https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Xmidnightfirex&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-19T21:44:11Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_23&diff=134153 2017 AMC 10B Problems/Problem 23 2020-09-27T18:09:21Z <p>Xmidnightfirex: /* Solution 3 */</p> <hr /> <div>==Problem 23==<br /> Let &lt;math&gt;N=123456789101112\dots4344&lt;/math&gt; be the &lt;math&gt;79&lt;/math&gt;-digit number that is formed by writing the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; in order, one after the other. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> We only need to find the remainders of N when divided by 5 and 9 to determine the answer.<br /> By inspection, &lt;math&gt;N \equiv 4 \text{ (mod 5)}&lt;/math&gt;.<br /> The remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4&lt;/math&gt;, but since &lt;math&gt;10 \equiv 1 \text{ (mod 9)}&lt;/math&gt;, we can also write this as &lt;math&gt;1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45&lt;/math&gt;, which has a remainder of 0 mod 9. Solving these modular congruence using CRT(Chinese Remainder Theorem) we get the remainder to be &lt;math&gt;9 \pmod{45}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) } 9}&lt;/math&gt;.<br /> <br /> ==Alternative Ending to Solution 1==<br /> Once we find our 2 modular congruences, we can narrow our options down to &lt;math&gt;{C}&lt;/math&gt; and &lt;math&gt;{D}&lt;/math&gt; because the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt; should be a multiple of 9 by our modular congruence that states &lt;math&gt;N&lt;/math&gt; has a remainder of &lt;math&gt;0&lt;/math&gt; when divided by &lt;math&gt;9&lt;/math&gt;. Also, our other modular congruence states that the remainder when divided by &lt;math&gt;45&lt;/math&gt; should have a remainder of &lt;math&gt;4&lt;/math&gt; when divided by &lt;math&gt;5&lt;/math&gt;. Out of options &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, only &lt;math&gt;\boxed{\textbf{(C) } 9}&lt;/math&gt; satisfies that the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 45 &lt;math&gt;\equiv 4 \text{ (mod 5)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Realize that &lt;math&gt;10 \equiv 10 \cdot 10 \equiv 10^{k} \pmod{45}&lt;/math&gt; for all positive integers &lt;math&gt;k&lt;/math&gt;.<br /> <br /> Apply this on the expanded form of &lt;math&gt;N&lt;/math&gt;:<br /> &lt;cmath&gt;N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv&lt;/cmath&gt;<br /> &lt;cmath&gt;10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv&lt;/cmath&gt;<br /> &lt;cmath&gt;10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_23&diff=134152 2017 AMC 10B Problems/Problem 23 2020-09-27T18:09:10Z <p>Xmidnightfirex: /* Solution 2 */</p> <hr /> <div>==Problem 23==<br /> Let &lt;math&gt;N=123456789101112\dots4344&lt;/math&gt; be the &lt;math&gt;79&lt;/math&gt;-digit number that is formed by writing the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; in order, one after the other. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> We only need to find the remainders of N when divided by 5 and 9 to determine the answer.<br /> By inspection, &lt;math&gt;N \equiv 4 \text{ (mod 5)}&lt;/math&gt;.<br /> The remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4&lt;/math&gt;, but since &lt;math&gt;10 \equiv 1 \text{ (mod 9)}&lt;/math&gt;, we can also write this as &lt;math&gt;1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45&lt;/math&gt;, which has a remainder of 0 mod 9. Solving these modular congruence using CRT(Chinese Remainder Theorem) we get the remainder to be &lt;math&gt;9 \pmod{45}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) } 9}&lt;/math&gt;.<br /> <br /> ==Alternative Ending to Solution 1==<br /> Once we find our 2 modular congruences, we can narrow our options down to &lt;math&gt;{C}&lt;/math&gt; and &lt;math&gt;{D}&lt;/math&gt; because the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt; should be a multiple of 9 by our modular congruence that states &lt;math&gt;N&lt;/math&gt; has a remainder of &lt;math&gt;0&lt;/math&gt; when divided by &lt;math&gt;9&lt;/math&gt;. Also, our other modular congruence states that the remainder when divided by &lt;math&gt;45&lt;/math&gt; should have a remainder of &lt;math&gt;4&lt;/math&gt; when divided by &lt;math&gt;5&lt;/math&gt;. Out of options &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, only &lt;math&gt;\boxed{\textbf{(C) } 9}&lt;/math&gt; satisfies that the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 45 &lt;math&gt;\equiv 4 \text{ (mod 5)}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Realize that &lt;math&gt;10 \equiv 10 \cdot 10 \equiv 10^{k} \pmod{45}&lt;/math&gt; for all positive integers &lt;math&gt;k&lt;/math&gt;.<br /> <br /> Apply this on the expanded form of &lt;math&gt;N&lt;/math&gt;:<br /> &lt;cmath&gt;N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv&lt;/cmath&gt;<br /> &lt;cmath&gt;10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv&lt;/cmath&gt;<br /> &lt;cmath&gt;10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_1&diff=132077 1992 AHSME Problems/Problem 1 2020-08-18T21:20:50Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;3(4x+5\pi)=P&lt;/math&gt; then &lt;math&gt;6(8x+10\pi)=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 2P\quad<br /> \text{(B) } 4P\quad<br /> \text{(C) } 6P\quad<br /> \text{(D) } 8P\quad<br /> \text{(E) } 18P&lt;/math&gt;<br /> == Solution ==<br /> We can see that &lt;math&gt;8x+10\pi&lt;/math&gt; is equal to &lt;math&gt;2(4x+5\pi),&lt;/math&gt; and we know that &lt;math&gt;2^2 = 4,&lt;/math&gt; so the answer is &lt;math&gt;\boxed{B}\, .&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=1|num-a=2}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_1&diff=132076 1992 AHSME Problems/Problem 1 2020-08-18T21:20:32Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;3(4x+5\pi)=P&lt;/math&gt; then &lt;math&gt;6(8x+10\pi)=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 2P\quad<br /> \text{(B) } 4P\quad<br /> \text{(C) } 6P\quad<br /> \text{(D) } 8P\quad<br /> \text{(E) } 18P&lt;/math&gt;<br /> == Solution ==<br /> We can see that &lt;math&gt;8x+10p&lt;/math&gt;i is equal to &lt;math&gt;2(4x+5pi),&lt;/math&gt; and we know that &lt;math&gt;2^2 = 4,&lt;/math&gt; so the answer is &lt;math&gt;\boxed{B}\, .&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=1|num-a=2}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_21&diff=132075 1992 AHSME Problems/Problem 21 2020-08-18T21:18:48Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For a finite sequence &lt;math&gt;A=(a_1,a_2,...,a_n)&lt;/math&gt; of numbers, the ''Cesáro sum'' of A is defined to be <br /> &lt;math&gt;\frac{S_1+\cdots+S_n}{n}&lt;/math&gt; , where &lt;math&gt;S_k=a_1+\cdots+a_k&lt;/math&gt; and &lt;math&gt;1\leq k\leq n&lt;/math&gt;. If the Cesáro sum of<br /> the 99-term sequence &lt;math&gt;(a_1,...,a_{99})&lt;/math&gt; is 1000, what is the Cesáro sum of the 100-term sequence <br /> &lt;math&gt;(1,a_1,...,a_{99})&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 991\quad<br /> \text{(B) } 999\quad<br /> \text{(C) } 1000\quad<br /> \text{(D) } 1001\quad<br /> \text{(E) } 1009&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is <br /> &lt;cmath&gt;S_1 + S_2 + S_3 + ... + S_n&lt;/cmath&gt;<br /> &lt;cmath&gt;= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)&lt;/cmath&gt;<br /> &lt;cmath&gt;= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.&lt;/cmath&gt;<br /> If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is &lt;math&gt;1000 * 99 = 99,000,&lt;/math&gt; than the Cesáro total of the second sequence is &lt;math&gt;n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.&lt;/math&gt; Thus the Cesáro sum of the second sequence is &lt;math&gt;\frac{99,100}{100} = \boxed{991, A}\, .&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=20|num-a=22}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_21&diff=132074 1992 AHSME Problems/Problem 21 2020-08-18T21:17:25Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For a finite sequence &lt;math&gt;A=(a_1,a_2,...,a_n)&lt;/math&gt; of numbers, the ''Cesáro sum'' of A is defined to be <br /> &lt;math&gt;\frac{S_1+\cdots+S_n}{n}&lt;/math&gt; , where &lt;math&gt;S_k=a_1+\cdots+a_k&lt;/math&gt; and &lt;math&gt;1\leq k\leq n&lt;/math&gt;. If the Cesáro sum of<br /> the 99-term sequence &lt;math&gt;(a_1,...,a_{99})&lt;/math&gt; is 1000, what is the Cesáro sum of the 100-term sequence <br /> &lt;math&gt;(1,a_1,...,a_{99})&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 991\quad<br /> \text{(B) } 999\quad<br /> \text{(C) } 1000\quad<br /> \text{(D) } 1001\quad<br /> \text{(E) } 1009&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is <br /> &lt;cmath&gt;S_1 + S_2 + S_3 + ... + S_n&lt;/cmath&gt;<br /> &lt;cmath&gt;= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)&lt;/cmath&gt;<br /> &lt;cmath&gt;= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.&lt;/cmath&gt;<br /> If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is &lt;math&gt;1000 * 99 = 99,000,&lt;/math&gt; than the Cesáro total of the second sequence is &lt;math&gt;n \cdot a_1 + 99,000 = 100 * 1 + 99,000 = 99,100.&lt;/math&gt; Thus the Cesáro sum of the second sequence is &lt;math&gt;\frac{99,100}{100} = \boxed{991, A}\, .&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=20|num-a=22}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_21&diff=132073 1992 AHSME Problems/Problem 21 2020-08-18T21:17:01Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For a finite sequence &lt;math&gt;A=(a_1,a_2,...,a_n)&lt;/math&gt; of numbers, the ''Cesáro sum'' of A is defined to be <br /> &lt;math&gt;\frac{S_1+\cdots+S_n}{n}&lt;/math&gt; , where &lt;math&gt;S_k=a_1+\cdots+a_k&lt;/math&gt; and &lt;math&gt;1\leq k\leq n&lt;/math&gt;. If the Cesáro sum of<br /> the 99-term sequence &lt;math&gt;(a_1,...,a_{99})&lt;/math&gt; is 1000, what is the Cesáro sum of the 100-term sequence <br /> &lt;math&gt;(1,a_1,...,a_{99})&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 991\quad<br /> \text{(B) } 999\quad<br /> \text{(C) } 1000\quad<br /> \text{(D) } 1001\quad<br /> \text{(E) } 1009&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is <br /> &lt;cmath&gt;S_1 + S_2 + S_3 + ... + S_n<br /> = a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n) <br /> = n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.&lt;/cmath&gt;<br /> If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is &lt;math&gt;1000 * 99 = 99,000,&lt;/math&gt; than the Cesáro total of the second sequence is &lt;math&gt;n \cdot a_1 + 99,000 = 100 * 1 + 99,000 = 99,100.&lt;/math&gt; Thus the Cesáro sum of the second sequence is &lt;math&gt;\frac{99,100}{100} = \boxed{991, A}\, .&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=20|num-a=22}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_21&diff=132072 1992 AHSME Problems/Problem 21 2020-08-18T21:16:31Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For a finite sequence &lt;math&gt;A=(a_1,a_2,...,a_n)&lt;/math&gt; of numbers, the ''Cesáro sum'' of A is defined to be <br /> &lt;math&gt;\frac{S_1+\cdots+S_n}{n}&lt;/math&gt; , where &lt;math&gt;S_k=a_1+\cdots+a_k&lt;/math&gt; and &lt;math&gt;1\leq k\leq n&lt;/math&gt;. If the Cesáro sum of<br /> the 99-term sequence &lt;math&gt;(a_1,...,a_{99})&lt;/math&gt; is 1000, what is the Cesáro sum of the 100-term sequence <br /> &lt;math&gt;(1,a_1,...,a_{99})&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 991\quad<br /> \text{(B) } 999\quad<br /> \text{(C) } 1000\quad<br /> \text{(D) } 1001\quad<br /> \text{(E) } 1009&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is <br /> \begin{align*}<br /> S_1 + S_2 + S_3 + ... + S_n \\<br /> &amp;= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n) \\<br /> &amp;= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.<br /> \end{align*}<br /> If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is &lt;math&gt;1000 * 99 = 99,000,&lt;/math&gt; than the Cesáro total of the second sequence is &lt;math&gt;n \cdot a_1 + 99,000 = 100 * 1 + 99,000 = 99,100.&lt;/math&gt; Thus the Cesáro sum of the second sequence is &lt;math&gt;\frac{99,100}{100} = \boxed{991, A}\, .&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=20|num-a=22}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=User:Xmidnightfirex&diff=130389 User:Xmidnightfirex 2020-08-04T02:40:32Z <p>Xmidnightfirex: Created page with &quot;I mean, hi? I don't really know what I'm doing here right now and what the point of this is, and I'm also pretty sure they've got my name on the wiki wrong. I'm xMidnightFirex...&quot;</p> <hr /> <div>I mean, hi? I don't really know what I'm doing here right now and what the point of this is, and I'm also pretty sure they've got my name on the wiki wrong. I'm xMidnightFirex, not Xmidnightfirex. That just looks plain weird in my opinion.</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1993_UNCO_Math_Contest_II_Problems/Problem_8&diff=117780 1993 UNCO Math Contest II Problems/Problem 8 2020-02-14T19:22:10Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For what integer value of &lt;math&gt;n&lt;/math&gt; is the expression<br /> &lt;cmath&gt;\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{n}+\sqrt{n+1}}&lt;/cmath&gt;<br /> equal to &lt;math&gt;7&lt;/math&gt; ? (Hint: &lt;math&gt;(1+\sqrt{2})(1-\sqrt{2})=-1.&lt;/math&gt;)<br /> <br /> <br /> == Solution ==<br /> <br /> Looking at the first fraction, we get that &lt;math&gt;\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}&lt;/math&gt;. Moving on, we see an interesting pattern in which the fraction &lt;math&gt;\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}&lt;/math&gt;. This means that we can rewrite the fractions as &lt;math&gt;(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n})&lt;/math&gt;. We can cancel out most of the terms in that sequence and get &lt;math&gt;\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1&lt;/math&gt;. However, we need to solve for &lt;math&gt;n&lt;/math&gt; now. Setting the previous expression equal to &lt;math&gt;7&lt;/math&gt;, we get that &lt;math&gt;\sqrt{n+1} = 8&lt;/math&gt;. Squaring both sides, we get that &lt;math&gt;n+1 = 64&lt;/math&gt;. Hence, &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;\boxed{63}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=1993|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1993_UNCO_Math_Contest_II_Problems/Problem_8&diff=117779 1993 UNCO Math Contest II Problems/Problem 8 2020-02-14T19:21:46Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For what integer value of &lt;math&gt;n&lt;/math&gt; is the expression<br /> &lt;cmath&gt;\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{n}+\sqrt{n+1}}&lt;/cmath&gt;<br /> equal to &lt;math&gt;7&lt;/math&gt; ? (Hint: &lt;math&gt;(1+\sqrt{2})(1-\sqrt{2})=-1.