https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ssding&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-27T15:30:01Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_24&diff=164464 2017 AMC 10A Problems/Problem 24 2021-11-02T13:46:52Z <p>Ssding: /* Solution 1 */ Changed the grammar and formatting of a sentence</p> <hr /> <div>==Problem==<br /> For certain real numbers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, the polynomial &lt;cmath&gt;g(x) = x^3 + ax^2 + x + 10&lt;/cmath&gt;has three distinct roots, and each root of &lt;math&gt;g(x)&lt;/math&gt; is also a root of the polynomial &lt;cmath&gt;f(x) = x^4 + x^3 + bx^2 + 100x + c.&lt;/cmath&gt;What is &lt;math&gt;f(1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;f(x)&lt;/math&gt; must have four roots, three of which are roots of &lt;math&gt;g(x)&lt;/math&gt;. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of &lt;math&gt;f(x)&lt;/math&gt; and &lt;math&gt;g(x)&lt;/math&gt; are the same, we know that<br /> <br /> &lt;cmath&gt;f(x)=g(x)(x-r)&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;r\in\mathbb{C}&lt;/math&gt; is the fourth root of &lt;math&gt;f(x)&lt;/math&gt;. Substituting &lt;math&gt;g(x)&lt;/math&gt; and expanding, we find that<br /> <br /> &lt;cmath&gt;\begin{align*}f(x)&amp;=(x^3+ax^2+x+10)(x-r)\\<br /> &amp;=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}&lt;/cmath&gt;<br /> <br /> Comparing coefficients with &lt;math&gt;f(x)&lt;/math&gt;, we see that<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> a-r&amp;=1\\<br /> 1-ar&amp;=b\\<br /> 10-r&amp;=100\\<br /> -10r&amp;=c.\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> (Solution 1.1 picks up here.)<br /> <br /> Let's solve for &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt;. Since &lt;math&gt;10-r=100&lt;/math&gt;, &lt;math&gt;r=-90&lt;/math&gt;, so &lt;math&gt;c=(-10)(-90)=900&lt;/math&gt;. Since &lt;math&gt;a-r=1&lt;/math&gt;, &lt;math&gt;a=-89&lt;/math&gt;. Then, since &lt;math&gt;b=1-ar&lt;/math&gt;, &lt;math&gt;b=-8009&lt;/math&gt;. Thus, we know that<br /> <br /> &lt;cmath&gt;f(x)=x^4+x^3-8009x^2+100x+900.&lt;/cmath&gt;<br /> <br /> Taking &lt;math&gt;f(1)&lt;/math&gt;, we find that<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> f(1)&amp;=1^4+1^3-8009(1)^2+100(1)+900\\<br /> &amp;=1+1-8009+100+900\\<br /> &amp;=\boxed{\bold{(C)}\, -7007}.\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 1.1==<br /> A faster ending to Solution 1 is as follows.<br /> We shall solve for only &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt;. Since &lt;math&gt;10-r=100&lt;/math&gt;, &lt;math&gt;r=-90&lt;/math&gt;, and since &lt;math&gt;a-r=1&lt;/math&gt;, &lt;math&gt;a=-89&lt;/math&gt;. Then,<br /> &lt;cmath&gt;\begin{align*}<br /> f(1)&amp;=(1-r)(1^3+a\cdot1^2+1+10)\\<br /> &amp;=(91)(-77)\\<br /> &amp;=\boxed{\bold{(C)}\, -7007}.\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 1.2 (Fast)==<br /> Let the term &lt;math&gt;q(x)&lt;/math&gt; be the linear term that we are solving for in the equation &lt;math&gt;f(x)=g(x)\cdot q(x)&lt;/math&gt;. Now, we know that &lt;math&gt;q(x)=mx-r&lt;/math&gt; must have &lt;math&gt;m=1&lt;/math&gt;, because only &lt;math&gt;x \cdot x^3=x^4&lt;/math&gt;. In addition, we know that, by distributing, &lt;math&gt;10x-rx=100x&lt;/math&gt;. Therefore, &lt;math&gt;r=-90&lt;/math&gt;, and all the other variables are quickly solved for.<br /> <br /> ==Solution 2==<br /> We notice that the constant term of &lt;math&gt;f(x)=c&lt;/math&gt; and the constant term in &lt;math&gt;g(x)=10&lt;/math&gt;. Because &lt;math&gt;f(x)&lt;/math&gt; can be factored as &lt;math&gt;g(x) \cdot (x- r)&lt;/math&gt; (where &lt;math&gt;r&lt;/math&gt; is the unshared root of &lt;math&gt;f(x)&lt;/math&gt;, we see that using the constant term, &lt;math&gt;-10 \cdot r = c&lt;/math&gt; and therefore &lt;math&gt;r = -\frac{c}{10}&lt;/math&gt;.<br /> Now we once again write &lt;math&gt;f(x)&lt;/math&gt; out in factored form: <br /> <br /> &lt;cmath&gt;f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)(x+\frac{c}{10})&lt;/cmath&gt;.<br /> <br /> We can expand the expression on the right-hand side to get:<br /> <br /> &lt;cmath&gt;f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c&lt;/cmath&gt;<br /> <br /> Now we have &lt;math&gt;f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c=x^4+x^3+bx^2+100x+c&lt;/math&gt;.<br /> <br /> Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations:<br /> &lt;cmath&gt;10+\frac{c}{10}=100 \Rightarrow c=900&lt;/cmath&gt;<br /> &lt;cmath&gt;a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89&lt;/cmath&gt;<br /> <br /> and finally,<br /> <br /> &lt;cmath&gt;1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009&lt;/cmath&gt;.<br /> <br /> We know that &lt;math&gt;f(1)&lt;/math&gt; is the sum of its coefficients, hence &lt;math&gt;1+1+b+100+c&lt;/math&gt;. We substitute the values we obtained for &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; into this expression to get &lt;math&gt;f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{\textbf{(C)}\,-7007}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;r_1,r_2,&lt;/math&gt; and &lt;math&gt;r_3&lt;/math&gt; be the roots of &lt;math&gt;g(x)&lt;/math&gt;. Let &lt;math&gt;r_4&lt;/math&gt; be the additional root of &lt;math&gt;f(x)&lt;/math&gt;. Then from Vieta's formulas on the quadratic term of &lt;math&gt;g(x)&lt;/math&gt; and the cubic term of &lt;math&gt;f(x)&lt;/math&gt;, we obtain the following:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> r_1+r_2+r_3&amp;=-a \\ <br /> r_1+r_2+r_3+r_4&amp;=-1<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;r_4=a-1&lt;/math&gt;.<br /> <br /> Now applying Vieta's formulas on the constant term of &lt;math&gt;g(x)&lt;/math&gt;, the linear term of &lt;math&gt;g(x)&lt;/math&gt;, and the linear term of &lt;math&gt;f(x)&lt;/math&gt;, we obtain:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> r_1r_2r_3 &amp; = -10\\<br /> r_1r_2+r_2r_3+r_3r_1 &amp;= 1\\ <br /> r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 &amp; = -100\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Substituting for &lt;math&gt;r_1r_2r_3&lt;/math&gt; in the bottom equation and factoring the remainder of the expression, we obtain:<br /> <br /> &lt;cmath&gt;-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100&lt;/cmath&gt;<br /> <br /> It follows that &lt;math&gt;r_4=-90&lt;/math&gt;. But &lt;math&gt;r_4=a-1&lt;/math&gt; so &lt;math&gt;a=-89&lt;/math&gt;<br /> <br /> Now we can factor &lt;math&gt;f(x)&lt;/math&gt; in terms of &lt;math&gt;g(x)&lt;/math&gt; as<br /> <br /> &lt;cmath&gt;f(x)=(x-r_4)g(x)=(x+90)g(x)&lt;/cmath&gt;<br /> <br /> Then &lt;math&gt;f(1)=91g(1)&lt;/math&gt; and<br /> <br /> &lt;cmath&gt;g(1)=1^3-89\cdot 1^2+1+10=-77&lt;/cmath&gt;<br /> <br /> Hence &lt;math&gt;f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Risky)==<br /> Let the roots of &lt;math&gt;g(x)&lt;/math&gt; be &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, and &lt;math&gt;r_3&lt;/math&gt;. Let the roots of &lt;math&gt;f(x)&lt;/math&gt; be &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt;, and &lt;math&gt;r_4&lt;/math&gt;. From Vieta's, we have:<br /> &lt;cmath&gt;\begin{align*}<br /> r_1+r_2+r_3=-a \\<br /> r_1+r_2+r_3+r_4=-1 \\<br /> r_4=a-1<br /> \end{align*}&lt;/cmath&gt;<br /> The fourth root is &lt;math&gt;a-1&lt;/math&gt;. Since &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, and &lt;math&gt;r_3&lt;/math&gt; are common roots, we have:<br /> &lt;cmath&gt;\begin{align*}<br /> f(x)=g(x)(x-(a-1)) \\<br /> f(1)=g(1)(1-(a-1)) \\<br /> f(1)=(a+12)(2-a) \\<br /> f(1)=-(a+12)(a-2) \\<br /> \end{align*}&lt;/cmath&gt;<br /> Let &lt;math&gt;a-2=k&lt;/math&gt;:<br /> &lt;cmath&gt;\begin{align*}<br /> f(1)=-k(k+14)<br /> \end{align*}&lt;/cmath&gt;<br /> Note that &lt;math&gt;-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)&lt;/math&gt;<br /> This gives us a pretty good guess of &lt;math&gt;\boxed{\textbf{(C)}\, -7007}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> First off, let's get rid of the &lt;math&gt;x^4&lt;/math&gt; term by finding &lt;math&gt;h(x)=f(x)-xg(x)&lt;/math&gt;. This polynomial consists of the difference of two polynomials with &lt;math&gt;3&lt;/math&gt; common factors, so it must also have these factors. The polynomial is &lt;math&gt;h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c&lt;/math&gt;, and must be equal to &lt;math&gt;(1-a)g(x)&lt;/math&gt;. Equating the coefficients, we get &lt;math&gt;3&lt;/math&gt; equations. We will tackle the situation one equation at a time, starting the &lt;math&gt;x&lt;/math&gt; terms. Looking at the coefficients, we get &lt;math&gt;\dfrac{90}{1-a} = 1&lt;/math&gt;. &lt;cmath&gt;\therefore 90=1-a.&lt;/cmath&gt; The solution to the previous is obviously &lt;math&gt;a=-89&lt;/math&gt;. We can now find &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. &lt;math&gt;\dfrac{b-1}{1-a} = a&lt;/math&gt;, &lt;cmath&gt;\therefore b-1=a(1-a)=-89*90=-8010&lt;/cmath&gt; and &lt;math&gt;b=-8009&lt;/math&gt;. Finally &lt;math&gt;\dfrac{c}{1-a} = 10&lt;/math&gt;, &lt;cmath&gt;\therefore c=10(1-a)=10*90=900&lt;/cmath&gt; Solving the original problem, &lt;math&gt;f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{\textbf{(C)}\, -7007}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of &lt;math&gt;\frac{f(x)}{g(x)}&lt;/math&gt; eventually brings us the final step &lt;math&gt;(1-a)x^3 + (b-1)x^2 + 90x + c&lt;/math&gt; minus &lt;math&gt;(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)&lt;/math&gt; after we multiply &lt;math&gt;f(x)&lt;/math&gt; by &lt;math&gt;(1-a)&lt;/math&gt;. Now we equate coefficients of same-degree &lt;math&gt;x&lt;/math&gt; terms. This gives us &lt;math&gt; 10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009&lt;/math&gt;. We are interested in finding &lt;math&gt;f(1)&lt;/math&gt;, which equals &lt;math&gt;1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{\textbf{(C)}\,-7007}&lt;/math&gt;. ~skyscraper<br /> <br /> ==General Note==<br /> <br /> Note that &lt;math&gt;f(1)&lt;/math&gt; for any polynomial is simply the sum of the coefficients of the polynomial.<br /> <br /> ==Video Solution==<br /> https://youtu.be/wXvRNC-48Lk<br /> <br /> https://www.youtube.com/watch?v=MBIiz0mroqk (By Richard Rusczyk)<br /> <br /> https://youtu.be/3dfbWzOfJAI?t=4412<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2017|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems&diff=163588 2013 AMC 10A Problems 2021-10-16T12:41:49Z <p>Ssding: CHanged formatting</p> <hr /> <div>{{AMC10 Problems|year=2013|ab=A}}<br /> ==Problem 1==<br /> <br /> A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Alice is making a batch of cookies and needs &lt;math&gt;2\frac{1}{2}&lt;/math&gt; cups of sugar. Unfortunately, her measuring cup holds only &lt;math&gt;\frac{1}{4}&lt;/math&gt; cup of sugar. How many times must she fill that cup to get the correct amount of sugar?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt;10&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt;, and the area of &lt;math&gt; \triangle ABE &lt;/math&gt;<br /> is &lt;math&gt;40&lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &lt;math&gt;\$105&lt;/math&gt;, Dorothy paid &lt;math&gt;\$125&lt;/math&gt;, and Sammy paid &lt;math&gt;\$175&lt;/math&gt;. In order to share costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?&lt;/math&gt; <br /> <br /> &lt;math&gt; \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many three-digit numbers are not divisible by &lt;math&gt;5&lt;/math&gt;, have digits that sum to less than &lt;math&gt;20&lt;/math&gt;, and have the first digit equal to the third digit?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A solid cube of side length &lt;math&gt;1&lt;/math&gt; is removed from each corner of a solid cube of side length &lt;math&gt;3&lt;/math&gt;. How many edges does the remaining solid have?<br /> <br /> &lt;math&gt; \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Two sides of a triangle have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A triangle with vertices &lt;math&gt;(6, 5)&lt;/math&gt;, &lt;math&gt;(8, -3)&lt;/math&gt;, and &lt;math&gt;(9, 1)&lt;/math&gt; is reflected about the line &lt;math&gt;x=8&lt;/math&gt; to create a second triangle. What is the area of the union of the two triangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let points &lt;math&gt;A = (0, 0)&lt;/math&gt;, &lt;math&gt;B = (1, 2)&lt;/math&gt;, &lt;math&gt;C=(3, 3)&lt;/math&gt;, and &lt;math&gt;D = (4, 0)&lt;/math&gt;. Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt;A&lt;/math&gt;. This line intersects &lt;math&gt;\overline{CD}&lt;/math&gt; at point &lt;math&gt;\bigg(\frac{p}{q}, \frac{r}{s}\bigg)&lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt;p+q+r+s&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In base &lt;math&gt;10&lt;/math&gt;, the number &lt;math&gt;2013&lt;/math&gt; ends in the digit &lt;math&gt;3&lt;/math&gt;. In base &lt;math&gt;9&lt;/math&gt;, on the other hand, the same number is written as &lt;math&gt;(2676)_{9}&lt;/math&gt; and ends in the digit &lt;math&gt;6&lt;/math&gt;. For how many positive integers &lt;math&gt;b&lt;/math&gt; does the base-&lt;math&gt;b&lt;/math&gt;-representation of &lt;math&gt;2013&lt;/math&gt; end in the digit &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A group of &lt;math&gt;12&lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt;k^{\text{th}}&lt;/math&gt; pirate to take a share takes &lt;math&gt;\frac{k}{12}&lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt;12^{\text{th}}&lt;/math&gt; pirate receive?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2013|ab=A|before=[[2012 AMC 10B Problems]]|after=[[2013 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2013 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems&diff=163587 2013 AMC 10A Problems 2021-10-16T12:38:58Z <p>Ssding: /* Problem 4 */ Changed formatting</p> <hr /> <div>{{AMC10 Problems|year=2013|ab=A}}<br /> ==Problem 1==<br /> <br /> A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Alice is making a batch of cookies and needs &lt;math&gt;2\frac{1}{2}&lt;/math&gt; cups of sugar. Unfortunately, her measuring cup holds only &lt;math&gt;\frac{1}{4}&lt;/math&gt; cup of sugar. How many times must she fill that cup to get the correct amount of sugar?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt;10&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt;, and the area of &lt;math&gt; \triangle ABE &lt;/math&gt;<br /> is &lt;math&gt;40&lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &lt;math&gt;\$105&lt;/math&gt;, Dorothy paid &lt;math&gt;\$125&lt;/math&gt;, and Sammy paid &lt;math&gt;\$175&lt;/math&gt;. In order to share costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?&lt;/math&gt; <br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many three-digit numbers are not divisible by &lt;math&gt;5&lt;/math&gt;, have digits that sum to less than &lt;math&gt;20&lt;/math&gt;, and have the first digit equal to the third digit?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A solid cube of side length &lt;math&gt;1&lt;/math&gt; is removed from each corner of a solid cube of side length &lt;math&gt;3&lt;/math&gt;. How many edges does the remaining solid have?<br /> <br /> <br /> &lt;math&gt; \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Two sides of a triangle have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A triangle with vertices &lt;math&gt;(6, 5)&lt;/math&gt;, &lt;math&gt;(8, -3)&lt;/math&gt;, and &lt;math&gt;(9, 1)&lt;/math&gt; is reflected about the line &lt;math&gt;x=8&lt;/math&gt; to create a second triangle. What is the area of the union of the two triangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let points &lt;math&gt;A = (0, 0)&lt;/math&gt;, &lt;math&gt;B = (1, 2)&lt;/math&gt;, &lt;math&gt;C=(3, 3)&lt;/math&gt;, and &lt;math&gt;D = (4, 0)&lt;/math&gt;. Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt;A&lt;/math&gt;. This line intersects &lt;math&gt;\overline{CD}&lt;/math&gt; at point &lt;math&gt;\bigg(\frac{p}{q}, \frac{r}{s}\bigg)&lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt;p+q+r+s&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In base &lt;math&gt;10&lt;/math&gt;, the number &lt;math&gt;2013&lt;/math&gt; ends in the digit &lt;math&gt;3&lt;/math&gt;. In base &lt;math&gt;9&lt;/math&gt;, on the other hand, the same number is written as &lt;math&gt;(2676)_{9}&lt;/math&gt; and ends in the digit &lt;math&gt;6&lt;/math&gt;. For how many positive integers &lt;math&gt;b&lt;/math&gt; does the base-&lt;math&gt;b&lt;/math&gt;-representation of &lt;math&gt;2013&lt;/math&gt; end in the digit &lt;math&gt;3&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A group of &lt;math&gt;12&lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt;k^{\text{th}}&lt;/math&gt; pirate to take a share takes &lt;math&gt;\frac{k}{12}&lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt;12^{\text{th}}&lt;/math&gt; pirate receive?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 23|Solution]]<br /> ==Problem 24==<br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 24|Solution]]<br /> ==Problem 25==<br /> <br /> All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2013|ab=A|before=[[2012 AMC 10B Problems]]|after=[[2013 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2013 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems&diff=163586 2013 AMC 10A Problems 2021-10-16T12:38:33Z <p>Ssding: /* Problem 3 */ Changed formatting</p> <hr /> <div>{{AMC10 Problems|year=2013|ab=A}}<br /> ==Problem 1==<br /> <br /> A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Alice is making a batch of cookies and needs &lt;math&gt;2\frac{1}{2}&lt;/math&gt; cups of sugar. Unfortunately, her measuring cup holds only &lt;math&gt;\frac{1}{4}&lt;/math&gt; cup of sugar. How many times must she fill that cup to get the correct amount of sugar?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt;10&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt;, and the area of &lt;math&gt; \triangle ABE &lt;/math&gt;<br /> is &lt;math&gt;40&lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &lt;math&gt;\$105&lt;/math&gt;, Dorothy paid &lt;math&gt;\$125&lt;/math&gt;, and Sammy paid &lt;math&gt;\$175&lt;/math&gt;. In order to share costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?&lt;/math&gt; <br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many three-digit numbers are not divisible by &lt;math&gt;5&lt;/math&gt;, have digits that sum to less than &lt;math&gt;20&lt;/math&gt;, and have the first digit equal to the third digit?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A solid cube of side length &lt;math&gt;1&lt;/math&gt; is removed from each corner of a solid cube of side length &lt;math&gt;3&lt;/math&gt;. How many edges does the remaining solid have?<br /> <br /> <br /> &lt;math&gt; \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Two sides of a triangle have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A triangle with vertices &lt;math&gt;(6, 5)&lt;/math&gt;, &lt;math&gt;(8, -3)&lt;/math&gt;, and &lt;math&gt;(9, 1)&lt;/math&gt; is reflected about the line &lt;math&gt;x=8&lt;/math&gt; to create a second triangle. What is the area of the union of the two triangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let points &lt;math&gt;A = (0, 0)&lt;/math&gt;, &lt;math&gt;B = (1, 2)&lt;/math&gt;, &lt;math&gt;C=(3, 3)&lt;/math&gt;, and &lt;math&gt;D = (4, 0)&lt;/math&gt;. Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt;A&lt;/math&gt;. This line intersects &lt;math&gt;\overline{CD}&lt;/math&gt; at point &lt;math&gt;\bigg(\frac{p}{q}, \frac{r}{s}\bigg)&lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt;p+q+r+s&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In base &lt;math&gt;10&lt;/math&gt;, the number &lt;math&gt;2013&lt;/math&gt; ends in the digit &lt;math&gt;3&lt;/math&gt;. In base &lt;math&gt;9&lt;/math&gt;, on the other hand, the same number is written as &lt;math&gt;(2676)_{9}&lt;/math&gt; and ends in the digit &lt;math&gt;6&lt;/math&gt;. For how many positive integers &lt;math&gt;b&lt;/math&gt; does the base-&lt;math&gt;b&lt;/math&gt;-representation of &lt;math&gt;2013&lt;/math&gt; end in the digit &lt;math&gt;3&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A group of &lt;math&gt;12&lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt;k^{\text{th}}&lt;/math&gt; pirate to take a share takes &lt;math&gt;\frac{k}{12}&lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt;12^{\text{th}}&lt;/math&gt; pirate receive?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 23|Solution]]<br /> ==Problem 24==<br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 24|Solution]]<br /> ==Problem 25==<br /> <br /> All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2013|ab=A|before=[[2012 AMC 10B Problems]]|after=[[2013 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2013 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems&diff=163585 2013 AMC 10A Problems 2021-10-16T12:38:10Z <p>Ssding: /* Problem 2 */ Changed formatting</p> <hr /> <div>{{AMC10 Problems|year=2013|ab=A}}<br /> ==Problem 1==<br /> <br /> A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Alice is making a batch of cookies and needs &lt;math&gt;2\frac{1}{2}&lt;/math&gt; cups of sugar. Unfortunately, her measuring cup holds only &lt;math&gt;\frac{1}{4}&lt;/math&gt; cup of sugar. How many times must she fill that cup to get the correct amount of sugar?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt;10&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt;, and the area of &lt;math&gt; \triangle ABE &lt;/math&gt;<br /> is &lt;math&gt;40&lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &lt;math&gt;\$105&lt;/math&gt;, Dorothy paid &lt;math&gt;\$125&lt;/math&gt;, and Sammy paid &lt;math&gt;\$175&lt;/math&gt;. In order to share costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?&lt;/math&gt; <br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many three-digit numbers are not divisible by &lt;math&gt;5&lt;/math&gt;, have digits that sum to less than &lt;math&gt;20&lt;/math&gt;, and have the first digit equal to the third digit?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A solid cube of side length &lt;math&gt;1&lt;/math&gt; is removed from each corner of a solid cube of side length &lt;math&gt;3&lt;/math&gt;. How many edges does the remaining solid have?<br /> <br /> <br /> &lt;math&gt; \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Two sides of a triangle have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A triangle with vertices &lt;math&gt;(6, 5)&lt;/math&gt;, &lt;math&gt;(8, -3)&lt;/math&gt;, and &lt;math&gt;(9, 1)&lt;/math&gt; is reflected about the line &lt;math&gt;x=8&lt;/math&gt; to create a second triangle. What is the area of the union of the two triangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let points &lt;math&gt;A = (0, 0)&lt;/math&gt;, &lt;math&gt;B = (1, 2)&lt;/math&gt;, &lt;math&gt;C=(3, 3)&lt;/math&gt;, and &lt;math&gt;D = (4, 0)&lt;/math&gt;. Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt;A&lt;/math&gt;. This line intersects &lt;math&gt;\overline{CD}&lt;/math&gt; at point &lt;math&gt;\bigg(\frac{p}{q}, \frac{r}{s}\bigg)&lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt;p+q+r+s&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In base &lt;math&gt;10&lt;/math&gt;, the number &lt;math&gt;2013&lt;/math&gt; ends in the digit &lt;math&gt;3&lt;/math&gt;. In base &lt;math&gt;9&lt;/math&gt;, on the other hand, the same number is written as &lt;math&gt;(2676)_{9}&lt;/math&gt; and ends in the digit &lt;math&gt;6&lt;/math&gt;. For how many positive integers &lt;math&gt;b&lt;/math&gt; does the base-&lt;math&gt;b&lt;/math&gt;-representation of &lt;math&gt;2013&lt;/math&gt; end in the digit &lt;math&gt;3&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A group of &lt;math&gt;12&lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt;k^{\text{th}}&lt;/math&gt; pirate to take a share takes &lt;math&gt;\frac{k}{12}&lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt;12^{\text{th}}&lt;/math&gt; pirate receive?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 23|Solution]]<br /> ==Problem 24==<br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 24|Solution]]<br /> ==Problem 25==<br /> <br /> All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2013|ab=A|before=[[2012 AMC 10B Problems]]|after=[[2013 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2013 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems&diff=163584 2013 AMC 10A Problems 2021-10-16T12:37:49Z <p>Ssding: /* Problem 1 */ Changed formatting</p> <hr /> <div>{{AMC10 Problems|year=2013|ab=A}}<br /> ==Problem 1==<br /> <br /> A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75 &lt;/math&gt;<br /> <br /> [[2013 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Alice is making a batch of cookies and needs &lt;math&gt;2\frac{1}{2}&lt;/math&gt; cups of sugar. Unfortunately, her measuring cup holds only &lt;math&gt;\frac{1}{4}&lt;/math&gt; cup of sugar. How many times must she fill that cup to get the correct amount of sugar?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt;10&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt;, and the area of &lt;math&gt; \triangle ABE &lt;/math&gt;<br /> is &lt;math&gt;40&lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid &lt;math&gt;\$105&lt;/math&gt;, Dorothy paid &lt;math&gt;\$125&lt;/math&gt;, and Sammy paid &lt;math&gt;\$175&lt;/math&gt;. In order to share costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?&lt;/math&gt; <br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many three-digit numbers are not divisible by &lt;math&gt;5&lt;/math&gt;, have digits that sum to less than &lt;math&gt;20&lt;/math&gt;, and have the first digit equal to the third digit?