https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=PiDragon&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-13T05:56:27Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_21&diff=76039 2016 AMC 10A Problems/Problem 21 2016-02-16T01:44:00Z <p>PiDragon: /* Solution 2 */</p> <hr /> <div>Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> ==Solution== <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(linewidth(0.4));<br /> //Variable Declarations<br /> pair P,Q,R,Pp,Qp,Rp;<br /> pair A,B;<br /> <br /> //Variable Definitions<br /> A=(-5, 0);<br /> B=(8, 0);<br /> P=(-2.828,1);<br /> Q=(0,2);<br /> R=(4.899,3);<br /> Pp=foot(P,A,B);<br /> Qp=foot(Q,A,B);<br /> Rp=foot(R,A,B);<br /> path PQR = P--Q--R--cycle;<br /> //Initial Diagram<br /> dot(P);<br /> dot(Q);<br /> dot(R);<br /> dot(Pp);<br /> dot(Qp);<br /> dot(Rp);<br /> draw(Circle(P, 1), linewidth(0.8));<br /> draw(Circle(Q, 2), linewidth(0.8));<br /> draw(Circle(R, 3), linewidth(0.8));<br /> draw(A--B,Arrows);<br /> label(&quot;$P$&quot;,P,N);<br /> label(&quot;$Q$&quot;,Q,N);<br /> label(&quot;$R$&quot;,R,N);<br /> label(&quot;$P'$&quot;,Pp,S);<br /> label(&quot;$Q'$&quot;,Qp,S);<br /> label(&quot;$R'$&quot;,Rp,S);<br /> label(&quot;$l$&quot;,B,E);<br /> <br /> //Added lines<br /> draw(PQR);<br /> draw(P--Pp);<br /> draw(Q--Qp);<br /> draw(R--Rp);<br /> <br /> //Angle marks<br /> draw(rightanglemark(P,Pp,B));<br /> draw(rightanglemark(Q,Qp,B));<br /> draw(rightanglemark(R,Rp,B));<br /> &lt;/asy&gt;<br /> Notice that we can find &lt;math&gt;[P'PQRR']&lt;/math&gt; in two different ways: &lt;math&gt;[P'PQQ']+[Q'QRR']&lt;/math&gt; and &lt;math&gt;[PQR]+[P'PRR']&lt;/math&gt;, so &lt;math&gt;[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']&lt;/math&gt; <br /> &lt;math&gt;\break&lt;/math&gt;<br /> <br /> &lt;math&gt;P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}&lt;/math&gt;. Additionally, &lt;math&gt;Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}&lt;/math&gt;. Therefore, &lt;math&gt;[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}&lt;/math&gt;. Similarly, &lt;math&gt;[Q'QRR']=5\sqrt6&lt;/math&gt;. We can calculate &lt;math&gt;[P'PRR']&lt;/math&gt; easily because &lt;math&gt;P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}&lt;/math&gt;. &lt;math&gt;[P'PRR']=4\sqrt{2}+4\sqrt{6}&lt;/math&gt;. &lt;math&gt;\newline&lt;/math&gt;<br /> <br /> Plugging into first equation, the two sums of areas, &lt;math&gt;3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]&lt;/math&gt;. &lt;math&gt;\newline&lt;/math&gt;<br /> <br /> &lt;math&gt;[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}&lt;/math&gt;.<br /> <br /> ==Solution 2== <br /> <br /> Use the Shoelace Thm!<br /> <br /> Let the center of the first circle of radius 1 be at (0, 1). <br /> <br /> Draw the trapezoid &lt;math&gt;PQQ'P'&lt;/math&gt; and using Pythagorean Thm., we get that &lt;math&gt;P'Q' = 2\sqrt{2}&lt;/math&gt; so the center of the second circle of radius 2 is at &lt;math&gt;(2\sqrt{2}, 2)&lt;/math&gt;.<br /> <br /> Draw the trapezoid &lt;math&gt;QRR'Q'&lt;/math&gt; and using Pythagorean Thm., we get that &lt;math&gt;Q'R' = 2\sqrt{2} + 2\sqrt{6}&lt;/math&gt; so the center of the third circle of radius 3 is at &lt;math&gt;(2\sqrt{2}+2\sqrt{6}, 3)&lt;/math&gt;.