&lt;/math&gt;)<br /> <br /> <br /> == Solution ==<br /> <br /> Looking at the first fraction, we get that &lt;math&gt;\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}&lt;/math&gt;. Moving on, we see an interesting pattern in which the fraction &lt;math&gt;\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}&lt;/math&gt;. This means that we can rewrite the fractions as &lt;math&gt;(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n}&lt;/math&gt;. We can cancel out most of the terms in that sequence and get &lt;math&gt;\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1&lt;/math&gt;. However, we need to solve for &lt;math&gt;n&lt;/math&gt; now. Setting the previous expression equal to &lt;math&gt;7&lt;/math&gt;, we get that &lt;math&gt;\sqrt{n+1} = 8&lt;/math&gt;. Squaring both sides, we get that &lt;math&gt;n+1 = 64&lt;/math&gt;. Hence, &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;\boxed{63}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=1993|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_24&diff=116208 2020 AMC 10A Problems/Problem 24 2020-02-01T14:02:30Z <p>Xmidnightfirex: /* Solution 2 (bashing) */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer greater than &lt;math&gt;1000&lt;/math&gt; for which&lt;cmath&gt;\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> We know that &lt;math&gt;gcd(63, n+120)=21&lt;/math&gt;, so we can write &lt;math&gt;n+120\equiv0\pmod {21}&lt;/math&gt;. Simplifying, we get &lt;math&gt;n\equiv6\pmod {21}&lt;/math&gt;. Similarly, we can write &lt;math&gt;n+63\equiv0\pmod {60}&lt;/math&gt;, or &lt;math&gt;n\equiv-3\pmod {60}&lt;/math&gt;. Solving these two modular congruences, &lt;math&gt;n\equiv237\pmod {420}&lt;/math&gt; which we know is the only solution by CRT (Chinese Remainder Theorem). Now, since the problem is asking for the least positive integer greater than &lt;math&gt;1000&lt;/math&gt;, we find the least solution is &lt;math&gt;n=1077&lt;/math&gt;. However, we are have not considered cases where &lt;math&gt;gcd(63, n+120) =63&lt;/math&gt; or &lt;math&gt;gcd(n+63, 120) =120&lt;/math&gt;. &lt;math&gt;{1077+120}\equiv0\pmod {63}&lt;/math&gt; so we try &lt;math&gt;n=1077+420=1497&lt;/math&gt;. &lt;math&gt;{1497+63}\equiv0\pmod {120}&lt;/math&gt; so again we add &lt;math&gt;420&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. It turns out that &lt;math&gt;n=1497+420=1917&lt;/math&gt; does indeed satisfy the original conditions, so our answer is &lt;math&gt;1+9+1+7=\boxed{\textsf{(C) } 18}&lt;/math&gt;.<br /> <br /> ==Solution 2 (bashing)==<br /> <br /> We are given that &lt;math&gt;\gcd(63, n+120)=21&lt;/math&gt; and &lt;math&gt;\gcd(n+63,120) = 60&lt;/math&gt;. This tells us that &lt;math&gt;n+120&lt;/math&gt; is divisible by &lt;math&gt;21&lt;/math&gt; but not &lt;math&gt;63&lt;/math&gt;. It also tells us that &lt;math&gt;n+63&lt;/math&gt; is divisible by 60 but not 120. Starting, we find the least value of &lt;math&gt;n+120&lt;/math&gt; which is divisible by &lt;math&gt;21&lt;/math&gt; which satisfies the conditions for &lt;math&gt;n&lt;/math&gt;, which is &lt;math&gt;1134&lt;/math&gt;, making &lt;math&gt;n=1014&lt;/math&gt;. We then now keep on adding &lt;math&gt;21&lt;/math&gt; until we get a number which satisfies the second equation. This number turns out to be &lt;math&gt;1917&lt;/math&gt;, whose digits add up to &lt;math&gt;\boxed{\text{C}, 18}&lt;/math&gt;.<br /> <br /> -Midnight<br /> <br /> == Video Solution ==<br /> <br /> https://youtu.be/tk3yOGG2K-s - &lt;math&gt;Phineas1500&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_24&diff=116207 2020 AMC 10A Problems/Problem 24 2020-02-01T14:02:09Z <p>Xmidnightfirex: /* Solution 2 (bashing) */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer greater than &lt;math&gt;1000&lt;/math&gt; for which&lt;cmath&gt;\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> We know that &lt;math&gt;gcd(63, n+120)=21&lt;/math&gt;, so we can write &lt;math&gt;n+120\equiv0\pmod {21}&lt;/math&gt;. Simplifying, we get &lt;math&gt;n\equiv6\pmod {21}&lt;/math&gt;. Similarly, we can write &lt;math&gt;n+63\equiv0\pmod {60}&lt;/math&gt;, or &lt;math&gt;n\equiv-3\pmod {60}&lt;/math&gt;. Solving these two modular congruences, &lt;math&gt;n\equiv237\pmod {420}&lt;/math&gt; which we know is the only solution by CRT (Chinese Remainder Theorem). Now, since the problem is asking for the least positive integer greater than &lt;math&gt;1000&lt;/math&gt;, we find the least solution is &lt;math&gt;n=1077&lt;/math&gt;. However, we are have not considered cases where &lt;math&gt;gcd(63, n+120) =63&lt;/math&gt; or &lt;math&gt;gcd(n+63, 120) =120&lt;/math&gt;. &lt;math&gt;{1077+120}\equiv0\pmod {63}&lt;/math&gt; so we try &lt;math&gt;n=1077+420=1497&lt;/math&gt;. &lt;math&gt;{1497+63}\equiv0\pmod {120}&lt;/math&gt; so again we add &lt;math&gt;420&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. It turns out that &lt;math&gt;n=1497+420=1917&lt;/math&gt; does indeed satisfy the original conditions, so our answer is &lt;math&gt;1+9+1+7=\boxed{\textsf{(C) } 18}&lt;/math&gt;.<br /> <br /> ==Solution 2 (bashing)==<br /> <br /> We are given that &lt;math&gt;\gcd(63, n+120)=21&lt;/math&gt; and &lt;math&gt;\gcd(n+63,120) = 60&lt;/math&gt;. This tells us that &lt;math&gt;n+120&lt;/math&gt; is divisible by &lt;math&gt;21&lt;/math&gt; but not &lt;math&gt;63&lt;/math&gt;. It also tells us that &lt;math&gt;n+63&lt;/math&gt; is divisible by 60 but not 120. Starting, we find the least value of &lt;math&gt;n+120&lt;/math&gt; which is divisible by &lt;math&gt;21&lt;/math&gt; which satisfies the conditions for &lt;math&gt;n&lt;/math&gt;, which is &lt;math&gt;1134&lt;/math&gt;, making &lt;math&gt;n&lt;/math&gt; &lt;math&gt;1014&lt;/math&gt;. We then now keep on adding &lt;math&gt;21&lt;/math&gt; until we get a number which satisfies the second equation. This number turns out to be &lt;math&gt;1917&lt;/math&gt;, whose digits add up to &lt;math&gt;\boxed{\text{C}, 18}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> <br /> https://youtu.be/tk3yOGG2K-s - &lt;math&gt;Phineas1500&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_24&diff=116206 2020 AMC 10A Problems/Problem 24 2020-02-01T13:54:58Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer greater than &lt;math&gt;1000&lt;/math&gt; for which&lt;cmath&gt;\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> We know that &lt;math&gt;gcd(63, n+120)=21&lt;/math&gt;, so we can write &lt;math&gt;n+120\equiv0\pmod {21}&lt;/math&gt;. Simplifying, we get &lt;math&gt;n\equiv6\pmod {21}&lt;/math&gt;. Similarly, we can write &lt;math&gt;n+63\equiv0\pmod {60}&lt;/math&gt;, or &lt;math&gt;n\equiv-3\pmod {60}&lt;/math&gt;. Solving these two modular congruences, &lt;math&gt;n\equiv237\pmod {420}&lt;/math&gt; which we know is the only solution by CRT (Chinese Remainder Theorem). Now, since the problem is asking for the least positive integer greater than &lt;math&gt;1000&lt;/math&gt;, we find the least solution is &lt;math&gt;n=1077&lt;/math&gt;. However, we are have not considered cases where &lt;math&gt;gcd(63, n+120) =63&lt;/math&gt; or &lt;math&gt;gcd(n+63, 120) =120&lt;/math&gt;. &lt;math&gt;{1077+120}\equiv0\pmod {63}&lt;/math&gt; so we try &lt;math&gt;n=1077+420=1497&lt;/math&gt;. &lt;math&gt;{1497+63}\equiv0\pmod {120}&lt;/math&gt; so again we add &lt;math&gt;420&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. It turns out that &lt;math&gt;n=1497+420=1917&lt;/math&gt; does indeed satisfy the original conditions, so our answer is &lt;math&gt;1+9+1+7=\boxed{\textsf{(C) } 18}&lt;/math&gt;.<br /> <br /> ==Solution 2 (bashing)==<br /> <br /> We are given that &lt;math&gt;gcd(63, n+120)=21&lt;/math&gt; and<br /> <br /> == Video Solution ==<br /> <br /> https://youtu.be/tk3yOGG2K-s - &lt;math&gt;Phineas1500&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_18&diff=116125 2020 AMC 10A Problems/Problem 18 2020-02-01T03:52:50Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;(a,b,c,d)&lt;/math&gt; be an ordered quadruple of not necessarily distinct integers, each one of them in the set &lt;math&gt;{0,1,2,3}.&lt;/math&gt; For how many such quadruples is it true that &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; is odd? (For example, &lt;math&gt;(0,3,1,1)&lt;/math&gt; is one such quadruple, because &lt;math&gt;0\cdot 1-3\cdot 1 = -3&lt;/math&gt; is odd.)<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> == Solution ==<br /> In order for &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are &lt;math&gt;2 \cdot 4 = 8&lt;/math&gt; ways to pick numbers to obtain a even product. There are &lt;math&gt;2 \cdot 2 = 4&lt;/math&gt; ways to obtain an odd product. Therefore, the total amount of ways to make &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; odd is &lt;math&gt;2 \cdot (8 \cdot 4) = \boxed{\text{B}, 64}&lt;/math&gt;.<br /> <br /> -Midnight<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_18&diff=116124 2020 AMC 10A Problems/Problem 18 2020-02-01T03:52:14Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;(a,b,c,d)&lt;/math&gt; be an ordered quadruple of not necessarily distinct integers, each one of them in the set &lt;math&gt;{0,1,2,3}.&lt;/math&gt; For how many such quadruples is it true that &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; is odd? (For example, &lt;math&gt;(0,3,1,1)&lt;/math&gt; is one such quadruple, because &lt;math&gt;0\cdot 1-3\cdot 1 = -3&lt;/math&gt; is odd.)<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> == Solution ==<br /> In order for &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are &lt;math&gt;2 \cdot 4 = 8&lt;/math&gt; ways to pick numbers to obtain a even product. There are &lt;math&gt;2 \cdot 2 = 4&lt;/math&gt; ways to obtain an odd product. Therefore, the total amount of ways to make &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; odd is &lt;math&gt;2 \cdot (8 \cdot 4) = \boxed{\text{B}, 64}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_18&diff=116122 2020 AMC 10A Problems/Problem 18 2020-02-01T03:51:29Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;(a,b,c,d)&lt;/math&gt; be an ordered quadruple of not necessarily distinct integers, each one of them in the set &lt;math&gt;{0,1,2,3}.&lt;/math&gt; For how many such quadruples is it true that &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; is odd? (For example, &lt;math&gt;(0,3,1,1)&lt;/math&gt; is one such quadruple, because &lt;math&gt;0\cdot 1-3\cdot 1 = -3&lt;/math&gt; is odd.)<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> == Solution ==<br /> In order for &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are &lt;math&gt;2 \cdot 4 = 8&lt;/math&gt; ways to pick numbers to obtain a even product. There are &lt;math&gt;2 \cdot 2 = 4&lt;/math&gt; ways to obtain an odd product. Therefore, the total amount of ways to make &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; odd is 2 \cdot (8 \cdot 4) = 64<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_25&diff=113962 2002 AMC 10A Problems/Problem 25 2020-01-01T18:29:19Z <p>Xmidnightfirex: /* Solution 2 but quicker */</p> <hr /> <div>== Problem ==<br /> In [[trapezoid]] &lt;math&gt;ABCD&lt;/math&gt; with bases &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, we have &lt;math&gt;AB = 52&lt;/math&gt;, &lt;math&gt;BC = 12&lt;/math&gt;, &lt;math&gt;CD = 39&lt;/math&gt;, and &lt;math&gt;DA = 5&lt;/math&gt;. The area of &lt;math&gt;ABCD&lt;/math&gt; is<br /> <br /> &lt;math&gt;\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; to meet at point &lt;math&gt;E&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(250);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);<br /> draw(A--B--C--D--cycle);<br /> draw(D--F--C,dashed);<br /> label(&quot;$$A$$&quot;,A,S);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,W);<br /> label(&quot;$$E$$&quot;,F,N);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,E);<br /> label(&quot;12&quot;,(B+C)/2,WSW);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since &lt;math&gt;\overline{AB} || \overline{CD}&lt;/math&gt; we have &lt;math&gt;\triangle AEB \sim \triangle DEC&lt;/math&gt;, with the ratio of [[proportion]]ality being &lt;math&gt;\frac {39}{52} = \frac {3}{4}&lt;/math&gt;. Thus<br /> &lt;cmath&gt;<br /> \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} &amp; \Longrightarrow CE = 36 \\<br /> \frac {DE}{DE + 5} = \frac {3}{4} &amp; \Longrightarrow DE = 15 \end{align*}<br /> &lt;/cmath&gt;<br /> So the sides of &lt;math&gt;\triangle CDE&lt;/math&gt; are &lt;math&gt;15,36,39&lt;/math&gt;, which we recognize to be a &lt;math&gt;5 - 12 - 13&lt;/math&gt; [[right triangle]]. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),<br /> &lt;cmath&gt;<br /> [ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}&lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> <br /> Draw altitudes from points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br /> draw(A--B--C--D--cycle);<br /> draw(C--E,dashed);<br /> draw(D--F,dashed);<br /> label(&quot;$$A$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,N);<br /> label(&quot;$$D'$$&quot;,F,SSE);<br /> label(&quot;$$C'$$&quot;,E,S);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Translate the triangle &lt;math&gt;ADD'&lt;/math&gt; so that &lt;math&gt;DD'&lt;/math&gt; coincides with &lt;math&gt;CC'&lt;/math&gt;. We get the following triangle:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0);<br /> draw(A--B--C--cycle);<br /> draw(C--F,dashed);<br /> label(&quot;$$A'$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,N);<br /> label(&quot;$$C'$$&quot;,F,SE);<br /> label(&quot;5&quot;,(A+C)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The length of &lt;math&gt;A'B&lt;/math&gt; in this triangle is equal to the length of the original &lt;math&gt;AB&lt;/math&gt;, minus the length of &lt;math&gt;CD&lt;/math&gt;.<br /> Thus &lt;math&gt;A'B = 52 - 39 = 13&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;A'BC&lt;/math&gt; is a well-known &lt;math&gt;(5,12,13)&lt;/math&gt; right triangle. Its area is &lt;math&gt;[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30&lt;/math&gt;, and therefore its altitude &lt;math&gt;CC'&lt;/math&gt; is &lt;math&gt;\frac{[A'BC]}{A'B} = \frac{60}{13}&lt;/math&gt;.<br /> <br /> Now the area of the original trapezoid is &lt;math&gt;\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Draw altitudes from points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br /> draw(A--B--C--D--cycle);<br /> draw(C--E,dashed);<br /> draw(D--F,dashed);<br /> label(&quot;$$A$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,N);<br /> label(&quot;$$D'$$&quot;,F,SSE);<br /> label(&quot;$$C'$$&quot;,E,S);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Call the length of &lt;math&gt;AD'&lt;/math&gt; to be &lt;math&gt;y&lt;/math&gt;, the length of &lt;math&gt;BC'&lt;/math&gt; to be &lt;math&gt;z&lt;/math&gt;, and the height of the trapezoid to be &lt;math&gt;x&lt;/math&gt;.