&gt;40\%&lt;/math&gt; of the seniors play a musical instrument, while &lt;math&gt;30\%&lt;/math&gt; of the non-seniors do not play a musical instrument. In all, &lt;math&gt;46.8\%&lt;/math&gt; of the students do not play a musical instrument. How many non-seniors play a musical instrument?<br /> <br /> &lt;math&gt;\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> 60% of seniors do not play a musical instrument. If we denote x as the number of seniors, then &lt;cmath&gt;\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{3}{5}x + 150 - \frac{3}{10}x = 234&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{3}{10}x = 84&lt;/cmath&gt;<br /> &lt;cmath&gt;x = 84\cdot\frac{10}{3} \Rightarrow 280&lt;/cmath&gt;<br /> <br /> Thus there are &lt;math&gt;500-x = 220&lt;/math&gt; non-seniors. Since 70% of the non-seniors play a musical instrument, &lt;math&gt;220 \cdot \frac{7}{10} = \boxed{B) 154}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let x be the number of seniors, and y be the number of non-seniors. Then &lt;cmath&gt;\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234&lt;/cmath&gt;<br /> <br /> Multiplying 10 to every term gives us<br /> &lt;cmath&gt;6x + 3y = 2340&lt;/cmath&gt;<br /> <br /> Also, &lt;math&gt;x + y = 500&lt;/math&gt; because there are 500 students in total.<br /> <br /> <br /> Solving these system of equations give us &lt;math&gt;x = 280&lt;/math&gt;, &lt;math&gt;y = 220&lt;/math&gt;<br /> <br /> <br /> Since 70% of the non-seniors play a musical instrument, we simply get 70% of 220, which gives us &lt;math&gt;\boxed{B) 154}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_4&diff=103094 2019 AMC 10B Problems/Problem 4 2019-02-17T00:38:43Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> <br /> All lines with equation &lt;math&gt;ax+by=c&lt;/math&gt; such that &lt;math&gt;a,b,c&lt;/math&gt; form an arithmetic progression pass through a common point. What are the coordinates of that point?<br /> <br /> &lt;math&gt;\textbf{(A) } (-1,2)<br /> \qquad\textbf{(B) } (0,1)<br /> \qquad\textbf{(C) } (1,-2)<br /> \qquad\textbf{(D) } (1,0)<br /> \qquad\textbf{(E) } (1,2)&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> If all lines satisfy the equation, then we can just plug in values for a, b, and c that form an arithmetic progression. Let's do a=1, b=2, c=3 and a=1, b=3, and c=5. Then the two lines we get are: &lt;cmath&gt;x+2y=3&lt;/cmath&gt; &lt;cmath&gt;x+3y=5&lt;/cmath&gt;<br /> Use elimination: &lt;cmath&gt;y = 2&lt;/cmath&gt; Plug this into one of the previous lines. &lt;cmath&gt;x+4 = 3 \Rightarrow x=-1&lt;/cmath&gt; Thus the common point is &lt;math&gt;\boxed{A) (-1,2)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We know that &lt;math&gt;a,b,c&lt;/math&gt; are an arithmetic progression, so if the common difference is &lt;math&gt;d&lt;/math&gt; we can say &lt;math&gt;a,b,c = a, a+d, a+2d.&lt;/math&gt; Now we have &lt;math&gt;ax+ (a+d)y = a+2d&lt;/math&gt;, and expanding gives &lt;math&gt;ax + ay + dy = a + 2d.&lt;/math&gt; Factoring gives &lt;math&gt;a(x+y-1)+d(y-2) = 0&lt;/math&gt;. Since this must always be true, we know that &lt;math&gt;x+y-1 = 0&lt;/math&gt; and &lt;math&gt;y-2 = 0&lt;/math&gt;, so &lt;math&gt;x,y = -1, 2,&lt;/math&gt; and the common point is &lt;math&gt;\boxed{A) (-1,2)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_5&diff=103093 2019 AMC 10B Problems/Problem 5 2019-02-17T00:38:20Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are reflected across the line &lt;math&gt;y=x&lt;/math&gt; to points &lt;math&gt;A'&lt;/math&gt;, &lt;math&gt;B'&lt;/math&gt;, and &lt;math&gt;C'&lt;/math&gt;, respectively. Assume that none of the vertices of the triangle lie on the line &lt;math&gt;y=x&lt;/math&gt;. Which of the following statements is &lt;i&gt;&lt;u&gt;not&lt;/u&gt;&lt;/i&gt; always true?<br /> <br /> &lt;math&gt;\textbf{(A) } &lt;/math&gt; Triangle &lt;math&gt;A'B'C'&lt;/math&gt; lies in the first quadrant.<br /> <br /> &lt;math&gt;\textbf{(B) } &lt;/math&gt; Triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'&lt;/math&gt; have the same area.<br /> <br /> &lt;math&gt;\textbf{(C) } &lt;/math&gt; The slope of line &lt;math&gt;AA'&lt;/math&gt; is &lt;math&gt;-1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) } &lt;/math&gt; The slopes of lines &lt;math&gt;AA'&lt;/math&gt; and &lt;math&gt;CC'&lt;/math&gt; are the same.<br /> <br /> &lt;math&gt;\textbf{(E) } &lt;/math&gt; Lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;A'B'&lt;/math&gt; are perpendicular to each other.<br /> <br /> ==Solution==<br /> Let's analyze all of the options separately. <br /> <br /> A: Clearly A is true, because a coordinate in the first quadrant will have (+,+), and its inverse would also have (+,+)<br /> <br /> B: The triangles have the same area, it's the same triangle.<br /> <br /> C: If coordinate A has (x,y), then its inverse will have (y,x). (x-y)/(y-x)=-1, so this is true.<br /> <br /> D: Likewise, if coordinate A has (x1,y1), and AA' has a slope of -1, then coordinate B, with (x2,y2), will also have a slope of -1. This is true. <br /> <br /> E: By process of elimination, this is the answer, but if coordinate A has (x1,y1) and coordinate B has (x2,y2), then their inverses will be (y1,x1), (y2,x2), and it is not necessarily true that (y2-y1)/(x2-x1)=-(y2-y1)/(x2-x1). (Negative inverses of each other). <br /> <br /> Clearly, the answer is &lt;math&gt;\boxed{\textbf{(E)}}&lt;/math&gt;.<br /> <br /> ==Counterexamples==<br /> If &lt;math&gt;(x_1,y_1) = (2,3)&lt;/math&gt; and &lt;math&gt;(x_2,y_2) = (7,1)&lt;/math&gt;, then the slope of &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;m_{AB}&lt;/math&gt;, is &lt;math&gt;\frac{1 - 3}{7 - 2} = -\frac{2}{5}&lt;/math&gt;, while the slope of &lt;math&gt;A'B'&lt;/math&gt;, &lt;math&gt;m_{A'B'}&lt;/math&gt;, is &lt;math&gt;\frac{7 - 2}{1 - 3} = -\frac{5}{2}&lt;/math&gt;. &lt;math&gt;m_{A'B'}&lt;/math&gt; is the &lt;math&gt;\textbf{reciprocal}&lt;/math&gt; of &lt;math&gt;m_{AB}&lt;/math&gt;, but it is not the negative reciprocal of &lt;math&gt;m_{AB}&lt;/math&gt;. To generalize, let &lt;math&gt;(x_1,y_1)&lt;/math&gt; denote the coordinates of point A, let &lt;math&gt;(x_2, y_2)&lt;/math&gt; denote the coordinates of point B, let &lt;math&gt;m_{AB}&lt;/math&gt; denote the slope of segment &lt;math&gt;\overline{AB}&lt;/math&gt;, and let &lt;math&gt;m_{A'B'}&lt;/math&gt; denote the slope of segment &lt;math&gt;\overline{A'B'}&lt;/math&gt;. Then, the coordinate pair for &lt;math&gt;A'&lt;/math&gt; is &lt;math&gt;(y_1, x_1)&lt;/math&gt;, and the pair for &lt;math&gt;B'&lt;/math&gt; is &lt;math&gt;(y_2, x_2)&lt;/math&gt;. Then, &lt;math&gt;m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}&lt;/math&gt;, and &lt;math&gt;m_{A'B'} = \frac{x_2 - x_1}{y_2 - y_1} = \frac{1}{m_{ab}}&lt;/math&gt;. If &lt;math&gt;y_1 \neq y_2&lt;/math&gt; and &lt;math&gt;x_1 \neq x_2&lt;/math&gt;, &lt;math&gt;\frac{1}{m_{AB}} \neq \frac{1}{m_{A'B'}} \Rightarrow m_{AB} \neq m_{A'B'}&lt;/math&gt;, and in these cases, the condition is false.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_8&diff=103092 2019 AMC 10B Problems/Problem 8 2019-02-17T00:37:49Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region? <br /> <br /> &lt;math&gt;(A)&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; &lt;math&gt;(B)&lt;/math&gt; &lt;math&gt;12 - 4\sqrt{3}&lt;/math&gt; &lt;math&gt;(C)&lt;/math&gt; &lt;math&gt;3\sqrt{3}&lt;/math&gt; &lt;math&gt;(D)&lt;/math&gt; &lt;math&gt;4\sqrt{3}&lt;/math&gt; &lt;math&gt;(E)&lt;/math&gt; &lt;math&gt;16 - \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We notice that the square can be split into &lt;math&gt;4&lt;/math&gt; congruent smaller squares with the altitude of the equilateral triangle being the side of the square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (Note that it has already been split in half). <br /> <br /> When we split an equilateral triangle in half, we get &lt;math&gt;2&lt;/math&gt; triangles with a &lt;math&gt;30-60-90&lt;/math&gt; relationship. Therefore, we get that the altitude and a side length of a square is &lt;math&gt;\sqrt{3}&lt;/math&gt;. <br /> <br /> We can then compute the area of the two triangles using the base-height-area relationship and get &lt;math&gt;2 \cdot \frac{1 \cdot \sqrt{3}}{2} = \sqrt{3}&lt;/math&gt;.<br /> <br /> The area of the small squares is the altitude squared which is &lt;math&gt;(\sqrt{3})^2 = 3&lt;/math&gt;. Therefore, the area of the shaded region in each of the four squares is &lt;math&gt;3 - \sqrt{3}&lt;/math&gt;<br /> <br /> Since there are four of these squares, we multiply this by &lt;math&gt;4&lt;/math&gt; to get &lt;math&gt;4(3 - \sqrt{3}) = 12 - 4 \sqrt{3}&lt;/math&gt; as our answer. This is choice &lt;math&gt;\boxed{ B) 12 - 4 \sqrt{3}}&lt;/math&gt;.<br /> <br /> ~Awesome2.1 and edited by greersc.