https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Vsamc&feedformat=atom AoPS Wiki - User contributions [en] 2020-07-09T08:39:33Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems&diff=125585 2013 AMC 12A Problems 2020-06-16T16:45:34Z <p>Vsamc: fixed latex</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 10A Problems/Problem 23|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 &lt;math&gt;12&lt;/math&gt;-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=A|before=[[2012 AMC 12B Problems]]|after=[[2013 AMC 12B Problems]]}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems&diff=125583 2013 AMC 12B Problems 2020-06-16T16:43:12Z <p>Vsamc: fixed latex again</p> <hr /> <div>{{AMC12 Problems|year=2013|ab=B}}<br /> <br /> ==Problem 1==<br /> On a particular January day, the high temperature in Lincoln, Nebraska, was &lt;math&gt;16&lt;/math&gt; degrees higher than the low temperature, and the average of the high and low temperatures was &lt;math&gt;3\textdegree&lt;/math&gt;. In degrees, what was the low temperature in Lincoln that day?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \textbf{(D)}\ -3 \qquad \textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Mr. Green measures his rectangular garden by walking two of the sides and finds that it is &lt;math&gt;15&lt;/math&gt; steps by &lt;math&gt;20&lt;/math&gt; steps. Each of Mr. Green’s steps is &lt;math&gt;2&lt;/math&gt; feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> When counting from &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;201&lt;/math&gt;, &lt;math&gt;53&lt;/math&gt; is the &lt;math&gt;51^{\text{st}}&lt;/math&gt; number counted. When counting backwards from &lt;math&gt;201&lt;/math&gt; to &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;53&lt;/math&gt; is the &lt;math&gt;n^{\text{th}}&lt;/math&gt; number counted. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The average age of &lt;math&gt;33&lt;/math&gt; fifth-graders is &lt;math&gt;11&lt;/math&gt;. The average age of &lt;math&gt;55&lt;/math&gt; of their parents is &lt;math&gt;33&lt;/math&gt;. What is the average age of all of these parents and fifth-graders?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy the equation &lt;math&gt;x^2 + y^2 = 10x - 6y - 34&lt;/math&gt;. What is &lt;math&gt;x + y&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Jo and Blair take turns counting from &lt;math&gt;1&lt;/math&gt; to one more than the last number said by the other person. Jo starts by saying &lt;math&gt;1&quot;&lt;/math&gt;, so Blair follows by saying &lt;math&gt;1, 2&quot;&lt;/math&gt;. Jo then says &lt;math&gt;1, 2, 3&quot;&lt;/math&gt;, and so on. What is the &lt;math&gt;53^{\text{rd}}&lt;/math&gt; number said?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Line &lt;math&gt;l_1&lt;/math&gt; has equation &lt;math&gt;3x - 2y = 1&lt;/math&gt; and goes through &lt;math&gt;A = (-1, -2)&lt;/math&gt;. Line &lt;math&gt;l_2&lt;/math&gt; has equation &lt;math&gt;y = 1&lt;/math&gt; and meets line &lt;math&gt;l_1&lt;/math&gt; at point &lt;math&gt;B&lt;/math&gt;. Line &lt;math&gt;l_3&lt;/math&gt; has positive slope, goes through point &lt;math&gt;A&lt;/math&gt;, and meets &lt;math&gt;l_2&lt;/math&gt; at point &lt;math&gt;C&lt;/math&gt;. The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. What is the slope of &lt;math&gt;l_3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{3}{4} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{4}{3} \qquad \textbf{(E)}\ \frac{3}{2}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides &lt;math&gt;12!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Alex has &lt;math&gt;75&lt;/math&gt; red tokens and &lt;math&gt;75&lt;/math&gt; blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Two bees start at the same spot and fly at the same rate in the following directions. Bee &lt;math&gt;A&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot north, then &lt;math&gt;1&lt;/math&gt; foot east, then &lt;math&gt;1&lt;/math&gt; foot upwards, and then continues to repeat this pattern. Bee &lt;math&gt;B&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot south, then &lt;math&gt;1&lt;/math&gt; foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly &lt;math&gt;10&lt;/math&gt; feet away from each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ A&lt;/math&gt; east, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt;\textbf{(B)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt;\textbf{(C)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt;\textbf{(D)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt;\textbf{(E)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Cities &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; are connected by roads &lt;math&gt;\widetilde{AB}&lt;/math&gt;, &lt;math&gt;\widetilde{AD}&lt;/math&gt;, &lt;math&gt;\widetilde{AE}&lt;/math&gt;, &lt;math&gt;\widetilde{BC}&lt;/math&gt;, &lt;math&gt;\widetilde{BD}&lt;/math&gt;, &lt;math&gt;\widetilde{CD}&lt;/math&gt;, and &lt;math&gt;\widetilde{DE}&lt;/math&gt;. How many different routes are there from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; that use each road exactly once? (Such a route will necessarily visit some cities more than once.)<br /> &lt;asy&gt;<br /> unitsize(10mm);<br /> defaultpen(linewidth(1.2pt)+fontsize(10pt));<br /> dotfactor=4;<br /> pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08);<br /> dot (A);<br /> dot (B);<br /> dot (C);<br /> dot (D);<br /> dot (E);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$E$&quot;,E,W);<br /> guide squiggly(path g, real stepsize, real slope=45)<br /> {<br /> real len = arclength(g);<br /> real step = len / round(len / stepsize);<br /> guide squig;<br /> for (real u = 0; u &lt; len; u += step){<br /> real a = arctime(g, u);<br /> real b = arctime(g, u + step / 2);<br /> pair p = point(g, a);<br /> pair q = point(g, b);<br /> pair np = unit( rotate(slope) * dir(g,a));<br /> pair nq = unit( rotate(0 - slope) * dir(g,b));<br /> squig = squig .. p{np} .. q{nq};<br /> }<br /> squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))};<br /> return squig;<br /> }<br /> pen pp = defaultpen + 2.718;<br /> draw(squiggly(A--B, 4.04, 30), pp);<br /> draw(squiggly(A--D, 7.777, 20), pp);<br /> draw(squiggly(A--E, 5.050, 15), pp);<br /> draw(squiggly(B--C, 5.050, 15), pp);<br /> draw(squiggly(B--D, 4.04, 20), pp);<br /> draw(squiggly(C--D, 2.718, 20), pp);<br /> draw(squiggly(D--E, 2.718, -60), pp);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> The internal angles of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; form an arithmetic progression. Triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;DCB&lt;/math&gt; are similar with &lt;math&gt;\angle DBA = \angle DCB&lt;/math&gt; and &lt;math&gt;\angle ADB = \angle CBD&lt;/math&gt;. Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is &lt;math&gt;N&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;N&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> The number &lt;math&gt;2013&lt;/math&gt; is expressed in the form &lt;br \&gt; &lt;center&gt; &lt;math&gt;2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}&lt;/math&gt;,&lt;/center&gt;&lt;br /&gt;where &lt;math&gt;a_1 \ge a_2 \ge ... \ge a_m&lt;/math&gt; and &lt;math&gt;b_1 \ge b_2 \ge ... \ge b_n&lt;/math&gt; are positive integers and &lt;math&gt;a_1 + b_1&lt;/math&gt; is as small as possible. What is &lt;math&gt;|a_1 - b_1|&lt;/math&gt;?<br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Let &lt;math&gt;ABCDE&lt;/math&gt; be an equiangular convex pentagon of perimeter &lt;math&gt;1&lt;/math&gt;. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let &lt;math&gt;s&lt;/math&gt; be the perimeter of this star. What is the difference between the maximum and the minimum possible values of &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> <br /> Let &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be real numbers such that <br /> <br /> &lt;cmath&gt;a+b+c=2, \text{ and} &lt;/cmath&gt;<br /> &lt;cmath&gt; a^2+b^2+c^2=12 &lt;/cmath&gt;<br /> <br /> What is the difference between the maximum and minimum possible values of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }2\qquad \textbf{ (B) }\frac{10}{3}\qquad \textbf{ (C) }4 \qquad \textbf{ (D) }\frac{16}{3}\qquad \textbf{ (E) }\frac{20}{3} &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt; coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with &lt;math&gt;2013&lt;/math&gt; coins and when the game starts with &lt;math&gt;2014&lt;/math&gt; coins?<br /> <br /> &lt;math&gt; \textbf{(A)}&lt;/math&gt; Barbara will win with &lt;math&gt;2013&lt;/math&gt; coins and Jenna will win with &lt;math&gt;2014&lt;/math&gt; coins. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; Jenna will win with &lt;math&gt;2013&lt;/math&gt; coins, and whoever goes first will win with &lt;math&gt;2014&lt;/math&gt; coins. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; Barbara will win with &lt;math&gt;2013&lt;/math&gt; coins, and whoever goes second will win with &lt;math&gt;2014&lt;/math&gt; coins.<br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; Jenna will win with &lt;math&gt;2013&lt;/math&gt; coins, and Barbara will win with &lt;math&gt;2014&lt;/math&gt; coins.<br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; Whoever goes first will win with &lt;math&gt;2013&lt;/math&gt; coins, and whoever goes second will win with &lt;math&gt;2014&lt;/math&gt; coins.<br /> <br /> [[2013 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=13&lt;/math&gt;, &lt;math&gt;BC=14&lt;/math&gt;, and &lt;math&gt;CA=15&lt;/math&gt;. Distinct points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; lie on segments &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt;, and &lt;math&gt;\overline{DE}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{AD}\perp\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{DE}\perp\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{AF}\perp\overline{BF}&lt;/math&gt;. The length of segment &lt;math&gt;\overline{DF}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> For &lt;math&gt;135^\circ &lt; x &lt; 180^\circ&lt;/math&gt;, points &lt;math&gt;P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)&lt;/math&gt; and &lt;math&gt;S =(\tan x, \tan^2 x)&lt;/math&gt; are the vertices of a trapezoid. What is &lt;math&gt;\sin(2x)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D)}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Consider the set of &lt;math&gt;30&lt;/math&gt; parabolas defined as follows: all parabolas have as focus the point &lt;math&gt;(0,0)&lt;/math&gt; and the directrix lines have the form &lt;math&gt;y=ax+b&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; integers such that &lt;math&gt;a\in \{-2,-1,0,1,2\}&lt;/math&gt; and &lt;math&gt;b\in \{-3,-2,-1,1,2,3\}&lt;/math&gt;. No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D)}\ 840\qquad\textbf{(E)}\ 870 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Let &lt;math&gt;m&gt;1&lt;/math&gt; and &lt;math&gt;n&gt;1&lt;/math&gt; be integers. Suppose that the product of the solutions for &lt;math&gt;x&lt;/math&gt; of the equation<br /> &lt;cmath&gt; 8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0 &lt;/cmath&gt;<br /> is the smallest possible integer. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Bernardo chooses a three-digit positive integer &lt;math&gt;N&lt;/math&gt; and writes both its base-&lt;math&gt;5&lt;/math&gt; and base-&lt;math&gt;6&lt;/math&gt; representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-&lt;math&gt;10&lt;/math&gt; integers, he adds them to obtain an integer &lt;math&gt;S&lt;/math&gt;. For example, if &lt;math&gt;N=749&lt;/math&gt;, Bernardo writes the numbers &lt;math&gt;10,444&lt;/math&gt; and &lt;math&gt;3,245&lt;/math&gt;, and LeRoy obtains the sum &lt;math&gt;S=13,689&lt;/math&gt;. For how many choices of &lt;math&gt;N&lt;/math&gt; are the two rightmost digits of &lt;math&gt;S&lt;/math&gt;, in order, the same as those of &lt;math&gt;2N&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> [[2013 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{CN}&lt;/math&gt; is the angle bisector of &lt;math&gt;\angle{ACB}&lt;/math&gt; with &lt;math&gt;N&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be the intersection of the median &lt;math&gt;\overline{BM}&lt;/math&gt; and the bisector &lt;math&gt;\overline{CN}&lt;/math&gt;. In addition &lt;math&gt;\triangle BXN&lt;/math&gt; is equilateral with &lt;math&gt;AC=2&lt;/math&gt;. What is &lt;math&gt;BN^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;G&lt;/math&gt; be the set of polynomials of the form<br /> &lt;cmath&gt; P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50, &lt;/cmath&gt;<br /> where &lt;math&gt; c_1,c_2,\cdots, c_{n-1} &lt;/math&gt; are integers and &lt;math&gt;P(z)&lt;/math&gt; has distinct roots of the form &lt;math&gt;a+ib&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; integers. How many polynomials are in &lt;math&gt;G&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|before=[[2013 AMC 12A Problems]]|after=[[2014 AMC 12A Problems]]}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems&diff=125582 2013 AMC 12B Problems 2020-06-16T16:41:35Z <p>Vsamc: LaTeXed some stuff</p> <hr /> <div>{{AMC12 Problems|year=2013|ab=B}}<br /> <br /> ==Problem 1==<br /> On a particular January day, the high temperature in Lincoln, Nebraska, was &lt;math&gt;16&lt;/math&gt; degrees higher than the low temperature, and the average of the high and low temperatures was &lt;math&gt;3\textdegree&lt;/math&gt;. In degrees, what was the low temperature in Lincoln that day?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \textbf{(D)}\ -3 \qquad \textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Mr. Green measures his rectangular garden by walking two of the sides and finds that it is &lt;math&gt;15&lt;/math&gt; steps by &lt;math&gt;20&lt;/math&gt; steps. Each of Mr. Green’s steps is &lt;math&gt;2&lt;/math&gt; feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> When counting from &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;201&lt;/math&gt;, &lt;math&gt;53&lt;/math&gt; is the &lt;math&gt;51^{\text{st}}&lt;/math&gt; number counted. When counting backwards from &lt;math&gt;201&lt;/math&gt; to &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;53&lt;/math&gt; is the &lt;math&gt;n^{\text{th}}&lt;/math&gt; number counted. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The average age of &lt;math&gt;33&lt;/math&gt; fifth-graders is &lt;math&gt;11&lt;/math&gt;. The average age of &lt;math&gt;55&lt;/math&gt; of their parents is &lt;math&gt;33&lt;/math&gt;. What is the average age of all of these parents and fifth-graders?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy the equation &lt;math&gt;x^2 + y^2 = 10x - 6y - 34&lt;/math&gt;. What is &lt;math&gt;x + y&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Jo and Blair take turns counting from &lt;math&gt;1&lt;/math&gt; to one more than the last number said by the other person. Jo starts by saying &lt;math&gt;1&quot;&lt;/math&gt;, so Blair follows by saying &lt;math&gt;1, 2&quot;&lt;/math&gt;. Jo then says &lt;math&gt;1, 2, 3&quot;&lt;/math&gt;, and so on. What is the &lt;math&gt;53^{\text{rd}}&lt;/math&gt; number said?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Line &lt;math&gt;l_1&lt;/math&gt; has equation &lt;math&gt;3x - 2y = 1&lt;/math&gt; and goes through &lt;math&gt;A = (-1, -2)&lt;/math&gt;. Line &lt;math&gt;l_2&lt;/math&gt; has equation &lt;math&gt;y = 1&lt;/math&gt; and meets line &lt;math&gt;l_1&lt;/math&gt; at point &lt;math&gt;B&lt;/math&gt;. Line &lt;math&gt;l_3&lt;/math&gt; has positive slope, goes through point &lt;math&gt;A&lt;/math&gt;, and meets &lt;math&gt;l_2&lt;/math&gt; at point &lt;math&gt;C&lt;/math&gt;. The area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. What is the slope of &lt;math&gt;l_3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{3}{4} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{4}{3} \qquad \textbf{(E)}\ \frac{3}{2}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides &lt;math&gt;12!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Alex has &lt;math&gt;75&lt;/math&gt; red tokens and &lt;math&gt;75&lt;/math&gt; blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Two bees start at the same spot and fly at the same rate in the following directions. Bee &lt;math&gt;A&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot north, then &lt;math&gt;1&lt;/math&gt; foot east, then &lt;math&gt;1&lt;/math&gt; foot upwards, and then continues to repeat this pattern. Bee &lt;math&gt;B&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot south, then &lt;math&gt;1&lt;/math&gt; foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly &lt;math&gt;10&lt;/math&gt; feet away from each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ A&lt;/math&gt; east, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt;\textbf{(B)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt;\textbf{(C)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt;\textbf{(D)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt;\textbf{(E)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Cities &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; are connected by roads &lt;math&gt;\widetilde{AB}&lt;/math&gt;, &lt;math&gt;\widetilde{AD}&lt;/math&gt;, &lt;math&gt;\widetilde{AE}&lt;/math&gt;, &lt;math&gt;\widetilde{BC}&lt;/math&gt;, &lt;math&gt;\widetilde{BD}&lt;/math&gt;, &lt;math&gt;\widetilde{CD}&lt;/math&gt;, and &lt;math&gt;\widetilde{DE}&lt;/math&gt;. How many different routes are there from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; that use each road exactly once? (Such a route will necessarily visit some cities more than once.)<br /> &lt;asy&gt;<br /> unitsize(10mm);<br /> defaultpen(linewidth(1.2pt)+fontsize(10pt));<br /> dotfactor=4;<br /> pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08);<br /> dot (A);<br /> dot (B);<br /> dot (C);<br /> dot (D);<br /> dot (E);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$E$&quot;,E,W);<br /> guide squiggly(path g, real stepsize, real slope=45)<br /> {<br /> real len = arclength(g);<br /> real step = len / round(len / stepsize);<br /> guide squig;<br /> for (real u = 0; u &lt; len; u += step){<br /> real a = arctime(g, u);<br /> real b = arctime(g, u + step / 2);<br /> pair p = point(g, a);<br /> pair q = point(g, b);<br /> pair np = unit( rotate(slope) * dir(g,a));<br /> pair nq = unit( rotate(0 - slope) * dir(g,b));<br /> squig = squig .. p{np} .. q{nq};<br /> }<br /> squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))};<br /> return squig;<br /> }<br /> pen pp = defaultpen + 2.718;<br /> draw(squiggly(A--B, 4.04, 30), pp);<br /> draw(squiggly(A--D, 7.777, 20), pp);<br /> draw(squiggly(A--E, 5.050, 15), pp);<br /> draw(squiggly(B--C, 5.050, 15), pp);<br /> draw(squiggly(B--D, 4.04, 20), pp);<br /> draw(squiggly(C--D, 2.718, 20), pp);<br /> draw(squiggly(D--E, 2.718, -60), pp);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> The internal angles of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; form an arithmetic progression. Triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;DCB&lt;/math&gt; are similar with &lt;math&gt;\angle DBA = \angle DCB&lt;/math&gt; and &lt;math&gt;\angle ADB = \angle CBD&lt;/math&gt;. Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is &lt;math&gt;N&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;N&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> The number &lt;math&gt;2013&lt;/math&gt; is expressed in the form &lt;br \&gt; &lt;center&gt; &lt;math&gt;2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}&lt;/math&gt;,&lt;/center&gt;&lt;br /&gt;where &lt;math&gt;a_1 \ge a_2 \ge ... \ge a_m&lt;/math&gt; and &lt;math&gt;b_1 \ge b_2 \ge ... \ge b_n&lt;/math&gt; are positive integers and &lt;math&gt;a_1 + b_1&lt;/math&gt; is as small as possible. What is &lt;math&gt;|a_1 - b_1|&lt;/math&gt;?<br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Let &lt;math&gt;ABCDE&lt;/math&gt; be an equiangular convex pentagon of perimeter &lt;math&gt;1&lt;/math&gt;. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let &lt;math&gt;s&lt;/math&gt; be the perimeter of this star. What is the difference between the maximum and the minimum possible values of &lt;math&gt;s&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> <br /> Let &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be real numbers such that <br /> <br /> &lt;cmath&gt;a+b+c=2, \text{ and} &lt;/cmath&gt;<br /> &lt;cmath&gt; a^2+b^2+c^2=12 &lt;/cmath&gt;<br /> <br /> What is the difference between the maximum and minimum possible values of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }2\qquad \textbf{ (B) }\frac{10}{3}\qquad \textbf{ (C) }4 \qquad \textbf{ (D) }\frac{16}{3}\qquad \textbf{ (E) }\frac{20}{3} &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt; coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with &lt;math&gt;2013&lt;/math&gt; coins and when the game starts with &lt;math&gt;2014&lt;/math&gt; coins?<br /> <br /> &lt;math&gt; \textbf{(A)}&lt;/math&gt; Barbara will win with &lt;math&gt;2013&lt;/math&gt; coins and Jenna will win with &lt;math&gt;2014&lt;/math&gt; coins. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; Jenna will win with &lt;math&gt;2013&lt;/math&gt; coins, and whoever goes first will win with &lt;math&gt;2014&lt;/math&gt; coins. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; Barbara will win with &lt;math&gt;2013&lt;/math&gt; coins, and whoever goes second will win with &lt;math&gt;2014&lt;/math&gt; coins.<br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; Jenna will win with &lt;math&gt;2013&lt;/math&gt; coins, and Barbara will win with &lt;math&gt;2014&lt;/math&gt; coins.<br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; Whoever goes first will win with &lt;math&gt;2013&lt;/math&gt; coins, and whoever goes second will win with &lt;math&gt;2014&lt;/math&gt; coins.<br /> <br /> [[2013 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=13&lt;/math&gt;, &lt;math&gt;BC=14&lt;/math&gt;, and &lt;math&gt;CA=15&lt;/math&gt;. Distinct points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; lie on segments &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt;, and &lt;math&gt;\overline{DE}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{AD}\perp\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{DE}\perp\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{AF}\perp\overline{BF}&lt;/math&gt;. The length of segment &lt;math&gt;\overline{DF}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> For &lt;math&gt;135^\circ &lt; x &lt; 180^\circ&lt;/math&gt;, points &lt;math&gt;P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)&lt;/math&gt; and &lt;math&gt;S =(\tan x, \tan^2 x)&lt;/math&gt; are the vertices of a trapezoid. What is &lt;math&gt;\sin(2x)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D)}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point &lt;math&gt;(0,0)&lt;/math&gt; and the directrix lines have the form &lt;math&gt;y=ax+b&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; integers such that &lt;math&gt;a\in \{-2,-1,0,1,2\}&lt;/math&gt; and &lt;math&gt;b\in \{-3,-2,-1,1,2,3\}&lt;/math&gt;. No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D)}\ 840\qquad\textbf{(E)}\ 870 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Let &lt;math&gt;m&gt;1&lt;/math&gt; and &lt;math&gt;n&gt;1&lt;/math&gt; be integers. Suppose that the product of the solutions for &lt;math&gt;x&lt;/math&gt; of the equation<br /> &lt;cmath&gt; 8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0 &lt;/cmath&gt;<br /> is the smallest possible integer. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Bernardo chooses a three-digit positive integer &lt;math&gt;N&lt;/math&gt; and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer &lt;math&gt;S&lt;/math&gt;. For example, if &lt;math&gt;N=749&lt;/math&gt;, Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains the sum &lt;math&gt;S=13,689&lt;/math&gt;. For how many choices of &lt;math&gt;N&lt;/math&gt; are the two rightmost digits of &lt;math&gt;S&lt;/math&gt;, in order, the same as those of &lt;math&gt;2N&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> [[2013 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{CN}&lt;/math&gt; is the angle bisector of &lt;math&gt;\angle{ACB}&lt;/math&gt; with &lt;math&gt;N&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be the intersection of the median &lt;math&gt;\overline{BM}&lt;/math&gt; and the bisector &lt;math&gt;\overline{CN}&lt;/math&gt;. In addition &lt;math&gt;\triangle BXN&lt;/math&gt; is equilateral with &lt;math&gt;AC=2&lt;/math&gt;. What is &lt;math&gt;BN^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}&lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;G&lt;/math&gt; be the set of polynomials of the form<br /> &lt;cmath&gt; P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50, &lt;/cmath&gt;<br /> where &lt;math&gt; c_1,c_2,\cdots, c_{n-1} &lt;/math&gt; are integers and &lt;math&gt;P(z)&lt;/math&gt; has distinct roots of the form &lt;math&gt;a+ib&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; integers. How many polynomials are in &lt;math&gt;G&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056 &lt;/math&gt;<br /> <br /> [[2013 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|before=[[2013 AMC 12A Problems]]|after=[[2014 AMC 12A Problems]]}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17&diff=124626 2013 AMC 12B Problems/Problem 17 2020-06-09T20:05:08Z <p>Vsamc: </p> <hr /> <div>==Problem==<br /> <br /> <br /> Let &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be real numbers such that <br /> <br /> &lt;cmath&gt;a+b+c=2, \text{ and} &lt;/cmath&gt;<br /> &lt;cmath&gt; a^2+b^2+c^2=12 &lt;/cmath&gt;<br /> <br /> What is the difference between the maximum and minimum possible values of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;a+b= 2-c&lt;/math&gt;. Now, by Cauchy-Schwarz, we have that &lt;math&gt;(a^2+b^2) \ge \frac{(2-c)^2}{2}&lt;/math&gt;. Therefore, we have that &lt;math&gt;\frac{(2-c)^2}{2}+c^2 \le 12&lt;/math&gt;. We then find the roots of &lt;math&gt;c&lt;/math&gt; that satisfy equality and find the difference of the roots. This gives the answer, &lt;math&gt;\boxed{\textbf{(D)} \ \frac{16}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> This is similar to the first solution but is far more intuitive. From the given, we have &lt;cmath&gt; a + b = 2 - c&lt;/cmath&gt; &lt;cmath&gt;a^2 + b^2 = 12 - c^2 &lt;/cmath&gt; This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have &lt;cmath&gt; 2\,(a^2 + b^2) \geq (a + b)^2&lt;/cmath&gt; Substitution of the above results and some algebra yields &lt;cmath&gt; 3c^2 - 4c - 20 \leq 0 &lt;/cmath&gt; This quadratic inequality is easily solved, and it is seen that equality holds for &lt;math&gt;c = -2&lt;/math&gt; and &lt;math&gt;c = \frac{10}{3}&lt;/math&gt;. <br /> <br /> The difference between these two values is &lt;math&gt;\boxed{\textbf{(D)} \ \frac{16}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> (no Cauchy-Schwarz)<br /> <br /> From the first equation, we know that &lt;math&gt;c=2-a-b&lt;/math&gt;. We substitute this into the second equation to find that<br /> &lt;cmath&gt;a^2+b^2+(2-a-b)^2=12.&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;2a^2+2b^2-4a-4b+2ab=8&lt;/math&gt;, which we can write as the quadratic &lt;math&gt;a^2+(b-2)a+(b^2-2b-4)=0&lt;/math&gt;. We wish to find real values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy this equation. Therefore, the discriminant is nonnegative. Hence,<br /> &lt;cmath&gt;(b-2)^2-4(b^2-2b-4)\ge0,&lt;/cmath&gt;<br /> or &lt;math&gt;-3b^2+4b+20\ge 0&lt;/math&gt;. This factors as &lt;math&gt;-(3b-10)(b+2)\ge 0&lt;/math&gt;. Therefore, &lt;math&gt;-2\le b\le \frac{10}{3}&lt;/math&gt;, and by symmetry this must be true for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; as well.<br /> <br /> Now &lt;math&gt;a=b=2&lt;/math&gt; and &lt;math&gt;c=-2&lt;/math&gt; satisfy both equations, so we see that &lt;math&gt;c=-2&lt;/math&gt; must be the minimum possible value of &lt;math&gt;c&lt;/math&gt;. Also, &lt;math&gt;c=\frac{10}{3}&lt;/math&gt; and &lt;math&gt;a=b=-\frac{2}{3}&lt;/math&gt; satisfy both equations, so we see that &lt;math&gt;c=\frac{10}{3}&lt;/math&gt; is the maximum possible value of &lt;math&gt;c&lt;/math&gt;. The difference between these is &lt;math&gt;\frac{10}{3}-(-2)=\frac{16}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We take a geometrical approach.<br /> <br /> From the given, we have &lt;math&gt; a + b = 2 - c&lt;/math&gt; and &lt;math&gt;a^2 + b^2 = 12 - c^2 &lt;/math&gt;. The first equation is a line with x and y intercepts of &lt;math&gt;2-c&lt;/math&gt; and the second equation is a circle centered at the origin with radius &lt;math&gt;\sqrt{12-c^2}&lt;/math&gt;. Intuitively, if we want to find the minimum / maximum &lt;math&gt;c&lt;/math&gt; such that there still exist real solutions, the two graphs of the equations should be tangent.<br /> <br /> Thus, we have that &lt;math&gt;\sqrt{2} \cdot \sqrt{12-c^2} = 2-c&lt;/math&gt;, which simplifies to &lt;math&gt;3c^2-4c-20=0&lt;/math&gt;. Solving the quadratic, we get that the values of &lt;math&gt;c&lt;/math&gt; for which the two graphs are tangent are &lt;math&gt;c=-2&lt;/math&gt; and &lt;math&gt;c=\frac{10}{3}&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{\frac{16}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Draw the sphere and the plane represented by the two equations in Cartesian space, with the &lt;math&gt;z&lt;/math&gt;-axis representing &lt;math&gt;c&lt;/math&gt;. The intersection between the sphere and plane is a circle. We wish to find the point on the circle where &lt;math&gt;z&lt;/math&gt; is minimized and the point where &lt;math&gt;z&lt;/math&gt; is maximized. <br /> Looking at the graph, it is clear by symmetry that &lt;math&gt;x = y&lt;/math&gt; when &lt;math&gt;z&lt;/math&gt; is maximized or minimized. Thus, we can set &lt;math&gt;a = b&lt;/math&gt;. This gives us the following system of equations:<br /> <br /> &lt;math&gt;2a + c = 2&lt;/math&gt;<br /> <br /> &lt;math&gt;2a^2 + c^2 = 12&lt;/math&gt;<br /> <br /> Solving gives &lt;math&gt;c = \frac{10}{3}, -2&lt;/math&gt;, which are the maximum and minimum values of &lt;math&gt;c&lt;/math&gt; respectively. The answer follows from here.