<br /> By the Pythagorean Theorem, we have:<br /> &lt;cmath&gt;z^2 + x^2 = 144&lt;/cmath&gt;<br /> &lt;cmath&gt;y^2 + x^2 = 25&lt;/cmath&gt;<br /> <br /> Subtracting these two equation yields:<br /> &lt;cmath&gt;z^2-y^2=119 \implies (z+y)(z-y)=119&lt;/cmath&gt;<br /> <br /> We also have: &lt;math&gt;z+y=52-39=13&lt;/math&gt;.<br /> <br /> We can substitute the value of &lt;math&gt;z+y&lt;/math&gt; into the equation we just obtained, so we now have:<br /> <br /> &lt;cmath&gt;(13) (z-y)=119 \implies z-y=\frac{119}{13}&lt;/cmath&gt;.<br /> <br /> We can add the &lt;math&gt;z+y&lt;/math&gt; and the &lt;math&gt;z-y&lt;/math&gt; equation to find the value of &lt;math&gt;z&lt;/math&gt;, which simplifies down to be &lt;math&gt;\frac{144}{13}&lt;/math&gt;. Finally, we can plug in &lt;math&gt;z&lt;/math&gt; and use the Pythagorean theorem to find the height of the trapezoid.<br /> <br /> &lt;cmath&gt;\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}&lt;/cmath&gt;<br /> <br /> Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.<br /> <br /> The median of the trapezoid is &lt;math&gt;\frac{39+52}{2} = \frac{91}{2}&lt;/math&gt;, and multiplying this and the height of the trapezoid gets us:<br /> <br /> &lt;cmath&gt;\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}&lt;/cmath&gt;<br /> <br /> === Solution 4 ===<br /> <br /> We construct a line segment parallel to &lt;math&gt;\overline{AD}&lt;/math&gt; from point &lt;math&gt;C&lt;/math&gt; to line &lt;math&gt;\overline{AB},&lt;/math&gt; and label the intersection of this segment with line &lt;math&gt;\overline{AB}&lt;/math&gt; as point &lt;math&gt;E.&lt;/math&gt; Then quadrilateral &lt;math&gt;AECD&lt;/math&gt; is a parallelogram, so &lt;math&gt;CE=5, AE=39,&lt;/math&gt; and &lt;math&gt;EB=13.&lt;/math&gt; Triangle &lt;math&gt;EBC&lt;/math&gt; is therefore a right triangle, with area &lt;math&gt;\frac12 \cdot 5 \cdot 12 = 30.&lt;/math&gt;<br /> <br /> By continuing to split &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; into segments of length &lt;math&gt;13,&lt;/math&gt; we can connect these vertices in a &quot;zig-zag,&quot; creating seven congruent right triangles, each with sides &lt;math&gt;5,12,&lt;/math&gt; and &lt;math&gt;13,&lt;/math&gt; and each with area &lt;math&gt;30.&lt;/math&gt; The total area is therefore &lt;math&gt;7 \cdot 30 = \boxed{\textbf{(C)} 210}.&lt;/math&gt;<br /> <br /> === Solution 2 but quicker ===<br /> From Solution &lt;math&gt;2&lt;/math&gt; we know that the the altitude of the trapezoid is &lt;math&gt;\frac{60}{13}&lt;/math&gt; and the triangle's area is &lt;math&gt;30&lt;/math&gt;.<br /> Note that once we remove the triangle we get a rectangle with length &lt;math&gt;39&lt;/math&gt; and height &lt;math&gt;\frac{60}{13}&lt;/math&gt;.<br /> The numbers multiply nicely to get &lt;math&gt;180+30=\boxed{(C) 210}&lt;/math&gt;<br /> -harsha12345<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|num-b=24|after=Last Problem|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_25&diff=113961 2002 AMC 10A Problems/Problem 25 2020-01-01T18:28:47Z <p>Xmidnightfirex: /* See also */</p> <hr /> <div>== Problem ==<br /> In [[trapezoid]] &lt;math&gt;ABCD&lt;/math&gt; with bases &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, we have &lt;math&gt;AB = 52&lt;/math&gt;, &lt;math&gt;BC = 12&lt;/math&gt;, &lt;math&gt;CD = 39&lt;/math&gt;, and &lt;math&gt;DA = 5&lt;/math&gt;. The area of &lt;math&gt;ABCD&lt;/math&gt; is<br /> <br /> &lt;math&gt;\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; to meet at point &lt;math&gt;E&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(250);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);<br /> draw(A--B--C--D--cycle);<br /> draw(D--F--C,dashed);<br /> label(&quot;$$A$$&quot;,A,S);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,W);<br /> label(&quot;$$E$$&quot;,F,N);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,E);<br /> label(&quot;12&quot;,(B+C)/2,WSW);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since &lt;math&gt;\overline{AB} || \overline{CD}&lt;/math&gt; we have &lt;math&gt;\triangle AEB \sim \triangle DEC&lt;/math&gt;, with the ratio of [[proportion]]ality being &lt;math&gt;\frac {39}{52} = \frac {3}{4}&lt;/math&gt;. Thus<br /> &lt;cmath&gt;<br /> \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} &amp; \Longrightarrow CE = 36 \\<br /> \frac {DE}{DE + 5} = \frac {3}{4} &amp; \Longrightarrow DE = 15 \end{align*}<br /> &lt;/cmath&gt;<br /> So the sides of &lt;math&gt;\triangle CDE&lt;/math&gt; are &lt;math&gt;15,36,39&lt;/math&gt;, which we recognize to be a &lt;math&gt;5 - 12 - 13&lt;/math&gt; [[right triangle]]. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),<br /> &lt;cmath&gt;<br /> [ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}&lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> <br /> Draw altitudes from points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br /> draw(A--B--C--D--cycle);<br /> draw(C--E,dashed);<br /> draw(D--F,dashed);<br /> label(&quot;$$A$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,N);<br /> label(&quot;$$D'$$&quot;,F,SSE);<br /> label(&quot;$$C'$$&quot;,E,S);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Translate the triangle &lt;math&gt;ADD'&lt;/math&gt; so that &lt;math&gt;DD'&lt;/math&gt; coincides with &lt;math&gt;CC'&lt;/math&gt;. We get the following triangle:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0);<br /> draw(A--B--C--cycle);<br /> draw(C--F,dashed);<br /> label(&quot;$$A'$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,N);<br /> label(&quot;$$C'$$&quot;,F,SE);<br /> label(&quot;5&quot;,(A+C)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The length of &lt;math&gt;A'B&lt;/math&gt; in this triangle is equal to the length of the original &lt;math&gt;AB&lt;/math&gt;, minus the length of &lt;math&gt;CD&lt;/math&gt;.<br /> Thus &lt;math&gt;A'B = 52 - 39 = 13&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;A'BC&lt;/math&gt; is a well-known &lt;math&gt;(5,12,13)&lt;/math&gt; right triangle. Its area is &lt;math&gt;[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30&lt;/math&gt;, and therefore its altitude &lt;math&gt;CC'&lt;/math&gt; is &lt;math&gt;\frac{[A'BC]}{A'B} = \frac{60}{13}&lt;/math&gt;.<br /> <br /> Now the area of the original trapezoid is &lt;math&gt;\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Draw altitudes from points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br /> draw(A--B--C--D--cycle);<br /> draw(C--E,dashed);<br /> draw(D--F,dashed);<br /> label(&quot;$$A$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,N);<br /> label(&quot;$$D'$$&quot;,F,SSE);<br /> label(&quot;$$C'$$&quot;,E,S);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Call the length of &lt;math&gt;AD'&lt;/math&gt; to be &lt;math&gt;y&lt;/math&gt;, the length of &lt;math&gt;BC'&lt;/math&gt; to be &lt;math&gt;z&lt;/math&gt;, and the height of the trapezoid to be &lt;math&gt;x&lt;/math&gt;.<br /> By the Pythagorean Theorem, we have:<br /> &lt;cmath&gt;z^2 + x^2 = 144&lt;/cmath&gt;<br /> &lt;cmath&gt;y^2 + x^2 = 25&lt;/cmath&gt;<br /> <br /> Subtracting these two equation yields:<br /> &lt;cmath&gt;z^2-y^2=119 \implies (z+y)(z-y)=119&lt;/cmath&gt;<br /> <br /> We also have: &lt;math&gt;z+y=52-39=13&lt;/math&gt;.<br /> <br /> We can substitute the value of &lt;math&gt;z+y&lt;/math&gt; into the equation we just obtained, so we now have:<br /> <br /> &lt;cmath&gt;(13) (z-y)=119 \implies z-y=\frac{119}{13}&lt;/cmath&gt;.<br /> <br /> We can add the &lt;math&gt;z+y&lt;/math&gt; and the &lt;math&gt;z-y&lt;/math&gt; equation to find the value of &lt;math&gt;z&lt;/math&gt;, which simplifies down to be &lt;math&gt;\frac{144}{13}&lt;/math&gt;. Finally, we can plug in &lt;math&gt;z&lt;/math&gt; and use the Pythagorean theorem to find the height of the trapezoid.<br /> <br /> &lt;cmath&gt;\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}&lt;/cmath&gt;<br /> <br /> Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.<br /> <br /> The median of the trapezoid is &lt;math&gt;\frac{39+52}{2} = \frac{91}{2}&lt;/math&gt;, and multiplying this and the height of the trapezoid gets us:<br /> <br /> &lt;cmath&gt;\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}&lt;/cmath&gt;<br /> <br /> === Solution 4 ===<br /> <br /> We construct a line segment parallel to &lt;math&gt;\overline{AD}&lt;/math&gt; from point &lt;math&gt;C&lt;/math&gt; to line &lt;math&gt;\overline{AB},&lt;/math&gt; and label the intersection of this segment with line &lt;math&gt;\overline{AB}&lt;/math&gt; as point &lt;math&gt;E.&lt;/math&gt; Then quadrilateral &lt;math&gt;AECD&lt;/math&gt; is a parallelogram, so &lt;math&gt;CE=5, AE=39,&lt;/math&gt; and &lt;math&gt;EB=13.&lt;/math&gt; Triangle &lt;math&gt;EBC&lt;/math&gt; is therefore a right triangle, with area &lt;math&gt;\frac12 \cdot 5 \cdot 12 = 30.&lt;/math&gt;<br /> <br /> By continuing to split &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; into segments of length &lt;math&gt;13,&lt;/math&gt; we can connect these vertices in a &quot;zig-zag,&quot; creating seven congruent right triangles, each with sides &lt;math&gt;5,12,&lt;/math&gt; and &lt;math&gt;13,&lt;/math&gt; and each with area &lt;math&gt;30.&lt;/math&gt; The total area is therefore &lt;math&gt;7 \cdot 30 = \boxed{\textbf{(C)} 210}.&lt;/math&gt;<br /> <br /> === Solution 2 but quicker ===<br /> From solution &lt;math&gt;2&lt;/math&gt; we know that the the altitude of the trapezoid is &lt;math&gt;\frac{60}{13}&lt;/math&gt; and the triangle's area is &lt;math&gt;30&lt;/math&gt;.<br /> Note that once we remove the triangle we get a rectangle with length &lt;math&gt;39&lt;/math&gt; and height &lt;math&gt;\frac{60}{13}&lt;/math&gt;<br /> the number multiply nicely to get &lt;math&gt;180+30=\boxed{(C) 210}&lt;/math&gt;<br /> -harsha12345<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|num-b=24|after=Last Problem|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=113467 2011 AMC 10A Problems 2019-12-26T19:48:01Z <p>Xmidnightfirex: /* Problem 12 */</p> <hr /> <div>{{AMC10 Problems|year=2011|ab=A}}<br /> <br /> == Problem 1 ==<br /> <br /> A cell phone plan costs &lt;math&gt;\textdollar 20&lt;/math&gt; each month, plus &lt;math&gt;5&lt;/math&gt;¢ per text message sent, plus &lt;math&gt;10&lt;/math&gt;¢ for each minute used over &lt;math&gt;30&lt;/math&gt; hours. In January Juan sent &lt;math&gt;100&lt;/math&gt; text messages and talked for &lt;math&gt;30.5&lt;/math&gt; hours. How much did he have to pay?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \textdollar 24.00 \qquad\textbf{(B)}\ \textdollar 24.50 \qquad\textbf{(C)}\ \textdollar 25.50\qquad\textbf{(D)}\ \textdollar 28.00\qquad\textbf{(E)}\ \textdollar 30.00 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Suppose &lt;math&gt;[a\ b]&lt;/math&gt; denotes the average of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;\{a\ b\ c\}&lt;/math&gt; denotes the average of &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. What is &lt;math&gt;\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Let &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; be the following sums of arithmetic sequences: &lt;cmath&gt; \begin{eqnarray*} X &amp;=&amp; 10 + 12 + 14 + \cdots + 100, \\ Y &amp;=&amp; 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} &lt;/cmath&gt; What is the value of &lt;math&gt;Y - X&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of &lt;math&gt;12&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt; minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> Set &lt;math&gt;A &lt;/math&gt; has 20 elements, and set &lt;math&gt;B &lt;/math&gt; has 15 elements. What is the smallest possible number of elements in &lt;math&gt;A \cup B &lt;/math&gt;, the union of &lt;math&gt;A &lt;/math&gt; and &lt;math&gt;B &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Which of the following equations does NOT have a solution?<br /> <br /> &lt;math&gt;\textbf{(A)}\:(x+7)^2=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(B)}\:\left|-3x\right|+5=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(C)}\:\sqrt{-x}-2=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\:\sqrt{x}-8=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\:\left|-3x\right|-4=0 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> A rectangular region is bounded by the graphs of the equations &lt;math&gt;y=a, y=-b, x=-c,&lt;/math&gt; and &lt;math&gt;x=d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are all positive numbers. Which of the following represents the area of this region?<br /> <br /> &lt;math&gt; \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad&lt;/math&gt; &lt;math&gt;+bc-bd\qquad\textbf{(C)}\ ac+ad&lt;/math&gt; &lt;math&gt;-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad&lt;/math&gt; &lt;math&gt;+bc+bd\qquad\textbf{(E)}\ ac-ad&lt;/math&gt; &lt;math&gt; -bc+bd &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt; \textdollar 17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;\textbf{(A)}\,7 \qquad\textbf{(B)}\,11 \qquad\textbf{(C)}\,17 \qquad\textbf{(D)}\,23 \qquad\textbf{(E)}\,77&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Square &lt;math&gt;EFGH&lt;/math&gt; has one vertex on each side of square &lt;math&gt;ABCD&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AE=7\cdot EB&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;EFGH&lt;/math&gt; to the area of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\,\frac{49}{64} \qquad\textbf{(B)}\,\frac{25}{32} \qquad\textbf{(C)}\,\frac78 \qquad\textbf{(D)}\,\frac{5\sqrt{2}}{8} \qquad\textbf{(E)}\,\frac{\sqrt{14}}{4} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br /> <br /> &lt;math&gt;\textbf{(A)}\,13 \qquad\textbf{(B)}\,14 \qquad\textbf{(C)}\,15 \qquad\textbf{(D)}\,16 \qquad\textbf{(E)}\,17&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br /> <br /> &lt;math&gt;\textbf{(A)}\,12 \qquad\textbf{(B)}\,20 \qquad\textbf{(C)}\,72 \qquad\textbf{(D)}\,120 \qquad\textbf{(E)}\,200&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br /> <br /> &lt;math&gt;\textbf{(A)}\,\frac{1}{36} \qquad\textbf{(B)}\,\frac{1}{12} \qquad\textbf{(C)}\,\frac{1}{6} \qquad\textbf{(D)}\,\frac{1}{4} \qquad\textbf{(E)}\,\frac{5}{18}&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br /> <br /> &lt;math&gt;\textbf{(A)}\,140 \qquad\textbf{(B)}\,240 \qquad\textbf{(C)}\,440 \qquad\textbf{(D)}\,640 \qquad\textbf{(E)}\,840&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Which of the following is equal to &lt;math&gt;\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\,3\sqrt2 \qquad\textbf{(B)}\,2\sqrt6 \qquad\textbf{(C)}\,\frac{7\sqrt2}{2} \qquad\textbf{(D)}\,3\sqrt3 \qquad\textbf{(E)}\,6&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> In the eight-term sequence &lt;math&gt;A,B,C,D,E,F,G,H&lt;/math&gt;, the value of &lt;math&gt;C&lt;/math&gt; is 5 and the sum of any three consecutive terms is 30. What is &lt;math&gt;A+H&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\,17 \qquad\textbf{(B)}\,18 \qquad\textbf{(C)}\,25 \qquad\textbf{(D)}\,26 \qquad\textbf{(E)}\,43&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Circles &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; each have radius 1. Circles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share one point of tangency. Circle &lt;math&gt;C&lt;/math&gt; has a point of tangency with the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;. What is the area inside Circle &lt;math&gt;C&lt;/math&gt; but outside circle &lt;math&gt;A&lt;/math&gt; and circle &lt;math&gt;B&lt;/math&gt; ?