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A solid cube of side length &lt;math&gt;1&lt;/math&gt; is removed from each corner of a solid cube of side length &lt;math&gt;3&lt;/math&gt;. How many edges does the remaining solid have?<br /> <br /> <br /> &lt;math&gt; \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Two sides of a triangle have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A triangle with vertices &lt;math&gt;(6, 5)&lt;/math&gt;, &lt;math&gt;(8, -3)&lt;/math&gt;, and &lt;math&gt;(9, 1)&lt;/math&gt; is reflected about the line &lt;math&gt;x=8&lt;/math&gt; to create a second triangle. What is the area of the union of the two triangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let points &lt;math&gt;A = (0, 0)&lt;/math&gt;, &lt;math&gt;B = (1, 2)&lt;/math&gt;, &lt;math&gt;C=(3, 3)&lt;/math&gt;, and &lt;math&gt;D = (4, 0)&lt;/math&gt;. Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt;A&lt;/math&gt;. This line intersects &lt;math&gt;\overline{CD}&lt;/math&gt; at point &lt;math&gt;\bigg(\frac{p}{q}, \frac{r}{s}\bigg)&lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt;p+q+r+s&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In base &lt;math&gt;10&lt;/math&gt;, the number &lt;math&gt;2013&lt;/math&gt; ends in the digit &lt;math&gt;3&lt;/math&gt;. In base &lt;math&gt;9&lt;/math&gt;, on the other hand, the same number is written as &lt;math&gt;(2676)_{9}&lt;/math&gt; and ends in the digit &lt;math&gt;6&lt;/math&gt;. For how many positive integers &lt;math&gt;b&lt;/math&gt; does the base-&lt;math&gt;b&lt;/math&gt;-representation of &lt;math&gt;2013&lt;/math&gt; end in the digit &lt;math&gt;3&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A group of &lt;math&gt;12&lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt;k^{\text{th}}&lt;/math&gt; pirate to take a share takes &lt;math&gt;\frac{k}{12}&lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt;12^{\text{th}}&lt;/math&gt; pirate receive?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 23|Solution]]<br /> ==Problem 24==<br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 24|Solution]]<br /> ==Problem 25==<br /> <br /> All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2013|ab=A|before=[[2012 AMC 10B Problems]]|after=[[2013 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2013 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems&diff=163422 2002 AMC 10A Problems 2021-10-12T15:04:14Z <p>Ssding: /* Problem 25 */ Removed link to trapezoid page</p> <hr /> <div>{{AMC10 Problems|year=2002|ab=A}}<br /> ==Problem 1==<br /> <br /> The ratio &lt;math&gt;\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}&lt;/math&gt; is closest to which of the following numbers?<br /> <br /> &lt;math&gt;\text{(A)}\ 0.1 \qquad \text{(B)}\ 0.2 \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Given that &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are non-zero real numbers, define &lt;math&gt;(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}&lt;/math&gt;. Find &lt;math&gt;(2, 12, 9)&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> According to the standard convention for exponentiation, <br /> <br /> &lt;math&gt;2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65,536&lt;/math&gt;.<br /> <br /> If the order in which the exponentiations are performed is changed, how many other values are possible?<br /> <br /> <br /> &lt;math&gt;\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> For how many positive integers &lt;math&gt;m&lt;/math&gt; is there at least 1 positive integer &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;mn \le m + n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}&lt;/math&gt; infinitely many<br /> <br /> <br /> [[2002 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.<br /> <br /> &lt;asy&gt;<br /> unitsize(.3cm);<br /> path c=Circle((0,2),1);<br /> filldraw(Circle((0,0),3),grey,black);<br /> filldraw(Circle((0,0),1),white,black);<br /> filldraw(c,white,black);<br /> filldraw(rotate(60)*c,white,black);<br /> filldraw(rotate(120)*c,white,black);<br /> filldraw(rotate(180)*c,white,black);<br /> filldraw(rotate(240)*c,white,black);<br /> filldraw(rotate(300)*c,white,black);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ \pi \qquad \text{(B)}\ 1.5\pi \qquad \text{(C)}\ 2\pi \qquad \text{(D)}\ 3\pi \qquad \text{(E)}\ 3.5\pi&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> From a starting number, Cindy was supposed to subtract 3, and then divide by 9, but instead, Cindy subtracted 9, then divided by 3, getting 43. If the correct instructions were followed, what would the result be?<br /> <br /> &lt;math&gt;\text{(A)}\ 15 \qquad \text{(B)}\ 34 \qquad \text{(C)}\ 43 \qquad \text{(D)}\ 51 \qquad \text{(E)} 138&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> A &lt;math&gt;45^\circ&lt;/math&gt; arc of circle A is equal in length to a &lt;math&gt;30^\circ&lt;/math&gt; arc of circle B. What is the ratio of circle A's area and circle B's area?<br /> <br /> &lt;math&gt;\text{(A)}\ 4/9 \qquad \text{(B)}\ 2/3 \qquad \text{(C)}\ 5/6 \qquad \text{(D)}\ 3/2 \qquad \text{(E)}\ 9/4&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let &lt;math&gt;B&lt;/math&gt; be the total area of the blue triangles, &lt;math&gt;W&lt;/math&gt; the total area of the white squares, and &lt;math&gt;R&lt;/math&gt; the area of the red square. Which of the following is correct?<br /> <br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);<br /> fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red);<br /> path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle;<br /> path divider=(-2,2)--(-3,3)--cycle;<br /> fill(onewhite,white);<br /> fill(rotate(90)*onewhite,white);<br /> fill(rotate(180)*onewhite,white);<br /> fill(rotate(270)*onewhite,white);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ B = W \qquad \text{(B)}\ W = R \qquad \text{(C)}\ B = R \qquad \text{(D)}\ 3B = 2R \qquad \text{(E)}\ 2R = W&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> There are 3 numbers A, B, and C, such that &lt;math&gt;1001C - 2002A = 4004&lt;/math&gt;, and &lt;math&gt;1001B + 3003A = 5005&lt;/math&gt;. What is the average of A, B, and C?<br /> <br /> &lt;math&gt;\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}&lt;/math&gt; Not uniquely determined<br /> <br /> <br /> [[2002 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> What is the sum of all of the roots of &lt;math&gt;(2x + 3) (x - 4) + (2x + 3) (x - 6) = 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ 7/2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 13&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB each, 12 of the files take up 0.7 MB each, and the rest take up 0.4 MB each. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?<br /> <br /> &lt;math&gt;\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 15 \qquad \text{(E)} 16&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> Mr. Earl E. Bird leaves home every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?<br /> <br /> &lt;math&gt;\text{(A)}\ 45 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 55 \qquad \text{(E)} 58&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> Given a triangle with side lengths 15, 20, and 25, find the triangle's smallest altitude.<br /> <br /> &lt;math&gt;\text{(A)}\ 6 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> Both roots of the quadratic equation &lt;math&gt;x^2 - 63x + k = 0&lt;/math&gt; are prime numbers. The number of possible values of &lt;math&gt;k&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?<br /> <br /> &lt;math&gt;\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> Let &lt;math&gt;a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5&lt;/math&gt;. What is &lt;math&gt;a + b + c + d&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> Sarah pours 4 ounces of coffee into a cup that can hold 8 ounces. Then she pours 4 ounces of cream into a second cup that can also hold 8 ounces. She then pours half of the contents of the first cup into the second cup, completely mixes the contents of the second cup, then pours half of the contents of the second cup back into the first cup. What fraction of the contents in the first cup is cream?<br /> <br /> &lt;math&gt;\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/3 \qquad \text{(C)}\ 3/8 \qquad \text{(D)}\ 2/5 \qquad \text{(E)} 1/2 &lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> A 3x3x3 cube is made of 27 normal dice. Each die's opposite sides sum to 7. What is the smallest possible sum of all of the values visible on the 6 faces of the large cube?<br /> <br /> &lt;math&gt;\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?<br /> <br /> &lt;math&gt;\text{(A)}\ 2\pi/3 \qquad \text{(B)}\ 2\pi \qquad \text{(C)}\ 5\pi/2 \qquad \text{(D)}\ 8\pi/3 \qquad \text{(E)}\ 3\pi&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> Points &lt;math&gt;A,B,C,D,E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie, in that order, on &lt;math&gt;\overline{AF}&lt;/math&gt;, dividing it into five segments, each of length 1. Point &lt;math&gt;G&lt;/math&gt; is not on line &lt;math&gt;AF&lt;/math&gt;. Point &lt;math&gt;H&lt;/math&gt; lies on &lt;math&gt;\overline{GD}&lt;/math&gt;, and point &lt;math&gt;J&lt;/math&gt; lies on &lt;math&gt;\overline{GF}&lt;/math&gt;. The line segments &lt;math&gt;\overline{HC}, \overline{JE},&lt;/math&gt; and &lt;math&gt;\overline{AG}&lt;/math&gt; are parallel. Find &lt;math&gt;HC/JE&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is<br /> <br /> &lt;math&gt;\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a [[perfect square]], and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?<br /> <br /> &lt;math&gt;\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> Points &lt;math&gt;A,B,C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on a line, in that order, with &lt;math&gt;AB = CD&lt;/math&gt; and &lt;math&gt;BC = 12&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is not on the line, and &lt;math&gt;BE = CE = 10&lt;/math&gt;. The perimeter of &lt;math&gt;\triangle AED&lt;/math&gt; is twice the perimeter of &lt;math&gt;\triangle BEC&lt;/math&gt;. Find &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina? <br /> <br /> &lt;math&gt;\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25&lt;/math&gt;<br /> <br /> [[2002 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> &lt;asy&gt;<br /> pair A,B,C,D;<br /> A=(0,0);<br /> B=(52,0);<br /> C=(38,20);<br /> D=(5,20);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> draw(A--B--C--D--cycle);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D&quot;,D,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;12&quot;,(B+C)/2,E);<br /> label(&quot;5&quot;,(D+A)/2,W);<br /> &lt;/asy&gt;<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; (diagram not to scale). 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 /> [[2002 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|ab=A|before=[[2001 AMC 10 Problems]]|after=[[2002 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_23&diff=163259 2020 AMC 10A Problems/Problem 23 2021-10-07T18:42:07Z <p>Ssding: /* Solution 2 (Rewording Solution 1) */ Added &quot;of&quot;</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #20]] and [[2020 AMC 10A Problems|2020 AMC 10A #23]]}}<br /> <br /> == Problem ==<br /> Let &lt;math&gt;T&lt;/math&gt; be the triangle in the coordinate plane with vertices &lt;math&gt;(0,0), (4,0),&lt;/math&gt; and &lt;math&gt;(0,3).&lt;/math&gt; Consider the following five isometries (rigid transformations) of the plane: rotations of &lt;math&gt;90^{\circ}, 180^{\circ},&lt;/math&gt; and &lt;math&gt;270^{\circ}&lt;/math&gt; counterclockwise around the origin, reflection across the &lt;math&gt;x&lt;/math&gt;-axis, and reflection across the &lt;math&gt;y&lt;/math&gt;-axis. How many of the &lt;math&gt;125&lt;/math&gt; sequences of three of these transformations (not necessarily distinct) will return &lt;math&gt;T&lt;/math&gt; to its original position? (For example, a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by a reflection across the &lt;math&gt;y&lt;/math&gt;-axis will return &lt;math&gt;T&lt;/math&gt; to its original position, but a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by another reflection across the &lt;math&gt;x&lt;/math&gt;-axis will not return &lt;math&gt;T&lt;/math&gt; to its original position.)<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> &lt;asy&gt;<br /> size(10cm);<br /> Label f; <br /> f.p=fontsize(6); <br /> xaxis(-6,6,Ticks(f, 2.0)); <br /> yaxis(-6,6,Ticks(f, 2.0));<br /> <br /> filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1));<br /> &lt;/asy&gt;<br /> <br /> First, any combination of motions we can make must reflect &lt;math&gt;T&lt;/math&gt; an even number of times. This is because every time we reflect &lt;math&gt;T&lt;/math&gt;, it changes orientation. Once &lt;math&gt;T&lt;/math&gt; has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed &lt;math&gt;3&lt;/math&gt; transformations and an even number of them must be reflections, we either reflect &lt;math&gt;T&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; times or &lt;math&gt;2&lt;/math&gt; times.