<br /> <br /> ==Solution 6==<br /> We can consider &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; to be solutions to a cubic equation. Then, given our information, we have &lt;math&gt;a+b+c=2&lt;/math&gt; and &lt;math&gt;ab+ac+bc=-4&lt;/math&gt;, so our cubic equation looks like this: &lt;math&gt;x^3-2x^2-4x+m&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; can be any real number. <br /> <br /> Since this cubic has &lt;math&gt;3&lt;/math&gt; real solutions, it must have both a maximum and a minimum (noticed by looking at the graph). Then the greatest solution is maximized when the other two solutions are the same and is minimized when the other two solutions are the same. <br /> <br /> Thus, equating &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we have &lt;cmath&gt;2a+c=2&lt;/cmath&gt;and&lt;cmath&gt;2a^2+c^2=12&lt;/cmath&gt;Solving this, we get the quadratic &lt;cmath&gt;3c^2-4c+4=24 \Rightarrow c=\frac{10}{3}, -2&lt;/cmath&gt;implying the answer.<br /> <br /> ==Solution 7==<br /> Subtracting &lt;math&gt;c&lt;/math&gt; from the first equation yields &lt;math&gt;a+b=2-c&lt;/math&gt;. Subtracting &lt;math&gt;c^2&lt;/math&gt; from the second equation yields &lt;math&gt;a^2+b^2=12-c^2&lt;/math&gt;. Thus we have the equations <br /> &lt;cmath&gt;\begin{cases} <br /> a+b=2-c\\<br /> a^2+b^2=12-c^2\\<br /> \end{cases}&lt;/cmath&gt;<br /> Squaring the first equation yields &lt;cmath&gt;a^2+2ab+b^2=4-4c+c^2&lt;/cmath&gt; Subtracting the second equation from this one yields &lt;cmath&gt;2ab=2c^2-4c-8&lt;/cmath&gt; &lt;cmath&gt;\iff ab=c^2-2c-4&lt;/cmath&gt; Thus we have the system of equations <br /> &lt;cmath&gt;\begin{cases} <br /> a+b=2-c\\<br /> ab=c^2-2c-4\\<br /> \end{cases}&lt;/cmath&gt;<br /> We can reverse [[Vieta's Formulas]], to get that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are roots of the quadratic equation(in &lt;math&gt;x&lt;/math&gt;) &lt;cmath&gt;x^2-(2-c)x+(c^2-2c-4).&lt;/cmath&gt; Because we have &lt;math&gt;a, b\in \mathbb{R}&lt;/math&gt;, we have that the discriminant of this quadratic equation must be nonnegative. The discriminant is &lt;cmath&gt;(2-c)^2-4(c^2-2c-4)=-3c^2+4c+20,&lt;/cmath&gt; and it can be factored as &lt;math&gt;-3(c+2)(c-\frac{10}{2})&lt;/math&gt;. Since we have &lt;math&gt;-3(c+2)(c-\frac{10}{2})\geq 0&lt;/math&gt;, we must have &lt;math&gt;(c+2)(c-\frac{10}{2})\leq 0&lt;/math&gt;. If &lt;math&gt;c&lt;-2&lt;/math&gt;, then we have that &lt;math&gt;(c+2)(c-\frac{10}{2})&gt;0&lt;/math&gt;. If &lt;math&gt;c=-2&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})=0&lt;/math&gt;. If &lt;math&gt;c\in (-2, \frac{10}{3})&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})&lt;0&lt;/math&gt;. If &lt;math&gt;c=\frac{10}{3}&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})=0&lt;/math&gt;. If &lt;math&gt;c&gt;\frac{10}{3}&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})&gt;0&lt;/math&gt;. Thus our inequality holds if and only if &lt;math&gt;c\in \left [-2, \frac{10}{3}\right]&lt;/math&gt;, and the maxmimum value is &lt;math&gt;\frac{10}{3}&lt;/math&gt;, whilst the minimum value is &lt;math&gt;-2&lt;/math&gt;. Thus the difference between the maximum and minimum values is &lt;math&gt;\frac{10}{3}-(-2)=\boxed{\textbf{(D)} \frac{16}{3}}&lt;/math&gt;<br /> <br /> -vsamc<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17&diff=124625 2013 AMC 12B Problems/Problem 17 2020-06-09T20:04:49Z <p>Vsamc: </p> <hr /> <div>==Problem==<br /> <br /> <br /> Let &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be real numbers such that <br /> <br /> &lt;cmath&gt;a+b+c=2, \text{ and} &lt;/cmath&gt;<br /> &lt;cmath&gt; a^2+b^2+c^2=12 &lt;/cmath&gt;<br /> <br /> What is the difference between the maximum and minimum possible values of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;a+b= 2-c&lt;/math&gt;. Now, by Cauchy-Schwarz, we have that &lt;math&gt;(a^2+b^2) \ge \frac{(2-c)^2}{2}&lt;/math&gt;. Therefore, we have that &lt;math&gt;\frac{(2-c)^2}{2}+c^2 \le 12&lt;/math&gt;. We then find the roots of &lt;math&gt;c&lt;/math&gt; that satisfy equality and find the difference of the roots. This gives the answer, &lt;math&gt;\boxed{\textbf{(D)} \ \frac{16}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> This is similar to the first solution but is far more intuitive. From the given, we have &lt;cmath&gt; a + b = 2 - c&lt;/cmath&gt; &lt;cmath&gt;a^2 + b^2 = 12 - c^2 &lt;/cmath&gt; This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have &lt;cmath&gt; 2\,(a^2 + b^2) \geq (a + b)^2&lt;/cmath&gt; Substitution of the above results and some algebra yields &lt;cmath&gt; 3c^2 - 4c - 20 \leq 0 &lt;/cmath&gt; This quadratic inequality is easily solved, and it is seen that equality holds for &lt;math&gt;c = -2&lt;/math&gt; and &lt;math&gt;c = \frac{10}{3}&lt;/math&gt;. <br /> <br /> The difference between these two values is &lt;math&gt;\boxed{\textbf{(D)} \ \frac{16}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> (no Cauchy-Schwarz)<br /> <br /> From the first equation, we know that &lt;math&gt;c=2-a-b&lt;/math&gt;. We substitute this into the second equation to find that<br /> &lt;cmath&gt;a^2+b^2+(2-a-b)^2=12.&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;2a^2+2b^2-4a-4b+2ab=8&lt;/math&gt;, which we can write as the quadratic &lt;math&gt;a^2+(b-2)a+(b^2-2b-4)=0&lt;/math&gt;. We wish to find real values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy this equation. Therefore, the discriminant is nonnegative. Hence,<br /> &lt;cmath&gt;(b-2)^2-4(b^2-2b-4)\ge0,&lt;/cmath&gt;<br /> or &lt;math&gt;-3b^2+4b+20\ge 0&lt;/math&gt;. This factors as &lt;math&gt;-(3b-10)(b+2)\ge 0&lt;/math&gt;. Therefore, &lt;math&gt;-2\le b\le \frac{10}{3}&lt;/math&gt;, and by symmetry this must be true for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; as well.<br /> <br /> Now &lt;math&gt;a=b=2&lt;/math&gt; and &lt;math&gt;c=-2&lt;/math&gt; satisfy both equations, so we see that &lt;math&gt;c=-2&lt;/math&gt; must be the minimum possible value of &lt;math&gt;c&lt;/math&gt;. Also, &lt;math&gt;c=\frac{10}{3}&lt;/math&gt; and &lt;math&gt;a=b=-\frac{2}{3}&lt;/math&gt; satisfy both equations, so we see that &lt;math&gt;c=\frac{10}{3}&lt;/math&gt; is the maximum possible value of &lt;math&gt;c&lt;/math&gt;. The difference between these is &lt;math&gt;\frac{10}{3}-(-2)=\frac{16}{3}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We take a geometrical approach.<br /> <br /> From the given, we have &lt;math&gt; a + b = 2 - c&lt;/math&gt; and &lt;math&gt;a^2 + b^2 = 12 - c^2 &lt;/math&gt;. The first equation is a line with x and y intercepts of &lt;math&gt;2-c&lt;/math&gt; and the second equation is a circle centered at the origin with radius &lt;math&gt;\sqrt{12-c^2}&lt;/math&gt;. Intuitively, if we want to find the minimum / maximum &lt;math&gt;c&lt;/math&gt; such that there still exist real solutions, the two graphs of the equations should be tangent.<br /> <br /> Thus, we have that &lt;math&gt;\sqrt{2} \cdot \sqrt{12-c^2} = 2-c&lt;/math&gt;, which simplifies to &lt;math&gt;3c^2-4c-20=0&lt;/math&gt;. Solving the quadratic, we get that the values of &lt;math&gt;c&lt;/math&gt; for which the two graphs are tangent are &lt;math&gt;c=-2&lt;/math&gt; and &lt;math&gt;c=\frac{10}{3}&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{\frac{16}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Draw the sphere and the plane represented by the two equations in Cartesian space, with the &lt;math&gt;z&lt;/math&gt;-axis representing &lt;math&gt;c&lt;/math&gt;. The intersection between the sphere and plane is a circle. We wish to find the point on the circle where &lt;math&gt;z&lt;/math&gt; is minimized and the point where &lt;math&gt;z&lt;/math&gt; is maximized. <br /> Looking at the graph, it is clear by symmetry that &lt;math&gt;x = y&lt;/math&gt; when &lt;math&gt;z&lt;/math&gt; is maximized or minimized. Thus, we can set &lt;math&gt;a = b&lt;/math&gt;. This gives us the following system of equations:<br /> <br /> &lt;math&gt;2a + c = 2&lt;/math&gt;<br /> <br /> &lt;math&gt;2a^2 + c^2 = 12&lt;/math&gt;<br /> <br /> Solving gives &lt;math&gt;c = \frac{10}{3}, -2&lt;/math&gt;, which are the maximum and minimum values of &lt;math&gt;c&lt;/math&gt; respectively. The answer follows from here.<br /> <br /> ==Solution 6==<br /> We can consider &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; to be solutions to a cubic equation. Then, given our information, we have &lt;math&gt;a+b+c=2&lt;/math&gt; and &lt;math&gt;ab+ac+bc=-4&lt;/math&gt;, so our cubic equation looks like this: &lt;math&gt;x^3-2x^2-4x+m&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; can be any real number. <br /> <br /> Since this cubic has &lt;math&gt;3&lt;/math&gt; real solutions, it must have both a maximum and a minimum (noticed by looking at the graph). Then the greatest solution is maximized when the other two solutions are the same and is minimized when the other two solutions are the same. <br /> <br /> Thus, equating &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we have &lt;cmath&gt;2a+c=2&lt;/cmath&gt;and&lt;cmath&gt;2a^2+c^2=12&lt;/cmath&gt;Solving this, we get the quadratic &lt;cmath&gt;3c^2-4c+4=24 \Rightarrow c=\frac{10}{3}, -2&lt;/cmath&gt;implying the answer.<br /> <br /> ==Solution 7==<br /> Subtracting &lt;math&gt;c&lt;/math&gt; from the first equation yields &lt;math&gt;a+b=2-c&lt;/math&gt;. Subtracting &lt;math&gt;c^2&lt;/math&gt; from the second equation yields &lt;math&gt;a^2+b^2=12-c^2&lt;/math&gt;. Thus we have the equations <br /> &lt;cmath&gt;\begin{cases} <br /> a+b=2-c\\<br /> a^2+b^2=12-c^2\\<br /> \end{cases}&lt;/cmath&gt;<br /> Squaring the first equation yields &lt;cmath&gt;a^2+2ab+b^2=4-4c+c^2&lt;/cmath&gt; Subtracting the second equation from this one yields &lt;cmath&gt;2ab=2c^2-4c-8&lt;/cmath&gt; &lt;cmath&gt;\iff ab=c^2-2c-4&lt;/cmath&gt; Thus we have the system of equations <br /> &lt;cmath&gt;\begin{cases} <br /> a+b=2-c\\<br /> ab=c^2-2c-4\\<br /> \end{cases}&lt;/cmath&gt;<br /> We can reverse [[Vieta's Formula's]], to get that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are roots of the quadratic equation(in &lt;math&gt;x&lt;/math&gt;) &lt;cmath&gt;x^2-(2-c)x+(c^2-2c-4).&lt;/cmath&gt; Because we have &lt;math&gt;a, b\in \mathbb{R}&lt;/math&gt;, we have that the discriminant of this quadratic equation must be nonnegative. The discriminant is &lt;cmath&gt;(2-c)^2-4(c^2-2c-4)=-3c^2+4c+20,&lt;/cmath&gt; and it can be factored as &lt;math&gt;-3(c+2)(c-\frac{10}{2})&lt;/math&gt;. Since we have &lt;math&gt;-3(c+2)(c-\frac{10}{2})\geq 0&lt;/math&gt;, we must have &lt;math&gt;(c+2)(c-\frac{10}{2})\leq 0&lt;/math&gt;. If &lt;math&gt;c&lt;-2&lt;/math&gt;, then we have that &lt;math&gt;(c+2)(c-\frac{10}{2})&gt;0&lt;/math&gt;. If &lt;math&gt;c=-2&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})=0&lt;/math&gt;. If &lt;math&gt;c\in (-2, \frac{10}{3})&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})&lt;0&lt;/math&gt;. If &lt;math&gt;c=\frac{10}{3}&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})=0&lt;/math&gt;. If &lt;math&gt;c&gt;\frac{10}{3}&lt;/math&gt;, then &lt;math&gt;(c+2)(c-\frac{10}{2})&gt;0&lt;/math&gt;. Thus our inequality holds if and only if &lt;math&gt;c\in \left [-2, \frac{10}{3}\right]&lt;/math&gt;, and the maxmimum value is &lt;math&gt;\frac{10}{3}&lt;/math&gt;, whilst the minimum value is &lt;math&gt;-2&lt;/math&gt;. Thus the difference between the maximum and minimum values is &lt;math&gt;\frac{10}{3}-(-2)=\boxed{\textbf{(D)} \frac{16}{3}}&lt;/math&gt;<br /> <br /> -vsamc<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=120860 2018 AIME I Problems/Problem 5 2020-04-12T18:52:54Z <p>Vsamc: </p> <hr /> <div>==Problem 5==<br /> <br /> For each ordered pair of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfying &lt;cmath&gt;\log_2(2x+y) = \log_4(x^2+xy+7y^2)&lt;/cmath&gt;there is a real number &lt;math&gt;K&lt;/math&gt; such that &lt;cmath&gt;\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).&lt;/cmath&gt;Find the product of all possible values of &lt;math&gt;K&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Using the logarithmic property &lt;math&gt;\log_{a^n}b^n = \log_{a}b&lt;/math&gt;, we note that &lt;math&gt;(2x+y)^2 = 4x^2+4xy+y^2&lt;/math&gt;. <br /> That gives &lt;math&gt;x^2+xy-2y^2=0&lt;/math&gt; upon simplification and division by &lt;math&gt;3&lt;/math&gt;. Then, &lt;math&gt;x=y&lt;/math&gt; or &lt;math&gt;x=-2y&lt;/math&gt;.<br /> From the second equation, &lt;math&gt;9x^2+6xy+y^2=3x^2+4xy+Ky^2&lt;/math&gt;. If we take &lt;math&gt;x=y&lt;/math&gt;, we see that &lt;math&gt;K=9&lt;/math&gt;. If we take &lt;math&gt;x=-2y&lt;/math&gt;, we see that &lt;math&gt;K=21&lt;/math&gt;. The product is &lt;math&gt;\boxed{189}&lt;/math&gt;.<br /> <br /> -expiLnCalc<br /> <br /> ==Note==<br /> <br /> The cases &lt;math&gt;x=y&lt;/math&gt; and &lt;math&gt;x=-2y&lt;/math&gt; can be found by SFFT ([https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick]) from &lt;math&gt;x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0&lt;/math&gt;.<br /> ==Solution 2==<br /> Do as done in Solution 1 to get &lt;math&gt;x^2+xy-2y^2=0\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2&lt;/math&gt;. Do as done in Solution 1 to get &lt;math&gt;9x^2+6xy+y^2=3x^2+4xy+Ky^2\implies 6x^2+2xy+(1-K)y^2=0\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\implies \frac{x}{y}=&lt;/math&gt; &lt;math&gt;\frac{-2\pm \sqrt{4-24(1-K)}}{12}&lt;/math&gt; &lt;math&gt;\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}&lt;/math&gt;. If &lt;math&gt;\frac{x}{y}=1&lt;/math&gt;, then &lt;math&gt;1=\frac{-1\pm \sqrt{6K-5}}{6}\implies 6=-1\pm \sqrt{6K-5}\implies 7=\pm \sqrt{6K-5}\implies 49=6K-5\implies K=9&lt;/math&gt;. If &lt;math&gt;\frac{x}{y}=-2&lt;/math&gt;, then &lt;math&gt;-2=\frac{-1\pm \sqrt{6K-5}}{6}\implies -12=-1\pm \sqrt{6K-5}\implies -11=\sqrt{6K-5}\implies 121=6K-5\implies 126=6K\implies K=21&lt;/math&gt;. Hence our final answer is &lt;math&gt;21\cdot 6=\boxed{126}&lt;/math&gt;<br /> -vsamc<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=120859 2018 AIME I Problems/Problem 5 2020-04-12T18:52:17Z <p>Vsamc: </p> <hr /> <div>==Problem 5==<br /> <br /> For each ordered pair of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfying &lt;cmath&gt;\log_2(2x+y) = \log_4(x^2+xy+7y^2)&lt;/cmath&gt;there is a real number &lt;math&gt;K&lt;/math&gt; such that &lt;cmath&gt;\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).&lt;/cmath&gt;Find the product of all possible values of &lt;math&gt;K&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Using the logarithmic property &lt;math&gt;\log_{a^n}b^n = \log_{a}b&lt;/math&gt;, we note that &lt;math&gt;(2x+y)^2 = 4x^2+4xy+y^2&lt;/math&gt;. <br /> That gives &lt;math&gt;x^2+xy-2y^2=0&lt;/math&gt; upon simplification and division by &lt;math&gt;3&lt;/math&gt;. Then, &lt;math&gt;x=y&lt;/math&gt; or &lt;math&gt;x=-2y&lt;/math&gt;.<br /> From the second equation, &lt;math&gt;9x^2+6xy+y^2=3x^2+4xy+Ky^2&lt;/math&gt;. If we take &lt;math&gt;x=y&lt;/math&gt;, we see that &lt;math&gt;K=9&lt;/math&gt;. If we take &lt;math&gt;x=-2y&lt;/math&gt;, we see that &lt;math&gt;K=21&lt;/math&gt;. The product is &lt;math&gt;\boxed{189}&lt;/math&gt;.<br /> <br /> -expiLnCalc<br /> <br /> ==Note==<br /> <br /> The cases &lt;math&gt;x=y&lt;/math&gt; and &lt;math&gt;x=-2y&lt;/math&gt; can be found by SFFT ([https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick]) from &lt;math&gt;x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0&lt;/math&gt;.<br /> ==Solution 2==<br /> Do as done in Solution 1 to get &lt;math&gt;x^2+xy-2y^2=0\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2&lt;/math&gt;. Do as done in Solution 1 to get &lt;math&gt;9x^2+6xy+y^2=3x^2+4xy+Ky^2\implies 6x^2+2xy+(1-K)y^2=0\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}&lt;/math&gt; &lt;math&gt;\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}&lt;/math&gt;. If &lt;math&gt;\frac{x}{y}=1&lt;/math&gt;, then &lt;math&gt;1=\frac{-1\pm \sqrt{6K-5}}{6}\implies 6=-1\pm \sqrt{6K-5}\implies 7=\pm \sqrt{6K-5}\implies 49=6K-5\implies K=9&lt;/math&gt;. If &lt;math&gt;\frac{x}{y}=-2&lt;/math&gt;, then &lt;math&gt;-2=\frac{-1\pm \sqrt{6K-5}}{6}\implies -12=-1\pm \sqrt{6K-5}\implies -11=\sqrt{6K-5}\implies 121=6K-5\implies 126=6K\implies K=21&lt;/math&gt;. Hence our final answer is &lt;math&gt;21\cdot 6=\boxed{126}&lt;/math&gt;<br /> -vsamc<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=120858 2018 AIME I Problems/Problem 5 2020-04-12T18:51:49Z <p>Vsamc: </p> <hr /> <div>==Problem 5==<br /> <br /> For each ordered pair of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfying &lt;cmath&gt;\log_2(2x+y) = \log_4(x^2+xy+7y^2)&lt;/cmath&gt;there is a real number &lt;math&gt;K&lt;/math&gt; such that &lt;cmath&gt;\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).&lt;/cmath&gt;Find the product of all possible values of &lt;math&gt;K&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Using the logarithmic property &lt;math&gt;\log_{a^n}b^n = \log_{a}b&lt;/math&gt;, we note that &lt;math&gt;(2x+y)^2 = 4x^2+4xy+y^2&lt;/math&gt;. <br /> That gives &lt;math&gt;x^2+xy-2y^2=0&lt;/math&gt; upon simplification and division by &lt;math&gt;3&lt;/math&gt;. Then, &lt;math&gt;x=y&lt;/math&gt; or &lt;math&gt;x=-2y&lt;/math&gt;.<br /> From the second equation, &lt;math&gt;9x^2+6xy+y^2=3x^2+4xy+Ky^2&lt;/math&gt;. If we take &lt;math&gt;x=y&lt;/math&gt;, we see that &lt;math&gt;K=9&lt;/math&gt;. If we take &lt;math&gt;x=-2y&lt;/math&gt;, we see that &lt;math&gt;K=21&lt;/math&gt;. The product is &lt;math&gt;\boxed{189}&lt;/math&gt;.<br /> <br /> -expiLnCalc<br /> <br /> ==Note==<br /> <br /> The cases &lt;math&gt;x=y&lt;/math&gt; and &lt;math&gt;x=-2y&lt;/math&gt; can be found by SFFT ([https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick]) from &lt;math&gt;x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0&lt;/math&gt;.<br /> ==Solution 2==<br /> Do as done in Solution 1 to get &lt;math&gt;x^2+xy-2y^2=0\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2&lt;/math&gt;. Do as done in Solution 1 to get &lt;math&gt;9x^2+6xy+y^2=3x^2+4xy+Ky^2\implies 6x^2+2xy+(1-K)y^2=0\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}&lt;/math&gt;. If &lt;math&gt;\frac{x}{y}=1&lt;/math&gt;, then &lt;math&gt;1=\frac{-1\pm \sqrt{6K-5}}{6}\implies 6=-1\pm \sqrt{6K-5}\implies 7=\pm \sqrt{6K-5}\implies 49=6K-5\implies K=9&lt;/math&gt;. If &lt;math&gt;\frac{x}{y}=-2&lt;/math&gt;, then &lt;math&gt;-2=\frac{-1\pm \sqrt{6K-5}}{6}\implies -12=-1\pm \sqrt{6K-5}\implies -11=\sqrt{6K-5}\implies 121=6K-5\implies 126=6K\implies K=21&lt;/math&gt;. Hence our final answer is &lt;math&gt;21\cdot 6=\boxed{126}&lt;/math&gt;<br /> -vsamc<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=120631 2018 AMC 12A Problems/Problem 19 2020-04-06T18:52:56Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;A&lt;/math&gt; be the set of positive integers that have no prime factors other than &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;5&lt;/math&gt;. The infinite sum &lt;cmath&gt;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots&lt;/cmath&gt;of the reciprocals of the elements of &lt;math&gt;A&lt;/math&gt; can be expressed as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}&lt;/math&gt;<br /> <br /> == Solution ==<br /> It's just &lt;cmath&gt;<br /> \sum_{a=0}^\infty\frac1{2^a}\sum_{b=0}^\infty\frac1{3^b}\sum_{c=0}^\infty\frac{1}{5^c} =\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty\frac1{2^a3^b5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br /> &lt;/cmath&gt; since this represents all the numbers in the denominator.<br /> (athens2016)<br /> <br /> == Solution 2==<br /> Separate into 7 separate infinite series's so we can calculate each and find the original sum: <br /> <br /> The first infinite sequence shall be all the reciprocals of the powers of &lt;math&gt;2&lt;/math&gt;, the second shall be reciprocals of the powers of &lt;math&gt;3&lt;/math&gt;, and the third will consist of reciprocals of the powers of 5. We can easily calculate these to be &lt;math&gt;1, \frac{1}{2}, \frac{1}{4}&lt;/math&gt; respectively. <br /> <br /> The fourth infinite series shall be all real numbers in the form &lt;math&gt; \frac{1}{2^a3^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The fifth is all real numbers in the form &lt;math&gt; \frac{1}{2^a5^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The sixth is all real numbers in the form &lt;math&gt; \frac{1}{3^a5^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The seventh infinite series is all real numbers in the form &lt;math&gt; \frac{1}{2^a3^b5^c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are greater than or equal to 1. <br /> <br /> Let us denote the first sequence as &lt;math&gt;a_{1}&lt;/math&gt;, the second as &lt;math&gt;a_{2}&lt;/math&gt;, etc. We know &lt;math&gt;a_{1}=1&lt;/math&gt;, &lt;math&gt;a_{2}=\frac{1}{2}&lt;/math&gt;, &lt;math&gt;a_{3}=\frac{1}{4}&lt;/math&gt;, let us find &lt;math&gt;a_{4}&lt;/math&gt;. factoring out &lt;math&gt;\frac{1}{6}&lt;/math&gt; from the terms in this subsequence, we would get &lt;math&gt;a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})&lt;/math&gt;. <br /> <br /> Knowing &lt;math&gt;a_{1}&lt;/math&gt; and &lt;math&gt;a_{2}&lt;/math&gt;, we can substitute and solve for &lt;math&gt;a_{4}&lt;/math&gt;, and we get &lt;math&gt;\frac{1}{2}&lt;/math&gt;. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them &lt;math&gt;\frac{1}{4}&lt;/math&gt; and &lt;math&gt;\frac{1}{8}&lt;/math&gt;. <br /> <br /> Finally, for the seventh sequence, we see &lt;math&gt;a_{7}=\frac{a_{8}}{30}&lt;/math&gt;, where &lt;math&gt;a_{8}&lt;/math&gt; is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get &lt;math&gt;1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}&lt;/math&gt;, but when we separated the sequence into its parts, we ignored the &lt;math&gt;1/1&lt;/math&gt;, so adding in the &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}&lt;/math&gt;, which when we solve for, we get &lt;math&gt;\frac{29}{8}=\frac{29a_{8}}{30}&lt;/math&gt;, &lt;math&gt;\frac{1}{8}=\frac{a_{8}}{30}&lt;/math&gt;, &lt;math&gt;\frac{30}{8}=(a_{8})&lt;/math&gt;, &lt;math&gt;\frac{15}{4}=(a_{8})&lt;/math&gt;. So our answer is &lt;math&gt;\frac{15}{4}&lt;/math&gt;, but we are asked to add the numerator and denominator, which sums up to &lt;math&gt;19&lt;/math&gt;, which is the answer.<br /> <br /> ~~Edited by mprincess0229~~<br /> ==Solution 3==<br /> Clearly this is just summing over the reciprocals of the numbers of the form &lt;math&gt;2^i3^j5^k&lt;/math&gt;, where &lt;math&gt;i,j,k\in [0,\infty)&lt;/math&gt;. SO our desired sum is &lt;math&gt;\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}&lt;/math&gt;. By the infinite geometric series formula, &lt;math&gt;\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}&lt;/math&gt; is just &lt;math&gt;\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}&lt;/math&gt;. Applying the infinite geometric series formula again gives that &lt;math&gt;\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}&lt;/math&gt;. Applying the infinite geometric series formula again yields &lt;math&gt;\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}&lt;/math&gt;. Hence our final answer is &lt;math&gt;15+4=\boxed{19\textbf{(C)}}&lt;/math&gt;.<br /> <br /> -vsamc<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=120630 2018 AMC 12A Problems/Problem 19 2020-04-06T18:52:31Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;A&lt;/math&gt; be the set of positive integers that have no prime factors other than &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;5&lt;/math&gt;. The infinite sum &lt;cmath&gt;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots&lt;/cmath&gt;of the reciprocals of the elements of &lt;math&gt;A&lt;/math&gt; can be expressed as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}&lt;/math&gt;<br /> <br /> == Solution ==<br /> It's just &lt;cmath&gt;<br /> \sum_{a=0}^\infty\frac1{2^a}\sum_{b=0}^\infty\frac1{3^b}\sum_{c=0}^\infty\frac{1}{5^c} =\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty\frac1{2^a3^b5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br /> &lt;/cmath&gt; since this represents all the numbers in the denominator.<br /> (athens2016)<br /> <br /> == Solution 2==<br /> Separate into 7 separate infinite series's so we can calculate each and find the original sum: <br /> <br /> The first infinite sequence shall be all the reciprocals of the powers of &lt;math&gt;2&lt;/math&gt;, the second shall be reciprocals of the powers of &lt;math&gt;3&lt;/math&gt;, and the third will consist of reciprocals of the powers of 5. We can easily calculate these to be &lt;math&gt;1, \frac{1}{2}, \frac{1}{4}&lt;/math&gt; respectively. <br /> <br /> The fourth infinite series shall be all real numbers in the form &lt;math&gt; \frac{1}{2^a3^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The fifth is all real numbers in the form &lt;math&gt; \frac{1}{2^a5^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The sixth is all real numbers in the form &lt;math&gt; \frac{1}{3^a5^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The seventh infinite series is all real numbers in the form &lt;math&gt; \frac{1}{2^a3^b5^c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are greater than or equal to 1. <br /> <br /> Let us denote the first sequence as &lt;math&gt;a_{1}&lt;/math&gt;, the second as &lt;math&gt;a_{2}&lt;/math&gt;, etc. We know &lt;math&gt;a_{1}=1&lt;/math&gt;, &lt;math&gt;a_{2}=\frac{1}{2}&lt;/math&gt;, &lt;math&gt;a_{3}=\frac{1}{4}&lt;/math&gt;, let us find &lt;math&gt;a_{4}&lt;/math&gt;. factoring out &lt;math&gt;\frac{1}{6}&lt;/math&gt; from the terms in this subsequence, we would get &lt;math&gt;a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})&lt;/math&gt;. <br /> <br /> Knowing &lt;math&gt;a_{1}&lt;/math&gt; and &lt;math&gt;a_{2}&lt;/math&gt;, we can substitute and solve for &lt;math&gt;a_{4}&lt;/math&gt;, and we get &lt;math&gt;\frac{1}{2}&lt;/math&gt;. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them &lt;math&gt;\frac{1}{4}&lt;/math&gt; and &lt;math&gt;\frac{1}{8}&lt;/math&gt;. <br /> <br /> Finally, for the seventh sequence, we see &lt;math&gt;a_{7}=\frac{a_{8}}{30}&lt;/math&gt;, where &lt;math&gt;a_{8}&lt;/math&gt; is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get &lt;math&gt;1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}&lt;/math&gt;, but when we separated the sequence into its parts, we ignored the &lt;math&gt;1/1&lt;/math&gt;, so adding in the &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}&lt;/math&gt;, which when we solve for, we get &lt;math&gt;\frac{29}{8}=\frac{29a_{8}}{30}&lt;/math&gt;, &lt;math&gt;\frac{1}{8}=\frac{a_{8}}{30}&lt;/math&gt;, &lt;math&gt;\frac{30}{8}=(a_{8})&lt;/math&gt;, &lt;math&gt;\frac{15}{4}=(a_{8})&lt;/math&gt;. So our answer is &lt;math&gt;\frac{15}{4}&lt;/math&gt;, but we are asked to add the numerator and denominator, which sums up to &lt;math&gt;19&lt;/math&gt;, which is the answer.<br /> <br /> ~~Edited by mprincess0229~~<br /> ==Solution 3==<br /> Clearly this is just summing over the reciprocals of the numbers of the form &lt;math&gt;2^i3^j5^k&lt;/math&gt;, where &lt;math&gt;i,j,k\in [0,\infty)&lt;/math&gt;. SO our desired sum is &lt;math&gt;\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}&lt;/math&gt;. By the infinite geometric series formula, &lt;math&gt;\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}&lt;/math&gt; is just &lt;math&gt;\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}&lt;/math&gt;. Applying the infinite geometric series formula again gives that &lt;math&gt;\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}&lt;/math&gt;. Applying the infinite geometric series formula again yields &lt;math&gt;\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}&lt;/math&gt;. Hence ur final answer is &lt;math&gt;15+4=\boxed{19\textbf{(C)}}&lt;/math&gt;.<br /> <br /> -vsamc<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=120629 2018 AMC 12A Problems/Problem 19 2020-04-06T18:51:58Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;A&lt;/math&gt; be the set of positive integers that have no prime factors other than &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;5&lt;/math&gt;. The infinite sum &lt;cmath&gt;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots&lt;/cmath&gt;of the reciprocals of the elements of &lt;math&gt;A&lt;/math&gt; can be expressed as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}&lt;/math&gt;<br /> <br /> == Solution ==<br /> It's just &lt;cmath&gt;<br /> \sum_{a=0}^\infty\frac1{2^a}\sum_{b=0}^\infty\frac1{3^b}\sum_{c=0}^\infty\frac{1}{5^c} =\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty\frac1{2^a3^b5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br /> &lt;/cmath&gt; since this represents all the numbers in the denominator.