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad<br /> \textbf{(B)}\ \frac{\pi}{2} \qquad<br /> \textbf{(C)}\ 2 \qquad<br /> \textbf{(D)}\ \frac{3\pi}{4} \qquad<br /> \textbf{(E)}\ 1+\frac{\pi}{2} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> Two points on the circumference of a circle of radius &lt;math&gt;r&lt;/math&gt; are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> Each vertex of convex pentagon &lt;math&gt;ABCDE&lt;/math&gt; is to be assigned a color. There are &lt;math&gt;6&lt;/math&gt; colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> Seven students count from 1 to 1000 as follows:<br /> <br /> •Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br /> <br /> •Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> •Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> •Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br /> <br /> •Finally, George says the only number that no one else says.<br /> <br /> What number does George say?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> Let &lt;math&gt;R&lt;/math&gt; be a square region and &lt;math&gt;n\ge4&lt;/math&gt; an integer. A point &lt;math&gt;X&lt;/math&gt; in the interior of &lt;math&gt;R&lt;/math&gt; is called &lt;math&gt;n\text{-}ray&lt;/math&gt; partitional if there are &lt;math&gt;n&lt;/math&gt; rays emanating from &lt;math&gt;X&lt;/math&gt; that divide &lt;math&gt;R&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br /> <br /> &lt;math&gt;\textbf{(A)}\,1500 \qquad\textbf{(B)}\,1560 \qquad\textbf{(C)}\,2320 \qquad\textbf{(D)}\,2480 \qquad\textbf{(E)}\,2500&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2011|ab=A|before=[[2010 AMC 10B Problems]]|after=[[2011 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=113466 2011 AMC 10A Problems 2019-12-26T19:39:52Z <p>Xmidnightfirex: /* Problem 1 */</p> <hr /> <div>{{AMC10 Problems|year=2011|ab=A}}<br /> <br /> == Problem 1 ==<br /> <br /> A cell phone plan costs &lt;math&gt;\textdollar 20&lt;/math&gt; each month, plus &lt;math&gt;5&lt;/math&gt;¢ per text message sent, plus &lt;math&gt;10&lt;/math&gt;¢ for each minute used over &lt;math&gt;30&lt;/math&gt; hours. In January Juan sent &lt;math&gt;100&lt;/math&gt; text messages and talked for &lt;math&gt;30.5&lt;/math&gt; hours. How much did he have to pay?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \textdollar 24.00 \qquad\textbf{(B)}\ \textdollar 24.50 \qquad\textbf{(C)}\ \textdollar 25.50\qquad\textbf{(D)}\ \textdollar 28.00\qquad\textbf{(E)}\ \textdollar 30.00 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Suppose &lt;math&gt;[a\ b]&lt;/math&gt; denotes the average of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;\{a\ b\ c\}&lt;/math&gt; denotes the average of &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. What is &lt;math&gt;\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Let &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; be the following sums of arithmetic sequences: &lt;cmath&gt; \begin{eqnarray*} X &amp;=&amp; 10 + 12 + 14 + \cdots + 100, \\ Y &amp;=&amp; 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} &lt;/cmath&gt; What is the value of &lt;math&gt;Y - X&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 &lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of &lt;math&gt;12&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt; minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> Set &lt;math&gt;A &lt;/math&gt; has 20 elements, and set &lt;math&gt;B &lt;/math&gt; has 15 elements. What is the smallest possible number of elements in &lt;math&gt;A \cup B &lt;/math&gt;, the union of &lt;math&gt;A &lt;/math&gt; and &lt;math&gt;B &lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Which of the following equations does NOT have a solution?<br /> <br /> &lt;math&gt;\textbf{(A)}\:(x+7)^2=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(B)}\:\left|-3x\right|+5=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(C)}\:\sqrt{-x}-2=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\:\sqrt{x}-8=0&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\:\left|-3x\right|-4=0 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> A rectangular region is bounded by the graphs of the equations &lt;math&gt;y=a, y=-b, x=-c,&lt;/math&gt; and &lt;math&gt;x=d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are all positive numbers. Which of the following represents the area of this region?<br /> <br /> &lt;math&gt; \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad&lt;/math&gt; &lt;math&gt;+bc-bd\qquad\textbf{(C)}\ ac+ad&lt;/math&gt; &lt;math&gt;-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad&lt;/math&gt; &lt;math&gt;+bc+bd\qquad\textbf{(E)}\ ac-ad&lt;/math&gt; &lt;math&gt; -bc+bd &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt; \textdollar 17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;\textbf{(A)}\,7 \qquad\textbf{(B)}\,11 \qquad\textbf{(C)}\,17 \qquad\textbf{(D)}\,23 \qquad\textbf{(E)}\,77&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Square &lt;math&gt;EFGH&lt;/math&gt; has one vertex on each side of square &lt;math&gt;ABCD&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AE=7\cdot EB&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;EFGH&lt;/math&gt; to the area of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\,\frac{49}{64} \qquad\textbf{(B)}\,\frac{25}{32} \qquad\textbf{(C)}\,\frac78 \qquad\textbf{(D)}\,\frac{5\sqrt{2}}{8} \qquad\textbf{(E)}\,\frac{\sqrt{14}}{4} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br /> <br /> &lt;math&gt;\textbf{(A)}\,13 \qquad\textbf{(B)}\,14 \qquad\textbf{(C)}\,15 \qquad\textbf{(D)}\,16 \qquad\textbf{(E)}\,17&lt;/math&gt;<br /> <br /> <br /> [[2011 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br /> <br /> &lt;math&gt;\textbf{(A)}\,12 \qquad\textbf{(B)}\,20 \qquad\textbf{(C)}\,72 \qquad\textbf{(D)}\,120 \qquad\textbf{(E)}\,200&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br /> <br /> &lt;math&gt;\textbf{(A)}\,\frac{1}{36} \qquad\textbf{(B)}\,\frac{1}{12} \qquad\textbf{(C)}\,\frac{1}{6} \qquad\textbf{(D)}\,\frac{1}{4} \qquad\textbf{(E)}\,\frac{5}{18}&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br /> <br /> &lt;math&gt;\textbf{(A)}\,140 \qquad\textbf{(B)}\,240 \qquad\textbf{(C)}\,440 \qquad\textbf{(D)}\,640 \qquad\textbf{(E)}\,840&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Which of the following is equal to &lt;math&gt;\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\,3\sqrt2 \qquad\textbf{(B)}\,2\sqrt6 \qquad\textbf{(C)}\,\frac{7\sqrt2}{2} \qquad\textbf{(D)}\,3\sqrt3 \qquad\textbf{(E)}\,6&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> In the eight-term sequence &lt;math&gt;A,B,C,D,E,F,G,H&lt;/math&gt;, the value of &lt;math&gt;C&lt;/math&gt; is 5 and the sum of any three consecutive terms is 30. What is &lt;math&gt;A+H&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\,17 \qquad\textbf{(B)}\,18 \qquad\textbf{(C)}\,25 \qquad\textbf{(D)}\,26 \qquad\textbf{(E)}\,43&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Circles &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; each have radius 1. Circles &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share one point of tangency. Circle &lt;math&gt;C&lt;/math&gt; has a point of tangency with the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;. What is the area inside Circle &lt;math&gt;C&lt;/math&gt; but outside circle &lt;math&gt;A&lt;/math&gt; and circle &lt;math&gt;B&lt;/math&gt; ?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad<br /> \textbf{(B)}\ \frac{\pi}{2} \qquad<br /> \textbf{(C)}\ 2 \qquad<br /> \textbf{(D)}\ \frac{3\pi}{4} \qquad<br /> \textbf{(E)}\ 1+\frac{\pi}{2} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> Two points on the circumference of a circle of radius &lt;math&gt;r&lt;/math&gt; are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> Each vertex of convex pentagon &lt;math&gt;ABCDE&lt;/math&gt; is to be assigned a color. There are &lt;math&gt;6&lt;/math&gt; colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> Seven students count from 1 to 1000 as follows:<br /> <br /> •Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br /> <br /> •Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> •Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br /> <br /> •Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br /> <br /> •Finally, George says the only number that no one else says.<br /> <br /> What number does George say?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} &lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> Let &lt;math&gt;R&lt;/math&gt; be a square region and &lt;math&gt;n\ge4&lt;/math&gt; an integer. A point &lt;math&gt;X&lt;/math&gt; in the interior of &lt;math&gt;R&lt;/math&gt; is called &lt;math&gt;n\text{-}ray&lt;/math&gt; partitional if there are &lt;math&gt;n&lt;/math&gt; rays emanating from &lt;math&gt;X&lt;/math&gt; that divide &lt;math&gt;R&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br /> <br /> &lt;math&gt;\textbf{(A)}\,1500 \qquad\textbf{(B)}\,1560 \qquad\textbf{(C)}\,2320 \qquad\textbf{(D)}\,2480 \qquad\textbf{(E)}\,2500&lt;/math&gt;<br /> <br /> [[2011 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2011|ab=A|before=[[2010 AMC 10B Problems]]|after=[[2011 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_25&diff=113461 2010 AMC 10A Problems/Problem 25 2019-12-26T18:10:37Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Jim starts with a positive integer &lt;math&gt;n&lt;/math&gt; and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with &lt;math&gt;n = 55&lt;/math&gt;, then his sequence contains &lt;math&gt;5&lt;/math&gt; numbers:<br /> <br /> <br /> &lt;cmath&gt;\begin{array}{ccccc}<br /> {}&amp;{}&amp;{}&amp;{}&amp;55\\<br /> 55&amp;-&amp;7^2&amp;=&amp;6\\<br /> 6&amp;-&amp;2^2&amp;=&amp;2\\<br /> 2&amp;-&amp;1^2&amp;=&amp;1\\<br /> 1&amp;-&amp;1^2&amp;=&amp;0\\<br /> \end{array}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;N&lt;/math&gt; be the smallest number for which Jim’s sequence has &lt;math&gt;8&lt;/math&gt; numbers. What is the units digit of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 1<br /> \qquad<br /> \mathrm{(B)}\ 3<br /> \qquad<br /> \mathrm{(C)}\ 5<br /> \qquad<br /> \mathrm{(D)}\ 7<br /> \qquad<br /> \mathrm{(E)}\ 9<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can find the answer by working backwards. We begin with &lt;math&gt;1-1^2=0&lt;/math&gt; on the bottom row, then the &lt;math&gt;1&lt;/math&gt; goes to the right of the equal's sign in the row above. We find the smallest value &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;x-1^2=1&lt;/math&gt; and &lt;math&gt;x&gt;1^2&lt;/math&gt;, which is &lt;math&gt;x=2&lt;/math&gt;.<br /> <br /> We repeat the same procedure except with &lt;math&gt;x-1^2=1&lt;/math&gt; for the next row and &lt;math&gt;x-1^2=2&lt;/math&gt; for the row after that. However, at the fourth row, we see that solving &lt;math&gt;x-1^2=3&lt;/math&gt; yields &lt;math&gt;x=4&lt;/math&gt;, in which case it would be incorrect since &lt;math&gt;1^2=1&lt;/math&gt; is not the greatest perfect square less than or equal to &lt;math&gt;x&lt;/math&gt; . So we make it a &lt;math&gt;2^2&lt;/math&gt; and solve &lt;math&gt;x-2^2=3&lt;/math&gt;. We continue on using this same method where we increase the perfect square until &lt;math&gt;x&lt;/math&gt; can be made bigger than it. When we repeat this until we have &lt;math&gt;8&lt;/math&gt; rows, we get:<br /> <br /> &lt;cmath&gt; \begin{array}{ccccc}{}&amp;{}&amp;{}&amp;{}&amp;7223\\ 7223&amp;-&amp;84^{2}&amp;=&amp;167\\ 167&amp;-&amp;12^{2}&amp;=&amp;23\\ 23&amp;-&amp;4^{2}&amp;=&amp;7\\ 7&amp;-&amp;2^{2}&amp;=&amp;3\\ 3&amp;-&amp;1^{2}&amp;=&amp;2\\ 2&amp;-&amp;1^{2}&amp;=&amp;1\\ 1&amp;-&amp;1^{2}&amp;=&amp;0\\ \end{array} &lt;/cmath&gt;<br /> <br /> Hence the solution is the last digit of &lt;math&gt;7223&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)}\ 3}&lt;/math&gt;.<br /> <br /> Note: We can go up to &lt;math&gt;167&lt;/math&gt;, and then notice the pattern of units digits alternating between &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;, so we do not need to calculate &lt;math&gt;7223&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Notice that to get the previous term, we must add the smallest square number, (let's call it &lt;math&gt;n^2&lt;/math&gt;) such that the sum is less than &lt;math&gt;(n+1)^2&lt;/math&gt;. Otherwise, instead of subtracting &lt;math&gt;n^2&lt;/math&gt; from the previous term, we're subtracting a greater square number. <br /> <br /> <br /> Remember that &lt;math&gt;(x+1)^2 = x^2 + x + (x+1)&lt;/math&gt;. Recall that to find the previous term, we must add a square number such that it is less than the next square number. &lt;math&gt;a + n^2 &lt; (n+1)^2&lt;/math&gt;. For this to be true, &lt;math&gt;a &lt; n + (n+1)&lt;/math&gt;. What that means is that given a term &lt;math&gt;a&lt;/math&gt;, we can find the previous term by adding &lt;math&gt;n^2&lt;/math&gt; where &lt;math&gt;n &gt; \frac {a-1}{2}&lt;/math&gt;. <br /> <br /> <br /> For example, to find the term that precedes &lt;math&gt;167&lt;/math&gt;, we know that &lt;math&gt;n&gt;166/2 = 83&lt;/math&gt;. Therefore, &lt;math&gt;n=84&lt;/math&gt; and the previous term is &lt;math&gt;167 + 84^2 = 7223&lt;/math&gt;. The last digit of &lt;math&gt;7223&lt;/math&gt; is &lt;math&gt;3 \Rightarrow \boxed{\textbf{(B)}\ 3}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=24|after=Last Question|ab=A}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=113322 2019 AMC 8 Problems/Problem 24 2019-12-24T00:40:12Z <p>Xmidnightfirex: /* Problem 24 */</p> <hr /> <div><br /> <br /> <br /> ==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{(B) 30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of 3. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of 3 also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)&lt;/math&gt;<br /> -Brudder<br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{(B) 30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6(Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{(B) 30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;Area&lt;/math&gt; &lt;math&gt;of&lt;/math&gt; &lt;math&gt;Triangle&lt;/math&gt; &lt;math&gt;XYZ&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = 30&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> The diagram is very inaccurate. <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M);<br /> label(&quot;$M$&quot;,M,S);<br /> label(&quot;$1$&quot;,A--DD,N);<br /> label(&quot;$2$&quot;,DD--C,N);<br /> label(&quot;$1$&quot;,EE--M,N);<br /> &lt;/asy&gt;<br /> Note: All numbers above &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EM}&lt;/math&gt; are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated. &lt;br /&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{B} \, 30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(F--D);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$x$&quot;, (D+E+F)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120+2x$&quot;, (D+F+C)/3);<br /> &lt;/asy&gt;<br /> Labeling the areas in the diagram, we have:&lt;br /&gt;<br /> &lt;math&gt;[DBC]=240=[BFE]+[FED]+[FDC]=x+x+120+2x=120+4x&lt;/math&gt; so &lt;math&gt;240=120+4x, 120=4x, 30=x&lt;/math&gt;.&lt;br /&gt;<br /> So our answer is &lt;math&gt;\boxed{\textbf{(B)} 30}&lt;/math&gt;. ~~RWhite<br /> ==Solution 10 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=111978 2019 AMC 8 Problems/Problem 3 2019-11-21T21:48:16Z <p>Xmidnightfirex: </p> <hr /> <div>==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider subtracting 1 from each of the fractions. Our new fractions would then be &lt;math&gt;\frac{4}{11}, \frac{4}{15},&lt;/math&gt; and &lt;math&gt;\frac{4}{13}&lt;/math&gt;. Since &lt;math&gt;\frac{4}{15}&lt;\frac{4}{13}&lt;\frac{4}{11}&lt;/math&gt;, it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;<br /> <br /> -will3145<br /> <br /> ==Solution 2==<br /> You could change everything to a common denominator, which eventually gives us an answer of &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8&diff=111977 2019 AMC 8 2019-11-21T21:45:48Z <p>Xmidnightfirex: </p> <hr /> <div>Can someone please add link to AMC 8 home page, thanks!<br /> <br /> *[[2019 AMC 8 Problems]]<br /> * [[2019 AMC 8 Answer Key]]<br /> ** [[2019 AMC 8 Problems/Problem 1]]<br /> ** [[2019 AMC 8 Problems/Problem 2]]<br /> ** [[2019 AMC 8 Problems/Problem 3]]<br /> ** [[2019 AMC 8 Problems/Problem 4]]<br /> ** [[2019 AMC 8 Problems/Problem 5]]<br /> ** [[2019 AMC 8 Problems/Problem 6]]<br /> ** [[2019 AMC 8 Problems/Problem 7]]<br /> ** [[2019 AMC 8 Problems/Problem 8]]<br /> ** [[2019 AMC 8 Problems/Problem 9]]<br /> ** [[2019 AMC 8 Problems/Problem 10]]<br /> ** [[2019 AMC 8 Problems/Problem 11]]<br /> ** [[2019 AMC 8 Problems/Problem 12]]<br /> ** [[2019 AMC 8 Problems/Problem 13]]<br /> ** [[2019 AMC 8 Problems/Problem 14]]<br /> ** [[2019 AMC 8 Problems/Problem 15]]<br /> ** [[2019 AMC 8 Problems/Problem 16]]<br /> ** [[2019 AMC 8 Problems/Problem 17]]<br /> ** [[2019 AMC 8 Problems/Problem 18]]<br /> ** [[2019 AMC 8 Problems/Problem 19]]<br /> ** [[2019 AMC 8 Problems/Problem 20]]<br /> ** [[2019 AMC 8 Problems/Problem 21]]<br /> ** [[2019 AMC 8 Problems/Problem 22]]<br /> ** [[2019 AMC 8 Problems/Problem 23]]<br /> ** [[2019 AMC 8 Problems/Problem 24]]<br /> ** [[2019 AMC 8 Problems/Problem 25]]<br /> <br /> == See also ==<br /> {{Succession box|<br /> |header=2019 AMC 8<br /> |before=[[2018 AMC 8]]<br /> |title=[[AMC 8]]<br /> |after=[[2020 AMC 8]]<br /> }}<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8&diff=111567 2019 AMC 8 2019-11-20T02:13:28Z <p>Xmidnightfirex: </p> <hr /> <div>PROBLEMS RELEASE<br /> <br /> Please?<br /> <br /> Hello?<br /> == See also ==<br /> {{Succession box|<br /> |header=2019 AMC 8<br /> |before=[[2018 AMC 8]]<br /> |title=[[AMC 8]]<br /> |after=[[2020 AMC 8]]<br /> }}<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8&diff=111558 2019 AMC 8 2019-11-19T12:41:59Z <p>Xmidnightfirex: </p> <hr /> <div>PROBLEMS RELEASE<br /> <br /> == See also ==<br /> {{Succession box|<br /> |header=2019 AMC 8<br /> |before=[[2018 AMC 8]]<br /> |title=[[AMC 8]]<br /> |after=[[2020 AMC 8]]<br /> }}<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8&diff=111513 2019 AMC 8 2019-11-18T00:24:46Z <p>Xmidnightfirex: </p> <hr /> <div>Test not posted yet.<br /> <br /> Do not discuss the 2019 AMC 8 until November 19, 2019.<br /> <br /> Almost there! Problems will be posted on the date above ^ ~:)<br /> |<br /> <br /> == See also ==<br /> {{Succession box|<br /> |header=2019 AMC 8<br /> |before=[[2018 AMC 8]]<br /> |title=[[AMC 8]]<br /> |after=[[2020 AMC 8]]<br /> }}<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8&diff=111487 2019 AMC 8 2019-11-17T14:16:55Z <p>Xmidnightfirex: </p> <hr /> <div>Test not posted yet.<br /> <br /> Do not discuss the 2019 AMC 8 until November 19, 2019.<br /> <br /> Two more days!<br /> <br /> == See also ==<br /> {{Succession box|<br /> |header=2019 AMC 8<br /> |before=[[2018 AMC 8]]<br /> |title=[[AMC 8]]<br /> |after=[[2020 AMC 8]]<br /> }}<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1993_UNCO_Math_Contest_II_Problems/Problem_4&diff=110605 1993 UNCO Math Contest II Problems/Problem 4 2019-10-26T14:43:40Z <p>Xmidnightfirex: /* Problem */</p> <hr /> <div>== Problem ==<br /> <br /> The table gives some of the straight line distances between certain pairs of cities. for example the distance<br /> between city &lt;math&gt;A&lt;/math&gt; and city &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;34.&lt;/math&gt; Use the given data to determine the distance between city &lt;math&gt;A&lt;/math&gt; and city &lt;math&gt;C&lt;/math&gt;.<br /> (Hint: a problem in the first round was similar in spirit to this one.)<br /> &lt;cmath&gt;\begin{tabular}{c|cccc}<br /> &amp; A &amp; B &amp; C &amp; D \\ \hline<br /> A &amp; 34 &amp; &amp; 16 \\ <br /> B &amp; &amp; 42 &amp; \\ <br /> C &amp; &amp; &amp; 12\\ <br /> D &amp; 30 &amp; &amp; \\ <br /> \end{tabular}<br /> &lt;/cmath&gt;<br /> <br /> == Solution ==<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=1993|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1993_UNCO_Math_Contest_II_Problems/Problem_8&diff=110580 1993 UNCO Math Contest II Problems/Problem 8 2019-10-25T02:30:32Z <p>Xmidnightfirex: </p> <hr /> <div>== Problem ==<br /> <br /> For what integer value of &lt;math&gt;n&lt;/math&gt; is the expression<br /> &lt;cmath&gt;\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{n}+\sqrt{n+1}}&lt;/cmath&gt;<br /> equal to &lt;math&gt;7&lt;/math&gt; ? (Hint: &lt;math&gt;(1+\sqrt{2})(1-\sqrt{2})=-1.&lt;/math&gt;)<br /> <br /> <br /> == Solution ==<br /> <br /> Looking at the first fraction, we get that &lt;math&gt;\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}&lt;/math&gt;. Moving on, we see an interesting pattern in which the fraction &lt;math&gt;\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n})&lt;/math&gt;. This means that we can rewrite the fractions as &lt;math&gt;(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n}&lt;/math&gt;. We can cancel out most of the terms in that sequence and get &lt;math&gt;\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1&lt;/math&gt;. However, we need to solve for &lt;math&gt;n&lt;/math&gt; now. Setting the previous expression equal to &lt;math&gt;7&lt;/math&gt;, we get that &lt;math&gt;\sqrt{n+1} = 8&lt;/math&gt;. Squaring both sides, we get that &lt;math&gt;n+1 = 64&lt;/math&gt;. Hence, &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;\boxed{63}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=1993|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=1993_UNCO_Math_Contest_II_Problems/Problem_8&diff=110579 1993 UNCO Math Contest II Problems/Problem 8 2019-10-25T02:29:23Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For what integer value of &lt;math&gt;n&lt;/math&gt; is the expression<br /> &lt;cmath&gt;\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{n}+\sqrt{n+1}}&lt;/cmath&gt;<br /> equal to &lt;math&gt;7&lt;/math&gt; ? (Hint: &lt;math&gt;(1+\sqrt{2})(1-\sqrt{2})=-1.&lt;/math&gt;)<br /> <br /> <br /> == Solution ==<br /> <br /> Looking at the first fraction, we get that &lt;math&gt;\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}&lt;/math&gt;. Moving on, we see an interesting pattern in which the fraction &lt;math&gt;\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}&lt;/math&gt;. This means that we can rewrite the fractions as &lt;math&gt;(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n}&lt;/math&gt;. We can cancel out most of the terms in that sequence and get &lt;math&gt;\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1&lt;/math&gt;. However, we need to solve for &lt;math&gt;n&lt;/math&gt; now. Setting the previous expression equal to &lt;math&gt;7&lt;/math&gt;, we get that &lt;math&gt;\sqrt{n+1} = 8&lt;/math&gt;. Squaring both sides, we get that &lt;math&gt;n+1 = 64&lt;/math&gt;. Hence, &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;\boxed{63}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=1993|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=LaTeX&diff=110489 LaTeX 2019-10-21T01:51:17Z <p>Xmidnightfirex: </p> <hr /> <div>{{Latex}}<br /> Admins, I figured out a way to edit anything in the wiki page, please pm me so I can explain to you how so this bug can hopefully be fixed. It is a very big bug ~Firebolt360<br /> <br /> Yes, it's quite a big bug. -owo<br /> <br /> The &lt;math&gt;\LaTeX&lt;/math&gt; typesetting system (pronounced &quot;Lay-Tek&quot; by most, or &quot;Lah-Tek&quot; by some) is widely used to produce well-formatted [[math|mathematical]] and scientific writing. With &lt;math&gt;\LaTeX&lt;/math&gt;, it is very easy to produce expressions like <br /> &lt;cmath&gt;<br /> \sqrt{\frac {a^2+b^2+c^2}3} \geq \frac {a+b+c}3 \geq \sqrt{abc} \geq \frac 3 { \frac 1a + \frac 1b + \frac 1c } .<br /> &lt;/cmath&gt; Nearly every serious student of math and science will use &lt;math&gt;\LaTeX&lt;/math&gt; frequently. Through these web pages, you will learn much of what you'll need to express math and science like a pro.<br /> <br /> * [http://www.artofproblemsolving.com/wiki/index.php/LaTeX:LaTeX_on_AoPS Click here] to start learning how to use &lt;math&gt;\LaTeX&lt;/math&gt; on AoPS<br /> {{Asymptote}}<br /> {{main|Asymptote}}<br /> '''Asymptote''' is a powerful vector graphics language designed for creating mathematical diagrams and figures. It can output images in either eps or pdf format and is compatible with the standard mathematics typesetting language, [[LaTeX]]. It is also a complete programming language and has cleaner syntax than its predecessor, [http://netlib.bell-labs.com/who/hobby/MetaPost.html MetaPost], which was a language used only for two-dimensional graphics.<br /> <br /> Here is an example of an image that can be produced using Asymptote:<br /> <br /> &lt;center&gt;[[Image:Figure1.jpg]]&lt;/center&gt;<br /> <br /> In a sense, Asymptote is the ruler and compass of typesetting.<br /> <br /> <br /> You can use Asymptote on the AoPSWiki right now, by enclosing the Asymptote code within &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;...&lt;/asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tags. For example, the following code<br /> &lt;pre&gt;&lt;nowiki&gt;&lt;asy&gt;<br /> draw((0,0)--(3,7),red);<br /> dot((0,0));<br /> dot((3,7));<br /> label(&quot;Produced with Asymptote &quot;+version.VERSION,point(S),2S);<br /> &lt;/asy&gt;&lt;/nowiki&gt;&lt;/pre&gt;<br /> created the picture <br /> &lt;center&gt;&lt;asy&gt;<br /> draw((0,0)--(3,7),red);<br /> dot((0,0));<br /> dot((3,7));<br /> label(&quot;Produced with Asymptote &quot;+version.VERSION,point(S),2S);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> And on the AoPS forums you can use &lt;tt&gt;&lt;nowiki&gt;[asy]..[/asy]&lt;/nowiki&gt;&lt;/tt&gt;<br /> <br /> Another example:<br /> <br /> &lt;pre&gt;&lt;nowiki&gt;[asy]<br /> pair A,B,C,X,Y,Z; <br /> A = (0,0);<br /> B = (1,0);<br /> C = (0.3,0.8);<br /> draw(A--B--C--A);<br /> X = (B+C)/2;<br /> Y = (A+C)/2;<br /> Z = (A+B)/2;<br /> draw(A--X, red);<br /> draw(B--Y,red);<br /> draw(C--Z,red);<br /> [/asy]&lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,X,Y,Z;<br /> A = (0,0);<br /> B = (1,0);<br /> C = (0.3,0.8);<br /> draw(A--B--C--A);<br /> X = (B+C)/2;<br /> Y = (A+C)/2;<br /> Z = (A+B)/2;<br /> draw(A--X, red);<br /> draw(B--Y,red);<br /> draw(C--Z,red);&lt;/asy&gt;<br /> <br /> Admins, I figured out a way to edit anything in the wiki page, please pm mem so I can explain to you how so this bug can hopefully be fixed. It is a very big bug ~Firebolt360</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2010_Problems/Problem_6&diff=105070 Mock AIME 2 2010 Problems/Problem 6 2019-04-01T02:27:10Z <p>Xmidnightfirex: Created page with &quot;...&quot;</p> <hr /> <div>...</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2011&diff=105069 Mock AIME 1 2011 2019-04-01T02:26:57Z <p>Xmidnightfirex: Created page with &quot;Why are you here?&quot;</p> <hr /> <div>Why are you here?</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&diff=103287 2019 AMC 10B Problems/Problem 20 2019-02-18T02:32:21Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}<br /> ==Problem==<br /> As shown in the figure, line segment &lt;math&gt;\overline{AD}&lt;/math&gt; is trisected by points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AB=BC=CD=2.&lt;/math&gt; Three semicircles of radius &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;\overarc{AEB},\overarc{BFC},&lt;/math&gt; and &lt;math&gt;\overarc{CGD},&lt;/math&gt; have their diameters on &lt;math&gt;\overline{AD},&lt;/math&gt; and are tangent to line &lt;math&gt;EG&lt;/math&gt; at &lt;math&gt;E,F,&lt;/math&gt; and &lt;math&gt;G,&lt;/math&gt; respectively. A circle of radius &lt;math&gt;2&lt;/math&gt; has its center on &lt;math&gt;F. &lt;/math&gt; The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form <br /> &lt;cmath&gt;\frac{a}{b}\cdot\pi-\sqrt{c}+d,&lt;/cmath&gt;<br /> where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), grey);<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17&lt;/math&gt;<br /> ==Solution==<br /> Divide the circle into four parts: The top semicircle: (A), the bottom sector with arc length 120 degrees: (B), the triangle formed by the radii of (A) and the chord: (C), and the four parts which are the corners of a circle inscribed in a square (D). The area is just (A) + (B) - (C) + (D).