<br /> <br /> Now, we may use the Shoelace Thm!<br /> <br /> &lt;math&gt;(0,1)&lt;/math&gt;<br /> <br /> &lt;math&gt;(2\sqrt{2}, 2)&lt;/math&gt;<br /> <br /> &lt;math&gt;(2\sqrt{2}+2\sqrt{6}, 3)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|&lt;/math&gt;<br /> <br /> &lt;math&gt;= \sqrt{6}-\sqrt{2}&lt;/math&gt; &lt;math&gt;\fbox{D}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2016|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> PiDragon https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_21&diff=76038 2016 AMC 10A Problems/Problem 21 2016-02-16T01:43:34Z <p>PiDragon: </p> <hr /> <div>Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> ==Solution== <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(linewidth(0.4));<br /> //Variable Declarations<br /> pair P,Q,R,Pp,Qp,Rp;<br /> pair A,B;<br /> <br /> //Variable Definitions<br /> A=(-5, 0);<br /> B=(8, 0);<br /> P=(-2.828,1);<br /> Q=(0,2);<br /> R=(4.899,3);<br /> Pp=foot(P,A,B);<br /> Qp=foot(Q,A,B);<br /> Rp=foot(R,A,B);<br /> path PQR = P--Q--R--cycle;<br /> //Initial Diagram<br /> dot(P);<br /> dot(Q);<br /> dot(R);<br /> dot(Pp);<br /> dot(Qp);<br /> dot(Rp);<br /> draw(Circle(P, 1), linewidth(0.8));<br /> draw(Circle(Q, 2), linewidth(0.8));<br /> draw(Circle(R, 3), linewidth(0.8));<br /> draw(A--B,Arrows);<br /> label(&quot;$P$&quot;,P,N);<br /> label(&quot;$Q$&quot;,Q,N);<br /> label(&quot;$R$&quot;,R,N);<br /> label(&quot;$P'$&quot;,Pp,S);<br /> label(&quot;$Q'$&quot;,Qp,S);<br /> label(&quot;$R'$&quot;,Rp,S);<br /> label(&quot;$l$&quot;,B,E);<br /> <br /> //Added lines<br /> draw(PQR);<br /> draw(P--Pp);<br /> draw(Q--Qp);<br /> draw(R--Rp);<br /> <br /> //Angle marks<br /> draw(rightanglemark(P,Pp,B));<br /> draw(rightanglemark(Q,Qp,B));<br /> draw(rightanglemark(R,Rp,B));<br /> &lt;/asy&gt;<br /> Notice that we can find &lt;math&gt;[P'PQRR']&lt;/math&gt; in two different ways: &lt;math&gt;[P'PQQ']+[Q'QRR']&lt;/math&gt; and &lt;math&gt;[PQR]+[P'PRR']&lt;/math&gt;, so &lt;math&gt;[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']&lt;/math&gt; <br /> &lt;math&gt;\break&lt;/math&gt;<br /> <br /> &lt;math&gt;P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}&lt;/math&gt;. Additionally, &lt;math&gt;Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}&lt;/math&gt;. Therefore, &lt;math&gt;[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}&lt;/math&gt;. Similarly, &lt;math&gt;[Q'QRR']=5\sqrt6&lt;/math&gt;. We can calculate &lt;math&gt;[P'PRR']&lt;/math&gt; easily because &lt;math&gt;P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}&lt;/math&gt;. &lt;math&gt;[P'PRR']=4\sqrt{2}+4\sqrt{6}&lt;/math&gt;. &lt;math&gt;\newline&lt;/math&gt;<br /> <br /> Plugging into first equation, the two sums of areas, &lt;math&gt;3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]&lt;/math&gt;. &lt;math&gt;\newline&lt;/math&gt;<br /> <br /> &lt;math&gt;[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}&lt;/math&gt;.<br /> <br /> ==Solution 2== <br /> <br /> Use the Shoelace Thm!<br /> <br /> Let the center of the first circle of radius 1 be at (0, 1). <br /> <br /> Draw the trapezoid &lt;math&gt;PQQ'P'&lt;/math&gt; and using Pythagorean Thm., we get that &lt;math&gt;P'Q' = 2\sqrt{2}&lt;/math&gt; so the center of the second circle of radius 2 is at &lt;math&gt;(2\sqrt{2}, 2)&lt;/math&gt;.