<br /> (athens2016)<br /> <br /> == Solution 2==<br /> Separate into 7 separate infinite series's so we can calculate each and find the original sum: <br /> <br /> The first infinite sequence shall be all the reciprocals of the powers of &lt;math&gt;2&lt;/math&gt;, the second shall be reciprocals of the powers of &lt;math&gt;3&lt;/math&gt;, and the third will consist of reciprocals of the powers of 5. We can easily calculate these to be &lt;math&gt;1, \frac{1}{2}, \frac{1}{4}&lt;/math&gt; respectively. <br /> <br /> The fourth infinite series shall be all real numbers in the form &lt;math&gt; \frac{1}{2^a3^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The fifth is all real numbers in the form &lt;math&gt; \frac{1}{2^a5^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The sixth is all real numbers in the form &lt;math&gt; \frac{1}{3^a5^b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are greater than or equal to 1. <br /> <br /> The seventh infinite series is all real numbers in the form &lt;math&gt; \frac{1}{2^a3^b5^c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are greater than or equal to 1. <br /> <br /> Let us denote the first sequence as &lt;math&gt;a_{1}&lt;/math&gt;, the second as &lt;math&gt;a_{2}&lt;/math&gt;, etc. We know &lt;math&gt;a_{1}=1&lt;/math&gt;, &lt;math&gt;a_{2}=\frac{1}{2}&lt;/math&gt;, &lt;math&gt;a_{3}=\frac{1}{4}&lt;/math&gt;, let us find &lt;math&gt;a_{4}&lt;/math&gt;. factoring out &lt;math&gt;\frac{1}{6}&lt;/math&gt; from the terms in this subsequence, we would get &lt;math&gt;a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})&lt;/math&gt;. <br /> <br /> Knowing &lt;math&gt;a_{1}&lt;/math&gt; and &lt;math&gt;a_{2}&lt;/math&gt;, we can substitute and solve for &lt;math&gt;a_{4}&lt;/math&gt;, and we get &lt;math&gt;\frac{1}{2}&lt;/math&gt;. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them &lt;math&gt;\frac{1}{4}&lt;/math&gt; and &lt;math&gt;\frac{1}{8}&lt;/math&gt;. <br /> <br /> Finally, for the seventh sequence, we see &lt;math&gt;a_{7}=\frac{a_{8}}{30}&lt;/math&gt;, where &lt;math&gt;a_{8}&lt;/math&gt; is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get &lt;math&gt;1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}&lt;/math&gt;, but when we separated the sequence into its parts, we ignored the &lt;math&gt;1/1&lt;/math&gt;, so adding in the &lt;math&gt;1&lt;/math&gt;, we get &lt;math&gt;1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}&lt;/math&gt;, which when we solve for, we get &lt;math&gt;\frac{29}{8}=\frac{29a_{8}}{30}&lt;/math&gt;, &lt;math&gt;\frac{1}{8}=\frac{a_{8}}{30}&lt;/math&gt;, &lt;math&gt;\frac{30}{8}=(a_{8})&lt;/math&gt;, &lt;math&gt;\frac{15}{4}=(a_{8})&lt;/math&gt;. So our answer is &lt;math&gt;\frac{15}{4}&lt;/math&gt;, but we are asked to add the numerator and denominator, which sums up to &lt;math&gt;19&lt;/math&gt;, which is the answer.<br /> <br /> ~~Edited by mprincess0229~~<br /> ==Solution 3==<br /> Clearly this is just summing over the reciprocals of the numbers of the form &lt;math&gt;2^i3^j5^k&lt;/math&gt;, where &lt;math&gt;i,j,k\in [0,\infty)&lt;/math&gt;. SO our desired sum is &lt;math&gt;\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}&lt;/math&gt;. By the infinite geometric series formula, &lt;math&gt;\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}&lt;/math&gt; is just &lt;math&gt;\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}&lt;/math&gt;. Applying the infinite geometric series formula again gives that &lt;math&gt;\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}&lt;/math&gt;. Applying the infinite geometric series formula again yields &lt;math&gt;\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}&lt;/math&gt;. Hence ur final answer is &lt;math&gt;15+4=\boxed{19\textbf{(D)}}&lt;/math&gt;.<br /> <br /> -vsamc<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_20&diff=120628 2018 AMC 12A Problems/Problem 20 2020-04-06T18:41:11Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is an isosceles right triangle with &lt;math&gt;AB=AC=3&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of hypotenuse &lt;math&gt;\overline{BC}&lt;/math&gt;. Points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively, so that &lt;math&gt;AI&gt;AE&lt;/math&gt; and &lt;math&gt;AIME&lt;/math&gt; is a cyclic quadrilateral. Given that triangle &lt;math&gt;EMI&lt;/math&gt; has area &lt;math&gt;2&lt;/math&gt;, the length &lt;math&gt;CI&lt;/math&gt; can be written as &lt;math&gt;\frac{a-\sqrt{b}}{c}&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;b&lt;/math&gt; is not divisible by the square of any prime. What is the value of &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }9 \qquad<br /> \textbf{(B) }10 \qquad<br /> \textbf{(C) }11 \qquad<br /> \textbf{(D) }12 \qquad<br /> \textbf{(E) }13 \qquad<br /> &lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> import olympiad;<br /> <br /> size(200);<br /> <br /> pair A, B, C, I, M, E;<br /> <br /> A = (0, 0);<br /> B = (3, 0);<br /> C = (0, 3);<br /> M = (1.5, 1.5);<br /> I = (0, 1.5 + sqrt(2) / 2);<br /> E = (1.5 - sqrt(2) / 2, 0);<br /> <br /> draw(A -- B -- C -- cycle);<br /> draw(I -- M -- E -- cycle);<br /> draw(rightanglemark(I, A, E, 4));<br /> <br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(I);<br /> dot(M);<br /> dot(E);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$I$&quot;, I, NE);<br /> label(&quot;$M$&quot;, M, NE);<br /> label(&quot;$E$&quot;, E + (0.1, 0.04), NE);<br /> label(&quot;$3$&quot;, (A + C) / 2, W);<br /> label(&quot;$3$&quot;, (A + B) / 2, S);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> == Solution 1==<br /> <br /> Observe that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles right (&lt;math&gt;M&lt;/math&gt; is the midpoint of diameter arc &lt;math&gt;EI&lt;/math&gt;), so &lt;math&gt;MI=2,MC=\frac{3}{\sqrt{2}}&lt;/math&gt;. With &lt;math&gt;\angle{MCI}=45^\circ&lt;/math&gt;, we can use Law of Cosines to determine that &lt;math&gt;CI=\frac{3\pm\sqrt{7}}{2}&lt;/math&gt;. The same calculations hold for &lt;math&gt;BE&lt;/math&gt; also, and since &lt;math&gt;CI&lt;BE&lt;/math&gt;, we deduce that &lt;math&gt;CI&lt;/math&gt; is the smaller root, giving the answer of &lt;math&gt;\boxed{12}&lt;/math&gt;.<br /> <br /> == Solution 2 (Using Ptolemy) ==<br /> <br /> We first claim that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles and right.<br /> <br /> Proof: Construct &lt;math&gt;\overline{MF}\perp\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{MG}\perp\overline{AC}&lt;/math&gt;. Since &lt;math&gt;\overline{AM}&lt;/math&gt; bisects &lt;math&gt;\angle{BAC}&lt;/math&gt;, one can deduce that &lt;math&gt;MF=MG&lt;/math&gt;. Then by AAS it is clear that &lt;math&gt;MI=ME&lt;/math&gt; and therefore &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles. Since quadrilateral &lt;math&gt;AIME&lt;/math&gt; is cyclic, one can deduce that &lt;math&gt;\angle{EMI}=90^\circ&lt;/math&gt;. Q.E.D.<br /> <br /> Since the area of &lt;math&gt;\triangle{EMI}&lt;/math&gt; is 2, we can find that &lt;math&gt;MI=ME=2&lt;/math&gt;, &lt;math&gt;EI=2\sqrt{2}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;M&lt;/math&gt; is the mid-point of &lt;math&gt;\overline{BC}&lt;/math&gt;, it is clear that &lt;math&gt;AM=\frac{3\sqrt{2}}{2}&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;AE=a&lt;/math&gt; and &lt;math&gt;AI=b&lt;/math&gt;. By Ptolemy's Theorem, in cyclic quadrilateral &lt;math&gt;AIME&lt;/math&gt;, we have &lt;math&gt;2a+2b=6&lt;/math&gt;. By Pythagorean Theorem, we have &lt;math&gt;a^2+b^2=8&lt;/math&gt;. One can solve the simultaneous system and find &lt;math&gt;b=\frac{3+\sqrt{7}}{2}&lt;/math&gt;. Then by deducting the length of &lt;math&gt;\overline{AI}&lt;/math&gt; from 3 we get &lt;math&gt;CI=\frac{3-\sqrt{7}}{2}&lt;/math&gt;, giving the answer of &lt;math&gt;\boxed{12}&lt;/math&gt;. (Surefire2019)<br /> <br /> == Solution 3 (More Elementary) ==<br /> <br /> Like above, notice that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles and right, which means that &lt;math&gt;\dfrac{ME \cdot MI}{2} = 2&lt;/math&gt;, so &lt;math&gt;MI^2=4&lt;/math&gt; and &lt;math&gt;MI = 2&lt;/math&gt;. Then construct &lt;math&gt;\overline{MF}\perp\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{MG}\perp\overline{AC}&lt;/math&gt; as well as &lt;math&gt;\overline{MI}&lt;/math&gt;. It's clear that &lt;math&gt;MG^2+GI^2 = MI^2&lt;/math&gt; by Pythagorean, so knowing that &lt;math&gt;MG = \dfrac{AB}{2} = \dfrac{3}{2}&lt;/math&gt; allows one to solve to get &lt;math&gt;GI = \dfrac{\sqrt{7}}{2}&lt;/math&gt;. By just looking at the diagram, &lt;math&gt;CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}&lt;/math&gt;. The answer is thus &lt;math&gt;3+7+2=12&lt;/math&gt;.<br /> <br /> == Solution 4 (Coordinate Geometry) ==<br /> Let &lt;math&gt;A&lt;/math&gt; lie on &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; on &lt;math&gt;(0,y)&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt; on &lt;math&gt;(x,0)&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;(\frac{3}{2},\frac{3}{2})&lt;/math&gt;. Since &lt;math&gt;{AIME}&lt;/math&gt; is cyclic, &lt;math&gt;\angle EMI&lt;/math&gt; (which is opposite of another right angle) must be a right angle; therefore, &lt;math&gt;\vec{ME} \cdot \vec{MI} = &lt;\frac{-3}{2}, y-\frac{3}{2}&gt; \cdot &lt;x-\frac{3}{2}, -\frac{3}{2}&gt; = 0&lt;/math&gt;. Compute the dot product to arrive at the relation &lt;math&gt;y=3-x&lt;/math&gt;. We can set up another equation involving the area of &lt;math&gt;\triangle EMI&lt;/math&gt; using the [[Shoelace Theorem]]. This is &lt;math&gt;2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]&lt;/math&gt;. Multiplying, substituting &lt;math&gt;3-x&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;, and simplifying, we get &lt;math&gt;x^2 -3x + \frac{1}{2}=0&lt;/math&gt;. Thus, &lt;math&gt;(x,y)=&lt;/math&gt; &lt;math&gt;(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})&lt;/math&gt;. But &lt;math&gt;AI&gt;AE&lt;/math&gt;, meaning &lt;math&gt;x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}&lt;/math&gt;, and the final answer is &lt;math&gt;3+7+2=\boxed{12}&lt;/math&gt;.<br /> <br /> == Solution 5 (Quick) ==<br /> From &lt;math&gt;AIME&lt;/math&gt; cyclic we get &lt;math&gt;\angle{MEI} = \angle{MAI} = 45^\circ&lt;/math&gt; and &lt;math&gt;\angle{MIE} = \angle{MAE} = 45^\circ&lt;/math&gt;, so &lt;math&gt;\triangle{EMI}&lt;/math&gt; is an isosceles right triangle. <br /> <br /> From &lt;math&gt;[EMI]=2&lt;/math&gt; we get &lt;math&gt;EM=MI=2&lt;/math&gt;. <br /> <br /> Notice &lt;math&gt;\triangle{AEM} \cong \triangle{CIM}&lt;/math&gt;, because &lt;math&gt;\angle{AEM}=180-\angle{AIM}=\angle{CIM}&lt;/math&gt;, &lt;math&gt;EM=IM&lt;/math&gt;, and &lt;math&gt;\angle{EAM}=\angle{ICM}=45^\circ&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;CI=AE=x&lt;/math&gt;, so &lt;math&gt;AI=3-x&lt;/math&gt;. <br /> <br /> By Pythagoras on &lt;math&gt;\triangle{EAI}&lt;/math&gt; we have &lt;math&gt;x^2+(3-x)^2=EI^2=8&lt;/math&gt;, and solve this to get &lt;math&gt;x=CI=\dfrac{3-\sqrt{7}}{2}&lt;/math&gt; for a final answer of &lt;math&gt;3+7+2=\boxed{12}&lt;/math&gt;.<br /> ==Solution 6(bash)==<br /> Let &lt;math&gt;CI=a&lt;/math&gt;, &lt;math&gt;BE=b&lt;/math&gt;. Because opposite angles in a cyclic quadrilateral are supplementary, we have &lt;math&gt;\angle EMI=90^{\circ}&lt;/math&gt;. By the law of cosines, we have &lt;math&gt;MI^2=a^2+\frac{9}{4}-\frac{3}{2}a&lt;/math&gt;, and &lt;math&gt;ME^2=b^2+\frac{9}{4}-\frac{3}{2}b&lt;/math&gt;. Notice that &lt;math&gt;EI=2MO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the origin of the circle mentioned in the problem. Thus &lt;math&gt;\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8&lt;/math&gt;. Thus we have &lt;math&gt;18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}&lt;/math&gt; &lt;math&gt;=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a&lt;/math&gt;. We know that &lt;math&gt;(3-a)^2+(3-b)^2=8\implies (3-a)^2+a^2=8\implies 2a^2-6a+9=8\implies 2a^2-6a+1=0\implies a=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}&lt;/math&gt;. We take the smaller solution because we have &lt;math&gt;AI&gt;AE\implies 3-AI&lt;3-AE\implies CI&lt;CE&lt;/math&gt;, and we want &lt;math&gt;CI&lt;/math&gt;, not &lt;math&gt;CE&lt;/math&gt;, thus &lt;math&gt;CI=\frac{3-\sqrt{7}}{2}&lt;/math&gt;. Thus our final answer is &lt;math&gt;3+7+2=\boxed{12\textbf{(D)}}&lt;/math&gt;<br /> <br /> -vsamc<br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_20&diff=120627 2018 AMC 12A Problems/Problem 20 2020-04-06T18:40:52Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is an isosceles right triangle with &lt;math&gt;AB=AC=3&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of hypotenuse &lt;math&gt;\overline{BC}&lt;/math&gt;. Points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively, so that &lt;math&gt;AI&gt;AE&lt;/math&gt; and &lt;math&gt;AIME&lt;/math&gt; is a cyclic quadrilateral. Given that triangle &lt;math&gt;EMI&lt;/math&gt; has area &lt;math&gt;2&lt;/math&gt;, the length &lt;math&gt;CI&lt;/math&gt; can be written as &lt;math&gt;\frac{a-\sqrt{b}}{c}&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;b&lt;/math&gt; is not divisible by the square of any prime. What is the value of &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }9 \qquad<br /> \textbf{(B) }10 \qquad<br /> \textbf{(C) }11 \qquad<br /> \textbf{(D) }12 \qquad<br /> \textbf{(E) }13 \qquad<br /> &lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> import olympiad;<br /> <br /> size(200);<br /> <br /> pair A, B, C, I, M, E;<br /> <br /> A = (0, 0);<br /> B = (3, 0);<br /> C = (0, 3);<br /> M = (1.5, 1.5);<br /> I = (0, 1.5 + sqrt(2) / 2);<br /> E = (1.5 - sqrt(2) / 2, 0);<br /> <br /> draw(A -- B -- C -- cycle);<br /> draw(I -- M -- E -- cycle);<br /> draw(rightanglemark(I, A, E, 4));<br /> <br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(I);<br /> dot(M);<br /> dot(E);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$I$&quot;, I, NE);<br /> label(&quot;$M$&quot;, M, NE);<br /> label(&quot;$E$&quot;, E + (0.1, 0.04), NE);<br /> label(&quot;$3$&quot;, (A + C) / 2, W);<br /> label(&quot;$3$&quot;, (A + B) / 2, S);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> == Solution 1==<br /> <br /> Observe that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles right (&lt;math&gt;M&lt;/math&gt; is the midpoint of diameter arc &lt;math&gt;EI&lt;/math&gt;), so &lt;math&gt;MI=2,MC=\frac{3}{\sqrt{2}}&lt;/math&gt;. With &lt;math&gt;\angle{MCI}=45^\circ&lt;/math&gt;, we can use Law of Cosines to determine that &lt;math&gt;CI=\frac{3\pm\sqrt{7}}{2}&lt;/math&gt;. The same calculations hold for &lt;math&gt;BE&lt;/math&gt; also, and since &lt;math&gt;CI&lt;BE&lt;/math&gt;, we deduce that &lt;math&gt;CI&lt;/math&gt; is the smaller root, giving the answer of &lt;math&gt;\boxed{12}&lt;/math&gt;.<br /> <br /> == Solution 2 (Using Ptolemy) ==<br /> <br /> We first claim that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles and right.<br /> <br /> Proof: Construct &lt;math&gt;\overline{MF}\perp\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{MG}\perp\overline{AC}&lt;/math&gt;. Since &lt;math&gt;\overline{AM}&lt;/math&gt; bisects &lt;math&gt;\angle{BAC}&lt;/math&gt;, one can deduce that &lt;math&gt;MF=MG&lt;/math&gt;. Then by AAS it is clear that &lt;math&gt;MI=ME&lt;/math&gt; and therefore &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles. Since quadrilateral &lt;math&gt;AIME&lt;/math&gt; is cyclic, one can deduce that &lt;math&gt;\angle{EMI}=90^\circ&lt;/math&gt;. Q.E.D.<br /> <br /> Since the area of &lt;math&gt;\triangle{EMI}&lt;/math&gt; is 2, we can find that &lt;math&gt;MI=ME=2&lt;/math&gt;, &lt;math&gt;EI=2\sqrt{2}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;M&lt;/math&gt; is the mid-point of &lt;math&gt;\overline{BC}&lt;/math&gt;, it is clear that &lt;math&gt;AM=\frac{3\sqrt{2}}{2}&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;AE=a&lt;/math&gt; and &lt;math&gt;AI=b&lt;/math&gt;. By Ptolemy's Theorem, in cyclic quadrilateral &lt;math&gt;AIME&lt;/math&gt;, we have &lt;math&gt;2a+2b=6&lt;/math&gt;. By Pythagorean Theorem, we have &lt;math&gt;a^2+b^2=8&lt;/math&gt;. One can solve the simultaneous system and find &lt;math&gt;b=\frac{3+\sqrt{7}}{2}&lt;/math&gt;. Then by deducting the length of &lt;math&gt;\overline{AI}&lt;/math&gt; from 3 we get &lt;math&gt;CI=\frac{3-\sqrt{7}}{2}&lt;/math&gt;, giving the answer of &lt;math&gt;\boxed{12}&lt;/math&gt;. (Surefire2019)<br /> <br /> == Solution 3 (More Elementary) ==<br /> <br /> Like above, notice that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles and right, which means that &lt;math&gt;\dfrac{ME \cdot MI}{2} = 2&lt;/math&gt;, so &lt;math&gt;MI^2=4&lt;/math&gt; and &lt;math&gt;MI = 2&lt;/math&gt;. Then construct &lt;math&gt;\overline{MF}\perp\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{MG}\perp\overline{AC}&lt;/math&gt; as well as &lt;math&gt;\overline{MI}&lt;/math&gt;. It's clear that &lt;math&gt;MG^2+GI^2 = MI^2&lt;/math&gt; by Pythagorean, so knowing that &lt;math&gt;MG = \dfrac{AB}{2} = \dfrac{3}{2}&lt;/math&gt; allows one to solve to get &lt;math&gt;GI = \dfrac{\sqrt{7}}{2}&lt;/math&gt;. By just looking at the diagram, &lt;math&gt;CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}&lt;/math&gt;. The answer is thus &lt;math&gt;3+7+2=12&lt;/math&gt;.<br /> <br /> == Solution 4 (Coordinate Geometry) ==<br /> Let &lt;math&gt;A&lt;/math&gt; lie on &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; on &lt;math&gt;(0,y)&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt; on &lt;math&gt;(x,0)&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;(\frac{3}{2},\frac{3}{2})&lt;/math&gt;. Since &lt;math&gt;{AIME}&lt;/math&gt; is cyclic, &lt;math&gt;\angle EMI&lt;/math&gt; (which is opposite of another right angle) must be a right angle; therefore, &lt;math&gt;\vec{ME} \cdot \vec{MI} = &lt;\frac{-3}{2}, y-\frac{3}{2}&gt; \cdot &lt;x-\frac{3}{2}, -\frac{3}{2}&gt; = 0&lt;/math&gt;. Compute the dot product to arrive at the relation &lt;math&gt;y=3-x&lt;/math&gt;. We can set up another equation involving the area of &lt;math&gt;\triangle EMI&lt;/math&gt; using the [[Shoelace Theorem]]. This is &lt;math&gt;2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]&lt;/math&gt;. Multiplying, substituting &lt;math&gt;3-x&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;, and simplifying, we get &lt;math&gt;x^2 -3x + \frac{1}{2}=0&lt;/math&gt;. Thus, &lt;math&gt;(x,y)=&lt;/math&gt; &lt;math&gt;(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})&lt;/math&gt;. But &lt;math&gt;AI&gt;AE&lt;/math&gt;, meaning &lt;math&gt;x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}&lt;/math&gt;, and the final answer is &lt;math&gt;3+7+2=\boxed{12}&lt;/math&gt;.<br /> <br /> == Solution 5 (Quick) ==<br /> From &lt;math&gt;AIME&lt;/math&gt; cyclic we get &lt;math&gt;\angle{MEI} = \angle{MAI} = 45^\circ&lt;/math&gt; and &lt;math&gt;\angle{MIE} = \angle{MAE} = 45^\circ&lt;/math&gt;, so &lt;math&gt;\triangle{EMI}&lt;/math&gt; is an isosceles right triangle. <br /> <br /> From &lt;math&gt;[EMI]=2&lt;/math&gt; we get &lt;math&gt;EM=MI=2&lt;/math&gt;. <br /> <br /> Notice &lt;math&gt;\triangle{AEM} \cong \triangle{CIM}&lt;/math&gt;, because &lt;math&gt;\angle{AEM}=180-\angle{AIM}=\angle{CIM}&lt;/math&gt;, &lt;math&gt;EM=IM&lt;/math&gt;, and &lt;math&gt;\angle{EAM}=\angle{ICM}=45^\circ&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;CI=AE=x&lt;/math&gt;, so &lt;math&gt;AI=3-x&lt;/math&gt;. <br /> <br /> By Pythagoras on &lt;math&gt;\triangle{EAI}&lt;/math&gt; we have &lt;math&gt;x^2+(3-x)^2=EI^2=8&lt;/math&gt;, and solve this to get &lt;math&gt;x=CI=\dfrac{3-\sqrt{7}}{2}&lt;/math&gt; for a final answer of &lt;math&gt;3+7+2=\boxed{12}&lt;/math&gt;.<br /> ==Solution 6(bash)==<br /> Let &lt;math&gt;CI=a&lt;/math&gt;, &lt;math&gt;BE=b&lt;/math&gt;. Because opposite angles in a cyclic quadrilateral are supplementary, we have &lt;math&gt;\angle EMI=90^{\circ}&lt;/math&gt;. By the law of cosines, we have &lt;math&gt;MI^2=a^2+\frac{9}{4}-\frac{3}{2}a&lt;/math&gt;, and &lt;math&gt;ME^2=b^2+\frac{9}{4}-\frac{3}{2}b&lt;/math&gt;. Notice that &lt;math&gt;EI=2MO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the origin of the circle mentioned in the problem. Thus &lt;math&gt;\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8&lt;/math&gt;. Thus we have &lt;math&gt;18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}&lt;/math&gt; &lt;math&gt;=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a&lt;/math&gt;. We know that &lt;math&gt;(3-a)^2+(3-b)^2=8\implies (3-a)^2+a^2=8\implies 2a^2-6a+9=8\implies 2a^2-6a+1=0\implies a=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}&lt;/math&gt;. We take the smaller solution because we have &lt;math&gt;AI&gt;AE\implies 3-AI&lt;3-AE\implies CI&lt;CE&lt;/math&gt;, and we want &lt;math&gt;CI&lt;/math&gt;, not &lt;math&gt;CE&lt;/math&gt;, thus &lt;math&gt;CI=\frac{3-\sqrt{7}}{2}&lt;/math&gt;. Thus our final answer is &lt;math&gt;3+7+2=\boxed{12\textbf{(D)}}&lt;/math&gt;<br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_20&diff=120626 2018 AMC 12A Problems/Problem 20 2020-04-06T18:40:21Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is an isosceles right triangle with &lt;math&gt;AB=AC=3&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of hypotenuse &lt;math&gt;\overline{BC}&lt;/math&gt;. Points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively, so that &lt;math&gt;AI&gt;AE&lt;/math&gt; and &lt;math&gt;AIME&lt;/math&gt; is a cyclic quadrilateral. Given that triangle &lt;math&gt;EMI&lt;/math&gt; has area &lt;math&gt;2&lt;/math&gt;, the length &lt;math&gt;CI&lt;/math&gt; can be written as &lt;math&gt;\frac{a-\sqrt{b}}{c}&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;b&lt;/math&gt; is not divisible by the square of any prime. What is the value of &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }9 \qquad<br /> \textbf{(B) }10 \qquad<br /> \textbf{(C) }11 \qquad<br /> \textbf{(D) }12 \qquad<br /> \textbf{(E) }13 \qquad<br /> &lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> import olympiad;<br /> <br /> size(200);<br /> <br /> pair A, B, C, I, M, E;<br /> <br /> A = (0, 0);<br /> B = (3, 0);<br /> C = (0, 3);<br /> M = (1.5, 1.5);<br /> I = (0, 1.5 + sqrt(2) / 2);<br /> E = (1.5 - sqrt(2) / 2, 0);<br /> <br /> draw(A -- B -- C -- cycle);<br /> draw(I -- M -- E -- cycle);<br /> draw(rightanglemark(I, A, E, 4));<br /> <br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(I);<br /> dot(M);<br /> dot(E);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$I$&quot;, I, NE);<br /> label(&quot;$M$&quot;, M, NE);<br /> label(&quot;$E$&quot;, E + (0.1, 0.04), NE);<br /> label(&quot;$3$&quot;, (A + C) / 2, W);<br /> label(&quot;$3$&quot;, (A + B) / 2, S);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> == Solution 1==<br /> <br /> Observe that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles right (&lt;math&gt;M&lt;/math&gt; is the midpoint of diameter arc &lt;math&gt;EI&lt;/math&gt;), so &lt;math&gt;MI=2,MC=\frac{3}{\sqrt{2}}&lt;/math&gt;. With &lt;math&gt;\angle{MCI}=45^\circ&lt;/math&gt;, we can use Law of Cosines to determine that &lt;math&gt;CI=\frac{3\pm\sqrt{7}}{2}&lt;/math&gt;. The same calculations hold for &lt;math&gt;BE&lt;/math&gt; also, and since &lt;math&gt;CI&lt;BE&lt;/math&gt;, we deduce that &lt;math&gt;CI&lt;/math&gt; is the smaller root, giving the answer of &lt;math&gt;\boxed{12}&lt;/math&gt;.<br /> <br /> == Solution 2 (Using Ptolemy) ==<br /> <br /> We first claim that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles and right.<br /> <br /> Proof: Construct &lt;math&gt;\overline{MF}\perp\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{MG}\perp\overline{AC}&lt;/math&gt;. Since &lt;math&gt;\overline{AM}&lt;/math&gt; bisects &lt;math&gt;\angle{BAC}&lt;/math&gt;, one can deduce that &lt;math&gt;MF=MG&lt;/math&gt;. Then by AAS it is clear that &lt;math&gt;MI=ME&lt;/math&gt; and therefore &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles. Since quadrilateral &lt;math&gt;AIME&lt;/math&gt; is cyclic, one can deduce that &lt;math&gt;\angle{EMI}=90^\circ&lt;/math&gt;. Q.E.D.<br /> <br /> Since the area of &lt;math&gt;\triangle{EMI}&lt;/math&gt; is 2, we can find that &lt;math&gt;MI=ME=2&lt;/math&gt;, &lt;math&gt;EI=2\sqrt{2}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;M&lt;/math&gt; is the mid-point of &lt;math&gt;\overline{BC}&lt;/math&gt;, it is clear that &lt;math&gt;AM=\frac{3\sqrt{2}}{2}&lt;/math&gt;.<br /> <br /> Now let &lt;math&gt;AE=a&lt;/math&gt; and &lt;math&gt;AI=b&lt;/math&gt;. By Ptolemy's Theorem, in cyclic quadrilateral &lt;math&gt;AIME&lt;/math&gt;, we have &lt;math&gt;2a+2b=6&lt;/math&gt;. By Pythagorean Theorem, we have &lt;math&gt;a^2+b^2=8&lt;/math&gt;. One can solve the simultaneous system and find &lt;math&gt;b=\frac{3+\sqrt{7}}{2}&lt;/math&gt;. Then by deducting the length of &lt;math&gt;\overline{AI}&lt;/math&gt; from 3 we get &lt;math&gt;CI=\frac{3-\sqrt{7}}{2}&lt;/math&gt;, giving the answer of &lt;math&gt;\boxed{12}&lt;/math&gt;. (Surefire2019)<br /> <br /> == Solution 3 (More Elementary) ==<br /> <br /> Like above, notice that &lt;math&gt;\triangle{EMI}&lt;/math&gt; is isosceles and right, which means that &lt;math&gt;\dfrac{ME \cdot MI}{2} = 2&lt;/math&gt;, so &lt;math&gt;MI^2=4&lt;/math&gt; and &lt;math&gt;MI = 2&lt;/math&gt;. Then construct &lt;math&gt;\overline{MF}\perp\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{MG}\perp\overline{AC}&lt;/math&gt; as well as &lt;math&gt;\overline{MI}&lt;/math&gt;. It's clear that &lt;math&gt;MG^2+GI^2 = MI^2&lt;/math&gt; by Pythagorean, so knowing that &lt;math&gt;MG = \dfrac{AB}{2} = \dfrac{3}{2}&lt;/math&gt; allows one to solve to get &lt;math&gt;GI = \dfrac{\sqrt{7}}{2}&lt;/math&gt;. By just looking at the diagram, &lt;math&gt;CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}&lt;/math&gt;. The answer is thus &lt;math&gt;3+7+2=12&lt;/math&gt;.<br /> <br /> == Solution 4 (Coordinate Geometry) ==<br /> Let &lt;math&gt;A&lt;/math&gt; lie on &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; on &lt;math&gt;(0,y)&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt; on &lt;math&gt;(x,0)&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;(\frac{3}{2},\frac{3}{2})&lt;/math&gt;. Since &lt;math&gt;{AIME}&lt;/math&gt; is cyclic, &lt;math&gt;\angle EMI&lt;/math&gt; (which is opposite of another right angle) must be a right angle; therefore, &lt;math&gt;\vec{ME} \cdot \vec{MI} = &lt;\frac{-3}{2}, y-\frac{3}{2}&gt; \cdot &lt;x-\frac{3}{2}, -\frac{3}{2}&gt; = 0&lt;/math&gt;. Compute the dot product to arrive at the relation &lt;math&gt;y=3-x&lt;/math&gt;. We can set up another equation involving the area of &lt;math&gt;\triangle EMI&lt;/math&gt; using the [[Shoelace Theorem]]. This is &lt;math&gt;2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]&lt;/math&gt;. Multiplying, substituting &lt;math&gt;3-x&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;, and simplifying, we get &lt;math&gt;x^2 -3x + \frac{1}{2}=0&lt;/math&gt;. Thus, &lt;math&gt;(x,y)=&lt;/math&gt; &lt;math&gt;(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})&lt;/math&gt;. But &lt;math&gt;AI&gt;AE&lt;/math&gt;, meaning &lt;math&gt;x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}&lt;/math&gt;, and the final answer is &lt;math&gt;3+7+2=\boxed{12}&lt;/math&gt;.<br /> <br /> == Solution 5 (Quick) ==<br /> From &lt;math&gt;AIME&lt;/math&gt; cyclic we get &lt;math&gt;\angle{MEI} = \angle{MAI} = 45^\circ&lt;/math&gt; and &lt;math&gt;\angle{MIE} = \angle{MAE} = 45^\circ&lt;/math&gt;, so &lt;math&gt;\triangle{EMI}&lt;/math&gt; is an isosceles right triangle. <br /> <br /> From &lt;math&gt;[EMI]=2&lt;/math&gt; we get &lt;math&gt;EM=MI=2&lt;/math&gt;. <br /> <br /> Notice &lt;math&gt;\triangle{AEM} \cong \triangle{CIM}&lt;/math&gt;, because &lt;math&gt;\angle{AEM}=180-\angle{AIM}=\angle{CIM}&lt;/math&gt;, &lt;math&gt;EM=IM&lt;/math&gt;, and &lt;math&gt;\angle{EAM}=\angle{ICM}=45^\circ&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;CI=AE=x&lt;/math&gt;, so &lt;math&gt;AI=3-x&lt;/math&gt;. <br /> <br /> By Pythagoras on &lt;math&gt;\triangle{EAI}&lt;/math&gt; we have &lt;math&gt;x^2+(3-x)^2=EI^2=8&lt;/math&gt;, and solve this to get &lt;math&gt;x=CI=\dfrac{3-\sqrt{7}}{2}&lt;/math&gt; for a final answer of &lt;math&gt;3+7+2=\boxed{12}&lt;/math&gt;.<br /> ==Solution 6(bash)==<br /> Let &lt;math&gt;CI=a&lt;/math&gt;, &lt;math&gt;BE=b&lt;/math&gt;. Because opposite angles in a cyclic quadrilateral are supplementary, we have &lt;math&gt;\angle EMI=90^{\circ}&lt;/math&gt;. By the law of cosines, we have &lt;math&gt;MI^2=a^2+\frac{9}{4}-\frac{3}{2}a&lt;/math&gt;, and &lt;math&gt;ME^2=b^2+\frac{9}{4}-\frac{3}{2}b&lt;/math&gt;. Notice that &lt;math&gt;EI=2MO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the origin of the circle mentioned in the problem. Thus &lt;math&gt;\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2(\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8&lt;/math&gt;. Thus we have &lt;math&gt;18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}&lt;/math&gt; &lt;math&gt;=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a&lt;/math&gt;. We know that &lt;math&gt;(3-a)^2+(3-b)^2=8\implies (3-a)^2+a^2=8\implies 2a^2-6a+9=8\implies 2a^2-6a+1=0\implies a=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}&lt;/math&gt;. We take the smaller solution because we have &lt;math&gt;AI&gt;AE\implies 3-AI&lt;3-AE\implies CI&lt;CE&lt;/math&gt;, and we want &lt;math&gt;CI&lt;/math&gt;, not &lt;math&gt;CE&lt;/math&gt;, thus &lt;math&gt;CI=\frac{3-\sqrt{7}}{2}&lt;/math&gt;. Thus our final answer is &lt;math&gt;3+7+2=\boxed{12\textbf{(D)}}&lt;/math&gt;<br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_14&diff=120577 2018 AMC 12A Problems/Problem 14 2020-04-05T23:20:21Z <p>Vsamc: </p> <hr /> <div>==Problem==<br /> <br /> The solutions to the equation &lt;math&gt;\log_{3x} 4 = \log_{2x} 8&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is a positive real number other than &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; or &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;, can be written as &lt;math&gt;\tfrac {p}{q}&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p + q&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 5 \qquad <br /> \textbf{(B) } 13 \qquad <br /> \textbf{(C) } 17 \qquad <br /> \textbf{(D) } 31 \qquad <br /> \textbf{(E) } 35 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> Base switch to log 2 and you have &lt;math&gt;\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}&lt;/math&gt; .