<br /> <br /> Area of (A): &lt;math&gt;2\pi&lt;/math&gt;<br /> <br /> Area of (B): &lt;math&gt;\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is &lt;math&gt;2\sqrt{3}*1/2=\sqrt{3}&lt;/math&gt;<br /> <br /> Area of (D): &lt;math&gt;4*1-1/4*\pi*4=4-\pi&lt;/math&gt;<br /> <br /> Total sum: &lt;math&gt;\frac{7\pi}{3}-\sqrt{3}+4&lt;/math&gt;<br /> <br /> &lt;math&gt;7+3+3+4=\boxed{17}&lt;/math&gt;<br /> <br /> For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region. <br /> - edited by IronicNinja<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=103077 2019 AMC 10B Problems/Problem 23 2019-02-16T21:32:06Z <p>Xmidnightfirex: /* See Also */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br /> <br /> ==Problem==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was &lt;math&gt;(x, 0)&lt;/math&gt;. Using Pythagorean Theorem gives &lt;math&gt;x=5&lt;/math&gt;.<br /> <br /> Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;(5, 0)&lt;/math&gt; form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br /> <br /> &lt;math&gt;2\sqrt{170}x = d * \sqrt{40}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; represents the distance between circle center and &lt;math&gt;(5, 0)&lt;/math&gt;. Therefore, &lt;math&gt;d = \sqrt{17}x&lt;/math&gt;. Using Pythagorean Theorem on &lt;math&gt;(5, 0)&lt;/math&gt;, either one of &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;, and the circle center, we realize that &lt;math&gt;170 + x^2 = 17x^2&lt;/math&gt;, at which point &lt;math&gt;x^2 = \frac{85}{8}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, follow solution 1 and obtain &lt;math&gt;x=5&lt;/math&gt;. Label the point &lt;math&gt;(5,0)&lt;/math&gt; as point &lt;math&gt;C&lt;/math&gt;. The midpoint &lt;math&gt;M&lt;/math&gt; of segment &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;(9, 12)&lt;/math&gt;. Notice that the center of the circle must lie on the line that goes through the points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt;. Thus, the center of the circle lies on the line &lt;math&gt;y=3x-15&lt;/math&gt;. <br /> <br /> Line &lt;math&gt;AC&lt;/math&gt; is &lt;math&gt;y=13x-65&lt;/math&gt;. The perpendicular line must pass through &lt;math&gt;A(6, 13)&lt;/math&gt; and &lt;math&gt;(x, 3x-15)&lt;/math&gt;. The slope of the perpendicular line is &lt;math&gt;-\frac{1}{13}&lt;/math&gt;. The line is hence &lt;math&gt;y=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. The point &lt;math&gt;(x, 3x-15)&lt;/math&gt; lies on this line. Therefore, &lt;math&gt;3x-15=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. Solving this equation tells us that &lt;math&gt;x=\frac{37}{4}&lt;/math&gt;. So the center of the circle is &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;. The distance between the center, &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;, and point A is &lt;math&gt;\frac{\sqrt{170}}{4}&lt;/math&gt;. Hence, the area is &lt;math&gt;\frac{85}{8}\pi&lt;/math&gt;. The answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The mid point of &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;D(9,12)&lt;/math&gt;. Let the tangent lines at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at &lt;math&gt;C(a,0)&lt;/math&gt; on the &lt;math&gt;X&lt;/math&gt; axis. Then &lt;math&gt;CD&lt;/math&gt; would be the perpendicular bisector of &lt;math&gt;AB&lt;/math&gt;. Let the center of circle be O. Then &lt;math&gt;\triangle AOC&lt;/math&gt; is similar to &lt;math&gt;\triangle DAC&lt;/math&gt;, that is &lt;math&gt;\frac{OA}{AC} = \frac{AD}{DC}.&lt;/math&gt;<br /> The slope of &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;\frac{13-11}{6-12}=\frac{-1}{3}&lt;/math&gt;, therefore the slope of CD will be 3. The equation of &lt;math&gt;CD&lt;/math&gt; is &lt;math&gt;y-12=3*(x-9)&lt;/math&gt;, that is &lt;math&gt;y=3x-15&lt;/math&gt;. Let &lt;math&gt;y=0&lt;/math&gt;. Then we have &lt;math&gt;x=5&lt;/math&gt;, which is the &lt;math&gt;x&lt;/math&gt; coordinate of &lt;math&gt;C(5,0)&lt;/math&gt;.<br /> <br /> &lt;math&gt;AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}&lt;/math&gt;,<br /> &lt;math&gt;AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}&lt;/math&gt;,<br /> &lt;math&gt;DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}&lt;/math&gt;,<br /> Therefore &lt;math&gt;OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}&lt;/math&gt;,<br /> Consequently, the area of the circle is &lt;math&gt;\pi\cdot OA^2 = \pi\cdot\frac{85}{8}&lt;/math&gt;.<br /> (by Zhen Qin)<br /> (P.S. Will someone please Latex this?)<br /> (&lt;math&gt;\LaTeX&lt;/math&gt;ed by a pewdiepie subscriber)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=102930 2019 AMC 10B Problems/Problem 25 2019-02-15T16:28:02Z <p>Xmidnightfirex: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br /> <br /> ==Problem==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> ==Solution 1 (Recursion)==<br /> We can deduce that any valid sequence of length &lt;math&gt;n&lt;/math&gt; will start with a 0 followed by either &quot;10&quot; or &quot;110&quot;.<br /> Because of this, we can define a recursive function:<br /> <br /> &lt;math&gt;f(n) = f(n-3) + f(n-2)&lt;/math&gt;<br /> <br /> This is because for any valid sequence of length &lt;math&gt;n&lt;/math&gt;, you can append either &quot;10&quot; or &quot;110&quot; and the resulting sequence would still satisfy the given conditions.<br /> <br /> &lt;math&gt;f(5) = 1&lt;/math&gt; and &lt;math&gt;f(6) = 2&lt;/math&gt;, so you follow the recursion up until &lt;math&gt;f(19) = 65 \quad \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 2 (Casework)==<br /> After any given zero, the next zero must appear exactly two or three spots down the line. And we started at position 1 and ended at position 19, so we moved over 18. Therefore, we must add a series of 2's and 3's to get 18. How can we do this?<br /> <br /> Option 1: nine 2's (there is only 1 way to arrange this).<br /> <br /> Option 2: two 3's and six 2's (&lt;math&gt;{8\choose2} =28&lt;/math&gt; ways to arrange this).<br /> <br /> Option 3: four 3's and three 2's (&lt;math&gt;{7\choose3}=35&lt;/math&gt; ways to arrange this).<br /> <br /> Option 4: six 3's (there is only 1 way to arrange this).<br /> <br /> Sum the four numbers given above: 1+28+35+1=65<br /> <br /> ==Solution3==<br /> first have 1 choose<br /> second have 1 choose<br /> 2 2 3 4 5 7 9 12 16 21 28 37 49 86 65 last is 0<br /> at the last we see the answer is 65(C)<br /> (Can someone please edit this?)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=102929 2019 AMC 10B Problems/Problem 25 2019-02-15T16:27:17Z <p>Xmidnightfirex: /* Solution 2 (Casework) */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br /> <br /> ==Problem==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> ==Solution 1 (Recursion)==<br /> We can deduce that any valid sequence of length &lt;math&gt;n&lt;/math&gt; will start with a 0 followed by either &quot;10&quot; or &quot;110&quot;.<br /> Because of this, we can define a recursive function:<br /> <br /> &lt;math&gt;f(n) = f(n-3) + f(n-2)&lt;/math&gt;<br /> <br /> This is because for any valid sequence of length &lt;math&gt;n&lt;/math&gt;, you can append either &quot;10&quot; or &quot;110&quot; and the resulting sequence would still satisfy the given conditions.<br /> <br /> &lt;math&gt;f(5) = 1&lt;/math&gt; and &lt;math&gt;f(6) = 2&lt;/math&gt;, so you follow the recursion up until &lt;math&gt;f(19) = 65 \quad \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 2 (Casework)==<br /> After any given zero, the next zero must appear exactly two or three spots down the line. And we started at position 1 and ended at position 19, so we moved over 18. Therefore, we must add a series of 2's and 3's to get 18. How can we do this?<br /> <br /> Option 1: nine 2's (there is only 1 way to arrange this).<br /> <br /> Option 2: two 3's and six 2's (&lt;math&gt;{8\choose2} =28&lt;/math&gt; ways to arrange this).<br /> <br /> Option 3: four 3's and three 2's (&lt;math&gt;{7\choose3}=35&lt;/math&gt; ways to arrange this).<br /> <br /> Option 4: six 3's (there is only 1 way to arrange this).<br /> <br /> Sum the four numbers given above: 1+28+35+1=65<br /> <br /> ==Solution3==<br /> first have 1 choose<br /> second have 1 choose<br /> 2 2 3 4 5 7 9 12 16 21 28 37 49 86 65 last is 0<br /> at the last we see the answer is 65(C)<br /> ~Solution by rayfunmath<br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=102928 2019 AMC 10B Problems/Problem 25 2019-02-15T16:26:38Z <p>Xmidnightfirex: /* Solution 1 (Recursion) */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br /> <br /> ==Problem==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> ==Solution 1 (Recursion)==<br /> We can deduce that any valid sequence of length &lt;math&gt;n&lt;/math&gt; will start with a 0 followed by either &quot;10&quot; or &quot;110&quot;.<br /> Because of this, we can define a recursive function:<br /> <br /> &lt;math&gt;f(n) = f(n-3) + f(n-2)&lt;/math&gt;<br /> <br /> This is because for any valid sequence of length &lt;math&gt;n&lt;/math&gt;, you can append either &quot;10&quot; or &quot;110&quot; and the resulting sequence would still satisfy the given conditions.<br /> <br /> &lt;math&gt;f(5) = 1&lt;/math&gt; and &lt;math&gt;f(6) = 2&lt;/math&gt;, so you follow the recursion up until &lt;math&gt;f(19) = 65 \quad \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 2 (Casework)==<br /> After any given zero, the next zero must appear exactly two or three spots down the line. And we started at position 1 and ended at position 19, so we moved over 18. Therefore, we must add a series of 2's and 3's to get 18. How can we do this?<br /> <br /> Option 1: nine 2's (there is only 1 way to arrange this).<br /> <br /> Option 2: two 3's and six 2's (&lt;math&gt;{8\choose2} =28&lt;/math&gt; ways to arrange this).<br /> <br /> Option 3: four 3's and three 2's (&lt;math&gt;{7\choose3}=35&lt;/math&gt; ways to arrange this).<br /> <br /> Option 4: six 3's (there is only 1 way to arrange this).<br /> <br /> Sum the four numbers given above: 1+28+35+1=65<br /> <br /> ~Solution by mxnxn<br /> ==Solution3==<br /> first have 1 choose<br /> second have 1 choose<br /> 2 2 3 4 5 7 9 12 16 21 28 37 49 86 65 last is 0<br /> at the last we see the answer is 65(C)<br /> ~Solution by rayfunmath<br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_24&diff=102927 2019 AMC 10B Problems/Problem 24 2019-02-15T16:25:56Z <p>Xmidnightfirex: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #24]] and [[2019 AMC 12B Problems|2019 AMC 12B #22]]}}<br /> <br /> ==Problem==<br /> <br /> Define a sequence recursively by &lt;math&gt;x_0=5&lt;/math&gt; and &lt;cmath&gt;x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}&lt;/cmath&gt; for all nonnegative integers &lt;math&gt;n.&lt;/math&gt; Let &lt;math&gt;m&lt;/math&gt; be the least positive integer such that<br /> &lt;cmath&gt;x_m\leq 4+\frac{1}{2^{20}}.&lt;/cmath&gt;In which of the following intervals does &lt;math&gt;m&lt;/math&gt; lie?<br /> <br /> &lt;math&gt;\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We first prove that &lt;math&gt;x_n &gt; 4&lt;/math&gt; for all &lt;math&gt;n \ge 0&lt;/math&gt; by induction from <br /> &lt;cmath&gt;<br /> x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}<br /> &lt;/cmath&gt;<br /> and then prove &lt;math&gt;x_n&lt;/math&gt;'s are decreasing by<br /> &lt;cmath&gt;<br /> x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} &lt; 0<br /> &lt;/cmath&gt;<br /> Now we need to estimate the value of &lt;math&gt;x_{n+1}-4&lt;/math&gt; by<br /> &lt;cmath&gt;<br /> x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6} <br /> &lt;/cmath&gt;<br /> since &lt;math&gt;x_n&lt;/math&gt;'s are decreasing, &lt;math&gt;\frac{x_n + 5}{x_n+6}&lt;/math&gt; are also decreasing, so we have<br /> &lt;cmath&gt;<br /> \frac{9}{10} &lt; \frac{x_n + 5}{x_n+6} \le \frac{10}{11}<br /> &lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;<br /> \frac{9}{10}(x_n-4) &lt; x_{n+1} - 4 \le \frac{10}{11}(x_n-4)<br /> &lt;/cmath&gt;<br /> which leads to<br /> &lt;cmath&gt;<br /> (\frac{9}{10})^n = (\frac{9}{10})^n (x_0-4) &lt; x_{n} - 4 \le (\frac{10}{11})^n (x_0-4) = (\frac{10}{11})^n<br /> &lt;/cmath&gt;<br /> The problem requires us to find the value of &lt;math&gt;n&lt;/math&gt; such that<br /> &lt;cmath&gt;<br /> (\frac{9}{10})^n &lt; x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } <br /> (\frac{10}{11})^{n-1} &gt; x_{n-1} - 4 &gt; \frac{1}{2^{20}}<br /> &lt;/cmath&gt;<br /> using natural logarithm, we need<br /> &lt;math&gt;n \ln \frac{9}{10} &lt; -20 \ln 2&lt;/math&gt; and &lt;math&gt;(n-1)\ln \frac{10}{11} &gt; -20 \ln 2&lt;/math&gt;, or<br /> <br /> &lt;cmath&gt;<br /> n &gt; \frac{20\ln 2}{\ln\frac{10}{9}} \text{ and } n-1 &lt; \frac{20\ln 2}{\ln\frac{11}{10}}<br /> &lt;/cmath&gt;<br /> <br /> As estimations, &lt;math&gt;\ln\frac{10}{9} \approx 1/9&lt;/math&gt; and &lt;math&gt;\ln\frac{11}{10} \approx 1/10&lt;/math&gt;, &lt;math&gt;\ln 2\approx 0.7&lt;/math&gt;<br /> we can estimate that<br /> &lt;cmath&gt;<br /> 126 &lt; n &lt; 141<br /> &lt;/cmath&gt;<br /> Choose &lt;math&gt;\boxed{C}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> Making the reasonable assumption that &lt;math&gt;x_n&lt;/math&gt; approaches &lt;math&gt;4&lt;/math&gt;, we can translate &lt;math&gt;x&lt;/math&gt; down by &lt;math&gt;4&lt;/math&gt; to obtain a more simple sequence &lt;math&gt;a_n=x_n-4&lt;/math&gt; that should approach &lt;math&gt;0&lt;/math&gt;. <br /> <br /> <br /> Substitution of &lt;math&gt;(a_{n}+4)&lt;/math&gt; for &lt;math&gt;(x_{n})&lt;/math&gt; and &lt;math&gt;(a_{n+1}+4)&lt;/math&gt; for &lt;math&gt;(x_{n+1})&lt;/math&gt; in the definition of &lt;math&gt;x_{n+1}&lt;/math&gt; leads to<br /> <br /> <br /> &lt;math&gt;a_{n+1}+4 = \frac{(a_{n} + 8)(a_{n}+5)}{a_{n}+10} = \frac{a_{n}^2 + 13a_{n}+40}{a_{n}+10} \implies a_{n+1} = a_{n}(\frac{a_{n}+9}{a_{n}+10}) \implies \frac{a_{n+1}}{a_{n}} = \frac{a_{n}+9}{a_{n}+10}&lt;/math&gt;<br /> <br /> <br /> The ratio of consecutive terms is thus always positive and less than 1 (because &lt;math&gt;a_0&lt;/math&gt; is positive). This means that the largest possible value for &lt;math&gt;a_n&lt;/math&gt; is 1 and that no value of &lt;math&gt;a_n&lt;/math&gt; can be less than or equal to 0.