https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Warrenwangtennis&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-27T14:16:54Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_15&diff=84740 1987 AIME Problems/Problem 15 2017-03-17T17:57:45Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Problem ==<br /> Squares &lt;math&gt;S_1&lt;/math&gt; and &lt;math&gt;S_2&lt;/math&gt; are [[inscribe]]d in [[right triangle]] &lt;math&gt;ABC&lt;/math&gt;, as shown in the figures below. Find &lt;math&gt;AC + CB&lt;/math&gt; if area &lt;math&gt;(S_1) = 441&lt;/math&gt; and area &lt;math&gt;(S_2) = 440&lt;/math&gt;.<br /> <br /> [[Image:AIME_1987_Problem_15.png]]<br /> <br /> == Solution ==<br /> [[Image:1987 AIME-15a.png|360px]]<br /> <br /> Because all the [[triangle]]s in the figure are [[similar]] to triangle &lt;math&gt;ABC&lt;/math&gt;, it's a good idea to use [[area ratios]]. In the diagram above, &lt;math&gt;\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.&lt;/math&gt; Hence, &lt;math&gt;T_3 = \frac {440}{441}T_1&lt;/math&gt; and &lt;math&gt;T_4 = \frac {440}{441}T_2&lt;/math&gt;. Additionally, the area of triangle &lt;math&gt;ABC&lt;/math&gt; is equal to both &lt;math&gt;T_1 + T_2 + 441&lt;/math&gt; and &lt;math&gt;T_3 + T_4 + T_5 + 440.&lt;/math&gt; <br /> <br /> Setting the equations equal and solving for &lt;math&gt;T_5&lt;/math&gt;, &lt;math&gt;T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}&lt;/math&gt;. Therefore, &lt;math&gt;441T_5 = 441 + T_1 + T_2&lt;/math&gt;. However, &lt;math&gt;441 + T_1 + T_2&lt;/math&gt; is equal to the area of triangle &lt;math&gt;ABC&lt;/math&gt;! This means that the ratio between the areas &lt;math&gt;T_5&lt;/math&gt; and &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;441&lt;/math&gt;, and the ratio between the sides is &lt;math&gt;\sqrt {441} = 21&lt;/math&gt;. As a result, &lt;math&gt;AB = 21\sqrt {440} = \sqrt {AC^2 + BC^2}&lt;/math&gt;. We now need &lt;math&gt;(AC)(BC)&lt;/math&gt; to find the value of &lt;math&gt;AC + BC&lt;/math&gt;, because &lt;math&gt;AB^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;h&lt;/math&gt; denote the height to the [[hypotenuse]] of triangle &lt;math&gt;ABC&lt;/math&gt;. Notice that &lt;math&gt;h - \frac {1}{21}h = \sqrt {440}&lt;/math&gt;. (The height of &lt;math&gt;ABC&lt;/math&gt; decreased by the corresponding height of &lt;math&gt;T_5&lt;/math&gt;) Thus, &lt;math&gt;(AB)(h) = (AC)(BC) = 22\cdot 21^2&lt;/math&gt;. Because &lt;math&gt;AB^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2&lt;/math&gt;, &lt;math&gt;AC + BC = (21)(22) = \boxed{462}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1987|num-b=14|after=Last&lt;br /&gt;Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=Cauchy-Schwarz_Inequality&diff=84734 Cauchy-Schwarz Inequality 2017-03-17T04:10:41Z <p>Warrenwangtennis: /* Elementary Form */</p> <hr /> <div>The '''Cauchy-Schwarz Inequality''' (which is known by other names, including '''Cauchy's Inequality''', '''Schwarz's Inequality''', and the '''Cauchy-Bunyakovsky-Schwarz Inequality''') is a well-known [[inequality]] with many elegant applications. It has an elementary form, a complex form, and a general form. <br /> <br /> [[Augustin Louis Cauchy]] wrote the first paper about the elementary form in 1821. The general form was discovered by [[Viktor Bunyakovsky]] in 1849 and independently by [[Hermann Schwarz]] in 1888.<br /> <br /> == Elementary Form ==<br /> <br /> For any real numbers &lt;math&gt; a_1, \ldots, a_n &lt;/math&gt; and &lt;math&gt; b_1, \ldots, b_n &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \left( \sum_{i=1}^{n}a_ib_i \right)^2 \le \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right)<br /> &lt;/cmath&gt;<br /> with equality when there exist constants &lt;math&gt;\mu, \lambda &lt;/math&gt; not both zero such that for all &lt;math&gt; 1 \le i \le n &lt;/math&gt;, &lt;math&gt;\mu a_i = \lambda b_i &lt;/math&gt;.<br /> <br /> === Discussion ===<br /> <br /> Consider the vectors &lt;math&gt; \mathbf{a} = \langle a_1, \ldots a_n \rangle &lt;/math&gt; and &lt;math&gt; {} \mathbf{b} = \langle b_1, \ldots b_n \rangle &lt;/math&gt;. If &lt;math&gt;\theta &lt;/math&gt; is the [[angle]] formed by &lt;math&gt; \mathbf{a} &lt;/math&gt; and &lt;math&gt; \mathbf{b} &lt;/math&gt;, then the left-hand side of the inequality is equal to the square of the [[dot product]] of &lt;math&gt;\mathbf{a} &lt;/math&gt; and &lt;math&gt; \mathbf{b} &lt;/math&gt;, or &lt;math&gt;(\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2 (\cos\theta) ^2&lt;/math&gt; .The right hand side of the inequality is equal to &lt;math&gt; \left( ||\mathbf{a}|| * ||\mathbf{b}|| \right)^2 = a^2b^2&lt;/math&gt;. The inequality then follows from &lt;math&gt; |\cos\theta | \le 1 &lt;/math&gt;, with equality when one of &lt;math&gt; \mathbf{a,b} &lt;/math&gt; is a multiple of the other, as desired.<br /> <br /> === Complex Form ===<br /> <br /> The inequality sometimes appears in the following form.<br /> <br /> Let &lt;math&gt; a_1, \ldots, a_n &lt;/math&gt; and &lt;math&gt; b_1, \ldots, b_n &lt;/math&gt; be [[complex numbers]]. Then<br /> &lt;cmath&gt;<br /> \left| \sum_{i=1}^na_ib_i \right|^2 \le \left(\sum_{i=1}^{n}|a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right)<br /> &lt;/cmath&gt;<br /> This appears to be more powerful, but it follows from<br /> &lt;cmath&gt;<br /> \left| \sum_{i=1}^n a_ib_i \right| ^2 \le \left( \sum_{i=1}^n |a_i| \cdot |b_i| \right)^2 \le \left(\sum_{i=1}^n |a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right)<br /> &lt;/cmath&gt;<br /> <br /> == Upper Bound on (Σa)(Σb) ==<br /> <br /> Let &lt;math&gt;a_1, a_2, \ldots, a_n&lt;/math&gt; and &lt;math&gt;b_1, b_2, \ldots, b_n&lt;/math&gt; be two sequences of positive real numbers with<br /> &lt;cmath&gt;<br /> 0 &lt; m \le \frac{a_i}{b_i} \le M<br /> &lt;/cmath&gt;<br /> for &lt;math&gt;1 \le i \le n&lt;/math&gt;. Then<br /> &lt;cmath&gt;<br /> \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right) \le \frac{(M+m)^2}{4Mm} \left( \sum_{i=1}^{n}a_ib_i \right)^2,<br /> &lt;/cmath&gt;<br /> with equality if and only if, for some ordering of the pairs &lt;math&gt;(a_i,b_i) \mapsto (a_{\sigma(i)},b_{\sigma(i)})&lt;/math&gt;, some &lt;math&gt;0 \le j \le n&lt;/math&gt; exists such that &lt;math&gt;a_{\sigma(i)}=mb_{\sigma(i)}&lt;/math&gt; for &lt;math&gt;1 \le \sigma(i) \le j&lt;/math&gt; and &lt;math&gt;a_{\sigma(i)}=Mb_{\sigma(i)}&lt;/math&gt; for &lt;math&gt;j+1 \le \sigma(i) \le n&lt;/math&gt;, and<br /> &lt;cmath&gt;<br /> m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2.<br /> &lt;/cmath&gt;<br /> If we restrict that &lt;math&gt;m_1 \le a_i \le M_1&lt;/math&gt; and &lt;math&gt;m_2 \le b_i \le M_2&lt;/math&gt; for all &lt;math&gt;i&lt;/math&gt;, then it's clear that for &lt;math&gt;a_i/b_i&lt;/math&gt; to be &lt;math&gt;m=m_1/M_2&lt;/math&gt; or &lt;math&gt;M=M_1/m_2&lt;/math&gt; for all &lt;math&gt;i&lt;/math&gt;, then &lt;math&gt;a_i=m_1 \Longleftrightarrow b_i=M_2&lt;/math&gt; and &lt;math&gt;a_i=M_1 \Longleftrightarrow b_i=m_2&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2<br /> &lt;/cmath&gt;<br /> is equivalent to<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> m(jM_2^2) = M((n-j)m_2^2) &amp;\Longleftrightarrow m_1M_2j = M_1m_2(n-j)\\<br /> &amp;\Longleftrightarrow j = \left(\frac{M_1m_2}{M_1m_2+m_1M_2}\right) n.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> (When this is not an integer, the maximum occurs when &lt;math&gt;j&lt;/math&gt; is either the ceiling or floor of the right-hand side.) In the special case that &lt;math&gt;a_ib_i = k &gt; 0&lt;/math&gt; is constant for all &lt;math&gt;i&lt;/math&gt;, we have &lt;math&gt;M_1=1/m_2&lt;/math&gt; and &lt;math&gt;m_1=1/M_2&lt;/math&gt;, so here &lt;math&gt;j&lt;/math&gt; must be &lt;math&gt;n/2&lt;/math&gt;.<br /> <br /> === Proof ===<br /> <br /> Note that for all &lt;math&gt;i&lt;/math&gt;, we have<br /> &lt;cmath&gt;<br /> 0 \le \left(\frac{a_i}{b_i}-m\right)\left(M-\frac{a_i}{b_i}\right) = \frac{1}{b_i^2}(a_ib_iM-a_i^2-b_i^2Mm+a_ib_im)<br /> &lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt;<br /> (M+m)a_ib_i \ge a_i^2+(Mm)b_i^2,<br /> &lt;/cmath&gt;<br /> with equality if and only if &lt;math&gt;a_i=mb_i&lt;/math&gt; or &lt;math&gt;a_i=Mb_i&lt;/math&gt;. Summing up these inequalities over &lt;math&gt;1 \le i \le n&lt;/math&gt;, we obtain from AM-GM that<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> (M+m)\sum_{i=1}^{n}a_ib_i &amp;\ge \sum_{i=1}^{n}a_i^2 + (Mm)\sum_{i=1}^{n}b_i^2\\<br /> &amp;\ge 2\sqrt{Mm \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right)},<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> and squaring gives us the desired bound. For equality to occur, we must have &lt;math&gt;a_i=mb_i&lt;/math&gt; or &lt;math&gt;a_i=Mb_i&lt;/math&gt; for all &lt;math&gt;i&lt;/math&gt;. If, without loss of generality, &lt;math&gt;a_i=mb_i&lt;/math&gt; for &lt;math&gt;1 \le i \le j&lt;/math&gt; and &lt;math&gt;a_i=Mb_i&lt;/math&gt; for &lt;math&gt;j+1 \le i \le n&lt;/math&gt; for some &lt;math&gt;0 \le j \le n&lt;/math&gt;, then for the AM-GM to reach equality we must have (assume &lt;math&gt;M&gt;m&lt;/math&gt; since &lt;math&gt;M=m&lt;/math&gt; is trivial)<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \sum_{i=1}^{n}a_i^2 &amp;= Mm\sum_{i=1}^{n}b_i^2\\<br /> m^2\sum_{i=1}^{j}b_i^2 + M^2\sum_{i=j+1}^{n}b_i^2 &amp;= Mm\sum_{i=1}^{j}b_i^2 + Mm\sum_{i=j+1}^{n}b_i^2\\<br /> (m-M)m\sum_{i=1}^{j}b_i^2 &amp;= (m-M)M\sum_{i=j+1}^{n}b_i^2\\<br /> m\sum_{i=1}^{j}b_i^2 &amp;= M\sum_{i=j+1}^{n}b_i^2.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> == General Form ==<br /> <br /> Let &lt;math&gt;V &lt;/math&gt; be a [[vector space]], and let &lt;math&gt; \langle \cdot, \cdot \rangle : V \times V \to \mathbb{R} &lt;/math&gt; be an [[inner product]]. Then for any &lt;math&gt; \mathbf{a,b} \in V &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle ,<br /> &lt;/cmath&gt;<br /> with equality if and only if there exist constants &lt;math&gt;\mu, \lambda &lt;/math&gt; not both zero such that &lt;math&gt; \mu\mathbf{a} = \lambda\mathbf{b} &lt;/math&gt;.<br /> <br /> === Proof 1 ===<br /> <br /> Consider the polynomial of &lt;math&gt; t &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \langle t\mathbf{a + b}, t\mathbf{a + b} \rangle = t^2\langle \mathbf{a,a} \rangle + 2t\langle \mathbf{a,b} \rangle + \langle \mathbf{b,b} \rangle .<br /> &lt;/cmath&gt;<br /> This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., &lt;math&gt; \langle \mathbf{a,b} \rangle^2 &lt;/math&gt; must be less than or equal to &lt;math&gt; \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle &lt;/math&gt;, with equality when &lt;math&gt; \mathbf{a = 0} &lt;/math&gt; or when there exists some scalar &lt;math&gt;-t &lt;/math&gt; such that &lt;math&gt; -t\mathbf{a} = \mathbf{b} &lt;/math&gt;, as desired.<br /> <br /> === Proof 2 ===<br /> <br /> We consider<br /> &lt;cmath&gt;<br /> \langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle .<br /> &lt;/cmath&gt;<br /> Since this is always greater than or equal to zero, we have<br /> &lt;cmath&gt;<br /> \langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle .<br /> &lt;/cmath&gt;<br /> Now, if either &lt;math&gt; \mathbf{a} &lt;/math&gt; or &lt;math&gt; \mathbf{b} &lt;/math&gt; is equal to &lt;math&gt; \mathbf{0} &lt;/math&gt;, then &lt;math&gt; \langle \mathbf{a,b} \rangle^2 = \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle = 0 &lt;/math&gt;. Otherwise, we may [[normalize]] so that &lt;math&gt; \langle \mathbf {a,a} \rangle = \langle \mathbf{b,b} \rangle = 1 &lt;/math&gt;, and we have<br /> &lt;cmath&gt;<br /> \langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} ,<br /> &lt;/cmath&gt;<br /> with equality when &lt;math&gt;\mathbf{a} &lt;/math&gt; and &lt;math&gt; \mathbf{b} &lt;/math&gt; may be scaled to each other, as desired.<br /> <br /> === Examples ===<br /> <br /> The elementary form of the Cauchy-Schwarz inequality is a special case of the general form, as is the '''Cauchy-Schwarz Inequality for Integrals''': for integrable functions &lt;math&gt; f,g : [a,b] \mapsto \mathbb{R} &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \biggl( \int_{a}^b f(x)g(x)dx \biggr)^2 \le \int_{a}^b \bigl[ f(x) \bigr]^2dx \cdot \int_a^b \bigl[ g(x) \bigr]^2 dx<br /> &lt;/cmath&gt;<br /> with equality when there exist constants &lt;math&gt; \mu, \lambda &lt;/math&gt; not both equal to zero such that for &lt;math&gt; t \in [a,b] &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx .<br /> &lt;/cmath&gt;<br /> <br /> ==Problems==<br /> <br /> ===Introductory===<br /> <br /> *Consider the function &lt;math&gt;f(x)=\frac{(x+k)^2}{x^2+1},x\in (-\infty,\infty)&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is a positive integer. Show that &lt;math&gt;f(x)\le k^2+1&lt;/math&gt;. ([[User:Temperal/The_Problem_Solver's Resource Competition|Source]])<br /> * [http://www.mathlinks.ro/Forum/viewtopic.php?t=78687 (APMO 1991 #3)] Let &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_n&lt;/math&gt;, &lt;math&gt;b_1&lt;/math&gt;, &lt;math&gt;b_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;b_n&lt;/math&gt; be positive real numbers such that &lt;math&gt;a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n&lt;/math&gt;. Show that<br /> &lt;cmath&gt;\frac {a_1^2}{a_1 + b_1} + \frac {a_2^2}{a_2 + b_2} + \cdots + \frac {a_n^2}{a_n + b_n} \geq \frac {a_1 + a_2 + \cdots + a_n}{2}&lt;/cmath&gt;<br /> <br /> ===Intermediate===<br /> <br /> *Let &lt;math&gt;ABC&lt;/math&gt; be a triangle such that<br /> &lt;cmath&gt;<br /> \left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2 ,<br /> &lt;/cmath&gt;<br /> where &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; denote its [[semiperimeter]] and [[inradius]], respectively. Prove that triangle &lt;math&gt;ABC &lt;/math&gt; is similar to a triangle &lt;math&gt;T &lt;/math&gt; whose side lengths are all positive integers with no common divisor and determine those integers.<br /> ([[2002 USAMO Problems/Problem 2|Source]])<br /> <br /> ===Olympiad===<br /> <br /> *&lt;math&gt;P&lt;/math&gt; is a point inside a given triangle &lt;math&gt;ABC&lt;/math&gt;. &lt;math&gt;D, E, F&lt;/math&gt; are the feet of the perpendiculars from &lt;math&gt;P&lt;/math&gt; to the lines &lt;math&gt;BC, CA, AB&lt;/math&gt;, respectively. Find all &lt;math&gt;P&lt;/math&gt; for which<br /> &lt;cmath&gt;<br /> \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}<br /> &lt;/cmath&gt;<br /> is least.<br /> <br /> ([[1981 IMO Problems/Problem 1|Source]])<br /> <br /> == Other Resources ==<br /> * [http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality Wikipedia entry]<br /> <br /> ===Books===<br /> * [http://www.amazon.com/exec/obidos/ASIN/052154677X/artofproblems-20 The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities] by J. Michael Steele.<br /> * [http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20 Problem Solving Strategies] by Arthur Engel contains significant material on inequalities.<br /> <br /> <br /> [[Category:Algebra]]<br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=Cauchy-Schwarz_Inequality&diff=84733 Cauchy-Schwarz Inequality 2017-03-17T04:10:12Z <p>Warrenwangtennis: /* Complex Form */</p> <hr /> <div>The '''Cauchy-Schwarz Inequality''' (which is known by other names, including '''Cauchy's Inequality''', '''Schwarz's Inequality''', and the '''Cauchy-Bunyakovsky-Schwarz Inequality''') is a well-known [[inequality]] with many elegant applications. It has an elementary form, a complex form, and a general form. <br /> <br /> [[Augustin Louis Cauchy]] wrote the first paper about the elementary form in 1821. The general form was discovered by [[Viktor Bunyakovsky]] in 1849 and independently by [[Hermann Schwarz]] in 1888.<br /> <br /> == Elementary Form ==<br /> <br /> For any real numbers &lt;math&gt; a_1, \ldots, a_n &lt;/math&gt; and &lt;math&gt; b_1, \ldots, b_n &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \left( \sum_{i=1}^{n}a_ib_i \right)^2 \le \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right),<br /> &lt;/cmath&gt;<br /> with equality when there exist constants &lt;math&gt;\mu, \lambda &lt;/math&gt; not both zero such that for all &lt;math&gt; 1 \le i \le n &lt;/math&gt;, &lt;math&gt;\mu a_i = \lambda b_i &lt;/math&gt;.<br /> <br /> === Discussion ===<br /> <br /> Consider the vectors &lt;math&gt; \mathbf{a} = \langle a_1, \ldots a_n \rangle &lt;/math&gt; and &lt;math&gt; {} \mathbf{b} = \langle b_1, \ldots b_n \rangle &lt;/math&gt;. If &lt;math&gt;\theta &lt;/math&gt; is the [[angle]] formed by &lt;math&gt; \mathbf{a} &lt;/math&gt; and &lt;math&gt; \mathbf{b} &lt;/math&gt;, then the left-hand side of the inequality is equal to the square of the [[dot product]] of &lt;math&gt;\mathbf{a} &lt;/math&gt; and &lt;math&gt; \mathbf{b} &lt;/math&gt;, or &lt;math&gt;(\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2 (\cos\theta) ^2&lt;/math&gt; .The right hand side of the inequality is equal to &lt;math&gt; \left( ||\mathbf{a}|| * ||\mathbf{b}|| \right)^2 = a^2b^2&lt;/math&gt;. The inequality then follows from &lt;math&gt; |\cos\theta | \le 1 &lt;/math&gt;, with equality when one of &lt;math&gt; \mathbf{a,b} &lt;/math&gt; is a multiple of the other, as desired.<br /> <br /> === Complex Form ===<br /> <br /> The inequality sometimes appears in the following form.