https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Kj2002&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-23T15:31:01Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=82099 2012 AMC 12B Problems/Problem 17 2016-12-31T19:20:50Z <p>Kj2002: /* Solutions */</p> <hr /> <div>==Problem==<br /> <br /> Square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;(3,0), (5,0), (7,0),&lt;/math&gt; and &lt;math&gt;(13,0)&lt;/math&gt; lie on lines &lt;math&gt;SP, RQ, PQ&lt;/math&gt;, and &lt;math&gt;SR&lt;/math&gt;, respectively. What is the sum of the coordinates of the center of the square &lt;math&gt;PQRS&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> &lt;asy&gt; size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);&lt;/asy&gt;<br /> <br /> (diagram by [i] MSTang [/i])<br /> ===Solution 1===<br /> <br /> &lt;asy&gt; size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);&lt;/asy&gt;<br /> <br /> Let the four points be labeled &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, &lt;math&gt;P_3&lt;/math&gt;, and &lt;math&gt;P_4&lt;/math&gt;, respectively. Let the lines that go through each point be labeled &lt;math&gt;L_1&lt;/math&gt;, &lt;math&gt;L_2&lt;/math&gt;, &lt;math&gt;L_3&lt;/math&gt;, and &lt;math&gt;L_4&lt;/math&gt;, respectively. Since &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; go through &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt;, respectively, and &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; are opposite sides of the square, we can say that &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; are parallel with slope &lt;math&gt;m&lt;/math&gt;. Similarly, &lt;math&gt;L_3&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; have slope &lt;math&gt;-\frac{1}{m}&lt;/math&gt;. Also, note that since square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant, &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: &lt;math&gt;L_1: y = m(x-3)&lt;/math&gt;, &lt;math&gt;L_2: y = m(x-5)&lt;/math&gt;, &lt;math&gt;L_3: y = -\frac{1}{m}(x-7)&lt;/math&gt;, &lt;math&gt;L_4: y = -\frac{1}{m}(x-13)&lt;/math&gt;.<br /> <br /> <br /> Since &lt;math&gt;PQRS&lt;/math&gt; is a square, it follows that &lt;math&gt;\Delta x&lt;/math&gt; between points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is equal to &lt;math&gt;\Delta y&lt;/math&gt; between points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. Our approach will be to find &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt; in terms of &lt;math&gt;m&lt;/math&gt; and equate the two to solve for &lt;math&gt;m&lt;/math&gt;. &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;P&lt;/math&gt;. Setting the equations for &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; equal to each other and solving for &lt;math&gt;x&lt;/math&gt;, we find that they intersect at &lt;math&gt;x = \frac{3m^2 + 7}{m^2 + 1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_3&lt;/math&gt; intersect at point &lt;math&gt;Q&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;x&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is found to be &lt;math&gt;x = \frac{5m^2 + 7}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta x = \frac{2m^2}{m^2 + 1}&lt;/math&gt;. Substituting the &lt;math&gt;x&lt;/math&gt;-coordinate for point &lt;math&gt;Q&lt;/math&gt; found above into the equation for &lt;math&gt;L_2&lt;/math&gt;, we find that the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;y = \frac{2m}{m^2+1}&lt;/math&gt;. &lt;math&gt;L_2&lt;/math&gt; and &lt;math&gt;L_4&lt;/math&gt; intersect at point &lt;math&gt;R&lt;/math&gt;. Intersecting the two equations, the &lt;math&gt;y&lt;/math&gt;-coordinate of point &lt;math&gt;R&lt;/math&gt; is found to be &lt;math&gt;y = \frac{8m}{m^2 + 1}&lt;/math&gt;. Subtracting the two, we get &lt;math&gt;\Delta y = \frac{6m}{m^2 + 1}&lt;/math&gt;. Equating &lt;math&gt;\Delta x&lt;/math&gt; and &lt;math&gt;\Delta y&lt;/math&gt;, we get &lt;math&gt;2m^2 = 6m&lt;/math&gt; which gives us &lt;math&gt;m = 3&lt;/math&gt;. Finally, note that the line which goes though the midpoint of &lt;math&gt;P_1&lt;/math&gt; and &lt;math&gt;P_2&lt;/math&gt; with slope &lt;math&gt;3&lt;/math&gt; and the line which goes through the midpoint of &lt;math&gt;P_3&lt;/math&gt; and &lt;math&gt;P_4&lt;/math&gt; with slope &lt;math&gt;-\frac{1}{3}&lt;/math&gt; must intersect at at the center of the square. The equation of the line going through &lt;math&gt;(4,0)&lt;/math&gt; is given by &lt;math&gt;y = 3(x-4)&lt;/math&gt; and the equation of the line going through &lt;math&gt;(10,0)&lt;/math&gt; is &lt;math&gt;y = -\frac{1}{3}(x-10)&lt;/math&gt;. Equating the two, we find that they intersect at &lt;math&gt;(4.6, 1.8)&lt;/math&gt;. Adding the &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;-coordinates, we get &lt;math&gt;6.4&lt;/math&gt;. Thus, answer choice &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt; is correct.<br /> <br /> ===Solution 2===<br /> <br /> Note that the center of the square lies along a line that has an &lt;math&gt;x-&lt;/math&gt;intercept of &lt;math&gt;\frac{3+5}{2}=4&lt;/math&gt;, and also along another line with &lt;math&gt;x-&lt;/math&gt;intercept &lt;math&gt;\frac{7+13}{2}=10&lt;/math&gt;. Since these 2 lines are parallel to the sides of the square, they are perpendicular (since the sides of a square are). Let &lt;math&gt;m&lt;/math&gt; be the slope of the first line. Then &lt;math&gt;-\frac{1}{m}&lt;/math&gt; is the slope of the second line. We may use the point-slope form for the equation of a line to write &lt;math&gt;l_1:y=m(x-4)&lt;/math&gt; and &lt;math&gt;l_2:y=-\frac{1}{m}(x-10)&lt;/math&gt;. We easily calculate the intersection of these lines using substitution or elimination to obtain &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; as the center or the square. Let &lt;math&gt;\theta&lt;/math&gt; denote the (acute) angle formed by &lt;math&gt;l_1&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis. Note that &lt;math&gt;\tan\theta=m&lt;/math&gt;. Let &lt;math&gt;s&lt;/math&gt; denote the side length of the square. Then &lt;math&gt;\sin\theta=s/2&lt;/math&gt;. On the other hand the acute angle formed by &lt;math&gt;l_2&lt;/math&gt; and the &lt;math&gt;x-&lt;/math&gt;axis is &lt;math&gt;90-\theta&lt;/math&gt; so that &lt;math&gt;\cos\theta=\sin(90-\theta)=s/6&lt;/math&gt;. Then &lt;math&gt;m=\tan\theta=3&lt;/math&gt;. Substituting into &lt;math&gt;\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)&lt;/math&gt; we obtain &lt;math&gt;\left(\frac{23}{5},\frac{9}{5}\right)&lt;/math&gt; so that the sum of the coordinates is &lt;math&gt;\frac{32}{5}=6.4&lt;/math&gt;. Hence the answer is &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ===Solution 3 (Fast)===<br /> Suppose<br /> <br /> &lt;cmath&gt;SP: y=m(x-3)&lt;/cmath&gt;<br /> &lt;cmath&gt;RQ: y=m(x-5)&lt;/cmath&gt;<br /> &lt;cmath&gt;PQ: -my=x-7&lt;/cmath&gt;<br /> &lt;cmath&gt;SR: -my=x-13&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;m &gt;0&lt;/math&gt;.<br /> <br /> Recall that the distance between two parallel lines &lt;math&gt;Ax+By+C=0&lt;/math&gt; and &lt;math&gt;Ax+By+C_1=0&lt;/math&gt; is &lt;math&gt;|C-C_1|/\sqrt{A^2+B^2}&lt;/math&gt;, we have distance between &lt;math&gt;SP&lt;/math&gt; and &lt;math&gt;RQ&lt;/math&gt; equals to &lt;math&gt;2m/\sqrt{1+m^2}&lt;/math&gt;, and the distance between &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;SR&lt;/math&gt; equals to &lt;math&gt;6/\sqrt{1+m^2}&lt;/math&gt;. Equating them, we get &lt;math&gt;m=3&lt;/math&gt;.<br /> <br /> Then, the center of the square is just the intersection between the following two &quot;mid&quot; lines:<br /> <br /> &lt;cmath&gt;L_1: y=3(x-4)&lt;/cmath&gt;<br /> &lt;cmath&gt;L_2: -3y = x-10&lt;/cmath&gt;<br /> <br /> The solution is &lt;math&gt;(4.6,1.8)&lt;/math&gt;, so we get the answer &lt;math&gt;4.6+1.8=6.4&lt;/math&gt;. &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_14&diff=81733 2014 AMC 12A Problems/Problem 14 2016-11-30T00:18:20Z <p>Kj2002: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;b&lt;c&lt;/math&gt; be three integers such that &lt;math&gt;a,b,c&lt;/math&gt; is an arithmetic progression and &lt;math&gt;a,c,b&lt;/math&gt; is a geometric progression. What is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-2\qquad<br /> \textbf{(B) }1\qquad<br /> \textbf{(C) }2\qquad<br /> \textbf{(D) }4\qquad<br /> \textbf{(E) }6\qquad&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We have &lt;math&gt;b-a=c-b&lt;/math&gt;, so &lt;math&gt;a=2b-c&lt;/math&gt;. Since &lt;math&gt;a,c,b&lt;/math&gt; is geometric, &lt;math&gt;c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0&lt;/math&gt;. Since &lt;math&gt;a&lt;b&lt;c&lt;/math&gt;, we can't have &lt;math&gt;b=c&lt;/math&gt; and thus &lt;math&gt;c=-2b&lt;/math&gt;. Then our arithmetic progression is &lt;math&gt;4b,b,-2b&lt;/math&gt;. Since &lt;math&gt;4b &lt; b &lt; -2b&lt;/math&gt;, &lt;math&gt;b &lt; 0&lt;/math&gt;. The smallest possible value of &lt;math&gt;c=-2b&lt;/math&gt; is &lt;math&gt;(-2)(-1)=2&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> (Solution by AwesomeToad)<br /> <br /> ==Solution 2==<br /> <br /> Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us &lt;math&gt;(b-a) = (c-b)&lt;/math&gt;. From this, we can obtain the expression &lt;math&gt;a = 2b-c&lt;/math&gt;. Again, by taking the definition of a geometric progression, we can obtain the expression, &lt;math&gt;c=ar&lt;/math&gt; and &lt;math&gt;b=ar^2&lt;/math&gt;, where r serves as a value for the ratio between two terms in the progression. By substituting &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, &lt;math&gt;a=2ar^2-ar&lt;/math&gt; which can be simplified to &lt;math&gt;(r-1)(2r+1)=0&lt;/math&gt; giving us &lt;math&gt;r=1&lt;/math&gt; of &lt;math&gt;r=-1/2&lt;/math&gt;. Thus, from the geometric progression, &lt;math&gt;a=a&lt;/math&gt;, &lt;math&gt;b=1/4a&lt;/math&gt; and &lt;math&gt;c=-1/2a&lt;/math&gt;. Looking at the initial conditions of &lt;math&gt;a&lt;b&lt;c&lt;/math&gt; we can see that the lowest integer value that would satisfy the above expressions is if &lt;math&gt;a = -4&lt;/math&gt;, thus making &lt;math&gt;c=2&lt;/math&gt; or or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> (Solution by thatuser)<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2014|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_14&diff=81732 2014 AMC 12A Problems/Problem 14 2016-11-30T00:17:40Z <p>Kj2002: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;b&lt;c&lt;/math&gt; be three integers such that &lt;math&gt;a,b,c&lt;/math&gt; is an arithmetic progression and &lt;math&gt;a,c,b&lt;/math&gt; is a geometric progression. What is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-2\qquad<br /> \textbf{(B) }1\qquad<br /> \textbf{(C) }2\qquad<br /> \textbf{(D) }4\qquad<br /> \textbf{(E) }6\qquad&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We have &lt;math&gt;b-a=c-b&lt;/math&gt;, so &lt;math&gt;a=2b-c&lt;/math&gt;. Since &lt;math&gt;a,c,b&lt;/math&gt; is