<br /> <br /> &lt;math&gt;\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}&lt;/math&gt;<br /> <br /> &lt;math&gt;2*\log_2 2x = 3*\log_2 3x&lt;/math&gt;<br /> <br /> Then &lt;math&gt;\log_2 (2x)^2 = \log_2 (3x)^3&lt;/math&gt;. so &lt;math&gt;4x^2=27x^3&lt;/math&gt; and we have &lt;math&gt;x=\frac{4}{27}&lt;/math&gt; leading to &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt; (jeremylu)<br /> <br /> ==Solution 2==<br /> If you multiply both sides by &lt;math&gt;\log_2 (3x)&lt;/math&gt;<br /> <br /> then it should come out to &lt;math&gt;\log_2 (3x)&lt;/math&gt; * &lt;math&gt;\log_{3x} (4)&lt;/math&gt; = &lt;math&gt;\log_2 {3x}&lt;/math&gt; * &lt;math&gt;\log_{2x} (8)&lt;/math&gt;<br /> <br /> that then becomes &lt;math&gt;\log_2 (4)&lt;/math&gt; * &lt;math&gt;\log_{3x} (3x)&lt;/math&gt; = &lt;math&gt;\log_2 (8)&lt;/math&gt; * &lt;math&gt;\log_{2x} (3x)&lt;/math&gt;<br /> <br /> which simplifies to &lt;math&gt;2*1 = 3\log_{2x} (3x)&lt;/math&gt;<br /> <br /> so now &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\log_{2x} (3x)&lt;/math&gt; putting in exponent form gets<br /> <br /> &lt;math&gt;(2x)^2&lt;/math&gt; = &lt;math&gt;(3x)^3&lt;/math&gt;<br /> <br /> so &lt;math&gt;4x^2&lt;/math&gt; = &lt;math&gt;27x^3&lt;/math&gt;<br /> <br /> dividing yields &lt;math&gt;x = 4/27&lt;/math&gt; and <br /> <br /> &lt;math&gt;4+27 =&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt;<br /> - Pikachu13307<br /> <br /> ==Solution 3==<br /> <br /> We can convert both &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; into &lt;math&gt;2^2&lt;/math&gt; and &lt;math&gt;2^3&lt;/math&gt;, respectively, giving:<br /> <br /> &lt;math&gt;2\log_{3x} (2) = 3\log_{2x} (2)&lt;/math&gt;<br /> <br /> Converting the bases of the right side, we get &lt;math&gt;\log_{2x} 2 = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3}*\log_{3x} (2) = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;2^\frac{2}{3} = (3x)^\frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3} * \ln 2 = \frac{\ln 2}{\ln (2x)} * \ln (3x)&lt;/math&gt;<br /> <br /> Dividing both sides by &lt;math&gt;\ln 2&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)}&lt;/math&gt;<br /> <br /> Which simplifies to<br /> <br /> &lt;math&gt;2\ln (2x) = 3\ln (3x)&lt;/math&gt;<br /> <br /> Log expansion allows us to see that<br /> <br /> &lt;math&gt;2\ln 2 + 2\ln (x) = 3\ln 3 + 3\ln (x)&lt;/math&gt;, which then simplifies to<br /> <br /> &lt;math&gt;\ln (x) = 2\ln 2 - 3\ln 3&lt;/math&gt;<br /> <br /> Thus,<br /> <br /> &lt;math&gt;x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}}&lt;/math&gt;<br /> <br /> And<br /> <br /> &lt;math&gt;x = \frac{2^2}{3^3} = \frac{4}{27} = \boxed{\textbf{(D)}31}&lt;/math&gt; <br /> <br /> -lepetitmoulin<br /> <br /> ==Solution 4==<br /> &lt;math&gt;\log_{3x} 4=\log_{2x} 8&lt;/math&gt; is the same as &lt;math&gt;2\log_{3x} 2=3\log_{2x} 2&lt;/math&gt;<br /> <br /> Using Reciprocal law, we get &lt;math&gt;\log_{(3x)^\frac{1}{2}} 2=\log_{(2x)^\frac{1}{3}} 2&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow (3x)^\frac{1}{2}=(2x)^\frac{1}{3}&lt;/math&gt; &lt;math&gt;\Rightarrow 27x^3=4x^2&lt;/math&gt; &lt;math&gt;\Rightarrow \frac{x^3}{x^2}=\frac{4}{27}=x&lt;/math&gt;<br /> <br /> &lt;math&gt;\therefore \frac{p}{q}=\frac{4}{27}&lt;/math&gt; &lt;math&gt;\Rightarrow p+q=4+27=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) } 31}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> ==Solution 5==<br /> &lt;math&gt;\log_{3x} 4=\log_{2x} 8\implies 2\log_{3x} 2=3\log_{2x} 2 \implies \frac{2}{3}=\frac{\log_{2x}2}{\log_{2x}3}&lt;/math&gt;. We know that &lt;math&gt;\log_a{b}=\frac{\log_{c}b}{\log_{c}a}=\frac{\frac{1}{\log_b{c}}}{\frac{1}{\log_a{c}}}=\frac{\log_a{c}}{\log_b{c}}&lt;/math&gt;. Thus &lt;math&gt;\frac{2}{3}=\frac{\log_{2x}2}{\log_{2x}3}\implies \frac{2}{3}=\log_{2x}{3x}\implies (2x)^{\frac{2}{3}}=3x\implies 2^{\frac{2}{3}}x^{\frac{2}{3}}=3x\implies 2^{\frac{2}{3}}=3x^{\frac{1}{3}}\implies x^{\frac{1}{3}}=\frac{2^{\frac{2}{3}}}{3}\implies x=\frac{2^2}{3^3}=\frac{4}{27}&lt;/math&gt;. &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;27&lt;/math&gt; are indeed relatively prime thus our final answer is &lt;math&gt;4+27=31 \text{which is }\boxed{\textbf{(D)}}&lt;/math&gt;<br /> <br /> -vsamc<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_23&diff=119623 2004 AMC 12A Problems/Problem 23 2020-03-17T14:14:37Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> A [[polynomial]]<br /> <br /> &lt;cmath&gt;P(x) = c_{2004}x^{2004} + c_{2003}x^{2003} + ... + c_1x + c_0&lt;/cmath&gt;<br /> <br /> has [[real]] [[coefficient]]s with &lt;math&gt;c_{2004}\not = 0&lt;/math&gt; and &lt;math&gt;2004&lt;/math&gt; distinct complex [[zero]]es &lt;math&gt;z_k = a_k + b_ki&lt;/math&gt;, &lt;math&gt;1\leq k\leq 2004&lt;/math&gt; with &lt;math&gt;a_k&lt;/math&gt; and &lt;math&gt;b_k&lt;/math&gt; real, &lt;math&gt;a_1 = b_1 = 0&lt;/math&gt;, and<br /> <br /> &lt;cmath&gt;\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}.&lt;/cmath&gt;<br /> <br /> Which of the following quantities can be a nonzero number?<br /> <br /> &lt;math&gt;\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> We have to evaluate the answer choices and use process of elimination:<br /> <br /> *&lt;math&gt;\mathrm{(A)}&lt;/math&gt;: We are given that &lt;math&gt;a_1 = b_1 = 0&lt;/math&gt;, so &lt;math&gt;z_1 = 0&lt;/math&gt;. If one of the roots is zero, then &lt;math&gt;P(0) = c_0 = 0&lt;/math&gt;. <br /> *&lt;math&gt;\mathrm{(B)}&lt;/math&gt;: By [[Vieta's formulas]], we know that &lt;math&gt;-\frac{c_{2003}}{c_{2004}}&lt;/math&gt; is the sum of all of the roots of &lt;math&gt;P(x)&lt;/math&gt;. Since that is real, &lt;math&gt; \sum_{k = 1}^{2004}{b_k}=0=\sum_{k = 1}^{2004}{a_k}&lt;/math&gt;, and &lt;math&gt;\frac{c_{2003}}{c_{2004}}=0&lt;/math&gt;, so &lt;math&gt;c_{2003}=0&lt;/math&gt;.<br /> *&lt;math&gt;\mathrm{(C)}&lt;/math&gt;: All of the coefficients are real. For sake of contradiction suppose none of &lt;math&gt;b_{2\ldots 2004}&lt;/math&gt; are zero. Then for each complex root &lt;math&gt;z_i&lt;/math&gt;, its [[complex conjugate]] &lt;math&gt;\overline{z_i} = a_i - b_ik&lt;/math&gt; is also a root. So the roots should pair up, but we have an odd number of imaginary roots! (Remember that &lt;math&gt;b_1 = 0&lt;/math&gt;.) This gives us the contradiction, and therefore the product is equal to zero.<br /> *&lt;math&gt;\mathrm{(D)}&lt;/math&gt;: We are given that &lt;math&gt;\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}&lt;/math&gt;. Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.<br /> <br /> There is, however, no reason to believe that &lt;math&gt;\boxed{\mathrm{E}}&lt;/math&gt; should be zero (in fact, that quantity is &lt;math&gt;P(1)&lt;/math&gt;, and there is no evidence that &lt;math&gt;1&lt;/math&gt; is a root of &lt;math&gt;P(x)&lt;/math&gt;).<br /> ==Solution 2(cheap method using answer choices)==<br /> Rule out answer choices &lt;math&gt;A, B, D&lt;/math&gt; as done in solution 1. Assume for the sake of contradiction that &lt;math&gt;\mathrm{(E)}&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt;. Then &lt;math&gt;P(1)=0&lt;/math&gt;, so there is sum &lt;math&gt;m\neq 1&lt;/math&gt; such that &lt;math&gt;z_m=1+0i&lt;/math&gt;, which implies at least one &lt;math&gt;b_i&lt;/math&gt;, &lt;math&gt;i\neq 1&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt; SO if &lt;math&gt;\sum_{k=1}^{2004}c_k=0&lt;/math&gt;, then &lt;math&gt;\prod_{i=12}^{2004}b_i=0&lt;/math&gt;. So we have that &lt;math&gt;\sum_{k=1}^{2004}c_k\neq 0&lt;/math&gt;, because then that would rule out all the answer choices. Hence &lt;math&gt;\boxed{\mathrm{E}}&lt;/math&gt; must be nonzero, so the answer is &lt;math&gt;\mathrm{(E)}&lt;/math&gt;.<br /> -vsamc<br /> <br /> == See also ==<br /> {{AMC12 box|year=2004|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119335 2020 AMC 12A Problems/Problem 22 2020-03-13T20:20:13Z <p>Vsamc: Whoops Fixed My Silly Mistake</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(B)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is B<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2i}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4i(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4i}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4i}\sum_{n\ge 0}\tfrac{(3+4i)^n-(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}-\tfrac{1}{1-(\tfrac{3-4i}{7})}]=\tfrac{1}{4i}[\tfrac{7}{7-(3+4i)}-\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}-\tfrac{7}{4+4i}]=\tfrac{1}{4i}[\tfrac{7(4+4i)}{32}-\tfrac{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(B)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119333 2020 AMC 12A Problems/Problem 22 2020-03-13T20:14:34Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(B)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is B<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{(3+4i)^n-(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}-\tfrac{1}{1-(\tfrac{3-4i}{7})}]=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}-\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}-\tfrac{7}{4+4i}]=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}-\tfrac{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(B)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119332 2020 AMC 12A Problems/Problem 22 2020-03-13T19:39:19Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(B)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is B<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{(3+4i)^n+(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}+\tfrac{1}{1-(\tfrac{3-4i}{7})}]=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}]=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(B)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119331 2020 AMC 12A Problems/Problem 22 2020-03-13T19:36:13Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(C)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is C<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{(3+4i)^n+(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}+\tfrac{1}{1-(\tfrac{3-4i}{7})}]=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}]=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(C)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119329 2020 AMC 12A Problems/Problem 22 2020-03-13T19:33:25Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(C)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is C<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{(3+4i)^n+(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}+\tfrac{1}{1-(\tfrac{3-4i}{7})}=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}]=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}]=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(C)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119328 2020 AMC 12A Problems/Problem 22 2020-03-13T19:29:12Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(C)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is C<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{(3+4i)^n+(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}+\tfrac{1}{1-(\tfrac{3-4i}{7})}=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i)}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}]=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(C)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119327 2020 AMC 12A Problems/Problem 22 2020-03-13T19:25:10Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(C)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamc stating that the answer choice is C<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Expanding the summand yields &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{(3+4i)^n+(3-4i)^n}{7^n}&lt;/math&gt;=&lt;math&gt;\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}+\tfrac{1}{1-(\tfrac{3-4i}{7})}&lt;/math&gt;&lt;math&gt;=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i))}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}\=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}}\textbf{(C)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=119326 2020 AMC 12A Problems/Problem 22 2020-03-13T19:19:04Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> == Solution 2 (DeMoivre's Formula) ==<br /> Note that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}i \right)&lt;/math&gt;. Let &lt;math&gt;\theta = \arctan (1/2)&lt;/math&gt;, then, we know that &lt;math&gt;(2+i) = \sqrt{5} \cdot \left( \cos \theta + i\sin \theta \right)&lt;/math&gt;, so &lt;math&gt;(2+i)^n = (\cos (n \theta) + i\sin (n\theta))(\sqrt{5})^n&lt;/math&gt;. Therefore, &lt;math&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n} = \sum_{n=0}^\infty\frac{\cos(n\theta)\sin(n\theta) (5)^n}{7^n} =&lt;/math&gt; &lt;math&gt;\frac{1}{2}\sum_{n=0}^\infty \left( \frac{5}{7}\right)^n \sin (2n\theta) = \frac{1}{2} \text{im} \left( \sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} \right)&lt;/math&gt;. Aha &lt;math&gt;\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} &lt;/math&gt; is a geometric sequence that evaluates to &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}}&lt;/math&gt;. We can quickly see that &lt;math&gt;\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5}&lt;/math&gt;. &lt;math&gt;\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i&lt;/math&gt;. The imaginary part is &lt;math&gt;\frac{7}{8}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}}&lt;/math&gt;.(Which is answer choice &lt;math&gt;\textbf{(C)}&lt;/math&gt;<br /> <br /> ~AopsUser101, minor edit by vsamcstating that the answer choice is C<br /> ==Solution 3==<br /> Clearly &lt;math&gt;a_n=\tfrac{(2+i)^n+(2-i)^n}{2}, b_n=\tfrac{(2+i)^n-(2-i)^n}{2}&lt;/math&gt;. So we have &lt;math&gt;\sum_{n\ge 0}\tfrac{a_nb_n}{7^n}=\sum_{n\ge 0}\tfrac{((2+i)^n+(2-i)^n))((2+i)^n-(2-i)^n))}{4(7^n)}&lt;/math&gt;. By linearity, we have the latter is equivalent to &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{[(2+i)^n+(2-i)^n][(2+i)^n-(2-i)^n]}{7^n}&lt;/math&gt;. Exapanding the summand yields &lt;math&gt;\tfrac{1}{4}\sum_{n\ge 0}\tfrac{(3+4i)^n+(3-4i)^n}{7^n}=\tfrac{1}{4}[\tfrac{1}{1-(\tfrac{3+4i}{7})}+\tfrac{1}{1-(\tfrac{3-4i}{7})}=\tfrac{1}{4}[\tfrac{7}{7-(3+4i)}+\tfrac{7}{7-(3-4i))}]=\tfrac{1}{4}[\tfrac{7}{4-4i}+\tfrac{7}{4+4i}\=\tfrac{1}{4}[\tfrac{7(4+4i)}{32}+{7(4-4i)}{32}=\tfrac{1}{4}\cdot \tfrac{56}{32}=\boxed{\tfrac{7}{16}} \textbf{(C)}&lt;/math&gt;<br /> -vsamc<br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=119319 2000 AMC 12 Problems/Problem 8 2020-03-13T18:42:07Z <p>Vsamc: </p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> ==Problem==<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ==Video:==<br /> https://www.youtube.com/watch?v=HVP6qjKAkjA&amp;t=2s<br /> <br /> ==Solution 1==<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;f_n&lt;/math&gt; denote the number of unit cubes in a figure. We have<br /> &lt;cmath&gt;f_0=1&lt;/cmath&gt;<br /> &lt;cmath&gt;f_1=5&lt;/cmath&gt;<br /> &lt;cmath&gt;f_2=13&lt;/cmath&gt;<br /> &lt;cmath&gt;f_3=25&lt;/cmath&gt;<br /> &lt;cmath&gt;f_4=41&lt;/cmath&gt;<br /> &lt;cmath&gt;...&lt;/cmath&gt;<br /> <br /> Computing the difference between the number of cubes in each figure yields<br /> &lt;cmath&gt;4,8,12,16,...&lt;/cmath&gt;<br /> It is easy to notice that this is an arithmetic sequence, with the first term being &lt;math&gt;4&lt;/math&gt; and the difference being &lt;math&gt;4&lt;/math&gt;. Let this sequence be &lt;math&gt;a_n&lt;/math&gt;<br /> <br /> From &lt;math&gt;f_0&lt;/math&gt; to &lt;math&gt;f_{100}&lt;/math&gt;, the sequence will have &lt;math&gt;100&lt;/math&gt; terms. Using the arithmetic sum formula yields<br /> <br /> &lt;cmath&gt;S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;=50(2\cdot 4+99\cdot 4)&lt;/cmath&gt;<br /> &lt;cmath&gt;=50(101\cdot 4)&lt;/cmath&gt;<br /> &lt;cmath&gt;=200\cdot 101&lt;/cmath&gt;<br /> &lt;cmath&gt;=20200&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;f_{100}=1+20200=\boxed{\textbf{(C) }20201}&lt;/math&gt; unit cubes.<br /> <br /> ~ljlbox<br /> <br /> == Solution 6(Newton's Forward Differences) ==<br /> We know that 1 and 5 differ by 4, 5 and 13 differ by 8, and 13 and 25 differ by 12. Hence the differences are 4, 8, and 12, resp. And the differences of the differences area all 4. So by Newton's Forward Difference Formula, we get the 100th figure is(because Figure 0 exists) &lt;math&gt;\dbinom{101-1}{0}+4\dbinom{101-1}{1}+4\dbinom{101-1}{2}=20201&lt;/math&gt; or &lt;math&gt;\textbf{(C)}&lt;/math&gt;<br /> -vsamc<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_19&diff=119053 2010 AMC 12A Problems/Problem 19 2020-03-11T15:07:27Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Each of &lt;math&gt;2010&lt;/math&gt; boxes in a line contains a single red marble, and for &lt;math&gt;1 \le k \le 2010&lt;/math&gt;, the box in the &lt;math&gt;k\text{th}&lt;/math&gt; position also contains &lt;math&gt;k&lt;/math&gt; white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let &lt;math&gt;P(n)&lt;/math&gt; be the probability that Isabella stops after drawing exactly &lt;math&gt;n&lt;/math&gt; marbles. What is the smallest value of &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;P(n) &lt; \frac{1}{2010}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005&lt;/math&gt;<br /> <br /> == Solution 1==<br /> The probability of drawing a white marble from box &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;\frac{k}{k + 1}&lt;/math&gt;, and the probability of drawing a red marble from box &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;\frac{1}{k+1}&lt;/math&gt;.<br /> <br /> To stop after drawing &lt;math&gt;n&lt;/math&gt; marbles, we must draw a white marble from boxes &lt;math&gt;1, 2, \ldots, n-1,&lt;/math&gt; and draw a red marble from box &lt;math&gt;n.&lt;/math&gt; Thus, &lt;cmath&gt;P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.&lt;/cmath&gt;<br /> <br /> So, we must have &lt;math&gt;\frac{1}{n(n + 1)} &lt; \frac{1}{2010}&lt;/math&gt; or &lt;math&gt;n(n+1) &gt; 2010.&lt;/math&gt;<br /> <br /> Since &lt;math&gt;n(n+1)&lt;/math&gt; increases as &lt;math&gt;n&lt;/math&gt; increases, we can simply test values of &lt;math&gt;n&lt;/math&gt;; after some trial and error, we get that the minimum value of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A) }45}&lt;/math&gt;, since &lt;math&gt;45(46) = 2070&lt;/math&gt; but &lt;math&gt;44(45) = 1980.&lt;/math&gt;<br /> <br /> == Solution 2(cheap) ==<br /> Do the same thing as Solution 1, but when we get to &lt;math&gt;n(n+1)&gt;2010&lt;/math&gt; just test all the answer choices in ascending order(from A to E), and stop when one of the answer choices is greater than &lt;math&gt;2010&lt;/math&gt;. We get &lt;math&gt;45(46)=2070&lt;/math&gt;, which is greater than &lt;math&gt;2010&lt;/math&gt;, so we are done. The answer is &lt;math&gt;\textbf{(A)}&lt;/math&gt;<br /> <br /> -vsamc<br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_18&diff=119052 2010 AMC 12A Problems/Problem 18 2020-03-11T15:04:32Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> A 16-step path is to go from &lt;math&gt;(-4,-4)&lt;/math&gt; to &lt;math&gt;(4,4)&lt;/math&gt; with each step increasing either the &lt;math&gt;x&lt;/math&gt;-coordinate or the &lt;math&gt;y&lt;/math&gt;-coordinate by 1. How many such paths stay outside or on the boundary of the square &lt;math&gt;-2 \le x \le 2&lt;/math&gt;, &lt;math&gt;-2 \le y \le 2&lt;/math&gt; at each step?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800&lt;/math&gt;<br /> <br /> == Solution 1==<br /> Each path must go through either the second or the fourth quadrant.<br /> Each path that goes through the second quadrant must pass through exactly one of the points &lt;math&gt;(-4,4)&lt;/math&gt;, &lt;math&gt;(-3,3)&lt;/math&gt;, and &lt;math&gt;(-2,2)&lt;/math&gt;.<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; path of the first kind, &lt;math&gt;{8\choose 1}^2=64&lt;/math&gt; paths of the second kind, and &lt;math&gt;{8\choose 2}^2=28^2=784&lt;/math&gt; paths of the third type. <br /> Each path that goes through the fourth quadrant must pass through exactly one of the points &lt;math&gt;(4,-4)&lt;/math&gt;, &lt;math&gt;(3,-3)&lt;/math&gt;, and &lt;math&gt;(2,-2)&lt;/math&gt;.<br /> Again, there is &lt;math&gt;1&lt;/math&gt; path of the first kind, &lt;math&gt;{8\choose 1}^2=64&lt;/math&gt; paths of the second kind, and &lt;math&gt;{8\choose 2}^2=28^2=784&lt;/math&gt; paths of the third type. <br /> <br /> Hence the total number of paths is &lt;math&gt;2(1+64+784) = \boxed{1698}&lt;/math&gt;.<br /> == Solution 2==<br /> We will use the concept of complimentary counting. If you go in the square you have to go out by these labeled points(click the link) https://imgur.com/VysX4P0 and go out through the borders because if you didnt, you would touch another point in one of those points in the set of points in the link(Call it &lt;math&gt;S&lt;/math&gt;). There is symmetry about &lt;math&gt;y=x&lt;/math&gt;, so we only have to consider &lt;math&gt;(1,-1), (1,0)&lt;/math&gt;, and &lt;math&gt;(1,1)&lt;/math&gt;. &lt;math&gt;(1,1)&lt;/math&gt; can go on the boundary in two ways, so we can only consider one case and multiply it by two. For &lt;math&gt;(1,0)&lt;/math&gt; and &lt;math&gt;(1,-1)&lt;/math&gt; we can just multiply by two. So we count paths from &lt;math&gt;(-4,4)&lt;/math&gt; to each of these points, and then multiply that by the number of ways to get from the point one unit right of that to &lt;math&gt;(4,4)&lt;/math&gt;, and all in all, we get the answer is &lt;math&gt;\dbinom{16}{8}-2[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}]=1698&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> -vsamc<br /> Note: Sorry if this was rushed.<br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=17|num-a=19|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_18&diff=119051 2010 AMC 12A Problems/Problem 18 2020-03-11T15:04:14Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> A 16-step path is to go from &lt;math&gt;(-4,-4)&lt;/math&gt; to &lt;math&gt;(4,4)&lt;/math&gt; with each step increasing either the &lt;math&gt;x&lt;/math&gt;-coordinate or the &lt;math&gt;y&lt;/math&gt;-coordinate by 1. How many such paths stay outside or on the boundary of the square &lt;math&gt;-2 \le x \le 2&lt;/math&gt;, &lt;math&gt;-2 \le y \le 2&lt;/math&gt; at each step?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800&lt;/math&gt;<br /> <br /> == Solution ==<br /> Each path must go through either the second or the fourth quadrant.<br /> Each path that goes through the second quadrant must pass through exactly one of the points &lt;math&gt;(-4,4)&lt;/math&gt;, &lt;math&gt;(-3,3)&lt;/math&gt;, and &lt;math&gt;(-2,2)&lt;/math&gt;.<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; path of the first kind, &lt;math&gt;{8\choose 1}^2=64&lt;/math&gt; paths of the second kind, and &lt;math&gt;{8\choose 2}^2=28^2=784&lt;/math&gt; paths of the third type. <br /> Each path that goes through the fourth quadrant must pass through exactly one of the points &lt;math&gt;(4,-4)&lt;/math&gt;, &lt;math&gt;(3,-3)&lt;/math&gt;, and &lt;math&gt;(2,-2)&lt;/math&gt;.<br /> Again, there is &lt;math&gt;1&lt;/math&gt; path of the first kind, &lt;math&gt;{8\choose 1}^2=64&lt;/math&gt; paths of the second kind, and &lt;math&gt;{8\choose 2}^2=28^2=784&lt;/math&gt; paths of the third type. <br /> <br /> Hence the total number of paths is &lt;math&gt;2(1+64+784) = \boxed{1698}&lt;/math&gt;.<br /> == Solution 2==<br /> We will use the concept of complimentary counting. If you go in the square you have to go out by these labeled points(click the link) https://imgur.com/VysX4P0 and go out through the borders because if you didnt, you would touch another point in one of those points in the set of points in the link(Call it &lt;math&gt;S&lt;/math&gt;). There is symmetry about &lt;math&gt;y=x&lt;/math&gt;, so we only have to consider &lt;math&gt;(1,-1), (1,0)&lt;/math&gt;, and &lt;math&gt;(1,1)&lt;/math&gt;. &lt;math&gt;(1,1)&lt;/math&gt; can go on the boundary in two ways, so we can only consider one case and multiply it by two. For &lt;math&gt;(1,0)&lt;/math&gt; and &lt;math&gt;(1,-1)&lt;/math&gt; we can just multiply by two. So we count paths from &lt;math&gt;(-4,4)&lt;/math&gt; to each of these points, and then multiply that by the number of ways to get from the point one unit right of that to &lt;math&gt;(4,4)&lt;/math&gt;, and all in all, we get the answer is &lt;math&gt;\dbinom{16}{8}-2[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}]=1698&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> -vsamc<br /> Note: Sorry if this was rushed.<br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=17|num-a=19|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_18&diff=119050 2010 AMC 12A Problems/Problem 18 2020-03-11T15:03:29Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> A 16-step path is to go from &lt;math&gt;(-4,-4)&lt;/math&gt; to &lt;math&gt;(4,4)&lt;/math&gt; with each step increasing either the &lt;math&gt;x&lt;/math&gt;-coordinate or the &lt;math&gt;y&lt;/math&gt;-coordinate by 1. How many such paths stay outside or on the boundary of the square &lt;math&gt;-2 \le x \le 2&lt;/math&gt;, &lt;math&gt;-2 \le y \le 2&lt;/math&gt; at each step?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800&lt;/math&gt;<br /> <br /> == Solution ==<br /> Each path must go through either the second or the fourth quadrant.<br /> Each path that goes through the second quadrant must pass through exactly one of the points &lt;math&gt;(-4,4)&lt;/math&gt;, &lt;math&gt;(-3,3)&lt;/math&gt;, and &lt;math&gt;(-2,2)&lt;/math&gt;.<br /> <br /> There is &lt;math&gt;1&lt;/math&gt; path of the first kind, &lt;math&gt;{8\choose 1}^2=64&lt;/math&gt; paths of the second kind, and &lt;math&gt;{8\choose 2}^2=28^2=784&lt;/math&gt; paths of the third type. <br /> Each path that goes through the fourth quadrant must pass through exactly one of the points &lt;math&gt;(4,-4)&lt;/math&gt;, &lt;math&gt;(3,-3)&lt;/math&gt;, and &lt;math&gt;(2,-2)&lt;/math&gt;.<br /> Again, there is &lt;math&gt;1&lt;/math&gt; path of the first kind, &lt;math&gt;{8\choose 1}^2=64&lt;/math&gt; paths of the second kind, and &lt;math&gt;{8\choose 2}^2=28^2=784&lt;/math&gt; paths of the third type. <br /> <br /> Hence the total number of paths is &lt;math&gt;2(1+64+784) = \boxed{1698}&lt;/math&gt;.<br /> == Solution 2==<br /> We will use the concept of complimentary counting. If you go in the square you have to go out by these labeled points(click the link) https://imgur.com/VysX4P0 and go out through the borders because if you didnt, you would touch another point in one of those points in the set of points in the link(Call it &lt;math&gt;S&lt;/math&gt;). There is symmetry about &lt;math&gt;y=x&lt;/math&gt;, so we only have to consider &lt;math&gt;(1,-1), (1,0)&lt;/math&gt;, and &lt;math&gt;(1,1)&lt;/math&gt;. &lt;math&gt;(1,1)&lt;/math&gt; can go on the boundary in two ways, so we can only consider one case and multiply it by two. For &lt;math&gt;(1,0)&lt;/math&gt; and &lt;math&gt;(1,-1)&lt;/math&gt; we can just multiply by two. So we count paths from &lt;math&gt;(-4,4)&lt;/math&gt; to each of these points, and then multiply that by the number of ways to get from the point one unit right of that to &lt;math&gt;(4,4)&lt;/math&gt;, and all in all, we get the answer is &lt;math&gt;\dbinom{16}{8}-2[\dbinom{8}{3}\dbinom{7}{2}+\dbinom{9}{4}\dbinom{6}{2}+\dbinom{10}{5}\dbinom{5}{2}]=1698&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> -vsamc(sorry this was rushed)<br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=17|num-a=19|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_17&diff=119023 2010 AMC 12A Problems/Problem 17 2020-03-10T22:39:01Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; has side lengths &lt;math&gt;AB=CD=EF=1&lt;/math&gt; and &lt;math&gt;BC=DE=FA=r&lt;/math&gt;. The area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;70\%&lt;/math&gt; of the area of the hexagon. What is the sum of all possible values of &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> == Solution 1==<br /> It is clear that &lt;math&gt;\triangle ACE&lt;/math&gt; is an equilateral triangle. From the [[Law of Cosines]] on &lt;math&gt;\triangle ABC&lt;/math&gt;, we get that &lt;math&gt;AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1&lt;/math&gt;. Therefore, the area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{4}(r^2+r+1)&lt;/math&gt;.<br /> <br /> If we extend &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; so that &lt;math&gt;FA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; meet at &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;DE&lt;/math&gt; meet at &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; meet at &lt;math&gt;Z&lt;/math&gt;, we find that hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is formed by taking equilateral triangle &lt;math&gt;XYZ&lt;/math&gt; of side length &lt;math&gt;r+2&lt;/math&gt; and removing three equilateral triangles, &lt;math&gt;ABX&lt;/math&gt;, &lt;math&gt;CDY&lt;/math&gt; and &lt;math&gt;EFZ&lt;/math&gt;, of side length &lt;math&gt;1&lt;/math&gt;. The area of &lt;math&gt;ABCDEF&lt;/math&gt; is therefore<br /> <br /> &lt;math&gt;\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)&lt;/math&gt;.<br /> <br /> <br /> Based on the initial conditions,<br /> <br /> &lt;cmath&gt;\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)&lt;/cmath&gt;<br /> <br /> Simplifying this gives us &lt;math&gt;r^2-6r+1 = 0&lt;/math&gt;. By [[Vieta's Formulas]] we know that the sum of the possible value of &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(E)}\ 6}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Step 1: Use [[Law of Cosines]] in the same manner as the previous solution to get &lt;math&gt;AC=\sqrt{r^2+r+1}&lt;/math&gt;. <br /> <br /> Step 2: &lt;math&gt;\triangle{ABC}&lt;/math&gt;~&lt;math&gt;\triangle{CDE}&lt;/math&gt;~&lt;math&gt;\triangle{EFA}&lt;/math&gt; via SAS congruency. Using the formula &lt;math&gt;[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}&lt;/math&gt;. The area of the hexagon is equal to &lt;math&gt;[ACE] + 3[ABC]&lt;/math&gt;. We are given that &lt;math&gt;70\%&lt;/math&gt; of this area is equal to &lt;math&gt;[ACE]&lt;/math&gt;; solving for &lt;math&gt;AC&lt;/math&gt; in terms of &lt;math&gt;r&lt;/math&gt; gives &lt;math&gt;AC=\sqrt{7r}&lt;/math&gt;.