<br /> <br /> Let &lt;math&gt; a_1, \ldots, a_n &lt;/math&gt; and &lt;math&gt; b_1, \ldots, b_n &lt;/math&gt; be [[complex numbers]]. Then<br /> &lt;cmath&gt;<br /> \left| \sum_{i=1}^na_ib_i \right|^2 \le \left(\sum_{i=1}^{n}|a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right)<br /> &lt;/cmath&gt;<br /> This appears to be more powerful, but it follows from<br /> &lt;cmath&gt;<br /> \left| \sum_{i=1}^n a_ib_i \right| ^2 \le \left( \sum_{i=1}^n |a_i| \cdot |b_i| \right)^2 \le \left(\sum_{i=1}^n |a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right)<br /> &lt;/cmath&gt;<br /> <br /> == Upper Bound on (Σa)(Σb) ==<br /> <br /> Let &lt;math&gt;a_1, a_2, \ldots, a_n&lt;/math&gt; and &lt;math&gt;b_1, b_2, \ldots, b_n&lt;/math&gt; be two sequences of positive real numbers with<br /> &lt;cmath&gt;<br /> 0 &lt; m \le \frac{a_i}{b_i} \le M<br /> &lt;/cmath&gt;<br /> for &lt;math&gt;1 \le i \le n&lt;/math&gt;. Then<br /> &lt;cmath&gt;<br /> \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right) \le \frac{(M+m)^2}{4Mm} \left( \sum_{i=1}^{n}a_ib_i \right)^2,<br /> &lt;/cmath&gt;<br /> with equality if and only if, for some ordering of the pairs &lt;math&gt;(a_i,b_i) \mapsto (a_{\sigma(i)},b_{\sigma(i)})&lt;/math&gt;, some &lt;math&gt;0 \le j \le n&lt;/math&gt; exists such that &lt;math&gt;a_{\sigma(i)}=mb_{\sigma(i)}&lt;/math&gt; for &lt;math&gt;1 \le \sigma(i) \le j&lt;/math&gt; and &lt;math&gt;a_{\sigma(i)}=Mb_{\sigma(i)}&lt;/math&gt; for &lt;math&gt;j+1 \le \sigma(i) \le n&lt;/math&gt;, and<br /> &lt;cmath&gt;<br /> m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2.<br /> &lt;/cmath&gt;<br /> If we restrict that &lt;math&gt;m_1 \le a_i \le M_1&lt;/math&gt; and &lt;math&gt;m_2 \le b_i \le M_2&lt;/math&gt; for all &lt;math&gt;i&lt;/math&gt;, then it's clear that for &lt;math&gt;a_i/b_i&lt;/math&gt; to be &lt;math&gt;m=m_1/M_2&lt;/math&gt; or &lt;math&gt;M=M_1/m_2&lt;/math&gt; for all &lt;math&gt;i&lt;/math&gt;, then &lt;math&gt;a_i=m_1 \Longleftrightarrow b_i=M_2&lt;/math&gt; and &lt;math&gt;a_i=M_1 \Longleftrightarrow b_i=m_2&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2<br /> &lt;/cmath&gt;<br /> is equivalent to<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> m(jM_2^2) = M((n-j)m_2^2) &amp;\Longleftrightarrow m_1M_2j = M_1m_2(n-j)\\<br /> &amp;\Longleftrightarrow j = \left(\frac{M_1m_2}{M_1m_2+m_1M_2}\right) n.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> (When this is not an integer, the maximum occurs when &lt;math&gt;j&lt;/math&gt; is either the ceiling or floor of the right-hand side.) In the special case that &lt;math&gt;a_ib_i = k &gt; 0&lt;/math&gt; is constant for all &lt;math&gt;i&lt;/math&gt;, we have &lt;math&gt;M_1=1/m_2&lt;/math&gt; and &lt;math&gt;m_1=1/M_2&lt;/math&gt;, so here &lt;math&gt;j&lt;/math&gt; must be &lt;math&gt;n/2&lt;/math&gt;.<br /> <br /> === Proof ===<br /> <br /> Note that for all &lt;math&gt;i&lt;/math&gt;, we have<br /> &lt;cmath&gt;<br /> 0 \le \left(\frac{a_i}{b_i}-m\right)\left(M-\frac{a_i}{b_i}\right) = \frac{1}{b_i^2}(a_ib_iM-a_i^2-b_i^2Mm+a_ib_im)<br /> &lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt;<br /> (M+m)a_ib_i \ge a_i^2+(Mm)b_i^2,<br /> &lt;/cmath&gt;<br /> with equality if and only if &lt;math&gt;a_i=mb_i&lt;/math&gt; or &lt;math&gt;a_i=Mb_i&lt;/math&gt;. Summing up these inequalities over &lt;math&gt;1 \le i \le n&lt;/math&gt;, we obtain from AM-GM that<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> (M+m)\sum_{i=1}^{n}a_ib_i &amp;\ge \sum_{i=1}^{n}a_i^2 + (Mm)\sum_{i=1}^{n}b_i^2\\<br /> &amp;\ge 2\sqrt{Mm \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right)},<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> and squaring gives us the desired bound. For equality to occur, we must have &lt;math&gt;a_i=mb_i&lt;/math&gt; or &lt;math&gt;a_i=Mb_i&lt;/math&gt; for all &lt;math&gt;i&lt;/math&gt;. If, without loss of generality, &lt;math&gt;a_i=mb_i&lt;/math&gt; for &lt;math&gt;1 \le i \le j&lt;/math&gt; and &lt;math&gt;a_i=Mb_i&lt;/math&gt; for &lt;math&gt;j+1 \le i \le n&lt;/math&gt; for some &lt;math&gt;0 \le j \le n&lt;/math&gt;, then for the AM-GM to reach equality we must have (assume &lt;math&gt;M&gt;m&lt;/math&gt; since &lt;math&gt;M=m&lt;/math&gt; is trivial)<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \sum_{i=1}^{n}a_i^2 &amp;= Mm\sum_{i=1}^{n}b_i^2\\<br /> m^2\sum_{i=1}^{j}b_i^2 + M^2\sum_{i=j+1}^{n}b_i^2 &amp;= Mm\sum_{i=1}^{j}b_i^2 + Mm\sum_{i=j+1}^{n}b_i^2\\<br /> (m-M)m\sum_{i=1}^{j}b_i^2 &amp;= (m-M)M\sum_{i=j+1}^{n}b_i^2\\<br /> m\sum_{i=1}^{j}b_i^2 &amp;= M\sum_{i=j+1}^{n}b_i^2.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> == General Form ==<br /> <br /> Let &lt;math&gt;V &lt;/math&gt; be a [[vector space]], and let &lt;math&gt; \langle \cdot, \cdot \rangle : V \times V \to \mathbb{R} &lt;/math&gt; be an [[inner product]]. Then for any &lt;math&gt; \mathbf{a,b} \in V &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle ,<br /> &lt;/cmath&gt;<br /> with equality if and only if there exist constants &lt;math&gt;\mu, \lambda &lt;/math&gt; not both zero such that &lt;math&gt; \mu\mathbf{a} = \lambda\mathbf{b} &lt;/math&gt;.<br /> <br /> === Proof 1 ===<br /> <br /> Consider the polynomial of &lt;math&gt; t &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \langle t\mathbf{a + b}, t\mathbf{a + b} \rangle = t^2\langle \mathbf{a,a} \rangle + 2t\langle \mathbf{a,b} \rangle + \langle \mathbf{b,b} \rangle .<br /> &lt;/cmath&gt;<br /> This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., &lt;math&gt; \langle \mathbf{a,b} \rangle^2 &lt;/math&gt; must be less than or equal to &lt;math&gt; \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle &lt;/math&gt;, with equality when &lt;math&gt; \mathbf{a = 0} &lt;/math&gt; or when there exists some scalar &lt;math&gt;-t &lt;/math&gt; such that &lt;math&gt; -t\mathbf{a} = \mathbf{b} &lt;/math&gt;, as desired.<br /> <br /> === Proof 2 ===<br /> <br /> We consider<br /> &lt;cmath&gt;<br /> \langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle .<br /> &lt;/cmath&gt;<br /> Since this is always greater than or equal to zero, we have<br /> &lt;cmath&gt;<br /> \langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle .<br /> &lt;/cmath&gt;<br /> Now, if either &lt;math&gt; \mathbf{a} &lt;/math&gt; or &lt;math&gt; \mathbf{b} &lt;/math&gt; is equal to &lt;math&gt; \mathbf{0} &lt;/math&gt;, then &lt;math&gt; \langle \mathbf{a,b} \rangle^2 = \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle = 0 &lt;/math&gt;. Otherwise, we may [[normalize]] so that &lt;math&gt; \langle \mathbf {a,a} \rangle = \langle \mathbf{b,b} \rangle = 1 &lt;/math&gt;, and we have<br /> &lt;cmath&gt;<br /> \langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} ,<br /> &lt;/cmath&gt;<br /> with equality when &lt;math&gt;\mathbf{a} &lt;/math&gt; and &lt;math&gt; \mathbf{b} &lt;/math&gt; may be scaled to each other, as desired.<br /> <br /> === Examples ===<br /> <br /> The elementary form of the Cauchy-Schwarz inequality is a special case of the general form, as is the '''Cauchy-Schwarz Inequality for Integrals''': for integrable functions &lt;math&gt; f,g : [a,b] \mapsto \mathbb{R} &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \biggl( \int_{a}^b f(x)g(x)dx \biggr)^2 \le \int_{a}^b \bigl[ f(x) \bigr]^2dx \cdot \int_a^b \bigl[ g(x) \bigr]^2 dx<br /> &lt;/cmath&gt;<br /> with equality when there exist constants &lt;math&gt; \mu, \lambda &lt;/math&gt; not both equal to zero such that for &lt;math&gt; t \in [a,b] &lt;/math&gt;,<br /> &lt;cmath&gt;<br /> \mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx .<br /> &lt;/cmath&gt;<br /> <br /> ==Problems==<br /> <br /> ===Introductory===<br /> <br /> *Consider the function &lt;math&gt;f(x)=\frac{(x+k)^2}{x^2+1},x\in (-\infty,\infty)&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is a positive integer. Show that &lt;math&gt;f(x)\le k^2+1&lt;/math&gt;. ([[User:Temperal/The_Problem_Solver's Resource Competition|Source]])<br /> * [http://www.mathlinks.ro/Forum/viewtopic.php?t=78687 (APMO 1991 #3)] Let &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;a_n&lt;/math&gt;, &lt;math&gt;b_1&lt;/math&gt;, &lt;math&gt;b_2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;b_n&lt;/math&gt; be positive real numbers such that &lt;math&gt;a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n&lt;/math&gt;. Show that<br /> &lt;cmath&gt;\frac {a_1^2}{a_1 + b_1} + \frac {a_2^2}{a_2 + b_2} + \cdots + \frac {a_n^2}{a_n + b_n} \geq \frac {a_1 + a_2 + \cdots + a_n}{2}&lt;/cmath&gt;<br /> <br /> ===Intermediate===<br /> <br /> *Let &lt;math&gt;ABC&lt;/math&gt; be a triangle such that<br /> &lt;cmath&gt;<br /> \left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2 ,<br /> &lt;/cmath&gt;<br /> where &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; denote its [[semiperimeter]] and [[inradius]], respectively. Prove that triangle &lt;math&gt;ABC &lt;/math&gt; is similar to a triangle &lt;math&gt;T &lt;/math&gt; whose side lengths are all positive integers with no common divisor and determine those integers.<br /> ([[2002 USAMO Problems/Problem 2|Source]])<br /> <br /> ===Olympiad===<br /> <br /> *&lt;math&gt;P&lt;/math&gt; is a point inside a given triangle &lt;math&gt;ABC&lt;/math&gt;. &lt;math&gt;D, E, F&lt;/math&gt; are the feet of the perpendiculars from &lt;math&gt;P&lt;/math&gt; to the lines &lt;math&gt;BC, CA, AB&lt;/math&gt;, respectively. Find all &lt;math&gt;P&lt;/math&gt; for which<br /> &lt;cmath&gt;<br /> \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}<br /> &lt;/cmath&gt;<br /> is least.<br /> <br /> ([[1981 IMO Problems/Problem 1|Source]])<br /> <br /> == Other Resources ==<br /> * [http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality Wikipedia entry]<br /> <br /> ===Books===<br /> * [http://www.amazon.com/exec/obidos/ASIN/052154677X/artofproblems-20 The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities] by J. Michael Steele.<br /> * [http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20 Problem Solving Strategies] by Arthur Engel contains significant material on inequalities.<br /> <br /> <br /> [[Category:Algebra]]<br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_10&diff=84731 1987 AIME Problems/Problem 10 2017-03-17T02:23:44Z <p>Warrenwangtennis: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) <br /> == Solutions ==<br /> === Solution 1 ===<br /> Let the total number of steps be &lt;math&gt;x&lt;/math&gt;, the speed of the escalator be &lt;math&gt;e&lt;/math&gt; and the speed of Bob be &lt;math&gt;b&lt;/math&gt;.<br /> <br /> In the time it took Bob to climb up the escalator he saw 75 steps and also climbed the entire escalator. Thus the contribution of the escalator must have been an additional &lt;math&gt;x - 75&lt;/math&gt; steps. Since Bob and the escalator were both moving at a constant speed over the time it took Bob to climb, the [[ratio]] of their distances covered is the same as the ratio of their speeds, so &lt;math&gt;\frac{e}{b} = \frac{x - 75}{75}&lt;/math&gt;. <br /> <br /> Similarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved &lt;math&gt;150 - x&lt;/math&gt; steps in that time. Thus &lt;math&gt;\frac{e}{3b} = \frac{150 - x}{150}&lt;/math&gt; or &lt;math&gt;\frac{e}{b} = \frac{150 - x}{50}&lt;/math&gt;.<br /> <br /> Equating the two values of &lt;math&gt;\frac{e}{b}&lt;/math&gt; we have &lt;math&gt;\frac{x - 75}{75} = \frac{150 - x}{50}&lt;/math&gt; and so &lt;math&gt;2x - 150 = 450 - 3x&lt;/math&gt; and &lt;math&gt;5x = 600&lt;/math&gt; and &lt;math&gt;x = 120&lt;/math&gt;, the answer.<br /> <br /> === Solution 2 ===<br /> Again, let the total number of steps be &lt;math&gt;x&lt;/math&gt;, the speed of the escalator be &lt;math&gt;e&lt;/math&gt; and the speed of Bob be &lt;math&gt;b&lt;/math&gt; (all &quot;per unit time&quot;).<br /> <br /> Then this can be interpreted as a classic chasing problem: Bob is &quot;behind&quot; by &lt;math&gt;x&lt;/math&gt; steps, and since he moves at a pace of &lt;math&gt;b+e&lt;/math&gt; relative to the escalator, it will take &lt;math&gt;\frac{x}{b+e}=\frac{75}{e}&lt;/math&gt; time to get to the top.<br /> <br /> Similarly, Al will take &lt;math&gt;\frac{x}{3b-e}=\frac{150}{e}&lt;/math&gt; time to get to the bottom.<br /> <br /> From these two equations, we arrive at &lt;math&gt;150=\frac{ex}{3b-e}=2\cdot75=\frac{2ex}{b+e}=\frac{6ex}{3b+3e}=\frac{(ex)-(6ex)}{(3b-e)-(3b+3e)}=\frac{5x}{4}&lt;/math&gt;<br /> &lt;math&gt;\implies600=5x\implies x=\boxed{120}&lt;/math&gt;, where we have used the fact that &lt;math&gt;\frac{a}{b}=\frac{c}{d}=\frac{a\pm c}{b\pm d}&lt;/math&gt; (the proportion manipulations are motivated by the desire to isolate &lt;math&gt;x&lt;/math&gt;, prompting the isolation of the &lt;math&gt;150&lt;/math&gt; on one side, and the fact that if we could cancel out the &lt;math&gt;b&lt;/math&gt;'s, then the &lt;math&gt;e&lt;/math&gt;'s in the numerator and denominator would cancel out, resulting in an equation with &lt;math&gt;x&lt;/math&gt; by itself).<br /> <br /> == See also ==<br /> {{AIME box|year=1987|num-b=9|num-a=11}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems&diff=84730 1987 AIME Problems 2017-03-17T02:14:54Z <p>Warrenwangtennis: /* Problem 12 */</p> <hr /> <div>{{AIME Problems|year=1987}}<br /> <br /> == Problem 1 ==<br /> An ordered pair &lt;math&gt;(m,n)&lt;/math&gt; of non-negative integers is called &quot;simple&quot; if the addition &lt;math&gt;m+n&lt;/math&gt; in base &lt;math&gt;10&lt;/math&gt; requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to &lt;math&gt;1492&lt;/math&gt;. <br /> <br /> [[1987 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> What is the largest possible distance between two points, one on the sphere of radius 19 with center &lt;math&gt;(-2,-10,5),&lt;/math&gt; and the other on the sphere of radius 87 with center &lt;math&gt;(12,8,-16)&lt;/math&gt;?<br /> <br /> [[1987 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called &quot;nice&quot; if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?<br /> <br /> [[1987 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> Find the area of the region enclosed by the graph of &lt;math&gt;|x-60|+|y|=|x/4|.&lt;/math&gt;<br /> <br /> [[1987 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Find &lt;math&gt;3x^2 y^2&lt;/math&gt; if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are integers such that &lt;math&gt;y^2 + 3x^2 y^2 = 30x^2 + 517&lt;/math&gt;.<br /> <br /> [[1987 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is divided into four parts of equal area by five segments as shown in the figure, where &lt;math&gt;XY = YB + BC + CZ = ZW = WD + DA + AX&lt;/math&gt;, and &lt;math&gt;PQ&lt;/math&gt; is parallel to &lt;math&gt;AB&lt;/math&gt;. Find the length of &lt;math&gt;AB&lt;/math&gt; (in cm) if &lt;math&gt;BC = 19&lt;/math&gt; cm and &lt;math&gt;PQ = 87&lt;/math&gt; cm.<br /> <br /> [[Image:AIME_1987_Problem_6.png]]<br /> <br /> [[1987 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Let &lt;math&gt;[r,s]&lt;/math&gt; denote the least common multiple of positive integers &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt;. Find the number of ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers for which &lt;math&gt;[a,b] = 1000&lt;/math&gt;, &lt;math&gt;[b,c] = 2000&lt;/math&gt;, and &lt;math&gt;[c,a] = 2000&lt;/math&gt;.<br /> <br /> [[1987 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> What is the largest positive integer &lt;math&gt;n&lt;/math&gt; for which there is a unique integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;\frac{8}{15} &lt; \frac{n}{n + k} &lt; \frac{7}{13}&lt;/math&gt;?<br /> <br /> [[1987 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has right angle at &lt;math&gt;B&lt;/math&gt;, and contains a point &lt;math&gt;P&lt;/math&gt; for which &lt;math&gt;PA = 10&lt;/math&gt;, &lt;math&gt;PB = 6&lt;/math&gt;, and &lt;math&gt;\angle APB = \angle BPC = \angle CPA&lt;/math&gt;. Find &lt;math&gt;PC&lt;/math&gt;.<br /> <br /> [[Image:AIME_1987_Problem_9.png]]<br /> <br /> [[1987 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) <br /> <br /> [[1987 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Find the largest possible value of &lt;math&gt;k&lt;/math&gt; for which &lt;math&gt;3^{11}&lt;/math&gt; is expressible as the sum of &lt;math&gt;k&lt;/math&gt; consecutive positive integers.<br /> <br /> [[1987 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Let &lt;math&gt;m&lt;/math&gt; be the smallest integer whose cube root is of the form &lt;math&gt;n+r&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer and &lt;math&gt;r&lt;/math&gt; is a positive real number less than &lt;math&gt;1/1000&lt;/math&gt;. Find &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[1987 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A given sequence &lt;math&gt;r_1, r_2, \dots, r_n&lt;/math&gt; of distinct real numbers can be put in ascending order by means of one or more &quot;bubble passes&quot;. A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, &lt;math&gt;r_n&lt;/math&gt;, with its current predecessor and exchanging them if and only if the last term is smaller. <br /> <br /> The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.<br /> &lt;center&gt;&lt;math&gt;\underline{1 \quad 9} \quad 8 \quad 7&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad {}\underline{9 \quad 8} \quad 7&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad 8 \quad \underline{9 \quad 7}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;1 \quad 8 \quad 7 \quad 9&lt;/math&gt;&lt;/center&gt;<br /> Suppose that &lt;math&gt;n = 40&lt;/math&gt;, and that the terms of the initial sequence &lt;math&gt;r_1, r_2, \dots, r_{40}&lt;/math&gt; are distinct from one another and are in random order. Let &lt;math&gt;p/q&lt;/math&gt;, in lowest terms, be the probability that the number that begins as &lt;math&gt;r_{20}&lt;/math&gt; will end up, after one bubble pass, in the &lt;math&gt;30^{\mbox{th}}&lt;/math&gt; place. Find &lt;math&gt;p + q&lt;/math&gt;.<br /> <br /> [[1987 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Compute<br /> &lt;center&gt;&lt;math&gt;\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}&lt;/math&gt;&lt;/center&gt;.