https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Npip99&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-13T22:53:37Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_24&diff=105767 1950 AHSME Problems/Problem 24 2019-05-11T21:05:24Z <p>Npip99: /* Solution 2 */</p> <hr /> <div>== Problem==<br /> <br /> The equation &lt;math&gt;x + \sqrt{x-2} = 4&lt;/math&gt; has:<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;x + \sqrt{x-2} = 4&lt;/math&gt; Original Equation<br /> <br /> &lt;math&gt;\sqrt{x-2} = 4 - x&lt;/math&gt; Subtract x from both sides<br /> <br /> &lt;math&gt;x-2 = 16 - 8x + x^2&lt;/math&gt; Square both sides<br /> <br /> &lt;math&gt;x^2 - 9x + 18 = 0&lt;/math&gt; Get all terms on one side<br /> <br /> &lt;math&gt;(x-6)(x-3) = 0&lt;/math&gt; Factor<br /> <br /> &lt;math&gt;x = \{6, 3\}&lt;/math&gt;<br /> <br /> If you put down A as your answer, it's wrong. You need to check for extraneous roots.<br /> <br /> &lt;math&gt;6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark&lt;/math&gt;<br /> <br /> There is &lt;math&gt;\boxed{\textbf{(E)} \text{1 real root}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> It's not hard to note that &lt;math&gt;x=3&lt;/math&gt; simply works, as &lt;math&gt;3 + \sqrt{1} = 4&lt;/math&gt;. But, &lt;math&gt;x&lt;/math&gt; is increasing, and &lt;math&gt;\sqrt{x-2}&lt;/math&gt; is increasing, so &lt;math&gt;3&lt;/math&gt; is the only root. If &lt;math&gt;x &lt; 3&lt;/math&gt;, &lt;math&gt;x + \sqrt{x-2} &lt; 4&lt;/math&gt;, and similarly if &lt;math&gt;x &gt; 3&lt;/math&gt;, then &lt;math&gt;x + \sqrt{x-2} &gt; 4&lt;/math&gt;. Imaginary roots don't have to be considered because squaring the equation gives a quadratic, and quadratics either have only imaginary or only real roots.<br /> <br /> == See Also ==<br /> {{AHSME 50p box|year=1950|num-b=23|num-a=25}}<br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_24&diff=105766 1950 AHSME Problems/Problem 24 2019-05-11T21:04:57Z <p>Npip99: /* Solution 2 */</p> <hr /> <div>== Problem==<br /> <br /> The equation &lt;math&gt;x + \sqrt{x-2} = 4&lt;/math&gt; has:<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;x + \sqrt{x-2} = 4&lt;/math&gt; Original Equation<br /> <br /> &lt;math&gt;\sqrt{x-2} = 4 - x&lt;/math&gt; Subtract x from both sides<br /> <br /> &lt;math&gt;x-2 = 16 - 8x + x^2&lt;/math&gt; Square both sides<br /> <br /> &lt;math&gt;x^2 - 9x + 18 = 0&lt;/math&gt; Get all terms on one side<br /> <br /> &lt;math&gt;(x-6)(x-3) = 0&lt;/math&gt; Factor<br /> <br /> &lt;math&gt;x = \{6, 3\}&lt;/math&gt;<br /> <br /> If you put down A as your answer, it's wrong. You need to check for extraneous roots.<br /> <br /> &lt;math&gt;6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark&lt;/math&gt;<br /> <br /> There is &lt;math&gt;\boxed{\textbf{(E)} \text{1 real root}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> It's not hard to note that &lt;math&gt;x=3&lt;/math&gt; simply works, as &lt;math&gt;3 + \sqrt{1} = 4&lt;/math&gt;. But, &lt;math&gt;x&lt;/math&gt; is increasing, and &lt;math&gt;\sqrt{x-2}&lt;/math&gt; is increasing, so &lt;math&gt;3&lt;/math&gt; is the only root. If &lt;math&gt;x&lt;/math&gt; is smaller than &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x + \sqrt{x-2} &lt; 4&lt;/math&gt;, and similarly if &lt;math&gt;x &gt; 3&lt;/math&gt;, then &lt;math&gt;x + \sqrt{x-2} &gt; 4&lt;/math&gt;. Imaginary roots don't have to be considered because squaring the equation gives a quadratic, and quadratics either have only imaginary or only real roots.