geometric, &lt;math&gt;c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0&lt;/math&gt;. Since &lt;math&gt;a&lt;b&lt;c&lt;/math&gt;, we can't have &lt;math&gt;b=c&lt;/math&gt; and thus &lt;math&gt;c=-2b&lt;/math&gt;. Then our arithmetic progression is &lt;math&gt;4b,b,-2b&lt;/math&gt;. Since &lt;math&gt;4b &lt; b &lt; -2b&lt;/math&gt;, &lt;math&gt;b &lt; 0&lt;/math&gt;. The smallest possible value of &lt;math&gt;c=-2b&lt;/math&gt; is &lt;math&gt;(-2)(-1)=2&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> (Solution by AwesomeToad)<br /> <br /> ==Solution 2==<br /> <br /> Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us &lt;math&gt;(b-a) = (c-b)&lt;/math&gt;. From this, we can obtain the expression &lt;math&gt;a = 2b-c&lt;/math&gt;. Again, by taking the definition of a geometric progression, we can obtain the expression, &lt;math&gt;c=ar&lt;/math&gt; and &lt;math&gt;b=ar^2&lt;/math&gt;, where r serves as a value for the ratio between two terms in the progression. By substituting &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, &lt;math&gt;a=2ar^2-ar&lt;/math&gt; which can be simplified to &lt;math&gt;(r-1)(2r+1)=0&lt;/math&gt; giving us &lt;math&gt;r=1&lt;/math&gt; of &lt;math&gt;r=-1/2&lt;/math&gt;. Thus, from the geometric progression, &lt;math&gt;a=a&lt;/math&gt;, &lt;math&gt;b=1/4a&lt;/math&gt; and =&lt;math&gt;c=-1/2a&lt;/math&gt;. Looking at the initial conditions of &lt;math&gt;a&lt;b&lt;c&lt;/math&gt; we can see that the lowest integer value that would satisfy the above expressions is if &lt;math&gt;a = -4&lt;/math&gt;, thus making &lt;math&gt;c=2&lt;/math&gt; or or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> (Solution by thatuser)<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2014|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_14&diff=81731 2014 AMC 12A Problems/Problem 14 2016-11-30T00:16:39Z <p>Kj2002: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;b&lt;c&lt;/math&gt; be three integers such that &lt;math&gt;a,b,c&lt;/math&gt; is an arithmetic progression and &lt;math&gt;a,c,b&lt;/math&gt; is a geometric progression. What is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-2\qquad<br /> \textbf{(B) }1\qquad<br /> \textbf{(C) }2\qquad<br /> \textbf{(D) }4\qquad<br /> \textbf{(E) }6\qquad&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We have &lt;math&gt;b-a=c-b&lt;/math&gt;, so &lt;math&gt;a=2b-c&lt;/math&gt;. Since &lt;math&gt;a,c,b&lt;/math&gt; is geometric, &lt;math&gt;c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0&lt;/math&gt;. Since &lt;math&gt;a&lt;b&lt;c&lt;/math&gt;, we can't have &lt;math&gt;b=c&lt;/math&gt; and thus &lt;math&gt;c=-2b&lt;/math&gt;. Then our arithmetic progression is &lt;math&gt;4b,b,-2b&lt;/math&gt;. Since &lt;math&gt;4b &lt; b &lt; -2b&lt;/math&gt;, &lt;math&gt;b &lt; 0&lt;/math&gt;. The smallest possible value of &lt;math&gt;c=-2b&lt;/math&gt; is &lt;math&gt;(-2)(-1)=2&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> (Solution by AwesomeToad)<br /> <br /> ==Solution 2==<br /> <br /> Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us &lt;math&gt;(b-a) = (c-b)&lt;/math&gt;. From this, we can obtain the expression &lt;math&gt;a = 2b-c&lt;/math&gt;. Again, by taking the definition of a geometric progression, we can obtain the expression, &lt;math&gt;c=ar&lt;/math&gt; and &lt;math&gt;b=ar^2&lt;/math&gt;, where r serves as a value for the ratio between two terms in the progression. By substituting &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, &lt;math&gt;a=2ar^2-ar&lt;/math&gt; which can be simplified to &lt;math&gt;(r-1)(2r+1)=0&lt;/math&gt; giving us &lt;math&gt;r=1&lt;/math&gt; of &lt;math&gt;r=-1/2&lt;/math&gt;. Thus, from the geometric progression, &lt;math&gt;a=a&lt;/math&gt;, &lt;math&gt;b=1/4a&lt;/math&gt; and &lt;math&gt;c=-1/2a&lt;/math&gt;. Looking at the initial conditions of &lt;math&gt;a&lt;b&lt;c&lt;/math&gt; we can see that the lowest integer value that would satisfy the above expressions is if &lt;math&gt;a = -4&lt;/math&gt;, thus making &lt;math&gt;c=2&lt;/math&gt; or or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> (Solution by thatuser)<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2014|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=User:Kj2002&diff=69967 User:Kj2002 2015-04-19T14:56:31Z <p>Kj2002: Blanked the page</p> <hr /> <div></div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=User:Kj2002&diff=61293 User:Kj2002 2014-03-30T17:38:28Z <p>Kj2002: Created page with &quot;Hi my name is kj2002. In 6th grade, I got 5th in MC state, 6th in chapter. i play FTW!&quot;</p> <hr /> <div>Hi my name is kj2002.<br /> <br /> In 6th grade, I got 5th in MC state, 6th in chapter.<br /> <br /> i play FTW!</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_25&diff=53215 2013 AMC 12A Problems/Problem 25 2013-07-03T20:03:24Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Let &lt;math&gt;f : \mathbb{C} \to \mathbb{C} &lt;/math&gt; be defined by &lt;math&gt; f(z) = z^2 + iz + 1 &lt;/math&gt;. How many complex numbers &lt;math&gt;z &lt;/math&gt; are there such that &lt;math&gt; \text{Im}(z) &gt; 0 &lt;/math&gt; and both the real and the imaginary parts of &lt;math&gt;f(z)&lt;/math&gt; are integers with absolute value at most &lt;math&gt; 10 &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;f(z)=z^2+iz+1=c=a+bi&lt;/math&gt;. We look for &lt;math&gt;z&lt;/math&gt; with &lt;math&gt;\text{Im}(z)&gt;0&lt;/math&gt; such that &lt;math&gt;a,b&lt;/math&gt; are integers where &lt;math&gt;|a|, |b|\leq 10&lt;/math&gt;.<br /> <br /> First, use the quadratic formula:<br /> <br /> &lt;math&gt; z = \frac{1}{2} (-i \pm \sqrt{-1-4(1-c)}) = -\frac{i}{2} \pm \sqrt{ -\frac{5}{4} + c }&lt;/math&gt;<br /> <br /> Generally, consider the imaginary part of a radical of a complex number: &lt;math&gt;\sqrt{u}&lt;/math&gt;, where &lt;math&gt;u = v+wi = r e^{i\theta}&lt;/math&gt;.<br /> <br /> &lt;math&gt;Im (\sqrt{u}) = Im(\pm \sqrt{r} e^{i\theta/2}) = \pm \sqrt{r} \sin(i\theta/2) = \pm \sqrt{r}\sqrt{\frac{1-\cos\theta}{2}} = \pm \sqrt{\frac{r-v}{2}}&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;u= -5/4 + c&lt;/math&gt;, then &lt;math&gt;v = -5/4 + a&lt;/math&gt;, &lt;math&gt;w=b&lt;/math&gt;, &lt;math&gt;r=\sqrt{v^2 + w^2}&lt;/math&gt;.<br /> <br /> Note that &lt;math&gt;Im(z)&gt;0&lt;/math&gt; if and only if &lt;math&gt;\pm \sqrt{\frac{r-v}{2}}&gt;\frac{1}{2}&lt;/math&gt;. The latter is true only when we take the positive sign, and that &lt;math&gt;r-v &gt; 1/2&lt;/math&gt;,<br /> <br /> or &lt;math&gt;v^2 + w^2 &gt; (1/2 + v)^2 = 1/4 + v + v^2&lt;/math&gt;, &lt;math&gt;w^2 &gt; 1/4 + v&lt;/math&gt;, or &lt;math&gt;b^2 &gt; a-1&lt;/math&gt;.<br /> <br /> In other words, for all &lt;math&gt;z&lt;/math&gt;, &lt;math&gt;f(z)=a+bi&lt;/math&gt; satisfies &lt;math&gt;b^2 &gt; a-1&lt;/math&gt;, and there is one and only one &lt;math&gt;z&lt;/math&gt; that makes it true. Therefore we are just going to count the number of ordered pairs &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; are integers of magnitude no greater than &lt;math&gt;10&lt;/math&gt;, and that &lt;math&gt;b^2 \geq a&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a\leq 0&lt;/math&gt;, there is no restriction on &lt;math&gt;b&lt;/math&gt; so there are &lt;math&gt;11\cdot 21 = 231&lt;/math&gt; pairs;<br /> <br /> when &lt;math&gt;a &gt; 0&lt;/math&gt;, there are &lt;math&gt;2(1+4+9+10+10+10+10+10+10+10)=2(84)=168&lt;/math&gt; pairs.<br /> <br /> So there are &lt;math&gt;231+168=399&lt;/math&gt; in total.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=24|after=Last Question}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&diff=53214 2013 AMC 12A Problems/Problem 24 2013-07-03T20:03:11Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Suppose &lt;math&gt;p&lt;/math&gt; is the answer. We calculate &lt;math&gt;1-p&lt;/math&gt;.<br /> <br /> Assume that the circumradius of the 12-gon is &lt;math&gt;1&lt;/math&gt;, and the 6 different lengths are &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_6&lt;/math&gt;, in increasing order. Then<br /> <br /> &lt;math&gt;a_k = 2\sin ( \frac{k\pi}{12} )&lt;/math&gt;. <br /> <br /> So &lt;math&gt;a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_2=1&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_3=\sqrt{2}\approx 1.4&lt;/math&gt;, <br /> <br /> &lt;math&gt;a_4=\sqrt{3}\approx 1.7&lt;/math&gt;,<br /> <br /> &lt;math&gt;a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3 &lt;/math&gt;,<br /> <br /> &lt;math&gt;a_6 = 2&lt;/math&gt;.<br /> <br /> Note that there are &lt;math&gt;12&lt;/math&gt; segments of each length of &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_5&lt;/math&gt;, respectively, and &lt;math&gt;6&lt;/math&gt; segments of length &lt;math&gt;a_6&lt;/math&gt;. There are &lt;math&gt;66&lt;/math&gt; segments in total.<br /> <br /> Now, Consider the following inequalities:<br /> <br /> - &lt;math&gt;a_1 + a_1 &gt; a_2&lt;/math&gt;: Since &lt;math&gt;6 &gt; (1+\sqrt{2})^2&lt;/math&gt; <br /> <br /> - &lt;math&gt;a_1 + a_1 &lt; a_3&lt;/math&gt;.<br /> <br /> - &lt;math&gt;a_1 + a_2&lt;/math&gt; is greater than &lt;math&gt;a_3&lt;/math&gt; but less than &lt;math&gt;a_4&lt;/math&gt;.<br /> <br /> - &lt;math&gt;a_1 + a_3&lt;/math&gt; is greater than &lt;math&gt;a_4&lt;/math&gt; but equal to &lt;math&gt;a_5&lt;/math&gt;.<br /> <br /> - &lt;math&gt;a_1 + a_4&lt;/math&gt; is greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> - &lt;math&gt;a_2+a_2 = 2 = a_6&lt;/math&gt;. Then obviously any two segments with at least one them longer than &lt;math&gt;a_2&lt;/math&gt; have a sum greater than &lt;math&gt;a_6&lt;/math&gt;.<br /> <br /> Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means &lt;math&gt;(a_x, a_y, a_z)&lt;/math&gt;:<br /> <br /> 1-1-3, 1-1-4, 1-1-5, 1-1-6,<br /> 1-2-4, 1-2-5, 1-2-6,<br /> 1-3-5, 1-3-6,<br /> 2-2-6<br /> <br /> In the above list there are &lt;math&gt;3&lt;/math&gt; triples of the type a-a-b without ''6'', &lt;math&gt;2&lt;/math&gt; triples of a-a-6 where a is not ''6'', &lt;math&gt;3&lt;/math&gt; triples of a-b-c without ''6'', and &lt;math&gt;2&lt;/math&gt; triples of a-b-6 where a, b are not ''6''. So,<br /> <br /> &lt;cmath&gt;1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} &lt;/cmath&gt;<br /> <br /> So &lt;math&gt;p = 223/286&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_23&diff=53213 2013 AMC 12A Problems/Problem 23 2013-07-03T20:02:59Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> &lt;math&gt; ABCD&lt;/math&gt; is a square of side length &lt;math&gt; \sqrt{3} + 1 &lt;/math&gt;. Point &lt;math&gt; P &lt;/math&gt; is on &lt;math&gt; \overline{AC} &lt;/math&gt; such that &lt;math&gt; AP = \sqrt{2} &lt;/math&gt;. The square region bounded by &lt;math&gt; ABCD &lt;/math&gt; is rotated &lt;math&gt; 90^{\circ} &lt;/math&gt; counterclockwise with center &lt;math&gt; P &lt;/math&gt;, sweeping out a region whose area is &lt;math&gt; \frac{1}{c} (a \pi + b) &lt;/math&gt;, where &lt;math&gt;a &lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt; c &lt;/math&gt; are positive integers and &lt;math&gt; \text{gcd}(a,b,c) = 1 &lt;/math&gt;. What is &lt;math&gt; a + b + c &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We first note that diagonal &lt;math&gt; \overline{AC} &lt;/math&gt; is of length &lt;math&gt; \sqrt{6} + \sqrt{2} &lt;/math&gt;. It must be that &lt;math&gt; \overline{AP} &lt;/math&gt; divides the diagonal into two segments in the ratio &lt;math&gt;\sqrt{3}&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt;. It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;\sqrt{3} + 1&lt;/math&gt;. The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or &lt;math&gt; 2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3} &lt;/math&gt;.