<br /> <br /> <br /> Plugging the extrema of &lt;math&gt;a_n&lt;/math&gt; back into the ratio shows that &lt;math&gt;\frac{9}{10}&lt;\frac{a_{n+1}}{a_{n}}\leq\frac{10}{11}&lt;/math&gt; for all &lt;math&gt;n&lt;/math&gt;. <br /> <br /> <br /> For &lt;math&gt;(n&gt;0)&lt;/math&gt;, we can bound &lt;math&gt;a_{n}&lt;/math&gt; by applying this rule recursively : &lt;math&gt;(\frac{9}{10})^n &lt; a_{n} \leq (\frac{10}{11})^n&lt;/math&gt; <br /> <br /> <br /> <br /> Therefore, &lt;math&gt;a_{n}&lt;/math&gt; is always less than &lt;math&gt;(\frac{1}{2^{20}})&lt;/math&gt; when &lt;math&gt;(\frac{10}{11})^n&lt;\frac{1}{2^{20}}\implies n&gt;20log_{\frac{11}{10}}{2}&lt;/math&gt;<br /> <br /> and &lt;math&gt;a_{n}&lt;/math&gt; is never less than &lt;math&gt;(\frac{1}{2^{20}})&lt;/math&gt; when &lt;math&gt;(\frac{9}{10})^n&gt;\frac{1}{2^{20}}\implies n&lt;20log_{\frac{10}{9}}{2}&lt;/math&gt;<br /> <br /> <br /> <br /> The first integer &lt;math&gt;m&lt;/math&gt; such that &lt;math&gt;a_{m}\leq\frac{1}{2^{20}}&lt;/math&gt; must therefore lie in the interval <br /> &lt;math&gt;\left[\left\lfloor 20log_{\frac{10}{9}}{2}\right\rfloor+1, \left\lceil 20log_{\frac{11}{10}}{2}\right\rceil\right]&lt;/math&gt;<br /> <br /> Both of these can be quickly estimated at c. &lt;math&gt;140&lt;/math&gt;, so the answer must be &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> (actual values are &lt;math&gt;132&lt;/math&gt; and &lt;math&gt;147&lt;/math&gt;)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2019|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=102926 2019 AMC 10B Problems/Problem 23 2019-02-15T16:25:24Z <p>Xmidnightfirex: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br /> <br /> ==Problem==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was &lt;math&gt;(x, 0)&lt;/math&gt;. Using Pythagorean Theorem gives &lt;math&gt;x=5&lt;/math&gt;.<br /> <br /> Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;(5, 0)&lt;/math&gt; form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br /> <br /> &lt;math&gt;2\sqrt{170}x = d * \sqrt{40}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; represents the distance between circle center and &lt;math&gt;(5, 0)&lt;/math&gt;. Therefore, &lt;math&gt;d = \sqrt{17}x&lt;/math&gt;. Using Pythagorean Theorem on &lt;math&gt;(5, 0)&lt;/math&gt;, either one of &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;, and the circle center, we realize that &lt;math&gt;170 + x^2 = 17x^2&lt;/math&gt;, at which point &lt;math&gt;x^2 = \frac{85}{8}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, follow solution 1 and obtain &lt;math&gt;x=5&lt;/math&gt;. Label the point &lt;math&gt;(5,0)&lt;/math&gt; as point &lt;math&gt;C&lt;/math&gt;. The midpoint &lt;math&gt;M&lt;/math&gt; of segment &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;(9, 12)&lt;/math&gt;. Notice that the center of the circle must lie on the line that goes through the points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt;. Thus, the center of the circle lies on the line &lt;math&gt;y=3x-15&lt;/math&gt;. <br /> <br /> Line &lt;math&gt;AC&lt;/math&gt; is &lt;math&gt;y=13x-65&lt;/math&gt;. The perpendicular line must pass through &lt;math&gt;A(6, 13)&lt;/math&gt; and &lt;math&gt;(x, 3x-15)&lt;/math&gt;. The slope of the perpendicular line is &lt;math&gt;-\frac{1}{13}&lt;/math&gt;. The line is hence &lt;math&gt;y=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. The point &lt;math&gt;(x, 3x-15)&lt;/math&gt; lies on this line. Therefore, &lt;math&gt;3x-15=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. Solving this equation tells us that &lt;math&gt;x=\frac{37}{4}&lt;/math&gt;. So the center of the circle is &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;. The distance between the center, &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;, and point A is &lt;math&gt;\frac{\sqrt{170}}{4}&lt;/math&gt;. Hence, the area is &lt;math&gt;\frac{85}{8}\pi&lt;/math&gt;. The answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC.<br /> The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)<br /> <br /> AC=sqrt((6-5)^2+(13-0)^2)=sqrt(170)<br /> AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10)<br /> DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160)<br /> Therefore OA=AC*AD/DC=sqrt(85/5)<br /> Consequently, the area of the circle is pi*OA^2=pi*85/5<br /> (by Zhen Qin)<br /> (P.S. Will someone please Latex this?)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=102925 2019 AMC 10B Problems/Problem 23 2019-02-15T16:24:31Z <p>Xmidnightfirex: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #23]] and [[2019 AMC 12B Problems|2019 AMC 12B #20]]}}<br /> <br /> ==Problem==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was &lt;math&gt;(x, 0)&lt;/math&gt;. Using Pythagorean Theorem gives &lt;math&gt;x=5&lt;/math&gt;.<br /> <br /> Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;(5, 0)&lt;/math&gt; form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:<br /> <br /> &lt;math&gt;2\sqrt{170}x = d * \sqrt{40}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; represents the distance between circle center and &lt;math&gt;(5, 0)&lt;/math&gt;. Therefore, &lt;math&gt;d = \sqrt{17}x&lt;/math&gt;. Using Pythagorean Theorem on &lt;math&gt;(5, 0)&lt;/math&gt;, either one of &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt;, and the circle center, we realize that &lt;math&gt;170 + x^2 = 17x^2&lt;/math&gt;, at which point &lt;math&gt;x^2 = \frac{85}{8}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First, follow solution 1 and obtain &lt;math&gt;x=5&lt;/math&gt;. Label the point &lt;math&gt;(5,0)&lt;/math&gt; as point &lt;math&gt;C&lt;/math&gt;. The midpoint &lt;math&gt;M&lt;/math&gt; of segment &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;(9, 12)&lt;/math&gt;. Notice that the center of the circle must lie on the line that goes through the points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt;. Thus, the center of the circle lies on the line &lt;math&gt;y=3x-15&lt;/math&gt;. <br /> <br /> Line &lt;math&gt;AC&lt;/math&gt; is &lt;math&gt;y=13x-65&lt;/math&gt;. The perpendicular line must pass through &lt;math&gt;A(6, 13)&lt;/math&gt; and &lt;math&gt;(x, 3x-15)&lt;/math&gt;. The slope of the perpendicular line is &lt;math&gt;-\frac{1}{13}&lt;/math&gt;. The line is hence &lt;math&gt;y=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. The point &lt;math&gt;(x, 3x-15)&lt;/math&gt; lies on this line. Therefore, &lt;math&gt;3x-15=-\frac{x}{13}+\frac{175}{13}&lt;/math&gt;. Solving this equation tells us that &lt;math&gt;x=\frac{37}{4}&lt;/math&gt;. So the center of the circle is &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;. The distance between the center, &lt;math&gt;(\frac{37}{4}, \frac{51}{4})&lt;/math&gt;, and point A is &lt;math&gt;\frac{\sqrt{170}}{4}&lt;/math&gt;. Hence, the area is &lt;math&gt;\frac{85}{8}\pi&lt;/math&gt;. The answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{85}{8}\pi}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC is similiar to DAC, that is OA/AC=AD/DC.<br /> The slope of AB is (13-11)/(6-12)=-1/3, therefore the slope of CD will be 3. the equation of CD is y-12=3*(x-9), that is y=3x-15, let y=0, we have x=5, which is the x coordiante of C(5,0)<br /> <br /> AC=sqrt((6-5)^2+(13-0)^2)=sqrt(170)<br /> AD=sqrt((6-9)^2)+(13-12)^2)=sqrt(10)<br /> DC=sqrt((9-5)^2+(12-0)^2)=aqrt(160)<br /> Therefore OA=AC*AD/DC=sqrt(85/5)<br /> Consequently, the area of the circle is pi*OA^2=pi*85/5<br /> (by Zhen Qin)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_22&diff=102924 2019 AMC 10B Problems/Problem 22 2019-02-15T16:24:05Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #22]] and [[2019 AMC 12B Problems|2019 AMC 12B #19]]}}<br /> <br /> ==Problem==<br /> <br /> Raashan, Sylvia, and Ted play the following game. Each starts with &lt;math&gt; \$1&lt;/math&gt;. A bell rings every &lt;math&gt;15&lt;/math&gt; seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives &lt;math&gt;\$1&lt;/math&gt; to that player. What is the probability that after the bell has rung &lt;math&gt;2019&lt;/math&gt; times, each player will have &lt;math&gt;\$1&lt;/math&gt;? (For example, Raashan and Ted may each decide to give &lt;math&gt;\$1&lt;/math&gt; to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have &lt;math&gt;\$0&lt;/math&gt;, Sylvia will have &lt;math&gt;\$2&lt;/math&gt;, and Ted will have &lt;math&gt;\$1&lt;/math&gt;, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their &lt;math&gt; \$1&lt;/math&gt; to, and the holdings will be the same at the end of the second round.)<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> On the first turn, each player starts off with &lt;math&gt;\text{\$1}&lt;/math&gt; each. Each turn after that, there are only two situations possible: either everyone stays at &lt;math&gt;\text{\$1}&lt;/math&gt; &lt;math&gt;\text{(1-1-1)}&lt;/math&gt;, or the distribution of money becomes &lt;math&gt;\text{\$2-\$1-\$0}&lt;/math&gt;, in any order &lt;math&gt;\text{(2-1-0)}&lt;/math&gt;.<br /> <br /> (Note: &lt;math&gt;\text{S-T-R}&lt;/math&gt; means that &lt;math&gt;\text{R}&lt;/math&gt; gives his money to &lt;math&gt;\text{S}&lt;/math&gt;, &lt;math&gt;\text{S}&lt;/math&gt; gives her money to &lt;math&gt;\text{T}&lt;/math&gt;, and &lt;math&gt;\text{T}&lt;/math&gt; gives his money to &lt;math&gt;\text{R}&lt;/math&gt;.)<br /> <br /> From the &lt;math&gt;\text{1-1-1}&lt;/math&gt; state, there are two ways to distribute the money so that it stays in a &lt;math&gt;\text{1-1-1}&lt;/math&gt; state: &lt;math&gt;\text{S-T-R}&lt;/math&gt; and &lt;math&gt;\text{T-R-S}&lt;/math&gt;. There are 6 ways to change the state to &lt;math&gt;\text{2-1-0}&lt;/math&gt;: &lt;math&gt;\text{S-R-R}&lt;/math&gt;, &lt;math&gt;\text{T-R-R}&lt;/math&gt;, &lt;math&gt;\text{S-R-S}&lt;/math&gt;, &lt;math&gt;\text{S-T-S}&lt;/math&gt;, &lt;math&gt;\text{T-T-R}&lt;/math&gt;, and &lt;math&gt;\text{T-T-S}&lt;/math&gt;. This means that the probability that the state stays &lt;math&gt;\text{1-1-1}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{2}{8}=\frac{1}{4}&lt;/math&gt;, and the probability that the state changes to &lt;math&gt;\text{2-1-0}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{6}{8}=\frac{3}{4}&lt;/math&gt;.<br /> <br /> From the &lt;math&gt;\text{2-1-0}&lt;/math&gt; state, there is one way to change the state back to &lt;math&gt;\text{1-1-1}&lt;/math&gt;: &lt;math&gt;\text{S-T-0}&lt;/math&gt;. (We can assume that &lt;math&gt;\text{R}&lt;/math&gt; has &lt;math&gt;\text{\$2}&lt;/math&gt;, &lt;math&gt;\text{S}&lt;/math&gt; has &lt;math&gt;\text{\$1}&lt;/math&gt;, and &lt;math&gt;\text{T}&lt;/math&gt; has &lt;math&gt;\text{\$0}&lt;/math&gt; since only the distribution of money matters, not the specific people.) There are three ways to keep the &lt;math&gt;\text{2-1-0}&lt;/math&gt; state: &lt;math&gt;\text{S-R-0}&lt;/math&gt;, &lt;math&gt;\text{T-R-0}&lt;/math&gt;, &lt;math&gt;\text{T-T-0}&lt;/math&gt;. This means that the probability that the state changes to &lt;math&gt;\text{1-1-1}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{1}{4}&lt;/math&gt;, and the probability that the state stays &lt;math&gt;\text{2-1-0}&lt;/math&gt; is &lt;math&gt;\textstyle\frac{3}{4}&lt;/math&gt;.<br /> <br /> We can see that there will always be a &lt;math&gt;\textstyle\frac{1}{4}&lt;/math&gt; chance that the money is distributed &lt;math&gt;\text{1-1-1}&lt;/math&gt; (as long as the bell rings once), so the answer is &lt;math&gt;\boxed{\textbf{(B) }\frac{1}{4}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&diff=102923 2019 AMC 10B Problems/Problem 20 2019-02-15T16:23:42Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}<br /> ==Problem==<br /> As shown in the figure, line segment &lt;math&gt;\overline{AD}&lt;/math&gt; is trisected by points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AB=BC=CD=2.&lt;/math&gt; Three semicircles of radius &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;\overarc{AEB},\overarc{BFC},&lt;/math&gt; and &lt;math&gt;\overarc{CGD},&lt;/math&gt; have their diameters on &lt;math&gt;\overline{AD},&lt;/math&gt; and are tangent to line &lt;math&gt;EG&lt;/math&gt; at &lt;math&gt;E,F,&lt;/math&gt; and &lt;math&gt;G,&lt;/math&gt; respectively. A circle of radius &lt;math&gt;2&lt;/math&gt; has its center on &lt;math&gt;F. &lt;/math&gt; The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form <br /> &lt;cmath&gt;\frac{a}{b}\cdot\pi-\sqrt{c}+d,&lt;/cmath&gt;<br /> where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), grey);<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D\$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17&lt;/math&gt;<br /> ==Solution==<br /> Divide the circle into four parts: The top semicircle: (A), the bottom sector with arc length 120 degrees: (B), the triangle formed by the radii of (A) and the chord: (C), and the four parts which are the corners of a circle inscribed in a square (D). The area is just (A) + (B) - (C) + (D).<br /> <br /> Area of (A): &lt;math&gt;2\pi&lt;/math&gt;<br /> <br /> Area of (B): &lt;math&gt;\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> Area of (C): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is &lt;math&gt;2\sqrt{3}*1/2=\sqrt{3}&lt;/math&gt;<br /> <br /> Area of (D): &lt;math&gt;4*1-1/4*\pi*4=4-\pi&lt;/math&gt;<br /> <br /> Total sum: &lt;math&gt;\frac{7\pi}{3}-\sqrt{3}+4&lt;/math&gt;<br /> <br /> &lt;math&gt;7+3+3+4=\boxed{17}&lt;/math&gt;<br /> <br /> For this solution to be a tad more clear, we are finding the area of the sector in B of 120 degrees because the large circle radius is 2, and the short length (the radius of the semicircle) is 1, and so the triangle is a 30-60-90 triangle. In A, we find the top semicircle part, in B minus C, we find the area of the shaded region above the semicircles but below the diameter, and in D we find the bottom shaded region.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=102922 2019 AMC 10B Problems/Problem 19 2019-02-15T16:23:19Z <p>Xmidnightfirex: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br /> <br /> ==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all positive integer divisors of &lt;math&gt;100,000.&lt;/math&gt; How many numbers are the product of two distinct elements of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> To find the number of numbers that are the product of two distinct elements of &lt;math&gt;S&lt;/math&gt;, we first square &lt;math&gt;100,000&lt;/math&gt; and factor it. Factoring, we find &lt;math&gt;100,000^2 = 2^{10} \cdot 5^{10}&lt;/math&gt;. Therefore, &lt;math&gt;100,000^2&lt;/math&gt; has &lt;math&gt;(10 + 1)(10 + 1) = 121&lt;/math&gt; distinct factors. Each of these can be achieved by multiplying two factors of &lt;math&gt;S&lt;/math&gt;. However, the factors must be distinct, so we eliminate &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100,000^2&lt;/math&gt;, as well as &lt;math&gt;2^{10}&lt;/math&gt; and &lt;math&gt;5^{10}&lt;/math&gt;, so the answer is &lt;math&gt;121 - 4 = 117&lt;/math&gt;. &lt;math&gt;2^{10}&lt;/math&gt; and &lt;math&gt;5^{10}&lt;/math&gt; do not work since the factors chosen must be distinct, and those require &lt;math&gt;2^5 \cdot 2^5&lt;/math&gt; or &lt;math&gt;5^5 \cdot 5^5&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> The prime factorization of 100,000 is &lt;math&gt;2^5 \cdot 5^5&lt;/math&gt;. Thus, we choose two numbers &lt;math&gt;2^a5^b&lt;/math&gt; and &lt;math&gt;2^c5^d&lt;/math&gt; where &lt;math&gt;0 \le a,b,c,d \le 5&lt;/math&gt; and &lt;math&gt;(a,b) \neq (c,d)&lt;/math&gt;, whose product is &lt;math&gt;2^{a+c}5^{b+d}&lt;/math&gt;, where &lt;math&gt;0 \le a+c \le 10&lt;/math&gt; and &lt;math&gt;0 \le b+d \le 10&lt;/math&gt;.