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_13&diff=102973 2019 AMC 12B Problems/Problem 13 2019-02-15T21:14:07Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #17]] and [[2019 AMC 12B Problems|2019 AMC 12B #13]]}}<br /> ==Problem==<br /> <br /> A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;2^{-k}&lt;/math&gt; for &lt;math&gt;k = 1,2,3....&lt;/math&gt; What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?&lt;br&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher numbered bin. The probability of both landing in the same bin is &lt;math&gt;\sum_{k=1}^{\infty}2^{-2k}&lt;/math&gt;. The sum is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt;. Therefore the other two probabilities have to both be &lt;math&gt;\textbf{(C) } \frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 1 Variant==<br /> We solve for the probability by doing &lt;math&gt;\frac{1-(\text{Probability of Equality})}{2}&lt;/math&gt;.<br /> <br /> We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to &lt;math&gt;(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...&lt;/math&gt; <br /> <br /> The summation of this expression is equal to &lt;cmath&gt;\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1&lt;/cmath&gt;. Using the geometric sum formula, we obtain the summation of this expression to be &lt;math&gt;\frac{1}{\frac{3}{4}}-1&lt;/math&gt; or &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 2 (variant)==<br /> Suppose the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, for some &lt;math&gt;i \ge 1&lt;/math&gt;. The probability of this occurring is &lt;math&gt;\frac{1}{2^i}&lt;/math&gt;. Given this occurs, the probability that the red ball goes in a higher-numbered bin is &lt;math&gt;\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}&lt;/math&gt;. Thus the probability that the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, and the red ball goes in a bin greater than &lt;math&gt;i&lt;/math&gt;, is &lt;math&gt;\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}&lt;/math&gt;. Summing from &lt;math&gt;i=1&lt;/math&gt; to infinity, we get<br /> <br /> &lt;cmath&gt;\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/cmath&gt;<br /> <br /> (Note: to find this sum, we use the formula &lt;math&gt;\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}&lt;/math&gt;. Since in this case &lt;math&gt;r = \frac{1}{4}&lt;/math&gt;, the answer is &lt;math&gt;\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}&lt;/math&gt;. If you don't know this formula, you may instead note that if you multiply the sum by &lt;math&gt;4&lt;/math&gt;, it is equivalent to adding &lt;math&gt;1&lt;/math&gt;. Thus: &lt;math&gt;4n = n+1&lt;/math&gt;, which clearly simplifies to &lt;math&gt;n = \frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 3 (infinite geometric series)==<br /> The probability that the two balls will go into adjacent bins is &lt;math&gt;\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} = \frac{1}{6}&lt;/math&gt;. The probability that the two balls will go into bins that have a distance of 2 from each other is &lt;math&gt;\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} = \frac{1}{12}&lt;/math&gt;. We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is &lt;math&gt;\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + ...&lt;/math&gt;, which converges into &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 4 (quickest)==<br /> Define a win as a ball appearing in higher numbered box.<br /> <br /> Start from the first box. <br /> <br /> There are 4 possible results in the box: Red, Green, Red and Green, none, with probability of &lt;math&gt;\frac{1}{4}&lt;/math&gt; for each. Red win, Green win, Tie all have the same probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt;. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution.<br /> <br /> So finally, Red win, Green win and Tie all have a probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt;<br /> <br /> The answer is &lt;math&gt;\boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 5(easiest)==<br /> Write out the infinite geometric series as &lt;math&gt;\frac{1}{2}&lt;/math&gt;, &lt;math&gt;\frac{1}{4}&lt;/math&gt;, &lt;math&gt;\frac{1}{8}&lt;/math&gt;, &lt;math&gt;\frac{1}{16}&lt;/math&gt;... To find the probablilty that red scores after, we can simply remove all odd number terms(i.e term 1, term 3...), and then sum the remaining. This works for a similar reason to solution 2 variant(so look at that if you don't understand why I can do this). Writing this out as another infinite geometric sequence,&lt;math&gt;\frac{1}{4}&lt;/math&gt;, &lt;math&gt;\frac{1}{16}&lt;/math&gt;, &lt;math&gt;\frac{1}{64}&lt;/math&gt;..., Summing, we get &lt;cmath&gt;\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/cmath&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=102972 2019 AMC 10B Problems/Problem 16 2019-02-15T21:13:53Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; with a right angle at &lt;math&gt;C&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;AC=CD,&lt;/math&gt; &lt;math&gt;DE=EB,&lt;/math&gt; and the ratio &lt;math&gt;AC:DE=4:3&lt;/math&gt;. What is the ratio &lt;math&gt;AD:DB?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2&lt;/math&gt;<br /> <br /> ==Solution==<br /> Without loss of generality, let &lt;math&gt;AC = CD = 4&lt;/math&gt; and &lt;math&gt;DE = EB = 3&lt;/math&gt;. Let &lt;math&gt;\angle A = \alpha&lt;/math&gt; and &lt;math&gt;\angle B = \beta = 90^{\circ} - \alpha&lt;/math&gt;. As &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle DEB&lt;/math&gt; are isosceles, &lt;math&gt;\angle ADC = \alpha&lt;/math&gt; and &lt;math&gt;\angle BDE = \beta&lt;/math&gt;. Then &lt;math&gt;\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}&lt;/math&gt;, so &lt;math&gt;\triangle CDE&lt;/math&gt; is a 3-4-5 triangle with &lt;math&gt;CE = 5&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;CB = 5+3 = 8&lt;/math&gt;, and &lt;math&gt;\triangle ABC&lt;/math&gt; is a 1-2-&lt;math&gt;\sqrt{5}&lt;/math&gt; triangle.<br /> <br /> On isosceles triangles &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle DEB&lt;/math&gt;, drop altitudes from &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; onto &lt;math&gt;AB&lt;/math&gt;; denote the feet of these altitudes by &lt;math&gt;P_C&lt;/math&gt; and &lt;math&gt;P_E&lt;/math&gt; respectively. Then &lt;math&gt;\triangle ACP_C \sim \triangle ABC&lt;/math&gt; by AAA similarity, so we get that &lt;math&gt;AP_C = P_CD = \frac{4}{\sqrt{5}}&lt;/math&gt;, and &lt;math&gt;AD = 2 \times \frac{4}{\sqrt{5}}&lt;/math&gt;. Similarly we get &lt;math&gt;BD = 2 \times \frac{6}{\sqrt{5}}&lt;/math&gt;, and &lt;math&gt;AD:DB = \boxed{\textbf{(A) } 2:3}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;AC=CD=4x&lt;/math&gt;, and &lt;math&gt;DE=EB=3x&lt;/math&gt;. (For this solution, A is above C, and B is to the right of C). Denote the angle of point A as &quot;t&quot;. Then &lt;math&gt;&lt;ACD&lt;/math&gt; is &lt;math&gt;180-2t&lt;/math&gt; degrees, which implies that &lt;math&gt;&lt;DCB&lt;/math&gt; is &lt;math&gt;2t - 90&lt;/math&gt; degrees. Similarly, the angle of point B is &lt;math&gt;90 - t&lt;/math&gt; degrees, which implies that &lt;math&gt;&lt;BED&lt;/math&gt; is &lt;math&gt;2t&lt;/math&gt; degrees. This further implies that &lt;math&gt;&lt;DEC&lt;/math&gt; is &lt;math&gt;180 - 2t&lt;/math&gt; degrees. <br /> <br /> This may seem strange, but if you draw the diagram, the solution will work itself out like this.<br /> <br /> Now we see that &lt;math&gt;&lt;CDE = 180 - &lt;ECD - &lt;CED \Rightarrow 180 - 2x + 90 - 180 + 2x \Rightarrow 90&lt;/math&gt;. Thus triangle CDE is a right triangle, with side lengths of 3x, 4x, and by the pythaogrean theorem, 5x. Now we see that AC is 4x (by definition), BC is 5x+3x = 8x, and AB is &lt;math&gt;4\sqrt{5}&lt;/math&gt;x. Now, we find the cosine of 2y - this is &lt;math&gt;2cos^2x - 1&lt;/math&gt;. which is &lt;math&gt;2*(\frac{1}{\sqrt{5}})^2 - 1 \Rightarrow \frac{-3}{5}&lt;/math&gt; Using law of cosines on triangle BED, and denoting the length of BD as &quot;d&quot;, we get &lt;cmath&gt;d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x)&lt;/cmath&gt; &lt;cmath&gt;d^2 = 18x^2 + \frac{54x^2}{5} \Rightarrow {144x^2}{5}&lt;/cmath&gt; &lt;cmath&gt;d = \frac{12x}{\sqrt{5}}&lt;/cmath&gt; Since this is DB, and we know AB, to find the ratio we find AD, which is &lt;math&gt;\frac{4x}{\sqrt{5}} - \frac{12x}{\sqrt{5}}&lt;/math&gt;, which is &lt;math&gt;\frac{8x}{\sqrt{5}}&lt;/math&gt;. Thus the answer is &lt;math&gt;\frac{\frac{8x}{\sqrt{5}}}{\frac{12x}{\sqrt{5}}} \Rightarrow \frac{8x}{\sqrt{5}}\cdot\frac{\sqrt{5}}{12x} \Rightarrow \boxed {A)2:3}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_14&diff=102971 2019 AMC 10B Problems/Problem 14 2019-02-15T21:13:36Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> The base-ten representation for &lt;math&gt;19!