<br /> <br /> <br /> Apart from the rectangular region of the initial and final squares, the swept out area includes &lt;math&gt;4&lt;/math&gt; circular segments. Two of them are quarter-segments of circles centered at &lt;math&gt;P&lt;/math&gt; of radii &lt;math&gt;\sqrt{2}&lt;/math&gt; (the segment bounded by &lt;math&gt; \overline{PA} &lt;/math&gt; and &lt;math&gt; \overline{PA'}&lt;/math&gt;) and &lt;math&gt;\sqrt{6}&lt;/math&gt; (that bounded by &lt;math&gt;\overline{PC}&lt;/math&gt; and &lt;math&gt;\overline{PC'}&lt;/math&gt;). Assuming &lt;math&gt;A&lt;/math&gt; is the bottom-left vertex and &lt;math&gt;B&lt;/math&gt; is the bottom-right one, it is clear that the third segment is formed as &lt;math&gt;B&lt;/math&gt; swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when &lt;math&gt;D&lt;/math&gt; overshoots the final square's left edge. To find the areas of these segments, consider the perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt;. Call the point of intersection &lt;math&gt;E&lt;/math&gt;. From the previous paragraph, it is clear that &lt;math&gt;PE = \sqrt{3}&lt;/math&gt; and &lt;math&gt;BE = 1&lt;/math&gt;. This means &lt;math&gt;PB = 2&lt;/math&gt;, and &lt;math&gt;B&lt;/math&gt; swings back inside edge &lt;math&gt;\overline{BC}&lt;/math&gt; at a point &lt;math&gt;1&lt;/math&gt; unit above &lt;math&gt;E&lt;/math&gt; (since it left the edge &lt;math&gt;1&lt;/math&gt; unit below). The triangle of the circular sector is therefore an equilateral triangle of side length &lt;math&gt;2&lt;/math&gt;, and so the angle of the segment is &lt;math&gt;60^{\circ}&lt;/math&gt;. Imagining the process in reverse, it is clear that the situation is the same with point &lt;math&gt;D&lt;/math&gt;.<br /> <br /> <br /> To see that no other area is included, note that the four vertices represent local maxima of distance from point &lt;math&gt;P&lt;/math&gt;, so the path of any other point in or on the square lies in one of the four circular segments if it does not lie entirely within the initial or final squares.<br /> <br /> <br /> The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas &lt;math&gt; \frac{1}{4} \pi (\sqrt{2})^2 - \frac{1}{2} \sqrt{2} \sqrt{2} = \frac{\pi}{2} - 1 &lt;/math&gt; and &lt;math&gt; \frac{1}{4} \pi (\sqrt{6})^2 - \frac{1}{2} \sqrt{6} \sqrt{6} = \frac{3 \pi}{2} - 3 &lt;/math&gt;. The other two segments both have area &lt;math&gt;\frac{1}{6} \pi (2)^2 - \frac{(2)^2 \sqrt{3}}{4} = \frac{2 \pi}{3} - \sqrt{3} &lt;/math&gt;.<br /> <br /> <br /> The total area is therefore &lt;cmath&gt;(6 + 2 \sqrt{3}) + (\frac{\pi}{2} - 1) + (\frac{3 \pi}{2} - 3) + 2 (\frac{2 \pi}{3} - \sqrt{3})&lt;/cmath&gt; &lt;cmath&gt;= 2 + 2 \sqrt{3} + 2 \pi + \frac{4 \pi}{3} - 2 \sqrt{3}&lt;/cmath&gt; &lt;cmath&gt;= \frac{10 \pi}{3} + 2&lt;/cmath&gt; &lt;cmath&gt;= \frac{1}{3} (10 \pi + 6) &lt;/cmath&gt;<br /> <br /> <br /> This means &lt;math&gt;a = 10&lt;/math&gt;, &lt;math&gt;b = 6&lt;/math&gt;, and &lt;math&gt;c = 3&lt;/math&gt;. So &lt;math&gt;a + b + c = 10 + 6 + 3 = 19&lt;/math&gt;; answer choice &lt;math&gt;\boxed{\textbf{(C)} \ 19}&lt;/math&gt; is correct.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_22&diff=53212 2013 AMC 12A Problems/Problem 22 2013-07-03T20:02:22Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome &lt;math&gt;n&lt;/math&gt; is chosen uniformly at random. What is the probability that &lt;math&gt;\frac{n}{11}&lt;/math&gt; is also a palindrome?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Working backwards, we can multiply 5-digit palindromes &lt;math&gt;ABCBA&lt;/math&gt; by &lt;math&gt;11&lt;/math&gt;, giving a 6-digit palindrome:<br /> <br /> &lt;math&gt;A (A+B) (B+C) (B+C) (A+B) A&lt;/math&gt;<br /> <br /> Note that if &lt;math&gt;A + B &gt; 10&lt;/math&gt; or &lt;math&gt;B + C &gt; 10&lt;/math&gt;, then the symmetry will be broken by carried 1s<br /> <br /> Simply count the combinations of &lt;math&gt;(A, B, C)&lt;/math&gt; for which &lt;math&gt;A + B &lt; 10&lt;/math&gt; and &lt;math&gt;B + C &lt; 10&lt;/math&gt;<br /> <br /> &lt;math&gt;A = 1&lt;/math&gt; implies &lt;math&gt;9&lt;/math&gt; possible &lt;math&gt;B&lt;/math&gt; (0 through 8), for each of which there are &lt;math&gt;10, 9, 8, 7, 6, 5, 4, 3, 2&lt;/math&gt; possible C, respectively. There are &lt;math&gt;54&lt;/math&gt; valid palindromes when &lt;math&gt;A = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;A = 2&lt;/math&gt; implies &lt;math&gt;8&lt;/math&gt; possible &lt;math&gt;B&lt;/math&gt; (0 through 7), for each of which there are &lt;math&gt;10, 9, 8, 7, 6, 5, 4, 3&lt;/math&gt; possible C, respectively. There are &lt;math&gt;52&lt;/math&gt; valid palindromes when &lt;math&gt;A = 2&lt;/math&gt;<br /> <br /> Following this pattern, the total is<br /> <br /> &lt;math&gt;54 + 52 + 49 + 45 + 40 + 34 + 27 + 19 + 10 = 330&lt;/math&gt;<br /> <br /> 6-digit palindromes are of the form &lt;math&gt;XYZZYX&lt;/math&gt;, and the first digit cannot be a zero, so there are &lt;math&gt;9 * 10 * 10 = 900&lt;/math&gt; combinations of &lt;math&gt;(X, Y, Z)&lt;/math&gt;<br /> <br /> So, the probability is &lt;math&gt;\frac{330}{900} = \frac{11}{30}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_21&diff=53211 2013 AMC 12A Problems/Problem 21 2013-07-03T20:02:11Z <p>Kj2002: /* See Also */</p> <hr /> <div>==Problem==<br /> Consider &lt;math&gt; A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots )))) &lt;/math&gt;. Which of the following intervals contains &lt;math&gt; A &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ (\log 2016, \log 2017) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(B)} \ (\log 2017, \log 2018) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(C)} \ (\log 2018, \log 2019) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(D)} \ (\log 2019, \log 2020) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(E)} \ (\log 2020, \log 2021) &lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;f(x) = \log(x + f(x-1))&lt;/math&gt; and &lt;math&gt;f(2) = log(2)&lt;/math&gt;, and from the problem description, &lt;math&gt;A = f(2013)&lt;/math&gt;<br /> <br /> We can reason out an approximation, by ignoring the &lt;math&gt;f(x-1)&lt;/math&gt;:<br /> <br /> &lt;math&gt;f_{0}(x) \approx \log x&lt;/math&gt;<br /> <br /> And a better approximation, by plugging in our first approximation for &lt;math&gt;f(x-1)&lt;/math&gt; in our original definition for &lt;math&gt;f(x)&lt;/math&gt;:<br /> <br /> &lt;math&gt;f_{1}(x) \approx \log(x + \log(x-1))&lt;/math&gt;<br /> <br /> And an even better approximation:<br /> <br /> &lt;math&gt;f_{2}(x) \approx \log(x + \log(x-1 + \log(x-2)))&lt;/math&gt;<br /> <br /> Continuing this pattern, obviously, will eventually terminate at &lt;math&gt;f_{x-1}(x)&lt;/math&gt;, in other words our original definition of &lt;math&gt;f(x)&lt;/math&gt;.<br /> <br /> However, at &lt;math&gt;x = 2013&lt;/math&gt;, going further than &lt;math&gt;f_{1}(x)&lt;/math&gt; will not distinguish between our answer choices. &lt;math&gt;\log(2012 + \log(2011))&lt;/math&gt; is nearly indistinguishable from &lt;math&gt;\log(2012)&lt;/math&gt;.<br /> <br /> So we take &lt;math&gt;f_{1}(x)&lt;/math&gt; and plug in.<br /> <br /> &lt;math&gt;f(2013) \approx \log(2013 + \log 2012)&lt;/math&gt;<br /> <br /> Since &lt;math&gt;1000 &lt; 2012 &lt; 10000&lt;/math&gt;, we know &lt;math&gt;3 &lt; log(2012) &lt; 4&lt;/math&gt;. This gives us our answer range:<br /> <br /> &lt;math&gt;\log 2016 &lt; \log(2013 + \log 2012) &lt; \log(2017)&lt;/math&gt;<br /> <br /> &lt;math&gt;(\log 2016, \log 2017)&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_20&diff=53210 2013 AMC 12A Problems/Problem 20 2013-07-03T20:01:28Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem 20 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set &lt;math&gt;\{1,2,3,...,19\}&lt;/math&gt;. For &lt;math&gt;a,b \in S&lt;/math&gt;, define &lt;math&gt;a \succ b&lt;/math&gt; to mean that either &lt;math&gt;0 &lt; a - b \le 9&lt;/math&gt; or &lt;math&gt;b - a &gt; 9&lt;/math&gt;. How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of elements of &lt;math&gt;S&lt;/math&gt; have the property that &lt;math&gt;x \succ y&lt;/math&gt;, &lt;math&gt;y \succ z&lt;/math&gt;, and &lt;math&gt;z \succ x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Imagine 19 numbers are just 19 persons sitting evenly around a circle &lt;math&gt;C&lt;/math&gt;; each of them is facing to the center. <br /> <br /> One may check that &lt;math&gt;x \succ y&lt;/math&gt; iff &lt;math&gt;y&lt;/math&gt; is one of the 9 persons on the left of &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;y \succ x&lt;/math&gt; iff &lt;math&gt;y&lt;/math&gt; is one of the 9 persons on the right of &lt;math&gt;x&lt;/math&gt;. Therefore, &quot;&lt;math&gt;x \succ y&lt;/math&gt; and &lt;math&gt;y \succ z&lt;/math&gt; and &lt;math&gt;z \succ x&lt;/math&gt;&quot; implies that &lt;math&gt;x, y, z&lt;/math&gt; cuts the circumference of &lt;math&gt;C&lt;/math&gt; into three arcs, each of which has no more than &lt;math&gt;10&lt;/math&gt; numbers sitting on it (inclusive).<br /> <br /> We count the complement: where the cut generated by &lt;math&gt;(x,y,z)&lt;/math&gt; has ONE arc that has more than &lt;math&gt;10&lt;/math&gt; persons sitting on. Note that there can only be one such arc because there are only &lt;math&gt;19&lt;/math&gt; persons in total.<br /> <br /> Suppose the number of persons on the longest arc is &lt;math&gt;k&gt;10&lt;/math&gt;. Then two places of &lt;math&gt;x,y,z&lt;/math&gt; are just chosen from the two end-points of the arc, and there are &lt;math&gt;19-k&lt;/math&gt; possible places for the third person. Once the three places of &lt;math&gt;x,y,z&lt;/math&gt; is chosen, there are three possible ways to put &lt;math&gt;x,y,z&lt;/math&gt; into them clockwise. Also, note that for any &lt;math&gt;k&gt;10&lt;/math&gt;, there are &lt;math&gt;19&lt;/math&gt; ways to choose an arc of length &lt;math&gt;k&lt;/math&gt;. Therefore the total number of ways (of the complement) is <br /> <br /> &lt;cmath&gt;\sum_{k=11}^{18} 3\cdot 19 \cdot (19-k) = 3\cdot 19 \cdot (1+\cdots+8) = 3\cdot 19\cdot 36&lt;/cmath&gt;<br /> <br /> So the answer is<br /> <br /> &lt;cmath&gt; 3\cdot \binom{19}{3} - 3\cdot 19\cdot 36 = 3\cdot 19 \cdot (51 - 36) = 855 &lt;/cmath&gt;<br /> <br /> NOTE: this multiple choice problem can be done even faster -- after we realized the fact that each choice of the three places of &lt;math&gt;x,y,z&lt;/math&gt; corresponds to &lt;math&gt;3&lt;/math&gt; possible ways to put them in, and that each arc of length &lt;math&gt;k&gt;10&lt;/math&gt; has &lt;math&gt;19&lt;/math&gt; equitable positions, it is evident that the answer should be divisible by &lt;math&gt;3\cdot 19&lt;/math&gt;, which can only be &lt;math&gt;855&lt;/math&gt; from the five choices.