<br /> <br /> Draw the trapezoid &lt;math&gt;QRR'Q'&lt;/math&gt; and using Pythagorean Thm., we get that &lt;math&gt;Q'R' = 2\sqrt{2} + 2\sqrt{6}&lt;/math&gt; so the center of the third circle of radius 3 is at &lt;math&gt;(2\sqrt{2}+2\sqrt{6}, 3)&lt;/math&gt;.<br /> <br /> Now, we may use the Shoelace Thm!<br /> <br /> &lt;math&gt;(0,1)&lt;/math&gt;<br /> <br /> &lt;math&gt;(2\sqrt{2}, 2)&lt;/math&gt;<br /> <br /> &lt;math&gt;(2\sqrt{2}+2\sqrt{6}, 3)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|&lt;/math&gt;<br /> <br /> &lt;math&gt;= \sqrt{6}-\sqrt{2} \fbox{D}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2016|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> PiDragon https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_24&diff=74982 2015 AMC 10B Problems/Problem 24 2016-01-31T22:01:28Z <p>PiDragon: </p> <hr /> <div>==Problem==<br /> Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin &lt;math&gt;p_0=(0,0)&lt;/math&gt; facing to the east and walks one unit, arriving at &lt;math&gt;p_1=(1,0)&lt;/math&gt;. For &lt;math&gt;n=1,2,3,\dots&lt;/math&gt;, right after arriving at the point &lt;math&gt;p_n&lt;/math&gt;, if Aaron can turn &lt;math&gt;90^\circ&lt;/math&gt; left and walk one unit to an unvisited point &lt;math&gt;p_{n+1}&lt;/math&gt;, he does that. Otherwise, he walks one unit straight ahead to reach &lt;math&gt;p_{n+1}&lt;/math&gt;. Thus the sequence of points continues &lt;math&gt;p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)&lt;/math&gt;, and so on in a counterclockwise spiral pattern. What is &lt;math&gt;p_{2015}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) &lt;/math&gt;<br /> <br /> ==Solution==<br /> (Used from 2015 AMC 10/12 B Math Jam)<br /> <br /> The first thing we would do is track Aaron's footsteps:<br /> <br /> He starts by taking &lt;math&gt;1&lt;/math&gt; step East and &lt;math&gt;1&lt;/math&gt; step North, ending at &lt;math&gt;(1,1)&lt;/math&gt; after &lt;math&gt;2&lt;/math&gt; steps and about to head West.<br /> <br /> Then he takes &lt;math&gt;2&lt;/math&gt; steps West and &lt;math&gt;2&lt;/math&gt; steps South, ending at &lt;math&gt;(-1,-1&lt;/math&gt;) after &lt;math&gt;2+4&lt;/math&gt; steps, and about to head East.<br /> <br /> Then he takes &lt;math&gt;3&lt;/math&gt; steps East and &lt;math&gt;3&lt;/math&gt; steps North, ending at &lt;math&gt;(2,2)&lt;/math&gt; after &lt;math&gt;2+4+6&lt;/math&gt; steps, and about to head West.<br /> <br /> Then he takes &lt;math&gt;4&lt;/math&gt; steps West and &lt;math&gt;4&lt;/math&gt; steps South, ending at &lt;math&gt;(-2,-2)&lt;/math&gt; after &lt;math&gt;2+4+6+8&lt;/math&gt; steps, and about to head East.<br /> <br /> From this pattern, we can notice that for any integer &lt;math&gt;k \ge 1&lt;/math&gt; he's at &lt;math&gt;(-k, -k)&lt;/math&gt; after &lt;math&gt;2 + 4 + 6 + ... + 4k&lt;/math&gt; steps, and about to head East. There are &lt;math&gt;2k&lt;/math&gt; terms in the sum, with an average value of &lt;math&gt;(2 + 4k)/2 = 2k + 1&lt;/math&gt;, so:<br /> <br /> &lt;cmath&gt;2 + 4 + 6 + ... + 4k = 2k(2k + 1)&lt;/cmath&gt;<br /> <br /> If we substitute &lt;math&gt;k = 22&lt;/math&gt; into the equation: &lt;math&gt;44(45) = 1980 &lt; 2015&lt;/math&gt;. So he has &lt;math&gt;35&lt;/math&gt; moves to go. This makes him end up at &lt;math&gt;(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}&lt;/math&gt;<br /> <br /> ==Alternate Solution==<br /> We are given that Aaron starts at &lt;math&gt;(0, 0)&lt;/math&gt;, and we note that his net steps follow the pattern of &lt;math&gt;+1&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction, &lt;math&gt;+1&lt;/math&gt; in the &lt;math&gt;y&lt;/math&gt;-direction, &lt;math&gt;-2&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction, &lt;math&gt;-2&lt;/math&gt; in the &lt;math&gt;y&lt;/math&gt;-direction, &lt;math&gt;+3&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction, &lt;math&gt;+3&lt;/math&gt; in the &lt;math&gt;y&lt;/math&gt;-direction, and so on, where we add odd and subtract even.<br /> <br /> We want &lt;math&gt;2 + 4 + 6 + 8 + ... + 2n = 2015&lt;/math&gt;, but it does not work out cleanly. Instead, we get that &lt;math&gt;2 + 4 + 6 + ... + 2(44) = 1980&lt;/math&gt;, which means that there are &lt;math&gt;35&lt;/math&gt; extra steps past adding &lt;math&gt;-44&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction (and the final number we add in the &lt;math&gt;y&lt;/math&gt;-direction is &lt;math&gt;-44&lt;/math&gt;).<br /> <br /> So &lt;math&gt;p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)&lt;/math&gt;.<br /> <br /> We can group &lt;math&gt;1-2+3-4+5...-44&lt;/math&gt; as &lt;math&gt;(1-2)+(3-4)+(5-6)+...+(43-44) = -22&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;p_{2015} = (13, -22)&lt;/math&gt; &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> PiDragon https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_20&diff=74962 2015 AMC 12A Problems/Problem 20 2016-01-31T15:53:02Z <p>PiDragon: </p> <hr /> <div>==Problem==<br /> <br /> Isosceles triangles &lt;math&gt;T&lt;/math&gt; and &lt;math&gt;T'&lt;/math&gt; are not congruent but have the same area and the same perimeter. The sides of &lt;math&gt;T&lt;/math&gt; have lengths &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;, while those of &lt;math&gt;T'&lt;/math&gt; have lengths &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt;. Which of the following numbers is closest to &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1 ===<br /> The area of &lt;math&gt;T&lt;/math&gt; is &lt;math&gt;\dfrac{1}{2} \cdot 8 \cdot 3 = 12&lt;/math&gt; and the perimeter is 18.<br /> <br /> The area of &lt;math&gt;T'&lt;/math&gt; is &lt;math&gt;\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}&lt;/math&gt; and the perimeter is &lt;math&gt;2a + b&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;2a + b = 18&lt;/math&gt;, so &lt;math&gt;2a = 18 - b&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}&lt;/math&gt;, so &lt;math&gt;48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}&lt;/math&gt;.<br /> <br /> We square and divide 36 from both sides to obtain &lt;math&gt;64 = b^2 (9 - b)&lt;/math&gt;, so &lt;math&gt;b^3 - 9b^2 + 64 = 0&lt;/math&gt;. This factors as &lt;math&gt;(b - 8)(b^2 - b - 8) = 0&lt;/math&gt;. Because clearly &lt;math&gt;b \neq 8&lt;/math&gt; but &lt;math&gt;b &gt; 0&lt;/math&gt;, we have &lt;math&gt;b = \dfrac{1 + \sqrt{33}}{2} &lt; \dfrac{1 + 6}{2} = 3.5.