<br /> <br /> [[1987 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Squares &lt;math&gt;S_1&lt;/math&gt; and &lt;math&gt;S_2&lt;/math&gt; are inscribed in right triangle &lt;math&gt;ABC&lt;/math&gt;, as shown in the figures below. Find &lt;math&gt;AC + CB&lt;/math&gt; if area &lt;math&gt;(S_1) = 441&lt;/math&gt; and area &lt;math&gt;(S_2) = 440&lt;/math&gt;.<br /> <br /> [[Image:AIME_1987_Problem_15.png]]<br /> <br /> [[1987 AIME Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:AIME Problems|1987]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_5&diff=84729 1987 AIME Problems/Problem 5 2017-03-17T01:39:55Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;3x^2 y^2&lt;/math&gt; if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are [[integer]]s such that &lt;math&gt;y^2 + 3x^2 y^2 = 30x^2 + 517&lt;/math&gt;.<br /> == Solution ==<br /> If we move the &lt;math&gt;x^2&lt;/math&gt; term to the left side, it is [[SFFT|factorable]]:<br /> <br /> &lt;cmath&gt;(3x^2 + 1)(y^2 - 10) = 517 - 10&lt;/cmath&gt;<br /> <br /> &lt;math&gt;507&lt;/math&gt; is equal to &lt;math&gt;3 * 13^2&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are integers, &lt;math&gt;3x^2 + 1&lt;/math&gt; cannot equal a multiple of three. &lt;math&gt;169&lt;/math&gt; doesn't work either, so &lt;math&gt;3x^2 + 1 = 13&lt;/math&gt;, and &lt;math&gt;x^2 = 4&lt;/math&gt;. This leaves &lt;math&gt;y^2 - 10 = 39&lt;/math&gt;, so &lt;math&gt;y^2 = 49&lt;/math&gt;. Thus, &lt;math&gt;3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1987|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_13&diff=84728 1988 AIME Problems/Problem 13 2017-03-17T00:28:34Z <p>Warrenwangtennis: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a&lt;/math&gt; if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are [[integer]]s such that &lt;math&gt;x^2 - x - 1&lt;/math&gt; is a factor of &lt;math&gt;ax^{17} + bx^{16} + 1&lt;/math&gt;.<br /> <br /> __TOC__<br /> === Solution 1 ===<br /> Let's work backwards! Let &lt;math&gt;F(x) = ax^{17} + bx^{16} + 1&lt;/math&gt; and let &lt;math&gt;P(x)&lt;/math&gt; be the [[polynomial]] such that &lt;math&gt;P(x)(x^2 - x - 1) = F(x)&lt;/math&gt;.<br /> <br /> Clearly, the [[constant]] term of &lt;math&gt;P(x)&lt;/math&gt; must be &lt;math&gt;- 1&lt;/math&gt;. Now, we have &lt;math&gt;(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1)&lt;/math&gt;, where &lt;math&gt;c_{i}&lt;/math&gt; is some [[coefficient]]. However, since &lt;math&gt;F(x)&lt;/math&gt; has no &lt;math&gt;x&lt;/math&gt; term, it must be true that &lt;math&gt;c_{15} = 1&lt;/math&gt;. <br /> <br /> Let's find &lt;math&gt;c_{14}&lt;/math&gt; now. Notice that all we care about in finding &lt;math&gt;c_{14}&lt;/math&gt; is that &lt;math&gt;(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}&lt;/math&gt;. Therefore, &lt;math&gt;c_{14} = - 2&lt;/math&gt;. Undergoing a similar process, &lt;math&gt;c_{13} = 3&lt;/math&gt;, &lt;math&gt;c_{12} = - 5&lt;/math&gt;, &lt;math&gt;c_{11} = 8&lt;/math&gt;, and we see a nice pattern. The coefficients of &lt;math&gt;P(x)&lt;/math&gt; are just the [[Fibonacci sequence]] with alternating signs! Therefore, &lt;math&gt;a = c_1 = F_{16}&lt;/math&gt;, where &lt;math&gt;F_{16}&lt;/math&gt; denotes the 16th Fibonnaci number and &lt;math&gt;a = 987&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;F_n&lt;/math&gt; represent the &lt;math&gt;n&lt;/math&gt;th number in the Fibonacci sequence. Therefore,<br /> <br /> &lt;math&gt;x^2 - x - 1 = 0\Longrightarrow x^n = F_n(x),\ n\in N\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N\ .&lt;/math&gt;<br /> <br /> The above uses the similarity between the Fibonacci [[recursion|recursive]] definition, &lt;math&gt;F_{n+2} - F_{n+1} - F_n = 0&lt;/math&gt;, and the polynomial &lt;math&gt;x^2 - x - 1 = 0&lt;/math&gt;. <br /> <br /> &lt;math&gt;0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1\Longrightarrow&lt;/math&gt;<br /> <br /> &lt;math&gt;(aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q\Longrightarrow&lt;/math&gt;<br /> <br /> &lt;math&gt;aF_{17} + bF_{16} = 0&lt;/math&gt; and &lt;math&gt;aF_{16} + bF_{15} + 1 = 0\Longrightarrow&lt;/math&gt;<br /> <br /> &lt;math&gt;a = F_{16},\ b = - F_{17}\Longrightarrow \boxed {a = 987}\ .&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> We can long divide and search for a pattern; then the remainder would be set to zero to solve for &lt;math&gt;a&lt;/math&gt;. Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is &lt;math&gt;(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0&lt;/math&gt;. Since the coefficient of &lt;math&gt;x&lt;/math&gt; must be zero, this gives us two equations, &lt;math&gt;F_{16}b + F_{17}a = 0&lt;/math&gt; and &lt;math&gt;F_{15}b + F_{16}a + 1 = 0&lt;/math&gt;. Solving these two as above, we get that &lt;math&gt;a = 987&lt;/math&gt;.<br /> <br /> There are various similar solutions which yield the same pattern, such as repeated substitution of &lt;math&gt;x^2 = x + 1&lt;/math&gt; into the larger polynomial.<br /> <br /> === Solution 4 ===<br /> The roots of &lt;math&gt;x^2-x-1&lt;/math&gt; are &lt;math&gt;\phi&lt;/math&gt; (the [[Golden Ratio]]) and &lt;math&gt;1-\phi&lt;/math&gt;. These two must also be roots of &lt;math&gt;ax^{17}+bx^{16}+1&lt;/math&gt;. Thus, we have two equations: &lt;math&gt;a\phi^{17}+b\phi^{16}+1=0&lt;/math&gt; and &lt;math&gt;a(1-\phi)^{17}+b(1-\phi)^{16}+1=0&lt;/math&gt;. Subtract these two and divide by &lt;math&gt;\sqrt{5}&lt;/math&gt; to get &lt;math&gt;\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0&lt;/math&gt;. Noting that the formula for the &lt;math&gt;n&lt;/math&gt;th [[Fibonacci number]] is &lt;math&gt;\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}&lt;/math&gt;, we have &lt;math&gt;1597a+987b=0&lt;/math&gt;. Since &lt;math&gt;1597&lt;/math&gt; and &lt;math&gt;987&lt;/math&gt; are coprime, the solutions to this equation under the integers are of the form &lt;math&gt;a=987k&lt;/math&gt; and &lt;math&gt;b=-1597k&lt;/math&gt;, of which the only integral solutions for &lt;math&gt;a&lt;/math&gt; on &lt;math&gt;[0,999]&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;987&lt;/math&gt;. &lt;math&gt;(a,b)=(0,0)&lt;/math&gt; cannot work since &lt;math&gt;x^2-x-1&lt;/math&gt; does not divide &lt;math&gt;1&lt;/math&gt;, so the answer must be &lt;math&gt;\boxed{987}&lt;/math&gt;. (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between &lt;math&gt;000&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;).<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=84727 Ceva's Theorem 2017-03-16T23:38:39Z <p>Warrenwangtennis: /* Statement */</p> <hr /> <div>'''Ceva's Theorem''' is a criterion for the [[concurrence]] of [[cevian]]s in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> <br /> [[Image:Ceva1.PNG|thumb|right]]<br /> Let &lt;math&gt;ABC &lt;/math&gt; be a triangle, and let &lt;math&gt;D, E, F &lt;/math&gt; be points on lines &lt;math&gt;BC, CA, AB &lt;/math&gt;, respectively. Lines &lt;math&gt;AD, BE, CF &lt;/math&gt; are [[concurrent]] if and only if<br /> &lt;br&gt;&lt;center&gt;<br /> &lt;math&gt;\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 &lt;/math&gt;,<br /> &lt;/center&gt;&lt;br&gt;<br /> where lengths are [[directed segments | directed]]. This also works for the [[reciprocal]] of each of the ratios, as the reciprocal of &lt;math&gt;1&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt;.<br /> <br /> <br /> (Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)<br /> <br /> <br /> The proof using [[Routh's Theorem]] is extremely trivial, so we will not include it.<br /> <br /> == Proof ==<br /> <br /> We will use the notation &lt;math&gt;[ABC] &lt;/math&gt; to denote the area of a triangle with vertices &lt;math&gt;A,B,C &lt;/math&gt;.<br /> <br /> First, suppose &lt;math&gt;AD, BE, CF &lt;/math&gt; meet at a point &lt;math&gt;X &lt;/math&gt;. We note that triangles &lt;math&gt;ABD, ADC &lt;/math&gt; have the same altitude to line &lt;math&gt;BC &lt;/math&gt;, but bases &lt;math&gt;BD &lt;/math&gt; and &lt;math&gt;DC &lt;/math&gt;. It follows that &lt;math&gt; \frac {BD}{DC} = \frac{[ABD]}{[ADC]} &lt;/math&gt;. The same is true for triangles &lt;math&gt;XBD, XDC &lt;/math&gt;, so <br /> <br /> &lt;center&gt;&lt;math&gt; \frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]} &lt;/math&gt;. &lt;/center&gt;<br /> Similarly, &lt;math&gt; \frac{CE}{EA} = \frac{[BCX]}{[BXA]} &lt;/math&gt; and &lt;math&gt; \frac{AF}{FB} = \frac{[CAX]}{[CXB]} &lt;/math&gt;,<br /> so<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Now, suppose &lt;math&gt;D, E,F &lt;/math&gt; satisfy Ceva's criterion, and suppose &lt;math&gt;AD, BE &lt;/math&gt; intersect at &lt;math&gt;X &lt;/math&gt;. Suppose the line &lt;math&gt;CX &lt;/math&gt; intersects line &lt;math&gt;AB &lt;/math&gt; at &lt;math&gt;F' &lt;/math&gt;. We have proven that &lt;math&gt;F' &lt;/math&gt; must satisfy Ceva's criterion. This means that &lt;center&gt;&lt;math&gt; \frac{AF'}{F'B} = \frac{AF}{FB} &lt;/math&gt;, &lt;/center&gt; so &lt;center&gt;&lt;math&gt;F' = F &lt;/math&gt;, &lt;/center&gt; and line &lt;math&gt;CF &lt;/math&gt; concurrs with &lt;math&gt;AD &lt;/math&gt; and &lt;math&gt;BE &lt;/math&gt;. {{Halmos}}<br /> <br /> ==Proof by [[Barycentric coordinates]]==<br /> <br /> Since &lt;math&gt;D\in BC&lt;/math&gt;, we can write its coordinates as &lt;math&gt;(0,d,1-d)&lt;/math&gt;. The equation of line &lt;math&gt;AD&lt;/math&gt; is then &lt;math&gt;z=\frac{1-d}{d}y&lt;/math&gt;. <br /> <br /> Similarly, since &lt;math&gt;E=(1-e,0,e)&lt;/math&gt;, and &lt;math&gt;F=(f,1-f,0)&lt;/math&gt;, we can see that the equations of &lt;math&gt;BE&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt; respectively are &lt;math&gt;x=\frac{1-e}{e}z&lt;/math&gt; and &lt;math&gt;y=\frac{1-f}{f}x&lt;/math&gt;<br /> <br /> Multiplying the three together yields the solution to the equation:<br /> <br /> &lt;math&gt;xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y&lt;/math&gt;<br /> <br /> Dividing by &lt;math&gt;xyz&lt;/math&gt; yields:<br /> <br /> <br /> &lt;math&gt;1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}&lt;/math&gt;, which is equivalent to Ceva's theorem<br /> <br /> QED<br /> <br /> == Trigonometric Form ==<br /> <br /> The [[trig | trigonometric]] form of Ceva's Theorem (Trig Ceva) states that cevians &lt;math&gt;AD,BE,CF&lt;/math&gt; concur if and only if<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> === Proof ===<br /> <br /> First, suppose &lt;math&gt;AD, BE, CF &lt;/math&gt; concur at a point &lt;math&gt;X &lt;/math&gt;. We note that<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} &lt;/math&gt;, &lt;/center&gt;<br /> and similarly,<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} &lt;/math&gt;. &lt;/center&gt;<br /> It follows that<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} &lt;/math&gt; &lt;br&gt; &lt;br&gt; &lt;math&gt; \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod &lt;math&gt;\pi &lt;/math&gt; to be either positive or negative.<br /> <br /> The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. {{Halmos}}<br /> <br /> == Problems ==<br /> ===Introductory===<br /> *Suppose &lt;math&gt;AB, AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; have lengths &lt;math&gt;13, 14&lt;/math&gt;, and &lt;math&gt;15&lt;/math&gt;, respectively. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;, find &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;DC&lt;/math&gt;. ([[Ceva's Theorem/Problems|Source]])<br /> <br /> ===Intermediate===<br /> *In &lt;math&gt;\Delta ABC, AD, BE, CF&lt;/math&gt; are concurrent lines. &lt;math&gt;P, Q, R&lt;/math&gt; are points on &lt;math&gt;EF, FD, DE&lt;/math&gt; such that &lt;math&gt;DP, EQ, FR&lt;/math&gt; are concurrent. Prove that (using ''plane geometry'') &lt;math&gt;AP, BQ, CR&lt;/math&gt; are concurrent. (&lt;url&gt;viewtopic.php?f=151&amp;t=543574 &lt;/url&gt;)</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_12&diff=84726 1988 AIME Problems/Problem 12 2017-03-16T23:32:53Z <p>Warrenwangtennis: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P&lt;/math&gt; be an interior point of triangle &lt;math&gt;ABC&lt;/math&gt; and extend lines from the vertices through &lt;math&gt;P&lt;/math&gt; to the opposite sides. Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; denote the lengths of the segments indicated in the figure. Find the product &lt;math&gt;abc&lt;/math&gt; if &lt;math&gt;a + b + c = 43&lt;/math&gt; and &lt;math&gt;d = 3&lt;/math&gt;.<br /> <br /> [[Image:1988_AIME-12.png]]<br /> <br /> == Solution 1 ==<br /> Call the [[cevian]]s AD, BE, and CF. Using area ratios (&lt;math&gt;\triangle PBC&lt;/math&gt; and &lt;math&gt;\triangle ABC&lt;/math&gt; have the same base), we have: <br /> <br /> &lt;math&gt;\frac {d}{a + d} = \frac {[PBC]}{[ABC]}&lt;/math&gt;<br /> <br /> Similarily, &lt;math&gt;\frac {d}{b + d} = \frac {[PCA]}{[ABC]}&lt;/math&gt; and &lt;math&gt;\frac {d}{c + d} = \frac {[PAB]}{[ABC]}&lt;/math&gt;.<br /> <br /> Then,<br /> &lt;math&gt;\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} = \frac {[ABC]}{[ABC]} = 1&lt;/math&gt;<br /> <br /> The [[identity]] &lt;math&gt;\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1&lt;/math&gt; is a form of [[Ceva's Theorem]]. <br /> <br /> Plugging in &lt;math&gt;d = 3&lt;/math&gt;, we get <br /> <br /> &lt;cmath&gt;\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)&lt;/cmath&gt;<br /> &lt;cmath&gt;3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27&lt;/cmath&gt;<br /> &lt;cmath&gt;9(a + b + c) + 54 = abc=\boxed{441}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;A,B,C&lt;/math&gt; be the weights of the respective vertices. We see that the weights of the feet of the cevians are &lt;math&gt;A+B,B+C,C+A&lt;/math&gt;. By [[mass points]], we have that: &lt;cmath&gt;\dfrac{a}{3}=\dfrac{B+C}{A}&lt;/cmath&gt; &lt;cmath&gt;\dfrac{b}{3}=\dfrac{C+A}{B}&lt;/cmath&gt; &lt;cmath&gt;\dfrac{c}{3}=\dfrac{A+B}{C}&lt;/cmath&gt;<br /> <br /> If we add the equations together, we get &lt;math&gt;\frac{a+b+c}{3}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\frac{43}{3}&lt;/math&gt;<br /> <br /> If we multiply them together, we get &lt;math&gt;\frac{abc}{27}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\frac{49}{3} \implies abc=\boxed{441}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_13&diff=84723 1990 AIME Problems/Problem 13 2017-03-15T23:37:11Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}&lt;/math&gt;. Given that &lt;math&gt;9^{4000}_{}&lt;/math&gt; has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of &lt;math&gt;T_{}^{}&lt;/math&gt; have 9 as their leftmost digit?<br /> <br /> == Solution ==<br /> When a number is multiplied by &lt;math&gt;9&lt;/math&gt;, it gains a digit unless the new number starts with a 9. <br /> <br /> Since &lt;math&gt;9^{4000}&lt;/math&gt; has 3816 digits more than &lt;math&gt;9^1&lt;/math&gt;, &lt;math&gt;4000 - 3816 = 184&lt;/math&gt; numbers have 9 as their leftmost digits.<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_13&diff=84722 1990 AIME Problems/Problem 13 2017-03-15T23:36:18Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}&lt;/math&gt;. Given that &lt;math&gt;9^{4000}_{}&lt;/math&gt; has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of &lt;math&gt;T_{}^{}&lt;/math&gt; have 9 as their leftmost digit?<br /> <br /> == Solution ==<br /> When a number is multiplied by &lt;math&gt;9&lt;/math&gt;, the number will gain a digit, except when the new number starts with a 9. <br /> Since &lt;math&gt;9^{4000}&lt;/math&gt; has 3816 digits more than &lt;math&gt;9^1&lt;/math&gt;, &lt;math&gt;4000 - (3817 - 1) = 184&lt;/math&gt; numbers have 9 as their leftmost digits.<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_13&diff=84721 1990 AIME Problems/Problem 13 2017-03-15T23:35:38Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}&lt;/math&gt;. Given that &lt;math&gt;9^{4000}_{}&lt;/math&gt; has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of &lt;math&gt;T_{}^{}&lt;/math&gt; have 9 as their leftmost digit?<br /> <br /> == Solution ==<br /> When a number is multiplied by &lt;math&gt;9&lt;/math&gt;, the number will gain a digit, except when the new number starts with a &lt;math&gt;9&lt;/math&gt;. Since &lt;math&gt;9^{4000}&lt;/math&gt; has 3816 digits more than &lt;math&gt;9^1&lt;/math&gt;, exactly &lt;math&gt;4000 - (3817 - 1) = 184&lt;/math&gt; numbers have 9 as their leftmost digits.<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_13&diff=84720 1990 AIME Problems/Problem 13 2017-03-15T23:35:03Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}&lt;/math&gt;. Given that &lt;math&gt;9^{4000}_{}&lt;/math&gt; has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of &lt;math&gt;T_{}^{}&lt;/math&gt; have 9 as their leftmost digit?<br /> <br /> == Solution ==<br /> When a number is multiplied by &lt;math&gt;9&lt;/math&gt;, the number will gain a digit, except when the new number starts with a &lt;math&gt;9&lt;/math&gt;, when the number of digits remain the same. Since &lt;math&gt;9^{4000}&lt;/math&gt; has 3816 digits more than &lt;math&gt;9^1&lt;/math&gt;, exactly &lt;math&gt;4000 - (3817 - 1) = 184&lt;/math&gt; numbers have 9 as their leftmost digits.<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_11&diff=84718 1990 AIME Problems/Problem 11 2017-03-15T22:55:49Z <p>Warrenwangtennis: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Someone observed that &lt;math&gt;6! = 8 \cdot 9 \cdot 10&lt;/math&gt;. Find the largest [[positive]] [[integer]] &lt;math&gt;n^{}_{}&lt;/math&gt; for which &lt;math&gt;n^{}_{}!&lt;/math&gt; can be expressed as the [[product]] of &lt;math&gt;n - 3_{}^{}&lt;/math&gt; [[consecutive]] positive integers.<br /> <br /> == Solution 1 ==<br /> The product of &lt;math&gt;n - 3&lt;/math&gt; consecutive integers can be written as &lt;math&gt;\frac{(n - 3 + a)!}{a!