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_19&diff=53209 2013 AMC 12A Problems/Problem 19 2013-07-03T20:01:14Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> In &lt;math&gt; \bigtriangleup ABC &lt;/math&gt;, &lt;math&gt; AB = 86 &lt;/math&gt;, and &lt;math&gt; AC = 97 &lt;/math&gt;. A circle with center &lt;math&gt; A &lt;/math&gt; and radius &lt;math&gt; AB &lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at points &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; X &lt;/math&gt;. Moreover &lt;math&gt; \overline{BX} &lt;/math&gt; and &lt;math&gt; \overline{CX} &lt;/math&gt; have integer lengths. What is &lt;math&gt; BC &lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> <br /> Let CX=x, BX=y. Let the circle intersect AC at D and the diameter including AD intersect the circle again at E.<br /> Use power of a point on point C to the circle centered at A.<br /> <br /> So CX*CB=CD*CE<br /> x(x+y)=(97-86)(97+86)<br /> x(x+y)=3*11*61.<br /> <br /> Obviously x+y&gt;x so we have three solution pairs for (x,x+y)=(1,2013),(3,671),(11,183),(33,61).<br /> By the Triangle Inequality, only x+y=61 yields a possible length of BX+CX=BC.<br /> <br /> Therefore, the answer is '''D) 61.'''<br /> <br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;h&lt;/math&gt; be the perpendicular from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;AX=x&lt;/math&gt;, &lt;math&gt;XC=y&lt;/math&gt;, then by Pythagorean Theorem,<br /> <br /> &lt;math&gt;h^2 + (x/2)^2 = 86^2&lt;/math&gt;<br /> <br /> &lt;math&gt;h^2 + (x/2 + y)^2 = 97^2&lt;/math&gt;<br /> <br /> Subtracting the two equations, we get &lt;math&gt;(x+y)y = (97-86)(97+86)&lt;/math&gt;,<br /> <br /> then the rest is similar to the above solution by power of points.<br /> <br /> ===Solution 3===<br /> Let &lt;math&gt;x&lt;/math&gt; represent &lt;math&gt;CX&lt;/math&gt;, and let &lt;math&gt;y&lt;/math&gt; represent &lt;math&gt;BX&lt;/math&gt;. Since the circle goes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;AX&lt;/math&gt; = 86.<br /> Then by Stewart's Theorem,<br /> <br /> &lt;math&gt;xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 + xy + 86^2 = 97^2&lt;/math&gt;<br /> <br /> (Since &lt;math&gt;y&lt;/math&gt; cannot be equal to 0, dividing both sides of the equation by &lt;math&gt;y&lt;/math&gt; is allowed.)<br /> <br /> &lt;math&gt;x(x+y) = (97+86)(97-86)&lt;/math&gt;<br /> <br /> &lt;math&gt;x(x+y) = 2013&lt;/math&gt;<br /> <br /> The prime factors of 2013 are 3, 11, and 61. Obviously, &lt;math&gt;x &lt; x+y&lt;/math&gt;. In addition, by the Triangle Inequality, &lt;math&gt;BC &lt; AB + AC&lt;/math&gt;, so &lt;math&gt;x+y &lt; 183&lt;/math&gt;. Therefore, &lt;math&gt;x&lt;/math&gt; must equal 33, and &lt;math&gt;x+y&lt;/math&gt; must equal 61. &lt;math&gt;\boxed{D}&lt;/math&gt;<br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_18&diff=53208 2013 AMC 12A Problems/Problem 18 2013-07-03T20:00:51Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> You have a regular hexagon with side lengths 2 and six spheres on each vertice radius 1 that are internally tangent, therefore drawing radius's through them all would create this regular hexagon.<br /> <br /> There is a larger sphere the 6 spheres are internally tangent to, with center in the center of the hexagon. To find the radius of the larger sphere we must first, either by prior knowledge or by deducing from the angle sum that the hexagon can be split into 6 equilateral triangles from it's vertices and then the radius is &lt;math&gt;2+1=3&lt;/math&gt;<br /> <br /> The 8th sphere is now, when thinking about it in 3D sitting on top of the 6 spheres which is the only possibility for it to tangent all the 6 small spheres externally and the larger sphere internally. The ring of the 6 small spheres is symmetrical and the 8th sphere will be resting with it's center alligned with the diameter of the large sphere.<br /> <br /> We can therefore now create a triangle with the horizontal component 2, as it is from the vertice of the hexagon to the center of the hexagon. <br /> The vertical component is from the center of the large sphere to the center of the 8th sphere. This length equals 3, the radius of the large sphere, take away the radius of the 8th sphere, we can call it r, since the radius of the large sphere will include the diameter of the 8th sphere if we subtract radius we will reach the center.<br /> The last component is the hypotenuse of the right angled triangle. This consists of the radius of the small sphere - 1 - and the radius of the 8th sphere - r -.<br /> <br /> We therefore now have a right angled triangle which when applied pythagoras states &lt;math&gt;2^2+(3-r)^2=(1+r)^2&lt;/math&gt;<br /> Expanding brackets gives us &lt;math&gt;4+9-6r+r^2=1+2r+r^2&lt;/math&gt; here we can cancel out &lt;math&gt;r^2&lt;/math&gt;<br /> Isolating the r's &lt;math&gt;12=8r&lt;/math&gt;<br /> and then finally we have the answer: &lt;math&gt;r=\frac{12}{8}=\frac{3}{2}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=17|num-a=19}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_17&diff=53207 2013 AMC 12A Problems/Problem 17 2013-07-03T20:00:36Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The first pirate takes &lt;math&gt;\frac{1}{12}&lt;/math&gt; of the &lt;math&gt;x&lt;/math&gt; coins, leaving &lt;math&gt;\frac{11}{12} x&lt;/math&gt;.<br /> <br /> The second pirate takes &lt;math&gt;\frac{2}{12}&lt;/math&gt; of the remaining coins, leaving &lt;math&gt;\frac{10}{12}*\frac{11}{12}*x&lt;/math&gt;.<br /> <br /> Continuing this pattern, the eleventh pirate must take &lt;math&gt;\frac{11}{12}&lt;/math&gt; of the remaining coins after the first ten pirates have taken their share, which leaves &lt;math&gt;\frac{11!}{12^{12}}*x&lt;/math&gt;. The twelfth pirate takes all of this.<br /> <br /> <br /> Note that<br /> <br /> &lt;math&gt;12^{12} = (2^2 * 3)^{12} = 2^{24} * 3^{12}&lt;/math&gt;<br /> <br /> &lt;math&gt;11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2&lt;/math&gt;<br /> <br /> <br /> All the 2s and 3s cancel out of &lt;math&gt;11!&lt;/math&gt;, leaving<br /> <br /> &lt;math&gt;11 * 5 * 7 * 5 = 1925&lt;/math&gt;<br /> <br /> in the numerator.<br /> <br /> <br /> We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, &lt;math&gt;x&lt;/math&gt; is the denominator, leaving &lt;math&gt;1925&lt;/math&gt; coins for the twelfth pirate.<br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_16&diff=53206 2013 AMC 12A Problems/Problem 16 2013-07-03T20:00:20Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are three piles of rocks. The mean weight of the rocks in &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;40&lt;/math&gt; pounds, the mean weight of the rocks in &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; pounds, the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;43&lt;/math&gt; pounds, and the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;44&lt;/math&gt; pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let pile &lt;math&gt;A&lt;/math&gt; have &lt;math&gt;A&lt;/math&gt; rocks, and so on.<br /> <br /> The mean weight of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; together is &lt;math&gt;44&lt;/math&gt;, so the total weight of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;44(A + C)&lt;/math&gt;<br /> <br /> To get the total weight of &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, we need to add the total weight of &lt;math&gt;B&lt;/math&gt; and subtract the total weight of &lt;math&gt;A&lt;/math&gt;<br /> <br /> &lt;math&gt;44A + 44C + 50B - 40A = 4A + 44C + 50B&lt;/math&gt;<br /> <br /> And then dividing by the number of rocks &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; together, to get the mean of &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;,<br /> <br /> &lt;math&gt;\frac{4A + 44C + 50B}{B + C}&lt;/math&gt;<br /> <br /> Simplifying,<br /> <br /> &lt;math&gt;\frac{4A + 44C + 44B + 6B}{B + C}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;44 + \frac{4A + 6B}{B + C}&lt;/math&gt;<br /> <br /> Now, to get rid of the &lt;math&gt;A&lt;/math&gt; in the numerator, we use two definitions of the total weight of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;<br /> <br /> &lt;math&gt;40A + 50B = 43A + 43B&lt;/math&gt;<br /> <br /> &lt;math&gt;3A = 7B&lt;/math&gt;<br /> <br /> &lt;math&gt;A = \frac{7}{3}B&lt;/math&gt;<br /> <br /> Substituting back in,<br /> <br /> &lt;math&gt;44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;44 + \frac{46}{3}*\frac{B}{B + C}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;\frac{B}{B + C} &lt; 1&lt;/math&gt;, and the maximal value of this factor occurs when &lt;math&gt;C = 1&lt;/math&gt;<br /> <br /> Also note that &lt;math&gt;\frac{46}{3}&lt;/math&gt; must cancel to give an integer value, and the only fraction that satisfies both these conditions is &lt;math&gt;\frac{45}{46}&lt;/math&gt;<br /> <br /> Plugging in, we get<br /> <br /> &lt;math&gt;44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Suppose there are &lt;math&gt;A,B,C&lt;/math&gt; rocks in the three piles, and that the mean of pile C is &lt;math&gt;x&lt;/math&gt;, and that the mean of the combination of &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;y&lt;/math&gt;. We are going to maximize &lt;math&gt;y&lt;/math&gt;, subject to the following conditions:<br /> <br /> &lt;cmath&gt;40A+50B=43(A+B)&lt;/cmath&gt;<br /> &lt;cmath&gt;40A+xC=44(A+C)&lt;/cmath&gt;<br /> &lt;cmath&gt;50B+xC=y(B+C)&lt;/cmath&gt;<br /> <br /> which can be rearranged as:<br /> <br /> &lt;cmath&gt;7B=3A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-44)C=4A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-y)C=(y-50)B&lt;/cmath&gt;<br /> <br /> Let us test &lt;math&gt;y=59&lt;/math&gt; is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes <br /> <br /> &lt;cmath&gt;(x-59)C=9B.&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;15C = (x-44)C - (x-59)C = 4A - 9B&lt;/math&gt;, &lt;math&gt;45C=4(3A)-27B=28B-27B&lt;/math&gt;, &lt;math&gt;105C=28A-9(7B)=A&lt;/math&gt;, therefore,<br /> <br /> &lt;math&gt;A=105C, B=45C, x=4(105)+44=464&lt;/math&gt;, which gives us a consistent solution. Therefore &lt;math&gt;y=59&lt;/math&gt; is the answer.<br /> <br /> (Note: To further illustrate the idea, let us look at &lt;math&gt;y=60&lt;/math&gt; and see what happens. We then get &lt;math&gt;7\cdot 16C = 4A-30A&lt;0&lt;/math&gt;, which is a contradiction!)<br /> <br /> ===Solution 3===<br /> Obtain the 3 equations as in solution 2.