<br /> <br /> <br /> <br /> Case 1: &lt;math&gt;0&lt;/math&gt; reflections on &lt;math&gt;T&lt;/math&gt;.<br /> <br /> <br /> <br /> In this case, we must use &lt;math&gt;3&lt;/math&gt; rotations to return &lt;math&gt;T&lt;/math&gt; to its original position. Notice that our set of rotations, &lt;math&gt;\{90^\circ,180^\circ,270^\circ\}&lt;/math&gt;, contains every multiple of &lt;math&gt;90^\circ&lt;/math&gt; except for &lt;math&gt;0^\circ&lt;/math&gt;. We can start with any two rotations &lt;math&gt;a,b&lt;/math&gt; in &lt;math&gt;\{90^\circ,180^\circ,270^\circ\}&lt;/math&gt; and there must be exactly one &lt;math&gt;c \equiv -a - b \pmod{360^\circ}&lt;/math&gt; such that we can use the three rotations &lt;math&gt;(a,b,c)&lt;/math&gt; which ensures that &lt;math&gt;a + b + c \equiv 0^\circ \pmod{360^\circ}&lt;/math&gt;. That way, the composition of rotations &lt;math&gt;a,b,c&lt;/math&gt; yields a full rotation. For example, if &lt;math&gt;a = b = 90^\circ&lt;/math&gt;, then &lt;math&gt;c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}&lt;/math&gt;, so &lt;math&gt;c = 180^\circ&lt;/math&gt; and the rotations &lt;math&gt;(90^\circ,90^\circ,180^\circ)&lt;/math&gt; yields a full rotation.<br /> <br /> The only case in which this fails is when &lt;math&gt;c&lt;/math&gt; would have to equal &lt;math&gt;0^\circ&lt;/math&gt;. This happens when &lt;math&gt;(a,b)&lt;/math&gt; is already a full rotation, namely, &lt;math&gt;(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),&lt;/math&gt; or &lt;math&gt;(270^\circ,90^\circ)&lt;/math&gt;. However, we can simply subtract these three cases from the total. Selecting &lt;math&gt;(a,b)&lt;/math&gt; from &lt;math&gt;\{90^\circ,180^\circ,270^\circ\}&lt;/math&gt; yields &lt;math&gt;3 \cdot 3 = 9&lt;/math&gt; choices, and with &lt;math&gt;3&lt;/math&gt; that fail, we are left with &lt;math&gt;6&lt;/math&gt; combinations for case &lt;math&gt;1&lt;/math&gt;.<br /> <br /> <br /> <br /> Case 2: &lt;math&gt;2&lt;/math&gt; reflections on &lt;math&gt;T&lt;/math&gt;.<br /> <br /> <br /> <br /> In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps &lt;math&gt;T&lt;/math&gt; back to itself, inserting a rotation before, between, or after these two reflections would change &lt;math&gt;T&lt;/math&gt;'s final location, meaning that any combination involving two reflections across the x-axis would not map &lt;math&gt;T&lt;/math&gt; back to itself. The same applies to two reflections across the y-axis.<br /> <br /> Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a &lt;math&gt;180^\circ&lt;/math&gt; rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us &lt;math&gt;3! = 6&lt;/math&gt; combinations for case 2.<br /> <br /> Combining both cases we get &lt;math&gt;6+6=\boxed{\textbf{(A)} 12}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Rewording of Solution 1)==<br /> <br /> As in the previous solution, note that we must have either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt; reflections because of orientation since reflection changes orientation that is impossible to fix by rotation. We also know we can't have the same reflection twice, since that would give a net of no change and would require an identity rotation.<br /> <br /> Suppose there are no reflections. Denote &lt;math&gt;90^{\circ}&lt;/math&gt; as &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;180^{\circ}&lt;/math&gt; as &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;270^{\circ}&lt;/math&gt; as &lt;math&gt;3&lt;/math&gt;, just for simplification purposes. We want a combination of &lt;math&gt;3&lt;/math&gt; of these that will sum to either &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt; (&lt;math&gt;0&lt;/math&gt; and &lt;math&gt;12&lt;/math&gt; are impossible since the minimum is &lt;math&gt;3&lt;/math&gt; and the max is &lt;math&gt;9&lt;/math&gt;). &lt;math&gt;4&lt;/math&gt; can be achieved with any permutation of &lt;math&gt;(1-1-2)&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; can be achieved with any permutation of &lt;math&gt;(2-3-3)&lt;/math&gt;. This case can be done in &lt;math&gt;3+3=6&lt;/math&gt; ways.<br /> <br /> Suppose there are two reflections. As noted already, they must be different, and as a result will take the triangle to the opposite side of the origin if we don't do any rotation. We have &lt;math&gt;1&lt;/math&gt; rotation left that we can do though, and the only one that will return to the original position is &lt;math&gt;2&lt;/math&gt;, which is &lt;math&gt;180^{\circ}&lt;/math&gt; AKA reflection across origin. Therefore, since all &lt;math&gt;3&lt;/math&gt; transformations are distinct. The three transformations can be applied anywhere since they are commutative (think quadrants). This gives &lt;math&gt;6&lt;/math&gt; ways.<br /> <br /> &lt;math&gt;6+6=\boxed{(A) 12}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Group Theory)==<br /> Define &lt;math&gt;s&lt;/math&gt; as a reflection, and &lt;math&gt;r&lt;/math&gt; as a &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise rotation. Thus, &lt;math&gt;r^4=s^2=e&lt;/math&gt;, and the five transformations can be represented as &lt;math&gt;{r, r^2, r^3, r^2s, s}&lt;/math&gt;, and &lt;math&gt;rs=sr^{-1}&lt;/math&gt;.<br /> <br /> Now either &lt;math&gt;s&lt;/math&gt; doesn't appear at all or appears twice. For the former case, it's easy to see that only &lt;math&gt;r, r, r^2&lt;/math&gt; and &lt;math&gt;r^2, r^3, r^3&lt;/math&gt; will work. Both can be permuted in &lt;math&gt;3&lt;/math&gt; ways, giving &lt;math&gt;6&lt;/math&gt; ways in total.<br /> <br /> For the latter case, note that &lt;math&gt;s&lt;/math&gt; can't appear twice, neither does &lt;math&gt;r^2s&lt;/math&gt;, else we need to get &lt;math&gt;e&lt;/math&gt; from &lt;math&gt;{r, r^2, r^3}&lt;/math&gt;, which is not possible. So &lt;math&gt;r^2s&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; must appear once each. The last transformation must be &lt;math&gt;r^2&lt;/math&gt;. A quick check shows that &lt;math&gt;{r^2, r^2s, s}&lt;/math&gt; is permutable, since &lt;math&gt;r^2s=sr^{-2}=sr^2&lt;/math&gt; (since &lt;math&gt;r^4=e&lt;/math&gt;). This gives &lt;math&gt;6&lt;/math&gt; ways.<br /> <br /> Thus the answer is &lt;math&gt;\boxed{(A) 12}&lt;/math&gt;.<br /> <br /> ~Xrider100<br /> <br /> ==Video Solutions==<br /> ===Video Solution 1===<br /> https://youtu.be/yAkj_5YMhhQ - Happytwin<br /> <br /> ===Video Solution 2 (Quick &amp; Simple)===<br /> Education, The Study of Everything<br /> <br /> https://youtu.be/SBhkM2frTUA<br /> <br /> ===Video Solution 3 (Richard Rusczyk)===<br /> https://artofproblemsolving.com/videos/amc/2020amc10a/513<br /> <br /> ===Video Solution 4===<br /> https://www.youtube.com/watch?v=iXwvTmFvo0c ~ MathEx<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_15&diff=163195 2021 AMC 10B Problems/Problem 15 2021-10-05T16:14:29Z <p>Ssding: /* Solution 3 */ Fixed mentions of x^3</p> <hr /> <div>==Problem==<br /> <br /> The real number &lt;math&gt;x&lt;/math&gt; satisfies the equation &lt;math&gt;x+\frac{1}{x} = \sqrt{5}&lt;/math&gt;. What is the value of &lt;math&gt;x^{11}-7x^{7}+x^3?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We square &lt;math&gt;x+\frac{1}{x}=\sqrt5&lt;/math&gt; to get &lt;math&gt;x^2+2+\frac{1}{x^2}=5&lt;/math&gt;. We subtract 2 on both sides for &lt;math&gt;x^2+\frac{1}{x^2}=3&lt;/math&gt; and square again, and see that &lt;math&gt;x^4+2+\frac{1}{x^4}=9&lt;/math&gt; so &lt;math&gt;x^4+\frac{1}{x^4}=7&lt;/math&gt;. We can divide our original expression of &lt;math&gt;x^{11}-7x^7+x^3&lt;/math&gt; by &lt;math&gt;x^7&lt;/math&gt; to get that it is equal to &lt;math&gt;x^7(x^4-7+\frac{1}{x^4})&lt;/math&gt;. Therefore because &lt;math&gt;x^4+\frac{1}{x^4}&lt;/math&gt; is 7, it is equal to &lt;math&gt;x^7(0)=\boxed{(B) 0}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt; and using the quadratic formula, we get &lt;math&gt;\frac{\sqrt{5} \pm 1}{2}&lt;/math&gt;. We can assume that it is &lt;math&gt;\frac{\sqrt{5}+1}{2}&lt;/math&gt;, and notice that this is also a solution the equation &lt;math&gt;x^2-x-1=0&lt;/math&gt;, i.e. we have &lt;math&gt;x^2=x+1&lt;/math&gt;. Repeatedly using this on the given (you can also just note Fibonacci numbers), <br /> &lt;cmath&gt; \begin{align*} <br /> (x^{11})-7x^7+x^3 &amp;= (x^{10}+x^9)-7x^7+x^3 \\<br /> &amp;=(2x^9+x^8)-7x^7+x^3 \\<br /> &amp;=(3x^8+2x^7)-7x^7+x^3 \\<br /> &amp;=(3x^8-5x^7)+x^3 \\<br /> &amp;=(-2x^7+3x^6)+x^3 \\<br /> &amp;=(x^6-2x^5)+x^3 \\<br /> &amp;=(-x^5+x^4+x^3) \\<br /> &amp;=-x^3(x^2-x-1) = \boxed{(\textbf{B}) 0}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ~Lcz<br /> <br /> ==Solution 3==<br /> We can immediately note that the exponents of &lt;math&gt;x^{11}-7x^7+x^3&lt;/math&gt; are an arithmetic sequence, so they are symmetric around the middle term. So, &lt;math&gt;x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})&lt;/math&gt;. We can see that since &lt;math&gt;x+\frac{1}{x} = \sqrt{5}&lt;/math&gt;, &lt;math&gt;x^2+2+\frac{1}{x^2} = 5&lt;/math&gt; and therefore &lt;math&gt;x^2+\frac{1}{x^2} = 3&lt;/math&gt;. Continuing from here, we get &lt;math&gt;x^4+2+\frac{1}{x^4} = 9&lt;/math&gt;, so &lt;math&gt;x^4-7+\frac{1}{x^4} = 0&lt;/math&gt;. We don't even need to find what &lt;math&gt;x^7&lt;/math&gt; is! This is since &lt;math&gt;x^7\cdot0&lt;/math&gt; is evidently &lt;math&gt;\boxed{(B) 0}&lt;/math&gt;, which is our answer.<br /> <br /> ~sosiaops<br /> <br /> ==Solution 4==<br /> We begin by multiplying &lt;math&gt;x+\frac{1}{x} = \sqrt{5}&lt;/math&gt; by &lt;math&gt;x&lt;/math&gt;, resulting in &lt;math&gt;x^2+1 = \sqrt{5}x&lt;/math&gt;. Now we see this equation: &lt;math&gt;x^{11}-7x^{7}+x^3&lt;/math&gt;. The terms all have &lt;math&gt;x^3&lt;/math&gt; in common, so we can factor that out, and what we're looking for becomes &lt;math&gt;x^3(x^8-7x^4+1)&lt;/math&gt;. Looking back to our original equation, we have &lt;math&gt;x^2+1 = \sqrt{5}x&lt;/math&gt;, which is equal to &lt;math&gt;x^2 = \sqrt{5}x-1&lt;/math&gt;. Using this, we can evaluate &lt;math&gt;x^4&lt;/math&gt; to be &lt;math&gt;5x^2-2\sqrt{5}x+1&lt;/math&gt;, and we see that there is another &lt;math&gt;x^2&lt;/math&gt;, so we put substitute it in again, resulting in &lt;math&gt;3\sqrt{5}x-4&lt;/math&gt;. Using the same way, we find that &lt;math&gt;x^8&lt;/math&gt; is &lt;math&gt;21\sqrt{5}x-29&lt;/math&gt;. We put this into &lt;math&gt;x^3(x^8-7x^4+1)&lt;/math&gt;, resulting in &lt;math&gt;x^3(0)&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{(B)~0}&lt;/math&gt;.<br /> <br /> ~purplepenguin2<br /> <br /> ==Solution 5==<br /> The equation we are given is &lt;math&gt;x+\tfrac{1}{x}=\sqrt{5}...&lt;/math&gt; Yuck. Fractions and radicals! We multiply both sides by &lt;math&gt;x,&lt;/math&gt; square, and re-arrange to get &lt;cmath&gt;x^2+1=\sqrt{5}x \implies x^4+2x^2+1=5x^2 \implies x^4-3x^2+1=0.&lt;/cmath&gt; Now, let us consider the expression we wish to acquire. Factoring out &lt;math&gt;x^3,&lt;/math&gt; we have &lt;cmath&gt;x^3\left(x^8-7x^4+1\right) = x^3\left(x^8+2x^4+1-9x^4\right).&lt;/cmath&gt; Then, we notice that &lt;math&gt;x^8+2x^4+1=\left(x^4+1\right)^2.&lt;/math&gt; Furthermore, &lt;cmath&gt;x^4+1=3x^2 \implies \left(x^4+1\right)^2=x^8+2x^4+1=9x^4.&lt;/cmath&gt; Thus, our answer is &lt;cmath&gt;x^3\left(9x^4-9x^4\right) = x^3 \cdot 0 = \boxed{\textbf{(B)}} ~ 0.&lt;/cmath&gt;<br /> ~peace09<br /> <br /> == Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials) ==<br /> https://youtu.be/hzcSPVGFbC8<br /> <br /> ~ pi_is_3.14<br /> <br /> == Video Solution by Interstigation (Simple Silly Bashing) ==<br /> https://youtu.be/Hdk2SDOcw7c<br /> <br /> ~ Interstigation<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> Not the most efficient method, but gets the job done.<br /> <br /> https://youtu.be/L1iW94Ue3eI?t=1468<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2021|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_22&diff=163194 2021 AMC 10A Problems/Problem 22 2021-10-05T15:51:17Z <p>Ssding: /* Solution 1 */ Changed &quot;pages&quot; to &quot;sheets&quot;</p> <hr /> <div>==Problem==<br /> Hiram's algebra notes are &lt;math&gt;50&lt;/math&gt; pages long and are printed on &lt;math&gt;25&lt;/math&gt; sheets of paper; the first sheet contains pages &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;, the second sheet contains pages &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly &lt;math&gt;19&lt;/math&gt;. How many sheets were borrowed?<br /> <br /> &lt;math&gt;\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose the roommate took sheets &lt;math&gt;a&lt;/math&gt; through &lt;math&gt;b&lt;/math&gt;, or equivalently, page numbers &lt;math&gt;2a-1&lt;/math&gt; through &lt;math&gt;2b&lt;/math&gt;. Because there are &lt;math&gt;(2b-2a+2)&lt;/math&gt; numbers taken, &lt;cmath&gt;\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.