}&lt;/math&gt; for some integer &lt;math&gt;a&lt;/math&gt;. Thus, &lt;math&gt;n! = \frac{(n - 3 + a)!}{a!}&lt;/math&gt;, from which it becomes evident that &lt;math&gt;a \ge 3&lt;/math&gt;. Since &lt;math&gt;(n - 3 + a)! &gt; n!&lt;/math&gt;, we can rewrite this as &lt;math&gt;\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!&lt;/math&gt;. For &lt;math&gt;a = 4&lt;/math&gt;, we get &lt;math&gt;n + 1 = 4!&lt;/math&gt; so &lt;math&gt;n = 23&lt;/math&gt;. For greater values of &lt;math&gt;a&lt;/math&gt;, we need to find the product of &lt;math&gt;a-3&lt;/math&gt; consecutive integers that equals &lt;math&gt;a!&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; can be approximated as &lt;math&gt;^{a-3}\sqrt{a!}&lt;/math&gt;, which decreases as &lt;math&gt;a&lt;/math&gt; increases. Thus, &lt;math&gt;n = 23&lt;/math&gt; is the greatest possible value to satisfy the given conditions.<br /> <br /> == Solution 2 ==<br /> Let the largest of the &lt;math&gt;n-3&lt;/math&gt; consecutive positive integers be &lt;math&gt;k&lt;/math&gt;. Clearly &lt;math&gt;k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;n&lt;/math&gt;, else the product of &lt;math&gt;n-3&lt;/math&gt; consecutive positive integers will be less than &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Key observation:<br /> Now for &lt;math&gt;n&lt;/math&gt; to be maximum the smallest number (or starting number) of the &lt;math&gt;n-3&lt;/math&gt; consecutive positive integers must be minimum, implying that &lt;math&gt;k&lt;/math&gt; needs to be minimum. But the least &lt;math&gt;k &gt; n&lt;/math&gt; is &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> So the &lt;math&gt;n-3&lt;/math&gt; consecutive positive integers are &lt;math&gt;5, 6, 7…, n+1&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{(n+1)!}{4!} = n!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n+1 = 24&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n = 23&lt;/math&gt;<br /> <br /> Kris17<br /> <br /> == Generalization: ==<br /> Largest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;n!&lt;/math&gt; can be expressed as the product of &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers is &lt;math&gt;(a+1)! - 1&lt;/math&gt;<br /> <br /> For ex. largest &lt;math&gt;n&lt;/math&gt; such that product of &lt;math&gt;n-6&lt;/math&gt; consecutive positive integers is equal to &lt;math&gt;n!&lt;/math&gt; is &lt;math&gt;7!-1 = 5039&lt;/math&gt;<br /> <br /> Proof:<br /> Reasoning the same way as above, let the largest of the &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers be &lt;math&gt;k&lt;/math&gt;. Clearly &lt;math&gt;k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;n&lt;/math&gt;, else the product of &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers will be less than &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Now, observe that for &lt;math&gt;n&lt;/math&gt; to be maximum the smallest number (or starting number) of the &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers must be minimum, implying that &lt;math&gt;k&lt;/math&gt; needs to be minimum. But the least &lt;math&gt;k &gt; n&lt;/math&gt; is &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> So the &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers are &lt;math&gt;a+2, a+3, … n+1&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{(n+1)!}{(a+1)!} = n!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n+1 = (a+1)!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n = (a+1)! -1&lt;/math&gt;<br /> <br /> Kris17<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_11&diff=84717 1990 AIME Problems/Problem 11 2017-03-15T22:53:56Z <p>Warrenwangtennis: /* Generalization: */</p> <hr /> <div>== Problem ==<br /> Someone observed that &lt;math&gt;6! = 8 \cdot 9 \cdot 10&lt;/math&gt;. Find the largest [[positive]] [[integer]] &lt;math&gt;n^{}_{}&lt;/math&gt; for which &lt;math&gt;n^{}_{}!&lt;/math&gt; can be expressed as the [[product]] of &lt;math&gt;n - 3_{}^{}&lt;/math&gt; [[consecutive]] positive integers.<br /> <br /> == Solution 1 ==<br /> The product of &lt;math&gt;n - 3&lt;/math&gt; consecutive integers can be written as &lt;math&gt;\frac{(n - 3 + a)!}{a!}&lt;/math&gt; for some integer &lt;math&gt;a&lt;/math&gt;. Thus, &lt;math&gt;n! = \frac{(n - 3 + a)!}{a!}&lt;/math&gt;, from which it becomes evident that &lt;math&gt;a \ge 3&lt;/math&gt;. Since &lt;math&gt;(n - 3 + a)! &gt; n!&lt;/math&gt;, we can rewrite this as &lt;math&gt;\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!&lt;/math&gt;. For &lt;math&gt;a = 4&lt;/math&gt;, we get &lt;math&gt;n + 1 = 4!&lt;/math&gt; so &lt;math&gt;n = 23&lt;/math&gt;. For greater values of &lt;math&gt;a&lt;/math&gt;, we need to find the product of &lt;math&gt;a-3&lt;/math&gt; consecutive integers that equals &lt;math&gt;a!&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; can be approximated as &lt;math&gt;^{a-3}\sqrt{a!}&lt;/math&gt;, which decreases as &lt;math&gt;a&lt;/math&gt; increases. Thus, &lt;math&gt;n = 23&lt;/math&gt; is the greatest possible value to satisfy the given conditions.<br /> <br /> == Solution 2 ==<br /> Let the largest of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers be &lt;math&gt;k&lt;/math&gt;. Clearly &lt;math&gt;k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;n&lt;/math&gt;, else the product of &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers will be less than &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Key observation:<br /> Now for &lt;math&gt;n&lt;/math&gt; to be maximum the smallest number (or starting number) of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers must be minimum, implying that &lt;math&gt;k&lt;/math&gt; needs to be minimum. But the least &lt;math&gt;k &gt; n&lt;/math&gt; is &lt;math&gt;(n+1).&lt;/math&gt;<br /> <br /> So the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers are &lt;math&gt;(5, 6, 7…, n+1)&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;(n+1)! /4! = n!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n+1 = 24&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n = 23&lt;/math&gt;<br /> <br /> Kris17<br /> <br /> == Generalization: ==<br /> Largest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;n!&lt;/math&gt; can be expressed as the product of &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers is &lt;math&gt;(a+1)! - 1&lt;/math&gt;<br /> <br /> For ex. largest &lt;math&gt;n&lt;/math&gt; such that product of &lt;math&gt;n-6&lt;/math&gt; consecutive positive integers is equal to &lt;math&gt;n!&lt;/math&gt; is &lt;math&gt;7!-1 = 5039&lt;/math&gt;<br /> <br /> Proof:<br /> Reasoning the same way as above, let the largest of the &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers be &lt;math&gt;k&lt;/math&gt;. Clearly &lt;math&gt;k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;n&lt;/math&gt;, else the product of &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers will be less than &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Now, observe that for &lt;math&gt;n&lt;/math&gt; to be maximum the smallest number (or starting number) of the &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers must be minimum, implying that &lt;math&gt;k&lt;/math&gt; needs to be minimum. But the least &lt;math&gt;k &gt; n&lt;/math&gt; is &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> So the &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers are &lt;math&gt;a+2, a+3, … n+1&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{(n+1)!}{(a+1)!} = n!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n+1 = (a+1)!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n = (a+1)! -1&lt;/math&gt;<br /> <br /> Kris17<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_11&diff=84716 1990 AIME Problems/Problem 11 2017-03-15T22:52:49Z <p>Warrenwangtennis: /* Generalization: */</p> <hr /> <div>== Problem ==<br /> Someone observed that &lt;math&gt;6! = 8 \cdot 9 \cdot 10&lt;/math&gt;. Find the largest [[positive]] [[integer]] &lt;math&gt;n^{}_{}&lt;/math&gt; for which &lt;math&gt;n^{}_{}!&lt;/math&gt; can be expressed as the [[product]] of &lt;math&gt;n - 3_{}^{}&lt;/math&gt; [[consecutive]] positive integers.<br /> <br /> == Solution 1 ==<br /> The product of &lt;math&gt;n - 3&lt;/math&gt; consecutive integers can be written as &lt;math&gt;\frac{(n - 3 + a)!}{a!}&lt;/math&gt; for some integer &lt;math&gt;a&lt;/math&gt;. Thus, &lt;math&gt;n! = \frac{(n - 3 + a)!}{a!}&lt;/math&gt;, from which it becomes evident that &lt;math&gt;a \ge 3&lt;/math&gt;. Since &lt;math&gt;(n - 3 + a)! &gt; n!&lt;/math&gt;, we can rewrite this as &lt;math&gt;\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!&lt;/math&gt;. For &lt;math&gt;a = 4&lt;/math&gt;, we get &lt;math&gt;n + 1 = 4!&lt;/math&gt; so &lt;math&gt;n = 23&lt;/math&gt;. For greater values of &lt;math&gt;a&lt;/math&gt;, we need to find the product of &lt;math&gt;a-3&lt;/math&gt; consecutive integers that equals &lt;math&gt;a!&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; can be approximated as &lt;math&gt;^{a-3}\sqrt{a!}&lt;/math&gt;, which decreases as &lt;math&gt;a&lt;/math&gt; increases. Thus, &lt;math&gt;n = 23&lt;/math&gt; is the greatest possible value to satisfy the given conditions.<br /> <br /> == Solution 2 ==<br /> Let the largest of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers be &lt;math&gt;k&lt;/math&gt;. Clearly &lt;math&gt;k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;n&lt;/math&gt;, else the product of &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers will be less than &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Key observation:<br /> Now for &lt;math&gt;n&lt;/math&gt; to be maximum the smallest number (or starting number) of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers must be minimum, implying that &lt;math&gt;k&lt;/math&gt; needs to be minimum. But the least &lt;math&gt;k &gt; n&lt;/math&gt; is &lt;math&gt;(n+1).&lt;/math&gt;<br /> <br /> So the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers are &lt;math&gt;(5, 6, 7…, n+1)&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;(n+1)! /4! = n!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n+1 = 24&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n = 23&lt;/math&gt;<br /> <br /> Kris17<br /> <br /> == Generalization: ==<br /> Largest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;n!&lt;/math&gt; can be expressed as the product of &lt;math&gt;n-a&lt;/math&gt; consecutive positive integers is &lt;math&gt;(a+1)! – 1&lt;/math&gt;<br /> <br /> For ex. largest 4n&lt;math&gt; such that product of &lt;/math&gt;n-6&lt;math&gt; consecutive positive integers is equal to &lt;/math&gt;n!&lt;math&gt; is &lt;/math&gt;7!-1 = 5039&lt;math&gt;<br /> <br /> Proof:<br /> Reasoning the same way as above, let the largest of the &lt;/math&gt;n-a&lt;math&gt; consecutive positive integers be &lt;/math&gt;k&lt;math&gt;. Clearly &lt;/math&gt;k&lt;math&gt; cannot be less than or equal to &lt;/math&gt;n&lt;math&gt;, else the product of &lt;/math&gt;n-a&lt;math&gt; consecutive positive integers will be less than &lt;/math&gt;n!&lt;math&gt;.<br /> <br /> Now, observe that for &lt;/math&gt;n&lt;math&gt; to be maximum the smallest number (or starting number) of the &lt;/math&gt;n-a&lt;math&gt; consecutive positive integers must be minimum, implying that &lt;/math&gt;k&lt;math&gt; needs to be minimum. But the least &lt;/math&gt;k &gt; n&lt;math&gt; is &lt;/math&gt;n+1&lt;math&gt;.<br /> <br /> So the &lt;/math&gt;n-a&lt;math&gt; consecutive positive integers are &lt;/math&gt;a+2, a+3, … n+1&lt;math&gt;<br /> <br /> So we have &lt;/math&gt;\frac{(n+1)!}/{(a+1)!} = n!&lt;math&gt;<br /> &lt;/math&gt;\Longrightarrow n+1 = (a+1)!&lt;math&gt;<br /> &lt;/math&gt;\Longrightarrow n = (a+1)! -1$<br /> <br /> Kris17<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_11&diff=84715 1990 AIME Problems/Problem 11 2017-03-15T22:48:40Z <p>Warrenwangtennis: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Someone observed that &lt;math&gt;6! = 8 \cdot 9 \cdot 10&lt;/math&gt;. Find the largest [[positive]] [[integer]] &lt;math&gt;n^{}_{}&lt;/math&gt; for which &lt;math&gt;n^{}_{}!&lt;/math&gt; can be expressed as the [[product]] of &lt;math&gt;n - 3_{}^{}&lt;/math&gt; [[consecutive]] positive integers.<br /> <br /> == Solution 1 ==<br /> The product of &lt;math&gt;n - 3&lt;/math&gt; consecutive integers can be written as &lt;math&gt;\frac{(n - 3 + a)!}{a!}&lt;/math&gt; for some integer &lt;math&gt;a&lt;/math&gt;. Thus, &lt;math&gt;n! = \frac{(n - 3 + a)!}{a!}&lt;/math&gt;, from which it becomes evident that &lt;math&gt;a \ge 3&lt;/math&gt;. Since &lt;math&gt;(n - 3 + a)! &gt; n!&lt;/math&gt;, we can rewrite this as &lt;math&gt;\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!&lt;/math&gt;. For &lt;math&gt;a = 4&lt;/math&gt;, we get &lt;math&gt;n + 1 = 4!&lt;/math&gt; so &lt;math&gt;n = 23&lt;/math&gt;. For greater values of &lt;math&gt;a&lt;/math&gt;, we need to find the product of &lt;math&gt;a-3&lt;/math&gt; consecutive integers that equals &lt;math&gt;a!&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; can be approximated as &lt;math&gt;^{a-3}\sqrt{a!}&lt;/math&gt;, which decreases as &lt;math&gt;a&lt;/math&gt; increases. Thus, &lt;math&gt;n = 23&lt;/math&gt; is the greatest possible value to satisfy the given conditions.<br /> <br /> == Solution 2 ==<br /> Let the largest of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers be &lt;math&gt;k&lt;/math&gt;. Clearly &lt;math&gt;k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;n&lt;/math&gt;, else the product of &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers will be less than &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Key observation:<br /> Now for &lt;math&gt;n&lt;/math&gt; to be maximum the smallest number (or starting number) of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers must be minimum, implying that &lt;math&gt;k&lt;/math&gt; needs to be minimum. But the least &lt;math&gt;k &gt; n&lt;/math&gt; is &lt;math&gt;(n+1).&lt;/math&gt;<br /> <br /> So the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers are &lt;math&gt;(5, 6, 7…, n+1)&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;(n+1)! /4! = n!&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n+1 = 24&lt;/math&gt;<br /> &lt;math&gt;\Longrightarrow n = 23&lt;/math&gt;<br /> <br /> Kris17<br /> <br /> == Generalization: ==<br /> Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = '''(a+1)! – 1'''<br /> <br /> For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039<br /> <br /> Proof:<br /> Reasoning the same way as above, let the largest of the (n-a) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-a) consecutive positive integers will be less than n!.<br /> <br /> Now, observe that for n to be maximum the smallest number (or starting number) of the (n-a) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k &gt; n is (n+1).<br /> <br /> So the (n-a) consecutive positive integers are (a+2, a+3, … n+1)<br /> <br /> So we have (n+1)! / (a+1)! = n!<br /> =&gt; n+1 = (a+1)!<br /> =&gt; n = '''(a+1)! -1'''<br /> <br /> Kris17<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_11&diff=84714 1990 AIME Problems/Problem 11 2017-03-15T22:48:16Z <p>Warrenwangtennis: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Someone observed that &lt;math&gt;6! = 8 \cdot 9 \cdot 10&lt;/math&gt;. Find the largest [[positive]] [[integer]] &lt;math&gt;n^{}_{}&lt;/math&gt; for which &lt;math&gt;n^{}_{}!&lt;/math&gt; can be expressed as the [[product]] of &lt;math&gt;n - 3_{}^{}&lt;/math&gt; [[consecutive]] positive integers.<br /> <br /> == Solution 1 ==<br /> The product of &lt;math&gt;n - 3&lt;/math&gt; consecutive integers can be written as &lt;math&gt;\frac{(n - 3 + a)!}{a!}&lt;/math&gt; for some integer &lt;math&gt;a&lt;/math&gt;. Thus, &lt;math&gt;n! = \frac{(n - 3 + a)!}{a!}&lt;/math&gt;, from which it becomes evident that &lt;math&gt;a \ge 3&lt;/math&gt;. Since &lt;math&gt;(n - 3 + a)! &gt; n!&lt;/math&gt;, we can rewrite this as &lt;math&gt;\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!&lt;/math&gt;. For &lt;math&gt;a = 4&lt;/math&gt;, we get &lt;math&gt;n + 1 = 4!&lt;/math&gt; so &lt;math&gt;n = 23&lt;/math&gt;. For greater values of &lt;math&gt;a&lt;/math&gt;, we need to find the product of &lt;math&gt;a-3&lt;/math&gt; consecutive integers that equals &lt;math&gt;a!&lt;/math&gt;. &lt;math&gt;n&lt;/math&gt; can be approximated as &lt;math&gt;^{a-3}\sqrt{a!}&lt;/math&gt;, which decreases as &lt;math&gt;a&lt;/math&gt; increases. Thus, &lt;math&gt;n = 23&lt;/math&gt; is the greatest possible value to satisfy the given conditions.<br /> <br /> == Solution 2 ==<br /> Let the largest of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers be &lt;math&gt;k&lt;/math&gt;. Clearly &lt;math&gt;k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;n&lt;/math&gt;, else the product of &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers will be less than &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Key observation:<br /> Now for &lt;math&gt;n&lt;/math&gt; to be maximum the smallest number (or starting number) of the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers must be minimum, implying that &lt;math&gt;k&lt;/math&gt; needs to be minimum. But the least &lt;math&gt;k &gt; n&lt;/math&gt; is &lt;math&gt;(n+1).&lt;/math&gt;<br /> <br /> So the &lt;math&gt;(n-3)&lt;/math&gt; consecutive positive integers are &lt;math&gt;(5, 6, 7…, n+1)&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;(n+1)! /4! = n!&lt;/math&gt;<br /> &lt;math&gt;\longrightarrow n+1 = 24&lt;/math&gt;<br /> &lt;math&gt;\longrightarrow n = 23&lt;/math&gt;<br /> <br /> Kris17<br /> <br /> == Generalization: ==<br /> Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = '''(a+1)! – 1'''<br /> <br /> For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039<br /> <br /> Proof:<br /> Reasoning the same way as above, let the largest of the (n-a) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-a) consecutive positive integers will be less than n!.