<br /> <br /> Consider &lt;math&gt;100000^2 = 2^{10}5^{10}&lt;/math&gt;. The number of divisors is &lt;math&gt;(10+1)(10+1) = 121&lt;/math&gt;. However, some of the divisors of &lt;math&gt;2^{10}5^{10}&lt;/math&gt; cannot be written as a product of two distinct divisors of &lt;math&gt;2^5 \cdot 5^5&lt;/math&gt;, namely: &lt;math&gt;1 = 2^05^0&lt;/math&gt;, &lt;math&gt;2^{10}5^{10}&lt;/math&gt;, &lt;math&gt;2^{10}&lt;/math&gt;, and &lt;math&gt;5^{10}&lt;/math&gt;. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only &lt;math&gt;2^5&lt;/math&gt; or &lt;math&gt;5^5&lt;/math&gt;. Since the factors chosen must be distinct, the last two numbers cannot be created because those require &lt;math&gt;2^5 \cdot 2^5&lt;/math&gt; or &lt;math&gt;5^5 \cdot 5^5&lt;/math&gt;. This gives &lt;math&gt;121-4 = 117&lt;/math&gt; candidate numbers. It is not too hard to show that every number of the form &lt;math&gt;2^p5^q&lt;/math&gt; where &lt;math&gt;0 \le p, q \le 10&lt;/math&gt;, and &lt;math&gt;p,q&lt;/math&gt; are not both 0 or 10, can be written as a product of two distinct elements in &lt;math&gt;S&lt;/math&gt;. Hence the answer is &lt;math&gt;\boxed{\textbf{(C) } 117}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=102921 2019 AMC 10B Problems/Problem 19 2019-02-15T16:22:49Z <p>Xmidnightfirex: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br /> <br /> ==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all positive integer divisors of &lt;math&gt;100,000.&lt;/math&gt; How many numbers are the product of two distinct elements of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> To find the number of numbers that are the product of two distinct elements of &lt;math&gt;S&lt;/math&gt;, we first square &lt;math&gt;100,000&lt;/math&gt; and factor it. Factoring, we find &lt;math&gt;100,000^2 = 2^{10} \cdot 5^{10}&lt;/math&gt;. Therefore, &lt;math&gt;100,000^2&lt;/math&gt; has &lt;math&gt;(10 + 1)(10 + 1) = 121&lt;/math&gt; distinct factors. Each of these can be achieved by multiplying two factors of &lt;math&gt;S&lt;/math&gt;. However, the factors must be distinct, so we eliminate &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100,000^2&lt;/math&gt;, as well as &lt;math&gt;2^{10}&lt;/math&gt; and &lt;math&gt;5^{10}&lt;/math&gt;, so the answer is &lt;math&gt;121 - 4 = 117&lt;/math&gt;. &lt;math&gt;2^{10}&lt;/math&gt; and &lt;math&gt;5^{10}&lt;/math&gt; do not work since the factors chosen must be distinct, and those require &lt;math&gt;2^5 \cdot 2^5&lt;/math&gt; or &lt;math&gt;5^5 \cdot 5^5&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> The prime factorization of 100,000 is &lt;math&gt;2^5 \cdot 5^5&lt;/math&gt;. Thus, we choose two numbers &lt;math&gt;2^a5^b&lt;/math&gt; and &lt;math&gt;2^c5^d&lt;/math&gt; where &lt;math&gt;0 \le a,b,c,d \le 5&lt;/math&gt; and &lt;math&gt;(a,b) \neq (c,d)&lt;/math&gt;, whose product is &lt;math&gt;2^{a+c}5^{b+d}&lt;/math&gt;, where &lt;math&gt;0 \le a+c \le 10&lt;/math&gt; and &lt;math&gt;0 \le b+d \le 10&lt;/math&gt;.<br /> <br /> Consider &lt;math&gt;100000^2 = 2^{10}5^{10}&lt;/math&gt;. The number of divisors is &lt;math&gt;(10+1)(10+1) = 121&lt;/math&gt;. However, some of the divisors of &lt;math&gt;2^{10}5^{10}&lt;/math&gt; cannot be written as a product of two distinct divisors of &lt;math&gt;2^5 \cdot 5^5&lt;/math&gt;, namely: &lt;math&gt;1 = 2^05^0&lt;/math&gt;, &lt;math&gt;2^{10}5^{10}&lt;/math&gt;, &lt;math&gt;2^{10}&lt;/math&gt;, and &lt;math&gt;5^{10}&lt;/math&gt;. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only &lt;math&gt;2^5&lt;/math&gt; or &lt;math&gt;5^5&lt;/math&gt;. Since the factors chosen must be distinct, the last two numbers cannot be created because those require &lt;math&gt;2^5 \cdot 2^5&lt;/math&gt; or &lt;math&gt;5^5 \cdot 5^5&lt;/math&gt;. This gives &lt;math&gt;121-4 = 117&lt;/math&gt; candidate numbers. It is not too hard to show that every number of the form &lt;math&gt;2^p5^q&lt;/math&gt; where &lt;math&gt;0 \le p, q \le 10&lt;/math&gt;, and &lt;math&gt;p,q&lt;/math&gt; are not both 0 or 10, can be written as a product of two distinct elements in &lt;math&gt;S&lt;/math&gt;. Hence the answer is &lt;math&gt;\boxed{\textbf{(C) } 117}&lt;/math&gt;.<br /> <br /> -scrabbler94 (edited by mshell214)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_18&diff=102920 2019 AMC 10B Problems/Problem 18 2019-02-15T16:22:24Z <p>Xmidnightfirex: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Henry decides one morning to do a workout, and he walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the way from his home to his gym. The gym is &lt;math&gt;2&lt;/math&gt; kilometers away from Henry's home. At that point, he changes his mind and walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point &lt;math&gt;A&lt;/math&gt; kilometers from home and a point &lt;math&gt;B&lt;/math&gt; kilometers from home. What is &lt;math&gt;|A-B|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 1\frac{1}{5} \qquad \textbf{(D) } 1\frac{1}{4} \qquad \textbf{(E) } 1\frac{1}{2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let the two points that Henry walks in-between be &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, with &lt;math&gt;A&lt;/math&gt; being closer to home. In addition, let the distance that the points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are from his home be &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, respectively. By symmetry, the distance from point &lt;math&gt;B&lt;/math&gt; is from the gym is the same as the distance from home to point &lt;math&gt;A&lt;/math&gt;. Thus, &lt;math&gt;a = 2 - b&lt;/math&gt;. In addition, the distance that he walks as he repeatedly heads to home and then to the gym is &lt;math&gt;b - a&lt;/math&gt;. Therefore, we are looking for the value of &lt;math&gt;b - a&lt;/math&gt;. When he walks from point &lt;math&gt;B&lt;/math&gt; to home, he walks &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the distance, ending at point &lt;math&gt;A&lt;/math&gt;. Therefore, we know that &lt;math&gt;b - a = \frac{3}{4} \cdot b&lt;/math&gt;. Similarily, we know &lt;math&gt;b - a = \frac{3}{4} \cdot (2 - a)&lt;/math&gt;. Adding these equations, we get &lt;math&gt;2(b - a) = \frac{3}{4} \cdot (2 + b - a)&lt;/math&gt;. Multiplying by &lt;math&gt;4&lt;/math&gt;, we get &lt;math&gt;8(b - a) = 6 + 3(b - a)&lt;/math&gt;, so &lt;math&gt;b - a = \frac{6}{5} = 1 \frac{1}{5}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_13&diff=102919 2019 AMC 12B Problems/Problem 13 2019-02-15T16:22:04Z <p>Xmidnightfirex: /* Solution 4 (quickest) */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #17]] and [[2019 AMC 12B Problems|2019 AMC 12B #13]]}}<br /> ==Problem==<br /> <br /> A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;2^{-k}&lt;/math&gt; for &lt;math&gt;k = 1,2,3....&lt;/math&gt; What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?&lt;br&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher numbered bin. The probability of both landing in the same bin is &lt;math&gt;\sum_{k=1}^{\infty}2^{-2k}&lt;/math&gt;. The sum is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt;. Therefore the other two probabilities have to both be &lt;math&gt;\textbf{(C) } \frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 1 Variant==<br /> We solve for the probability by doing &lt;math&gt;\frac{1-(\text{Probability of Equality})}{2}&lt;/math&gt;.<br /> <br /> We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to &lt;math&gt;(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...&lt;/math&gt; <br /> <br /> The summation of this expression is equal to &lt;cmath&gt;\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1&lt;/cmath&gt;. Using the geometric sum formula, we obtain the summation of this expression to be &lt;math&gt;\frac{1}{\frac{3}{4}}-1&lt;/math&gt; or &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 2 (variant)==<br /> Suppose the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, for some &lt;math&gt;i \ge 1&lt;/math&gt;. The probability of this occurring is &lt;math&gt;\frac{1}{2^i}&lt;/math&gt;. Given this occurs, the probability that the red ball goes in a higher-numbered bin is &lt;math&gt;\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}&lt;/math&gt;. Thus the probability that the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, and the red ball goes in a bin greater than &lt;math&gt;i&lt;/math&gt;, is &lt;math&gt;\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}&lt;/math&gt;. Summing from &lt;math&gt;i=1&lt;/math&gt; to infinity, we get<br /> <br /> &lt;cmath&gt;\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/cmath&gt;<br /> <br /> (Note: to find this sum, we use the formula &lt;math&gt;\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}&lt;/math&gt;. Since in this case &lt;math&gt;r = \frac{1}{4}&lt;/math&gt;, the answer is &lt;math&gt;\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}&lt;/math&gt;. If you don't know this formula, you may instead note that if you multiply the sum by &lt;math&gt;4&lt;/math&gt;, it is equivalent to adding &lt;math&gt;1&lt;/math&gt;. Thus: &lt;math&gt;4n = n+1&lt;/math&gt;, which clearly simplifies to &lt;math&gt;n = \frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 3 (infinite geometric series)==<br /> The probability that the two balls will go into adjacent bins is &lt;math&gt;\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} = \frac{1}{6}&lt;/math&gt;. The probability that the two balls will go into bins that have a distance of 2 from each other is &lt;math&gt;\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} = \frac{1}{12}&lt;/math&gt;. We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is &lt;math&gt;\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + ...&lt;/math&gt;, which converges into &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 4 (quickest)==<br /> Define a win as a ball appearing in higher numbered box.<br /> <br /> Start from the first box. <br /> <br /> There are 4 possible results in the box: Red, Green, Red and Green, none, with probability of &lt;math&gt;\frac{1}{4}&lt;/math&gt; for each. Red win, Green win, Tie all have the same probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt;. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution.<br /> <br /> So finally, Red win, Green win and Tie all have a probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt;<br /> <br /> The answer is &lt;math&gt;\boxed{C}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Xmidnightfirex https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_13&diff=102918 2019 AMC 12B Problems/Problem 13 2019-02-15T16:21:23Z <p>Xmidnightfirex: /* Solution 2 (variant) */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #17]] and [[2019 AMC 12B Problems|2019 AMC 12B #13]]}}<br /> ==Problem==<br /> <br /> A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;2^{-k}&lt;/math&gt; for &lt;math&gt;k = 1,2,3....&lt;/math&gt; What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?&lt;br&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher numbered bin. The probability of both landing in the same bin is &lt;math&gt;\sum_{k=1}^{\infty}2^{-2k}&lt;/math&gt;. The sum is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt;. Therefore the other two probabilities have to both be &lt;math&gt;\textbf{(C) } \frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 1 Variant==<br /> We solve for the probability by doing &lt;math&gt;\frac{1-(\text{Probability of Equality})}{2}&lt;/math&gt;.<br /> <br /> We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to &lt;math&gt;(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...&lt;/math&gt; <br /> <br /> The summation of this expression is equal to &lt;cmath&gt;\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1&lt;/cmath&gt;. Using the geometric sum formula, we obtain the summation of this expression to be &lt;math&gt;\frac{1}{\frac{3}{4}}-1&lt;/math&gt; or &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 2 (variant)==<br /> Suppose the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, for some &lt;math&gt;i \ge 1&lt;/math&gt;. The probability of this occurring is &lt;math&gt;\frac{1}{2^i}&lt;/math&gt;. Given this occurs, the probability that the red ball goes in a higher-numbered bin is &lt;math&gt;\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}&lt;/math&gt;. Thus the probability that the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, and the red ball goes in a bin greater than &lt;math&gt;i&lt;/math&gt;, is &lt;math&gt;\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}&lt;/math&gt;. Summing from &lt;math&gt;i=1&lt;/math&gt; to infinity, we get<br /> <br /> &lt;cmath&gt;\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/cmath&gt;<br /> <br /> (Note: to find this sum, we use the formula &lt;math&gt;\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}&lt;/math&gt;. Since in this case &lt;math&gt;r = \frac{1}{4}&lt;/math&gt;, the answer is &lt;math&gt;\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}&lt;/math&gt;. If you don't know this formula, you may instead note that if you multiply the sum by &lt;math&gt;4&lt;/math&gt;, it is equivalent to adding &lt;math&gt;1&lt;/math&gt;. Thus: &lt;math&gt;4n = n+1&lt;/math&gt;, which clearly simplifies to &lt;math&gt;n = \frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 3 (infinite geometric series)==<br /> The probability that the two balls will go into adjacent bins is &lt;math&gt;\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} = \frac{1}{6}&lt;/math&gt;. The probability that the two balls will go into bins that have a distance of 2 from each other is &lt;math&gt;\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} = \frac{1}{12}&lt;/math&gt;. We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is &lt;math&gt;\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + ...&lt;/math&gt;, which converges into &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> ==Solution 4 (quickest)==<br /> Define a win as a ball appearing in higher numbered box.<br /> <br /> Start from the first box. <br /> <br /> There are 4 possible results in the box: Red, Green, Red and Green, none, with probability of &lt;math&gt;\frac{1}{4}&lt;/math&gt; for each. Red win, Green win, Tie all have the same probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt;. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution.<br /> <br /> So finally, Red win, Green win and Tie all have a probability of &lt;math&gt;\frac{1}{3}&lt;/math&gt;<br /> <br /> The answer is &lt;math&gt;\boxed{C}&lt;/math&gt;<br /> <br /> -- Solution by mathsuper(丹神)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Xmidnightfirex