&lt;/cmath&gt; The first possible solution that comes to mind is if &lt;math&gt;2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12&lt;/math&gt;, which indeed works, giving &lt;math&gt;b=22&lt;/math&gt; and &lt;math&gt;a=10&lt;/math&gt;. The answer is &lt;math&gt;22-10+1=\boxed{\textbf{(B)} ~13}&lt;/math&gt;.<br /> <br /> ~Lcz<br /> <br /> ==Solution 2==<br /> Suppose the smallest page number borrowed is &lt;math&gt;k,&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; pages are borrowed. It follows that the largest page number borrowed is &lt;math&gt;k+n-1.&lt;/math&gt;<br /> <br /> We have the following preconditions:<br /> &lt;ol style=&quot;margin-left: 1.5em;&quot;&gt;<br /> &lt;li&gt;&lt;math&gt;n&lt;/math&gt; pages are borrowed means that &lt;math&gt;\frac{n}{2}&lt;/math&gt; sheets are borrowed, from which &lt;math&gt;n&lt;/math&gt; must be even.&lt;/li&gt;&lt;p&gt;<br /> &lt;li&gt;&lt;math&gt;k&lt;/math&gt; must be odd, as the smallest page number borrowed is on the right side (odd-numbered).&lt;/li&gt;&lt;p&gt;<br /> &lt;li&gt;&lt;math&gt;1+2+3+\cdots+50=\frac{51(50)}{2}=1275.&lt;/math&gt;&lt;/li&gt;&lt;p&gt;<br /> &lt;li&gt;The sum of the page numbers borrowed is &lt;math&gt;\frac{(2k+n-1)n}{2}.&lt;/math&gt;&lt;/li&gt;&lt;p&gt;<br /> &lt;/ol&gt;<br /> Together, we have &lt;cmath&gt;\begin{align*}<br /> \frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&amp;=19 \\<br /> 1275-\frac{(2k+n-1)n}{2}&amp;=19(50-n) \\<br /> 2550-(2k+n-1)n&amp;=38(50-n) \\<br /> 2550-(2k+n-1)n&amp;=1900-38n \\<br /> 650&amp;=(2k+n-39)n.<br /> \end{align*}&lt;/cmath&gt;<br /> The factors of &lt;math&gt;650&lt;/math&gt; are &lt;cmath&gt;1,2,5,10,13,25,26,50,65,130,325,650.&lt;/cmath&gt; Since &lt;math&gt;n&lt;/math&gt; is even, we only have a few cases to consider:<br /> &lt;cmath&gt;\begin{array}{c|c|c}<br /> &amp; &amp; \\ [-2.25ex]<br /> \boldsymbol{n} &amp; \boldsymbol{2k+n-39} &amp; \boldsymbol{k} \\ [0.5ex]<br /> \hline <br /> &amp; &amp; \\ [-2ex]<br /> 2 &amp; 325 &amp; 181 \\ <br /> 10 &amp; 65 &amp; 47 \\<br /> 26 &amp; 25 &amp; 19 \\<br /> 50 &amp; 13 &amp; 1 \\<br /> 130 &amp; 5 &amp; -43 \\<br /> 650 &amp; 1 &amp; -305 \\<br /> \end{array}&lt;/cmath&gt;<br /> Since &lt;math&gt;1\leq k \leq 49,&lt;/math&gt; only &lt;math&gt;k=47,19,1&lt;/math&gt; are possible:<br /> <br /> * If &lt;math&gt;k=47,&lt;/math&gt; then there will not be sufficient pages when we take &lt;math&gt;10&lt;/math&gt; pages out starting from page &lt;math&gt;47.&lt;/math&gt;<br /> <br /> * If &lt;math&gt;k=1,&lt;/math&gt; then the average page number of all remaining sheets will be undefined, as there will be no sheets remaining after we take &lt;math&gt;50&lt;/math&gt; pages (&lt;math&gt;25&lt;/math&gt; sheets) out starting from page &lt;math&gt;1.&lt;/math&gt;<br /> <br /> Therefore, the only possibility is &lt;math&gt;k=19.&lt;/math&gt; We conclude that &lt;math&gt;n=26&lt;/math&gt; pages, or &lt;math&gt;\frac n2=\boxed{\textbf{(B)} ~13}&lt;/math&gt; sheets, are borrowed.<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 3== <br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the number of sheets borrowed, with an average page number &lt;math&gt;k+25.5&lt;/math&gt;. The remaining &lt;math&gt;25-n&lt;/math&gt; sheets have an average page number of &lt;math&gt;19&lt;/math&gt; which is less than &lt;math&gt;25.5&lt;/math&gt;, the average page number of all &lt;math&gt;50&lt;/math&gt; pages, therefore &lt;math&gt;k&gt;0&lt;/math&gt;. Since the borrowed sheets start with an odd page number and end with an even page number we have &lt;math&gt;k \in \mathbb N&lt;/math&gt;. We notice that &lt;math&gt;n &lt; 25&lt;/math&gt; and &lt;math&gt;k \le (49+50)/2-25.5=24&lt;25&lt;/math&gt;.<br /> <br /> The weighted increase of average page number from &lt;math&gt;25.5&lt;/math&gt; to &lt;math&gt;k+25.5&lt;/math&gt; should be equal to the weighted decrease of average page number from &lt;math&gt;25.5&lt;/math&gt; to &lt;math&gt;19&lt;/math&gt;, where the weights are the page number in each group (borrowed vs. remained), therefore<br /> <br /> &lt;cmath&gt;2nk=2(25-n)(25.5-19)=13(25-n) \implies 13 | n \text{ or } 13 | k&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;n, k &lt; 25&lt;/math&gt; we have either &lt;math&gt;n=13&lt;/math&gt; or &lt;math&gt;k=13&lt;/math&gt;. If &lt;math&gt;n=13&lt;/math&gt; then &lt;math&gt;k=6&lt;/math&gt;. If &lt;math&gt;k=13&lt;/math&gt; then &lt;math&gt;2n=25-n&lt;/math&gt; which is impossible. Therefore the answer should be &lt;math&gt;n=\boxed{\textbf{(B)} ~13}&lt;/math&gt;.<br /> <br /> ~asops<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;(2k-1)-2n&lt;/math&gt; be pages be borrowed, the sum of digits in those pages is &lt;math&gt;(2n+2k+1)(n-k)&lt;/math&gt; while the sum of the rest pages is &lt;math&gt;1275-(2n+2k+1)(n-k)&lt;/math&gt; and we know the average of the rest is &lt;math&gt;\frac{1275-(2n+2k+1)}{50-2n+2k}&lt;/math&gt; which equals to &lt;math&gt;19&lt;/math&gt;; multiply this out we got &lt;math&gt;950-38(n-k)=1275-(2n+2k+1)(n-k)&lt;/math&gt; and we got &lt;math&gt;(2n+2k-37)(n-k)=325&lt;/math&gt;. As &lt;math&gt;325=25\cdot13&lt;/math&gt;, we can see &lt;math&gt;n-k=13&lt;/math&gt; and that is desired &lt;math&gt;\boxed{\textbf{(B)} ~13}&lt;/math&gt;.<br /> <br /> ~bluesoul<br /> <br /> == Video Solution by OmegaLearn (Arithmetic Sequences and System of Equations) ==<br /> https://youtu.be/dWOLIdTxwa4<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by MRENTHUSIASM (English &amp; Chinese)==<br /> https://www.youtube.com/watch?v=28te8OUiVxE<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_21&diff=163193 2021 AMC 10A Problems/Problem 21 2021-10-05T15:45:25Z <p>Ssding: /* Solution */ Deleted the reference to the Segment Addition Postulate</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be an equiangular hexagon. The lines &lt;math&gt;AB, CD,&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt; determine a triangle with area &lt;math&gt;192\sqrt{3}&lt;/math&gt;, and the lines &lt;math&gt;BC, DE,&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; determine a triangle with area &lt;math&gt;324\sqrt{3}&lt;/math&gt;. The perimeter of hexagon &lt;math&gt;ABCDEF&lt;/math&gt; can be expressed as &lt;math&gt;m +n\sqrt{p}&lt;/math&gt;, where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. What is &lt;math&gt;m + n + p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63&lt;/math&gt;<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> /* Made by MRENTHUSIASM */<br /> size(250);<br /> path P1, P2;<br /> P1 = scale(16sqrt(3))*polygon(3);<br /> P2 = shift(3,3)*scale(36)*rotate(180)*polygon(3);<br /> draw(P1, dashed+black);<br /> draw(P2, dashed+black);<br /> pair A, B, C, D, E, F;<br /> E = intersectionpoints(P1,P2);<br /> F = intersectionpoints(P1,P2);<br /> A = intersectionpoints(P1,P2);<br /> B = intersectionpoints(P1,P2);<br /> C = intersectionpoints(P1,P2);<br /> D = intersectionpoints(P1,P2);<br /> filldraw(A--B--C--D--E--F--cycle,yellow);<br /> dot(&quot;E$&quot;,E,1.5*dir(0),linewidth(4));<br /> dot(&quot;$F$&quot;,F,1.5*dir(60),linewidth(4));<br /> dot(&quot;$A$&quot;,A,1.5*dir(120),linewidth(4));<br /> dot(&quot;$B$&quot;,B,1.5*dir(180),linewidth(4));<br /> dot(&quot;$C$&quot;,C,1.5*dir(-120),linewidth(4));<br /> dot(&quot;$D&quot;,D,1.5*dir(-60),linewidth(4));<br /> dot(16sqrt(3)*dir(90)^^16sqrt(3)*dir(210)^^16sqrt(3)*dir(330),linewidth(4));<br /> dot((3,3)+36*dir(30)^^(3,3)+36*dir(150)^^(3,3)+36*dir(270),linewidth(4));<br /> &lt;/asy&gt;<br /> ~MRENTHUSIASM<br /> <br /> ==Solution==<br /> Let &lt;math&gt;P,Q,R,X,Y,&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; be the intersections &lt;math&gt;\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{FA}\cap\overleftrightarrow{BC},&lt;/math&gt; respectively.<br /> <br /> The sum of the interior angles of any hexagon is &lt;math&gt;720^\circ.&lt;/math&gt; Since hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is equiangular, each of its interior angles is &lt;math&gt;720^\circ\div6=120^\circ.&lt;/math&gt; By angle chasing, we conclude that the interior angles of &lt;math&gt;\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,&lt;/math&gt; and &lt;math&gt;\triangle ZAB&lt;/math&gt; are all &lt;math&gt;60^\circ.&lt;/math&gt; Therefore, these triangles are all equilateral triangles, from which &lt;math&gt;\triangle PQR&lt;/math&gt; and &lt;math&gt;\triangle XYZ&lt;/math&gt; are both equilateral triangles.<br /> <br /> We are given that<br /> &lt;cmath&gt;\begin{alignat*}{8}<br /> [PQR]&amp;=\frac{\sqrt{3}}{4}\cdot PQ^2&amp;&amp;=192\sqrt3, \\<br /> [XYZ]&amp;=\frac{\sqrt{3}}{4}\cdot YZ^2&amp;&amp;=324\sqrt3,<br /> \end{alignat*}&lt;/cmath&gt;<br /> so we get &lt;math&gt;PQ=16\sqrt3&lt;/math&gt; and &lt;math&gt;YZ=36,&lt;/math&gt; respectively.<br /> <br /> By the equilateral triangles, we find the perimeter of hexagon &lt;math&gt;ABCDEF:&lt;/math&gt;<br /> &lt;cmath&gt;\begin{align*}<br /> AB+BC+CD+DE+EF+FA&amp;=AZ+PC+CD+DQ+YF+FA \\<br /> &amp;=(YF+FA+AZ)+(PC+CD+DQ) \\<br /> &amp;=YZ+PQ \\<br /> &amp;=36+16\sqrt{3}.<br /> \end{align*}&lt;/cmath&gt;<br /> Finally, the answer is &lt;math&gt;36+16+3=\boxed{\textbf{(C)} ~55}.&lt;/math&gt;<br /> <br /> ~sugar_rush (Fundamental Logic)<br /> <br /> ~MRENTHUSIASM (Reconstruction)<br /> <br /> == Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) ==<br /> https://youtu.be/ptBwDcmDaLA<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/8qcbZ8c7fHg<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution by MRENTHUSIASM (English &amp; Chinese)==<br /> https://www.youtube.com/watch?v=0n8EAu2VAiM<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2021|ab=A|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_25&diff=163048 2019 AMC 10A Problems/Problem 25 2021-10-02T16:13:10Z <p>Ssding: /* Solution 2 */ Fixed the link to Legendre's Formula</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #25]] and [[2019 AMC 12A Problems|2019 AMC 12A #24]]}}<br /> <br /> ==Problem==<br /> <br /> For how many integers &lt;math&gt;n&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt;, inclusive, is &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}&lt;/cmath&gt; an integer? (Recall that &lt;math&gt;0! = 1&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35&lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> The main insight is that <br /> <br /> &lt;cmath&gt;\frac{(n^2)!}{(n!)^{n+1}}&lt;/cmath&gt; <br /> <br /> is always an integer. This is true because it is precisely the number of ways to split up &lt;math&gt;n^2&lt;/math&gt; objects into &lt;math&gt;n&lt;/math&gt; unordered groups of size &lt;math&gt;n&lt;/math&gt;. Thus,<br /> <br /> &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}&lt;/cmath&gt;<br /> <br /> is an integer if &lt;math&gt;n! \mid n^2&lt;/math&gt;, or in other words, if &lt;math&gt;(n-1)! \mid n&lt;/math&gt;, is an integer. This condition is false precisely when &lt;math&gt;n=4&lt;/math&gt; or &lt;math&gt;n&lt;/math&gt; is prime, by [[Wilson's Theorem]]. There are &lt;math&gt;15&lt;/math&gt; primes between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt;, inclusive, so there are &lt;math&gt;15 + 1 = 16&lt;/math&gt; terms for which<br /> <br /> &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^{n}}&lt;/cmath&gt;<br /> <br /> is potentially not an integer. It can be easily verified that the above expression is not an integer for &lt;math&gt;n=4&lt;/math&gt; as there are more factors of &lt;math&gt;2&lt;/math&gt; in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime &lt;math&gt;n=p&lt;/math&gt;, as there are more factors of p in the denominator than the numerator. Thus all &lt;math&gt;16&lt;/math&gt; values of n make the expression not an integer and the answer is &lt;math&gt;50-16=\boxed{\textbf{(D)}\ 34}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We can use the P-Adic Valuation (more info could be found here: [[Mathematicial notation]]) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by &lt;math&gt;v_p (n)&lt;/math&gt; and is defined as the greatest power of some prime 'p' that divides n. For example, &lt;math&gt;v_2 (6)=1&lt;/math&gt; or &lt;math&gt;v_7 (245)=2&lt;/math&gt; .) Using Legendre's formula, we know that :<br /> <br /> &lt;cmath&gt; v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor &lt;/cmath&gt;<br /> <br /> Seeing factorials involved in the problem, this prompts us to use [[Legendre's Formula]] where n is a power of a prime.<br /> <br /> We also know that , &lt;math&gt;v_p (m^n) = n \cdot v_p (m)&lt;/math&gt; .<br /> Knowing that &lt;math&gt;a\mid b&lt;/math&gt; if &lt;math&gt;v_p (a) \le v_p (b)&lt;/math&gt; , we have that :<br /> <br /> &lt;cmath&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!) &lt;/cmath&gt; and we must find all n for which this is true.<br /> <br /> If we plug in &lt;math&gt;n=p&lt;/math&gt;, by Legendre's we get two equations:<br /> <br /> &lt;cmath&gt; v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1 &lt;/cmath&gt;<br /> <br /> And we also get :<br /> <br /> &lt;cmath&gt; v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p &lt;/cmath&gt;<br /> <br /> But we are asked to prove that &lt;math&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1 &lt;/math&gt; which is false for all 'n' where n is prime.