&lt;/math&gt; The answer is &lt;math&gt;\textbf{(A)}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Triangle &lt;math&gt;T&lt;/math&gt;, being isosceles, has an area of &lt;math&gt;\frac{1}{2}(8)\sqrt{5^2-4^2}=12&lt;/math&gt; and a perimeter of &lt;math&gt;5+5+8=18&lt;/math&gt;.<br /> Triangle &lt;math&gt;T'&lt;/math&gt; similarly has an area of &lt;math&gt;\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12&lt;/math&gt; and &lt;math&gt;2a+b=18&lt;/math&gt;.<br /> <br /> Now we apply our computational fortitude.<br /> <br /> &lt;cmath&gt;\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12&lt;/cmath&gt;<br /> &lt;cmath&gt;(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=24&lt;/cmath&gt;<br /> &lt;cmath&gt;(b)\sqrt{4a^2-b^2}=48&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2(4a^2-b^2)=48^2&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2(2a+b)(2a-b)=48^2&lt;/cmath&gt;<br /> Plug in &lt;math&gt;2a+b=18&lt;/math&gt; to obtain<br /> &lt;cmath&gt;18b^2(2a-b)=48^2&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2(2a-b)=128&lt;/cmath&gt;<br /> Plug in &lt;math&gt;2a=18-b&lt;/math&gt; to obtain<br /> &lt;cmath&gt;b^2(18-2b)=128&lt;/cmath&gt;<br /> &lt;cmath&gt;2b^3-18b^2+128=0&lt;/cmath&gt;<br /> &lt;cmath&gt;b^3-9b^2+64=0&lt;/cmath&gt;<br /> We know that &lt;math&gt;b=8&lt;/math&gt; is a valid solution by &lt;math&gt;T&lt;/math&gt;. Factoring out &lt;math&gt;b-8&lt;/math&gt;, we obtain<br /> &lt;cmath&gt;(b-8)(b^2-b-8)=0 \Rightarrow b^2-b-8=0&lt;/cmath&gt;<br /> Utilizing the quadratic formula gives<br /> &lt;cmath&gt;b=\frac{1\pm\sqrt{33}}{2}&lt;/cmath&gt;<br /> We clearly must pick the positive solution. Note that &lt;math&gt;5&lt;\sqrt{33}&lt;6&lt;/math&gt;, and so &lt;math&gt;{3&lt;\frac{1+\sqrt{33}}{2}&lt;\frac{7}{2}}&lt;/math&gt;, which clearly gives an answer of &lt;math&gt;\fbox{A}&lt;/math&gt;, as desired.<br /> <br /> ===Solution 3===<br /> Triangle T has perimeter &lt;math&gt;5 + 5 + 8 = 18&lt;/math&gt; so &lt;math&gt;18 = 2a + b&lt;/math&gt;.<br /> <br /> Using Heron's, we get &lt;math&gt;\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})(\frac{d}{2})^2(\frac{2a-b}{2})}&lt;/math&gt;.<br /> <br /> We know that &lt;math&gt;2a + b = 18&lt;/math&gt; from above so we plug that in, and we also know that then &lt;math&gt;2a - b = 18 - 2b&lt;/math&gt;.<br /> <br /> &lt;math&gt;12 = 3\frac{b}{2}\sqrt{9-b}&lt;/math&gt;<br /> <br /> &lt;math&gt;64 = 9b^2 - b^3&lt;/math&gt;<br /> <br /> We plug in 3 for &lt;math&gt;b&lt;/math&gt; in the LHS, and we get 54 which is too low. We plug in 4 for &lt;math&gt;b&lt;/math&gt; in the LHS, and we get 80 which is too high. We now know that b is some number between 3 and 4. <br /> <br /> If &lt;math&gt;b \geq 3.5&lt;/math&gt;, then we would round up to 4, but if &lt;math&gt;b &lt; 3.5&lt;/math&gt;, then we would round down to 3. So let us plug in 3.5 for b. <br /> <br /> We get 67.375 which is too high, so we know that &lt;math&gt;b &lt; 3.5&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;3&lt;/math&gt;. &lt;math&gt;\textbf{(A)}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=19|num-a=21}}</div> PiDragon https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_9&diff=74960 2015 AMC 12A Problems/Problem 9 2016-01-31T15:20:06Z <p>PiDragon: </p> <hr /> <div>==Problem==<br /> <br /> A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is &lt;math&gt;\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}&lt;/math&gt;. Since there are three different colors, our final probability is &lt;math&gt;3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is &lt;math&gt;\frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The total number of ways they can draw is &lt;math&gt;{6 \choose 2}&lt;/math&gt; &lt;math&gt;{4 \choose 2}&lt;/math&gt; &lt;math&gt;{2 \choose 2}&lt;/math&gt;. Let Cheryl draw first and since there are three colors, there are &lt;math&gt;{3 \choose 1}&lt;/math&gt; ways she can get 2 marbles of the same color. The other two pick two each, which leads to &lt;math&gt;{4 \choose 2}&lt;/math&gt; and &lt;math&gt;{2 \choose 2}&lt;/math&gt;, respectively.<br /> &lt;math&gt;\frac{{3 \choose 1}{4 \choose 2}{2 \choose 2}}{{6 \choose 2}{4 \choose 2}{2 \choose 2}} = \frac{1}{5} \textbf{ (C)}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}}</div> PiDragon https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_9&diff=74959 2015 AMC 12A Problems/Problem 9 2016-01-31T15:19:25Z <p>PiDragon: </p> <hr /> <div>==Problem==<br /> <br /> A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is &lt;math&gt;\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}&lt;/math&gt;. Since there are three different colors, our final probability is &lt;math&gt;3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is &lt;math&gt;\frac{1}{5} \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The total number of ways they can draw is &lt;math&gt;{6 \choose 2}&lt;/math&gt; &lt;math&gt;{4 \choose 2}&lt;/math&gt; &lt;math&gt;{2 \choose 2}&lt;/math&gt;. Let Cheryl draw first and since there are three colors, there are &lt;math&gt;{3 \choose 1}&lt;/math&gt; ways she can get 2 marbles of the same color. The other two pick two each, which leads to &lt;math&gt;{4 \choose 2}&lt;/math&gt; and &lt;math&gt;{2 \choose 2}&lt;/math&gt;, respectively.<br /> &lt;math&gt;\frac{{3 \choose 1}{4 \choose 2}{2 \choose 2}}{{6 \choose 2}{4 \choose 2}{2 \choose 2}} = \frac{1}{5}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}}</div> PiDragon https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_22&diff=74930 2015 AMC 10A Problems/Problem 22 2016-01-31T01:19:35Z <p>PiDragon: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #17]] and [[2015 AMC 10A Problems|2015 AMC 10A #22]]}}<br /> ==Problem==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt;\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. We casework on how many people are standing.<br /> <br /> Case &lt;math&gt;1:&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; people are standing. This yields &lt;math&gt;1&lt;/math&gt; arrangement.<br /> <br /> Case &lt;math&gt;2:&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; person is standing. This yields &lt;math&gt;8&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;3:&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; people are standing. This yields &lt;math&gt;\dbinom{8}{2} - 8 = 20&lt;/math&gt; arrangements, because the two people cannot be next to each other.<br /> <br /> Case &lt;math&gt;4:&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding &lt;math&gt;2&lt;/math&gt; possible arrangements.