&lt;/math&gt; is &lt;math&gt;121,6T5,100,40M,832,H00&lt;/math&gt;, where &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, and &lt;math&gt;H&lt;/math&gt; denote digits that are not given. What is &lt;math&gt;T+M+H&lt;/math&gt;?<br /> <br /> ==Solution 1==<br /> We can figure out &lt;math&gt;H = 0&lt;/math&gt; by noticing that &lt;math&gt;19!&lt;/math&gt; will end with &lt;math&gt;3&lt;/math&gt; zeroes, as there are three &lt;math&gt;5&lt;/math&gt;'s in its prime factorization. Next we use the fact that &lt;math&gt;19!&lt;/math&gt; is a multiple of both &lt;math&gt;11&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;. Since their divisibility rules gives us that &lt;math&gt;T + M&lt;/math&gt; is congruent to &lt;math&gt;3&lt;/math&gt; mod &lt;math&gt;9&lt;/math&gt; and that &lt;math&gt;T - M&lt;/math&gt; is congruent to &lt;math&gt;7&lt;/math&gt; mod &lt;math&gt;11&lt;/math&gt;. By inspection, we see that &lt;math&gt;T = 4, M = 8&lt;/math&gt; is a valid solution. Therefore the answer is &lt;math&gt;4 + 8 + 0 = 12&lt;/math&gt;, which is (C).<br /> <br /> ==Solution 2==<br /> With investing just a little bit of time, we can manually calculate 19!. If we prime factorize 19!, it becomes &lt;math&gt;2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19&lt;/math&gt;. This looks complicated, but we can use elimination methods to make it simpler. &lt;math&gt;2^3 \cdot 5^3 = 1000&lt;/math&gt;, and &lt;math&gt;7 \cdot 11 \cdot 13 \cdot = 1001&lt;/math&gt;. If we put these aside for a moment, we have &lt;math&gt;2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19&lt;/math&gt; left from the original 19!. &lt;math&gt;2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192&lt;/math&gt;, and &lt;math&gt;3^8 = (3^4)^2 = 81^2 = 6561&lt;/math&gt;. We have the 2's and 3's out of the way, and then we have &lt;math&gt;7 \cdot 17 \cdot 19 = 2261&lt;/math&gt;. Now if we multiply all the values calculated, we get &lt;math&gt;1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000&lt;/math&gt;. Thus &lt;math&gt;T = 4, M = 8, H = 0&lt;/math&gt;, and the answer &lt;math&gt;T + M + H = 12&lt;/math&gt;, thus (C).<br /> <br /> ==Solution 3==<br /> We know that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt; are both factors of &lt;math&gt;19!&lt;/math&gt;. Furthermore, we know that H is 0 because &lt;math&gt;19!&lt;/math&gt; ends in three zeroes. We can simply use the divisibility rules for &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt; for this problem to find T and M. For &lt;math&gt;19!&lt;/math&gt; to be divisible by &lt;math&gt;9&lt;/math&gt;, the sum of digits must be divisible by &lt;math&gt;9&lt;/math&gt;. Summing the digits, we get that T + M + &lt;math&gt;33&lt;/math&gt; must be divisible by &lt;math&gt;9&lt;/math&gt;. This leaves either A or C as our answer choice. Now we test for divisibility by &lt;math&gt;11&lt;/math&gt;. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. &lt;math&gt;2728&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;-&lt;math&gt;7&lt;/math&gt;+&lt;math&gt;2&lt;/math&gt;-&lt;math&gt;8&lt;/math&gt; = -11 so &lt;math&gt;2728&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;). Applying the alternating sum to this problem, we see that T-M-7 must be divisible by 11. By inspection, we can see that this holds if T is &lt;math&gt;4&lt;/math&gt; and M is &lt;math&gt;8&lt;/math&gt;. The sum is &lt;math&gt;8&lt;/math&gt; + &lt;math&gt;4&lt;/math&gt; + &lt;math&gt;0&lt;/math&gt; = 12 or (C). -- krishdhar<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_15&diff=102970 2019 AMC 10B Problems/Problem 15 2019-02-15T21:13:19Z <p>Bgav: </p> <hr /> <div><br /> ==Problem==<br /> <br /> Two right triangles, &lt;math&gt;T_1&lt;/math&gt; and &lt;math&gt;T_2&lt;/math&gt;, have areas of 1 and 2, respectively. One side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other. What is the product of the third side lengths of &lt;math&gt;T_1&lt;/math&gt; and &lt;math&gt;T_2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{28}{3}\qquad\textbf{(B) }10\qquad\textbf{(C) }\frac{32}{3}\qquad\textbf{(D) }\frac{34}{3}\qquad\textbf{(E) }12&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> First of all, name the two sides which are congruent to be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, where &lt;math&gt;y &gt; x&lt;/math&gt;. The only way that the conditions of the problem can be satisfied is if &lt;math&gt;x&lt;/math&gt; was the shorter leg of &lt;math&gt;T_{2}&lt;/math&gt; and the longer leg of &lt;math&gt;T_{1}&lt;/math&gt;, and &lt;math&gt;y&lt;/math&gt; is the longer leg of &lt;math&gt;T_{2}&lt;/math&gt; and the hypotenuse of &lt;math&gt;T_{1}&lt;/math&gt;.<br /> <br /> Notice that this means the value we are looking for is the square of &lt;math&gt;\sqrt{x^{2}+y^{2}} \cdot \sqrt{y^{2}-x^{2}} = \sqrt{y^{4}-x^{4}}&lt;/math&gt;, which is just &lt;math&gt;y^{4}-x^{4}&lt;/math&gt;.<br /> <br /> We have two equations: &lt;math&gt;\frac{xy}{2} = 2&lt;/math&gt; and &lt;math&gt;\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1&lt;/math&gt;.<br /> <br /> This means that &lt;math&gt;y = \frac{4}{x}&lt;/math&gt; and that &lt;math&gt;\frac{4}{x^{2}} = y^{2} - x^{2}&lt;/math&gt;.<br /> <br /> Taking the second equation, we get &lt;math&gt;x^{2}y^{2} - x^{4} = 4&lt;/math&gt;, so since &lt;math&gt;xy = 4&lt;/math&gt;, &lt;math&gt;x^{4} = 12&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;y = \frac{4}{x}&lt;/math&gt;, we get &lt;math&gt;y^{4} = \frac{256}{12} = \frac{64}{3}&lt;/math&gt;.<br /> <br /> The value we are looking for is just &lt;math&gt;y^{4}-x^{4} = \frac{64-36}{3} = \frac{28}{3}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> First, construct right triangles △ABC and △EDF, with △ABC being the smaller triangle. We are given that one side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other. <br /> <br /> So, &lt;math&gt;\overline{AB}&lt;/math&gt; ≅ &lt;math&gt;\overline{EF}&lt;/math&gt;, call this length &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt; ≅ &lt;math&gt;\overline{DF}&lt;/math&gt;, call this length &lt;math&gt;y&lt;/math&gt;<br /> <br /> Additionally, call the length &lt;math&gt;\overline{AC}&lt;/math&gt; &lt;math&gt;z&lt;/math&gt;, and call the length &lt;math&gt;\overline{DE}&lt;/math&gt; &lt;math&gt;w&lt;/math&gt;<br /> <br /> Recapping our variables, we have &lt;math&gt;\overline{AB}&lt;/math&gt; = &lt;math&gt;\overline{EF}&lt;/math&gt; = &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt; = &lt;math&gt;\overline{DF}&lt;/math&gt; = &lt;math&gt;y&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt; = &lt;math&gt;z&lt;/math&gt;, and &lt;math&gt;\overline{DE}&lt;/math&gt; = &lt;math&gt;w&lt;/math&gt;<br /> <br /> We are given that &lt;math&gt;[ABC] = 1&lt;/math&gt; and &lt;math&gt;[EDF] = 2&lt;/math&gt;<br /> <br /> Since area = &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, this gives &lt;math&gt;\frac{xy}{2} = 1&lt;/math&gt; and &lt;math&gt;\frac{xw}{2} = 2&lt;/math&gt;<br /> <br /> Dividing the two equations, we get &lt;math&gt;\frac{xy}{xw}&lt;/math&gt; = &lt;math&gt;\frac{y}{w} = 2&lt;/math&gt;<br /> <br /> From this, we get &lt;math&gt;y = 2w&lt;/math&gt;<br /> <br /> We see that △EDF is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle, meaning that &lt;math&gt;x = w\sqrt{3}&lt;/math&gt;<br /> <br /> In △ABC, &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the legs. By the Pythagorean Theorem,<br /> <br /> &lt;math&gt;(w\sqrt{3})^2 + (2w)^2 = z^2&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;3w^2 + 4w^2 = z^2&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;7w^2 = z^2&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;w\sqrt{7} = z&lt;/math&gt;<br /> <br /> The question asks for the square of the product of the third side lengths of each triangle, which is &lt;math&gt;(wz)^2&lt;/math&gt;<br /> <br /> Using substitution, we see that &lt;math&gt;wz&lt;/math&gt; = &lt;math&gt;(w)(w\sqrt{7}&lt;/math&gt;) = &lt;math&gt;w^2\sqrt{7}&lt;/math&gt;<br /> <br /> We know &lt;math&gt;\frac{xw}{2} = 1&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;\frac{(w)(w\sqrt{3})}{2} =1&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;(w)(w\sqrt{3}) = 2&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;(w^2\sqrt{3}) = 2&lt;/math&gt;<br /> <br /> Dividing both sides by &lt;math&gt;\sqrt{3}&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;w^2 = \frac{2}{\sqrt{3}}&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;w^2 = \frac{2\sqrt{3}}{3}&lt;/math&gt;<br /> <br /> Since we want &lt;math&gt;(w^2\sqrt{7})^2&lt;/math&gt;, multiplying both sides by &lt;math&gt;\sqrt{7}&lt;/math&gt; gets us<br /> <br /> &lt;math&gt;w^2\sqrt{7} = \frac{2\sqrt{21}}{3}&lt;/math&gt;<br /> <br /> Squaring this,<br /> <br /> &lt;math&gt;(\frac{2\sqrt{21}}{3})^2 = \frac{4*21}{9} = \frac{84}{9} = \frac{28}{3}&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=102965 2019 AMC 10B Problems/Problem 25 2019-02-15T21:08:28Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br /> <br /> ==Problem==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> ==Solution 1 (Recursion)==<br /> We can deduce that any valid sequence of length &lt;math&gt;n&lt;/math&gt; will start with a 0 followed by either &quot;10&quot; or &quot;110&quot;.