<br /> <br /> Step 3: &lt;math&gt;\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0&lt;/math&gt; and by [[Vieta's Formulas]] , we get &lt;math&gt;\boxed{\textbf{E}}&lt;/math&gt;.<br /> <br /> Note: Since &lt;math&gt;r&lt;/math&gt; has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.<br /> <br /> ==Solution 3==<br /> Find the area of the triangle &lt;math&gt;ACE&lt;/math&gt; as how it was done in solution 1. Find the sum of the areas of the congruent triangles &lt;math&gt;ABC, CDE, EFA&lt;/math&gt; as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles &lt;math&gt;ABC, CDE, EFA&lt;/math&gt; is &lt;math&gt;30\%&lt;/math&gt; of the area of the hexagon. Hence &lt;math&gt;\frac{7}{3}&lt;/math&gt; times the latter is equal to the triangle &lt;math&gt;ACE&lt;/math&gt;. Hence &lt;math&gt;\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)&lt;/math&gt;. We can simplify this to &lt;math&gt;7r=r^2+r+1\implies r^2-6r+1=0&lt;/math&gt;. By Vieta's, we get the sum of all possible values of &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;-\frac{-6}{1}=6\text{or} \textbf{(E)}&lt;/math&gt;.<br /> -vsamc<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_17&diff=119022 2010 AMC 12A Problems/Problem 17 2020-03-10T22:38:31Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; has side lengths &lt;math&gt;AB=CD=EF=1&lt;/math&gt; and &lt;math&gt;BC=DE=FA=r&lt;/math&gt;. The area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;70\%&lt;/math&gt; of the area of the hexagon. What is the sum of all possible values of &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> == Solution 1==<br /> It is clear that &lt;math&gt;\triangle ACE&lt;/math&gt; is an equilateral triangle. From the [[Law of Cosines]] on &lt;math&gt;\triangle ABC&lt;/math&gt;, we get that &lt;math&gt;AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1&lt;/math&gt;. Therefore, the area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{4}(r^2+r+1)&lt;/math&gt;.<br /> <br /> If we extend &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; so that &lt;math&gt;FA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; meet at &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;DE&lt;/math&gt; meet at &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; meet at &lt;math&gt;Z&lt;/math&gt;, we find that hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is formed by taking equilateral triangle &lt;math&gt;XYZ&lt;/math&gt; of side length &lt;math&gt;r+2&lt;/math&gt; and removing three equilateral triangles, &lt;math&gt;ABX&lt;/math&gt;, &lt;math&gt;CDY&lt;/math&gt; and &lt;math&gt;EFZ&lt;/math&gt;, of side length &lt;math&gt;1&lt;/math&gt;. The area of &lt;math&gt;ABCDEF&lt;/math&gt; is therefore<br /> <br /> &lt;math&gt;\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)&lt;/math&gt;.<br /> <br /> <br /> Based on the initial conditions,<br /> <br /> &lt;cmath&gt;\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)&lt;/cmath&gt;<br /> <br /> Simplifying this gives us &lt;math&gt;r^2-6r+1 = 0&lt;/math&gt;. By [[Vieta's Formulas]] we know that the sum of the possible value of &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(E)}\ 6}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Step 1: Use [[Law of Cosines]] in the same manner as the previous solution to get &lt;math&gt;AC=\sqrt{r^2+r+1}&lt;/math&gt;. <br /> <br /> Step 2: &lt;math&gt;\triangle{ABC}&lt;/math&gt;~&lt;math&gt;\triangle{CDE}&lt;/math&gt;~&lt;math&gt;\triangle{EFA}&lt;/math&gt; via SAS congruency. Using the formula &lt;math&gt;[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}&lt;/math&gt;. The area of the hexagon is equal to &lt;math&gt;[ACE] + 3[ABC]&lt;/math&gt;. We are given that &lt;math&gt;70\%&lt;/math&gt; of this area is equal to &lt;math&gt;[ACE]&lt;/math&gt;; solving for &lt;math&gt;AC&lt;/math&gt; in terms of &lt;math&gt;r&lt;/math&gt; gives &lt;math&gt;AC=\sqrt{7r}&lt;/math&gt;.<br /> <br /> Step 3: &lt;math&gt;\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0&lt;/math&gt; and by [[Vieta's Formulas]] , we get &lt;math&gt;\boxed{\textbf{E}}&lt;/math&gt;.<br /> <br /> Note: Since &lt;math&gt;r&lt;/math&gt; has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.<br /> <br /> ==Solution 3==<br /> Find the area of the triangle &lt;math&gt;ACE&lt;/math&gt; as how it was done in solution 1. Find the sum of the areas of the congruent triangles &lt;math&gt;ABC, CDE, EFA&lt;/math&gt; as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles &lt;math&gt;ABC, CDE, EFA&lt;/math&gt; is &lt;math&gt;30\%&lt;/math&gt; of the area of the hexagon. Hence &lt;math&gt;\frac{7}{3}&lt;/math&gt; times the latter is equal to the triangle &lt;math&gt;ACE&lt;/math&gt;. Hence &lt;math&gt;\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)&lt;/math&gt;. We can simplify this to &lt;math&gt;7r=r^2+r+1\implies r^2-6r+1=0&lt;/math&gt;. By Vieta's, we get the sum of all possible values of &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;-\frac{-7}{1}=7\text{or} \textbf{(E)}&lt;/math&gt;.<br /> -vsamc<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_17&diff=119021 2010 AMC 12A Problems/Problem 17 2020-03-10T22:38:06Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; has side lengths &lt;math&gt;AB=CD=EF=1&lt;/math&gt; and &lt;math&gt;BC=DE=FA=r&lt;/math&gt;. The area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;70\%&lt;/math&gt; of the area of the hexagon. What is the sum of all possible values of &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> == Solution 1==<br /> It is clear that &lt;math&gt;\triangle ACE&lt;/math&gt; is an equilateral triangle. From the [[Law of Cosines]] on &lt;math&gt;\triangle ABC&lt;/math&gt;, we get that &lt;math&gt;AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1&lt;/math&gt;. Therefore, the area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{4}(r^2+r+1)&lt;/math&gt;.<br /> <br /> If we extend &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; so that &lt;math&gt;FA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; meet at &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;DE&lt;/math&gt; meet at &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; meet at &lt;math&gt;Z&lt;/math&gt;, we find that hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is formed by taking equilateral triangle &lt;math&gt;XYZ&lt;/math&gt; of side length &lt;math&gt;r+2&lt;/math&gt; and removing three equilateral triangles, &lt;math&gt;ABX&lt;/math&gt;, &lt;math&gt;CDY&lt;/math&gt; and &lt;math&gt;EFZ&lt;/math&gt;, of side length &lt;math&gt;1&lt;/math&gt;. The area of &lt;math&gt;ABCDEF&lt;/math&gt; is therefore<br /> <br /> &lt;math&gt;\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)&lt;/math&gt;.<br /> <br /> <br /> Based on the initial conditions,<br /> <br /> &lt;cmath&gt;\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)&lt;/cmath&gt;<br /> <br /> Simplifying this gives us &lt;math&gt;r^2-6r+1 = 0&lt;/math&gt;. By [[Vieta's Formulas]] we know that the sum of the possible value of &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(E)}\ 6}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Step 1: Use [[Law of Cosines]] in the same manner as the previous solution to get &lt;math&gt;AC=\sqrt{r^2+r+1}&lt;/math&gt;. <br /> <br /> Step 2: &lt;math&gt;\triangle{ABC}&lt;/math&gt;~&lt;math&gt;\triangle{CDE}&lt;/math&gt;~&lt;math&gt;\triangle{EFA}&lt;/math&gt; via SAS congruency. Using the formula &lt;math&gt;[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}&lt;/math&gt;. The area of the hexagon is equal to &lt;math&gt;[ACE] + 3[ABC]&lt;/math&gt;. We are given that &lt;math&gt;70\%&lt;/math&gt; of this area is equal to &lt;math&gt;[ACE]&lt;/math&gt;; solving for &lt;math&gt;AC&lt;/math&gt; in terms of &lt;math&gt;r&lt;/math&gt; gives &lt;math&gt;AC=\sqrt{7r}&lt;/math&gt;.<br /> <br /> Step 3: &lt;math&gt;\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0&lt;/math&gt; and by [[Vieta's Formulas]] , we get &lt;math&gt;\boxed{\textbf{E}}&lt;/math&gt;.<br /> <br /> Note: Since &lt;math&gt;r&lt;/math&gt; has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.<br /> <br /> ==Solution 3==<br /> Find the area of the triangle &lt;math&gt;ACE&lt;/math&gt; as how it was done in solution 1. Find the sum of the areas of the congruent triangles &lt;math&gt;ABC, CDE, EFA&lt;/math&gt; as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles &lt;math&gt;ABC, CDE, EFA&lt;/math&gt; is &lt;math&gt;30\%&lt;/math&gt; of the area of the hexagon. Hence &lt;math&gt;\frac{7}{3}&lt;/math&gt; times the latter is equal to the triangle &lt;math&gt;ACE&lt;/math&gt;. Hence &lt;math&gt;\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)&lt;/math&gt;. We can simplify this to &lt;math&gt;7r=r^2+r+1\implies r^2-6r+1=0&lt;/math&gt;. By Vieta's, we get the sum of all possible values of &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;-\frac{-7}{1}=7\text{or} \textbf{E}&lt;/math&gt;.<br /> -vsamc<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_8&diff=119012 2010 AMC 12A Problems/Problem 8 2020-03-10T20:42:33Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=2 \cdot AC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, respectively, such that &lt;math&gt;\angle BAE = \angle ACD&lt;/math&gt;. Let &lt;math&gt;F&lt;/math&gt; be the intersection of segments &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, and suppose that &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral. What is &lt;math&gt;\angle ACB&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> &lt;center&gt;[[File:AMC 2010 12A Problem 8.png]]&lt;/center&gt;<br /> <br /> <br /> Let &lt;math&gt;\angle BAE = \angle ACD = x&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}\angle BCD &amp;= \angle AEC = 60^\circ\\<br /> \angle EAC + \angle FCA + \angle ECF + \angle AEC &amp;= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\<br /> \angle EAC &amp;= 60^\circ - x\\<br /> \angle BAC &amp;= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\frac{AC}{AB} = \frac{1}{2}&lt;/math&gt;, triangle &lt;math&gt;ABC&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, so &lt;math&gt;\angle BCA = \boxed{90^\circ\,\textbf{(C)}}&lt;/math&gt;.<br /> <br /> == Solution 2(Trig and Angle Chasing) ==<br /> Let &lt;math&gt;AB=2a, AC=a&lt;/math&gt;. Let &lt;math&gt;\angle BAE=\angle ACD=x&lt;/math&gt;. Because &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral, we get &lt;math&gt;\angle FCE=60&lt;/math&gt;, so &lt;math&gt;\angle ACB=60+x&lt;/math&gt;. Because &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral, we get &lt;math&gt;\angle CFE=60&lt;/math&gt;. Angles &lt;math&gt;AFD&lt;/math&gt; and &lt;math&gt;CFE&lt;/math&gt; are vertical, so &lt;math&gt;\angle AFD=60&lt;/math&gt;. By triangle &lt;math&gt;ADF&lt;/math&gt;, we have &lt;math&gt;\angle ADF=120-x&lt;/math&gt;, and because of line &lt;math&gt;AB&lt;/math&gt;, we have &lt;math&gt;\angle BDC=60+x&lt;/math&gt;. Because Of line &lt;math&gt;BC&lt;/math&gt;, we have &lt;math&gt;\angle AEB=120&lt;/math&gt;, and by line &lt;math&gt;CD&lt;/math&gt;, we have &lt;math&gt;\angle DFE=120&lt;/math&gt;. By quadrilateral &lt;math&gt;BDFE&lt;/math&gt;, we have &lt;math&gt;\angle ABC=60-x&lt;/math&gt;.<br /> <br /> By the Law of Sines, we have &lt;math&gt;\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies \sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)&lt;/math&gt;. By the sine addition formula(which states &lt;math&gt;\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)&lt;/math&gt; by the way), we have &lt;math&gt;2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)&lt;/math&gt;. Because cosine is an even function, and sine is an odd function, we have &lt;math&gt;2\sin(60)\cos(x)-2\cos(60)\sin(x)=\sin(60)\cos(x)+\cos(60)\sin(x) \implies \sin(60)\cos(x)=3\cos(60)\sin(x)&lt;/math&gt;. We know that &lt;math&gt;\sin(60)=\frac{\sqrt{3}}{2}&lt;/math&gt;, and &lt;math&gt;\cos(60)=\frac{1}{2}&lt;/math&gt;, hence &lt;math&gt;\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}&lt;/math&gt;. The only value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;60+x&lt;180&lt;/math&gt;(because &lt;math&gt;60+x&lt;/math&gt; is an angle of the triangle) is &lt;math&gt;x=30^{\circ}&lt;/math&gt;. We seek to find &lt;math&gt;\angle ACB&lt;/math&gt;, which as we found before is &lt;math&gt;60+x&lt;/math&gt;, which is &lt;math&gt;90&lt;/math&gt;. The answer is &lt;math&gt;90, \text{or} \textbf{(C)}&lt;/math&gt;<br /> <br /> -vsamc<br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_8&diff=119011 2010 AMC 12A Problems/Problem 8 2020-03-10T20:39:40Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=2 \cdot AC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, respectively, such that &lt;math&gt;\angle BAE = \angle ACD&lt;/math&gt;. Let &lt;math&gt;F&lt;/math&gt; be the intersection of segments &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, and suppose that &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral. What is &lt;math&gt;\angle ACB&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> &lt;center&gt;[[File:AMC 2010 12A Problem 8.png]]&lt;/center&gt;<br /> <br /> <br /> Let &lt;math&gt;\angle BAE = \angle ACD = x&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}\angle BCD &amp;= \angle AEC = 60^\circ\\<br /> \angle EAC + \angle FCA + \angle ECF + \angle AEC &amp;= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\<br /> \angle EAC &amp;= 60^\circ - x\\<br /> \angle BAC &amp;= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\frac{AC}{AB} = \frac{1}{2}&lt;/math&gt;, triangle &lt;math&gt;ABC&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, so &lt;math&gt;\angle BCA = \boxed{90^\circ\,\textbf{(C)}}&lt;/math&gt;.<br /> <br /> == Solution 2(Trig and Angle Chasing) ==<br /> Let &lt;math&gt;AB=2a, AC=a&lt;/math&gt;. Let &lt;math&gt;\angle BAE=\angle ACD=x&lt;/math&gt;. Because &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral, we get &lt;math&gt;\angle FCE=60&lt;/math&gt;, so &lt;math&gt;\angle ACB=60+x&lt;/math&gt;. Because &lt;math&gt;\triangle CFE&lt;/math&gt; is equilateral, we get &lt;math&gt;\angle CFE=60&lt;/math&gt;. Angles &lt;math&gt;AFD&lt;/math&gt; and &lt;math&gt;CFE&lt;/math&gt; are vertical, so &lt;math&gt;\angle AFD=60&lt;/math&gt;. By triangle &lt;math&gt;ADF&lt;/math&gt;, we have &lt;math&gt;\angle ADF=120-x&lt;/math&gt;, and because of line &lt;math&gt;AB&lt;/math&gt;, we have &lt;math&gt;\angle BDC=60+x&lt;/math&gt;. Because Of line &lt;math&gt;BC&lt;/math&gt;, we have &lt;math&gt;\angle AEB=120&lt;/math&gt;, and by line &lt;math&gt;CD&lt;/math&gt;, we have &lt;math&gt;\angle DFE=120&lt;/math&gt;. By quadrilateral &lt;math&gt;BDFE&lt;/math&gt;, we have &lt;math&gt;\angle ABC=60-x&lt;/math&gt;.<br /> <br /> By the Law of Sines, we have &lt;math&gt;\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies &lt;/math&gt;\sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)&lt;math&gt;. By the sine addition formula(which states &lt;/math&gt;\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)&lt;math&gt; by the way), we have &lt;/math&gt;2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)&lt;math&gt;. Because cosine is an even function, and sine is an odd function, we have &lt;/math&gt;2\sin(60)\cos(x)-2\cos(60)\sin(x)=\sin(60)\cos(x)+\cos(60)\sin(x) \implies \sin(60)\cos(x)=3\cos(60)\sin(x)&lt;math&gt;. We know that &lt;/math&gt;\sin(60)=\frac{\sqrt{3}}{2}&lt;math&gt;, and &lt;/math&gt;\cos(60)=\frac{1}{2}&lt;math&gt;, hence &lt;/math&gt;\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies &lt;math&gt;\tan(x)=\frac{\sqrt(3)}{3}&lt;/math&gt;. The only value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;60+x&lt;180&lt;/math&gt;(because &lt;math&gt;60+x&lt;/math&gt; is an angle of the triangle)&lt;math&gt; is &lt;/math&gt;x=30^{\circ}&lt;math&gt;. We seek to find &lt;/math&gt;\angle ACB&lt;math&gt;, which as we found before is &lt;/math&gt;60+x&lt;math&gt;, which is &lt;/math&gt;90$. The answer is (C)<br /> <br /> -vsamc<br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_13&diff=118152 2012 AIME I Problems/Problem 13 2020-02-20T14:16:25Z <p>Vsamc: </p> <hr /> <div>==Problem 13==<br /> Three concentric circles have radii &lt;math&gt;3,&lt;/math&gt; &lt;math&gt;4,&lt;/math&gt; and &lt;math&gt;5.&lt;/math&gt; An equilateral triangle with one vertex on each circle has side length &lt;math&gt;s.&lt;/math&gt; The largest possible area of the triangle can be written as &lt;math&gt;a + \tfrac{b}{c} \sqrt{d},&lt;/math&gt; where &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; &lt;math&gt;c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime, and &lt;math&gt;d&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;a+b+c+d.&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Reinterpret the problem in the following manner. Equilateral triangle &lt;math&gt;ABC&lt;/math&gt; has a point &lt;math&gt;X&lt;/math&gt; on the interior such that &lt;math&gt;AX = 5,&lt;/math&gt; &lt;math&gt;BX = 4,&lt;/math&gt; and &lt;math&gt;CX = 3.&lt;/math&gt; A &lt;math&gt;60^\circ&lt;/math&gt; counter-clockwise rotation about vertex &lt;math&gt;A&lt;/math&gt; maps &lt;math&gt;X&lt;/math&gt; to &lt;math&gt;X'&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;C'.&lt;/math&gt; Note that angle &lt;math&gt;XAX'&lt;/math&gt; is &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;XA = X'A = 5&lt;/math&gt; which tells us that triangle &lt;math&gt;XAX'&lt;/math&gt; is equilateral and that &lt;math&gt;XX' = 5.&lt;/math&gt; We now notice that &lt;math&gt;XC = 3&lt;/math&gt; and &lt;math&gt;X'C = 4&lt;/math&gt; which tells us that angle &lt;math&gt;XCX'&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt; because there is a &lt;math&gt;3&lt;/math&gt;-&lt;math&gt;4&lt;/math&gt;-&lt;math&gt;5&lt;/math&gt; Pythagorean triple. Now note that &lt;math&gt;\angle ABC + \angle ACB = 120^\circ&lt;/math&gt; and &lt;math&gt;\angle XCA + \angle XBA = 90^\circ,&lt;/math&gt; so &lt;math&gt;\angle XCB+\angle XBC = 30^\circ&lt;/math&gt; and &lt;math&gt;\angle BXC = 150^\circ.&lt;/math&gt; Applying the law of cosines on triangle &lt;math&gt;BXC&lt;/math&gt; yields<br /> <br /> &lt;cmath&gt;BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}&lt;/cmath&gt;<br /> <br /> and thus the area of &lt;math&gt;ABC&lt;/math&gt; equals &lt;cmath&gt;BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.&lt;/cmath&gt;<br /> <br /> so our final answer is &lt;math&gt;3+4+25+9 = \boxed{041}.&lt;/math&gt;<br /> <br /> Remark: The new figure (the rotations and the triangle) must be twice the original triangle's area. So it is simply <br /> &lt;cmath&gt;\frac{\frac{9\sqrt3+16\sqrt3+25\sqrt3}{4}+3(\frac{3*4}{2}}{2})=25\frac{\sqrt{3}}{4}+9&lt;/cmath&gt;<br /> <br /> ===Solution 2===<br /> Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties:<br /> We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center &lt;math&gt;O&lt;/math&gt; of the circles lies in the interior of triangle &lt;math&gt;ABC&lt;/math&gt; or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let &lt;math&gt;\theta&lt;/math&gt; be the measure of angle &lt;math&gt;XAC&lt;/math&gt; so that angle &lt;math&gt;BAX&lt;/math&gt; has measure &lt;math&gt;60-\theta&lt;/math&gt;. Let &lt;math&gt;AB=BC=AC=x&lt;/math&gt;. The law of cosines on triangles &lt;math&gt;BAX&lt;/math&gt; and &lt;math&gt;XAC&lt;/math&gt; yields &lt;math&gt;\cos(60-\theta)=\frac{x^2+9}{10x}&lt;/math&gt; and &lt;math&gt;\cos\theta=\frac{x^2+16}{10x}&lt;/math&gt;. Solving this system will yield the value of &lt;math&gt;x&lt;/math&gt;. Since &lt;math&gt;\cos\theta=\frac{x^2+16}{10x}&lt;/math&gt; we have that &lt;math&gt;\sin\theta=\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}&lt;/math&gt;. Substituting these into the equation &lt;math&gt;\frac{x^2+9}{10x}=\cos(60-\theta)=\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta&lt;/math&gt; we obtain &lt;math&gt;\frac{x^2+9}{10x}=\frac{1}{2}\frac{x^2+16}{10x}+\frac{\sqrt{3}}{2}\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}&lt;/math&gt;. After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain &lt;math&gt;x^4-50x^2+193=0&lt;/math&gt; so that by the quadratic formula &lt;math&gt;x^2=25\pm12\sqrt{3}&lt;/math&gt;. Under the hypothesis that &lt;math&gt;O&lt;/math&gt; lies in the interior of triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;x^2&lt;/math&gt; must be &lt;math&gt;25+12\sqrt{3}&lt;/math&gt;. To see this, note that the other value for &lt;math&gt;x^2&lt;/math&gt; is roughly &lt;math&gt;4.2&lt;/math&gt; so that &lt;math&gt;x\approx 2.05&lt;/math&gt;, but since &lt;math&gt;AX=5&lt;/math&gt; and &lt;math&gt;AX\leq x&lt;/math&gt; we have a contradiction. We then obtain the area as in Solution 1.<br /> <br /> Now, suppose &lt;math&gt;O&lt;/math&gt; does not lie in the interior of triangle &lt;math&gt;ABC&lt;/math&gt;. We then obtain convex quadrilateral &lt;math&gt;OBAC&lt;/math&gt; with diagonals &lt;math&gt;CB&lt;/math&gt; and &lt;math&gt;OA&lt;/math&gt; intersecting at &lt;math&gt;X&lt;/math&gt;. Here &lt;math&gt;AX=AB=AC=x&lt;/math&gt;. We may let &lt;math&gt;\alpha&lt;/math&gt; denote the measure of angle &lt;math&gt;CAX&lt;/math&gt; so that angle &lt;math&gt;XAB&lt;/math&gt; measures &lt;math&gt;60-\alpha&lt;/math&gt;. Note that the law of cosines on triangles &lt;math&gt;CXA&lt;/math&gt; and &lt;math&gt;BXA&lt;/math&gt; yield the same equations as in the first case with &lt;math&gt;\theta&lt;/math&gt; replaced with &lt;math&gt;\alpha&lt;/math&gt;. Thus we obtain again &lt;math&gt;x^2=25\pm12\sqrt{3}&lt;/math&gt;. If &lt;math&gt;x^2=25+12\sqrt{3}&lt;/math&gt; then &lt;math&gt;x\approx 6.8&lt;/math&gt;, but this is impossible since &lt;math&gt;AX\leq 5&lt;/math&gt; but the shortest possible distance from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;X&lt;/math&gt; is the height of equilateral triangle &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\approx6.8\sqrt{3}\approx5.8&lt;/math&gt;; a contradiction. Hence in this case &lt;math&gt;x^2=25-12\sqrt{3}&lt;/math&gt;. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).<br /> <br /> ===Solution 3===<br /> The problem basically asks for the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; such that it is equilateral and there is a point &lt;math&gt;O&lt;/math&gt; inside of the triangle which satisfies &lt;math&gt;AO = 3&lt;/math&gt;, &lt;math&gt;BO = 4&lt;/math&gt;, and &lt;math&gt;CO = 5&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;AB = BC = AC = s&lt;/math&gt;. We want the area of the triangle, which is just &lt;math&gt;[ABC] = \frac{s^{2}\sqrt{3}}{4}&lt;/math&gt;. Thus, we want to know &lt;math&gt;s^{2}&lt;/math&gt;, and then finding the area will be a matter of simple calculation.<br /> <br /> By law of cosines, &lt;math&gt;s^{2} = 3^{2} + 4^{2} - 2 \cdot 3 \cdot 4 \cdot \cos{\theta} = 25 - 24\cos{\theta}&lt;/math&gt;, where &lt;math&gt;\theta = \angle{AOB}&lt;/math&gt;.<br /> <br /> Now, if we plug this value in for &lt;math&gt;s^{2}&lt;/math&gt; in the area formula, we get &lt;math&gt;[ABC] = \frac{25}{4}\sqrt{3} - 6\sqrt{3}\cos{\theta}&lt;/math&gt;.<br /> <br /> Notice that, in order for this expression to be in the required answer form &lt;math&gt;a + \tfrac{b}{c} \sqrt{d}&lt;/math&gt;, &lt;math&gt;\cos{\theta}&lt;/math&gt; must involve &lt;math&gt;\sqrt{3}&lt;/math&gt;. Now, even minimal experience with simple trigonometric functions will instantly make you think of &lt;math&gt;\frac{\pi}{6}&lt;/math&gt; or &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt;. The former doesn't work, since &lt;math&gt;\angle{AOB}&lt;/math&gt; is obtuse, so &lt;math&gt;\theta = \frac{5\pi}{6}&lt;/math&gt;.<br /> <br /> Thus, our area must be &lt;math&gt;[ABC] = \frac{25}{4}\sqrt{3} + 9&lt;/math&gt;, and so our answer is &lt;math&gt;\boxed{041}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{NOTE:}&lt;/math&gt;<br /> <br /> When we were using law of cosines, we instantly went to the sub-triangle &lt;math&gt;\bigtriangleup ABO&lt;/math&gt;. Why?<br /> <br /> Notice that &lt;math&gt;\angle{ABO}&lt;/math&gt; is bigger than both &lt;math&gt;\angle{BOC}&lt;/math&gt; and &lt;math&gt;\angle{AOC}&lt;/math&gt; because the sides which meet to form the angle are shorter for &lt;math&gt;\angle{ABO}&lt;/math&gt; than the other two triangles.<br /> <br /> Also notice that the three angles add up to &lt;math&gt;2\pi&lt;/math&gt; radians. One of our angles is &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt; by our reasoning/guessing above. This angle then must be the greatest angle out of the three, since the three angles all must be obtuse, and if one angle was &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt;, and another was greater than &lt;math&gt;\frac{5\pi}{6}&lt;/math&gt;, then the third angle would be less than &lt;math&gt;\frac{\pi}{3}&lt;/math&gt;, leading to a contradiction, since we know that all of the three angles are obtuse.<br /> <br /> Thus, we know that our greatest angle &lt;math&gt;\angle{ABO} = \frac{5\pi}{6}&lt;/math&gt;. So, we instantly went to the sub-triangle &lt;math&gt;\bigtriangleup ABO&lt;/math&gt; when using law of cosines since the other sub-triangles will get us nowhere.<br /> <br /> ===Solution 4===<br /> &lt;center&gt;&lt;asy&gt;<br /> import olympiad;<br /> import cse5;<br /> import graph;<br /> <br /> dotfactor = 2;<br /> unitsize(0.3inch);<br /> <br /> pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4);<br /> pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B;<br /> pair X = extension(A,F,D,C);<br /> pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M;<br /> <br /> dot(&quot;$C$&quot;, C, dir(0)); dot(&quot;$A$&quot;, A, dir(90));dot(&quot;$B$&quot;, B, dir(180));<br /> dot(&quot;$D$&quot;, D, NE); dot(&quot;$E$&quot;, E, dir(90));dot(&quot;$F$&quot;, F, dir(270));<br /> dot(&quot;$M$&quot;, M, NE); dot(&quot;$N$&quot;, N, dir(270));dot(&quot;$L$&quot;, L, NW);<br /> dot(&quot;$X$&quot;, X, dir(250));<br /> draw(L--X); draw(M--X); draw(N--X); <br /> <br /> draw(A--B--C--cycle);<br /> draw(A--D--B); draw(B--F--C); draw(A--E--C);<br /> draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed);<br /> draw(L--M--N--cycle);<br /> <br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let's call the circle center &lt;math&gt;X&lt;/math&gt;. It has a distance of 3, 4, 5 to an equilateral triangle &lt;math&gt;LMN&lt;/math&gt;. Consider &lt;math&gt;X&lt;/math&gt;’s pedal triangle &lt;math&gt;ABC&lt;/math&gt;. Since &lt;math&gt;X&lt;/math&gt;’s antipedal triangle is equilateral, &lt;math&gt;X&lt;/math&gt; must be the one of the isogonic centers of &lt;math&gt;\triangle{ABC}&lt;/math&gt;. We’ll take the one inside &lt;math&gt;ABC&lt;/math&gt;, i.e., the Fermat point, because it leads to larger &lt;math&gt;\triangle LMN&lt;/math&gt;. Now we construct the three equilateral triangles &lt;math&gt;ABD&lt;/math&gt;, &lt;math&gt;ACE&lt;/math&gt;, and &lt;math&gt;BCF&lt;/math&gt;, the same way the Fermat point is constructed. Then we have &lt;math&gt;\angle DXE = \angle EXF = \angle FXE = 120&lt;/math&gt;. Since &lt;math&gt;AEMCX&lt;/math&gt; is concyclic with &lt;math&gt;XM&lt;/math&gt;=4 as diameter, we have &lt;math&gt;AC=4\sin(60)&lt;/math&gt;. Similarly, &lt;math&gt;AB=3\sin(60)&lt;/math&gt;, and &lt;math&gt;BC=5\sin(60)&lt;/math&gt;. So &lt;math&gt;\triangle ABC&lt;/math&gt; is a 3-4-5 right triangle with &lt;math&gt;\angle BAC=90&lt;/math&gt;. With some more angle chasing we get <br /> &lt;cmath&gt;\angle MXC+\angle LXB = \angle MAC + \angle LAB = 180 – \angle BAC = 90&lt;/cmath&gt;<br /> &lt;cmath&gt;\angle LXM = 360 – (\angle MXC + \angle LXB + \angle BXC) = 360 –(90+120)=150&lt;/cmath&gt;<br /> By Law of Cosines, we have <br /> &lt;cmath&gt;LM^2 = 3^2+4^2-2*3*4\cos(150)=25+12\sqrt 3&lt;/cmath&gt;<br /> And the area follows.<br /> &lt;cmath&gt;[LMN] = \frac{25}{4}\sqrt{3} + 9; \boxed{041}.&lt;/cmath&gt; <br /> By Mathdummy<br /> <br /> ===Solution 5===<br /> Let &lt;math&gt;ABC&lt;/math&gt; be the equilateral triangle with &lt;math&gt;AB=BC=CA=x.&lt;/math&gt; Assume the coordinates of the vertices are &lt;math&gt;A(-\dfrac{x}{2},0), B(\dfrac{x}{2},0)&lt;/math&gt; and &lt;math&gt;C(0,\dfrac{\sqrt{3}}{2}x).&lt;/math&gt; Let &lt;math&gt;P(a,b)&lt;/math&gt; be such that &lt;math&gt;PA=3, PB=4&lt;/math&gt; and &lt;math&gt;PC=5.&lt;/math&gt; Then<br /> &lt;cmath&gt;\left (a+\dfrac{x}{2}\right )^2 +b^2=3^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;\left (a-\dfrac{x}{2}\right )^2 +b^2=4^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+\left (b-\dfrac{\sqrt{3}}{2}x\right )^2 =5^2.&lt;/cmath&gt;<br /> Subtraction and addition of the first two equations yield &lt;math&gt;2ax=-7, 2a^2+\dfrac{1}{2}x^2+2b^2=25.&lt;/math&gt; The third equation gives &lt;math&gt;2a^2+2b^2-2\sqrt{3}bx+\dfrac{3}{2}x^2=50.&lt;/math&gt; Then &lt;math&gt;x^2-2\sqrt{3}bx=25.&lt;/math&gt; We can then solve for &lt;math&gt;a, b&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt; and have a substitution. We have <br /> &lt;cmath&gt;2\left (\dfrac{-7}{2x}\right )^2 + \dfrac{1}{2}x^2 + 2\left (\dfrac{x^2-25}{2\sqrt{3}x}\right )^2 =25.&lt;/cmath&gt;<br /> Simplify it we have a quadratic equation for &lt;math&gt;x^2: x^4-50x^2+193=0.&lt;/math&gt; So &lt;math&gt;x^2=25\pm 12\sqrt{3}.&lt;/math&gt; The larger one leads to the solution. The smaller one relates to another equilateral triangle, as indicated in Solution 2.