<br /> <br /> Now, observe that for n to be maximum the smallest number (or starting number) of the (n-a) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k &gt; n is (n+1).<br /> <br /> So the (n-a) consecutive positive integers are (a+2, a+3, … n+1)<br /> <br /> So we have (n+1)! / (a+1)! = n!<br /> =&gt; n+1 = (a+1)!<br /> =&gt; n = '''(a+1)! -1'''<br /> <br /> Kris17<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_7&diff=84698 1992 AIME Problems/Problem 7 2017-03-14T21:40:15Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Problem ==<br /> Faces &lt;math&gt;ABC^{}_{}&lt;/math&gt; and &lt;math&gt;BCD^{}_{}&lt;/math&gt; of tetrahedron &lt;math&gt;ABCD^{}_{}&lt;/math&gt; meet at an angle of &lt;math&gt;30^\circ&lt;/math&gt;. The area of face &lt;math&gt;ABC^{}_{}&lt;/math&gt; is &lt;math&gt;120^{}_{}&lt;/math&gt;, the area of face &lt;math&gt;BCD^{}_{}&lt;/math&gt; is &lt;math&gt;80^{}_{}&lt;/math&gt;, and &lt;math&gt;BC=10^{}_{}&lt;/math&gt;. Find the volume of the tetrahedron.<br /> <br /> == Solution ==<br /> Since the area &lt;math&gt;BCD=80=\frac{1}{2}\cdot10\cdot16&lt;/math&gt;, the perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; has length &lt;math&gt;16&lt;/math&gt;.<br /> <br /> The perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;16 \cdot \sin 30^\circ=8&lt;/math&gt;. Therefore, the volume is &lt;math&gt;\frac{8\cdot120}{3}=\boxed{320}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1992|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_8&diff=83861 2014 AIME II Problems/Problem 8 2017-02-16T23:06:17Z <p>Warrenwangtennis: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Circle &lt;math&gt;C&lt;/math&gt; with radius 2 has diameter &lt;math&gt;\overline{AB}&lt;/math&gt;. Circle D is internally tangent to circle &lt;math&gt;C&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt;. Circle &lt;math&gt;E&lt;/math&gt; is internally tangent to circle &lt;math&gt;C&lt;/math&gt;, externally tangent to circle &lt;math&gt;D&lt;/math&gt;, and tangent to &lt;math&gt;\overline{AB}&lt;/math&gt;. The radius of circle &lt;math&gt;D&lt;/math&gt; is three times the radius of circle &lt;math&gt;E&lt;/math&gt;, and can be written in the form &lt;math&gt;\sqrt{m}-n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution 1== <br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> import graph; size(7.99cm); <br /> real labelscalefactor = 0.5; <br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> pen dotstyle = black; <br /> real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079; <br /> draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000)); <br /> draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670)); <br /> draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416)); <br /> draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432)); <br /> draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666)); <br /> draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227)); <br /> draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002)); <br /> dot((7.780000000000009,-1.320000000000002),dotstyle); <br /> label(&quot;$C$&quot;, (7.707377218471464,-1.576266740352400), NE * labelscalefactor); <br /> dot((7.271934046987930,-1.319179731427737),dotstyle); <br /> label(&quot;$D$&quot;, (7.303064016111554,-1.276266740352400), NE * labelscalefactor); <br /> dot((9.198812158392690,-0.8249788848962227),dotstyle); <br /> label(&quot;$E$&quot;, (9.225301294671791,-0.7792624249832147), NE * labelscalefactor); <br /> dot((9.198009254448635,-1.322289365031666),dotstyle); <br /> label(&quot;$F$&quot;, (9.225301294671791,-1.276266740352400), NE * labelscalefactor); <br /> dot((9.779997393419690,-1.323228980404432),dotstyle); <br /> label(&quot;$B$&quot;, (9.810012253929656,-1.276266740352400), NE * labelscalefactor); <br /> dot((5.780002606580324,-1.316771019595571),dotstyle); <br /> label(&quot;$A$&quot;, (5.812051070003994,-1.276266740352400), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> Using the diagram above, let the radius of &lt;math&gt;D&lt;/math&gt; be &lt;math&gt;3r&lt;/math&gt;, and the radius of &lt;math&gt;E&lt;/math&gt; be &lt;math&gt;r&lt;/math&gt;. Then, &lt;math&gt;EF=r&lt;/math&gt;, and &lt;math&gt;CE=2-r&lt;/math&gt;, so the Pythagorean theorem in &lt;math&gt;\triangle CEF&lt;/math&gt; gives &lt;math&gt;CF=\sqrt{4-4r}&lt;/math&gt;. Also, &lt;math&gt;CD=CA-AD=2-3r&lt;/math&gt;, so &lt;cmath&gt;DF=DC+CF=2-3r+\sqrt{4-4r}.&lt;/cmath&gt; Noting that &lt;math&gt;DE=4r&lt;/math&gt;, we can now use the Pythagorean theorem in &lt;math&gt;\triangle DEF&lt;/math&gt; to get <br /> &lt;cmath&gt;(2-3r+\sqrt{4-4r})^2+r^2=16r^2.&lt;/cmath&gt;<br /> <br /> Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives &lt;math&gt;3r=\sqrt{240}-14&lt;/math&gt; for a final answer of &lt;math&gt;\boxed{254}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Consider a reflection of circle &lt;math&gt;E&lt;/math&gt; over diameter &lt;math&gt;\overline{AB}&lt;/math&gt;. By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;r&lt;/math&gt;, and &lt;math&gt;3r&lt;/math&gt;, and the big circle has radius &lt;math&gt;2&lt;/math&gt;. <br /> <br /> Descartes' Circle Theorem gives &lt;math&gt;(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12)^2 = 2((\frac{1}{r})^2+(\frac{1}{r})^2+(\frac{1}{3r})^2+(-\frac12)^2)&lt;/math&gt;<br /> <br /> Note that the big circle has curvature &lt;math&gt;-\frac12&lt;/math&gt; because it is internally tangent. <br /> Solving gives &lt;math&gt;3r=\sqrt{240}-14&lt;/math&gt; for a final answer of &lt;math&gt;\boxed{254}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_20&diff=82968 2017 AMC 12A Problems/Problem 20 2017-02-08T22:13:23Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> How many ordered pairs &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;a&lt;/math&gt; is a positive real number and &lt;math&gt;b&lt;/math&gt; is an integer between &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;200&lt;/math&gt;, inclusive, satisfy the equation &lt;math&gt;(log_b a)^{2017}=log_b(a^{2017})?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> By the properties of logarithms, we can rearrange the equation to read &lt;math&gt;2017 log_b a=(log_b a)^{2017}&lt;/math&gt;. Then, subtracting &lt;math&gt;2017log_b a&lt;/math&gt; from each side yields &lt;math&gt;(log_b a)^{2017}-2017log_b a=0&lt;/math&gt;. We then proceed to factor out the term &lt;math&gt;log_b a&lt;/math&gt; which results in &lt;math&gt;(log_b a)[(log_b a)2016-2017]=0&lt;/math&gt;. Then, we set both factors equal to zero and solve. <br /> <br /> &lt;math&gt;log_b a=0&lt;/math&gt; has exactly &lt;math&gt;199&lt;/math&gt; solutions with the restricted domain of &lt;math&gt;[2,200]&lt;/math&gt; since this equation will always have a solution in the form of &lt;math&gt;(1, b)&lt;/math&gt;, and there are &lt;math&gt;199&lt;/math&gt; possible values of &lt;math&gt;b&lt;/math&gt; since &lt;math&gt;200-2+1 = 199&lt;/math&gt;. <br /> <br /> We proceed to solve the other factor, &lt;math&gt;(log_b a)2016-2017&lt;/math&gt;. We add &lt;math&gt;2017&lt;/math&gt; to both sides, and take the &lt;math&gt;2016th&lt;/math&gt; root, this gives us &lt;math&gt;log_b a=\sqrt{2017}&lt;/math&gt; &lt;math&gt;\sqrt{2017}&lt;/math&gt; is a real number, and therefore &lt;math&gt;a=b^{\sqrt{2017}}&lt;/math&gt; Again, there are &lt;math&gt;199&lt;/math&gt; solutions, as <br /> &lt;math&gt;b^{\sqrt{2017}}&lt;/math&gt; must be a real number (It's a real number raised to a real number). <br /> <br /> Therefore, there are as many solutions as possible &lt;math&gt;b&lt;/math&gt; values, and as there is only one value of a for each &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;199 + 199 = 398&lt;/math&gt;, therefore the answer is &lt;math&gt;\textbf{D}&lt;/math&gt;.<br /> <br /> Note: this solution is incorrect because when we take the &lt;math&gt;2016th&lt;/math&gt; root, we must also consider the negative root which is valid because the taking the reciprocal of &lt;math&gt;a&lt;/math&gt; negates &lt;math&gt;log_ba&lt;/math&gt;. Therefore the answer is &lt;math&gt;199 \cdot 3&lt;/math&gt; or &lt;math&gt;\textbf{E}&lt;/math&gt;.</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_16&diff=75492 2016 AMC 12A Problems/Problem 16 2016-02-04T22:28:05Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem 16==<br /> <br /> The graphs of &lt;math&gt;y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,&lt;/math&gt; and &lt;math&gt;y=\log_x \dfrac{1}{3}&lt;/math&gt; are plotted on the same set of axes. How many points in the plane with positive &lt;math&gt;x&lt;/math&gt;-coordinates lie on two or more of the graphs? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Setting the first two equations equal to each other, &lt;math&gt;\log_3 x = \log_x 3&lt;/math&gt;.<br /> <br /> Solving this, we get &lt;math&gt;(3, 1)&lt;/math&gt; and &lt;math&gt;(\frac{1}{3}, 1)&lt;/math&gt;.<br /> <br /> Similarly with the last two equations, we get &lt;math&gt;(3, -1)&lt;/math&gt; and &lt;math&gt;(\frac{1}{3}, -1)&lt;/math&gt;.<br /> <br /> Now, by setting the first and third equations equal to each other, we get &lt;math&gt;(1, 0)&lt;/math&gt;.<br /> <br /> Pairing the first and fourth or second and third equations won't work because then &lt;math&gt;\log x \leq 0&lt;/math&gt;.<br /> <br /> Pairing the second and fourth equations will yield &lt;math&gt;x = 1&lt;/math&gt;, but since you can't divide by &lt;math&gt;\log 1 = 0&lt;/math&gt;, it doesn't work.<br /> <br /> After trying all pairs, we have a total of &lt;math&gt;5&lt;/math&gt; solutions &lt;math&gt;\rightarrow \boxed{\textbf{(D)} 5}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_16&diff=75491 2016 AMC 12A Problems/Problem 16 2016-02-04T22:27:27Z <p>Warrenwangtennis: Created page with &quot;==Problem 16== The graphs of &lt;math&gt;y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,&lt;/math&gt; and &lt;math&gt;y=\log_x \dfrac{1}{3}&lt;/math&gt; are plotted on the same set of axes. How many p...&quot;</p> <hr /> <div>==Problem 16==<br /> <br /> The graphs of &lt;math&gt;y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,&lt;/math&gt; and &lt;math&gt;y=\log_x \dfrac{1}{3}&lt;/math&gt; are plotted on the same set of axes. How many points in the plane with positive &lt;math&gt;x&lt;/math&gt;-coordinates lie on two or more of the graphs? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Setting the first two equations equal to each other, &lt;math&gt;\log_3 x = log_x 3&lt;/math&gt;.<br /> <br /> Solving this, we get &lt;math&gt;(3, 1)&lt;/math&gt; and &lt;math&gt;(\frac{1}{3}, 1)&lt;/math&gt;.<br /> <br /> Similarly with the last two equations, we get &lt;math&gt;(3, -1)&lt;/math&gt; and &lt;math&gt;(\frac{1}{3}, -1)&lt;/math&gt;.<br /> <br /> Now, by setting the first and third equations equal to each other, we get &lt;math&gt;(1, 0)&lt;/math&gt;.<br /> <br /> Pairing the first and fourth or second and third equations won't work because then &lt;math&gt;\log x \leq 0&lt;/math&gt;.<br /> <br /> Pairing the second and fourth equations will yield &lt;math&gt;x = 1&lt;/math&gt;, but since you can't divide by &lt;math&gt;\log 1 = 0&lt;/math&gt;, it doesn't work.<br /> <br /> After trying all pairs, we have a total of &lt;math&gt;5&lt;/math&gt; solutions &lt;math&gt;\rightarrow \boxed{\textbf{(D)} 5}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=75483 2016 AMC 10A Problems/Problem 22 2016-02-04T21:13:26Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^{10} \cdot 5 \cdot 8&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^{10} \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^{12} \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;110n^3&lt;/math&gt; clearly has at least three distinct prime factors, namely 2, 5, and 11.<br /> <br /> Further, since the number of factors of &lt;math&gt;p_1^{n_1}\cdots p_k^{n_k}&lt;/math&gt; is &lt;math&gt;(n_1+1)\cdots(n_k+1)&lt;/math&gt; when the &lt;math&gt;p&lt;/math&gt;'s are distinct primes, we see that there can be at most three distinct prime factors for a number with 110 factors.<br /> <br /> We conclude that &lt;math&gt;110n^3&lt;/math&gt; has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of &lt;math&gt;n&lt;/math&gt; all of the form &lt;math&gt;n=p_1\cdot p_2^3&lt;/math&gt;.<br /> <br /> &lt;math&gt;81n^4&lt;/math&gt; thus has prime factorization &lt;math&gt;81n^4=3^4\cdot p_1^4\cdot p_2^{12}&lt;/math&gt; and a factor count of &lt;math&gt;5\cdot5\cdot13=\boxed{\textbf{(D) }325}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_11&diff=75476 2016 AMC 12A Problems/Problem 11 2016-02-04T21:07:43Z <p>Warrenwangtennis: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> Each of the &lt;math&gt;100&lt;/math&gt; students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are &lt;math&gt;42&lt;/math&gt; students who cannot sing, &lt;math&gt;65&lt;/math&gt; students who cannot dance, and &lt;math&gt;29&lt;/math&gt; students who cannot act. How many students have two of these talents? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64&lt;/math&gt; <br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the number of students that can only sing, &lt;math&gt;b&lt;/math&gt; can only dance, and &lt;math&gt;c&lt;/math&gt; can only act.<br /> <br /> Let &lt;math&gt;ab&lt;/math&gt; be the number of students that can sing and dance, &lt;math&gt;ac&lt;/math&gt; can sing and act, and &lt;math&gt;bc&lt;/math&gt; can dance and act.<br /> <br /> From the information given in the problem, &lt;math&gt;a + ab + b = 29, b + bc + c = 42,&lt;/math&gt; and &lt;math&gt;a + ac + c = 65&lt;/math&gt;.<br /> <br /> Adding these equations together, we get &lt;math&gt;2(a + b + c) + ab + bc + ac = 136&lt;/math&gt;.<br /> <br /> Since there are a total of &lt;math&gt;100&lt;/math&gt; students, &lt;math&gt;a + b + c + ab + bc + ac = 100&lt;/math&gt;.<br /> <br /> Subtracting these equations, we get &lt;math&gt;a + b + c = 36&lt;/math&gt;.<br /> <br /> Our answer is &lt;math&gt;ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=24|num-a=26}}<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_11&diff=75475 2016 AMC 12A Problems/Problem 11 2016-02-04T21:06:58Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Each of the &lt;math&gt;100&lt;/math&gt; students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are &lt;math&gt;42&lt;/math&gt; students who cannot sing, &lt;math&gt;65&lt;/math&gt; students who cannot dance, and &lt;math&gt;29&lt;/math&gt; students who cannot act. How many students have two of these talents? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64&lt;/math&gt; <br /> <br /> ==Problem==<br /> <br /> Each of the &lt;math&gt;100&lt;/math&gt; students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are &lt;math&gt;42&lt;/math&gt; students who cannot sing, &lt;math&gt;65&lt;/math&gt; students who cannot dance, and &lt;math&gt;29&lt;/math&gt; students who cannot act. How many students have two of these talents? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the number of students that can only sing, &lt;math&gt;b&lt;/math&gt; can only dance, and &lt;math&gt;c&lt;/math&gt; can only act.<br /> <br /> Let &lt;math&gt;ab&lt;/math&gt; be the number of students that can sing and dance, &lt;math&gt;ac&lt;/math&gt; can sing and act, and &lt;math&gt;bc&lt;/math&gt; can dance and act.<br /> <br /> From the information given in the problem, &lt;math&gt;a + ab + b = 29, b + bc + c = 42,&lt;/math&gt; and &lt;math&gt;a + ac + c = 65&lt;/math&gt;.<br /> <br /> Adding these equations together, we get &lt;math&gt;2(a + b + c) + ab + bc + ac = 136&lt;/math&gt;.<br /> <br /> Since there are a total of &lt;math&gt;100&lt;/math&gt; students, &lt;math&gt;a + b + c + ab + bc + ac = 100&lt;/math&gt;.<br /> <br /> Subtracting these equations, we get &lt;math&gt;a + b + c = 36&lt;/math&gt;.<br /> <br /> Our answer is &lt;math&gt;ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=24|num-a=26}}<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_9&diff=75451 2016 AMC 12A Problems/Problem 9 2016-02-04T19:20:41Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem 9==<br /> <br /> The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is &lt;math&gt;\tfrac{a-\sqrt{2}}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt; ?<br /> <br /> &lt;asy&gt;<br /> real x=.369;<br /> draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br /> filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray);<br /> filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray);<br /> filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray);<br /> filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray);<br /> filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;s&lt;/math&gt; be the side length of the small squares.<br /> <br /> The diagonal of the big square can be written in two ways: &lt;math&gt;\sqrt{2}&lt;/math&gt; and &lt;math&gt;s \sqrt{2} + s + s \sqrt{2}&lt;/math&gt;.<br /> <br /> Solving for &lt;math&gt;s&lt;/math&gt;, we get &lt;math&gt;s = \frac{4 - \sqrt{2}}{7}&lt;/math&gt;, so our answer is &lt;math&gt;4 + 7 \Rightarrow \boxed{\textbf{(E)} 11}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_9&diff=75450 2016 AMC 12A Problems/Problem 9 2016-02-04T19:20:19Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem 9==<br /> <br /> The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is &lt;math&gt;\tfrac{a-\sqrt{2}}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt; ?