<br /> <br /> Now we try the same for &lt;math&gt;n=p^2&lt;/math&gt; , where p is a prime. By Legendre we arrive at:<br /> <br /> &lt;cmath&gt;v_p ((p^4 -1)!) = p^3 + p^2 + p -3&lt;/cmath&gt; and &lt;cmath&gt;p^2 \cdot v_p (p^2 !) = p^3 + p^2 &lt;/cmath&gt;<br /> <br /> Then we get:<br /> <br /> &lt;cmath&gt; p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3 &lt;/cmath&gt; Which is true for all primes except for 2, so &lt;math&gt;2^2 = 4&lt;/math&gt; doesn't work. It can easily be verified that for all &lt;math&gt;n=p^i&lt;/math&gt; where &lt;math&gt;i&lt;/math&gt; is an integer greater than 2, satisfies the inequality :&lt;cmath&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!).&lt;/cmath&gt;<br /> <br /> Therefore, there are 16 values that don't work and &lt;math&gt; 50-16 = \boxed{\mathbf{(D)}\ 34}&lt;/math&gt; values that work.<br /> <br /> ~qwertysri987<br /> <br /> ==See Also==<br /> Video Solution by Richard Rusczyk:<br /> https://www.youtube.com/watch?v=9klaWnZojq0<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_25&diff=163047 2019 AMC 10A Problems/Problem 25 2021-10-02T16:12:13Z <p>Ssding: /* Solution */ Changed Legendre's formula into a link, and changed the header from &quot;Solution&quot; to &quot;Solutions&quot;</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #25]] and [[2019 AMC 12A Problems|2019 AMC 12A #24]]}}<br /> <br /> ==Problem==<br /> <br /> For how many integers &lt;math&gt;n&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt;, inclusive, is &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}&lt;/cmath&gt; an integer? (Recall that &lt;math&gt;0! = 1&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35&lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> The main insight is that <br /> <br /> &lt;cmath&gt;\frac{(n^2)!}{(n!)^{n+1}}&lt;/cmath&gt; <br /> <br /> is always an integer. This is true because it is precisely the number of ways to split up &lt;math&gt;n^2&lt;/math&gt; objects into &lt;math&gt;n&lt;/math&gt; unordered groups of size &lt;math&gt;n&lt;/math&gt;. Thus,<br /> <br /> &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}&lt;/cmath&gt;<br /> <br /> is an integer if &lt;math&gt;n! \mid n^2&lt;/math&gt;, or in other words, if &lt;math&gt;(n-1)! \mid n&lt;/math&gt;, is an integer. This condition is false precisely when &lt;math&gt;n=4&lt;/math&gt; or &lt;math&gt;n&lt;/math&gt; is prime, by [[Wilson's Theorem]]. There are &lt;math&gt;15&lt;/math&gt; primes between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;50&lt;/math&gt;, inclusive, so there are &lt;math&gt;15 + 1 = 16&lt;/math&gt; terms for which<br /> <br /> &lt;cmath&gt;\frac{(n^2-1)!}{(n!)^{n}}&lt;/cmath&gt;<br /> <br /> is potentially not an integer. It can be easily verified that the above expression is not an integer for &lt;math&gt;n=4&lt;/math&gt; as there are more factors of &lt;math&gt;2&lt;/math&gt; in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime &lt;math&gt;n=p&lt;/math&gt;, as there are more factors of p in the denominator than the numerator. Thus all &lt;math&gt;16&lt;/math&gt; values of n make the expression not an integer and the answer is &lt;math&gt;50-16=\boxed{\textbf{(D)}\ 34}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We can use the P-Adic Valuation (more info could be found here: [[Mathematicial notation]]) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by &lt;math&gt;v_p (n)&lt;/math&gt; and is defined as the greatest power of some prime 'p' that divides n. For example, &lt;math&gt;v_2 (6)=1&lt;/math&gt; or &lt;math&gt;v_7 (245)=2&lt;/math&gt; .) Using Legendre's formula, we know that :<br /> <br /> &lt;cmath&gt; v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor &lt;/cmath&gt;<br /> <br /> Seeing factorials involved in the problem, this prompts us to use [[Legendre's formula]] where n is a power of a prime.<br /> <br /> We also know that , &lt;math&gt;v_p (m^n) = n \cdot v_p (m)&lt;/math&gt; .<br /> Knowing that &lt;math&gt;a\mid b&lt;/math&gt; if &lt;math&gt;v_p (a) \le v_p (b)&lt;/math&gt; , we have that :<br /> <br /> &lt;cmath&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!) &lt;/cmath&gt; and we must find all n for which this is true.<br /> <br /> If we plug in &lt;math&gt;n=p&lt;/math&gt;, by Legendre's we get two equations:<br /> <br /> &lt;cmath&gt; v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1 &lt;/cmath&gt;<br /> <br /> And we also get :<br /> <br /> &lt;cmath&gt; v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p &lt;/cmath&gt;<br /> <br /> But we are asked to prove that &lt;math&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1 &lt;/math&gt; which is false for all 'n' where n is prime.<br /> <br /> Now we try the same for &lt;math&gt;n=p^2&lt;/math&gt; , where p is a prime. By Legendre we arrive at:<br /> <br /> &lt;cmath&gt;v_p ((p^4 -1)!) = p^3 + p^2 + p -3&lt;/cmath&gt; and &lt;cmath&gt;p^2 \cdot v_p (p^2 !) = p^3 + p^2 &lt;/cmath&gt;<br /> <br /> Then we get:<br /> <br /> &lt;cmath&gt; p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3 &lt;/cmath&gt; Which is true for all primes except for 2, so &lt;math&gt;2^2 = 4&lt;/math&gt; doesn't work. It can easily be verified that for all &lt;math&gt;n=p^i&lt;/math&gt; where &lt;math&gt;i&lt;/math&gt; is an integer greater than 2, satisfies the inequality :&lt;cmath&gt; n \cdot v_p (n!) \le v_p ((n^2 -1 )!).&lt;/cmath&gt;<br /> <br /> Therefore, there are 16 values that don't work and &lt;math&gt; 50-16 = \boxed{\mathbf{(D)}\ 34}&lt;/math&gt; values that work.<br /> <br /> ~qwertysri987<br /> <br /> ==See Also==<br /> Video Solution by Richard Rusczyk:<br /> https://www.youtube.com/watch?v=9klaWnZojq0<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_24&diff=162730 2014 AMC 12B Problems/Problem 24 2021-09-25T13:58:45Z <p>Ssding: /* Solution 1 */ Changed two theorem names to links</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABCDE&lt;/math&gt; be a pentagon inscribed in a circle such that &lt;math&gt;AB = CD = 3&lt;/math&gt;, &lt;math&gt;BC = DE = 10&lt;/math&gt;, and &lt;math&gt;AE= 14&lt;/math&gt;. The sum of the lengths of all diagonals of &lt;math&gt;ABCDE&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }129\qquad<br /> \textbf{(B) }247\qquad<br /> \textbf{(C) }353\qquad<br /> \textbf{(D) }391\qquad<br /> \textbf{(E) }421\qquad&lt;/math&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;BE=a&lt;/math&gt;, &lt;math&gt;AD=b&lt;/math&gt;, and &lt;math&gt;AC=CE=BD=c&lt;/math&gt;. Let &lt;math&gt;F&lt;/math&gt; be on &lt;math&gt;AE&lt;/math&gt; such that &lt;math&gt;CF \perp AE&lt;/math&gt;. <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.4)+fontsize(10));<br /> pen s = linewidth(0.8)+fontsize(8);<br /> <br /> pair O,A,B,C,D,E0,F;<br /> O=origin;<br /> A= dir(198);<br /> path c = CR(O,1);<br /> real r = 0.13535;<br /> B = IP(c, CR(A,3*r));<br /> C = IP(c, CR(B,10*r));<br /> D = IP(c, CR(C,3*r));<br /> E0 = OP(c, CR(D,10*r));<br /> F = foot(C,A,E0);<br /> <br /> dot(&quot;A$&quot;, A, A-O);<br /> dot(&quot;$B$&quot;, B, B-O);<br /> dot(&quot;$C$&quot;, C, C-O);<br /> dot(&quot;$D$&quot;, D, D-O);<br /> dot(&quot;$E$&quot;, E0, E0-O);<br /> dot(&quot;$F$&quot;, F, F-C);<br /> label(&quot;$c$&quot;,A--C,S);<br /> label(&quot;$c$&quot;,E0--C,W);<br /> label(&quot;$7$&quot;,F--E0,S);<br /> label(&quot;$7$&quot;,F--A,S);<br /> label(&quot;$3$&quot;,A--B,2*W);<br /> label(&quot;$10$&quot;,B--C,2*N);<br /> label(&quot;$3$&quot;,C--D,2*NE);<br /> label(&quot;$10&quot;,D--E0,E);<br /> draw(A--B--C--D--E0--A, black+0.8);<br /> <br /> draw(CR(O,1), s);<br /> draw(A--C--E0, royalblue);<br /> draw(C--F, royalblue+dashed);<br /> draw(rightanglemark(E0,F,C,2));<br /> MA(&quot;\theta&quot;,A,B,C,0.075);<br /> MA(&quot;\pi-\theta&quot;,C,E0,A,0.1);<br /> &lt;/asy&gt;<br /> In &lt;math&gt;\triangle CFE&lt;/math&gt; we have &lt;math&gt;\cos\theta = -\cos(\pi-\theta)=-7/c&lt;/math&gt;. We use the [[Law of Cosines]] on &lt;math&gt;\triangle ABC&lt;/math&gt; to get &lt;math&gt;60\cos\theta = 109-c^2&lt;/math&gt;. Eliminating &lt;math&gt;\cos\theta&lt;/math&gt; we get &lt;math&gt;c^3-109c-420=0&lt;/math&gt; which factorizes as <br /> &lt;cmath&gt;(c+7)(c+5)(c-12)=0.&lt;/cmath&gt;Discarding the negative roots we have &lt;math&gt;c=12&lt;/math&gt;. Thus &lt;math&gt;BD=AC=CE=12&lt;/math&gt;. For &lt;math&gt;BE=a&lt;/math&gt;, we use Ptolemy's theorem on cyclic quadrilateral &lt;math&gt;ABCE&lt;/math&gt; to get &lt;math&gt;a=44/3&lt;/math&gt;. For &lt;math&gt;AD=b&lt;/math&gt;, we use [[Ptolemy's theorem]] on cyclic quadrilateral &lt;math&gt;ACDE&lt;/math&gt; to get &lt;math&gt;b=27/2&lt;/math&gt;.<br /> <br /> The sum of the lengths of the diagonals is &lt;math&gt;12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}&lt;/math&gt; so the answer is &lt;math&gt;385 + 6 = \fbox{\textbf{(D) }391}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;a&lt;/math&gt; denote the length of a diagonal opposite adjacent sides of length &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; for sides &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; for sides &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt;. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:<br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> c^2 &amp;= 3a+100 \\<br /> c^2 &amp;= 10b+9 \\<br /> ab &amp;= 30+14c \\<br /> ac &amp;= 3c+140\\<br /> bc &amp;= 10c+42<br /> \end{align}<br /> &lt;/cmath&gt;<br /> <br /> Using equations &lt;math&gt;(1)&lt;/math&gt; and &lt;math&gt;(2)&lt;/math&gt;, we obtain:<br /> <br /> &lt;cmath&gt;<br /> a = \frac{c^2-100}{3}<br /> &lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;<br /> b = \frac{c^2-9}{10}<br /> &lt;/cmath&gt;<br /> <br /> Plugging into equation &lt;math&gt;(4)&lt;/math&gt;, we find that:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{c^2-100}{3}c &amp;= 3c + 140\\<br /> \frac{c^3-100c}{3} &amp;= 3c + 140\\<br /> c^3-100c &amp;= 9c + 420\\<br /> c^3-109c-420 &amp;=0\\<br /> (c-12)(c+7)(c+5)&amp;=0<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Or similarly into equation &lt;math&gt;(5)&lt;/math&gt; to check:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{c^2-9}{10}c &amp;= 10c+42\\<br /> \frac{c^3-9c}{10} &amp;= 10c + 42\\<br /> c^3-9c &amp;= 100c + 420\\<br /> c^3-109c-420 &amp;=0\\<br /> (c-12)(c+7)(c+5)&amp;=0<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;c&lt;/math&gt;, being a length, must be positive, implying that &lt;math&gt;c=12&lt;/math&gt;. In fact, this is reasonable, since &lt;math&gt;10+3\approx 12&lt;/math&gt; in the pentagon with apparently obtuse angles. Plugging this back into equations &lt;math&gt;(1)&lt;/math&gt; and &lt;math&gt;(2)&lt;/math&gt; we find that &lt;math&gt;a = \frac{44}{3}&lt;/math&gt; and &lt;math&gt;b= \frac{135}{10}=\frac{27}{2}&lt;/math&gt;.<br /> <br /> We desire &lt;math&gt;3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{216+88+81}{6}=\frac{385}{6}&lt;/math&gt;, so it follows that the answer is &lt;math&gt;385 + 6 = \fbox{\textbf{(D) }391}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_24&diff=162728 2014 AMC 12B Problems/Problem 24 2021-09-25T13:50:17Z <p>Ssding: /* Solution 1 */ Replaced &quot;cosine formula&quot; with &quot;Law of Cosines&quot;</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABCDE&lt;/math&gt; be a pentagon inscribed in a circle such that &lt;math&gt;AB = CD = 3&lt;/math&gt;, &lt;math&gt;BC = DE = 10&lt;/math&gt;, and &lt;math&gt;AE= 14&lt;/math&gt;. The sum of the lengths of all diagonals of &lt;math&gt;ABCDE&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }129\qquad<br /> \textbf{(B) }247\qquad<br /> \textbf{(C) }353\qquad<br /> \textbf{(D) }391\qquad<br /> \textbf{(E) }421\qquad&lt;/math&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;BE=a&lt;/math&gt;, &lt;math&gt;AD=b&lt;/math&gt;, and &lt;math&gt;AC=CE=BD=c&lt;/math&gt;. Let &lt;math&gt;F&lt;/math&gt; be on &lt;math&gt;AE&lt;/math&gt; such that &lt;math&gt;CF \perp AE&lt;/math&gt;. <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.4)+fontsize(10));<br /> pen s = linewidth(0.8)+fontsize(8);<br /> <br /> pair O,A,B,C,D,E0,F;<br /> O=origin;<br /> A= dir(198);<br /> path c = CR(O,1);<br /> real r = 0.13535;<br /> B = IP(c, CR(A,3*r));<br /> C = IP(c, CR(B,10*r));<br /> D = IP(c, CR(C,3*r));<br /> E0 = OP(c, CR(D,10*r));<br /> F = foot(C,A,E0);<br /> <br /> dot(&quot;A$&quot;, A, A-O);<br /> dot(&quot;$B$&quot;, B, B-O);<br /> dot(&quot;$C$&quot;, C, C-O);<br /> dot(&quot;$D$&quot;, D, D-O);<br /> dot(&quot;$E$&quot;, E0, E0-O);<br /> dot(&quot;$F$&quot;, F, F-C);<br /> label(&quot;$c$&quot;,A--C,S);<br /> label(&quot;$c$&quot;,E0--C,W);<br /> label(&quot;$7$&quot;,F--E0,S);<br /> label(&quot;$7$&quot;,F--A,S);<br /> label(&quot;$3$&quot;,A--B,2*W);<br /> label(&quot;$10$&quot;,B--C,2*N);<br /> label(&quot;$3$&quot;,C--D,2*NE);<br /> label(&quot;$10\$&quot;,D--E0,E);<br /> draw(A--B--C--D--E0--A, black+0.8);<br /> <br /> draw(CR(O,1), s);<br /> draw(A--C--E0, royalblue);<br /> draw(C--F, royalblue+dashed);<br /> draw(rightanglemark(E0,F,C,2));<br /> MA(&quot;\theta&quot;,A,B,C,0.075);<br /> MA(&quot;\pi-\theta&quot;,C,E0,A,0.1);<br /> &lt;/asy&gt;<br /> In &lt;math&gt;\triangle CFE&lt;/math&gt; we have &lt;math&gt;\cos\theta = -\cos(\pi-\theta)=-7/c&lt;/math&gt;. We use the Law of Cosines on &lt;math&gt;\triangle ABC&lt;/math&gt; to get &lt;math&gt;60\cos\theta = 109-c^2&lt;/math&gt;. Eliminating &lt;math&gt;\cos\theta&lt;/math&gt; we get &lt;math&gt;c^3-109c-420=0&lt;/math&gt; which factorizes as <br /> &lt;cmath&gt;(c+7)(c+5)(c-12)=0.