<br /> <br /> == See Also ==<br /> {{AHSME 50p box|year=1950|num-b=23|num-a=25}}<br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_24&diff=105765 1950 AHSME Problems/Problem 24 2019-05-11T21:04:46Z <p>Npip99: /* Solution */</p> <hr /> <div>== Problem==<br /> <br /> The equation &lt;math&gt;x + \sqrt{x-2} = 4&lt;/math&gt; has:<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;x + \sqrt{x-2} = 4&lt;/math&gt; Original Equation<br /> <br /> &lt;math&gt;\sqrt{x-2} = 4 - x&lt;/math&gt; Subtract x from both sides<br /> <br /> &lt;math&gt;x-2 = 16 - 8x + x^2&lt;/math&gt; Square both sides<br /> <br /> &lt;math&gt;x^2 - 9x + 18 = 0&lt;/math&gt; Get all terms on one side<br /> <br /> &lt;math&gt;(x-6)(x-3) = 0&lt;/math&gt; Factor<br /> <br /> &lt;math&gt;x = \{6, 3\}&lt;/math&gt;<br /> <br /> If you put down A as your answer, it's wrong. You need to check for extraneous roots.<br /> <br /> &lt;math&gt;6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark&lt;/math&gt;<br /> <br /> There is &lt;math&gt;\boxed{\textbf{(E)} \text{1 real root}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> It's not hard to note that x=3 simply works, as &lt;math&gt;3 + \sqrt{1} = 4&lt;/math&gt;. But, &lt;math&gt;x&lt;/math&gt; is increasing, and &lt;math&gt;\sqrt{x-2}&lt;/math&gt; is increasing, so &lt;math&gt;3&lt;/math&gt; is the only root. If &lt;math&gt;x&lt;/math&gt; is smaller than &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x + \sqrt{x-2} &lt; 4&lt;/math&gt;, and similarly if &lt;math&gt;x &gt; 3&lt;/math&gt;, then &lt;math&gt;x + \sqrt{x-2} &gt; 4&lt;/math&gt;. Imaginary roots don't have to be considered because squaring the equation gives a quadratic, and quadratics either have only imaginary or only real roots.<br /> <br /> == See Also ==<br /> {{AHSME 50p box|year=1950|num-b=23|num-a=25}}<br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_13&diff=98508 2004 AMC 12A Problems/Problem 13 2018-11-05T04:10:28Z <p>Npip99: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of [[point]]s &lt;math&gt;(a,b)&lt;/math&gt; in the [[coordinate plane]], where each of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; may be &lt;math&gt;- 1&lt;/math&gt;, &lt;math&gt;0&lt;/math&gt;, or &lt;math&gt;1&lt;/math&gt;. How many distinct [[line]]s pass through at least two members of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text {(A)}\ 8 \qquad \text {(B)}\ 20 \qquad \text {(C)}\ 24 \qquad \text {(D)}\ 27\qquad \text {(E)}\ 36&lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> [[Image:2004_AMC12A-13.png|center]]<br /> === Solution 1 ===<br /> Let's count them by cases:<br /> <br /> [[Image:2004_AMC12A-13b.png|center]]<br /> <br /> *'''Case 1''': The line is horizontal or vertical, clearly &lt;math&gt;3 \cdot 2 = 6&lt;/math&gt;.<br /> *'''Case 2''': The line has slope &lt;math&gt;\pm 1&lt;/math&gt;, with &lt;math&gt;2&lt;/math&gt; through &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; additional ones one unit above or below those. These total &lt;math&gt;6&lt;/math&gt;. <br /> *'''Case 3''': The only remaining lines pass through two points, a vertex and a non-vertex point on the opposite side. Thus we have each vertex pairing up with two points on the two opposites sides, giving &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt; lines.<br /> These add up to &lt;math&gt;6+6+8=20\ \mathrm{(B)}&lt;/math&gt;. <br /> <br /> === Solution 2 ===<br /> There are &lt;math&gt;{9 \choose 2} = 36&lt;/math&gt; ways to pick two points, but we've clearly overcounted all of the lines which pass through three points. In fact, each line which passes through three points will have been counted &lt;math&gt;{3 \choose 2} = 3&lt;/math&gt; times, so we have to subtract &lt;math&gt;2&lt;/math&gt; for each of these lines. Quick counting yields &lt;math&gt;3&lt;/math&gt; horizontal, &lt;math&gt;3&lt;/math&gt; vertical, and &lt;math&gt;2&lt;/math&gt; diagonal lines, so the answer is &lt;math&gt;36 - 2(3+3+2) = 20&lt;/math&gt; distinct lines.