<br /> <br /> &lt;asy&gt;<br /> real x=.369;<br /> draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br /> filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray);<br /> filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray);<br /> filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray);<br /> filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray);<br /> filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;s&lt;/math&gt; be the side length of the small squares.<br /> <br /> The diagonal of the big square can be written in two ways: &lt;math&gt;\sqrt{2} and &lt;/math&gt;s \sqrt{2} + s + s \sqrt{2}&lt;math&gt;.<br /> <br /> Solving for &lt;/math&gt;s&lt;math&gt;, we get &lt;/math&gt;s = \frac{4 - \sqrt{2}}{7}&lt;math&gt;, so our answer is &lt;/math&gt;4 + 7 \Rightarrow \boxed{\textbf{(E)} 11}$</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_9&diff=75446 2016 AMC 12A Problems/Problem 9 2016-02-04T18:52:49Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem 9==<br /> <br /> The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is &lt;math&gt;\tfrac{a-\sqrt{2}}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt; ?<br /> <br /> &lt;asy&gt;<br /> real x=.369;<br /> draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br /> filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray);<br /> filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray);<br /> filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray);<br /> filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray);<br /> filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> ==Solution==</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_9&diff=75445 2016 AMC 12A Problems/Problem 9 2016-02-04T18:52:15Z <p>Warrenwangtennis: Created page with &quot;==Problem 9== The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with o...&quot;</p> <hr /> <div>==Problem 9==<br /> <br /> The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is &lt;math&gt;\tfrac{a-\sqrt{2}}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt; ?<br /> <br /> &lt;asy&gt;<br /> real x=.369;<br /> draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br /> filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray);<br /> filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray);<br /> filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray);<br /> filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray);<br /> filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;s&lt;/math&gt; be the side length of the small squares.<br /> <br /> The diagonal of the big square can be written in two ways: &lt;math&gt;\sqrt{2} and &lt;/math&gt;s \sqrt{2} + s + s \sqrt{2}&lt;math&gt;.<br /> <br /> Solving for &lt;/math&gt;s&lt;math&gt;, we get &lt;/math&gt;s = \frac{4 - \sqrt{2}}{4}&lt;math&gt;, so our answer is &lt;/math&gt;4 + 4 \Rightarrow \boxed{\textbf{(B)} 8}$</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=75438 2016 AMC 12A Problems/Problem 17 2016-02-04T18:33:16Z <p>Warrenwangtennis: Created page with &quot;==Problem 17== Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overlin...&quot;</p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems&diff=75435 2016 AMC 12A Problems 2016-02-04T18:18:26Z <p>Warrenwangtennis: /* Problem 17 */</p> <hr /> <div>{{AMC12 Problems|year=2016|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;math&gt;\frac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^x\cdot100^{2x}=1000^5&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, &lt;math&gt;2016=13+2003&lt;/math&gt;). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{an odd integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(B)}\ \text{an odd integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(C)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two numbers that are not prime}\\<br /> \qquad\textbf{(D)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(E)}\ \text{an even integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Which of these describes the graph of &lt;math&gt;x^2(x+y+1)=y^2(x+y+1)&lt;/math&gt; ?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{two parallel lines}\\<br /> \qquad\textbf{(B)}\ \text{two intersecting lines}\\<br /> \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\<br /> \qquad\textbf{(D)}\ \text{three lines that do not all pass through a comment point}\\<br /> \qquad\textbf{(E)}\ \text{a line and a parabola}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the area of the shaded region of the given &lt;math&gt;8\times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{4}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 5\dfrac{1}{4}\qquad\textbf{(D)}\ 6\dfrac{1}{2}\qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is &lt;math&gt;\tfrac{a-\sqrt{2}}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt; ?<br /> <br /> &lt;pre style=&quot;color: gray&quot;&gt;TODO: Diagram&lt;/pre&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Each of the 100 students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are 42 students who cannot sing, 65 students who cannot dance, and 29 students who cannot act. How many students have two of these talents?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 6&lt;/math&gt;, &lt;math&gt;BC = 7&lt;/math&gt;, and &lt;math&gt;CA = 8&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt; bisects &lt;math&gt;\angle BAC&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BE}&lt;/math&gt; bisects &lt;math&gt;\angle ABC&lt;/math&gt;. The bisectors intersect at &lt;math&gt;F&lt;/math&gt;. What is the ratio &lt;math&gt;AF&lt;/math&gt; : &lt;math&gt;FD&lt;/math&gt;?<br /> <br /> &lt;pre style=&quot;color: gray&quot;&gt;TODO: Diagram&lt;/pre&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Each vertex of a cube is to be labeled with an integer from 1 through 8, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> The graphs of &lt;math&gt;y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,&lt;/math&gt; and &lt;math&gt;y=\log_x \dfrac{1}{3}&lt;/math&gt; are plotted on the same set of axes. How many points in the plane with positive &lt;math&gt;x&lt;/math&gt;-coordinates lie on two or more of the graphs? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> For some positive integer &lt;math&gt;n,&lt;/math&gt; the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3.&lt;/math&gt; How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 110\qquad\textbf{(B)}\ 191\qquad\textbf{(C)}\ 261\qquad\textbf{(D)}\ 325\qquad\textbf{(E)}\ 425&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Jerry starts at &lt;math&gt;0&lt;/math&gt; on the real number line. He tosses a fair coin &lt;math&gt;8&lt;/math&gt; times. When he gets heads, he moves &lt;math&gt;1&lt;/math&gt; unit in the positive direction; when he gets tails, he moves &lt;math&gt;1&lt;/math&gt; unit in the negative direction. The probability that he reaches &lt;math&gt;4&lt;/math&gt; at some time during this process &lt;math&gt;\frac{a}{b},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b?&lt;/math&gt; (For example, he succeeds if his sequence of tosses is &lt;math&gt;HTHHHHHH.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> <br /> ==Problem 20==<br /> <br /> A binary operation &lt;math&gt;\ \diamondsuit &lt;/math&gt; has the properties that &lt;math&gt;a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\ \cdot\ c&lt;/math&gt; and that &lt;math&gt;a\ \diamondsuit\ a = 1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c.&lt;/math&gt; (Here the dot &lt;math&gt;\ \cdot&lt;/math&gt; represents the usual multiplication operation.) The solution to the equation &lt;math&gt;2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100&lt;/math&gt; can be written as &lt;math&gt;\frac{p}{q},&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relativelt prime positive integers. What is &lt;math&gt;p + q?&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}.&lt;/math&gt; Three of the sides of this quadrilateral have length &lt;math&gt;200.&lt;/math&gt; What is the length of its fourth side? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Problem text<br /> <br /> &lt;math&gt;\textbf{(A)}\ thing\qquad\textbf{(B)}\ thing\qquad\textbf{(C)}\ thng\qquad\textbf{(D)}\ thing\qquad\textbf{(E)}\ thing&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> <br /> ==Problem 24==<br /> <br /> Problem text<br /> <br /> &lt;math&gt;\textbf{(A)}\ thing\qquad\textbf{(B)}\ thing\qquad\textbf{(C)}\ thng\qquad\textbf{(D)}\ thing\qquad\textbf{(E)}\ thing&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;k&lt;/math&gt; be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with &lt;math&gt;k+1&lt;/math&gt; digits. Every time Bernardo writes a number, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it. Bernardo then writes the next perfect square, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let &lt;math&gt;f(k)&lt;/math&gt; be the smallest positive integer not written on the board. For example, if &lt;math&gt;k = 1&lt;/math&gt;, then the numbers that Bernardo writes are &lt;math&gt;16, 25, 36, 49, 64&lt;/math&gt;, and the numbers showing on the board after Silvia erases are &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;f(1) = 5&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;f(2) + f(4)+ f(6) + ... + f(2016)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems&diff=75434 2016 AMC 12A Problems 2016-02-04T18:17:41Z <p>Warrenwangtennis: /* Problem 17 */</p> <hr /> <div>{{AMC12 Problems|year=2016|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;math&gt;\frac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^x\cdot100^{2x}=1000^5&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, &lt;math&gt;2016=13+2003&lt;/math&gt;). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{an odd integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(B)}\ \text{an odd integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(C)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two numbers that are not prime}\\<br /> \qquad\textbf{(D)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(E)}\ \text{an even integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Which of these describes the graph of &lt;math&gt;x^2(x+y+1)=y^2(x+y+1)&lt;/math&gt; ?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{two parallel lines}\\<br /> \qquad\textbf{(B)}\ \text{two intersecting lines}\\<br /> \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\<br /> \qquad\textbf{(D)}\ \text{three lines that do not all pass through a comment point}\\<br /> \qquad\textbf{(E)}\ \text{a line and a parabola}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the area of the shaded region of the given &lt;math&gt;8\times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{4}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 5\dfrac{1}{4}\qquad\textbf{(D)}\ 6\dfrac{1}{2}\qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is &lt;math&gt;\tfrac{a-\sqrt{2}}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt; ?<br /> <br /> &lt;pre style=&quot;color: gray&quot;&gt;TODO: Diagram&lt;/pre&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Each of the 100 students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are 42 students who cannot sing, 65 students who cannot dance, and 29 students who cannot act. How many students have two of these talents?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 6&lt;/math&gt;, &lt;math&gt;BC = 7&lt;/math&gt;, and &lt;math&gt;CA = 8&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt; bisects &lt;math&gt;\angle BAC&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BE}&lt;/math&gt; bisects &lt;math&gt;\angle ABC&lt;/math&gt;. The bisectors intersect at &lt;math&gt;F&lt;/math&gt;. What is the ratio &lt;math&gt;AF&lt;/math&gt; : &lt;math&gt;FD&lt;/math&gt;?<br /> <br /> &lt;pre style=&quot;color: gray&quot;&gt;TODO: Diagram&lt;/pre&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Each vertex of a cube is to be labeled with an integer from 1 through 8, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> The graphs of &lt;math&gt;y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,&lt;/math&gt; and &lt;math&gt;y=\log_x \dfrac{1}{3}&lt;/math&gt; are plotted on the same set of axes. How many points in the plane with positive &lt;math&gt;x&lt;/math&gt;-coordinates lie on two or more of the graphs? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ration of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> For some positive integer &lt;math&gt;n,&lt;/math&gt; the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3.&lt;/math&gt; How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 110\qquad\textbf{(B)}\ 191\qquad\textbf{(C)}\ 261\qquad\textbf{(D)}\ 325\qquad\textbf{(E)}\ 425&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Jerry starts at &lt;math&gt;0&lt;/math&gt; on the real number line. He tosses a fair coin &lt;math&gt;8&lt;/math&gt; times. When he gets heads, he moves &lt;math&gt;1&lt;/math&gt; unit in the positive direction; when he gets tails, he moves &lt;math&gt;1&lt;/math&gt; unit in the negative direction. The probability that he reaches &lt;math&gt;4&lt;/math&gt; at some time during this process &lt;math&gt;\frac{a}{b},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b?&lt;/math&gt; (For example, he succeeds if his sequence of tosses is &lt;math&gt;HTHHHHHH.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> <br /> ==Problem 20==<br /> <br /> A binary operation &lt;math&gt;\ \diamondsuit &lt;/math&gt; has the properties that &lt;math&gt;a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\ \cdot\ c&lt;/math&gt; and that &lt;math&gt;a\ \diamondsuit\ a = 1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c.&lt;/math&gt; (Here the dot &lt;math&gt;\ \cdot&lt;/math&gt; represents the usual multiplication operation.) The solution to the equation &lt;math&gt;2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100&lt;/math&gt; can be written as &lt;math&gt;\frac{p}{q},&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relativelt prime positive integers. What is &lt;math&gt;p + q?&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}.&lt;/math&gt; Three of the sides of this quadrilateral have length &lt;math&gt;200.&lt;/math&gt; What is the length of its fourth side? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Problem text<br /> <br /> &lt;math&gt;\textbf{(A)}\ thing\qquad\textbf{(B)}\ thing\qquad\textbf{(C)}\ thng\qquad\textbf{(D)}\ thing\qquad\textbf{(E)}\ thing&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> <br /> ==Problem 24==<br /> <br /> Problem text<br /> <br /> &lt;math&gt;\textbf{(A)}\ thing\qquad\textbf{(B)}\ thing\qquad\textbf{(C)}\ thng\qquad\textbf{(D)}\ thing\qquad\textbf{(E)}\ thing&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;k&lt;/math&gt; be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with &lt;math&gt;k+1&lt;/math&gt; digits. Every time Bernardo writes a number, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it. Bernardo then writes the next perfect square, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let &lt;math&gt;f(k)&lt;/math&gt; be the smallest positive integer not written on the board. For example, if &lt;math&gt;k = 1&lt;/math&gt;, then the numbers that Bernardo writes are &lt;math&gt;16, 25, 36, 49, 64&lt;/math&gt;, and the numbers showing on the board after Silvia erases are &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;f(1) = 5&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;f(2) + f(4)+ f(6) + ... + f(2016)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=75433 2016 AMC 12A Problems/Problem 12 2016-02-04T18:16:13Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Problem 12==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 6&lt;/math&gt;, &lt;math&gt;BC = 7&lt;/math&gt;, and &lt;math&gt;CA = 8&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt; bisects &lt;math&gt;\angle BAC&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BE}&lt;/math&gt; bisects &lt;math&gt;\angle ABC&lt;/math&gt;. The bisectors intersect at &lt;math&gt;F&lt;/math&gt;. What is the ratio &lt;math&gt;AF&lt;/math&gt; : &lt;math&gt;FD&lt;/math&gt;?<br /> <br /> &lt;pre style=&quot;color: gray&quot;&gt;TODO: Diagram&lt;/pre&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2&lt;/math&gt;<br /> <br /> == Solution ==<br /> By the angle bisector theorem, &lt;math&gt;\frac{AB}{AE} = \frac{CB}{CE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{6}{AE} = \frac{7}{8 - AE}&lt;/math&gt; so &lt;math&gt;AE = \frac{48}{13}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;CD = 4&lt;/math&gt;.<br /> <br /> Now, we use mass points. Assign point &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;mC \cdot CD = mB \cdot DB&lt;/math&gt; , so &lt;math&gt;mB = \frac{4}{3}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;A&lt;/math&gt; will have a mass of &lt;math&gt;\frac{7}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}&lt;/math&gt;<br /> <br /> So &lt;math&gt;\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=75432 2016 AMC 12A Problems/Problem 12 2016-02-04T18:13:54Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Solution ==<br /> By the angle bisector theorem, &lt;math&gt;\frac{AB}{AE} = \frac{CB}{CE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{6}{AE} = \frac{7}{8 - AE}&lt;/math&gt; so &lt;math&gt;AE = \frac{48}{13}&lt;/math&gt;<br /> <br /> <br /> Similarly, &lt;math&gt;CD = 4&lt;/math&gt;.