<br /> <br /> &lt;cmath&gt;7B=3A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-44)C=4A&lt;/cmath&gt;<br /> &lt;cmath&gt;(x-y)C=(y-50)B&lt;/cmath&gt;<br /> <br /> Combining the 1st and 2nd equations, we see that<br /> <br /> &lt;cmath&gt;(x-44)C=4A=\frac{28}{3}B&lt;/cmath&gt;<br /> <br /> Subtracting equation 3 from equation 2, we have<br /> <br /> &lt;cmath&gt;(x-44)C-(x-y)C=\frac{28}{3}B-(y-50)B&lt;/cmath&gt;<br /> &lt;cmath&gt;(y-44)C=(50+\frac{28}{3}-y)B=(\frac{178}{3}-y)B&lt;/cmath&gt;<br /> <br /> In order for the coefficients to be positive, &lt;cmath&gt;44&lt;y&lt;\frac{178}{3}&lt;/cmath&gt;<br /> <br /> Thus, the greatest integer value is &lt;math&gt;y=59&lt;/math&gt;, choice &lt;math&gt;(E)&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_15&diff=53205 2013 AMC 12A Problems/Problem 15 2013-07-03T19:59:45Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> There are two possibilities regarding the parents.<br /> <br /> 1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are &lt;math&gt;4 * 3^3 = 108&lt;/math&gt; combinations.<br /> <br /> 2) The two are in different stores. In this case, one can go in any of the 4 stores, and the other can go in any of the 3 remaining stores. The 3 baby bunnies can each go in any of the remaining 2 stores. There are &lt;math&gt;4 * 3 * 2^3 = 96&lt;/math&gt; combinations.<br /> <br /> Adding up, we get &lt;math&gt;108 + 96 = 204&lt;/math&gt; combinations.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_14&diff=53204 2013 AMC 12A Problems/Problem 14 2013-07-03T19:59:28Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> The sequence<br /> <br /> &lt;math&gt;\log_{12}{162}&lt;/math&gt;, &lt;math&gt;\log_{12}{x}&lt;/math&gt;, &lt;math&gt;\log_{12}{y}&lt;/math&gt;, &lt;math&gt;\log_{12}{z}&lt;/math&gt;, &lt;math&gt;\log_{12}{1250}&lt;/math&gt;<br /> <br /> is an arithmetic progression. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since the sequence is arithmetic, <br /> <br /> &lt;math&gt;\log_{12}{162}&lt;/math&gt; + &lt;math&gt;4d&lt;/math&gt; = &lt;math&gt;\log_{12}{1250}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference.<br /> <br /> <br /> Therefore,<br /> <br /> &lt;math&gt;4d&lt;/math&gt; = &lt;math&gt;\log_{12}{1250}&lt;/math&gt; - &lt;math&gt;\log_{12}{162}&lt;/math&gt; = &lt;math&gt;\log_{12}{(1250/162)}&lt;/math&gt;, and<br /> <br /> &lt;math&gt;d&lt;/math&gt; = &lt;math&gt;\frac{1}{4}&lt;/math&gt;(&lt;math&gt;\log_{12}{(1250/162)}&lt;/math&gt;) = &lt;math&gt;\log_{12}{(1250/162)^{1/4}}&lt;/math&gt;<br /> <br /> <br /> <br /> Now that we found &lt;math&gt;d&lt;/math&gt;, we just add it to the first term to find &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;math&gt;\log_{12}{162}&lt;/math&gt; + &lt;math&gt;\log_{12}{(1250/162)^{1/4}}&lt;/math&gt; = &lt;math&gt;\log_{12}{((162)(1250/162)^{1/4})}&lt;/math&gt;<br /> <br /> &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;(162)&lt;/math&gt;&lt;math&gt;(1250/162)^{1/4}&lt;/math&gt; = &lt;math&gt;(162)&lt;/math&gt;&lt;math&gt;(625/81)^{1/4}&lt;/math&gt; = &lt;math&gt;(162)(5/3)&lt;/math&gt; = &lt;math&gt;270&lt;/math&gt;, which is &lt;math&gt;B&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_13&diff=53203 2013 AMC 12A Problems/Problem 13 2013-07-03T19:59:11Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Let points &lt;math&gt; A = (0,0) , \ B = (1,2), \ C = (3,3), &lt;/math&gt; and &lt;math&gt; D = (4,0) &lt;/math&gt;. Quadrilateral &lt;math&gt; ABCD &lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt; A &lt;/math&gt;. This line intersects &lt;math&gt; \overline{CD} &lt;/math&gt; at point &lt;math&gt; \left (\frac{p}{q}, \frac{r}{s} \right ) &lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt; p + q + r + s &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> <br /> If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.<br /> <br /> Pick's Theorem states that<br /> <br /> &lt;math&gt;A&lt;/math&gt; = &lt;math&gt;I&lt;/math&gt; &lt;math&gt;+&lt;/math&gt; &lt;math&gt;\frac{B}{2}&lt;/math&gt; - &lt;math&gt;1&lt;/math&gt;, where &lt;math&gt;I&lt;/math&gt; is the number of lattice points in the interior of the polygon, and &lt;math&gt;B&lt;/math&gt; is the number of lattice points on the boundary of the polygon.<br /> <br /> In this case, <br /> <br /> &lt;math&gt;A&lt;/math&gt; = &lt;math&gt;5&lt;/math&gt; &lt;math&gt;+&lt;/math&gt; &lt;math&gt;\frac{7}{2}&lt;/math&gt; - &lt;math&gt;1&lt;/math&gt; = &lt;math&gt;7.5&lt;/math&gt;<br /> <br /> so<br /> <br /> &lt;math&gt;\frac{A}{2}&lt;/math&gt; = &lt;math&gt;3.75&lt;/math&gt;<br /> <br /> The bottom half of the quadrilateral makes a triangle with base &lt;math&gt;4&lt;/math&gt; and half the total area, so we can deduce that the height of the triangle must be &lt;math&gt;\frac{15}{8}&lt;/math&gt; in order for its area to be &lt;math&gt;3.75&lt;/math&gt;. This height is the y coordinate of our desired intersection point.<br /> <br /> <br /> Note that segment CD lies on the line &lt;math&gt;y = -3x + 12&lt;/math&gt;. Substituting in &lt;math&gt;\frac{15}{8}&lt;/math&gt; for y, we can find that the x coordinate of our intersection point is &lt;math&gt;\frac{27}{8}&lt;/math&gt;.<br /> <br /> Therefore the point of intersection is (&lt;math&gt;\frac{27}{8}&lt;/math&gt;, &lt;math&gt;\frac{15}{8}&lt;/math&gt;), and our desired result is &lt;math&gt;27+8+15+8=58&lt;/math&gt;, which is &lt;math&gt;B&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_12&diff=53202 2013 AMC 12A Problems/Problem 12 2013-07-03T19:58:59Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> The angles in a particular triangle are in arithmetic progression, and the side lengths are &lt;math&gt;4,5,x&lt;/math&gt;. The sum of the possible values of x equals &lt;math&gt;a+\sqrt{b}+\sqrt{c}&lt;/math&gt; where &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Because the angles are in an arithmetic progression, and the angles add up to &lt;math&gt; 180^{\circ} &lt;/math&gt;, the second largest angle in the triangle must be &lt;math&gt; 60^{\circ} &lt;/math&gt;. Also, the side opposite of that angle must be the second longest because of the angle-side relationship. Any of the three sides, &lt;math&gt; 4 &lt;/math&gt;, &lt;math&gt; 5 &lt;/math&gt;, or &lt;math&gt; x &lt;/math&gt;, could be the second longest side of the triangle.<br /> <br /> The law of cosines can be applied to solve for &lt;math&gt; x &lt;/math&gt; in all three cases.<br /> <br /> When the second longest side is &lt;math&gt; 5 &lt;/math&gt;, we get that &lt;math&gt; 5^2 = 4^2 + x^2 - 2(4)(x)cos 60^{\circ} &lt;/math&gt;, therefore &lt;math&gt; x^2 - 4x - 9 = 0 &lt;/math&gt;. By using the quadratic formula, <br /> &lt;math&gt; x = \frac {4 + \sqrt{16 + 36}}{2} &lt;/math&gt;, therefore &lt;math&gt; x = 2 + \sqrt{13} &lt;/math&gt;.<br /> <br /> When the second longest side is &lt;math&gt; x &lt;/math&gt;, we get that &lt;math&gt; x^2 = 5^2 + 4^2 - 40cos 60^{\circ} &lt;/math&gt;, therefore &lt;math&gt; x = \sqrt{21} &lt;/math&gt;.<br /> <br /> When the second longest side is &lt;math&gt; 4 &lt;/math&gt;, we get that &lt;math&gt; 4^2 = 5^2 + x^2 - 2(5)(x)cos 60^{\circ} &lt;/math&gt;, therefore &lt;math&gt; x^2 - 5x + 9 = 0 &lt;/math&gt;. Using the quadratic formula, <br /> &lt;math&gt; x = \frac {5 + \sqrt{25 - 36}}{2} &lt;/math&gt;. However, &lt;math&gt; \sqrt{-11} &lt;/math&gt; is not real, therefore the second longest side cannot equal &lt;math&gt; 4 &lt;/math&gt;.<br /> <br /> Adding the two other possibilities gets &lt;math&gt; 2 + \sqrt{13} + \sqrt{21} &lt;/math&gt;, with &lt;math&gt; a = 2, b=13 &lt;/math&gt;, and &lt;math&gt; c=21 &lt;/math&gt;. &lt;math&gt; a + b + c = 36 &lt;/math&gt;, which is answer choice &lt;math&gt; A &lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_11&diff=53201 2013 AMC 12A Problems/Problem 11 2013-07-03T19:58:00Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is equilateral with &lt;math&gt;AB=1&lt;/math&gt;. Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are on &lt;math&gt;\overline{AC}&lt;/math&gt; and points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on &lt;math&gt;\overline{AB}&lt;/math&gt; such that both &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt; are parallel to &lt;math&gt;\overline{BC}&lt;/math&gt;. Furthermore, triangle &lt;math&gt;ADE&lt;/math&gt; and trapezoids &lt;math&gt;DFGE&lt;/math&gt; and &lt;math&gt;FBCG&lt;/math&gt; all have the same perimeter. What is &lt;math&gt;DE+FG&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real s=1/2,m=5/6,l=1;<br /> pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m;<br /> draw(A--B--C--cycle^^D--E^^F--G);<br /> dot(A^^B^^C^^D^^E^^F^^G);<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,S);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,S);<br /> label(&quot;$G$&quot;,G,NW);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad<br /> \textbf{(B) }\dfrac{3}{2}\qquad<br /> \textbf{(C) }\dfrac{21}{13}\qquad<br /> \textbf{(D) }\dfrac{13}{8}\qquad<br /> \textbf{(E) }\dfrac{5}{3}\qquad&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;AD = x&lt;/math&gt;, and &lt;math&gt;AG = y&lt;/math&gt;. We want to find &lt;math&gt;DE + FG&lt;/math&gt;, which is nothing but &lt;math&gt;x+y&lt;/math&gt;.<br /> <br /> Based on the fact that &lt;math&gt;ADE&lt;/math&gt;, &lt;math&gt;DEFG&lt;/math&gt;, and &lt;math&gt;BCFG&lt;/math&gt; have the same perimeters, we can say the following:<br /> <br /> &lt;math&gt;3x = x + 2(y-x) + y = y + 2(1-y) + 1&lt;/math&gt;<br /> <br /> Simplifying, we can find that<br /> <br /> &lt;math&gt;3x = 3y-x = 3-y&lt;/math&gt;<br /> <br /> Since &lt;math&gt;3-y = 3x&lt;/math&gt;, &lt;math&gt;y = 3-3x&lt;/math&gt;.<br /> <br /> After substitution, we find that &lt;math&gt;9-10x = 3x&lt;/math&gt;, and &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;\frac{9}{13}&lt;/math&gt;.<br /> <br /> Again substituting, we find &lt;math&gt;y&lt;/math&gt; = &lt;math&gt;\frac{12}{13}&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;x+y&lt;/math&gt; = &lt;math&gt;\frac{21}{13}&lt;/math&gt;, which is &lt;math&gt;C&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=10|num-a=12}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_10&diff=53200 2013 AMC 12A Problems/Problem 10 2013-07-03T19:57:44Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;\tfrac{1}{n}&lt;/math&gt; has the repeating decimal representation &lt;math&gt;0.\overline{ab} = 0.ababab\cdots,&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; different digits. What is the sum of the elements of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad &lt;/math&gt;<br /> <br /> ==Solution==<br /> ==Solution 1==<br /> Note that &lt;math&gt;\frac{1}{11} = 0.\overline{09}&lt;/math&gt;.<br /> <br /> Dividing by 3 gives &lt;math&gt;\frac{1}{33} = 0.\overline{03}&lt;/math&gt;, and dividing by 9 gives &lt;math&gt;\frac{1}{99} = 0.\overline{01}&lt;/math&gt;.