<br /> <br /> === Solution 3 ===<br /> First consider how many lines go through &lt;math&gt;(-1, -1)&lt;/math&gt; and hit two points in &lt;math&gt;S&lt;/math&gt;. You can see that there are &lt;math&gt;5&lt;/math&gt; such lines. Now, we cross out &lt;math&gt;(-1, -1)&lt;/math&gt; and make sure to never consider consider lines that go through it anymore (As doing so would be double counting). Repeat for &lt;math&gt;(-1, 0)&lt;/math&gt;, making sure not to count the vertical line as it goes through the crossed out &lt;math&gt;(-1, -1)&lt;/math&gt;. Then cross out &lt;math&gt;(-1, 0)&lt;/math&gt;. Repeat for the rest, and count &lt;math&gt;20&lt;/math&gt; lines in total.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2004|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> http://www.artofproblemsolving.com/Wiki/index.php?title=2004_AMC_12A_Problems/Problem_14&amp;action=edit&amp;section=3</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_1&diff=72547 1969 Canadian MO Problems/Problem 1 2015-10-18T19:06:19Z <p>Npip99: /* Solution */</p> <hr /> <div>== Problem ==<br /> Show that if &lt;math&gt;a_1/b_1=a_2/b_2=a_3/b_3&lt;/math&gt; and &lt;math&gt;p_1,p_2,p_3&lt;/math&gt; are not all zero, then &lt;math&gt;\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}&lt;/math&gt; for every positive integer &lt;math&gt;n.&lt;/math&gt;<br /> <br /> == Solution ==<br /> Instead of proving the two expressions equal, we prove that their difference equals zero.<br /> <br /> Subtracting the LHS from the RHS,<br /> &lt;math&gt;0=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}-\frac{a_1^n}{b_1^n}.&lt;/math&gt;<br /> <br /> Finding a common denominator, the numerator becomes<br /> &lt;math&gt;b_1^n(p_1a_1^n+p_2a_2^n+p_3a_3^n)-a_1^n(p_1b_1^n+p_2b_2^n+p_3b_3^n)=p_2(a_2^nb_1^n-a_1^nb_2^n)+p_3(a_3^nb_1^n-a_1^nb_3^n)=0.&lt;/math&gt;<br /> (The denominator is irrelevant since it never equals zero)<br /> <br /> From &lt;math&gt;a_1/b_1=a_2/b_2,&lt;/math&gt; &lt;math&gt;a_1^nb_2^n=a_2^nb_1^n.&lt;/math&gt; Similarly, &lt;math&gt;a_1^nb_3^n=a_3^nb_1^n&lt;/math&gt; from &lt;math&gt;a_1/b_1=a_3/b_3.&lt;/math&gt; <br /> <br /> Hence, &lt;math&gt;a_2^nb_1^n-a_1^nb_2^n=a_3^nb_1^n-a_1^nb_3^n=0&lt;/math&gt; and our proof is complete.<br /> {{Old CanadaMO box|before=First question|num-a=2|year=1969}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_14&diff=72066 1998 AIME Problems/Problem 14 2015-09-14T21:32:44Z <p>Npip99: /* Solution */</p> <hr /> <div>== Problem ==<br /> An &lt;math&gt;m\times n\times p&lt;/math&gt; rectangular box has half the volume of an &lt;math&gt;(m + 2)\times(n + 2)\times(p + 2)&lt;/math&gt; rectangular box, where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers, and &lt;math&gt;m\le n\le p.&lt;/math&gt; What is the largest possible value of &lt;math&gt;p&lt;/math&gt;?<br /> <br /> == Solution ==<br /> &lt;cmath&gt;2mnp = (m+2)(n+2)(p+2)&lt;/cmath&gt;<br /> <br /> Let’s solve for &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)&lt;/cmath&gt;<br /> &lt;cmath&gt;[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)&lt;/cmath&gt;<br /> &lt;cmath&gt;p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}&lt;/cmath&gt;<br /> <br /> Clearly, we want to minimize the denominator, so we test &lt;math&gt;(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9&lt;/math&gt;. The possible pairs of factors of &lt;math&gt;9&lt;/math&gt; are &lt;math&gt;(1,9)(3,3)&lt;/math&gt;. These give &lt;math&gt;m = 3, n = 11&lt;/math&gt; and &lt;math&gt;m = 5, n = 5&lt;/math&gt; respectively. Substituting into the numerator, we see that the first pair gives &lt;math&gt;130&lt;/math&gt;, while the second pair gives &lt;math&gt;98&lt;/math&gt;. We now check that &lt;math&gt;130&lt;/math&gt; is optimal, setting &lt;math&gt;a=m-2&lt;/math&gt;, &lt;math&gt;b=n-2&lt;/math&gt; in order to simplify calculations. Since<br /> &lt;cmath&gt;0 \le (a-1)(b-1) \implies a+b \le ab+1&lt;/cmath&gt;<br /> We have<br /> &lt;cmath&gt;p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130&lt;/cmath&gt;<br /> Where we see &lt;math&gt;(m,n)=(3,11)&lt;/math&gt; gives us our maximum value of &lt;math&gt;\boxed{130}&lt;/math&gt;.