<br /> Because of this, we can define a recursive function:<br /> <br /> &lt;math&gt;f(n) = f(n-3) + f(n-2)&lt;/math&gt;<br /> <br /> This is because for any valid sequence of length &lt;math&gt;n&lt;/math&gt;, you can append either &quot;10&quot; or &quot;110&quot; and the resulting sequence would still satisfy the given conditions.<br /> <br /> &lt;math&gt;f(5) = 1&lt;/math&gt; and &lt;math&gt;f(6) = 2&lt;/math&gt;, so you follow the recursion up until &lt;math&gt;f(19) = 65 \quad \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 2 (Casework)==<br /> After any given zero, the next zero must appear exactly two or three spots down the line. And we started at position 1 and ended at position 19, so we moved over 18. Therefore, we must add a series of 2's and 3's to get 18. How can we do this?<br /> <br /> Option 1: nine 2's (there is only 1 way to arrange this).<br /> <br /> Option 2: two 3's and six 2's (&lt;math&gt;{8\choose2} =28&lt;/math&gt; ways to arrange this).<br /> <br /> Option 3: four 3's and three 2's (&lt;math&gt;{7\choose3}=35&lt;/math&gt; ways to arrange this).<br /> <br /> Option 4: six 3's (there is only 1 way to arrange this).<br /> <br /> Sum the four numbers given above: 1+28+35+1=65<br /> <br /> ==Solution3==<br /> first have 1 choose<br /> second have 1 choose<br /> 2 2 3 4 5 7 9 12 16 21 28 37 49 86 65 last is 0<br /> at the last we see the answer is 65(C)<br /> (Can someone please edit this?)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_24&diff=102964 2019 AMC 10B Problems/Problem 24 2019-02-15T21:08:11Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #24]] and [[2019 AMC 12B Problems|2019 AMC 12B #22]]}}<br /> <br /> ==Problem==<br /> <br /> Define a sequence recursively by &lt;math&gt;x_0=5&lt;/math&gt; and &lt;cmath&gt;x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}&lt;/cmath&gt; for all nonnegative integers &lt;math&gt;n.&lt;/math&gt; Let &lt;math&gt;m&lt;/math&gt; be the least positive integer such that<br /> &lt;cmath&gt;x_m\leq 4+\frac{1}{2^{20}}.&lt;/cmath&gt;In which of the following intervals does &lt;math&gt;m&lt;/math&gt; lie?<br /> <br /> &lt;math&gt;\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We first prove that &lt;math&gt;x_n &gt; 4&lt;/math&gt; for all &lt;math&gt;n \ge 0&lt;/math&gt; by induction from <br /> &lt;cmath&gt;<br /> x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}<br /> &lt;/cmath&gt;<br /> and then prove &lt;math&gt;x_n&lt;/math&gt;'s are decreasing by<br /> &lt;cmath&gt;<br /> x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} &lt; 0<br /> &lt;/cmath&gt;<br /> Now we need to estimate the value of &lt;math&gt;x_{n+1}-4&lt;/math&gt; by<br /> &lt;cmath&gt;<br /> x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6} <br /> &lt;/cmath&gt;<br /> since &lt;math&gt;x_n&lt;/math&gt;'s are decreasing, &lt;math&gt;\frac{x_n + 5}{x_n+6}&lt;/math&gt; are also decreasing, so we have<br /> &lt;cmath&gt;<br /> \frac{9}{10} &lt; \frac{x_n + 5}{x_n+6} \le \frac{10}{11}<br /> &lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;<br /> \frac{9}{10}(x_n-4) &lt; x_{n+1} - 4 \le \frac{10}{11}(x_n-4)<br /> &lt;/cmath&gt;<br /> which leads to<br /> &lt;cmath&gt;<br /> (\frac{9}{10})^n = (\frac{9}{10})^n (x_0-4) &lt; x_{n} - 4 \le (\frac{10}{11})^n (x_0-4) = (\frac{10}{11})^n<br /> &lt;/cmath&gt;<br /> The problem requires us to find the value of &lt;math&gt;n&lt;/math&gt; such that<br /> &lt;cmath&gt;<br /> (\frac{9}{10})^n &lt; x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } <br /> (\frac{10}{11})^{n-1} &gt; x_{n-1} - 4 &gt; \frac{1}{2^{20}}<br /> &lt;/cmath&gt;<br /> using natural logarithm, we need<br /> &lt;math&gt;n \ln \frac{9}{10} &lt; -20 \ln 2&lt;/math&gt; and &lt;math&gt;(n-1)\ln \frac{10}{11} &gt; -20 \ln 2&lt;/math&gt;, or<br /> <br /> &lt;cmath&gt;<br /> n &gt; \frac{20\ln 2}{\ln\frac{10}{9}} \text{ and } n-1 &lt; \frac{20\ln 2}{\ln\frac{11}{10}}<br /> &lt;/cmath&gt;<br /> <br /> As estimations, &lt;math&gt;\ln\frac{10}{9} \approx 1/9&lt;/math&gt; and &lt;math&gt;\ln\frac{11}{10} \approx 1/10&lt;/math&gt;, &lt;math&gt;\ln 2\approx 0.7&lt;/math&gt;<br /> we can estimate that<br /> &lt;cmath&gt;<br /> 126 &lt; n &lt; 141<br /> &lt;/cmath&gt;<br /> Choose &lt;math&gt;\boxed{C}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> Making the reasonable assumption that &lt;math&gt;x_n&lt;/math&gt; approaches &lt;math&gt;4&lt;/math&gt;, we can translate &lt;math&gt;x&lt;/math&gt; down by &lt;math&gt;4&lt;/math&gt; to obtain a more simple sequence &lt;math&gt;a_n=x_n-4&lt;/math&gt; that should approach &lt;math&gt;0&lt;/math&gt;. <br /> <br /> <br /> Substitution of &lt;math&gt;(a_{n}+4)&lt;/math&gt; for &lt;math&gt;(x_{n})&lt;/math&gt; and &lt;math&gt;(a_{n+1}+4)&lt;/math&gt; for &lt;math&gt;(x_{n+1})&lt;/math&gt; in the definition of &lt;math&gt;x_{n+1}&lt;/math&gt; leads to<br /> <br /> <br /> &lt;math&gt;a_{n+1}+4 = \frac{(a_{n} + 8)(a_{n}+5)}{a_{n}+10} = \frac{a_{n}^2 + 13a_{n}+40}{a_{n}+10} \implies a_{n+1} = a_{n}(\frac{a_{n}+9}{a_{n}+10}) \implies \frac{a_{n+1}}{a_{n}} = \frac{a_{n}+9}{a_{n}+10}&lt;/math&gt;<br /> <br /> <br /> The ratio of consecutive terms is thus always positive and less than 1 (because &lt;math&gt;a_0&lt;/math&gt; is positive). This means that the largest possible value for &lt;math&gt;a_n&lt;/math&gt; is 1 and that no value of &lt;math&gt;a_n&lt;/math&gt; can be less than or equal to 0.<br /> <br /> <br /> Plugging the extrema of &lt;math&gt;a_n&lt;/math&gt; back into the ratio shows that &lt;math&gt;\frac{9}{10}&lt;\frac{a_{n+1}}{a_{n}}\leq\frac{10}{11}&lt;/math&gt; for all &lt;math&gt;n&lt;/math&gt;. <br /> <br /> <br /> For &lt;math&gt;(n&gt;0)&lt;/math&gt;, we can bound &lt;math&gt;a_{n}&lt;/math&gt; by applying this rule recursively : &lt;math&gt;(\frac{9}{10})^n &lt; a_{n} \leq (\frac{10}{11})^n&lt;/math&gt; <br /> <br /> <br /> <br /> Therefore, &lt;math&gt;a_{n}&lt;/math&gt; is always less than &lt;math&gt;(\frac{1}{2^{20}})&lt;/math&gt; when &lt;math&gt;(\frac{10}{11})^n&lt;\frac{1}{2^{20}}\implies n&gt;20log_{\frac{11}{10}}{2}&lt;/math&gt;<br /> <br /> and &lt;math&gt;a_{n}&lt;/math&gt; is never less than &lt;math&gt;(\frac{1}{2^{20}})&lt;/math&gt; when &lt;math&gt;(\frac{9}{10})^n&gt;\frac{1}{2^{20}}\implies n&lt;20log_{\frac{10}{9}}{2}&lt;/math&gt;<br /> <br /> <br /> <br /> The first integer &lt;math&gt;m&lt;/math&gt; such that &lt;math&gt;a_{m}\leq\frac{1}{2^{20}}&lt;/math&gt; must therefore lie in the interval <br /> &lt;math&gt;\left[\left\lfloor 20log_{\frac{10}{9}}{2}\right\rfloor+1, \left\lceil 20log_{\frac{11}{10}}{2}\right\rceil\right]&lt;/math&gt;<br /> <br /> Both of these can be quickly estimated at c. &lt;math&gt;140&lt;/math&gt;, so the answer must be &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> (actual values are &lt;math&gt;132&lt;/math&gt; and &lt;math&gt;147&lt;/math&gt;)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2019|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=102963 2019 AMC 10B Problems/Problem 23 2019-02-15T21:07:57Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br /> <br /> ==Problem==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was &lt;math&gt;(x, 0)&lt;/math&gt;. Using Pythagorean Theorem gives &lt;math&gt;x=5&lt;/math&gt;.<br /> <br /> Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;(5, 0)&lt;/math&gt; form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br /> <br /> &lt;math&gt;2\sqrt{170}x = d * \sqrt{40}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; represents the distance between circle center and &lt;math&gt;(5, 0)&lt;/math&gt;. Therefore, &lt;math&gt;d = \sqrt{17}x&lt;/math&gt;. Using Pythagorean Theorem on &lt;math&gt;(5, 0)&lt;/math&gt;, either one of &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;, and the circle center, we realize that &lt;math&gt;170 + x^2 = 17x^2&lt;/math&gt;, at which point &lt;math&gt;x^2 = \frac{85}{8}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, follow solution 1 and obtain &lt;math&gt;x=5&lt;/math&gt;. Label the point &lt;math&gt;(5,0)&lt;/math&gt; as point &lt;math&gt;C&lt;/math&gt;. The midpoint &lt;math&gt;M&lt;/math&gt; of segment &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;(9, 12)&lt;/math&gt;. Notice that the center of the circle must lie on the line that goes through the points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt;. Thus, the center of the circle lies on the line &lt;math&gt;y=3x-15&lt;/math&gt;. <br /> <br /> Line &lt;math&gt;AC&lt;/math&gt; is &lt;math&gt;y=13x-65&lt;/math&gt;. The perpendicular line must pass through &lt;math&gt;A(6, 13)&lt;/math&gt; and &lt;math&gt;(x, 3x-15)&lt;/math&gt;. The slope of the perpendicular line is &lt;math&gt;-\frac{1}{13}&lt;/math&gt;. The line is hence &lt;math&gt;y=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. The point &lt;math&gt;(x, 3x-15)&lt;/math&gt; lies on this line. Therefore, &lt;math&gt;3x-15=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. Solving this equation tells us that &lt;math&gt;x=\frac{37}{4}&lt;/math&gt;. So the center of the circle is &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;. The distance between the center, &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;, and point A is &lt;math&gt;\frac{\sqrt{170}}{4}&lt;/math&gt;. Hence, the area is &lt;math&gt;\frac{85}{8}\pi&lt;/math&gt;. The answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC.<br /> The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)<br /> <br /> AC=sqrt((6-5)^2+(13-0)^2)=sqrt(170)<br /> AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10)<br /> DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160)<br /> Therefore OA=AC*AD/DC=sqrt(85/5)<br /> Consequently, the area of the circle is pi*OA^2=pi*85/5<br /> (by Zhen Qin)<br /> (P.S. Will someone please Latex this?)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_22&diff=102962 2019 AMC 10B Problems/Problem 22 2019-02-15T21:07:40Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #22]] and [[2019 AMC 12B Problems|2019 AMC 12B #19]]}}<br /> <br /> ==Problem==<br /> <br /> Raashan, Sylvia, and Ted play the following game. Each starts with &lt;math&gt; \$1&lt;/math&gt;. A bell rings every &lt;math&gt;15&lt;/math&gt; seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives &lt;math&gt;\$1&lt;/math&gt; to that player. What is the probability that after the bell has rung &lt;math&gt;2019&lt;/math&gt; times, each player will have &lt;math&gt;\$1&lt;/math&gt;? (For example, Raashan and Ted may each decide to give &lt;math&gt;\$1&lt;/math&gt; to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have &lt;math&gt;\$0&lt;/math&gt;, Sylvia will have &lt;math&gt;\$2&lt;/math&gt;, and Ted will have &lt;math&gt;\$1&lt;/math&gt;, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their &lt;math&gt; \$1&lt;/math&gt; to, and the holdings will be the same at the end of the second round.)<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> On the first turn, each player starts off with &lt;math&gt;\text{\$1}&lt;/math&gt; each. Each turn after that, there are only two situations possible: either everyone stays at &lt;math&gt;\text{\$1}&lt;/math&gt; &lt;math&gt;\text{(1-1-1)}&lt;/math&gt;, or the distribution of money becomes &lt;math&gt;\text{\$2-\$1-\$0}&lt;/math&gt;, in any order &lt;math&gt;\text{(2-1-0)}&lt;/math&gt;.<br /> <br /> (Note: &lt;math&gt;\text{S-T-R}&lt;/math&gt; means that &lt;math&gt;\text{R}&lt;/math&gt; gives his money to &lt;math&gt;\text{S}&lt;/math&gt;, &lt;math&gt;\text{S}&lt;/math&gt; gives her money to &lt;math&gt;\text{T}&lt;/math&gt;, and &lt;math&gt;\text{T}&lt;/math&gt; gives his money to &lt;math&gt;\text{R}&lt;/math&gt;.)<br /> <br /> From the &lt;math&gt;\text{1-1-1}&lt;/math&gt; state, there are two ways to distribute the money so that it stays in a &lt;math&gt;\text{1-1-1}&lt;/math&gt; state: &lt;math&gt;\text{S-T-R}&lt;/math&gt; and &lt;math&gt;\text{T-R-S}&lt;/math&gt;. There are 6 ways to change the state to &lt;math&gt;\text{2-1-0}&lt;/math&gt;: &lt;math&gt;\text{S-R-R}&lt;/math&gt;, &lt;math&gt;\text{T-R-R}&lt;/math&gt;, &lt;math&gt;\text{S-R-S}&lt;/math&gt;, &lt;math&gt;\text{S-T-S}&lt;/math&gt;, &lt;math&gt;\text{T-T-R}&lt;/math&gt;, and &lt;math&gt;\text{T-T-S}&lt;/math&gt;. This means that the probability that the state stays &lt;math&gt;\text{1-1-1}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{2}{8}=\frac{1}{4}&lt;/math&gt;, and the probability that the state changes to &lt;math&gt;\text{2-1-0}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{6}{8}=\frac{3}{4}&lt;/math&gt;.<br /> <br /> From the &lt;math&gt;\text{2-1-0}&lt;/math&gt; state, there is one way to change the state back to &lt;math&gt;\text{1-1-1}&lt;/math&gt;: &lt;math&gt;\text{S-T-0}&lt;/math&gt;. (We can assume that &lt;math&gt;\text{R}&lt;/math&gt; has &lt;math&gt;\text{\$2}&lt;/math&gt;, &lt;math&gt;\text{S}&lt;/math&gt; has &lt;math&gt;\text{\$1}&lt;/math&gt;, and &lt;math&gt;\text{T}&lt;/math&gt; has &lt;math&gt;\text{\$0}&lt;/math&gt; since only the distribution of money matters, not the specific people.) There are three ways to keep the &lt;math&gt;\text{2-1-0}&lt;/math&gt; state: &lt;math&gt;\text{S-R-0}&lt;/math&gt;, &lt;math&gt;\text{T-R-0}&lt;/math&gt;, &lt;math&gt;\text{T-T-0}&lt;/math&gt;. This means that the probability that the state changes to &lt;math&gt;\text{1-1-1}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{1}{4}&lt;/math&gt;, and the probability that the state stays &lt;math&gt;\text{2-1-0}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{3}{4}&lt;/math&gt;.<br /> <br /> We can see that there will always be a &lt;math&gt;\textstyle\frac{1}{4}&lt;/math&gt; chance that the money is distributed &lt;math&gt;\text{1-1-1}&lt;/math&gt; (as long as the bell rings once), so the answer is &lt;math&gt;\boxed{\textbf{(B) }\frac{1}{4}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_21&diff=102961 2019 AMC 10B Problems/Problem 21 2019-02-15T21:07:24Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> <br /> Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We first want to find out the sequences of coin flips that satisfy this equation. Since Debra sees two tails before two heads, her first flip can't be heads, as that would mean she would either end at tails or see two heads before she sees two tails. Therefore, her first flip must be tails. If you calculate the shortest way she can get two heads in a row and see two tails before she sees two heads, it would be T H T H H, which would be &lt;math&gt;\frac{1}{2^5}&lt;/math&gt;, or &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Following this, she can prolong her coin flipping by adding an extra (T H), which is an extra &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance. Since she can do this indefinitely, this is an infinite geometric sequence, which means the answer is &lt;math&gt;\frac{\frac{1}{32}}{1-\frac{1}{4}}&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(B) }\frac{1}{24}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&diff=102960 2019 AMC 10B Problems/Problem 20 2019-02-15T21:07:10Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}<br /> ==Problem==<br /> As shown in the figure, line segment &lt;math&gt;\overline{AD}&lt;/math&gt; is trisected by points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AB=BC=CD=2.&lt;/math&gt; Three semicircles of radius &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;\overarc{AEB},\overarc{BFC},&lt;/math&gt; and &lt;math&gt;\overarc{CGD},&lt;/math&gt; have their diameters on &lt;math&gt;\overline{AD},&lt;/math&gt; and are tangent to line &lt;math&gt;EG&lt;/math&gt; at &lt;math&gt;E,F,&lt;/math&gt; and &lt;math&gt;G,&lt;/math&gt; respectively. A circle of radius &lt;math&gt;2&lt;/math&gt; has its center on &lt;math&gt;F. &lt;/math&gt; The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form <br /> &lt;cmath&gt;\frac{a}{b}\cdot\pi-\sqrt{c}+d,&lt;/cmath&gt;<br /> where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), grey);<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17&lt;/math&gt;<br /> ==Solution==<br /> Divide the circle into four parts: The top semicircle: (A), the bottom sector with arc length 120 degrees: (B), the triangle formed by the radii of (A) and the chord: (C), and the four parts which are the corners of a circle inscribed in a square (D). The area is just (A) + (B) - (C) + (D).<br /> <br /> Area of (A): &lt;math&gt;2\pi&lt;/math&gt;<br /> <br /> Area of (B): &lt;math&gt;\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is &lt;math&gt;2\sqrt{3}*1/2=\sqrt{3}&lt;/math&gt;<br /> <br /> Area of (D): &lt;math&gt;4*1-1/4*\pi*4=4-\pi&lt;/math&gt;<br /> <br /> Total sum: &lt;math&gt;\frac{7\pi}{3}-\sqrt{3}+4&lt;/math&gt;<br /> <br /> &lt;math&gt;7+3+3+4=\boxed{17}&lt;/math&gt;<br /> <br /> For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=102959 2019 AMC 10B Problems/Problem 19 2019-02-15T21:06:54Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br /> <br /> ==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all positive integer divisors of &lt;math&gt;100,000.