<br /> <br /> -JZ<br /> <br /> ==Partial Solution 6==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be the equilateral triangle with &lt;math&gt;AB=BC=CA=s&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; on the radius &lt;math&gt;3&lt;/math&gt; circle, &lt;math&gt;B&lt;/math&gt; on the radius &lt;math&gt;4&lt;/math&gt; circle, and &lt;math&gt;C&lt;/math&gt; on the radius &lt;math&gt;5&lt;/math&gt; circle. Let &lt;math&gt;O&lt;/math&gt; be the origin. Draw &lt;math&gt;AO, BO&lt;/math&gt;, and &lt;math&gt;CO&lt;/math&gt;. Then let &lt;math&gt;\angle AOB=\theta, \angle AOC=\gamma&lt;/math&gt;(And it follows that &lt;math&gt;\angle BOC=180-\theta-\gamma&lt;/math&gt;. From Law of Cosines on &lt;math&gt;\triangle AOB&lt;/math&gt; we have &lt;math&gt;s^2=25-24\cos(\theta)&lt;/math&gt;. From Law of Cosines on &lt;math&gt;\triangle AOC&lt;/math&gt; we have &lt;math&gt;s^2=34-30\cos(\gamma)&lt;/math&gt;. From Law of cosines and the cosine addition formula on &lt;math&gt;\triangle BOC&lt;/math&gt; we have &lt;math&gt;s^2=41-40(\cos(\theta)\cos(\gamma)-\sin(\theta)\sin(\gamma)&lt;/math&gt;, and I don't know a nice way to solve the equation because we can write &lt;math&gt;\cos(\gamma)&lt;/math&gt; in terms of &lt;math&gt;\cos(\theta)&lt;/math&gt; but we still have that &lt;math&gt;\sin(\theta)&lt;/math&gt; which could be plus or minus the square root of &lt;math&gt;1-cos^2(\theta)&lt;/math&gt; so ya if you could help that would be nice :) -vsamc<br /> <br /> ==See also==<br /> {{AIME box|year=2012|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_19&diff=116768 2018 AMC 10A Problems/Problem 19 2020-02-03T21:36:59Z <p>Vsamc: said that the solution was by me sorry for 3 edits in less then a minute</p> <hr /> <div>==Problem==<br /> <br /> A number &lt;math&gt;m&lt;/math&gt; is randomly selected from the set &lt;math&gt;\{11,13,15,17,19\}&lt;/math&gt;, and a number &lt;math&gt;n&lt;/math&gt; is randomly selected from &lt;math&gt;\{1999,2000,2001,\ldots,2018\}&lt;/math&gt;. What is the probability that &lt;math&gt;m^n&lt;/math&gt; has a units digit of &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since we only care about the unit digit, our set &lt;math&gt;\{11,13,15,17,19 \}&lt;/math&gt; can be turned into &lt;math&gt;\{1,3,5,7,9 \}&lt;/math&gt;. Call this set &lt;math&gt;A&lt;/math&gt; and call &lt;math&gt;\{1999, 2000, 2001, \cdots , 2018 \}&lt;/math&gt; set &lt;math&gt;B&lt;/math&gt;. Let's do casework on the element of &lt;math&gt;A&lt;/math&gt; that we choose. Since &lt;math&gt;1\cdot 1=1&lt;/math&gt;, any number from &lt;math&gt;B&lt;/math&gt; can be paired with &lt;math&gt;1&lt;/math&gt; to make &lt;math&gt;1^n&lt;/math&gt; have a units digit of &lt;math&gt;1&lt;/math&gt;. Therefore, the probability of this case happening is &lt;math&gt;\frac{1}{5}&lt;/math&gt; since there is a &lt;math&gt;\frac{1}{5}&lt;/math&gt; chance that the number &lt;math&gt;1&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let us consider the case where the number &lt;math&gt;3&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let's look at the unit digit when we repeatedly multiply the number &lt;math&gt;3&lt;/math&gt; by itself:<br /> &lt;cmath&gt;3\cdot 3=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 3=7&lt;/cmath&gt; <br /> &lt;cmath&gt;7\cdot 3=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 3=3&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;3^x&lt;/math&gt;, for some integer &lt;math&gt;x&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. Now, let's count how many numbers in &lt;math&gt;B&lt;/math&gt; are divisible by &lt;math&gt;4&lt;/math&gt;. This can be done by simply listing:<br /> &lt;cmath&gt;2000,2004,2008,2012,2016.&lt;/cmath&gt;<br /> There are &lt;math&gt;5&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt; divisible by &lt;math&gt;4&lt;/math&gt; out of the &lt;math&gt;2018-1999+1=20&lt;/math&gt; total numbers. Therefore, the probability that &lt;math&gt;3&lt;/math&gt; is picked from &lt;math&gt;A&lt;/math&gt; and a number divisible by &lt;math&gt;4&lt;/math&gt; is picked from &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}&lt;/math&gt;. <br /> Similarly, we can look at the repeating units digit for &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;7\cdot 7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 7=3&lt;/cmath&gt;<br /> &lt;cmath&gt;3\cdot 7=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 7=7&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;7^y&lt;/math&gt;, for some integer &lt;math&gt;y&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. This is exactly the same conditions as our last case with &lt;math&gt;3&lt;/math&gt; so the probability of this case is also &lt;math&gt;\frac{1}{20}&lt;/math&gt;. <br /> Since &lt;math&gt;5\cdot 5=25&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; ends in &lt;math&gt;5&lt;/math&gt;, the units digit of &lt;math&gt;5^w&lt;/math&gt;, for some integer, &lt;math&gt;w&lt;/math&gt; will always be &lt;math&gt;5&lt;/math&gt;. Thus, the probability in this case is &lt;math&gt;0&lt;/math&gt;.<br /> The last case we need to consider is when the number &lt;math&gt;9&lt;/math&gt; is chosen from &lt;math&gt;A&lt;/math&gt;. This happens with probability &lt;math&gt;\frac{1}{5}&lt;/math&gt;. We list out the repeating units digit for &lt;math&gt;9&lt;/math&gt; as we have done for &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;9\cdot 9=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 9=9&lt;/cmath&gt;<br /> We see that the units digit of &lt;math&gt;9^z&lt;/math&gt;, for some integer &lt;math&gt;z&lt;/math&gt;, is &lt;math&gt;1&lt;/math&gt; only when &lt;math&gt;z&lt;/math&gt; is an even number. From the &lt;math&gt;20&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt;, we see that exactly half of them are even. The probability in this case is &lt;math&gt;\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.&lt;/math&gt;<br /> Finally, we can add all of our probabilities together to get <br /> &lt;cmath&gt;\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.&lt;/cmath&gt;<br /> <br /> ~Nivek<br /> <br /> == Solution 2 ==<br /> Since only the units digit is relevant, we can turn the first set into &lt;math&gt;\{1,3,5,7,9\}&lt;/math&gt;. Note that &lt;math&gt;x^4 \equiv 1 \mod 10&lt;/math&gt; for all odd digits &lt;math&gt;x&lt;/math&gt;, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, &lt;math&gt;\mod 4&lt;/math&gt;, this set has 5 values which correspond to &lt;math&gt;\{0,1,2,3\}&lt;/math&gt;, making the probability equal for all of them. Next, check the values for which it is equal to &lt;math&gt;1 \mod 10&lt;/math&gt;. There are &lt;math&gt;4+1+0+1+2=8&lt;/math&gt; values for which it is equal to 1, remembering that &lt;math&gt;5^{4n} \equiv 1 \mod 10&lt;/math&gt; only if &lt;math&gt;n=0&lt;/math&gt;, which it is not. There are 20 values in total, and simplifying &lt;math&gt;\frac{8}{20}&lt;/math&gt; gives us &lt;math&gt;\boxed{\frac{2}{5}}&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> &lt;math&gt;QED\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> By Euler's Theorem, we have that &lt;math&gt;a^{4}=1(\mod 10)&lt;/math&gt;, iff &lt;math&gt;\gcd(a,10)=1&lt;/math&gt;. Hence &lt;math&gt;m=11,13,17,19&lt;/math&gt;, &lt;math&gt;n=2000,2004,2008,2012,2016&lt;/math&gt; work. Also note that &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1&lt;/math&gt;, and the latter mod 10 is clearly 1. So &lt;math&gt;m=11&lt;/math&gt;, &lt;math&gt;n=1999,2001,2002,2003,2005,...,2018&lt;/math&gt; work(not counting multiples of 4 as we would be double counting if we did). We can also note that &lt;math&gt;19^{2a}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;19^{2a}=361^{a}&lt;/math&gt;, and by the same logic as why &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt;, we are done. Hence &lt;math&gt;m=19&lt;/math&gt;, and &lt;math&gt;n=2002, 2006, 2010, 2014, 2018&lt;/math&gt; work(not counting any of the aforementioned cases as that would be double counting). We cannot make anymore observations that add more &lt;math&gt;m^n&lt;/math&gt; with units digit &lt;math&gt;1&lt;/math&gt;, hence the number of &lt;math&gt;m^n&lt;/math&gt; that have units digit one is &lt;math&gt;4\cdot 5+1\cdot 15+1\cdot 5=40&lt;/math&gt;. And the total number of combinations of an element of the set of all &lt;math&gt;m&lt;/math&gt; and an element of the set of all &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;5\cdot 20=100&lt;/math&gt;. Hence the desired probability is &lt;math&gt;\frac{40}{100}=\frac{2}{5}&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(E)}&lt;/math&gt;. <br /> ~vsamc<br /> <br /> ==Video Solution==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Probability Problems]]</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_19&diff=116767 2018 AMC 10A Problems/Problem 19 2020-02-03T21:35:51Z <p>Vsamc: (Phrased stuff better)</p> <hr /> <div>==Problem==<br /> <br /> A number &lt;math&gt;m&lt;/math&gt; is randomly selected from the set &lt;math&gt;\{11,13,15,17,19\}&lt;/math&gt;, and a number &lt;math&gt;n&lt;/math&gt; is randomly selected from &lt;math&gt;\{1999,2000,2001,\ldots,2018\}&lt;/math&gt;. What is the probability that &lt;math&gt;m^n&lt;/math&gt; has a units digit of &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since we only care about the unit digit, our set &lt;math&gt;\{11,13,15,17,19 \}&lt;/math&gt; can be turned into &lt;math&gt;\{1,3,5,7,9 \}&lt;/math&gt;. Call this set &lt;math&gt;A&lt;/math&gt; and call &lt;math&gt;\{1999, 2000, 2001, \cdots , 2018 \}&lt;/math&gt; set &lt;math&gt;B&lt;/math&gt;. Let's do casework on the element of &lt;math&gt;A&lt;/math&gt; that we choose. Since &lt;math&gt;1\cdot 1=1&lt;/math&gt;, any number from &lt;math&gt;B&lt;/math&gt; can be paired with &lt;math&gt;1&lt;/math&gt; to make &lt;math&gt;1^n&lt;/math&gt; have a units digit of &lt;math&gt;1&lt;/math&gt;. Therefore, the probability of this case happening is &lt;math&gt;\frac{1}{5}&lt;/math&gt; since there is a &lt;math&gt;\frac{1}{5}&lt;/math&gt; chance that the number &lt;math&gt;1&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let us consider the case where the number &lt;math&gt;3&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let's look at the unit digit when we repeatedly multiply the number &lt;math&gt;3&lt;/math&gt; by itself:<br /> &lt;cmath&gt;3\cdot 3=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 3=7&lt;/cmath&gt; <br /> &lt;cmath&gt;7\cdot 3=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 3=3&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;3^x&lt;/math&gt;, for some integer &lt;math&gt;x&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. Now, let's count how many numbers in &lt;math&gt;B&lt;/math&gt; are divisible by &lt;math&gt;4&lt;/math&gt;. This can be done by simply listing:<br /> &lt;cmath&gt;2000,2004,2008,2012,2016.&lt;/cmath&gt;<br /> There are &lt;math&gt;5&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt; divisible by &lt;math&gt;4&lt;/math&gt; out of the &lt;math&gt;2018-1999+1=20&lt;/math&gt; total numbers. Therefore, the probability that &lt;math&gt;3&lt;/math&gt; is picked from &lt;math&gt;A&lt;/math&gt; and a number divisible by &lt;math&gt;4&lt;/math&gt; is picked from &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}&lt;/math&gt;. <br /> Similarly, we can look at the repeating units digit for &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;7\cdot 7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 7=3&lt;/cmath&gt;<br /> &lt;cmath&gt;3\cdot 7=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 7=7&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;7^y&lt;/math&gt;, for some integer &lt;math&gt;y&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. This is exactly the same conditions as our last case with &lt;math&gt;3&lt;/math&gt; so the probability of this case is also &lt;math&gt;\frac{1}{20}&lt;/math&gt;. <br /> Since &lt;math&gt;5\cdot 5=25&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; ends in &lt;math&gt;5&lt;/math&gt;, the units digit of &lt;math&gt;5^w&lt;/math&gt;, for some integer, &lt;math&gt;w&lt;/math&gt; will always be &lt;math&gt;5&lt;/math&gt;. Thus, the probability in this case is &lt;math&gt;0&lt;/math&gt;.<br /> The last case we need to consider is when the number &lt;math&gt;9&lt;/math&gt; is chosen from &lt;math&gt;A&lt;/math&gt;. This happens with probability &lt;math&gt;\frac{1}{5}&lt;/math&gt;. We list out the repeating units digit for &lt;math&gt;9&lt;/math&gt; as we have done for &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;9\cdot 9=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 9=9&lt;/cmath&gt;<br /> We see that the units digit of &lt;math&gt;9^z&lt;/math&gt;, for some integer &lt;math&gt;z&lt;/math&gt;, is &lt;math&gt;1&lt;/math&gt; only when &lt;math&gt;z&lt;/math&gt; is an even number. From the &lt;math&gt;20&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt;, we see that exactly half of them are even. The probability in this case is &lt;math&gt;\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.&lt;/math&gt;<br /> Finally, we can add all of our probabilities together to get <br /> &lt;cmath&gt;\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.&lt;/cmath&gt;<br /> <br /> ~Nivek<br /> <br /> == Solution 2 ==<br /> Since only the units digit is relevant, we can turn the first set into &lt;math&gt;\{1,3,5,7,9\}&lt;/math&gt;. Note that &lt;math&gt;x^4 \equiv 1 \mod 10&lt;/math&gt; for all odd digits &lt;math&gt;x&lt;/math&gt;, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, &lt;math&gt;\mod 4&lt;/math&gt;, this set has 5 values which correspond to &lt;math&gt;\{0,1,2,3\}&lt;/math&gt;, making the probability equal for all of them. Next, check the values for which it is equal to &lt;math&gt;1 \mod 10&lt;/math&gt;. There are &lt;math&gt;4+1+0+1+2=8&lt;/math&gt; values for which it is equal to 1, remembering that &lt;math&gt;5^{4n} \equiv 1 \mod 10&lt;/math&gt; only if &lt;math&gt;n=0&lt;/math&gt;, which it is not. There are 20 values in total, and simplifying &lt;math&gt;\frac{8}{20}&lt;/math&gt; gives us &lt;math&gt;\boxed{\frac{2}{5}}&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> &lt;math&gt;QED\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> By Euler's Theorem, we have that &lt;math&gt;a^{4}=1(\mod 10)&lt;/math&gt;, iff &lt;math&gt;\gcd(a,10)=1&lt;/math&gt;. Hence &lt;math&gt;m=11,13,17,19&lt;/math&gt;, &lt;math&gt;n=2000,2004,2008,2012,2016&lt;/math&gt; work. Also note that &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1&lt;/math&gt;, and the latter mod 10 is clearly 1. So &lt;math&gt;m=11&lt;/math&gt;, &lt;math&gt;n=1999,2001,2002,2003,2005,...,2018&lt;/math&gt; work(not counting multiples of 4 as we would be double counting if we did). We can also note that &lt;math&gt;19^{2a}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;19^{2a}=361^{a}&lt;/math&gt;, and by the same logic as why &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt;, we are done. Hence &lt;math&gt;m=19&lt;/math&gt;, and &lt;math&gt;n=2002, 2006, 2010, 2014, 2018&lt;/math&gt; work(not counting any of the aforementioned cases as that would be double counting). We cannot make anymore observations that add more &lt;math&gt;m^n&lt;/math&gt; with units digit &lt;math&gt;1&lt;/math&gt;, hence the number of &lt;math&gt;m^n&lt;/math&gt; that have units digit one is &lt;math&gt;4\cdot 5+1\cdot 15+1\cdot 5=40&lt;/math&gt;. And the total number of combinations of an element of the set of all &lt;math&gt;m&lt;/math&gt; and an element of the set of all &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;5\cdot 20=100&lt;/math&gt;. Hence the desired probability is &lt;math&gt;\frac{40}{100}=\frac{2}{5}&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(E)}&lt;/math&gt;. <br /> <br /> ==Video Solution==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Probability Problems]]</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_19&diff=116766 2018 AMC 10A Problems/Problem 19 2020-02-03T21:34:38Z <p>Vsamc: </p> <hr /> <div>==Problem==<br /> <br /> A number &lt;math&gt;m&lt;/math&gt; is randomly selected from the set &lt;math&gt;\{11,13,15,17,19\}&lt;/math&gt;, and a number &lt;math&gt;n&lt;/math&gt; is randomly selected from &lt;math&gt;\{1999,2000,2001,\ldots,2018\}&lt;/math&gt;. What is the probability that &lt;math&gt;m^n&lt;/math&gt; has a units digit of &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since we only care about the unit digit, our set &lt;math&gt;\{11,13,15,17,19 \}&lt;/math&gt; can be turned into &lt;math&gt;\{1,3,5,7,9 \}&lt;/math&gt;. Call this set &lt;math&gt;A&lt;/math&gt; and call &lt;math&gt;\{1999, 2000, 2001, \cdots , 2018 \}&lt;/math&gt; set &lt;math&gt;B&lt;/math&gt;. Let's do casework on the element of &lt;math&gt;A&lt;/math&gt; that we choose. Since &lt;math&gt;1\cdot 1=1&lt;/math&gt;, any number from &lt;math&gt;B&lt;/math&gt; can be paired with &lt;math&gt;1&lt;/math&gt; to make &lt;math&gt;1^n&lt;/math&gt; have a units digit of &lt;math&gt;1&lt;/math&gt;. Therefore, the probability of this case happening is &lt;math&gt;\frac{1}{5}&lt;/math&gt; since there is a &lt;math&gt;\frac{1}{5}&lt;/math&gt; chance that the number &lt;math&gt;1&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let us consider the case where the number &lt;math&gt;3&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let's look at the unit digit when we repeatedly multiply the number &lt;math&gt;3&lt;/math&gt; by itself:<br /> &lt;cmath&gt;3\cdot 3=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 3=7&lt;/cmath&gt; <br /> &lt;cmath&gt;7\cdot 3=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 3=3&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;3^x&lt;/math&gt;, for some integer &lt;math&gt;x&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. Now, let's count how many numbers in &lt;math&gt;B&lt;/math&gt; are divisible by &lt;math&gt;4&lt;/math&gt;. This can be done by simply listing:<br /> &lt;cmath&gt;2000,2004,2008,2012,2016.&lt;/cmath&gt;<br /> There are &lt;math&gt;5&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt; divisible by &lt;math&gt;4&lt;/math&gt; out of the &lt;math&gt;2018-1999+1=20&lt;/math&gt; total numbers. Therefore, the probability that &lt;math&gt;3&lt;/math&gt; is picked from &lt;math&gt;A&lt;/math&gt; and a number divisible by &lt;math&gt;4&lt;/math&gt; is picked from &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}&lt;/math&gt;. <br /> Similarly, we can look at the repeating units digit for &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;7\cdot 7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 7=3&lt;/cmath&gt;<br /> &lt;cmath&gt;3\cdot 7=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 7=7&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;7^y&lt;/math&gt;, for some integer &lt;math&gt;y&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. This is exactly the same conditions as our last case with &lt;math&gt;3&lt;/math&gt; so the probability of this case is also &lt;math&gt;\frac{1}{20}&lt;/math&gt;. <br /> Since &lt;math&gt;5\cdot 5=25&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; ends in &lt;math&gt;5&lt;/math&gt;, the units digit of &lt;math&gt;5^w&lt;/math&gt;, for some integer, &lt;math&gt;w&lt;/math&gt; will always be &lt;math&gt;5&lt;/math&gt;. Thus, the probability in this case is &lt;math&gt;0&lt;/math&gt;.<br /> The last case we need to consider is when the number &lt;math&gt;9&lt;/math&gt; is chosen from &lt;math&gt;A&lt;/math&gt;. This happens with probability &lt;math&gt;\frac{1}{5}&lt;/math&gt;. We list out the repeating units digit for &lt;math&gt;9&lt;/math&gt; as we have done for &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;9\cdot 9=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 9=9&lt;/cmath&gt;<br /> We see that the units digit of &lt;math&gt;9^z&lt;/math&gt;, for some integer &lt;math&gt;z&lt;/math&gt;, is &lt;math&gt;1&lt;/math&gt; only when &lt;math&gt;z&lt;/math&gt; is an even number. From the &lt;math&gt;20&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt;, we see that exactly half of them are even. The probability in this case is &lt;math&gt;\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.&lt;/math&gt;<br /> Finally, we can add all of our probabilities together to get <br /> &lt;cmath&gt;\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.&lt;/cmath&gt;<br /> <br /> ~Nivek<br /> <br /> == Solution 2 ==<br /> Since only the units digit is relevant, we can turn the first set into &lt;math&gt;\{1,3,5,7,9\}&lt;/math&gt;. Note that &lt;math&gt;x^4 \equiv 1 \mod 10&lt;/math&gt; for all odd digits &lt;math&gt;x&lt;/math&gt;, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, &lt;math&gt;\mod 4&lt;/math&gt;, this set has 5 values which correspond to &lt;math&gt;\{0,1,2,3\}&lt;/math&gt;, making the probability equal for all of them. Next, check the values for which it is equal to &lt;math&gt;1 \mod 10&lt;/math&gt;. There are &lt;math&gt;4+1+0+1+2=8&lt;/math&gt; values for which it is equal to 1, remembering that &lt;math&gt;5^{4n} \equiv 1 \mod 10&lt;/math&gt; only if &lt;math&gt;n=0&lt;/math&gt;, which it is not. There are 20 values in total, and simplifying &lt;math&gt;\frac{8}{20}&lt;/math&gt; gives us &lt;math&gt;\boxed{\frac{2}{5}}&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> &lt;math&gt;QED\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> By Euler's Theorem, we have that &lt;math&gt;a^{4}=1(\mod 10)&lt;/math&gt;, iff &lt;math&gt;\gcd(a,10)=1&lt;/math&gt;. Hence &lt;math&gt;m=11,13,17,19&lt;/math&gt;, &lt;math&gt;n=2000,2004,2008,2012,2016&lt;/math&gt; work. Also note that &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1&lt;/math&gt;, and the latter mod 10 is clearly 1. So &lt;math&gt;m=11&lt;/math&gt;, &lt;math&gt;n=1999,2001,2002,2003,2005,...,2018&lt;/math&gt; work(not counting multiples of 4 because we already did that). We can also note that &lt;math&gt;19^{2a}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;19^{2a}=361^{a}&lt;/math&gt;, and by the same logic as why &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt;, we are done. Hence &lt;math&gt;m=19&lt;/math&gt;, and &lt;math&gt;n=2002, 2006, 2010, 2014, 2018&lt;/math&gt; work(not counting any of the aforementioned cases as that would be double counting). We cannot make anymore observations that add more &lt;math&gt;m^n&lt;/math&gt; with units digit &lt;math&gt;1&lt;/math&gt;, hence the number of &lt;math&gt;m^n&lt;/math&gt; that have units digit one is &lt;math&gt;4\cdot 5+1\cdot 15+1\cdot 5=40&lt;/math&gt;. And the total number of combinations of an element of the set of all &lt;math&gt;m&lt;/math&gt; and an element of the set of all &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;5\cdot 20=100&lt;/math&gt;. Hence the desired probability is &lt;math&gt;\frac{40}{100}=\frac{2}{5}&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(E)}&lt;/math&gt;. <br /> <br /> ==Video Solution==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Probability Problems]]</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_19&diff=116765 2018 AMC 10A Problems/Problem 19 2020-02-03T21:34:01Z <p>Vsamc: </p> <hr /> <div>==Problem==<br /> <br /> A number &lt;math&gt;m&lt;/math&gt; is randomly selected from the set &lt;math&gt;\{11,13,15,17,19\}&lt;/math&gt;, and a number &lt;math&gt;n&lt;/math&gt; is randomly selected from &lt;math&gt;\{1999,2000,2001,\ldots,2018\}&lt;/math&gt;. What is the probability that &lt;math&gt;m^n&lt;/math&gt; has a units digit of &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since we only care about the unit digit, our set &lt;math&gt;\{11,13,15,17,19 \}&lt;/math&gt; can be turned into &lt;math&gt;\{1,3,5,7,9 \}&lt;/math&gt;. Call this set &lt;math&gt;A&lt;/math&gt; and call &lt;math&gt;\{1999, 2000, 2001, \cdots , 2018 \}&lt;/math&gt; set &lt;math&gt;B&lt;/math&gt;. Let's do casework on the element of &lt;math&gt;A&lt;/math&gt; that we choose. Since &lt;math&gt;1\cdot 1=1&lt;/math&gt;, any number from &lt;math&gt;B&lt;/math&gt; can be paired with &lt;math&gt;1&lt;/math&gt; to make &lt;math&gt;1^n&lt;/math&gt; have a units digit of &lt;math&gt;1&lt;/math&gt;. Therefore, the probability of this case happening is &lt;math&gt;\frac{1}{5}&lt;/math&gt; since there is a &lt;math&gt;\frac{1}{5}&lt;/math&gt; chance that the number &lt;math&gt;1&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let us consider the case where the number &lt;math&gt;3&lt;/math&gt; is selected from &lt;math&gt;A&lt;/math&gt;. Let's look at the unit digit when we repeatedly multiply the number &lt;math&gt;3&lt;/math&gt; by itself:<br /> &lt;cmath&gt;3\cdot 3=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 3=7&lt;/cmath&gt; <br /> &lt;cmath&gt;7\cdot 3=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 3=3&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;3^x&lt;/math&gt;, for some integer &lt;math&gt;x&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. Now, let's count how many numbers in &lt;math&gt;B&lt;/math&gt; are divisible by &lt;math&gt;4&lt;/math&gt;. This can be done by simply listing:<br /> &lt;cmath&gt;2000,2004,2008,2012,2016.&lt;/cmath&gt;<br /> There are &lt;math&gt;5&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt; divisible by &lt;math&gt;4&lt;/math&gt; out of the &lt;math&gt;2018-1999+1=20&lt;/math&gt; total numbers. Therefore, the probability that &lt;math&gt;3&lt;/math&gt; is picked from &lt;math&gt;A&lt;/math&gt; and a number divisible by &lt;math&gt;4&lt;/math&gt; is picked from &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}&lt;/math&gt;. <br /> Similarly, we can look at the repeating units digit for &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;7\cdot 7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;9\cdot 7=3&lt;/cmath&gt;<br /> &lt;cmath&gt;3\cdot 7=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 7=7&lt;/cmath&gt;<br /> We see that the unit digit of &lt;math&gt;7^y&lt;/math&gt;, for some integer &lt;math&gt;y&lt;/math&gt;, will only be &lt;math&gt;1&lt;/math&gt; when &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt;. This is exactly the same conditions as our last case with &lt;math&gt;3&lt;/math&gt; so the probability of this case is also &lt;math&gt;\frac{1}{20}&lt;/math&gt;. <br /> Since &lt;math&gt;5\cdot 5=25&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; ends in &lt;math&gt;5&lt;/math&gt;, the units digit of &lt;math&gt;5^w&lt;/math&gt;, for some integer, &lt;math&gt;w&lt;/math&gt; will always be &lt;math&gt;5&lt;/math&gt;. Thus, the probability in this case is &lt;math&gt;0&lt;/math&gt;.<br /> The last case we need to consider is when the number &lt;math&gt;9&lt;/math&gt; is chosen from &lt;math&gt;A&lt;/math&gt;. This happens with probability &lt;math&gt;\frac{1}{5}&lt;/math&gt;. We list out the repeating units digit for &lt;math&gt;9&lt;/math&gt; as we have done for &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;:<br /> &lt;cmath&gt;9\cdot 9=1&lt;/cmath&gt;<br /> &lt;cmath&gt;1\cdot 9=9&lt;/cmath&gt;<br /> We see that the units digit of &lt;math&gt;9^z&lt;/math&gt;, for some integer &lt;math&gt;z&lt;/math&gt;, is &lt;math&gt;1&lt;/math&gt; only when &lt;math&gt;z&lt;/math&gt; is an even number. From the &lt;math&gt;20&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt;, we see that exactly half of them are even. The probability in this case is &lt;math&gt;\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.&lt;/math&gt;<br /> Finally, we can add all of our probabilities together to get <br /> &lt;cmath&gt;\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.&lt;/cmath&gt;<br /> <br /> ~Nivek<br /> <br /> == Solution 2 ==<br /> Since only the units digit is relevant, we can turn the first set into &lt;math&gt;\{1,3,5,7,9\}&lt;/math&gt;. Note that &lt;math&gt;x^4 \equiv 1 \mod 10&lt;/math&gt; for all odd digits &lt;math&gt;x&lt;/math&gt;, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, &lt;math&gt;\mod 4&lt;/math&gt;, this set has 5 values which correspond to &lt;math&gt;\{0,1,2,3\}&lt;/math&gt;, making the probability equal for all of them. Next, check the values for which it is equal to &lt;math&gt;1 \mod 10&lt;/math&gt;. There are &lt;math&gt;4+1+0+1+2=8&lt;/math&gt; values for which it is equal to 1, remembering that &lt;math&gt;5^{4n} \equiv 1 \mod 10&lt;/math&gt; only if &lt;math&gt;n=0&lt;/math&gt;, which it is not. There are 20 values in total, and simplifying &lt;math&gt;\frac{8}{20}&lt;/math&gt; gives us &lt;math&gt;\boxed{\frac{2}{5}}&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> &lt;math&gt;QED\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> By Euler's Theorem, we have that &lt;math&gt;a^{4}=1(\mod 10)&lt;/math&gt;, iff &lt;math&gt;\gcd(a,10)=1&lt;/math&gt;. Hence &lt;math&gt;m=11,13,17,19&lt;/math&gt;, &lt;math&gt;n=2000,2004,2008,2012,2016&lt;/math&gt; work. Also note that &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1&lt;/math&gt;, and the latter mod 10 is clearly 1. So &lt;math&gt;m=11&lt;/math&gt;, &lt;math&gt;n=1999,2001,2002,2003,2005,...,2018&lt;/math&gt; work(not counting multiples of 4 because we already did that). We can also note that &lt;math&gt;19^{2a}\equiv 1(\mod 10)&lt;/math&gt; because &lt;math&gt;19^{2a}=361^{a}&lt;/math&gt;, and by the same logic as why &lt;math&gt;11^{\text{any positive integer}}\equiv 1(\mod 10)&lt;/math&gt;, we are done. Hence &lt;math&gt;m=19&lt;/math&gt;, and &lt;math&gt;n=2002, 2006, 2010, 2014, 2018&lt;/math&gt; work(not counting any of the aforementioned cases as that would be double counting). We cannot mmake anymore observations tat add more &lt;math&gt;m^n&lt;/math&gt; with units digit &lt;math&gt;1&lt;/math&gt;, hence the number of &lt;math&gt;m^n&lt;/math&gt; that have units digit one is &lt;math&gt;4\cdot 5+1\cdot 15+1\cdot 5=40&lt;/math&gt;. And the total number of combinations of an element of the set of all &lt;math&gt;m&lt;/math&gt; and an element of the set of all &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;5\cdot 20=100&lt;/math&gt;. Hence the desired probability is &lt;math&gt;\frac{40}{100}=\frac{2}{5}&lt;/math&gt;, which is answer choice &lt;math&gt;\textbf{(E)}&lt;/math&gt;. <br /> <br /> ==Video Solution==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Probability Problems]]</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems&diff=115478 2014 AMC 10A Problems 2020-01-24T21:55:43Z <p>Vsamc: </p> <hr /> <div>==Problem 1==<br /> <br /> What is &lt;math&gt;10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{25}{2}\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Roy's cat eats &lt;math&gt;\frac{1}{3}&lt;/math&gt; of a can of cat food every morning and &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing &lt;math&gt;6&lt;/math&gt; cans of cat food. On what day of the week did Roy's cat finish eating all the cat food in the box?