<br /> <br /> *Note that &lt;math&gt;0 \le (a-1)(b-1)&lt;/math&gt; assumes &lt;math&gt;m,n \ge 3&lt;/math&gt;, but this is clear as &lt;math&gt;\frac{2m}{m+2} = \frac{(n+2)(p+2)}{np} &gt; 1&lt;/math&gt; and similarly for &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=Schur%27s_Inequality&diff=71978 Schur's Inequality 2015-09-12T01:38:06Z <p>Npip99: Repaired dead link</p> <hr /> <div>'''Schur's Inequality''' is an [[inequality]] that holds for [[positive number]]s. It is named for Issai Schur.<br /> <br /> <br /> == Theorem ==<br /> Schur's inequality states that for all non-negative &lt;math&gt;a,b,c \in \mathbb{R}&lt;/math&gt; and &lt;math&gt;r&gt;0&lt;/math&gt;:<br /> <br /> &lt;math&gt;{a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0}&lt;/math&gt;<br /> <br /> The four [[equality condition | equality cases]] occur when &lt;math&gt;a=b=c&lt;/math&gt; or when two of &lt;math&gt;a,b,c&lt;/math&gt; are equal and the third is &lt;math&gt;{0}&lt;/math&gt;.<br /> <br /> <br /> === Common Cases ===<br /> <br /> The &lt;math&gt;r=1&lt;/math&gt; case yields the well-known inequality:&lt;math&gt;a^3+b^3+c^3+3abc \geq a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b&lt;/math&gt;<br /> <br /> When &lt;math&gt;r=2&lt;/math&gt;, an equivalent form is:<br /> &lt;math&gt;a^4+b^4+c^4+abc(a+b+c) \geq a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b&lt;/math&gt;<br /> <br /> <br /> === Proof ===<br /> <br /> [[WLOG]], let &lt;math&gt;{a \geq b \geq c}&lt;/math&gt;. Note that &lt;math&gt;a^r(a-b)(a-c)+b^r(b-a)(b-c)&lt;/math&gt; &lt;math&gt;= a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))&lt;/math&gt;. Clearly, &lt;math&gt;a^r \geq b^r \geq 0&lt;/math&gt;, and &lt;math&gt;a-c \geq b-c \geq 0&lt;/math&gt;. Thus, &lt;math&gt;(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0&lt;/math&gt;. However, &lt;math&gt;c^r(c-a)(c-b) \geq 0&lt;/math&gt;, and thus the proof is complete.<br /> <br /> === Generalized Form ===<br /> <br /> It has been shown by [[Valentin Vornicu]] that a more general form of Schur's Inequality exists. Consider &lt;math&gt;a,b,c,x,y,z \in \mathbb{R}&lt;/math&gt;, where &lt;math&gt;{a \geq b \geq c}&lt;/math&gt;, and either &lt;math&gt;x \geq y \geq z&lt;/math&gt; or &lt;math&gt;z \geq y \geq x&lt;/math&gt;. Let &lt;math&gt;k \in \mathbb{Z}^{+}&lt;/math&gt;, and let &lt;math&gt;f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}&lt;/math&gt; be either convex or monotonic. Then,<br /> <br /> &lt;math&gt;{f(x)(a-b)^k(a-c)^k+f(y)(b-a)^k(b-c)^k+f(z)(c-a)^k(c-b)^k \geq 0}&lt;/math&gt;.<br /> <br /> The standard form of Schur's is the case of this inequality where &lt;math&gt;x=a,\ y=b,\ z=c,\ k=1,\ f(m)=m^r&lt;/math&gt;.<br /> <br /> == References ==<br /> <br /> * Mildorf, Thomas; ''Olympiad Inequalities''; January 20, 2006; &lt;http://artofproblemsolving.com/articles/files/MildorfInequalities.pdf&gt;<br /> <br /> * Vornicu, Valentin; ''Olimpiada de Matematica... de la provocare la experienta''; GIL Publishing House; Zalau, Romania.<br /> <br /> ==See Also==<br /> * [[Olympiad Mathematics]]<br /> * [[Inequalities]]<br /> * [[Number Theory]]<br /> <br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_12&diff=71883 1992 AIME Problems/Problem 12 2015-08-30T19:44:37Z <p>Npip99: /* Problem */</p> <hr /> <div>== Problem ==<br /> In a game of &lt;i&gt;Chomp&lt;/i&gt;, two players alternately take bites from a 5-by-7 grid of [[unit square]]s. To take a bite, a player chooses one of the remaining [[square (geometry) | squares]], then removes (&quot;eats&quot;) all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by &lt;math&gt;\times.