&lt;/math&gt; How many numbers are the product of two distinct elements of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> To find the number of numbers that are the product of two distinct elements of &lt;math&gt;S&lt;/math&gt;, we first square &lt;math&gt;100,000&lt;/math&gt; and factor it. Factoring, we find &lt;math&gt;100,000^2 = 2^{10} \cdot 5^{10}&lt;/math&gt;. Therefore, &lt;math&gt;100,000^2&lt;/math&gt; has &lt;math&gt;(10 + 1)(10 + 1) = 121&lt;/math&gt; distinct factors. Each of these can be achieved by multiplying two factors of &lt;math&gt;S&lt;/math&gt;. However, the factors must be distinct, so we eliminate &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100,000^2&lt;/math&gt;, as well as &lt;math&gt;2^{10}&lt;/math&gt; and &lt;math&gt;5^{10}&lt;/math&gt;, so the answer is &lt;math&gt;121 - 4 = 117&lt;/math&gt;. &lt;math&gt;2^{10}&lt;/math&gt; and &lt;math&gt;5^{10}&lt;/math&gt; do not work since the factors chosen must be distinct, and those require &lt;math&gt;2^5 \cdot 2^5&lt;/math&gt; or &lt;math&gt;5^5 \cdot 5^5&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> The prime factorization of 100,000 is &lt;math&gt;2^5 \cdot 5^5&lt;/math&gt;. Thus, we choose two numbers &lt;math&gt;2^a5^b&lt;/math&gt; and &lt;math&gt;2^c5^d&lt;/math&gt; where &lt;math&gt;0 \le a,b,c,d \le 5&lt;/math&gt; and &lt;math&gt;(a,b) \neq (c,d)&lt;/math&gt;, whose product is &lt;math&gt;2^{a+c}5^{b+d}&lt;/math&gt;, where &lt;math&gt;0 \le a+c \le 10&lt;/math&gt; and &lt;math&gt;0 \le b+d \le 10&lt;/math&gt;.<br /> <br /> Consider &lt;math&gt;100000^2 = 2^{10}5^{10}&lt;/math&gt;. The number of divisors is &lt;math&gt;(10+1)(10+1) = 121&lt;/math&gt;. However, some of the divisors of &lt;math&gt;2^{10}5^{10}&lt;/math&gt; cannot be written as a product of two distinct divisors of &lt;math&gt;2^5 \cdot 5^5&lt;/math&gt;, namely: &lt;math&gt;1 = 2^05^0&lt;/math&gt;, &lt;math&gt;2^{10}5^{10}&lt;/math&gt;, &lt;math&gt;2^{10}&lt;/math&gt;, and &lt;math&gt;5^{10}&lt;/math&gt;. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only &lt;math&gt;2^5&lt;/math&gt; or &lt;math&gt;5^5&lt;/math&gt;. Since the factors chosen must be distinct, the last two numbers cannot be created because those require &lt;math&gt;2^5 \cdot 2^5&lt;/math&gt; or &lt;math&gt;5^5 \cdot 5^5&lt;/math&gt;. This gives &lt;math&gt;121-4 = 117&lt;/math&gt; candidate numbers. It is not too hard to show that every number of the form &lt;math&gt;2^p5^q&lt;/math&gt; where &lt;math&gt;0 \le p, q \le 10&lt;/math&gt;, and &lt;math&gt;p,q&lt;/math&gt; are not both 0 or 10, can be written as a product of two distinct elements in &lt;math&gt;S&lt;/math&gt;. Hence the answer is &lt;math&gt;\boxed{\textbf{(C) } 117}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=102958 2019 AMC 12B Problems/Problem 8 2019-02-15T21:05:55Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;f(x) = x^{2}(1-x)^{2}&lt;/math&gt;. What is the value of the sum<br /> &lt;math&gt;f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots &lt;/math&gt;<br /> <br /> &lt;math&gt;+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that &lt;math&gt;f(x) = f(1-x)&lt;/math&gt;. We can see from this that the terms cancel and the answer is &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can first plug in a few numbers and see what happens. We get &lt;math&gt;(\frac{1}{2019})^2(\frac{2018}{2019})^2 + \frac{2}{2019})^2(\frac{2017}{2019})^2&lt;/math&gt; and so on. Then, we can skip to the end and see that the last term and the first term are equal, and cancel each other out because they have different signs. Therefore we see that every number cancels out. It might seem that there is some term in the middle, but if we use a smaller example to check, we see that that is not the case. Therefore, the answer is &lt;math&gt;\boxed{\text{(A) 0}}&lt;/math&gt;<br /> <br /> -- clara32356 (Claire)<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_13&diff=102957 2019 AMC 10B Problems/Problem 13 2019-02-15T21:05:41Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #13]] and [[2019 AMC 12B Problems|2019 AMC 12B #7]]}}<br /> <br /> ==Problem==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which the median of the numbers &lt;math&gt;4,6,8,17,&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; is equal to the mean of those five numbers?<br /> <br /> &lt;math&gt;\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are &lt;math&gt;3&lt;/math&gt; cases: &lt;math&gt;6&lt;/math&gt; is the median, &lt;math&gt;8&lt;/math&gt; is the median, and &lt;math&gt;x&lt;/math&gt; is the median. In all cases, the mean is &lt;math&gt;7+\frac{x}{5}&lt;/math&gt;.&lt;br&gt;<br /> For case 1, &lt;math&gt;x=-5&lt;/math&gt;. This allows 6 to be the median because the set is &lt;math&gt;-5,4,6,8,17&lt;/math&gt;.&lt;br&gt;<br /> For case 2, &lt;math&gt;x=5&lt;/math&gt;. This is impossible because the set is &lt;math&gt;4,5,6,8,17&lt;/math&gt;.&lt;br&gt;<br /> For case 3, &lt;math&gt;x=\frac{35}{4}&lt;/math&gt;. This is impossible because the set is &lt;math&gt;4,6,8,\frac{35}{4},17&lt;/math&gt;.&lt;br&gt;<br /> Only case 1 yields a solution, &lt;math&gt;x=-5&lt;/math&gt;, so the answer is &lt;math&gt;\textbf{(A) } -5&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> The mean is &lt;math&gt;\frac{4+6+8+17+x}{5}=\frac{35+x}{5}&lt;/math&gt;.<br /> <br /> There are 3 possibilities: either the median is 6, 8, or x.<br /> <br /> Let's start with 6.<br /> <br /> &lt;math&gt;\frac{35+x}{5}=6&lt;/math&gt; when &lt;math&gt;x=-5&lt;/math&gt; and the sequence is -5, 4, 6, 8, 17 which has 6 as the median so we're good.<br /> <br /> Now let the mean=8<br /> <br /> &lt;math&gt;\frac{35+x}{5}=8&lt;/math&gt; when &lt;math&gt;x=5&lt;/math&gt; and the sequence is 4, 5, 6, 8, 17 which has median 6 so no go.<br /> <br /> Finally we let the mean=x<br /> <br /> &lt;math&gt;\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75.&lt;/math&gt; and the sequence is 4, 6, 8, 8.75, 17 which has median 8 so no go.<br /> <br /> So the only option for x is &lt;math&gt;\boxed{-5}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2019|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_12&diff=102956 2019 AMC 10B Problems/Problem 12 2019-02-15T21:05:28Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> <br /> What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than &lt;math&gt;2019&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 11<br /> \qquad\textbf{(B) } 14<br /> \qquad\textbf{(C) } 22<br /> \qquad\textbf{(D) } 23<br /> \qquad\textbf{(E) } 27&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Convert &lt;math&gt;2019&lt;/math&gt; to base &lt;math&gt;7&lt;/math&gt;. This will get you &lt;math&gt;5613_7&lt;/math&gt;, which will be the upper bound. To maximize the sum of the digits, we want as many &lt;math&gt;6&lt;/math&gt;s as possible (which is the highest value in base &lt;math&gt;7&lt;/math&gt;), and this would be the number &lt;math&gt;4666_7&lt;/math&gt;. Thus, the answer is &lt;math&gt;4+6+6+6 = \boxed{\textbf{(C) }22}&lt;/math&gt;<br /> <br /> Note: the number can also be &lt;math&gt;5566_7&lt;/math&gt;, which will also give the answer of &lt;math&gt;22&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that all base 7 numbers with 5 digits or more is greater than 2019. Since the first answer that is possible using a 4 digit number is 23, we start with the smallest base 7 number that digits adds up to 23, 5666. 5666 in base 10 is greater than 2017, so we continue with trying 4666, which is less than 2019. So the answer is &lt;math&gt;\boxed{\textbf{(C) }22}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=102955 2019 AMC 10B Problems/Problem 11 2019-02-15T21:05:13Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> <br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?