<br /> <br /> Now, we use mass points. Assign point &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;mC \cdot CD = mB \cdot DB&lt;/math&gt; , so &lt;math&gt;mB = \frac{4}{3}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;A&lt;/math&gt; will have a mass of &lt;math&gt;\frac{7}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}&lt;/math&gt;<br /> <br /> So &lt;math&gt;\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=75431 2016 AMC 12A Problems/Problem 12 2016-02-04T18:11:13Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Solution ==<br /> By the angle bisector theorem, &lt;math&gt;\frac{AB}{AE} = \frac{CB}{CE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{6}{AE} = \frac{7}{8 - AE}&lt;/math&gt; so &lt;math&gt;AE = \frac{48}{13}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;CD = 4&lt;/math&gt;<br /> <br /> Now, we use mass points. Assign point &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;mC \cdot CD = mB \cdot DB&lt;/math&gt; , so &lt;math&gt;mB = \frac{4}{3}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;A&lt;/math&gt; will have a mass of &lt;math&gt;\frac{7}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}&lt;/math&gt;<br /> <br /> So &lt;math&gt;\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems&diff=75427 2016 AMC 12A Problems 2016-02-04T18:03:25Z <p>Warrenwangtennis: /* Problem 8 */</p> <hr /> <div>{{AMC12 Problems|year=2016|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;math&gt;\frac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^x\cdot100^{2x}=1000^5&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, &lt;math&gt;2016=13+2003&lt;/math&gt;). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{an odd integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(B)}\ \text{an odd integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(C)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two numbers that are not prime}\\<br /> \qquad\textbf{(D)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br /> \qquad\textbf{(E)}\ \text{an even integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Which of these describes the graph of &lt;math&gt;x^2(x+y+1)=y^2(x+y+1)&lt;/math&gt; ?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{two parallel lines}\\<br /> \qquad\textbf{(B)}\ \text{two intersecting lines}\\<br /> \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\<br /> \qquad\textbf{(D)}\ \text{three lines that do not all pass through a comment point}\\<br /> \qquad\textbf{(E)}\ \text{a line and a parabola}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the area of the shaded region of the given &lt;math&gt;8\times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{4}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 5\dfrac{1}{4}\qquad\textbf{(D)}\ 6\dfrac{1}{2}\qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is &lt;math&gt;\tfrac{a-\sqrt{2}}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt; ?<br /> <br /> &lt;pre style=&quot;color: gray&quot;&gt;TODO: Diagram&lt;/pre&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Each of the 100 students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are 42 students who cannot sing, 65 students who cannot dance, and 29 students who cannot act. How many students have two of these talents?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 6&lt;/math&gt;, &lt;math&gt;BC = 7&lt;/math&gt;, and &lt;math&gt;CA = 8&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt; bisects &lt;math&gt;\angle BAC&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BE}&lt;/math&gt; bisects &lt;math&gt;\angle ABC&lt;/math&gt;. The bisectors intersect at &lt;math&gt;F&lt;/math&gt;. What is the ratio &lt;math&gt;AF&lt;/math&gt; : &lt;math&gt;FD&lt;/math&gt;?<br /> <br /> &lt;pre style=&quot;color: gray&quot;&gt;TODO: Diagram&lt;/pre&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Each vertex of a cube is to be labeled with an integer from 1 through 8, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> The graphs of &lt;math&gt;y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,&lt;/math&gt; and &lt;math&gt;y=\log_x \dfrac{1}{3}&lt;/math&gt; are plotted on the same set of axes. How many points in the plane with positive &lt;math&gt;x&lt;/math&gt;-coordinates lie on two or more of the graphs? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is rge ration of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> For some positive integer &lt;math&gt;n,&lt;/math&gt; the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3.&lt;/math&gt; How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 110\qquad\textbf{(B)}\ 191\qquad\textbf{(C)}\ 261\qquad\textbf{(D)}\ 325\qquad\textbf{(E)}\ 425&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Jerry starts at &lt;math&gt;0&lt;/math&gt; on the real number line. He tosses a fair coin &lt;math&gt;8&lt;/math&gt; times. When he gets heads, he moves &lt;math&gt;1&lt;/math&gt; unit in the positive direction; when he gets tails, he moves &lt;math&gt;1&lt;/math&gt; unit in the negative direction. The probability that he reaches &lt;math&gt;4&lt;/math&gt; at some time during this process &lt;math&gt;\frac{a}{b},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b?&lt;/math&gt; (For example, he succeeds if his sequence of tosses is &lt;math&gt;HTHHHHHH.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> <br /> ==Problem 20==<br /> <br /> A binary operation &lt;math&gt;\ \diamondsuit &lt;/math&gt; has the properties that &lt;math&gt;a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\ \cdot\ c&lt;/math&gt; and that &lt;math&gt;a\ \diamondsuit\ a = 1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c.&lt;/math&gt; (Here the dot &lt;math&gt;\ \cdot&lt;/math&gt; represents the usual multiplication operation.) The solution to the equation &lt;math&gt;2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100&lt;/math&gt; can be written as &lt;math&gt;\frac{p}{q},&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relativelt prime positive integers. What is &lt;math&gt;p + q?&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}.&lt;/math&gt; Three of the sides of this quadrilateral have length &lt;math&gt;200.&lt;/math&gt; What is the length of its fourth side? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Problem text<br /> <br /> &lt;math&gt;\textbf{(A)}\ thing\qquad\textbf{(B)}\ thing\qquad\textbf{(C)}\ thng\qquad\textbf{(D)}\ thing\qquad\textbf{(E)}\ thing&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> <br /> ==Problem 24==<br /> <br /> Problem text<br /> <br /> &lt;math&gt;\textbf{(A)}\ thing\qquad\textbf{(B)}\ thing\qquad\textbf{(C)}\ thng\qquad\textbf{(D)}\ thing\qquad\textbf{(E)}\ thing&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;k&lt;/math&gt; be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with &lt;math&gt;k+1&lt;/math&gt; digits. Every time Bernardo writes a number, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it. Bernardo then writes the next perfect square, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let &lt;math&gt;f(k)&lt;/math&gt; be the smallest positive integer not written on the board. For example, if &lt;math&gt;k = 1&lt;/math&gt;, then the numbers that Bernardo writes are &lt;math&gt;16, 25, 36, 49, 64&lt;/math&gt;, and the numbers showing on the board after Silvia erases are &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;f(1) = 5&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;f(2) + f(4)+ f(6) + ... + f(2016)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064&lt;/math&gt;<br /> <br /> [[2016 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=75426 2016 AMC 12A Problems/Problem 12 2016-02-04T18:00:48Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Solution ==<br /> By the angle bisector theorem, &lt;math&gt;\frac{AB}{AE} = \frac{CB}{CE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{6}{AE} = \frac{7}{8 - AE}&lt;/math&gt; so &lt;math&gt;AE = \frac{48}{13}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;CD = 4&lt;/math&gt;<br /> <br /> Now, we use mass points.<br /> <br /> Assign point &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;1&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\frac{AE}{EC} = \frac{6}{7}, A&lt;/math&gt; will have a mass of &lt;math&gt;\frac{7}{6}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;B&lt;/math&gt; will have a mass of &lt;math&gt;\frac{4}{3}&lt;/math&gt;<br /> <br /> &lt;math&gt;mE = mA + mC = \frac{13}{6}&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;mD = mC + mB = \frac{7}{3}&lt;/math&gt;<br /> <br /> The mass of &lt;math&gt;F&lt;/math&gt; is the sum of the masses of &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> &lt;math&gt;mF = mE + mB = \frac{7}{2}&lt;/math&gt;<br /> <br /> This can be checked with &lt;math&gt;mD + mA&lt;/math&gt;, which is also &lt;math&gt;\frac{7}{2}&lt;/math&gt;<br /> <br /> So &lt;math&gt;\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=75388 2016 AMC 12A Problems/Problem 12 2016-02-04T16:59:56Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>== Solution ==<br /> By the angle bisector theorem, &lt;math&gt;\frac{AB}{AE} = \frac{CB}{CE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{6}{AE} = \frac{7}{8 - AE}&lt;/math&gt;, so &lt;math&gt;AE = \frac{48}{13}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;CD = 4&lt;/math&gt;<br /> <br /> Now, we use mass points.<br /> <br /> Assign point &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;1&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\frac{AE}{EC} = \frac{6}{7}, A&lt;/math&gt; will have a mass of &lt;math&gt;\frac{7}{6}&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;B&lt;/math&gt; will have a mass of &lt;math&gt;\frac{4}{3}&lt;/math&gt;.<br /> <br /> &lt;math&gt;mE = mA + mC = \frac{13}{6}&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;mD = mC + mB = \frac{7}{3}&lt;/math&gt;.<br /> <br /> The mass of &lt;math&gt;F&lt;/math&gt; is the sum of the masses of &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> &lt;math&gt;mF = mE + mB = \frac{7}{2}&lt;/math&gt;.<br /> <br /> This can be checked with &lt;math&gt;mD + mA&lt;/math&gt;, which is also &lt;math&gt;\frac{7}{2}&lt;/math&gt;.<br /> <br /> So &lt;math&gt;\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=75386 2016 AMC 12A Problems/Problem 12 2016-02-04T16:59:17Z <p>Warrenwangtennis: Created page with &quot;== Solution == By the angle bisector theorem, &lt;math&gt;\frac{AB}{AE} = \frac{CB}{CE}&lt;/math&gt; &lt;math&gt;\frac{6}{AE} = \frac{7}{8 - AE}&lt;/math&gt;, so &lt;math&gt;AE = \frac{48}{13}&lt;/math&gt; Sim...&quot;</p> <hr /> <div>== Solution ==<br /> By the angle bisector theorem, &lt;math&gt;\frac{AB}{AE} = \frac{CB}{CE}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{6}{AE} = \frac{7}{8 - AE}&lt;/math&gt;, so &lt;math&gt;AE = \frac{48}{13}&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;CD = 4&lt;/math&gt;<br /> <br /> Now, we use mass points.<br /> <br /> Assign point &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;1&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\frac{AE}/{EC} = \frac{6}{7}, A&lt;/math&gt; will have a mass of &lt;math&gt;\frac{7}{6}&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;B&lt;/math&gt; will have a mass of &lt;math&gt;\frac{4}{3}&lt;/math&gt;.<br /> <br /> &lt;math&gt;mE = mA + mC = \frac{13}{6}&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;mD = mC + mB = \frac{7}{3}&lt;/math&gt;.<br /> <br /> The mass of &lt;math&gt;F&lt;/math&gt; is the sum of the masses of &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> &lt;math&gt;mF = mE + mB = \frac{7}{2}&lt;/math&gt;.<br /> <br /> This can be checked with &lt;math&gt;mD + mA&lt;/math&gt;, which is also &lt;math&gt;\frac{7}{2}&lt;/math&gt;.<br /> <br /> So &lt;math&gt;\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=75186 2016 AMC 10A Problems 2016-02-04T01:27:36Z <p>Warrenwangtennis: /* Problem 19 */</p> <hr /> <div>==Problem 1==<br /> What is the value of &lt;math&gt;\dfrac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^{x}\cdot 100^{2x}=1000^{5}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> For every dollar Ben spent on bagels, David spent &lt;math&gt;25&lt;/math&gt; cents less. Ben paid &lt;math&gt;\$12.50&lt;/math&gt; more than David. How much did they spend in the bagel store together?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ximena lists the whole numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;30&lt;/math&gt; once. Emilio copies Ximena's numbers, replacing each occurrence of the digit &lt;math&gt;2&lt;/math&gt; by the digit &lt;math&gt;1&lt;/math&gt;. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays &lt;math&gt;40&lt;/math&gt; coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is &lt;math&gt;1&lt;/math&gt; foot wide on all four sides. What is the length in feet of the inner rectangle?<br /> &lt;asy&gt;<br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> path rectangle(pair X, pair Y){<br /> return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br /> }<br /> filldraw(rectangle((0,0),(7,5)),gray(0.5));<br /> filldraw(rectangle((1,1),(6,4)),gray(0.75));<br /> filldraw(rectangle((2,2),(5,3)),white);<br /> <br /> label(&quot;$1$&quot;,(0.5,2.5));<br /> draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br /> draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(1.5,2.5));<br /> draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br /> draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.5,2.5));<br /> draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br /> draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.1,1.5));<br /> draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br /> draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(3.7,0.5));<br /> draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br /> draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br /> &lt;/asy&gt;<br /> <br /> &lt;cmath&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8&lt;/cmath&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> What is the area of the shaded region of the given &lt;math&gt;8 \times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Three distinct integers are selected at random between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2016&lt;/math&gt;, inclusive. Which of the following is a correct statement about the probability &lt;math&gt;p&lt;/math&gt; that the product of the three integers is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ p&lt;\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}&lt;p&lt;\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p&gt;\dfrac{1}{3}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> How many ways are there to write &lt;math&gt;2016&lt;/math&gt; as the sum of twos and threes, ignoring order? (For example, &lt;math&gt;1008\cdot 2 + 0\cdot 3&lt;/math&gt; and &lt;math&gt;402\cdot 2 + 404\cdot 3&lt;/math&gt; are two such ways.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Seven cookies of radius &lt;math&gt;1&lt;/math&gt; inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),3));<br /> draw(circle((0,0),1));<br /> draw(circle((1,sqrt(3)),1));<br /> draw(circle((-1,sqrt(3)),1)); <br /> draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br /> draw(circle((2,0),1)); draw(circle((-2,0),1)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> A triangle with vertices &lt;math&gt;A(0, 2)&lt;/math&gt;, &lt;math&gt;B(-3, 2)&lt;/math&gt;, and &lt;math&gt;C(-3, 0)&lt;/math&gt; is reflected about the &lt;math&gt;x&lt;/math&gt;-axis, then the image &lt;math&gt;\triangle A'B'C'&lt;/math&gt; is rotated counterclockwise about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt; to produce &lt;math&gt;\triangle A''B''C''&lt;/math&gt;. Which of the following transformations will return &lt;math&gt;\triangle A''B''C''&lt;/math&gt; to &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; counterclockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; clockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; reflection about the &lt;math&gt;x&lt;/math&gt;-axis <br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; reflection about the line &lt;math&gt;y = x&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; reflection about the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> [[2016 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> In rectangle ABCD, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; between &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are such that &lt;math&gt;BE=EF=FC&lt;/math&gt;. Segments &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{AF}&lt;/math&gt; intersect &lt;math&gt;\overline{BD}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. The ration &lt;math&gt;BP:PQ:QD&lt;/math&gt; can be written as &lt;math&gt;r:s:t&lt;/math&gt; where the greatest common factor of &lt;math&gt;r,s&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; is 1. What is &lt;math&gt;r+s+t&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> <br /> [[2016 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> For some particular value of &lt;math&gt;N&lt;/math&gt;, when &lt;math&gt;(a+b+c+d+1)^N&lt;/math&gt; is expanded and like terms are combined, the resulting expression contains exactly &lt;math&gt;1001&lt;/math&gt; terms that include all four variables &lt;math&gt;a, b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, each to some positive power. What is &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 20|Solution]]<br /> ==Problem 21==<br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> A binary operation &lt;math&gt;\diamondsuit&lt;/math&gt; has the properties that &lt;math&gt;a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\,\diamondsuit \,a=1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. (Here &lt;math&gt;\cdot&lt;/math&gt; represents multiplication). The solution to the equation &lt;math&gt;2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}&lt;/math&gt;. Three of the sides of this quadrilateral have length &lt;math&gt;200&lt;/math&gt;. What is the length of the fourth side?