&lt;/math&gt; (The squares with two or more dotted edges have been removed form the original board in previous moves.) <br /> <br /> [[Image:AIME_1992_Problem_12.png]]<br /> <br /> The object of the game is to make one's opponent take the last bite. The diagram shows one of the many [[subset]]s of the [[set]] of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count.<br /> <br /> == Solution ==<br /> By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts.<br /> <br /> One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go down, and 7 that go across, with the shape on the right &quot;carved&quot; out by the path a possible subset.<br /> <br /> Therefore, the total number of such paths is &lt;math&gt;\binom{12}{5}=\boxed{792}&lt;/math&gt;<br /> <br /> == Links to Branches of Mathematics ==<br /> This question is one case of the problem of counting [[order ideal]]s or [[antichain]]s in [[poset]]s. Specifically, it ask for the number of order ideals of the [[product poset]] of the [[chain]] of length &lt;math&gt;4&lt;/math&gt; and the chain of length &lt;math&gt;6&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1992|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems&diff=71745 2015 AMC 10A Problems 2015-08-23T22:22:50Z <p>Npip99: /* Problem 3 */</p> <hr /> <div>==Problem 1==<br /> <br /> What is the value of &lt;math&gt;(2^0-1+5^2-0)^{-1}\times5?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{5}{24}\qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> A box contains a collection of triangular and square tiles. There are &lt;math&gt;25&lt;/math&gt; tiles in the box, containing &lt;math&gt;84&lt;/math&gt; edges total. How many square tiles are there in the box?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Ann made a 3-step staircase using 18 toothpicks as shown in the figure. How many toothpicks does she need to add to complete a 5-step staircase?<br /> <br /> &lt;asy&gt;<br /> unitsize(40);<br /> for(int i=0; i&lt;3; i+=1)<br /> {<br /> draw((0,i+0.05)--(0,i+0.95));<br /> draw((i+0.05,0)--(i+0.95,0));<br /> for(int j=0; j&lt;3-i; j+=1)<br /> {<br /> draw((i+1,j+0.05)--(i+1,j+0.95));<br /> draw((i+0.05,j+1)--(i+0.95,j+1));<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Mr. Patrick teaches math to &lt;math&gt; 15 &lt;/math&gt; students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was &lt;math&gt; 80 &lt;/math&gt;. After he graded Payton's test, the test average became &lt;math&gt; 81 &lt;/math&gt;. What was Payton's score on the test?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> The sum of two positive numbers is &lt;math&gt; 5 &lt;/math&gt; times their difference. What is the ratio of the larger number to the smaller number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> How many terms are there in the arithmetic sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, . . ., &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 61 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be &lt;math&gt;2&lt;/math&gt; : &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Two right circular cylinders have the same volume. The radius of the second cylinder is &lt;math&gt;10\%&lt;/math&gt; more than the radius of the first. What is the relationship between the heights of the two cylinders?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> How many rearrangements of &lt;math&gt;abcd&lt;/math&gt; are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either &lt;math&gt;ab&lt;/math&gt; or &lt;math&gt;ba&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> The ratio of the length to the width of a rectangle is &lt;math&gt;4&lt;/math&gt; : &lt;math&gt;3&lt;/math&gt;. If the rectangle has diagonal of length &lt;math&gt;d&lt;/math&gt;, then the area may be expressed as &lt;math&gt;kd^2&lt;/math&gt; for some constant &lt;math&gt;k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Points &lt;math&gt;(\sqrt{\pi}, a)&lt;/math&gt; and &lt;math&gt;(\sqrt{\pi}, b)&lt;/math&gt; are distinct points on the graph of &lt;math&gt;y^2+x^4=2x^2y+1&lt;/math&gt;. What is &lt;math&gt;|a-b|&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }1\qquad\textbf{(B) }\dfrac{\pi}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\sqrt{1+\pi}\qquad\textbf{(E) }1+\sqrt{\pi} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?