&lt;/cmath&gt;Discarding the negative roots we have &lt;math&gt;c=12&lt;/math&gt;. Thus &lt;math&gt;BD=AC=CE=12&lt;/math&gt;. For &lt;math&gt;BE=a&lt;/math&gt;, we use Ptolemy's theorem on cyclic quadrilateral &lt;math&gt;ABCE&lt;/math&gt; to get &lt;math&gt;a=44/3&lt;/math&gt;. For &lt;math&gt;AD=b&lt;/math&gt;, we use Ptolemy's theorem on cyclic quadrilateral &lt;math&gt;ACDE&lt;/math&gt; to get &lt;math&gt;b=27/2&lt;/math&gt;.<br /> <br /> The sum of the lengths of the diagonals is &lt;math&gt;12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}&lt;/math&gt; so the answer is &lt;math&gt;385 + 6 = \fbox{\textbf{(D) }391}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;a&lt;/math&gt; denote the length of a diagonal opposite adjacent sides of length &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; for sides &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; for sides &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt;. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:<br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> c^2 &amp;= 3a+100 \\<br /> c^2 &amp;= 10b+9 \\<br /> ab &amp;= 30+14c \\<br /> ac &amp;= 3c+140\\<br /> bc &amp;= 10c+42<br /> \end{align}<br /> &lt;/cmath&gt;<br /> <br /> Using equations &lt;math&gt;(1)&lt;/math&gt; and &lt;math&gt;(2)&lt;/math&gt;, we obtain:<br /> <br /> &lt;cmath&gt;<br /> a = \frac{c^2-100}{3}<br /> &lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;<br /> b = \frac{c^2-9}{10}<br /> &lt;/cmath&gt;<br /> <br /> Plugging into equation &lt;math&gt;(4)&lt;/math&gt;, we find that:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{c^2-100}{3}c &amp;= 3c + 140\\<br /> \frac{c^3-100c}{3} &amp;= 3c + 140\\<br /> c^3-100c &amp;= 9c + 420\\<br /> c^3-109c-420 &amp;=0\\<br /> (c-12)(c+7)(c+5)&amp;=0<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Or similarly into equation &lt;math&gt;(5)&lt;/math&gt; to check:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \frac{c^2-9}{10}c &amp;= 10c+42\\<br /> \frac{c^3-9c}{10} &amp;= 10c + 42\\<br /> c^3-9c &amp;= 100c + 420\\<br /> c^3-109c-420 &amp;=0\\<br /> (c-12)(c+7)(c+5)&amp;=0<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;c&lt;/math&gt;, being a length, must be positive, implying that &lt;math&gt;c=12&lt;/math&gt;. In fact, this is reasonable, since &lt;math&gt;10+3\approx 12&lt;/math&gt; in the pentagon with apparently obtuse angles. Plugging this back into equations &lt;math&gt;(1)&lt;/math&gt; and &lt;math&gt;(2)&lt;/math&gt; we find that &lt;math&gt;a = \frac{44}{3}&lt;/math&gt; and &lt;math&gt;b= \frac{135}{10}=\frac{27}{2}&lt;/math&gt;.<br /> <br /> We desire &lt;math&gt;3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{216+88+81}{6}=\frac{385}{6}&lt;/math&gt;, so it follows that the answer is &lt;math&gt;385 + 6 = \fbox{\textbf{(D) }391}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_25&diff=162691 2015 AMC 12A Problems/Problem 25 2021-09-24T17:47:59Z <p>Ssding: /* See Also */ Made the see also's after read &quot;Last Problem&quot; instead of &quot;Last Problem this year&quot;</p> <hr /> <div>==Problem==<br /> <br /> A collection of circles in the upper half-plane, all tangent to the &lt;math&gt;x&lt;/math&gt;-axis, is constructed in layers as follows. Layer &lt;math&gt;L_0&lt;/math&gt; consists of two circles of radii &lt;math&gt;70^2&lt;/math&gt; and &lt;math&gt;73^2&lt;/math&gt; that are externally tangent. For &lt;math&gt;k \ge 1&lt;/math&gt;, the circles in &lt;math&gt;\bigcup_{j=0}^{k-1}L_j&lt;/math&gt; are ordered according to their points of tangency with the &lt;math&gt;x&lt;/math&gt;-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer &lt;math&gt;L_k&lt;/math&gt; consists of the &lt;math&gt;2^{k-1}&lt;/math&gt; circles constructed in this way. Let &lt;math&gt;S=\bigcup_{j=0}^{6}L_j&lt;/math&gt;, and for every circle &lt;math&gt;C&lt;/math&gt; denote by &lt;math&gt;r(C)&lt;/math&gt; its radius. What is<br /> &lt;cmath&gt;\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?&lt;/cmath&gt;<br /> <br /> &lt;asy&gt;<br /> import olympiad;<br /> size(350);<br /> defaultpen(linewidth(0.7));<br /> // define a bunch of arrays and starting points<br /> pair[] coord = new pair;<br /> int[] trav = {32,16,8,4,2,1};<br /> coord = (0,73^2); coord = (2*73*70,70^2);<br /> // draw the big circles and the bottom line<br /> path arc1 = arc(coord,coord.y,260,360);<br /> path arc2 = arc(coord,coord.y,175,280);<br /> fill((coord.x-910,coord.y)--arc1--cycle,gray(0.75));<br /> fill((coord.x+870,coord.y+425)--arc2--cycle,gray(0.75));<br /> draw(arc1^^arc2);<br /> draw((-930,0)--(70^2+73^2+850,0));<br /> // We now apply the findCenter function 63 times to get<br /> // the location of the centers of all 63 constructed circles.<br /> // The complicated array setup ensures that all the circles<br /> // will be taken in the right order<br /> for(int i = 0;i&lt;=5;i=i+1)<br /> {<br /> int skip = trav[i];<br /> for(int k=skip;k&lt;=64 - skip; k = k + 2*skip)<br /> {<br /> pair cent1 = coord[k-skip], cent2 = coord[k+skip];<br /> real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);<br /> real shiftx = cent1.x + sqrt(4*r1*rn);<br /> coord[k] = (shiftx,rn);<br /> }<br /> // Draw the remaining 63 circles<br /> }<br /> for(int i=1;i&lt;=63;i=i+1)<br /> {<br /> filldraw(circle(coord[i],coord[i].y),gray(0.75));<br /> }&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let us start with the two circles in &lt;math&gt;L_0&lt;/math&gt; and the circle in &lt;math&gt;L_1&lt;/math&gt;. Let the larger circle in &lt;math&gt;L_0&lt;/math&gt; be named circle &lt;math&gt;X&lt;/math&gt; with radius &lt;math&gt;x&lt;/math&gt; and the smaller be named circle &lt;math&gt;Y&lt;/math&gt; with radius &lt;math&gt;y&lt;/math&gt;. Also let the single circle in &lt;math&gt;L_1&lt;/math&gt; be named circle &lt;math&gt;Z&lt;/math&gt; with radius &lt;math&gt;z&lt;/math&gt;. Draw radii &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; perpendicular to the x-axis. Drop altitudes &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; from the center of &lt;math&gt;Z&lt;/math&gt; to these radii &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, respectively, and drop altitude &lt;math&gt;c&lt;/math&gt; from the center of &lt;math&gt;Y&lt;/math&gt; to radius &lt;math&gt;x&lt;/math&gt; perpendicular to the x-axis. Connect the centers of circles &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; with their radii, and utilize the Pythagorean Theorem. We attain the following equations. &lt;cmath&gt;(x - z)^2 + a^2 = (x + z)^2 \implies a^2 = 4xz&lt;/cmath&gt; &lt;cmath&gt;(y - z)^2 + b^2 = (y + z)^2 \implies b^2 = 4yz&lt;/cmath&gt; &lt;cmath&gt;(x - y)^2 + c^2 = (x + y)^2 \implies c^2 = 4xy&lt;/cmath&gt;<br /> <br /> We see that &lt;math&gt;a = 2\sqrt{xz}&lt;/math&gt;, &lt;math&gt;b = 2\sqrt{yz}&lt;/math&gt;, and &lt;math&gt;c = 2\sqrt{xy}&lt;/math&gt;. Since &lt;math&gt;a + b = c&lt;/math&gt;, we have that &lt;math&gt;2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}&lt;/math&gt;. Divide this equation by &lt;math&gt;2\sqrt{xyz}&lt;/math&gt;, and this equation becomes the well-known relation of Descartes's Circle Theorem &lt;math&gt;\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.&lt;/math&gt; We can apply this relationship recursively with the circles in layers &lt;math&gt;L_2, L_3, \cdots, L_6&lt;/math&gt;.<br /> <br /> Here, let &lt;math&gt;S(n)&lt;/math&gt; denote the sum of the reciprocals of the square roots of all circles in layer &lt;math&gt;n&lt;/math&gt;. The notation in the problem asks us to find the sum of the reciprocals of the square roots of the radii in each circle in this collection, which is &lt;math&gt;\textstyle\sum_{n=0}^{6}S(n)&lt;/math&gt;. We already have that &lt;math&gt;S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}&lt;/math&gt;. Then, &lt;math&gt;S(2) = 2S(1) + S(0) = 3S(0)&lt;/math&gt;. Additionally, &lt;math&gt;S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)&lt;/math&gt;, and &lt;math&gt;S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)&lt;/math&gt;. Now, we notice that &lt;math&gt;S(n + 1) = 3S(n)&lt;/math&gt; because &lt;math&gt;S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)&lt;/math&gt;, which is a power of &lt;math&gt;3.&lt;/math&gt; Hence, our desired sum is &lt;math&gt;(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)&lt;/math&gt;. This simplifies to &lt;math&gt;365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{(D)}&lt;/math&gt;.<br /> <br /> Note that the circles in this question are known as Ford circles.<br /> <br /> ==Solution 2 (Pythagorean Theorem)==<br /> Let the two circles from &lt;math&gt;L_0&lt;/math&gt; be of radius &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt;, with &lt;math&gt;r_1&gt;r_2&lt;/math&gt;. Let the circle of radius &lt;math&gt;r_1&lt;/math&gt; be circle &lt;math&gt;A&lt;/math&gt; and the circle of radius &lt;math&gt;r_2&lt;/math&gt; be circle &lt;math&gt;B&lt;/math&gt;. Now, let the circle of &lt;math&gt;L_1&lt;/math&gt; have radius &lt;math&gt;r_3&lt;/math&gt;. Let this circle be circle &lt;math&gt;C&lt;/math&gt;. Draw the radii of the three circles down to the common tangential line and connect the radii. Draw two lines parallel to the common tangential line of the two layers intersecting the center point of circle &lt;math&gt;B&lt;/math&gt; and the center point of circle &lt;math&gt;C&lt;/math&gt;. Now, we have &lt;math&gt;3&lt;/math&gt; right triangles with a line of common length (The two parallel lines). Using the pythagorean theorem, we get the formula &lt;math&gt;\sqrt{(r_2+r_1)^2-(r_2-r_1)^2}=\sqrt{(r_1+r_3)^2-(r_1-r_3)^2}+\sqrt{(r_2+r_3)^2-(r_2-r_3)^2}&lt;/math&gt; Now we solve for &lt;math&gt;r_3&lt;/math&gt;. Square both sides, use the identity &lt;math&gt;(a^2-b^2)=(a+b)(a-b)&lt;/math&gt; and simplify: &lt;math&gt;(2r_2)(2r_1) = (2r_1)(2r_3)+2\sqrt{16r_1r_3r_2r_3}+(2r_2)(2r_3)=4(r_1r_3+r_2r_3+2r_3\sqrt{r_1r_2})=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \\ 4r_2r_1=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \implies r_3=\frac{r_2r_1}{r_1+r_2+2\sqrt{r_1r_2}}&lt;/math&gt;<br /> <br /> <br /> Now, let's change this into a function to clean things up: &lt;math&gt;f(x,y) = \frac{xy}{x+y+2\sqrt{xy}}=\frac{xy}{(\sqrt{x}+\sqrt{y})^2}&lt;/math&gt;<br /> Let's begin to rewrite the sum we want to find in terms of the radii of the circles, call this &lt;math&gt;g(x)&lt;/math&gt;: &lt;math&gt;g(x) = \frac{1}{\sqrt{r}}=\frac{1}{\sqrt{\frac{xy}{(\sqrt{x}+\sqrt{y})^2}}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}} = \frac{\sqrt{y}}{y}+\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}&lt;/math&gt; Using this, we can find the sum of some layers: &lt;math&gt;L_0&lt;/math&gt;, &lt;math&gt;\frac{1}{70}+\frac{1}{73}&lt;/math&gt;, &lt;math&gt;L_0&lt;/math&gt; and &lt;math&gt;L_1&lt;/math&gt;: &lt;math&gt;\frac{1}{70}+\frac{1}{73}+\frac{1}{70}+\frac{1}{73} = 2(\frac{1}{70}+\frac{1}{73})&lt;/math&gt; This is interesting, we have that the sum of Layer 0 and Layer 1 is equal to twice of Layer 0. If we continue and find the sum of layers 0, 1 and 2, we see it is equal to &lt;math&gt;5(L_0)&lt;/math&gt;. This is getting very interesting, there must be some pattern. First of all, we should observe that finding &lt;math&gt;g(x)&lt;/math&gt; of a circle is equivalent to adding up those of the 2 larger circles to construct the smaller one. Second, upon further observation, we can draw out the layers. When we're finding the next layer, we can split the current layers across the center, so that each half includes the center circle &lt;math&gt;L_1&lt;/math&gt;. Now, if we were to find &lt;math&gt;g(x)&lt;/math&gt;, we notice we are doubling the current sum and including the center circle twice. So, the recursive sum would be &lt;math&gt;a_n=3a_{n-1}-1&lt;/math&gt;. So, applying this new formula, we get &lt;math&gt;\sum_{C \in S}\frac{1}{\sqrt{r}} = (3(3(3(3(3(3-1)-1)-1)-1)-1)-1)(\frac{1}{70}+\frac{1}{73})=365\cdot(\frac{1}{70}+\frac{1}{73})=365\cdot\frac{143}{70\cdot73}=\boxed{\frac{143}{14}}&lt;/math&gt;<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2015amc12a/402<br /> <br /> ~ dolfin 7<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=24|after=Last Problem}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_25&diff=162690 2015 AMC 12A Problems/Problem 25 2021-09-24T17:45:52Z <p>Ssding: /* Problem */ Added a period and a space</p> <hr /> <div>==Problem==<br /> <br /> A collection of circles in the upper half-plane, all tangent to the &lt;math&gt;x&lt;/math&gt;-axis, is constructed in layers as follows. Layer &lt;math&gt;L_0&lt;/math&gt; consists of two circles of radii &lt;math&gt;70^2&lt;/math&gt; and &lt;math&gt;73^2&lt;/math&gt; that are externally tangent. For &lt;math&gt;k \ge 1&lt;/math&gt;, the circles in &lt;math&gt;\bigcup_{j=0}^{k-1}L_j&lt;/math&gt; are ordered according to their points of tangency with the &lt;math&gt;x&lt;/math&gt;-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer &lt;math&gt;L_k&lt;/math&gt; consists of the &lt;math&gt;2^{k-1}&lt;/math&gt; circles constructed in this way. Let &lt;math&gt;S=\bigcup_{j=0}^{6}L_j&lt;/math&gt;, and for every circle &lt;math&gt;C&lt;/math&gt; denote by &lt;math&gt;r(C)&lt;/math&gt; its radius. What is<br /> &lt;cmath&gt;\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?