<br /> <br /> &lt;math&gt;\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2016|ab=A|before=[[2015 AMC 10B Problems]]|after=[[2016 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2016 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=75185 2016 AMC 10A Problems 2016-02-04T01:26:58Z <p>Warrenwangtennis: /* Problem 19 */</p> <hr /> <div>==Problem 1==<br /> What is the value of &lt;math&gt;\dfrac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^{x}\cdot 100^{2x}=1000^{5}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> For every dollar Ben spent on bagels, David spent &lt;math&gt;25&lt;/math&gt; cents less. Ben paid &lt;math&gt;\$12.50&lt;/math&gt; more than David. How much did they spend in the bagel store together?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ximena lists the whole numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;30&lt;/math&gt; once. Emilio copies Ximena's numbers, replacing each occurrence of the digit &lt;math&gt;2&lt;/math&gt; by the digit &lt;math&gt;1&lt;/math&gt;. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays &lt;math&gt;40&lt;/math&gt; coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is &lt;math&gt;1&lt;/math&gt; foot wide on all four sides. What is the length in feet of the inner rectangle?<br /> &lt;asy&gt;<br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> path rectangle(pair X, pair Y){<br /> return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br /> }<br /> filldraw(rectangle((0,0),(7,5)),gray(0.5));<br /> filldraw(rectangle((1,1),(6,4)),gray(0.75));<br /> filldraw(rectangle((2,2),(5,3)),white);<br /> <br /> label(&quot;$1$&quot;,(0.5,2.5));<br /> draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br /> draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(1.5,2.5));<br /> draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br /> draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.5,2.5));<br /> draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br /> draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.1,1.5));<br /> draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br /> draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(3.7,0.5));<br /> draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br /> draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br /> &lt;/asy&gt;<br /> <br /> &lt;cmath&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8&lt;/cmath&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> What is the area of the shaded region of the given &lt;math&gt;8 \times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Three distinct integers are selected at random between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2016&lt;/math&gt;, inclusive. Which of the following is a correct statement about the probability &lt;math&gt;p&lt;/math&gt; that the product of the three integers is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ p&lt;\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}&lt;p&lt;\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p&gt;\dfrac{1}{3}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> How many ways are there to write &lt;math&gt;2016&lt;/math&gt; as the sum of twos and threes, ignoring order? (For example, &lt;math&gt;1008\cdot 2 + 0\cdot 3&lt;/math&gt; and &lt;math&gt;402\cdot 2 + 404\cdot 3&lt;/math&gt; are two such ways.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Seven cookies of radius &lt;math&gt;1&lt;/math&gt; inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),3));<br /> draw(circle((0,0),1));<br /> draw(circle((1,sqrt(3)),1));<br /> draw(circle((-1,sqrt(3)),1)); <br /> draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br /> draw(circle((2,0),1)); draw(circle((-2,0),1)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> A triangle with vertices &lt;math&gt;A(0, 2)&lt;/math&gt;, &lt;math&gt;B(-3, 2)&lt;/math&gt;, and &lt;math&gt;C(-3, 0)&lt;/math&gt; is reflected about the &lt;math&gt;x&lt;/math&gt;-axis, then the image &lt;math&gt;\triangle A'B'C'&lt;/math&gt; is rotated counterclockwise about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt; to produce &lt;math&gt;\triangle A''B''C''&lt;/math&gt;. Which of the following transformations will return &lt;math&gt;\triangle A''B''C''&lt;/math&gt; to &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; counterclockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; clockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; reflection about the &lt;math&gt;x&lt;/math&gt;-axis <br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; reflection about the line &lt;math&gt;y = x&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; reflection about the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> [[2016 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> In rectangle ABCD, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; between &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are such that &lt;math&gt;BE=EF=FC&lt;/math&gt;. Segments &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{AF}&lt;/math&gt; intersect &lt;math&gt;\overline{BD}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. The ration &lt;math&gt;BP:PQ:QD&lt;/math&gt; can be written as &lt;math&gt;r:s:t&lt;/math&gt; where the greatest common factor of &lt;math&gt;r,s&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; is 1. What is &lt;math&gt;r+s+t&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> <br /> [[2016 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> First, &lt;math&gt;\triangle APD&lt;/math&gt; and &lt;math&gt;\triangle EPB&lt;/math&gt; are similar by angle and parallel opposite side. Same for &lt;math&gt;\triangle AQD&lt;/math&gt; and &lt;math&gt;\triangle FQB&lt;/math&gt;.<br /> <br /> So &lt;math&gt;\frac {DP}}{AD}} = \frac{BP}{BE}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac {DP}{3} = \frac{\sqrt{6^2 + 3^2} - DP}{1}&lt;/math&gt;<br /> <br /> &lt;math&gt;DP = \frac{9\sqrt{5}}{4}&lt;/math&gt;<br /> <br /> Using the same process, &lt;math&gt;DQ = \frac{9\sqrt{5}}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;BP = 3\sqrt{5} - DP = \frac{3\sqrt{5}}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;PQ = DP - DQ = \frac{9\sqrt{5}}{20}&lt;/math&gt;<br /> <br /> &lt;math&gt;QD = DQ = \frac{9\sqrt{5}}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;BP : PQ : QD = 5 : 3 : 12. 5 + 3 + 12 = \boxed{\textbf{(E)} 20}&lt;/math&gt;.<br /> <br /> ==Problem 20==<br /> For some particular value of &lt;math&gt;N&lt;/math&gt;, when &lt;math&gt;(a+b+c+d+1)^N&lt;/math&gt; is expanded and like terms are combined, the resulting expression contains exactly &lt;math&gt;1001&lt;/math&gt; terms that include all four variables &lt;math&gt;a, b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, each to some positive power. What is &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 20|Solution]]<br /> ==Problem 21==<br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> A binary operation &lt;math&gt;\diamondsuit&lt;/math&gt; has the properties that &lt;math&gt;a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\,\diamondsuit \,a=1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. (Here &lt;math&gt;\cdot&lt;/math&gt; represents multiplication). The solution to the equation &lt;math&gt;2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}&lt;/math&gt;. Three of the sides of this quadrilateral have length &lt;math&gt;200&lt;/math&gt;. What is the length of the fourth side?<br /> <br /> &lt;math&gt;\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2016|ab=A|before=[[2015 AMC 10B Problems]]|after=[[2016 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2016 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_25&diff=75126 2016 AMC 10A Problems/Problem 25 2016-02-03T23:13:56Z <p>Warrenwangtennis: /* Problem 25 */</p> <hr /> <div>==Problem==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_25&diff=75121 2016 AMC 10A Problems/Problem 25 2016-02-03T23:10:52Z <p>Warrenwangtennis: /* Problem */</p> <hr /> <div>==Problem 25==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_25&diff=75119 2016 AMC 10A Problems/Problem 25 2016-02-03T23:10:15Z <p>Warrenwangtennis: /* Problem */</p> <hr /> <div>==Problem==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_14&diff=75099 2016 AMC 10A Problems/Problem 14 2016-02-03T22:59:24Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Solution==<br /> <br /> The amount of twos in our sum ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1008&lt;/math&gt;, with differences of &lt;math&gt;3&lt;/math&gt; because &lt;math&gt;2 \cdot 3 = lcm(2, 3)&lt;/math&gt;.<br /> <br /> The possible amount of twos is &lt;math&gt;\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{\textbf{(C)} 337}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_14&diff=75098 2016 AMC 10A Problems/Problem 14 2016-02-03T22:59:01Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Solution==<br /> <br /> The amount of twos in our sum ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1008&lt;/math&gt;, with differences of &lt;math&gt;3&lt;/math&gt; because &lt;math&gt;2 \cdot 3 = lcm(2, 3)&lt;/math&gt;.<br /> <br /> The possible amount of twos is &lt;math&gt;\frac{1008 - 0}{2} + 1 \Rightarrow \boxed{\textbf{(C)} 337}&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_14&diff=75094 2016 AMC 10A Problems/Problem 14 2016-02-03T22:57:43Z <p>Warrenwangtennis: /* Solution */</p> <hr /> <div>==Solution==<br /> <br /> The amount of twos in our sum ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1008&lt;/math&gt;, with differences of &lt;math&gt;3&lt;/math&gt; because &lt;math&gt;2 \cdot 3 = lcm(2, 3)&lt;/math&gt;.<br /> <br /> The possible amount of twos is &lt;math&gt;\frac{1008 - 0}{2} + 1 \Rightarrow \boxed{\textbf{(B)} 337&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_14&diff=75088 2016 AMC 10A Problems/Problem 14 2016-02-03T22:56:08Z <p>Warrenwangtennis: Created page with &quot;==Solution== The amount of twos in our sum ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1008&lt;/math&gt;, with differences of &lt;math&gt;3&lt;/math&gt; because &lt;math&gt;2 \cdot 3 = lcm(2, 3)&lt;/math&gt;. Th...&quot;</p> <hr /> <div>==Solution==<br /> <br /> The amount of twos in our sum ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1008&lt;/math&gt;, with differences of &lt;math&gt;3&lt;/math&gt; because &lt;math&gt;2 \cdot 3 = lcm(2, 3)&lt;/math&gt;.<br /> <br /> The possible amount of twos is &lt;math&gt;\frac{1008 - 0}{2} + 1 \rightarrow \boxed{\textbf{(B)} 337&lt;/math&gt;</div> Warrenwangtennis https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=75071 2016 AMC 10A Problems 2016-02-03T22:50:53Z <p>Warrenwangtennis: /* Problem 14 */</p> <hr /> <div>==Problem 1==<br /> What is the value of &lt;math&gt;\dfrac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^{x}\cdot 100^{2x}=1000^{5}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Problem 3==<br /> For every dollar Ben spent on bagels, David spent &lt;math&gt;25&lt;/math&gt; cents less. Ben paid &lt;math&gt;\$12.50&lt;/math&gt; more than David. How much did they spend in the bagel store together?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50&lt;/math&gt;<br /> ==Problem 4==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> ==Problem 5==<br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> ==Problem 6==<br /> Ximena lists the whole numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;30&lt;/math&gt; once. Emilio copies Ximena's numbers, replacing each occurrence of the digit &lt;math&gt;2&lt;/math&gt; by the digit &lt;math&gt;1&lt;/math&gt;. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110&lt;/math&gt;<br /> ==Problem 7==<br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> ==Problem 8==<br /> Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays &lt;math&gt;40&lt;/math&gt; coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45&lt;/math&gt;<br /> ==Problem 9==<br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> ==Problem 10==<br /> A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is &lt;math&gt;1&lt;/math&gt; foot wide on all four sides. What is the length in feet of the inner rectangle?<br /> &lt;asy&gt;<br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> path rectangle(pair X, pair Y){<br /> return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br /> }<br /> filldraw(rectangle((0,0),(7,5)),gray(0.5));<br /> filldraw(rectangle((1,1),(6,4)),gray(0.75));<br /> filldraw(rectangle((2,2),(5,3)),white);<br /> <br /> label(&quot;$1$&quot;,(0.5,2.5));<br /> draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br /> draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(1.5,2.5));<br /> draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br /> draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.5,2.5));<br /> draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br /> draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.1,1.5));<br /> draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br /> draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(3.7,0.5));<br /> draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br /> draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br /> &lt;/asy&gt;<br /> <br /> &lt;cmath&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8&lt;/cmath&gt;<br /> ==Problem 11==<br /> What is the area of the shaded region of the given &lt;math&gt;8 \times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4\$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> ==Problem 12==<br /> Three distinct integers are selected at random between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2016&lt;/math&gt;, inclusive. Which of the following is a correct statement about the probability &lt;math&gt;p&lt;/math&gt; that the product of the three integers is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ p&lt;\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}&lt;p&lt;\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p&gt;\dfrac{1}{3}&lt;/math&gt;<br /> ==Problem 13==<br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5&lt;/math&gt;<br /> ==Problem 14==<br /> How many ways are there to write &lt;math&gt;2016&lt;/math&gt; as the sum of twos and threes, ignoring order? (For example, &lt;math&gt;1008\cdot 2 + 0\cdot 3&lt;/math&gt; and &lt;math&gt;402\cdot 2 + 404\cdot 3&lt;/math&gt; are two such ways.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Seven cookies of radius &lt;math&gt;1&lt;/math&gt; inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),3));<br /> draw(circle((0,0),1));<br /> draw(circle((1,sqrt(3)),1));<br /> draw(circle((-1,sqrt(3)),1)); <br /> draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br /> draw(circle((2,0),1)); draw(circle((-2,0),1)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi&lt;/math&gt;<br /> <br /> ==Problem 16==<br /> A triangle with vertices &lt;math&gt;A(0, 2)&lt;/math&gt;, &lt;math&gt;B(-3, 2)&lt;/math&gt;, and &lt;math&gt;C(-3, 0)&lt;/math&gt; is reflected about the &lt;math&gt;x&lt;/math&gt;-axis, then the image &lt;math&gt;\triangle A'B'C'&lt;/math&gt; is rotated counterclockwise about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt; to produce &lt;math&gt;\triangle A''B''C''&lt;/math&gt;. Which of the following transformations will return &lt;math&gt;\triangle A''B''C''&lt;/math&gt; to &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; counterclockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; clockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; reflection about the &lt;math&gt;x&lt;/math&gt;-axis <br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; reflection about the line &lt;math&gt;y = x&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; reflection about the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> ==Problem 17==<br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> ==Problem 18==<br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> ==Problem 19==<br /> In rectangle ABCD, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; between &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are such that &lt;math&gt;BE=EF=FC&lt;/math&gt;. Segments &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{AF}&lt;/math&gt; intersect &lt;math&gt;\overline{BD}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. The ration &lt;math&gt;BP:PQ:QD&lt;/math&gt; can be written as &lt;math&gt;r:s:t&lt;/math&gt; where the greatest common factor of &lt;math&gt;r,s&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; is 1. What is &lt;math&gt;r+s+t&lt;/math&gt;?<br /> ==Problem 20==<br /> For some particular value of &lt;math&gt;N&lt;/math&gt;, when &lt;math&gt;(a+b+c+d+1)^N&lt;/math&gt; is expanded and like terms are combined, the resulting expression contains exactly &lt;math&gt;1001&lt;/math&gt; terms that include all four variables &lt;math&gt;a, b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, each to some positive power. What is &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19&lt;/math&gt;<br /> <br /> ==Problem 21==<br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> ==Problem 22==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> ==Problem 23==<br /> A binary operation &lt;math&gt;\diamondsuit&lt;/math&gt; has the properties that &lt;math&gt;a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\,\diamondsuit \,a=1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. (Here &lt;math&gt;\cdot&lt;/math&gt; represents multiplication). The solution to the equation &lt;math&gt;2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601&lt;/math&gt;<br /> <br /> ==Problem 24==<br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}&lt;/math&gt;. Three of the sides of this quadrilateral have length &lt;math&gt;200&lt;/math&gt;. What is the length of the fourth side?<br /> <br /> &lt;math&gt;\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500&lt;/math&gt;<br /> ==Problem 25==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;</div> Warrenwangtennis