<br /> <br /> &lt;math&gt; \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> The diagram below shows the circular face of a clock with radius &lt;math&gt;20&lt;/math&gt; cm and a circular disk with radius &lt;math&gt;10&lt;/math&gt; cm externally tangent to the clock face at &lt;math&gt;12&lt;/math&gt; o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br /> <br /> &lt;asy&gt;<br /> size(170);<br /> defaultpen(linewidth(0.9)+fontsize(13pt));<br /> draw(unitcircle^^circle((0,1.5),0.5));<br /> path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br /> for(int i=1;i&lt;=12;i=i+1)<br /> {<br /> draw(0.9*dir(90-30*i)--dir(90-30*i));<br /> label(&quot;$&quot;+(string) i+&quot;$&quot;,0.78*dir(90-30*i));<br /> }<br /> dot(origin);<br /> draw(shift((0,1.87))*arrow);<br /> draw(arc(origin,1.5,68,30),EndArrow(size=12));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Consider the set of all fractions &lt;math&gt;\tfrac{x}{y},&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by &lt;math&gt;1&lt;/math&gt;, the value of the fraction is increased by &lt;math&gt;10\%&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> If &lt;math&gt;y+4 = (x-2)^2, x+4 = (y-2)^2&lt;/math&gt;, and &lt;math&gt;x \neq y&lt;/math&gt;, what is the value of &lt;math&gt;x^2+y^2&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> A line that passes through the origin intersects both the line &lt;math&gt;x=1&lt;/math&gt; and the line &lt;math&gt;y=1+\frac{\sqrt{3}}{3}x&lt;/math&gt;. The three lines create an equilateral triangle. What is the perimeter of the triangle?<br /> <br /> &lt;math&gt; \textbf{(A) }2\sqrt{6}\qquad\textbf{(B) }2+2\sqrt{3}\qquad\textbf{(C) }6\qquad\textbf{(D) }3+2\sqrt{3}\qquad\textbf{(E) }6+\frac{\sqrt{3}}{3} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Hexadecimal (base-16) numbers are written using numeric digits &lt;math&gt;0&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt; as well as the letters &lt;math&gt;A&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; to represent &lt;math&gt;10&lt;/math&gt; through &lt;math&gt;15&lt;/math&gt;. Among the first &lt;math&gt;1000&lt;/math&gt; positive integers, there are &lt;math&gt;n&lt;/math&gt; whose hexadecimal representation contains only numeric digits. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> The isosceles right triangle &lt;math&gt;ABC&lt;/math&gt; has right angle at &lt;math&gt;C&lt;/math&gt; and area &lt;math&gt;12.5&lt;/math&gt;. The rays trisecting &lt;math&gt;\angle ACB&lt;/math&gt; intersect &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\bigtriangleup CDE&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\dfrac{25}{6} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> A rectangle with positive integer side lengths in &lt;math&gt;\text{cm}&lt;/math&gt; has area &lt;math&gt;A&lt;/math&gt; &lt;math&gt;\text{cm}^2&lt;/math&gt; and perimeter &lt;math&gt;P&lt;/math&gt; &lt;math&gt;\text{cm}&lt;/math&gt;. Which of the following numbers cannot equal &lt;math&gt;A+P&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 &lt;/math&gt;<br /> <br /> NOTE: As it originally appeared in the AMC 10, this problem was stated incorrectly and had no answer; it has been modified here to be solvable.<br /> <br /> [[2015 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Tetrahedron &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;AC=3&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;BD=4&lt;/math&gt;, &lt;math&gt;AD=3&lt;/math&gt;, and &lt;math&gt;CD=\tfrac{12}5\sqrt2&lt;/math&gt;. What is the volume of the tetrahedron?