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 2|solution]]<br /> <br /> ==Problem 3==<br /> <br /> Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for &lt;math&gt;\textdollar 2.50&lt;/math&gt; each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs &lt;math&gt;\textdollar 0.75&lt;/math&gt; for her to make. In dollars, what is her profit for the day?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> On an algebra quiz, &lt;math&gt;10\%&lt;/math&gt; of the students scored &lt;math&gt;0&lt;/math&gt; points, &lt;math&gt;35\%&lt;/math&gt; scored &lt;math&gt;10&lt;/math&gt; points, &lt;math&gt;30\%&lt;/math&gt; scored &lt;math&gt;15&lt;/math&gt; points, and the rest scored &lt;math&gt;30&lt;/math&gt; points. What is the difference between the mean and median score of the students' scores on this quiz?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Nonzero real numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; satisfy &lt;math&gt;x &lt; a&lt;/math&gt; and &lt;math&gt;y &lt; b&lt;/math&gt;. How many of the following inequalities must be true? <br /> <br /> &lt;math&gt;\textbf{(I)}\ x+y &lt; a+b\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(II)}\ x-y &lt; a-b\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(III)}\ xy &lt; ab\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(IV)}\ \frac{x}{y} &lt; \frac{a}{b}&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Which of the following numbers is a perfect square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The two legs of a right triangle, which are altitudes, have lengths &lt;math&gt;2\sqrt3&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;. How long is the third altitude of the triangle?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Five positive consecutive integers starting with &lt;math&gt;a&lt;/math&gt; have average &lt;math&gt;b&lt;/math&gt;. What is the average of &lt;math&gt;5&lt;/math&gt; consecutive integers that start with &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A customer who intends to purchase an appliance has three coupons, only one of which may be used:<br /> <br /> Coupon 1: &lt;math&gt;10\%&lt;/math&gt; off the listed price if the listed price is at least &lt;math&gt;\textdollar50&lt;/math&gt;<br /> <br /> Coupon 2: &lt;math&gt;\textdollar 20&lt;/math&gt; off the listed price if the listed price is at least &lt;math&gt;\textdollar100&lt;/math&gt;<br /> <br /> Coupon 3: &lt;math&gt;18\%&lt;/math&gt; off the amount by which the listed price exceeds &lt;math&gt;\textdollar100&lt;/math&gt;<br /> <br /> For which of the following listed prices will coupon &lt;math&gt;1&lt;/math&gt; offer a greater price reduction than either coupon &lt;math&gt;2&lt;/math&gt; or coupon &lt;math&gt;3&lt;/math&gt;?<br /> &lt;math&gt;\textbf{(A) }\textdollar179.95\qquad<br /> \textbf{(B) }\textdollar199.95\qquad<br /> \textbf{(C) }\textdollar219.95\qquad<br /> \textbf{(D) }\textdollar239.95\qquad<br /> \textbf{(E) }\textdollar259.95\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> size(125);<br /> defaultpen(linewidth(0.8));<br /> path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle;<br /> fill(hexagon,grey);<br /> for(int i=0;i&lt;=5;i=i+1)<br /> {<br /> path arc=2*dir(60*i)--arc(2*dir(60*i),1,120+60*i,240+60*i)--cycle;<br /> unfill(arc);<br /> draw(arc);<br /> }<br /> draw(hexagon,linewidth(1.8));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 27\sqrt{3}-9\pi\qquad\textbf{(B)}\ 27\sqrt{3}-6\pi\qquad\textbf{(C)}\ 54\sqrt{3}-18\pi\qquad\textbf{(D)}\ 54\sqrt{3}-12\pi\qquad\textbf{(E)}\ 108\sqrt{3}-9\pi &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;1&lt;/math&gt;, and squares &lt;math&gt;ABDE&lt;/math&gt;, &lt;math&gt;BCHI&lt;/math&gt;, &lt;math&gt;CAFG&lt;/math&gt; lie outside the triangle. What is the area of hexagon &lt;math&gt;DEFGHI&lt;/math&gt;?<br /> &lt;asy&gt;<br /> import graph;<br /> size(6cm);<br /> pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps);<br /> pair B = (0,0);<br /> pair C = (1,0);<br /> pair A = rotate(60,B)*C;<br /> <br /> pair E = rotate(270,A)*B;<br /> pair D = rotate(270,E)*A;<br /> <br /> pair F = rotate(90,A)*C;<br /> pair G = rotate(90,F)*A;<br /> <br /> pair I = rotate(270,B)*C;<br /> pair H = rotate(270,I)*B;<br /> <br /> draw(A--B--C--cycle);<br /> draw(A--E--D--B);<br /> draw(A--F--G--C);<br /> draw(B--I--H--C);<br /> <br /> draw(E--F);<br /> draw(D--I);<br /> draw(I--H);<br /> draw(H--G);<br /> <br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,W);<br /> label(&quot;$F$&quot;,F,E);<br /> label(&quot;$G$&quot;,G,E);<br /> label(&quot;$H$&quot;,H,SE);<br /> label(&quot;$I$&quot;,I,SW);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> The &lt;math&gt;y&lt;/math&gt;-intercepts, &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, of two perpendicular lines intersecting at the point &lt;math&gt;D(6,8)&lt;/math&gt; have a sum of zero. What is the area of &lt;math&gt;\triangle DAB&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> David drives from his home to the airport to catch a flight. He drives &lt;math&gt;35&lt;/math&gt; miles in the first hour, but realizes that he will be &lt;math&gt;1&lt;/math&gt; hour late if she continues at this speed. He increases his speed by &lt;math&gt;15&lt;/math&gt; miles per hour for the rest of the way to the airport and arrives &lt;math&gt;30&lt;/math&gt; minutes early. How many miles is the airport from his home?<br /> <br /> &lt;math&gt;\textbf{(A) }140\qquad<br /> \textbf{(B) }175\qquad<br /> \textbf{(C) }210\qquad<br /> \textbf{(D) }245\qquad<br /> \textbf{(E) }280\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=1&lt;/math&gt;, &lt;math&gt;BC=2&lt;/math&gt;, and points &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; are midpoints of &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt;, respectively. Point &lt;math&gt;H&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{GE}&lt;/math&gt;. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(9cm);<br /> pen dps = fontsize(10); defaultpen(dps);<br /> pair D = (0,0);<br /> pair F = (1/2,0);<br /> pair C = (1,0);<br /> pair G = (0,1);<br /> pair E = (1,1);<br /> pair A = (0,2);<br /> pair B = (1,2);<br /> pair H = (1/2,1);<br /> <br /> // do not look<br /> pair X = (1/3,2/3);<br /> pair Y = (2/3,2/3);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(G--E);<br /> draw(A--F--B);<br /> draw(D--H--C);<br /> filldraw(H--X--F--Y--cycle,grey);<br /> <br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E,E);<br /> label(&quot;$F$&quot;,F,S);<br /> label(&quot;$G$&quot;,G,W);<br /> label(&quot;$H$&quot;,H,N);<br /> <br /> label(&quot;$\frac12$&quot;,(0.25,0),S);<br /> label(&quot;$\frac12$&quot;,(0.75,0),S);<br /> label(&quot;$1$&quot;,(1,0.5),E);<br /> label(&quot;$1$&quot;,(1,1.5),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?‮<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> A square in the coordinate plane has vertices whose &lt;math&gt;y&lt;/math&gt;-coordinates are &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt;. What is the area of the square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Four cubes with edge lengths &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt; are stacked as shown. What is the length of the portion of &lt;math&gt;\overline{XY}&lt;/math&gt; contained in the cube with edge length &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{3\sqrt{33}}5\qquad\textbf{(B)}\ 2\sqrt3\qquad\textbf{(C)}\ \dfrac{2\sqrt{33}}3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 3\sqrt2 &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> dotfactor = 3;<br /> size(10cm);<br /> dot((0, 10));<br /> label(&quot;$X$&quot;, (0,10),W,fontsize(8pt));<br /> dot((6,2));<br /> label(&quot;$Y$&quot;, (6,2),E,fontsize(8pt));<br /> draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle);<br /> draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10));<br /> draw((1, 10)--(1.5,10.5));<br /> draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7));<br /> draw((2,9)--(3,10));<br /> draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4));<br /> draw((3,7)--(4.5,8.5));<br /> draw((4.5,6)--(6,6)--(6,2)--(4,0));<br /> draw((4,4)--(6,6));<br /> label(&quot;$1$&quot;, (1,9.5), W,fontsize(8pt));<br /> label(&quot;$2$&quot;, (2,8), W,fontsize(8pt));<br /> label(&quot;$3$&quot;, (3,5.5), W,fontsize(8pt));<br /> &lt;/asy&gt;<br /> [[2014 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> The product &lt;math&gt;(8)(888\dots8)&lt;/math&gt;, where the second factor has &lt;math&gt;991&lt;/math&gt; digits, is an integer whose digits have a sum of &lt;math&gt;1000&lt;/math&gt;. What is &lt;math&gt;991&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are such that the graphs of &lt;math&gt;y=ax+5&lt;/math&gt; and &lt;math&gt;y=3x+b&lt;/math&gt; intersect the &lt;math&gt;x&lt;/math&gt;-axis at the same point. What is the sum of all possible &lt;math&gt;x&lt;/math&gt;-coordinates of these points of intersection?<br /> <br /> &lt;math&gt; \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=10&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be a point on &lt;math&gt;\overline{CD}&lt;/math&gt; such that &lt;math&gt;\angle CBE=15^\circ&lt;/math&gt;. What is &lt;math&gt;AE&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> A rectangular piece of paper whose length is &lt;math&gt;\sqrt3&lt;/math&gt; times the width has area &lt;math&gt;A&lt;/math&gt;. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area &lt;math&gt;B&lt;/math&gt;. What is the ratio &lt;math&gt;B:A&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(6cm);<br /> <br /> real L = 0.05;<br /> <br /> pair A = (0,0);<br /> pair B = (sqrt(3),0);<br /> pair C = (sqrt(3),1);<br /> pair D = (0,1);<br /> <br /> pair X1 = (sqrt(3)/3,0);<br /> pair X2= (2*sqrt(3)/3,0);<br /> pair Y1 = (2*sqrt(3)/3,1);<br /> pair Y2 = (sqrt(3)/3,1);<br /> <br /> dot(X1);<br /> dot(Y1);<br /> <br /> draw(A--B--C--D--cycle, linewidth(2));<br /> draw(X1--Y1,dashed);<br /> <br /> draw(X2--(2*sqrt(3)/3,L));<br /> draw(Y2--(sqrt(3)/3,1-L));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 3:5\qquad\textbf{(C)}\ 2:3\qquad\textbf{(D)}\ 3:4\qquad\textbf{(E)}\ 4:5 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> A sequence of natural numbers is constructed by listing the first &lt;math&gt;4&lt;/math&gt;, then skipping one, listing the next &lt;math&gt;5&lt;/math&gt;, skipping &lt;math&gt;2&lt;/math&gt;, listing &lt;math&gt;6&lt;/math&gt;, skipping &lt;math&gt;3&lt;/math&gt;, and, on the &lt;math&gt;n&lt;/math&gt;th iteration, listing &lt;math&gt;n+3&lt;/math&gt; and skipping &lt;math&gt;n&lt;/math&gt;. The sequence begins &lt;math&gt;1,2,3,4,6,7,8,9,10,13&lt;/math&gt;. What is the &lt;math&gt;500,000&lt;/math&gt;th number in the sequence?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 996,506\qquad\textbf{(B)}\ 996,507\qquad\textbf{(C)}\ 996,508\qquad\textbf{(D)}\ 996,509\qquad\textbf{(E)}\ 996,510 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25==<br /> <br /> The number &lt;math&gt;5^{867}&lt;/math&gt; is between &lt;math&gt;2^{2013}&lt;/math&gt; and &lt;math&gt;2^{2014}&lt;/math&gt;. How many pairs of integers &lt;math&gt;(m,n)&lt;/math&gt; are there such that &lt;math&gt;1\leq m\leq 2012&lt;/math&gt; and &lt;cmath&gt;5^n&lt;2^m&lt;2^{m+2}&lt;5^{n+1}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }278\qquad\textbf{(B) }279\qquad\textbf{(C) }280\qquad\textbf{(D) }281\qquad\textbf{(E) }282\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2014|ab=A|before=[[2013 AMC 10B Problems]]|after=[[2014 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2014 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems&diff=115477 2014 AMC 10A Problems 2020-01-24T21:52:56Z <p>Vsamc: Undo revision 115476 by Arnold (talk)</p> <hr /> <div>==Problem 1==<br /> <br /> What is &lt;math&gt;10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{25}{2}\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Roy's cat eats &lt;math&gt;\frac{1}{3}&lt;/math&gt; of a can of cat food every morning and &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing &lt;math&gt;6&lt;/math&gt; cans of cat food. On what day of the week did Roy's cat finish eating all the cat food in the box?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}\qquad\textbf{(F)}\ \text{Option 6}&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 2|solution]]<br /> <br /> ==Problem 3==<br /> <br /> Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for &lt;math&gt;\textdollar 2.50&lt;/math&gt; each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs &lt;math&gt;\textdollar 0.75&lt;/math&gt; for her to make. In dollars, what is her profit for the day?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> On an algebra quiz, &lt;math&gt;10\%&lt;/math&gt; of the students scored &lt;math&gt;0&lt;/math&gt; points, &lt;math&gt;35\%&lt;/math&gt; scored &lt;math&gt;10&lt;/math&gt; points, &lt;math&gt;30\%&lt;/math&gt; scored &lt;math&gt;15&lt;/math&gt; points, and the rest scored &lt;math&gt;30&lt;/math&gt; points. What is the difference between the mean and median score of the students' scores on this quiz?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Nonzero real numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; satisfy &lt;math&gt;x &lt; a&lt;/math&gt; and &lt;math&gt;y &lt; b&lt;/math&gt;. How many of the following inequalities must be true? <br /> <br /> &lt;math&gt;\textbf{(I)}\ x+y &lt; a+b\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(II)}\ x-y &lt; a-b\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(III)}\ xy &lt; ab\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(IV)}\ \frac{x}{y} &lt; \frac{a}{b}&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Which of the following numbers is a perfect square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The two legs of a right triangle, which are altitudes, have lengths &lt;math&gt;2\sqrt3&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;. How long is the third altitude of the triangle?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Five positive consecutive integers starting with &lt;math&gt;a&lt;/math&gt; have average &lt;math&gt;b&lt;/math&gt;. What is the average of &lt;math&gt;5&lt;/math&gt; consecutive integers that start with &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A customer who intends to purchase an appliance has three coupons, only one of which may be used:<br /> <br /> Coupon 1: &lt;math&gt;10\%&lt;/math&gt; off the listed price if the listed price is at least &lt;math&gt;\textdollar50&lt;/math&gt;<br /> <br /> Coupon 2: &lt;math&gt;\textdollar 20&lt;/math&gt; off the listed price if the listed price is at least &lt;math&gt;\textdollar100&lt;/math&gt;<br /> <br /> Coupon 3: &lt;math&gt;18\%&lt;/math&gt; off the amount by which the listed price exceeds &lt;math&gt;\textdollar100&lt;/math&gt;<br /> <br /> For which of the following listed prices will coupon &lt;math&gt;1&lt;/math&gt; offer a greater price reduction than either coupon &lt;math&gt;2&lt;/math&gt; or coupon &lt;math&gt;3&lt;/math&gt;?<br /> &lt;math&gt;\textbf{(A) }\textdollar179.95\qquad<br /> \textbf{(B) }\textdollar199.95\qquad<br /> \textbf{(C) }\textdollar219.95\qquad<br /> \textbf{(D) }\textdollar239.95\qquad<br /> \textbf{(E) }\textdollar259.95\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> size(125);<br /> defaultpen(linewidth(0.8));<br /> path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle;<br /> fill(hexagon,grey);<br /> for(int i=0;i&lt;=5;i=i+1)<br /> {<br /> path arc=2*dir(60*i)--arc(2*dir(60*i),1,120+60*i,240+60*i)--cycle;<br /> unfill(arc);<br /> draw(arc);<br /> }<br /> draw(hexagon,linewidth(1.8));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 27\sqrt{3}-9\pi\qquad\textbf{(B)}\ 27\sqrt{3}-6\pi\qquad\textbf{(C)}\ 54\sqrt{3}-18\pi\qquad\textbf{(D)}\ 54\sqrt{3}-12\pi\qquad\textbf{(E)}\ 108\sqrt{3}-9\pi &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;1&lt;/math&gt;, and squares &lt;math&gt;ABDE&lt;/math&gt;, &lt;math&gt;BCHI&lt;/math&gt;, &lt;math&gt;CAFG&lt;/math&gt; lie outside the triangle. What is the area of hexagon &lt;math&gt;DEFGHI&lt;/math&gt;?<br /> &lt;asy&gt;<br /> import graph;<br /> size(6cm);<br /> pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps);<br /> pair B = (0,0);<br /> pair C = (1,0);<br /> pair A = rotate(60,B)*C;<br /> <br /> pair E = rotate(270,A)*B;<br /> pair D = rotate(270,E)*A;<br /> <br /> pair F = rotate(90,A)*C;<br /> pair G = rotate(90,F)*A;<br /> <br /> pair I = rotate(270,B)*C;<br /> pair H = rotate(270,I)*B;<br /> <br /> draw(A--B--C--cycle);<br /> draw(A--E--D--B);<br /> draw(A--F--G--C);<br /> draw(B--I--H--C);<br /> <br /> draw(E--F);<br /> draw(D--I);<br /> draw(I--H);<br /> draw(H--G);<br /> <br /> label(&quot;$A$&quot;,A,N);<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,W);<br /> label(&quot;$F$&quot;,F,E);<br /> label(&quot;$G$&quot;,G,E);<br /> label(&quot;$H$&quot;,H,SE);<br /> label(&quot;$I$&quot;,I,SW);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> The &lt;math&gt;y&lt;/math&gt;-intercepts, &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, of two perpendicular lines intersecting at the point &lt;math&gt;D(6,8)&lt;/math&gt; have a sum of zero. What is the area of &lt;math&gt;\triangle DAB&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> David drives from his home to the airport to catch a flight. He drives &lt;math&gt;35&lt;/math&gt; miles in the first hour, but realizes that he will be &lt;math&gt;1&lt;/math&gt; hour late if she continues at this speed. He increases his speed by &lt;math&gt;15&lt;/math&gt; miles per hour for the rest of the way to the airport and arrives &lt;math&gt;30&lt;/math&gt; minutes early. How many miles is the airport from his home?<br /> <br /> &lt;math&gt;\textbf{(A) }140\qquad<br /> \textbf{(B) }175\qquad<br /> \textbf{(C) }210\qquad<br /> \textbf{(D) }245\qquad<br /> \textbf{(E) }280\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=1&lt;/math&gt;, &lt;math&gt;BC=2&lt;/math&gt;, and points &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; are midpoints of &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt;, respectively. Point &lt;math&gt;H&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{GE}&lt;/math&gt;. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(9cm);<br /> pen dps = fontsize(10); defaultpen(dps);<br /> pair D = (0,0);<br /> pair F = (1/2,0);<br /> pair C = (1,0);<br /> pair G = (0,1);<br /> pair E = (1,1);<br /> pair A = (0,2);<br /> pair B = (1,2);<br /> pair H = (1/2,1);<br /> <br /> // do not look<br /> pair X = (1/3,2/3);<br /> pair Y = (2/3,2/3);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(G--E);<br /> draw(A--F--B);<br /> draw(D--H--C);<br /> filldraw(H--X--F--Y--cycle,grey);<br /> <br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E,E);<br /> label(&quot;$F$&quot;,F,S);<br /> label(&quot;$G$&quot;,G,W);<br /> label(&quot;$H$&quot;,H,N);<br /> <br /> label(&quot;$\frac12$&quot;,(0.25,0),S);<br /> label(&quot;$\frac12$&quot;,(0.75,0),S);<br /> label(&quot;$1$&quot;,(1,0.5),E);<br /> label(&quot;$1$&quot;,(1,1.5),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?‮<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> A square in the coordinate plane has vertices whose &lt;math&gt;y&lt;/math&gt;-coordinates are &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt;. What is the area of the square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Four cubes with edge lengths &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt; are stacked as shown. What is the length of the portion of &lt;math&gt;\overline{XY}&lt;/math&gt; contained in the cube with edge length &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{3\sqrt{33}}5\qquad\textbf{(B)}\ 2\sqrt3\qquad\textbf{(C)}\ \dfrac{2\sqrt{33}}3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 3\sqrt2 &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> dotfactor = 3;<br /> size(10cm);<br /> dot((0, 10));<br /> label(&quot;$X$&quot;, (0,10),W,fontsize(8pt));<br /> dot((6,2));<br /> label(&quot;$Y$&quot;, (6,2),E,fontsize(8pt));<br /> draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle);<br /> draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10));<br /> draw((1, 10)--(1.5,10.5));<br /> draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7));<br /> draw((2,9)--(3,10));<br /> draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4));<br /> draw((3,7)--(4.5,8.5));<br /> draw((4.5,6)--(6,6)--(6,2)--(4,0));<br /> draw((4,4)--(6,6));<br /> label(&quot;$1$&quot;, (1,9.5), W,fontsize(8pt));<br /> label(&quot;$2$&quot;, (2,8), W,fontsize(8pt));<br /> label(&quot;$3\$&quot;, (3,5.5), W,fontsize(8pt));<br /> &lt;/asy&gt;<br /> [[2014 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> The product &lt;math&gt;(8)(888\dots8)&lt;/math&gt;, where the second factor has &lt;math&gt;991&lt;/math&gt; digits, is an integer whose digits have a sum of &lt;math&gt;1000&lt;/math&gt;. What is &lt;math&gt;991&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are such that the graphs of &lt;math&gt;y=ax+5&lt;/math&gt; and &lt;math&gt;y=3x+b&lt;/math&gt; intersect the &lt;math&gt;x&lt;/math&gt;-axis at the same point. What is the sum of all possible &lt;math&gt;x&lt;/math&gt;-coordinates of these points of intersection?<br /> <br /> &lt;math&gt; \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=10&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be a point on &lt;math&gt;\overline{CD}&lt;/math&gt; such that &lt;math&gt;\angle CBE=15^\circ&lt;/math&gt;. What is &lt;math&gt;AE&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> A rectangular piece of paper whose length is &lt;math&gt;\sqrt3&lt;/math&gt; times the width has area &lt;math&gt;A&lt;/math&gt;. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area &lt;math&gt;B&lt;/math&gt;. What is the ratio &lt;math&gt;B:A&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(6cm);<br /> <br /> real L = 0.05;<br /> <br /> pair A = (0,0);<br /> pair B = (sqrt(3),0);<br /> pair C = (sqrt(3),1);<br /> pair D = (0,1);<br /> <br /> pair X1 = (sqrt(3)/3,0);<br /> pair X2= (2*sqrt(3)/3,0);<br /> pair Y1 = (2*sqrt(3)/3,1);<br /> pair Y2 = (sqrt(3)/3,1);<br /> <br /> dot(X1);<br /> dot(Y1);<br /> <br /> draw(A--B--C--D--cycle, linewidth(2));<br /> draw(X1--Y1,dashed);<br /> <br /> draw(X2--(2*sqrt(3)/3,L));<br /> draw(Y2--(sqrt(3)/3,1-L));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 3:5\qquad\textbf{(C)}\ 2:3\qquad\textbf{(D)}\ 3:4\qquad\textbf{(E)}\ 4:5 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> A sequence of natural numbers is constructed by listing the first &lt;math&gt;4&lt;/math&gt;, then skipping one, listing the next &lt;math&gt;5&lt;/math&gt;, skipping &lt;math&gt;2&lt;/math&gt;, listing &lt;math&gt;6&lt;/math&gt;, skipping &lt;math&gt;3&lt;/math&gt;, and, on the &lt;math&gt;n&lt;/math&gt;th iteration, listing &lt;math&gt;n+3&lt;/math&gt; and skipping &lt;math&gt;n&lt;/math&gt;. The sequence begins &lt;math&gt;1,2,3,4,6,7,8,9,10,13&lt;/math&gt;. What is the &lt;math&gt;500,000&lt;/math&gt;th number in the sequence?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 996,506\qquad\textbf{(B)}\ 996,507\qquad\textbf{(C)}\ 996,508\qquad\textbf{(D)}\ 996,509\qquad\textbf{(E)}\ 996,510 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25==<br /> <br /> The number &lt;math&gt;5^{867}&lt;/math&gt; is between &lt;math&gt;2^{2013}&lt;/math&gt; and &lt;math&gt;2^{2014}&lt;/math&gt;. How many pairs of integers &lt;math&gt;(m,n)&lt;/math&gt; are there such that &lt;math&gt;1\leq m\leq 2012&lt;/math&gt; and &lt;cmath&gt;5^n&lt;2^m&lt;2^{m+2}&lt;5^{n+1}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }278\qquad\textbf{(B) }279\qquad\textbf{(C) }280\qquad\textbf{(D) }281\qquad\textbf{(E) }282\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2014|ab=A|before=[[2013 AMC 10B Problems]]|after=[[2014 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2014 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_2&diff=115022 2019 Mock AMC 10B Problems/Problem 2 2020-01-20T15:43:24Z <p>Vsamc: </p> <hr /> <div>Simply choosing 3 of the boy scouts predetermines the other 3 boy scouts so we have &lt;math&gt;{6 \choose 3}=15&lt;/math&gt;<br /> &lt;cmath&gt;&lt;/cmath&gt;&lt;math&gt;\boxed{\bold{C}}&lt;/math&gt;<br /> &lt;math&gt;\bold{15}&lt;/math&gt;</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_15&diff=114262 1995 AIME Problems/Problem 15 2020-01-04T14:59:25Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_{}&lt;/math&gt; be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encounter a run of &lt;math&gt;5&lt;/math&gt; heads before one encounters a run of &lt;math&gt;2&lt;/math&gt; tails. Given that &lt;math&gt;p_{}&lt;/math&gt; can be written in the form &lt;math&gt;m/n&lt;/math&gt; where &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Think of the problem as a sequence of &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s. No two &lt;tt&gt;T&lt;/tt&gt;'s can occur in a row, so the sequence is blocks of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s separated by &lt;tt&gt;T&lt;/tt&gt;'s and ending in &lt;math&gt;5&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s. Since the first letter could be &lt;tt&gt;T&lt;/tt&gt; or the sequence could start with a block of &lt;tt&gt;H&lt;/tt&gt;'s, the total probability is that &lt;math&gt;3/2&lt;/math&gt; of it has to start with an &lt;tt&gt;H&lt;/tt&gt;. <br /> <br /> The answer to the problem is then the sum of all numbers of the form &lt;math&gt;\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5&lt;/math&gt;, where &lt;math&gt;a,b,c \ldots &lt;/math&gt; are all numbers &lt;math&gt;1-4&lt;/math&gt;, since the blocks of &lt;tt&gt;H&lt;/tt&gt;'s can range from &lt;math&gt;1-4&lt;/math&gt; in length. The sum of all numbers of the form &lt;math&gt;(1/2)^a&lt;/math&gt; is &lt;math&gt;1/2+1/4+1/8+1/16=15/16&lt;/math&gt;, so if there are n blocks of &lt;tt&gt;H&lt;/tt&gt;'s before the final five &lt;tt&gt;H&lt;/tt&gt;'s, the answer can be rewritten as the sum of all numbers of the form &lt;math&gt;\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;\infty&lt;/math&gt;, since that's how many blocks of &lt;tt&gt;H&lt;/tt&gt;'s there can be before the final five. This is an infinite geometric series whose sum is &lt;math&gt;\frac{3/64}{1-(15/32)}=\frac{3}{34}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;p_H, p_T&lt;/math&gt; respectively denote the probabilities that a string beginning with &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s are successful. Thus, <br /> &lt;center&gt;&lt;math&gt;p_T = \frac 12p_H.&lt;/math&gt;&lt;/center&gt;<br /> <br /> A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with &lt;tt&gt;H&lt;/tt&gt; (as there cannot be &lt;math&gt;2&lt;/math&gt; &lt;tt&gt;T&lt;/tt&gt;'s in a row, or be the string HHHHH.<br /> <br /> There is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll &lt;math&gt;4&lt;/math&gt; consecutive &lt;tt&gt;H&lt;/tt&gt;'s, and there is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll a &lt;tt&gt;T&lt;/tt&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;math&gt;p_H = \left(\frac{15}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> The answer is &lt;math&gt;p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> For simplicity, let's compute the complement, namely the probability of getting to &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads.<br /> <br /> Let &lt;math&gt;h_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive heads. Similarly, let &lt;math&gt;t_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive tails. Specifically, &lt;math&gt;h_{5} = 0&lt;/math&gt; and &lt;math&gt;t_{2} = 1&lt;/math&gt;. If we can solve for &lt;math&gt;h_{1}&lt;/math&gt; and &lt;math&gt;t_{1}&lt;/math&gt;, we are done; the answer is simply &lt;math&gt;1/2 * (h_{1} + t_{1})&lt;/math&gt;, since on our first roll, we have equal chances of getting a string with &quot;1 consecutive head&quot; or &quot;1 consecutive tail.&quot;<br /> <br /> Consider solving for &lt;math&gt;t_{1}&lt;/math&gt;. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} h_{1}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Applying similar logic, we get the equations:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> h_{1} &amp;= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\<br /> h_{2} &amp;= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\<br /> h_{3} &amp;= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\<br /> h_{4} &amp;= \frac{1}{2} h_{5} + \frac{1}{2} t_{1}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;h_{5} = 0&lt;/math&gt;, we get &lt;math&gt;h_{4} = \frac{1}{2} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{3} = \frac{3}{4} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{2} = \frac{7}{8} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}&lt;/math&gt;.<br /> <br /> So, the probability of reaching &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads is &lt;math&gt;\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}&lt;/math&gt;; we want the complement, &lt;math&gt;\frac{3}{34}&lt;/math&gt;, yielding an answer of &lt;math&gt;3 + 34 = \boxed{037}&lt;/math&gt;.<br /> <br /> Note: The same approach still works if we tried solving the original problem rather than the complement; we would have simply used &lt;math&gt;h_{5} = 1&lt;/math&gt; and &lt;math&gt;t_{2} = 0&lt;/math&gt; instead. The repeated back-substitution is cleaner because we used the complement.