<br /> <br /> &lt;math&gt;S = \{11, 33, 99\}&lt;/math&gt;<br /> <br /> &lt;math&gt;11 + 33 + 99 = 143&lt;/math&gt;<br /> <br /> The answer must be at least &lt;math&gt;143&lt;/math&gt;, but cannot be &lt;math&gt;155&lt;/math&gt; since no &lt;math&gt;n \le 12&lt;/math&gt; other than &lt;math&gt;11&lt;/math&gt; satisfies the conditions, so the answer is &lt;math&gt;143&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let us begin by working with the condition &lt;math&gt;0.\overline{ab} = 0.ababab\cdots,&lt;/math&gt;. Let &lt;math&gt;x = 0.ababab\cdots&lt;/math&gt;. So, &lt;math&gt;100x-x = ab \Rightarrow x = \frac{ab}{99}&lt;/math&gt;. In order for this fraction &lt;math&gt;x&lt;/math&gt; to be in the form &lt;math&gt;\frac{1}{n}&lt;/math&gt;, &lt;math&gt;99&lt;/math&gt; must be a multiple of &lt;math&gt;ab&lt;/math&gt;. Hence the possibilities of &lt;math&gt;ab&lt;/math&gt; are &lt;math&gt;1,3,9,11,33,99&lt;/math&gt;. Checking each of these, &lt;math&gt;\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},&lt;/math&gt; and &lt;math&gt;\frac{99}{99} = 1&lt;/math&gt;. So the only values of &lt;math&gt;n&lt;/math&gt; that have distinct &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are &lt;math&gt;11,33,&lt;/math&gt; and &lt;math&gt;99&lt;/math&gt;. So, &lt;math&gt;11+33+99= \boxed{\textbf{(D)} 143}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_9&diff=53199 2013 AMC 12A Problems/Problem 9 2013-07-03T19:57:26Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Note that because &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to the sides of &lt;math&gt;\triangle ABC&lt;/math&gt;, the internal triangles &lt;math&gt;\triangle BDE&lt;/math&gt; and &lt;math&gt;\triangle EFC&lt;/math&gt; are similar to &lt;math&gt;\triangle ABC&lt;/math&gt;, and are therefore also isosceles triangles.<br /> <br /> It follows that &lt;math&gt;BD = DE&lt;/math&gt;. Thus, &lt;math&gt;AD + DE = AD + DB = AB = 28&lt;/math&gt;.<br /> <br /> Since opposite sides of parallelograms are equal, the perimeter is &lt;math&gt;2 * (AD + DE) = 56&lt;/math&gt;.<br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_8&diff=53198 2013 AMC 12A Problems/Problem 8 2013-07-03T19:57:10Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> Given that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are distinct nonzero real numbers such that &lt;math&gt;x+\tfrac{2}{x} = y + \tfrac{2}{y}&lt;/math&gt;, what is &lt;math&gt;xy&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt; x+\tfrac{2}{x}= y+\tfrac{2}{y} &lt;/math&gt;<br /> <br /> Since &lt;math&gt;x\not=y&lt;/math&gt;, we may assume that &lt;math&gt;x=\frac{2}{y}&lt;/math&gt; and/or, equivalently, &lt;math&gt;y=\frac{2}{x}&lt;/math&gt;.<br /> <br /> Cross multiply in either equation, giving us &lt;math&gt;xy=2&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_7&diff=53197 2013 AMC 12A Problems/Problem 7 2013-07-03T19:56:54Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> The sequence &lt;math&gt;S_1, S_2, S_3, \cdots, S_{10}&lt;/math&gt; has the property that every term beginning with the third is the sum of the previous two. That is, &lt;cmath&gt; S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3. &lt;/cmath&gt; Suppose that &lt;math&gt;S_9 = 110&lt;/math&gt; and &lt;math&gt;S_7 = 42&lt;/math&gt;. What is &lt;math&gt;S_4&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;math&gt;S_9 = 110&lt;/math&gt;, &lt;math&gt;S_7 = 42&lt;/math&gt;<br /> <br /> &lt;math&gt;S_8 = S_9 - S_ 7 = 110 - 42 = 68&lt;/math&gt;<br /> <br /> &lt;math&gt;S_6 = S_8 - S_7 = 68 - 42 = 26&lt;/math&gt;<br /> <br /> &lt;math&gt;S_5 = S_7 - S_6 = 42 - 26 = 16&lt;/math&gt;<br /> <br /> &lt;math&gt;S_4 = S_6 - S_5 = 26 - 16 = 10&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_6&diff=53196 2013 AMC 12A Problems/Problem 6 2013-07-03T19:56:27Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let the number of 3-point shots attempted be &lt;math&gt;x&lt;/math&gt;. Since she attempted 30 shots, the number of 2-point shots attempted must be &lt;math&gt;30 - x&lt;/math&gt;.<br /> <br /> Since she was successful on &lt;math&gt;20\%&lt;/math&gt;, or &lt;math&gt;\frac{1}{5}&lt;/math&gt;, of her 3-pointers, and &lt;math&gt;30\%&lt;/math&gt;, or &lt;math&gt;\frac{3}{10}&lt;/math&gt;, of her 2-pointers, then her score must be<br /> <br /> &lt;math&gt;\frac{1}{5}*3x + \frac{3}{10}*2(30-x)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{3}{5}*x + \frac{3}{5}(30-x)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{3}{5}(x+30-x)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{3}{5}*30&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;18&lt;/math&gt;, which is &lt;math&gt;B&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_5&diff=53195 2013 AMC 12A Problems/Problem 5 2013-07-03T19:56:03Z <p>Kj2002: /* See also */</p> <hr /> <div>==Problem==<br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &amp;#36;&lt;math&gt;105&lt;/math&gt;, Dorothy paid &amp;#36;&lt;math&gt;125&lt;/math&gt;, and Sammy paid &amp;#36;&lt;math&gt;175&lt;/math&gt;. In order to share the costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Add up the amounts that Tom, Dorothy, and Sammy paid to get &amp;#036;&lt;math&gt;405&lt;/math&gt;, and divide by 3 to get &amp;#036;&lt;math&gt;135&lt;/math&gt;, the amount that each should have paid.<br /> <br /> Tom, having paid &amp;#036;&lt;math&gt;105&lt;/math&gt;, owes Sammy &amp;#036;&lt;math&gt;30&lt;/math&gt;, and Dorothy, having paid &amp;#036;&lt;math&gt;125&lt;/math&gt;, owes Sammy &amp;#036;&lt;math&gt;10&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;t - d = 30 - 10 = 20&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_4&diff=53194 2013 AMC 12A Problems/Problem 4 2013-07-03T19:55:35Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem ==<br /> <br /> What is the value of &lt;cmath&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}&lt;/math&gt;<br /> <br /> We can factor a &lt;math&gt;{2^{2012}}&lt;/math&gt; out of the numerator and denominator to obtain<br /> <br /> &lt;math&gt;\frac{2^{2012}*(2^2+1)}{2^{2012}*(2^2-1)}&lt;/math&gt;<br /> <br /> The &lt;math&gt;{2^{2012}}&lt;/math&gt; cancels, so we get <br /> <br /> &lt;math&gt;\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}&lt;/math&gt;, which is &lt;math&gt;C&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_3&diff=53193 2013 AMC 12A Problems/Problem 3 2013-07-03T19:55:20Z <p>Kj2002: /* See also */</p> <hr /> <div>==Problem==<br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We are given that &lt;math&gt;\frac{6}{10} = \frac{3}{5}&lt;/math&gt; of the flowers are pink, so we know &lt;math&gt;\frac{2}{5}&lt;/math&gt; of the flowers are red.<br /> <br /> Since &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the pink flowers are roses, &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the pink flowers are carnations.<br /> <br /> We are given that &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the red flowers are carnations.<br /> <br /> The number of carnations are <br /> <br /> &lt;math&gt;\frac{3}{5} * \frac{2}{3} + \frac{2}{5} * \frac{3}{4} = \frac{2}{5} + \frac{3}{10} = \frac{7}{10} = 70\%&lt;/math&gt;, which is &lt;math&gt;E&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_2&diff=53192 2013 AMC 12A Problems/Problem 2 2013-07-03T19:55:06Z <p>Kj2002: /* See also */</p> <hr /> <div>== Problem 2 ==<br /> <br /> A softball team played ten games, scoring &lt;math&gt;1,2,3,4,5,6,7,8,9&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt; runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score? <br /> <br /> &lt;math&gt; \textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> To score twice as many runs as their opponent, the softball team must have scored an even number.<br /> <br /> Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much.<br /> <br /> Therefore, the total runs by the opponent is &lt;math&gt;(2+4+6+8+10)+(1+2+3+4+5) = 45&lt;/math&gt;, which is &lt;math&gt;C&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_1&diff=53191 2013 AMC 12A Problems/Problem 1 2013-07-03T19:53:52Z <p>Kj2002: /* See also */</p> <hr /> <div>==Problem==<br /> <br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt; 10 &lt;/math&gt;. Point &lt;math&gt; E &lt;/math&gt; is on &lt;math&gt; \overline{BC} &lt;/math&gt;, and the area of &lt;math&gt; \bigtriangleup ABE &lt;/math&gt; is &lt;math&gt; 40 &lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> <br /> We are given that the area of &lt;math&gt;\triangle ABE&lt;/math&gt; is &lt;math&gt;40&lt;/math&gt;, and that &lt;math&gt;AB = 10&lt;/math&gt;.<br /> <br /> The area of a triangle:<br /> <br /> &lt;math&gt;A = \frac{bh}{2}&lt;/math&gt;<br /> <br /> Using &lt;math&gt;AB&lt;/math&gt; as the height of &lt;math&gt;\triangle ABE&lt;/math&gt;,<br /> <br /> &lt;math&gt;40 = \frac{10b}{2}&lt;/math&gt;<br /> <br /> and solving for b,<br /> <br /> &lt;math&gt;b = 8&lt;/math&gt;, which is &lt;math&gt;E&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems&diff=53190 2013 AMC 12A Problems 2013-07-03T19:53:00Z <p>Kj2002: /* Problem 1 */</p> <hr /> <div>{{AMC12 Problems|year=2013|ab=A}}<br /> <br /> == Problem 1 ==<br /> <br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt; 10 &lt;/math&gt;. Point &lt;math&gt; E &lt;/math&gt; is on &lt;math&gt; \overline{BC} &lt;/math&gt;, and the area of &lt;math&gt; \bigtriangleup ABE &lt;/math&gt; is &lt;math&gt; 40 &lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> A softball team played ten games, scoring &lt;math&gt;1,2,3,4,5,6,7,8,9&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt; runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score? <br /> <br /> &lt;math&gt; \textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> What is the value of &lt;cmath&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &amp;#36;&lt;math&gt;105&lt;/math&gt;, Dorothy paid &amp;#36;&lt;math&gt;125&lt;/math&gt;, and Sammy paid &amp;#36;&lt;math&gt;175&lt;/math&gt;. In order to share the costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> The sequence &lt;math&gt;S_1, S_2, S_3, \cdots, S_{10}&lt;/math&gt; has the property that every term beginning with the third is the sum of the previous two. That is, &lt;cmath&gt; S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3. &lt;/cmath&gt; Suppose that &lt;math&gt;S_9 = 110&lt;/math&gt; and &lt;math&gt;S_7 = 42&lt;/math&gt;. What is &lt;math&gt;S_4&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Given that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are distinct nonzero real numbers such that &lt;math&gt;x+\tfrac{2}{x} = y + \tfrac{2}{y}&lt;/math&gt;, what is &lt;math&gt;xy&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;\tfrac{1}{n}&lt;/math&gt; has the repeating decimal representation &lt;math&gt;0.\overline{ab} = 0.ababab\cdots,&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; different digits. What is the sum of the elements of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is equilateral with &lt;math&gt;AB=1&lt;/math&gt;. Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are on &lt;math&gt;\overline{AC}&lt;/math&gt; and points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on &lt;math&gt;\overline{AB}&lt;/math&gt; such that both &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt; are parallel to &lt;math&gt;\overline{BC}&lt;/math&gt;. Furthermore, triangle &lt;math&gt;ADE&lt;/math&gt; and trapezoids &lt;math&gt;DFGE&lt;/math&gt; and &lt;math&gt;FBCG&lt;/math&gt; all have the same perimeter. What is &lt;math&gt;DE+FG&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real s=1/2,m=5/6,l=1;<br /> pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m;<br /> draw(A--B--C--cycle^^D--E^^F--G);<br /> dot(A^^B^^C^^D^^E^^F^^G);<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,S);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,S);<br /> label(&quot;$G$&quot;,G,NW);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad<br /> \textbf{(B) }\dfrac{3}{2}\qquad<br /> \textbf{(C) }\dfrac{21}{13}\qquad<br /> \textbf{(D) }\dfrac{13}{8}\qquad<br /> \textbf{(E) }\dfrac{5}{3}\qquad&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> The angles in a particular triangle are in arithmetic progression, and the side lengths are &lt;math&gt;4,5,x&lt;/math&gt;. The sum of the possible values of x equals &lt;math&gt;a+\sqrt{b}+\sqrt{c}&lt;/math&gt; where &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> Let points &lt;math&gt; A = (0,0) , \ B = (1,2), \ C = (3,3), &lt;/math&gt; and &lt;math&gt; D = (4,0) &lt;/math&gt;. Quadrilateral &lt;math&gt; ABCD &lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt; A &lt;/math&gt;. This line intersects &lt;math&gt; \overline{CD} &lt;/math&gt; at point &lt;math&gt; \left (\frac{p}{q}, \frac{r}{s} \right ) &lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt; p + q + r + s &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The sequence<br /> <br /> &lt;math&gt;\log_{12}{162}&lt;/math&gt;, &lt;math&gt;\log_{12}{x}&lt;/math&gt;, &lt;math&gt;\log_{12}{y}&lt;/math&gt;, &lt;math&gt;\log_{12}{z}&lt;/math&gt;, &lt;math&gt;\log_{12}{1250}&lt;/math&gt;<br /> <br /> is an arithmetic progression. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are three piles of rocks. The mean weight of the rocks in &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;40&lt;/math&gt; pounds, the mean weight of the rocks in &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; pounds, the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;43&lt;/math&gt; pounds, and the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;44&lt;/math&gt; pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> In &lt;math&gt; \bigtriangleup ABC &lt;/math&gt;, &lt;math&gt; AB = 86 &lt;/math&gt;, and &lt;math&gt; AC = 97 &lt;/math&gt;. A circle with center &lt;math&gt; A &lt;/math&gt; and radius &lt;math&gt; AB &lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at points &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; X &lt;/math&gt;. Moreover &lt;math&gt; \overline{BX} &lt;/math&gt; and &lt;math&gt; \overline{CX} &lt;/math&gt; have integer lengths. What is &lt;math&gt; BC &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set &lt;math&gt;\{1,2,3,...,19\}&lt;/math&gt;. For &lt;math&gt;a,b \in S&lt;/math&gt;, define &lt;math&gt;a \succ b&lt;/math&gt; to mean that either &lt;math&gt;0 &lt; a - b \le 9&lt;/math&gt; or &lt;math&gt;b - a &gt; 9&lt;/math&gt;. How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of elements of &lt;math&gt;S&lt;/math&gt; have the property that &lt;math&gt;x \succ y&lt;/math&gt;, &lt;math&gt;y \succ z&lt;/math&gt;, and &lt;math&gt;z \succ x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Consider &lt;math&gt; A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots )))) &lt;/math&gt;. Which of the following intervals contains &lt;math&gt; A &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ (\log 2016, \log 2017) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(B)} \ (\log 2017, \log 2018) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(C)} \ (\log 2018, \log 2019) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(D)} \ (\log 2019, \log 2020) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(E)} \ (\log 2020, \log 2021) &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome &lt;math&gt;n&lt;/math&gt; is chosen uniformly at random. What is the probability that &lt;math&gt;\frac{n}{11}&lt;/math&gt; is also a palindrome?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> &lt;math&gt; ABCD&lt;/math&gt; is a square of side length &lt;math&gt; \sqrt{3} + 1 &lt;/math&gt;. Point &lt;math&gt; P &lt;/math&gt; is on &lt;math&gt; \overline{AC} &lt;/math&gt; such that &lt;math&gt; AP = \sqrt{2} &lt;/math&gt;. The square region bounded by &lt;math&gt; ABCD &lt;/math&gt; is rotated &lt;math&gt; 90^{\circ} &lt;/math&gt; counterclockwise with center &lt;math&gt; P &lt;/math&gt;, sweeping out a region whose area is &lt;math&gt; \frac{1}{c} (a \pi + b) &lt;/math&gt;, where &lt;math&gt;a &lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt; c &lt;/math&gt; are positive integers and &lt;math&gt; \text{gcd}(a,b,c) = 1 &lt;/math&gt;. What is &lt;math&gt; a + b + c &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> Let &lt;math&gt;f : \mathbb{C} \to \mathbb{C} &lt;/math&gt; be defined by &lt;math&gt; f(z) = z^2 + iz + 1 &lt;/math&gt;. How many complex numbers &lt;math&gt;z &lt;/math&gt; are there such that &lt;math&gt; \text{Im}(z) &gt; 0 &lt;/math&gt; and both the real and the imaginary parts of &lt;math&gt;f(z)&lt;/math&gt; are integers with absolute value at most &lt;math&gt; 10 &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D}} \ 431 \qquad \textbf{(E)} \ 441 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|before=[[2012 AMC 12B Problems]]|after=[[2013 AMC 12B Problems]]}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems&diff=53189 2013 AMC 12A Problems 2013-07-03T19:52:31Z <p>Kj2002: /* Problem 1 */</p> <hr /> <div>{{AMC12 Problems|year=2013|ab=A}}<br /> <br /> == Problem 1 ==<br /> <br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt; 10 &lt;/math&gt;. Point &lt;math&gt; E &lt;/math&gt; is on &lt;math&gt; \overline{BC} &lt;/math&gt;, and the area of &lt;math&gt; \bigtriangleup ABE &lt;/math&gt; is &lt;math&gt; 40 &lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 1|Solution]]<br /> {{MAA Notice}}<br /> <br /> == Problem 2 ==<br /> <br /> A softball team played ten games, scoring &lt;math&gt;1,2,3,4,5,6,7,8,9&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt; runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score? <br /> <br /> &lt;math&gt; \textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> What is the value of &lt;cmath&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &amp;#36;&lt;math&gt;105&lt;/math&gt;, Dorothy paid &amp;#36;&lt;math&gt;125&lt;/math&gt;, and Sammy paid &amp;#36;&lt;math&gt;175&lt;/math&gt;. In order to share the costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> The sequence &lt;math&gt;S_1, S_2, S_3, \cdots, S_{10}&lt;/math&gt; has the property that every term beginning with the third is the sum of the previous two. That is, &lt;cmath&gt; S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3. &lt;/cmath&gt; Suppose that &lt;math&gt;S_9 = 110&lt;/math&gt; and &lt;math&gt;S_7 = 42&lt;/math&gt;. What is &lt;math&gt;S_4&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Given that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are distinct nonzero real numbers such that &lt;math&gt;x+\tfrac{2}{x} = y + \tfrac{2}{y}&lt;/math&gt;, what is &lt;math&gt;xy&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;\tfrac{1}{n}&lt;/math&gt; has the repeating decimal representation &lt;math&gt;0.\overline{ab} = 0.ababab\cdots,&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; different digits. What is the sum of the elements of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is equilateral with &lt;math&gt;AB=1&lt;/math&gt;. Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are on &lt;math&gt;\overline{AC}&lt;/math&gt; and points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on &lt;math&gt;\overline{AB}&lt;/math&gt; such that both &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt; are parallel to &lt;math&gt;\overline{BC}&lt;/math&gt;. Furthermore, triangle &lt;math&gt;ADE&lt;/math&gt; and trapezoids &lt;math&gt;DFGE&lt;/math&gt; and &lt;math&gt;FBCG&lt;/math&gt; all have the same perimeter. What is &lt;math&gt;DE+FG&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real s=1/2,m=5/6,l=1;<br /> pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m;<br /> draw(A--B--C--cycle^^D--E^^F--G);<br /> dot(A^^B^^C^^D^^E^^F^^G);<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,S);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,S);<br /> label(&quot;$G$&quot;,G,NW);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad<br /> \textbf{(B) }\dfrac{3}{2}\qquad<br /> \textbf{(C) }\dfrac{21}{13}\qquad<br /> \textbf{(D) }\dfrac{13}{8}\qquad<br /> \textbf{(E) }\dfrac{5}{3}\qquad&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> The angles in a particular triangle are in arithmetic progression, and the side lengths are &lt;math&gt;4,5,x&lt;/math&gt;. The sum of the possible values of x equals &lt;math&gt;a+\sqrt{b}+\sqrt{c}&lt;/math&gt; where &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> Let points &lt;math&gt; A = (0,0) , \ B = (1,2), \ C = (3,3), &lt;/math&gt; and &lt;math&gt; D = (4,0) &lt;/math&gt;. Quadrilateral &lt;math&gt; ABCD &lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt; A &lt;/math&gt;. This line intersects &lt;math&gt; \overline{CD} &lt;/math&gt; at point &lt;math&gt; \left (\frac{p}{q}, \frac{r}{s} \right ) &lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt; p + q + r + s &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The sequence<br /> <br /> &lt;math&gt;\log_{12}{162}&lt;/math&gt;, &lt;math&gt;\log_{12}{x}&lt;/math&gt;, &lt;math&gt;\log_{12}{y}&lt;/math&gt;, &lt;math&gt;\log_{12}{z}&lt;/math&gt;, &lt;math&gt;\log_{12}{1250}&lt;/math&gt;<br /> <br /> is an arithmetic progression. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are three piles of rocks. The mean weight of the rocks in &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;40&lt;/math&gt; pounds, the mean weight of the rocks in &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; pounds, the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;43&lt;/math&gt; pounds, and the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;44&lt;/math&gt; pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> In &lt;math&gt; \bigtriangleup ABC &lt;/math&gt;, &lt;math&gt; AB = 86 &lt;/math&gt;, and &lt;math&gt; AC = 97 &lt;/math&gt;. A circle with center &lt;math&gt; A &lt;/math&gt; and radius &lt;math&gt; AB &lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at points &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; X &lt;/math&gt;. Moreover &lt;math&gt; \overline{BX} &lt;/math&gt; and &lt;math&gt; \overline{CX} &lt;/math&gt; have integer lengths. What is &lt;math&gt; BC &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set &lt;math&gt;\{1,2,3,...,19\}&lt;/math&gt;. For &lt;math&gt;a,b \in S&lt;/math&gt;, define &lt;math&gt;a \succ b&lt;/math&gt; to mean that either &lt;math&gt;0 &lt; a - b \le 9&lt;/math&gt; or &lt;math&gt;b - a &gt; 9&lt;/math&gt;. How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of elements of &lt;math&gt;S&lt;/math&gt; have the property that &lt;math&gt;x \succ y&lt;/math&gt;, &lt;math&gt;y \succ z&lt;/math&gt;, and &lt;math&gt;z \succ x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Consider &lt;math&gt; A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots )))) &lt;/math&gt;. Which of the following intervals contains &lt;math&gt; A &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ (\log 2016, \log 2017) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(B)} \ (\log 2017, \log 2018) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(C)} \ (\log 2018, \log 2019) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(D)} \ (\log 2019, \log 2020) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(E)} \ (\log 2020, \log 2021) &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome &lt;math&gt;n&lt;/math&gt; is chosen uniformly at random. What is the probability that &lt;math&gt;\frac{n}{11}&lt;/math&gt; is also a palindrome?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> &lt;math&gt; ABCD&lt;/math&gt; is a square of side length &lt;math&gt; \sqrt{3} + 1 &lt;/math&gt;. Point &lt;math&gt; P &lt;/math&gt; is on &lt;math&gt; \overline{AC} &lt;/math&gt; such that &lt;math&gt; AP = \sqrt{2} &lt;/math&gt;. The square region bounded by &lt;math&gt; ABCD &lt;/math&gt; is rotated &lt;math&gt; 90^{\circ} &lt;/math&gt; counterclockwise with center &lt;math&gt; P &lt;/math&gt;, sweeping out a region whose area is &lt;math&gt; \frac{1}{c} (a \pi + b) &lt;/math&gt;, where &lt;math&gt;a &lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt; c &lt;/math&gt; are positive integers and &lt;math&gt; \text{gcd}(a,b,c) = 1 &lt;/math&gt;. What is &lt;math&gt; a + b + c &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> Let &lt;math&gt;f : \mathbb{C} \to \mathbb{C} &lt;/math&gt; be defined by &lt;math&gt; f(z) = z^2 + iz + 1 &lt;/math&gt;. How many complex numbers &lt;math&gt;z &lt;/math&gt; are there such that &lt;math&gt; \text{Im}(z) &gt; 0 &lt;/math&gt; and both the real and the imaginary parts of &lt;math&gt;f(z)&lt;/math&gt; are integers with absolute value at most &lt;math&gt; 10 &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D}} \ 431 \qquad \textbf{(E)} \ 441 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|before=[[2012 AMC 12B Problems]]|after=[[2013 AMC 12B Problems]]}}<br /> {{MAA Notice}}</div> Kj2002 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems&diff=53188 2013 AMC 12A Problems 2013-07-03T19:50:50Z <p>Kj2002: /* See also */</p> <hr /> <div>{{AMC12 Problems|year=2013|ab=A}}<br /> <br /> == Problem 1 ==<br /> <br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt; 10 &lt;/math&gt;. Point &lt;math&gt; E &lt;/math&gt; is on &lt;math&gt; \overline{BC} &lt;/math&gt;, and the area of &lt;math&gt; \bigtriangleup ABE &lt;/math&gt; is &lt;math&gt; 40 &lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> A softball team played ten games, scoring &lt;math&gt;1,2,3,4,5,6,7,8,9&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt; runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score? <br /> <br /> &lt;math&gt; \textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> What is the value of &lt;cmath&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &amp;#36;&lt;math&gt;105&lt;/math&gt;, Dorothy paid &amp;#36;&lt;math&gt;125&lt;/math&gt;, and Sammy paid &amp;#36;&lt;math&gt;175&lt;/math&gt;. In order to share the costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> The sequence &lt;math&gt;S_1, S_2, S_3, \cdots, S_{10}&lt;/math&gt; has the property that every term beginning with the third is the sum of the previous two. That is, &lt;cmath&gt; S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3. &lt;/cmath&gt; Suppose that &lt;math&gt;S_9 = 110&lt;/math&gt; and &lt;math&gt;S_7 = 42&lt;/math&gt;. What is &lt;math&gt;S_4&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Given that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are distinct nonzero real numbers such that &lt;math&gt;x+\tfrac{2}{x} = y + \tfrac{2}{y}&lt;/math&gt;, what is &lt;math&gt;xy&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;\tfrac{1}{n}&lt;/math&gt; has the repeating decimal representation &lt;math&gt;0.\overline{ab} = 0.ababab\cdots,&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; different digits. What is the sum of the elements of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is equilateral with &lt;math&gt;AB=1&lt;/math&gt;. Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are on &lt;math&gt;\overline{AC}&lt;/math&gt; and points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on &lt;math&gt;\overline{AB}&lt;/math&gt; such that both &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt; are parallel to &lt;math&gt;\overline{BC}&lt;/math&gt;. Furthermore, triangle &lt;math&gt;ADE&lt;/math&gt; and trapezoids &lt;math&gt;DFGE&lt;/math&gt; and &lt;math&gt;FBCG&lt;/math&gt; all have the same perimeter. What is &lt;math&gt;DE+FG&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real s=1/2,m=5/6,l=1;<br /> pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m;<br /> draw(A--B--C--cycle^^D--E^^F--G);<br /> dot(A^^B^^C^^D^^E^^F^^G);<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,S);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,S);<br /> label(&quot;$G$&quot;,G,NW);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad<br /> \textbf{(B) }\dfrac{3}{2}\qquad<br /> \textbf{(C) }\dfrac{21}{13}\qquad<br /> \textbf{(D) }\dfrac{13}{8}\qquad<br /> \textbf{(E) }\dfrac{5}{3}\qquad&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> The angles in a particular triangle are in arithmetic progression, and the side lengths are &lt;math&gt;4,5,x&lt;/math&gt;. The sum of the possible values of x equals &lt;math&gt;a+\sqrt{b}+\sqrt{c}&lt;/math&gt; where &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> Let points &lt;math&gt; A = (0,0) , \ B = (1,2), \ C = (3,3), &lt;/math&gt; and &lt;math&gt; D = (4,0) &lt;/math&gt;. Quadrilateral &lt;math&gt; ABCD &lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt; A &lt;/math&gt;. This line intersects &lt;math&gt; \overline{CD} &lt;/math&gt; at point &lt;math&gt; \left (\frac{p}{q}, \frac{r}{s} \right ) &lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt; p + q + r + s &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The sequence<br /> <br /> &lt;math&gt;\log_{12}{162}&lt;/math&gt;, &lt;math&gt;\log_{12}{x}&lt;/math&gt;, &lt;math&gt;\log_{12}{y}&lt;/math&gt;, &lt;math&gt;\log_{12}{z}&lt;/math&gt;, &lt;math&gt;\log_{12}{1250}&lt;/math&gt;<br /> <br /> is an arithmetic progression. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are three piles of rocks. The mean weight of the rocks in &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;40&lt;/math&gt; pounds, the mean weight of the rocks in &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; pounds, the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;43&lt;/math&gt; pounds, and the mean weight of the rocks in the combined piles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;44&lt;/math&gt; pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> A group of &lt;math&gt; 12 &lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt; k^\text{th} &lt;/math&gt; pirate to take a share takes &lt;math&gt; \frac{k}{12} &lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt; 12^{\text{th}} &lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> In &lt;math&gt; \bigtriangleup ABC &lt;/math&gt;, &lt;math&gt; AB = 86 &lt;/math&gt;, and &lt;math&gt; AC = 97 &lt;/math&gt;. A circle with center &lt;math&gt; A &lt;/math&gt; and radius &lt;math&gt; AB &lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at points &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; X &lt;/math&gt;. Moreover &lt;math&gt; \overline{BX} &lt;/math&gt; and &lt;math&gt; \overline{CX} &lt;/math&gt; have integer lengths. What is &lt;math&gt; BC &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set &lt;math&gt;\{1,2,3,...,19\}&lt;/math&gt;. For &lt;math&gt;a,b \in S&lt;/math&gt;, define &lt;math&gt;a \succ b&lt;/math&gt; to mean that either &lt;math&gt;0 &lt; a - b \le 9&lt;/math&gt; or &lt;math&gt;b - a &gt; 9&lt;/math&gt;. How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of elements of &lt;math&gt;S&lt;/math&gt; have the property that &lt;math&gt;x \succ y&lt;/math&gt;, &lt;math&gt;y \succ z&lt;/math&gt;, and &lt;math&gt;z \succ x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Consider &lt;math&gt; A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots )))) &lt;/math&gt;. Which of the following intervals contains &lt;math&gt; A &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ (\log 2016, \log 2017) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(B)} \ (\log 2017, \log 2018) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(C)} \ (\log 2018, \log 2019) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(D)} \ (\log 2019, \log 2020) &lt;/math&gt;<br /> &lt;math&gt; \textbf{(E)} \ (\log 2020, \log 2021) &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome &lt;math&gt;n&lt;/math&gt; is chosen uniformly at random. What is the probability that &lt;math&gt;\frac{n}{11}&lt;/math&gt; is also a palindrome?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> &lt;math&gt; ABCD&lt;/math&gt; is a square of side length &lt;math&gt; \sqrt{3} + 1 &lt;/math&gt;. Point &lt;math&gt; P &lt;/math&gt; is on &lt;math&gt; \overline{AC} &lt;/math&gt; such that &lt;math&gt; AP = \sqrt{2} &lt;/math&gt;. The square region bounded by &lt;math&gt; ABCD &lt;/math&gt; is rotated &lt;math&gt; 90^{\circ} &lt;/math&gt; counterclockwise with center &lt;math&gt; P &lt;/math&gt;, sweeping out a region whose area is &lt;math&gt; \frac{1}{c} (a \pi + b) &lt;/math&gt;, where &lt;math&gt;a &lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt; c &lt;/math&gt; are positive integers and &lt;math&gt; \text{gcd}(a,b,c) = 1 &lt;/math&gt;. What is &lt;math&gt; a + b + c &lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt; \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}&lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> Let &lt;math&gt;f : \mathbb{C} \to \mathbb{C} &lt;/math&gt; be defined by &lt;math&gt; f(z) = z^2 + iz + 1 &lt;/math&gt;. How many complex numbers &lt;math&gt;z &lt;/math&gt; are there such that &lt;math&gt; \text{Im}(z) &gt; 0 &lt;/math&gt; and both the real and the imaginary parts of &lt;math&gt;f(z)&lt;/math&gt; are integers with absolute value at most &lt;math&gt; 10 &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D}} \ 431 \qquad \textbf{(E)} \ 441 &lt;/math&gt;<br /> <br /> [[2013 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|before=[[2012 AMC 12B Problems]]|after=[[2013 AMC 12B Problems]]}}<br /> {{MAA Notice}}</div> Kj2002