<br /> <br /> &lt;math&gt;\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt;\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> The zeroes of the function &lt;math&gt;f(x)=x^2-ax+2a&lt;/math&gt; are integers. What is the sum of the possible values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> For some positive integers &lt;math&gt;p&lt;/math&gt;, there is a quadrilateral &lt;math&gt;ABCD&lt;/math&gt; with positive integer side lengths, perimeter &lt;math&gt;p&lt;/math&gt;, right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;AB=2&lt;/math&gt;, and &lt;math&gt;CD=AD&lt;/math&gt;. How many different values of &lt;math&gt;p&lt;2015&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Let &lt;math&gt;S&lt;/math&gt; be a square of side length &lt;math&gt;1&lt;/math&gt;. Two points are chosen independently at random on the sides of &lt;math&gt;S&lt;/math&gt;. The probability that the straight-line distance between the points is at least &lt;math&gt;\tfrac12&lt;/math&gt; is &lt;math&gt;\tfrac{a-b\pi}c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers with &lt;math&gt;\gcd(a,b,c)=1&lt;/math&gt;. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2015|ab=A|before=[[2014 AMC 10B Problems]]|after=[[2015 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2015 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems&diff=71744 2015 AMC 10A Problems 2015-08-23T22:20:47Z <p>Npip99: /* Problem 3 */</p> <hr /> <div>==Problem 1==<br /> <br /> What is the value of &lt;math&gt;(2^0-1+5^2-0)^{-1}\times5?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{5}{24}\qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> A box contains a collection of triangular and square tiles. There are &lt;math&gt;25&lt;/math&gt; tiles in the box, containing &lt;math&gt;84&lt;/math&gt; edges total. How many square tiles are there in the box?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Ann made a 3-step staircase using 18 toothpicks as shown in the figure. How many toothpicks does she need to add to complete a 5-step staircase?<br /> <br /> &lt;asy&gt;<br /> for(int i=0; i&lt;3; i+=1)<br /> {<br /> draw((0,i+0.05)--(0,i+0.95));<br /> draw((i+0.05,0)--(i+0.95,0));<br /> for(int j=0; j&lt;3-i; j+=1)<br /> {<br /> draw((i+1,j+0.05)--(i+1,j+0.95));<br /> draw((i+0.05,j+1)--(i+0.95,j+1));<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Mr. Patrick teaches math to &lt;math&gt; 15 &lt;/math&gt; students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was &lt;math&gt; 80 &lt;/math&gt;. After he graded Payton's test, the test average became &lt;math&gt; 81 &lt;/math&gt;. What was Payton's score on the test?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> The sum of two positive numbers is &lt;math&gt; 5 &lt;/math&gt; times their difference. What is the ratio of the larger number to the smaller number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> How many terms are there in the arithmetic sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, . . ., &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 61 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be &lt;math&gt;2&lt;/math&gt; : &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Two right circular cylinders have the same volume. The radius of the second cylinder is &lt;math&gt;10\%&lt;/math&gt; more than the radius of the first. What is the relationship between the heights of the two cylinders?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> How many rearrangements of &lt;math&gt;abcd&lt;/math&gt; are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either &lt;math&gt;ab&lt;/math&gt; or &lt;math&gt;ba&lt;/math&gt;.