<br /> <br /> ==Solution 4(fast)==<br /> Consider what happens in the &quot;endgame&quot; or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are &lt;math&gt;\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}&lt;/math&gt; respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are &lt;math&gt; \frac{3}{64}&lt;/math&gt;. The sum of all these endgame outcomes are &lt;math&gt;\frac{34}{64}&lt;/math&gt;, hence the desired probability is &lt;math&gt;\frac{3}{34}&lt;/math&gt;, and in this case &lt;math&gt;m=3,n=34&lt;/math&gt; so we have &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;<br /> -vsamc<br /> <br /> == See also ==<br /> {{AIME box|year=1995|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_15&diff=114261 1995 AIME Problems/Problem 15 2020-01-04T14:58:57Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_{}&lt;/math&gt; be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encounter a run of &lt;math&gt;5&lt;/math&gt; heads before one encounters a run of &lt;math&gt;2&lt;/math&gt; tails. Given that &lt;math&gt;p_{}&lt;/math&gt; can be written in the form &lt;math&gt;m/n&lt;/math&gt; where &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Think of the problem as a sequence of &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s. No two &lt;tt&gt;T&lt;/tt&gt;'s can occur in a row, so the sequence is blocks of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s separated by &lt;tt&gt;T&lt;/tt&gt;'s and ending in &lt;math&gt;5&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s. Since the first letter could be &lt;tt&gt;T&lt;/tt&gt; or the sequence could start with a block of &lt;tt&gt;H&lt;/tt&gt;'s, the total probability is that &lt;math&gt;3/2&lt;/math&gt; of it has to start with an &lt;tt&gt;H&lt;/tt&gt;. <br /> <br /> The answer to the problem is then the sum of all numbers of the form &lt;math&gt;\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5&lt;/math&gt;, where &lt;math&gt;a,b,c \ldots &lt;/math&gt; are all numbers &lt;math&gt;1-4&lt;/math&gt;, since the blocks of &lt;tt&gt;H&lt;/tt&gt;'s can range from &lt;math&gt;1-4&lt;/math&gt; in length. The sum of all numbers of the form &lt;math&gt;(1/2)^a&lt;/math&gt; is &lt;math&gt;1/2+1/4+1/8+1/16=15/16&lt;/math&gt;, so if there are n blocks of &lt;tt&gt;H&lt;/tt&gt;'s before the final five &lt;tt&gt;H&lt;/tt&gt;'s, the answer can be rewritten as the sum of all numbers of the form &lt;math&gt;\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;\infty&lt;/math&gt;, since that's how many blocks of &lt;tt&gt;H&lt;/tt&gt;'s there can be before the final five. This is an infinite geometric series whose sum is &lt;math&gt;\frac{3/64}{1-(15/32)}=\frac{3}{34}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;p_H, p_T&lt;/math&gt; respectively denote the probabilities that a string beginning with &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s are successful. Thus, <br /> &lt;center&gt;&lt;math&gt;p_T = \frac 12p_H.&lt;/math&gt;&lt;/center&gt;<br /> <br /> A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with &lt;tt&gt;H&lt;/tt&gt; (as there cannot be &lt;math&gt;2&lt;/math&gt; &lt;tt&gt;T&lt;/tt&gt;'s in a row, or be the string HHHHH.<br /> <br /> There is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll &lt;math&gt;4&lt;/math&gt; consecutive &lt;tt&gt;H&lt;/tt&gt;'s, and there is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll a &lt;tt&gt;T&lt;/tt&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;math&gt;p_H = \left(\frac{15}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> The answer is &lt;math&gt;p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> For simplicity, let's compute the complement, namely the probability of getting to &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads.<br /> <br /> Let &lt;math&gt;h_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive heads. Similarly, let &lt;math&gt;t_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive tails. Specifically, &lt;math&gt;h_{5} = 0&lt;/math&gt; and &lt;math&gt;t_{2} = 1&lt;/math&gt;. If we can solve for &lt;math&gt;h_{1}&lt;/math&gt; and &lt;math&gt;t_{1}&lt;/math&gt;, we are done; the answer is simply &lt;math&gt;1/2 * (h_{1} + t_{1})&lt;/math&gt;, since on our first roll, we have equal chances of getting a string with &quot;1 consecutive head&quot; or &quot;1 consecutive tail.&quot;<br /> <br /> Consider solving for &lt;math&gt;t_{1}&lt;/math&gt;. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} h_{1}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Applying similar logic, we get the equations:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> h_{1} &amp;= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\<br /> h_{2} &amp;= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\<br /> h_{3} &amp;= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\<br /> h_{4} &amp;= \frac{1}{2} h_{5} + \frac{1}{2} t_{1}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;h_{5} = 0&lt;/math&gt;, we get &lt;math&gt;h_{4} = \frac{1}{2} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{3} = \frac{3}{4} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{2} = \frac{7}{8} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}&lt;/math&gt;.<br /> <br /> So, the probability of reaching &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads is &lt;math&gt;\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}&lt;/math&gt;; we want the complement, &lt;math&gt;\frac{3}{34}&lt;/math&gt;, yielding an answer of &lt;math&gt;3 + 34 = \boxed{037}&lt;/math&gt;.<br /> <br /> Note: The same approach still works if we tried solving the original problem rather than the complement; we would have simply used &lt;math&gt;h_{5} = 1&lt;/math&gt; and &lt;math&gt;t_{2} = 0&lt;/math&gt; instead. The repeated back-substitution is cleaner because we used the complement.<br /> <br /> ==Solution 4(fast)==<br /> Consider what happens in the &quot;endgame&quot; or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are &lt;math&gt;\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}&lt;/math&gt; respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are \frac{3}{64}&lt;math&gt;. The sum of all these endgame outcomes are &lt;/math&gt;\frac{34}{64}&lt;math&gt;, hence the desired probability is &lt;/math&gt;\frac{3}{34}&lt;math&gt;, and in this case &lt;/math&gt;m=3,n=34&lt;math&gt; so we have &lt;/math&gt;m+n=\boxed{037}<br /> -vsamc<br /> <br /> == See also ==<br /> {{AIME box|year=1995|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_15&diff=114260 1995 AIME Problems/Problem 15 2020-01-04T14:57:53Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_{}&lt;/math&gt; be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encounter a run of &lt;math&gt;5&lt;/math&gt; heads before one encounters a run of &lt;math&gt;2&lt;/math&gt; tails. Given that &lt;math&gt;p_{}&lt;/math&gt; can be written in the form &lt;math&gt;m/n&lt;/math&gt; where &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Think of the problem as a sequence of &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s. No two &lt;tt&gt;T&lt;/tt&gt;'s can occur in a row, so the sequence is blocks of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s separated by &lt;tt&gt;T&lt;/tt&gt;'s and ending in &lt;math&gt;5&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s. Since the first letter could be &lt;tt&gt;T&lt;/tt&gt; or the sequence could start with a block of &lt;tt&gt;H&lt;/tt&gt;'s, the total probability is that &lt;math&gt;3/2&lt;/math&gt; of it has to start with an &lt;tt&gt;H&lt;/tt&gt;. <br /> <br /> The answer to the problem is then the sum of all numbers of the form &lt;math&gt;\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5&lt;/math&gt;, where &lt;math&gt;a,b,c \ldots &lt;/math&gt; are all numbers &lt;math&gt;1-4&lt;/math&gt;, since the blocks of &lt;tt&gt;H&lt;/tt&gt;'s can range from &lt;math&gt;1-4&lt;/math&gt; in length. The sum of all numbers of the form &lt;math&gt;(1/2)^a&lt;/math&gt; is &lt;math&gt;1/2+1/4+1/8+1/16=15/16&lt;/math&gt;, so if there are n blocks of &lt;tt&gt;H&lt;/tt&gt;'s before the final five &lt;tt&gt;H&lt;/tt&gt;'s, the answer can be rewritten as the sum of all numbers of the form &lt;math&gt;\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;\infty&lt;/math&gt;, since that's how many blocks of &lt;tt&gt;H&lt;/tt&gt;'s there can be before the final five. This is an infinite geometric series whose sum is &lt;math&gt;\frac{3/64}{1-(15/32)}=\frac{3}{34}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;p_H, p_T&lt;/math&gt; respectively denote the probabilities that a string beginning with &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s are successful. Thus, <br /> &lt;center&gt;&lt;math&gt;p_T = \frac 12p_H.&lt;/math&gt;&lt;/center&gt;<br /> <br /> A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with &lt;tt&gt;H&lt;/tt&gt; (as there cannot be &lt;math&gt;2&lt;/math&gt; &lt;tt&gt;T&lt;/tt&gt;'s in a row, or be the string HHHHH.<br /> <br /> There is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll &lt;math&gt;4&lt;/math&gt; consecutive &lt;tt&gt;H&lt;/tt&gt;'s, and there is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll a &lt;tt&gt;T&lt;/tt&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;math&gt;p_H = \left(\frac{15}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> The answer is &lt;math&gt;p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> For simplicity, let's compute the complement, namely the probability of getting to &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads.<br /> <br /> Let &lt;math&gt;h_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive heads. Similarly, let &lt;math&gt;t_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive tails. Specifically, &lt;math&gt;h_{5} = 0&lt;/math&gt; and &lt;math&gt;t_{2} = 1&lt;/math&gt;. If we can solve for &lt;math&gt;h_{1}&lt;/math&gt; and &lt;math&gt;t_{1}&lt;/math&gt;, we are done; the answer is simply &lt;math&gt;1/2 * (h_{1} + t_{1})&lt;/math&gt;, since on our first roll, we have equal chances of getting a string with &quot;1 consecutive head&quot; or &quot;1 consecutive tail.&quot;<br /> <br /> Consider solving for &lt;math&gt;t_{1}&lt;/math&gt;. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} h_{1}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Applying similar logic, we get the equations:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> h_{1} &amp;= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\<br /> h_{2} &amp;= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\<br /> h_{3} &amp;= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\<br /> h_{4} &amp;= \frac{1}{2} h_{5} + \frac{1}{2} t_{1}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;h_{5} = 0&lt;/math&gt;, we get &lt;math&gt;h_{4} = \frac{1}{2} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{3} = \frac{3}{4} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{2} = \frac{7}{8} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}&lt;/math&gt;.<br /> <br /> So, the probability of reaching &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads is &lt;math&gt;\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}&lt;/math&gt;; we want the complement, &lt;math&gt;\frac{3}{34}&lt;/math&gt;, yielding an answer of &lt;math&gt;3 + 34 = \boxed{037}&lt;/math&gt;.<br /> <br /> Note: The same approach still works if we tried solving the original problem rather than the complement; we would have simply used &lt;math&gt;h_{5} = 1&lt;/math&gt; and &lt;math&gt;t_{2} = 0&lt;/math&gt; instead. The repeated back-substitution is cleaner because we used the complement.<br /> <br /> ==Solution 4(fast)==<br /> Consider what happens in the &quot;endgame&quot; or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are &lt;math&gt;\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64} respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are \frac{3}{64}&lt;/math&gt;. The sum of all these endgame outcomes are &lt;math&gt;\frac{34}{64}&lt;/math&gt;, hence the desired probability is &lt;math&gt;\frac{3}{34}&lt;/math&gt;, and in this case &lt;math&gt;m=3,n=34&lt;/math&gt; so we have &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;<br /> -vsamc<br /> <br /> == See also ==<br /> {{AIME box|year=1995|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_15&diff=114259 1995 AIME Problems/Problem 15 2020-01-04T14:57:23Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_{}&lt;/math&gt; be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encounter a run of &lt;math&gt;5&lt;/math&gt; heads before one encounters a run of &lt;math&gt;2&lt;/math&gt; tails. Given that &lt;math&gt;p_{}&lt;/math&gt; can be written in the form &lt;math&gt;m/n&lt;/math&gt; where &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Think of the problem as a sequence of &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s. No two &lt;tt&gt;T&lt;/tt&gt;'s can occur in a row, so the sequence is blocks of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s separated by &lt;tt&gt;T&lt;/tt&gt;'s and ending in &lt;math&gt;5&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s. Since the first letter could be &lt;tt&gt;T&lt;/tt&gt; or the sequence could start with a block of &lt;tt&gt;H&lt;/tt&gt;'s, the total probability is that &lt;math&gt;3/2&lt;/math&gt; of it has to start with an &lt;tt&gt;H&lt;/tt&gt;. <br /> <br /> The answer to the problem is then the sum of all numbers of the form &lt;math&gt;\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5&lt;/math&gt;, where &lt;math&gt;a,b,c \ldots &lt;/math&gt; are all numbers &lt;math&gt;1-4&lt;/math&gt;, since the blocks of &lt;tt&gt;H&lt;/tt&gt;'s can range from &lt;math&gt;1-4&lt;/math&gt; in length. The sum of all numbers of the form &lt;math&gt;(1/2)^a&lt;/math&gt; is &lt;math&gt;1/2+1/4+1/8+1/16=15/16&lt;/math&gt;, so if there are n blocks of &lt;tt&gt;H&lt;/tt&gt;'s before the final five &lt;tt&gt;H&lt;/tt&gt;'s, the answer can be rewritten as the sum of all numbers of the form &lt;math&gt;\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;\infty&lt;/math&gt;, since that's how many blocks of &lt;tt&gt;H&lt;/tt&gt;'s there can be before the final five. This is an infinite geometric series whose sum is &lt;math&gt;\frac{3/64}{1-(15/32)}=\frac{3}{34}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;p_H, p_T&lt;/math&gt; respectively denote the probabilities that a string beginning with &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s are successful. Thus, <br /> &lt;center&gt;&lt;math&gt;p_T = \frac 12p_H.&lt;/math&gt;&lt;/center&gt;<br /> <br /> A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with &lt;tt&gt;H&lt;/tt&gt; (as there cannot be &lt;math&gt;2&lt;/math&gt; &lt;tt&gt;T&lt;/tt&gt;'s in a row, or be the string HHHHH.<br /> <br /> There is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll &lt;math&gt;4&lt;/math&gt; consecutive &lt;tt&gt;H&lt;/tt&gt;'s, and there is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll a &lt;tt&gt;T&lt;/tt&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;math&gt;p_H = \left(\frac{15}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> The answer is &lt;math&gt;p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> For simplicity, let's compute the complement, namely the probability of getting to &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads.<br /> <br /> Let &lt;math&gt;h_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive heads. Similarly, let &lt;math&gt;t_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive tails. Specifically, &lt;math&gt;h_{5} = 0&lt;/math&gt; and &lt;math&gt;t_{2} = 1&lt;/math&gt;. If we can solve for &lt;math&gt;h_{1}&lt;/math&gt; and &lt;math&gt;t_{1}&lt;/math&gt;, we are done; the answer is simply &lt;math&gt;1/2 * (h_{1} + t_{1})&lt;/math&gt;, since on our first roll, we have equal chances of getting a string with &quot;1 consecutive head&quot; or &quot;1 consecutive tail.&quot;<br /> <br /> Consider solving for &lt;math&gt;t_{1}&lt;/math&gt;. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} h_{1}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Applying similar logic, we get the equations:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> h_{1} &amp;= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\<br /> h_{2} &amp;= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\<br /> h_{3} &amp;= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\<br /> h_{4} &amp;= \frac{1}{2} h_{5} + \frac{1}{2} t_{1}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;h_{5} = 0&lt;/math&gt;, we get &lt;math&gt;h_{4} = \frac{1}{2} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{3} = \frac{3}{4} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{2} = \frac{7}{8} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}&lt;/math&gt;.<br /> <br /> So, the probability of reaching &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads is &lt;math&gt;\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}&lt;/math&gt;; we want the complement, &lt;math&gt;\frac{3}{34}&lt;/math&gt;, yielding an answer of &lt;math&gt;3 + 34 = \boxed{037}&lt;/math&gt;.<br /> <br /> Note: The same approach still works if we tried solving the original problem rather than the complement; we would have simply used &lt;math&gt;h_{5} = 1&lt;/math&gt; and &lt;math&gt;t_{2} = 0&lt;/math&gt; instead. The repeated back-substitution is cleaner because we used the complement.<br /> <br /> ==Solution 4(fast)==<br /> Consider what happens in the &quot;endgame&quot; or what leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are &lt;math&gt;\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64} respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are \frac{3}{64}&lt;/math&gt;. The sum of all these endgame outcomes are &lt;math&gt;\frac{34}{64}&lt;/math&gt;, hence the desired probability is &lt;math&gt;\frac{3}{34}&lt;/math&gt;, and in this case &lt;math&gt;m=3,n=34&lt;/math&gt; so we have &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;<br /> -vsamc<br /> <br /> == See also ==<br /> {{AIME box|year=1995|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_15&diff=114258 1995 AIME Problems/Problem 15 2020-01-04T14:56:22Z <p>Vsamc: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_{}&lt;/math&gt; be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encounter a run of &lt;math&gt;5&lt;/math&gt; heads before one encounters a run of &lt;math&gt;2&lt;/math&gt; tails. Given that &lt;math&gt;p_{}&lt;/math&gt; can be written in the form &lt;math&gt;m/n&lt;/math&gt; where &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Think of the problem as a sequence of &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s. No two &lt;tt&gt;T&lt;/tt&gt;'s can occur in a row, so the sequence is blocks of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s separated by &lt;tt&gt;T&lt;/tt&gt;'s and ending in &lt;math&gt;5&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s. Since the first letter could be &lt;tt&gt;T&lt;/tt&gt; or the sequence could start with a block of &lt;tt&gt;H&lt;/tt&gt;'s, the total probability is that &lt;math&gt;3/2&lt;/math&gt; of it has to start with an &lt;tt&gt;H&lt;/tt&gt;. <br /> <br /> The answer to the problem is then the sum of all numbers of the form &lt;math&gt;\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5&lt;/math&gt;, where &lt;math&gt;a,b,c \ldots &lt;/math&gt; are all numbers &lt;math&gt;1-4&lt;/math&gt;, since the blocks of &lt;tt&gt;H&lt;/tt&gt;'s can range from &lt;math&gt;1-4&lt;/math&gt; in length. The sum of all numbers of the form &lt;math&gt;(1/2)^a&lt;/math&gt; is &lt;math&gt;1/2+1/4+1/8+1/16=15/16&lt;/math&gt;, so if there are n blocks of &lt;tt&gt;H&lt;/tt&gt;'s before the final five &lt;tt&gt;H&lt;/tt&gt;'s, the answer can be rewritten as the sum of all numbers of the form &lt;math&gt;\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;\infty&lt;/math&gt;, since that's how many blocks of &lt;tt&gt;H&lt;/tt&gt;'s there can be before the final five. This is an infinite geometric series whose sum is &lt;math&gt;\frac{3/64}{1-(15/32)}=\frac{3}{34}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;p_H, p_T&lt;/math&gt; respectively denote the probabilities that a string beginning with &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s are successful. Thus, <br /> &lt;center&gt;&lt;math&gt;p_T = \frac 12p_H.&lt;/math&gt;&lt;/center&gt;<br /> <br /> A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with &lt;tt&gt;H&lt;/tt&gt; (as there cannot be &lt;math&gt;2&lt;/math&gt; &lt;tt&gt;T&lt;/tt&gt;'s in a row, or be the string HHHHH.<br /> <br /> There is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll &lt;math&gt;4&lt;/math&gt; consecutive &lt;tt&gt;H&lt;/tt&gt;'s, and there is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll a &lt;tt&gt;T&lt;/tt&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;math&gt;p_H = \left(\frac{15}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> The answer is &lt;math&gt;p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> For simplicity, let's compute the complement, namely the probability of getting to &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads.<br /> <br /> Let &lt;math&gt;h_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive heads. Similarly, let &lt;math&gt;t_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive tails. Specifically, &lt;math&gt;h_{5} = 0&lt;/math&gt; and &lt;math&gt;t_{2} = 1&lt;/math&gt;. If we can solve for &lt;math&gt;h_{1}&lt;/math&gt; and &lt;math&gt;t_{1}&lt;/math&gt;, we are done; the answer is simply &lt;math&gt;1/2 * (h_{1} + t_{1})&lt;/math&gt;, since on our first roll, we have equal chances of getting a string with &quot;1 consecutive head&quot; or &quot;1 consecutive tail.&quot;<br /> <br /> Consider solving for &lt;math&gt;t_{1}&lt;/math&gt;. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} h_{1}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Applying similar logic, we get the equations:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> h_{1} &amp;= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\<br /> h_{2} &amp;= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\<br /> h_{3} &amp;= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\<br /> h_{4} &amp;= \frac{1}{2} h_{5} + \frac{1}{2} t_{1}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;h_{5} = 0&lt;/math&gt;, we get &lt;math&gt;h_{4} = \frac{1}{2} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{3} = \frac{3}{4} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{2} = \frac{7}{8} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}&lt;/math&gt;.<br /> <br /> So, the probability of reaching &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads is &lt;math&gt;\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}&lt;/math&gt;; we want the complement, &lt;math&gt;\frac{3}{34}&lt;/math&gt;, yielding an answer of &lt;math&gt;3 + 34 = \boxed{037}&lt;/math&gt;.<br /> <br /> Note: The same approach still works if we tried solving the original problem rather than the complement; we would have simply used &lt;math&gt;h_{5} = 1&lt;/math&gt; and &lt;math&gt;t_{2} = 0&lt;/math&gt; instead. The repeated back-substitution is cleaner because we used the complement.<br /> <br /> ==Solution 4(fast)==<br /> Consider what happens in the &quot;endgame&quot; or what leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are \(\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64} respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are $$\frac{3}{64}$$. The sum of all these endgame outcomes are $$\frac{34}{64}$$, hence the desired probability is $$\frac{3}{34}$$, and in this case $$m=3,n=34$$ so we have $$m+n=\boxed{037}$$<br /> -vsamc<br /> <br /> == See also ==<br /> {{AIME box|year=1995|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_11&diff=112238 2013 AMC 12B Problems/Problem 11 2019-11-27T19:47:59Z <p>Vsamc: </p> <hr /> <div>==Problem==<br /> Two bees start at the same spot and fly at the same rate in the following directions. Bee &lt;math&gt;A&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot north, then &lt;math&gt;1&lt;/math&gt; foot east, then &lt;math&gt;1&lt;/math&gt; foot upwards, and then continues to repeat this pattern. Bee &lt;math&gt;B&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot south, then &lt;math&gt;1&lt;/math&gt; foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly &lt;math&gt;10&lt;/math&gt; feet away from each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ A&lt;/math&gt; east, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt;\qquad \textbf{(B)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt; \qquad \textbf{(C)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt; \qquad \textbf{(D)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt; \qquad \textbf{(E)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;<br /> <br /> ==Solution==<br /> <br /> Let A and B begin at (0,0,0). In 6 steps, A will have done his route twice, ending up at (2,2,2), and B will have done his route three times, ending at (-3,-3,0). Their distance is &lt;math&gt;\sqrt{(2+3)^2+(2+3)^2+2^2}=\sqrt{54} &lt; 10&lt;/math&gt; We now move forward one step at a time until they are ten feet away:<br /> 7 steps: A moves north to (2,3,2), B moves south to (-3,-4,0), distance of &lt;math&gt;\sqrt{(2+3)^2+(3+4)^2+2^2}=\sqrt{78} &lt; 10&lt;/math&gt;<br /> 8 steps: A moves east to (3,3,2), B moves west to (-4,-4,0), distance of &lt;math&gt;\sqrt{(3+4)^2+(3+4)^2+2^2}=\sqrt{102}&gt;10&lt;/math&gt;<br /> <br /> Thus they reach 10 feet away when A is moving east and B is moving west, between moves 7 and 8. Thus the answer is &lt;math&gt;\textbf{(A)}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_11&diff=112237 2013 AMC 12B Problems/Problem 11 2019-11-27T19:38:58Z <p>Vsamc: </p> <hr /> <div>==Problem==<br /> Two bees start at the same spot and fly at the same rate in the following directions. Bee &lt;math&gt;A&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot north, then &lt;math&gt;1&lt;/math&gt; foot east, then &lt;math&gt;1&lt;/math&gt; foot upwards, and then continues to repeat this pattern. Bee &lt;math&gt;B&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot south, then &lt;math&gt;1&lt;/math&gt; foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly &lt;math&gt;10&lt;/math&gt; feet away from each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ A&lt;/math&gt; east, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt;\qquad \textbf{(B)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt; \qquad \textbf{(C)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt; \qquad \textbf{(D)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt; \qquad \textbf{(E)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;<br /> <br /> ==Solution==<br /> <br /> Let A and B begin at (0,0,0). In 6 steps, A will have done his route twice, ending up at (2,2,2), and B will have done his route three times, ending at (-3,-3,0). Their distance is &lt;math&gt;\sqrt{(2+3)^2+(2+3)^2+2^2}=\sqrt{54} &lt; 10&lt;/math&gt; We now move forward one step at a time until they are ten feet away:<br /> 7 steps: A moves north to (2,3,2), B moves south to (-3,-4,0), distance of &lt;math&gt;\sqrt{(2+3)^2+(3+4)^2+2^2}=\sqrt{78} &lt; 10&lt;/math&gt;<br /> 8 steps: A moves east to (3,3,2), B moves west to (-4,-4,0), distance of &lt;math&gt;\sqrt{(3+4)^2+(3+4)^2+2^2}=\sqrt{102}&gt;10&lt;/math&gt;<br /> <br /> Thus they reach 10 feet away when A is moving east and B is moving west, between moves 7 and 8. Thus the answer is &lt;math&gt;\textbf{(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Denote the origin of the two bees as A. Let the final positions of the bees be B and C respectively. Since they are moving an equal distance away from the origin with each of their respective steps, two right angled triangles can be constructed from their final positions. Since these triangles are congruent, they each have a hypotenuse of five units so they are both (3,4,5) triangles. 'Resolving' these triangles into the moves of each bee shows that they must have taken 7 steps to reach their final position at which point Bee B must move east next and Bee C is moving west. Hence &lt;math&gt;\textbf{(A)}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Vsamc https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_11&diff=112236 2013 AMC 12B Problems/Problem 11 2019-11-27T19:38:16Z <p>Vsamc: Undo revision 112235 by Vsamc (talk)</p> <hr /> <div>==Problem==<br /> Two bees start at the same spot and fly at the same rate in the following directions. Bee &lt;math&gt;A&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot north, then &lt;math&gt;1&lt;/math&gt; foot east, then &lt;math&gt;1&lt;/math&gt; foot upwards, and then continues to repeat this pattern. Bee &lt;math&gt;B&lt;/math&gt; travels &lt;math&gt;1&lt;/math&gt; foot south, then &lt;math&gt;1&lt;/math&gt; foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly &lt;math&gt;10&lt;/math&gt; feet away from each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ A&lt;/math&gt; east, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt;\qquad \textbf{(B)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt; \qquad \textbf{(C)}\ A&lt;/math&gt; north, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;&lt;math&gt; \qquad \textbf{(D)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; south&lt;br \&gt;&lt;math&gt; \qquad \textbf{(E)}\ A&lt;/math&gt; up, &lt;math&gt;B&lt;/math&gt; west&lt;br \&gt;<br /> <br /> ==Solution==<br /> <br /> Let A and B begin at (0,0,0). In 6 steps, A will have done his route twice, ending up at (2,2,2), and B will have done his route three times, ending at (-3,-3,0). Their distance is &lt;math&gt;\sqrt{(2+3)^2+(2+3)^2+2^2}=\sqrt{54} &lt; 10&lt;/math&gt; We now move forward one step at a time until they are ten feet away:<br /> 7 steps: A moves north to (2,3,2), B moves south to (-3,-4,0), distance of &lt;math&gt;\sqrt{(2+3)^2+(3+4)^2+2^2}=\sqrt{78} &lt; 10&lt;/math&gt;<br /> 8 steps: A moves east to (3,3,2), B moves west to (-4,-4,0), distance of &lt;math&gt;\sqrt{(3+4)^2+(3+4)^2+2^2}=\sqrt{102}&gt;10&lt;/math&gt;<br /> <br /> Thus they reach 10 feet away when A is moving east and B is moving west, between moves 7 and 8. Thus the answer is &lt;math&gt;\textbf{(A)}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Vsamc