<br /> <br /> &lt;math&gt;\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50&lt;/math&gt;<br /> <br /> ==Solution==<br /> Call the amount of marbles in each jar &lt;math&gt;x&lt;/math&gt;, because they are equivalent. Thus, &lt;math&gt;\frac{x}{10}&lt;/math&gt; is the amount of green marbles in &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{x}{9}&lt;/math&gt; is the amount of green marbles in &lt;math&gt;2&lt;/math&gt;. &lt;math&gt;\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}&lt;/math&gt;, &lt;math&gt;\frac{19x}{90}=95&lt;/math&gt;, and &lt;math&gt;x=450&lt;/math&gt; marbles in each jar. Because the &lt;math&gt;\frac{9x}{10}&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;\frac{8x}{9}&lt;/math&gt; is the amount of blue marbles in jar &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90}&lt;/math&gt;, so there must be &lt;math&gt;5&lt;/math&gt; more marbles in jar &lt;math&gt;1&lt;/math&gt; than jar &lt;math&gt;2&lt;/math&gt;. The answer is &lt;math&gt;\boxed{A}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_10&diff=102954 2019 AMC 10B Problems/Problem 10 2019-02-15T21:04:44Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #10]] and [[2019 AMC 12B Problems|2019 AMC 12B #6]]}}<br /> <br /> ==Problem==<br /> <br /> In a given plane, points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;10&lt;/math&gt; units apart. How many points &lt;math&gt;C&lt;/math&gt; are there in the plane such that the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; units and the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Notice that whatever point we pick for C, AB will be the base of the triangle. WLOG, points A and B are (0,0) and (0,10) [notice that for any other combination of points, we can just rotate the plane to be the same thing]. When we pick point C, we have to make sure the y value of C is 20, because that's the only way the area of the triangle can be 100. <br /> <br /> We figure that the one thing we need to test to see if there is such a triangle is when the perimeter is minimized, and the value of C is (x, 20). Thus, we put C in the middle, so point C is (5, 20). We can easily see that AC and BC will both be &lt;math&gt;\sqrt{20^2+5^2} \Rightarrow \sqrt{425}&lt;/math&gt;. The perimeter of this minimized triangle is &lt;math&gt;2\sqrt{425} + 10&lt;/math&gt;, which is larger than 50. Since the minimized perimeter is greater than 50, there is no triangle that satisfies the condition, giving us &lt;math&gt;\boxed{A) 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> WLOG let &lt;math&gt;AB&lt;/math&gt; be a horizontal segment of length 10. Now, realize &lt;math&gt;C&lt;/math&gt; has to lie on one of the lines parallel to &lt;math&gt;AB&lt;/math&gt; and vertically 20 units away from it. But 10+20+20 is already 50, and this doesn't form a triangle. Otherwise, WLOG &lt;math&gt;AC&lt;20&lt;/math&gt;. Dropping altitude &lt;math&gt;CD&lt;/math&gt; we have a right triangle &lt;math&gt;ACD&lt;/math&gt; with hypotenuse &lt;math&gt;AC&lt;20&lt;/math&gt; and leg &lt;math&gt;CD=20&lt;/math&gt; which is clearly impossible, giving us &lt;math&gt;\boxed{A) 0}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=9|num-a=11}}<br /> {{AMC12 box|year=2019|ab=B|num-b=5|num-a=7}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_9&diff=102953 2019 AMC 10B Problems/Problem 9 2019-02-15T21:04:30Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> <br /> The function &lt;math&gt;f&lt;/math&gt; is defined by &lt;cmath&gt;f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|&lt;/cmath&gt;for all real numbers &lt;math&gt;x&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to the real number &lt;math&gt;r&lt;/math&gt;. What is the range of &lt;math&gt;f&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\} \qquad\textbf{(E) } \text{The set of nonnegative integers} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> There are 4 cases we need to test here:<br /> <br /> Case 1: x is a positive integer. WLOG, assume x=1. Then f(1) = 1 - 1 = &lt;math&gt;0&lt;/math&gt;.<br /> <br /> Case 2: x is a positive fraction. WLOG, assume x=0.5. Then f(0.5) = 0 - 0 = &lt;math&gt;0&lt;/math&gt;.<br /> <br /> Case 3: x is a negative integer. WLOG, assume x=-1. Then f(-1) = 1 - 1 = &lt;math&gt;0&lt;/math&gt;.<br /> <br /> Case 4: x is a negative fraction. WLOG, assume x=-0.5. Then f(-0.5) = 0 - 1 = &lt;math&gt;-1&lt;/math&gt;.<br /> <br /> Thus the range of function f is &lt;math&gt;\boxed{\textbf{(A) } \{-1, 0\}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> It is easily verified that when &lt;math&gt;x&lt;/math&gt; is an integer, then &lt;math&gt;f(x)&lt;/math&gt; is zero. We need only to consider the case when &lt;math&gt;x&lt;/math&gt; is not.<br /> <br /> When &lt;math&gt;x&lt;/math&gt; is a positive number, &lt;math&gt;\lfloor x\rfloor \geq 0&lt;/math&gt;, so <br /> &lt;cmath&gt;f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|&lt;/cmath&gt;<br /> &lt;cmath&gt;=\lfloor x\rfloor-\lfloor x\rfloor&lt;/cmath&gt;<br /> &lt;cmath&gt;=\textbf{0}&lt;/cmath&gt;<br /> <br /> When &lt;math&gt;x&lt;/math&gt; is a negative number, let &lt;math&gt;x=-a-b&lt;/math&gt; be composed of integer part &lt;math&gt;a&lt;/math&gt; and decimal part &lt;math&gt;b&lt;/math&gt; (both &lt;math&gt;\geq 0&lt;/math&gt;):<br /> &lt;cmath&gt;f(x)=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor|&lt;/cmath&gt;<br /> &lt;cmath&gt;=\lfloor a+b\rfloor-|-a-1|&lt;/cmath&gt;<br /> &lt;cmath&gt;=a-(a+1)=\textbf{-1}&lt;/cmath&gt;<br /> <br /> Thus, the range of f is &lt;math&gt;\boxed{\textbf{(A) } \{-1, 0\}}&lt;/math&gt;<br /> <br /> <br /> Note: One could solve the case of &lt;math&gt;x&lt;/math&gt; as a negative non-integer this way:<br /> &lt;cmath&gt;f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|&lt;/cmath&gt;<br /> &lt;cmath&gt;=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1|&lt;/cmath&gt;<br /> &lt;cmath&gt;=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = \textbf{-1}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_8&diff=102952 2019 AMC 10B Problems/Problem 8 2019-02-15T21:04:15Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region? <br /> <br /> &lt;math&gt;(A)&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; &lt;math&gt;(B)&lt;/math&gt; &lt;math&gt;12 - 4\sqrt{3}&lt;/math&gt; &lt;math&gt;(C)&lt;/math&gt; &lt;math&gt;3\sqrt{3}&lt;/math&gt; &lt;math&gt;(D)&lt;/math&gt; &lt;math&gt;4\sqrt{3}&lt;/math&gt; &lt;math&gt;(E)&lt;/math&gt; &lt;math&gt;16 - \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We notice that the square can be split into &lt;math&gt;4&lt;/math&gt; congruent smaller squares with the altitude of the equilateral triangle being the side of the square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (Note that it has already been split in half). <br /> <br /> When we split an equilateral triangle in half, we get &lt;math&gt;2&lt;/math&gt; triangles with a &lt;math&gt;30-60-90&lt;/math&gt; relationship. Therefore, we get that the altitude and a side length of a square is &lt;math&gt;\sqrt{3}&lt;/math&gt;. <br /> <br /> We can then compute the area of the two triangles using the base-height-area relationship and get &lt;math&gt;\frac{2 \cdot 1 \cdot \sqrt{3}}{2} = \sqrt{3}&lt;/math&gt;.<br /> <br /> The area of the small squares is the altitude squared which is &lt;math&gt;(\sqrt{3})^2 = 3&lt;/math&gt;. Therefore, the area of the shaded region in each of the four squares is &lt;math&gt;3 - \sqrt{3}&lt;/math&gt;<br /> <br /> Since there are four of these squares, we multiply this by &lt;math&gt;4&lt;/math&gt; to get &lt;math&gt;4(3 - \sqrt{3}) = 12 - 4 \sqrt{3}&lt;/math&gt; as our answer. This is choice &lt;math&gt;\boxed{ B) 12 - 4 \sqrt{3}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_1&diff=102951 2019 AMC 10B Problems/Problem 1 2019-02-15T21:03:53Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #1]] and [[2019 AMC 12B Problems|2019 AMC 12B #1]]}}<br /> <br /> ==Problem==<br /> <br /> Alicia had two containers. The first was &lt;math&gt;\tfrac{5}{6}&lt;/math&gt; full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; full of water. What is the ratio of the volume of the first container to the volume of the second container?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the first jar's volume be &lt;math&gt;A&lt;/math&gt; and the second's be &lt;math&gt;B&lt;/math&gt;. It is given that &lt;math&gt;\frac{3}{4}A=\frac{5}{6}B&lt;/math&gt;. We find that &lt;math&gt;\frac{B}{A}=\frac{3/4}{5/6}=\boxed{\frac{9}{10}}.&lt;/math&gt; <br /> <br /> We already know that this is the ratio of smaller to larger volume because it is less than &lt;math&gt;1.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can set up a ratio to solve this problem. If x is the volume of the first container, and y is the volume of the second container, then:<br /> &lt;cmath&gt;\frac{5}{6}x = \frac{3}{4}y&lt;/cmath&gt;<br /> <br /> Cross Multiplying allows us to get &lt;math&gt;\frac{x}{y} = \frac{3}{4} \cdot \frac{6}{5} \Rightarrow \frac{18}{20} \Rightarrow \frac{9}{10}&lt;/math&gt;. Thus the ratio of the volume of the first container to the second container is &lt;math&gt;\boxed{(\text{D})\frac{9}{10}}&lt;/math&gt;<br /> <br /> An alternate solution is to plug in some maximum volume for the first container - let's say &lt;math&gt;72&lt;/math&gt;, so there was a volume of 60 in the first container, and then the second container also has a volume of &lt;math&gt;60&lt;/math&gt;, so you get &lt;math&gt;60 \cdot \frac{4}{3} \Rightarrow 80&lt;/math&gt;. Thus, &lt;math&gt;\frac{72}{80} \Rightarrow \frac{9}{10}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|before=First Problem|num-a=2}}<br /> {{AMC12 box|year=2019|ab=B|before=First Problem|num-a=2}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_3&diff=102950 2019 AMC 12B Problems/Problem 3 2019-02-15T21:03:24Z <p>Bgav: </p> <hr /> <div>==Problem==<br /> Which of the following rigid transformations (isometries) maps the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; onto the line segment &lt;math&gt;\overline{A'B'}&lt;/math&gt; so that the image of &lt;math&gt;A(-2, 1)&lt;/math&gt; is &lt;math&gt;A'(2, -1)&lt;/math&gt; and the image of &lt;math&gt;B(-1, 4)&lt;/math&gt; is &lt;math&gt;B'(1, -4)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } &lt;/math&gt; reflection in the &lt;math&gt;y&lt;/math&gt;-axis<br /> <br /> &lt;math&gt;\textbf{(B) } &lt;/math&gt; counterclockwise rotation around the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(C) } &lt;/math&gt; translation by 3 units to the right and 5 units down<br /> <br /> &lt;math&gt;\textbf{(D) } &lt;/math&gt; reflection in the &lt;math&gt;x&lt;/math&gt;-axis<br /> <br /> &lt;math&gt;\textbf{(E) } &lt;/math&gt; clockwise rotation about the origin by &lt;math&gt;180^{\circ}&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can simply graph or use coordinate rules to realize that both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are rotated &lt;math&gt;180^{\circ}&lt;/math&gt; about the origin, therefore &lt;math&gt;\overline{AB}&lt;/math&gt; is rotated &lt;math&gt;180^{\circ}&lt;/math&gt;, so &lt;math&gt;\boxed{(\text{E})}&lt;/math&gt; -Dodgers66<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_6&diff=102948 2019 AMC 10B Problems/Problem 6 2019-02-15T21:03:09Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #6]] and [[2019 AMC 12B Problems|2019 AMC 12B #4]]}}<br /> <br /> ==Problem==<br /> <br /> There is a real &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;(n+1)! + (n+2)! = n! \cdot 440&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;cmath&gt;(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!&lt;/cmath&gt;<br /> &lt;cmath&gt;n![n+1 + (n+2)(n+1)] = 440 \cdot n!&lt;/cmath&gt;<br /> &lt;cmath&gt;n + 1 + n^2 + 3n + 2 = 440&lt;/cmath&gt;<br /> &lt;cmath&gt;n^2 + 4n - 437 = 0&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 19&lt;/math&gt;. &lt;math&gt;1 + 9 = \boxed{\textbf{(C) }10}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Dividing both sides by &lt;math&gt;n!&lt;/math&gt; gives<br /> &lt;cmath&gt;(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.&lt;/cmath&gt;<br /> Since &lt;math&gt;n&lt;/math&gt; is positive, &lt;math&gt;n=19&lt;/math&gt;. The answer is &lt;math&gt;1 + 9 = \boxed{\textbf{(C) }10}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Divide both sides by &lt;math&gt;n!&lt;/math&gt;:<br /> <br /> <br /> &lt;math&gt;(n+1)+(n+1)(n+2)=440&lt;/math&gt;<br /> <br /> factor out &lt;math&gt;(n+1)&lt;/math&gt;:<br /> <br /> &lt;math&gt;(n+1)*(n+3)=440&lt;/math&gt;<br /> <br /> <br /> prime factorization of &lt;math&gt;440&lt;/math&gt; and a bit of experimentation gives us &lt;math&gt;n+1=20&lt;/math&gt; and &lt;math&gt;n+3=22&lt;/math&gt;, so &lt;math&gt;n=19&lt;/math&gt;, so the answer is &lt;math&gt;1 + 9 = \boxed{\textbf{(C) }10}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Obviously n must be very close to &lt;math&gt;\sqrt{440}&lt;/math&gt;. By quick inspection, &lt;math&gt;n = 19&lt;/math&gt; works. &lt;math&gt;1 + 9 = \boxed{\textbf{(C) }10}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_7&diff=102947 2019 AMC 10B Problems/Problem 7 2019-02-15T21:02:52Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #7]] and [[2019 AMC 12B Problems|2019 AMC 12B #5]]}}<br /> <br /> ==Problem==<br /> <br /> Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or &lt;math&gt;n&lt;/math&gt; pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> If he has enough money to buy 12 pieces of red candy, 14 pieces of green candy, and 15 pieces of blue candy, then the least money he can have is &lt;math&gt;lcm(12,14,15)&lt;/math&gt; = 420. Since a piece of purple candy costs 20 cents, the least value of n can be &lt;math&gt;\frac{420}{20} \implies \boxed{\textbf{(B)} 21}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We simply need to find a value of &lt;math&gt;20n&lt;/math&gt; that divides 12, 14, and 15. &lt;math&gt;20*18&lt;/math&gt; divides 12 and 15, but not 14. &lt;math&gt;20*21&lt;/math&gt; successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy, so the minimum value of &lt;math&gt;\boxed{n = 21}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> This problem is equivalent to finding the LCM of 12, 14, 15, and 20 (and then dividing it by 20). It is easy to see that the prime factorization of said LCM must be &lt;math&gt;7 \cdot 3 \cdot 5 \cdot 2^2&lt;/math&gt;. We can divide by 20 now, before we ever multiply it out, leaving us with &lt;math&gt;7 \cdot 3 = 21 = \boxed{B}&lt;/math&gt;. However, in this case multiplying it out nets us &lt;math&gt;420&lt;/math&gt;, which is worth the time it takes all on its own.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=6|num-a=8}}<br /> {{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_2&diff=102945 2019 AMC 10B Problems/Problem 2 2019-02-15T21:01:48Z <p>Bgav: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #2]] and [[2019 AMC 12B Problems|2019 AMC 12B #2]]}}<br /> <br /> ==Problem==<br /> <br /> Consider the statement, &quot;If &lt;math&gt;n&lt;/math&gt; is not prime, then &lt;math&gt;n-2&lt;/math&gt; is prime.&quot; Which of the following values of &lt;math&gt;n&lt;/math&gt; is a counterexample to this statement?<br /> <br /> &lt;math&gt;\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since a counterexample must be when n is not prime, n must be composite, so we eliminate A and C. Now we subtract 2 from the remaining answer choices, and we see that the only time &lt;math&gt;n-2&lt;/math&gt; is &lt;math&gt;\textbf{not}&lt;/math&gt; prime is when &lt;math&gt;n = 27&lt;/math&gt;, which is &lt;math&gt;\fbox {E}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=1|num-a=3}}<br /> {{AMC12 box|year=2019|ab=B|num-b=1|num-a=3}}<br /> {{MAA Notice}}<br /> SUB2PEWDS</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_25&diff=102096 2020 AMC 10A Problems/Problem 25 2019-02-14T08:47:26Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&diff=102092 2019 AMC 10B Problems/Problem 20 2019-02-14T08:45:46Z <p>Bgav: </p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems&diff=102065 2019 AMC 10B Problems 2019-02-14T05:23:16Z <p>Bgav: </p> <hr /> <div>T-series</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A&diff=102059 2020 AMC 10A 2019-02-14T05:05:27Z <p>Bgav: </p> <hr /> <div>SUBSCRIBE 2 PEWDIEPIE</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=102058 2019 AMC 10B Problems/Problem 25 2019-02-14T05:04:18Z <p>Bgav: </p> <hr /> <div>sub2pewds if u think this test was gg2ez</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_1&diff=102057 2019 AMC 10B Problems/Problem 1 2019-02-14T04:58:32Z <p>Bgav: </p> <hr /> <div>you feel like you're gonna have a bad time if u no sub to pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems&diff=102056 2019 AMC 10B Problems 2019-02-14T04:56:21Z <p>Bgav: </p> <hr /> <div>T-series</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Answer_Key&diff=102054 2019 AMC 10B Answer Key 2019-02-14T04:55:37Z <p>Bgav: </p> <hr /> <div>Subbing to pewdiepie is the only answer</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_22&diff=102051 2019 AMC 10B Problems/Problem 22 2019-02-14T04:52:57Z <p>Bgav: </p> <hr /> <div> totient<br /> <br /> sub2pewds pls this took a good bit of time</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_24&diff=102050 2019 AMC 10B Problems/Problem 24 2019-02-14T04:49:50Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=102049 2019 AMC 10B Problems/Problem 23 2019-02-14T04:49:34Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=102047 2019 AMC 10B Problems/Problem 19 2019-02-14T04:48:29Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_18&diff=102046 2019 AMC 10B Problems/Problem 18 2019-02-14T04:48:06Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_17&diff=102045 2019 AMC 10B Problems/Problem 17 2019-02-14T04:47:51Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=102044 2019 AMC 10B Problems/Problem 16 2019-02-14T04:47:39Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_15&diff=102043 2019 AMC 10B Problems/Problem 15 2019-02-14T04:47:19Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_14&diff=102042 2019 AMC 10B Problems/Problem 14 2019-02-14T04:47:06Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_13&diff=102040 2019 AMC 10B Problems/Problem 13 2019-02-14T04:46:50Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_12&diff=102039 2019 AMC 10B Problems/Problem 12 2019-02-14T04:46:25Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=102038 2019 AMC 10B Problems/Problem 11 2019-02-14T04:46:13Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_10&diff=102037 2019 AMC 10B Problems/Problem 10 2019-02-14T04:45:58Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_9&diff=102036 2019 AMC 10B Problems/Problem 9 2019-02-14T04:45:35Z <p>Bgav: Created page with &quot;sub2pewds&quot;</p> <hr /> <div>sub2pewds</div> Bgav