<br /> <br /> More difficult is:<br /> <br /> Case &lt;math&gt;5:&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; people are standing. First, choose the location of the first person standing (&lt;math&gt;8&lt;/math&gt; choices). Next, choose &lt;math&gt;2&lt;/math&gt; of the remaining people in the remaining &lt;math&gt;5&lt;/math&gt; legal seats to stand, amounting to &lt;math&gt;6&lt;/math&gt; arrangements considering that these two people cannot stand next to each other. However, we have to divide by &lt;math&gt;3,&lt;/math&gt; because there are &lt;math&gt;3&lt;/math&gt; ways to choose the first person given any three. This yields &lt;math&gt;\dfrac{8 \cdot 6}{3} = 16&lt;/math&gt; arrangements for Case &lt;math&gt;5.&lt;/math&gt;<br /> <br /> Alternate Case &lt;math&gt;5:&lt;/math&gt; Use complementary counting. Total number of ways to choose 3 people from 8 which is &lt;math&gt;\dbinom{8}{3}&lt;/math&gt;. Sub-case &lt;math&gt;1:&lt;/math&gt; three people are next to each other which is &lt;math&gt;\dbinom{8}{1}&lt;/math&gt;. Sub-case &lt;math&gt;2:&lt;/math&gt; two people are next to each other and the third person is not &lt;math&gt;\dbinom{8}{1}&lt;/math&gt; &lt;math&gt;\dbinom{4}{1}&lt;/math&gt;. This yields &lt;math&gt;\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16&lt;/math&gt; <br /> <br /> Summing gives &lt;math&gt;1 + 8 + 20 + 2 + 16 = 47,&lt;/math&gt; and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We will count how many valid standing arrangements there are counting rotations as distinct and divide by &lt;math&gt;256&lt;/math&gt; at the end.<br /> Line up all &lt;math&gt;8&lt;/math&gt; people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires &lt;math&gt;2&lt;/math&gt; spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka ball and urn.<br /> <br /> If there are &lt;math&gt;4&lt;/math&gt; standing, there are &lt;math&gt;{4 \choose 4}=1&lt;/math&gt; ways to place them.<br /> For &lt;math&gt;3,&lt;/math&gt; there are &lt;math&gt;{3+2 \choose 3}=10&lt;/math&gt; ways.<br /> etc.<br /> Summing, we get &lt;math&gt;{4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34&lt;/math&gt; ways.<br /> <br /> Now we consider that the far right person can be standing as well, so we have<br /> &lt;math&gt;{3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13&lt;/math&gt; ways<br /> <br /> Together we have &lt;math&gt;34+13=47&lt;/math&gt;, and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only &lt;math&gt;7&lt;/math&gt; people. If they stand, we count the arrangements with &lt;math&gt;6&lt;/math&gt; instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with &lt;math&gt;1&lt;/math&gt; person there are two ways and with &lt;math&gt;2&lt;/math&gt; people there are three ways. Carrying out the Fibonacci recursion until we get to &lt;math&gt;8&lt;/math&gt; people, we find there are &lt;math&gt;55&lt;/math&gt; standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for &lt;math&gt;4&lt;/math&gt; people to stand in a line, which is &lt;math&gt;8&lt;/math&gt; from our sequence. Therefore our probability is &lt;math&gt;\frac{55 - 8}{256} = \boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2015|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2015|ab=A|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]</div> PiDragon