&lt;/cmath&gt;<br /> <br /> &lt;asy&gt;<br /> import olympiad;<br /> size(350);<br /> defaultpen(linewidth(0.7));<br /> // define a bunch of arrays and starting points<br /> pair[] coord = new pair;<br /> int[] trav = {32,16,8,4,2,1};<br /> coord = (0,73^2); coord = (2*73*70,70^2);<br /> // draw the big circles and the bottom line<br /> path arc1 = arc(coord,coord.y,260,360);<br /> path arc2 = arc(coord,coord.y,175,280);<br /> fill((coord.x-910,coord.y)--arc1--cycle,gray(0.75));<br /> fill((coord.x+870,coord.y+425)--arc2--cycle,gray(0.75));<br /> draw(arc1^^arc2);<br /> draw((-930,0)--(70^2+73^2+850,0));<br /> // We now apply the findCenter function 63 times to get<br /> // the location of the centers of all 63 constructed circles.<br /> // The complicated array setup ensures that all the circles<br /> // will be taken in the right order<br /> for(int i = 0;i&lt;=5;i=i+1)<br /> {<br /> int skip = trav[i];<br /> for(int k=skip;k&lt;=64 - skip; k = k + 2*skip)<br /> {<br /> pair cent1 = coord[k-skip], cent2 = coord[k+skip];<br /> real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);<br /> real shiftx = cent1.x + sqrt(4*r1*rn);<br /> coord[k] = (shiftx,rn);<br /> }<br /> // Draw the remaining 63 circles<br /> }<br /> for(int i=1;i&lt;=63;i=i+1)<br /> {<br /> filldraw(circle(coord[i],coord[i].y),gray(0.75));<br /> }&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let us start with the two circles in &lt;math&gt;L_0&lt;/math&gt; and the circle in &lt;math&gt;L_1&lt;/math&gt;. Let the larger circle in &lt;math&gt;L_0&lt;/math&gt; be named circle &lt;math&gt;X&lt;/math&gt; with radius &lt;math&gt;x&lt;/math&gt; and the smaller be named circle &lt;math&gt;Y&lt;/math&gt; with radius &lt;math&gt;y&lt;/math&gt;. Also let the single circle in &lt;math&gt;L_1&lt;/math&gt; be named circle &lt;math&gt;Z&lt;/math&gt; with radius &lt;math&gt;z&lt;/math&gt;. Draw radii &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; perpendicular to the x-axis. Drop altitudes &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; from the center of &lt;math&gt;Z&lt;/math&gt; to these radii &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, respectively, and drop altitude &lt;math&gt;c&lt;/math&gt; from the center of &lt;math&gt;Y&lt;/math&gt; to radius &lt;math&gt;x&lt;/math&gt; perpendicular to the x-axis. Connect the centers of circles &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; with their radii, and utilize the Pythagorean Theorem. We attain the following equations. &lt;cmath&gt;(x - z)^2 + a^2 = (x + z)^2 \implies a^2 = 4xz&lt;/cmath&gt; &lt;cmath&gt;(y - z)^2 + b^2 = (y + z)^2 \implies b^2 = 4yz&lt;/cmath&gt; &lt;cmath&gt;(x - y)^2 + c^2 = (x + y)^2 \implies c^2 = 4xy&lt;/cmath&gt;<br /> <br /> We see that &lt;math&gt;a = 2\sqrt{xz}&lt;/math&gt;, &lt;math&gt;b = 2\sqrt{yz}&lt;/math&gt;, and &lt;math&gt;c = 2\sqrt{xy}&lt;/math&gt;. Since &lt;math&gt;a + b = c&lt;/math&gt;, we have that &lt;math&gt;2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}&lt;/math&gt;. Divide this equation by &lt;math&gt;2\sqrt{xyz}&lt;/math&gt;, and this equation becomes the well-known relation of Descartes's Circle Theorem &lt;math&gt;\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.&lt;/math&gt; We can apply this relationship recursively with the circles in layers &lt;math&gt;L_2, L_3, \cdots, L_6&lt;/math&gt;.<br /> <br /> Here, let &lt;math&gt;S(n)&lt;/math&gt; denote the sum of the reciprocals of the square roots of all circles in layer &lt;math&gt;n&lt;/math&gt;. The notation in the problem asks us to find the sum of the reciprocals of the square roots of the radii in each circle in this collection, which is &lt;math&gt;\textstyle\sum_{n=0}^{6}S(n)&lt;/math&gt;. We already have that &lt;math&gt;S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}&lt;/math&gt;. Then, &lt;math&gt;S(2) = 2S(1) + S(0) = 3S(0)&lt;/math&gt;. Additionally, &lt;math&gt;S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)&lt;/math&gt;, and &lt;math&gt;S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)&lt;/math&gt;. Now, we notice that &lt;math&gt;S(n + 1) = 3S(n)&lt;/math&gt; because &lt;math&gt;S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)&lt;/math&gt;, which is a power of &lt;math&gt;3.&lt;/math&gt; Hence, our desired sum is &lt;math&gt;(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)&lt;/math&gt;. This simplifies to &lt;math&gt;365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{(D)}&lt;/math&gt;.<br /> <br /> Note that the circles in this question are known as Ford circles.<br /> <br /> ==Solution 2 (Pythagorean Theorem)==<br /> Let the two circles from &lt;math&gt;L_0&lt;/math&gt; be of radius &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt;, with &lt;math&gt;r_1&gt;r_2&lt;/math&gt;. Let the circle of radius &lt;math&gt;r_1&lt;/math&gt; be circle &lt;math&gt;A&lt;/math&gt; and the circle of radius &lt;math&gt;r_2&lt;/math&gt; be circle &lt;math&gt;B&lt;/math&gt;. Now, let the circle of &lt;math&gt;L_1&lt;/math&gt; have radius &lt;math&gt;r_3&lt;/math&gt;. Let this circle be circle &lt;math&gt;C&lt;/math&gt;. Draw the radii of the three circles down to the common tangential line and connect the radii. Draw two lines parallel to the common tangential line of the two layers intersecting the center point of circle &lt;math&gt;B&lt;/math&gt; and the center point of circle &lt;math&gt;C&lt;/math&gt;. Now, we have &lt;math&gt;3&lt;/math&gt; right triangles with a line of common length (The two parallel lines). Using the pythagorean theorem, we get the formula &lt;math&gt;\sqrt{(r_2+r_1)^2-(r_2-r_1)^2}=\sqrt{(r_1+r_3)^2-(r_1-r_3)^2}+\sqrt{(r_2+r_3)^2-(r_2-r_3)^2}&lt;/math&gt; Now we solve for &lt;math&gt;r_3&lt;/math&gt;. Square both sides, use the identity &lt;math&gt;(a^2-b^2)=(a+b)(a-b)&lt;/math&gt; and simplify: &lt;math&gt;(2r_2)(2r_1) = (2r_1)(2r_3)+2\sqrt{16r_1r_3r_2r_3}+(2r_2)(2r_3)=4(r_1r_3+r_2r_3+2r_3\sqrt{r_1r_2})=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \\ 4r_2r_1=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \implies r_3=\frac{r_2r_1}{r_1+r_2+2\sqrt{r_1r_2}}&lt;/math&gt;<br /> <br /> <br /> Now, let's change this into a function to clean things up: &lt;math&gt;f(x,y) = \frac{xy}{x+y+2\sqrt{xy}}=\frac{xy}{(\sqrt{x}+\sqrt{y})^2}&lt;/math&gt;<br /> Let's begin to rewrite the sum we want to find in terms of the radii of the circles, call this &lt;math&gt;g(x)&lt;/math&gt;: &lt;math&gt;g(x) = \frac{1}{\sqrt{r}}=\frac{1}{\sqrt{\frac{xy}{(\sqrt{x}+\sqrt{y})^2}}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}} = \frac{\sqrt{y}}{y}+\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}&lt;/math&gt; Using this, we can find the sum of some layers: &lt;math&gt;L_0&lt;/math&gt;, &lt;math&gt;\frac{1}{70}+\frac{1}{73}&lt;/math&gt;, &lt;math&gt;L_0&lt;/math&gt; and &lt;math&gt;L_1&lt;/math&gt;: &lt;math&gt;\frac{1}{70}+\frac{1}{73}+\frac{1}{70}+\frac{1}{73} = 2(\frac{1}{70}+\frac{1}{73})&lt;/math&gt; This is interesting, we have that the sum of Layer 0 and Layer 1 is equal to twice of Layer 0. If we continue and find the sum of layers 0, 1 and 2, we see it is equal to &lt;math&gt;5(L_0)&lt;/math&gt;. This is getting very interesting, there must be some pattern. First of all, we should observe that finding &lt;math&gt;g(x)&lt;/math&gt; of a circle is equivalent to adding up those of the 2 larger circles to construct the smaller one. Second, upon further observation, we can draw out the layers. When we're finding the next layer, we can split the current layers across the center, so that each half includes the center circle &lt;math&gt;L_1&lt;/math&gt;. Now, if we were to find &lt;math&gt;g(x)&lt;/math&gt;, we notice we are doubling the current sum and including the center circle twice. So, the recursive sum would be &lt;math&gt;a_n=3a_{n-1}-1&lt;/math&gt;. So, applying this new formula, we get &lt;math&gt;\sum_{C \in S}\frac{1}{\sqrt{r}} = (3(3(3(3(3(3-1)-1)-1)-1)-1)-1)(\frac{1}{70}+\frac{1}{73})=365\cdot(\frac{1}{70}+\frac{1}{73})=365\cdot\frac{143}{70\cdot73}=\boxed{\frac{143}{14}}&lt;/math&gt;<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://artofproblemsolving.com/videos/amc/2015amc12a/402<br /> <br /> ~ dolfin 7<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=24|after=Last Problem this year}}</div> Ssding https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_23&diff=162348 2017 AMC 12B Problems/Problem 23 2021-09-17T17:40:55Z <p>Ssding: /* Solution 2 */ Neatened the formatting and fixed some grammar issues. Also made clearer some parts.</p> <hr /> <div>==Problem==<br /> The graph of &lt;math&gt;y=f(x)&lt;/math&gt;, where &lt;math&gt;f(x)&lt;/math&gt; is a polynomial of degree &lt;math&gt;3&lt;/math&gt;, contains points &lt;math&gt;A(2,4)&lt;/math&gt;, &lt;math&gt;B(3,9)&lt;/math&gt;, and &lt;math&gt;C(4,16)&lt;/math&gt;. Lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; intersect the graph again at points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively, and the sum of the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is 24. What is &lt;math&gt;f(0)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, we can define &lt;math&gt;f(x) = a(x-2)(x-3)(x-4) +x^2&lt;/math&gt;, which contains points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. Now we find that lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; are defined by the equations &lt;math&gt;y = 5x - 6&lt;/math&gt;, &lt;math&gt;y= 6x-8&lt;/math&gt;, and &lt;math&gt;y=7x-12&lt;/math&gt; respectively. Since we want to find the &lt;math&gt;x&lt;/math&gt;-coordinates of the intersections of these lines and &lt;math&gt;f(x)&lt;/math&gt;, we set each of them to &lt;math&gt;f(x)&lt;/math&gt;, and synthetically divide by the solutions we already know exist (eg. if we were looking at line &lt;math&gt;AB&lt;/math&gt;, we would synthetically divide by the solutions &lt;math&gt;x=2&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt;, because we already know &lt;math&gt;AB&lt;/math&gt; intersects the graph at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, which have &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;). After completing this process on all three lines, we get that the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; are &lt;math&gt;\frac{4a-1}{a}&lt;/math&gt;, &lt;math&gt;\frac{3a-1}{a}&lt;/math&gt;, and &lt;math&gt;\frac{2a-1}{a}&lt;/math&gt; respectively. Adding these together, we get &lt;math&gt;\frac{9a-3}{a} = 24&lt;/math&gt; which gives us &lt;math&gt;a = -\frac{1}{5}&lt;/math&gt;. Substituting this back into the original equation, we get &lt;math&gt;f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2&lt;/math&gt;, and &lt;math&gt;f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}&lt;/math&gt;<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 2==<br /> &lt;math&gt;\boxed{\textbf{No need to find the equations for the lines, really.}}&lt;/math&gt; First of all, &lt;math&gt;f(x) = a(x-2)(x-3)(x-4) +x^2&lt;/math&gt;. Let's say the line &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;y=bx+c&lt;/math&gt;, and &lt;math&gt;x_1&lt;/math&gt; is the &lt;math&gt;x&lt;/math&gt; coordinate of the third intersection, then &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;x_1&lt;/math&gt; are the three roots of &lt;math&gt;f(x) - bx-c&lt;/math&gt;. The values of &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; have no effect on the sum of the 3 roots, because the coefficient of the &lt;math&gt;x^2&lt;/math&gt; term is always &lt;math&gt;-9a+1&lt;/math&gt;. So we have <br /> &lt;cmath&gt; \frac{9a-1}{a} = 2 + 3 + x_1= 3 + 4 + x_2 = 2 + 4 + x_3&lt;/cmath&gt;<br /> Adding all three equations up, we get <br /> &lt;cmath&gt; 3(\frac{9a-1}{a}) = 18 + x_1 + x_2 + x_3 = 18 + 24&lt;/cmath&gt;<br /> Solving this equation, we get &lt;math&gt;a = -\frac{1}{5}&lt;/math&gt;. We finish as Solution 1 does.<br /> &lt;math&gt;\boxed{\textbf{(D)}\frac{24}{5}}&lt;/math&gt;.<br /> <br /> - Mathdummy<br /> <br /> Cleaned up by SSding<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Ssding