<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> The ratio of the length to the width of a rectangle is &lt;math&gt;4&lt;/math&gt; : &lt;math&gt;3&lt;/math&gt;. If the rectangle has diagonal of length &lt;math&gt;d&lt;/math&gt;, then the area may be expressed as &lt;math&gt;kd^2&lt;/math&gt; for some constant &lt;math&gt;k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{12}{25}\qquad\textbf{(D)}\ \frac{16}{25}\qquad\textbf{(E)}\ \frac{3}{4}&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Points &lt;math&gt;(\sqrt{\pi}, a)&lt;/math&gt; and &lt;math&gt;(\sqrt{\pi}, b)&lt;/math&gt; are distinct points on the graph of &lt;math&gt;y^2+x^4=2x^2y+1&lt;/math&gt;. What is &lt;math&gt;|a-b|&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }1\qquad\textbf{(B) }\dfrac{\pi}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\sqrt{1+\pi}\qquad\textbf{(E) }1+\sqrt{\pi} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?<br /> <br /> &lt;math&gt; \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> The diagram below shows the circular face of a clock with radius &lt;math&gt;20&lt;/math&gt; cm and a circular disk with radius &lt;math&gt;10&lt;/math&gt; cm externally tangent to the clock face at &lt;math&gt;12&lt;/math&gt; o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br /> <br /> &lt;asy&gt;<br /> size(170);<br /> defaultpen(linewidth(0.9)+fontsize(13pt));<br /> draw(unitcircle^^circle((0,1.5),0.5));<br /> path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br /> for(int i=1;i&lt;=12;i=i+1)<br /> {<br /> draw(0.9*dir(90-30*i)--dir(90-30*i));<br /> label(&quot;$&quot;+(string) i+&quot;$&quot;,0.78*dir(90-30*i));<br /> }<br /> dot(origin);<br /> draw(shift((0,1.87))*arrow);<br /> draw(arc(origin,1.5,68,30),EndArrow(size=12));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Consider the set of all fractions &lt;math&gt;\tfrac{x}{y},&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by &lt;math&gt;1&lt;/math&gt;, the value of the fraction is increased by &lt;math&gt;10\%&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> If &lt;math&gt;y+4 = (x-2)^2, x+4 = (y-2)^2&lt;/math&gt;, and &lt;math&gt;x \neq y&lt;/math&gt;, what is the value of &lt;math&gt;x^2+y^2&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> A line that passes through the origin intersects both the line &lt;math&gt;x=1&lt;/math&gt; and the line &lt;math&gt;y=1+\frac{\sqrt{3}}{3}x&lt;/math&gt;. The three lines create an equilateral triangle. What is the perimeter of the triangle?<br /> <br /> &lt;math&gt; \textbf{(A) }2\sqrt{6}\qquad\textbf{(B) }2+2\sqrt{3}\qquad\textbf{(C) }6\qquad\textbf{(D) }3+2\sqrt{3}\qquad\textbf{(E) }6+\frac{\sqrt{3}}{3} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Hexadecimal (base-16) numbers are written using numeric digits &lt;math&gt;0&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt; as well as the letters &lt;math&gt;A&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; to represent &lt;math&gt;10&lt;/math&gt; through &lt;math&gt;15&lt;/math&gt;. Among the first &lt;math&gt;1000&lt;/math&gt; positive integers, there are &lt;math&gt;n&lt;/math&gt; whose hexadecimal representation contains only numeric digits. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21 &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> The isosceles right triangle &lt;math&gt;ABC&lt;/math&gt; has right angle at &lt;math&gt;C&lt;/math&gt; and area &lt;math&gt;12.5&lt;/math&gt;. The rays trisecting &lt;math&gt;\angle ACB&lt;/math&gt; intersect &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\bigtriangleup CDE&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\dfrac{25}{6} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> A rectangle with positive integer side lengths in &lt;math&gt;\text{cm}&lt;/math&gt; has area &lt;math&gt;A&lt;/math&gt; &lt;math&gt;\text{cm}^2&lt;/math&gt; and perimeter &lt;math&gt;P&lt;/math&gt; &lt;math&gt;\text{cm}&lt;/math&gt;. Which of the following numbers cannot equal &lt;math&gt;A+P&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 &lt;/math&gt;<br /> <br /> NOTE: As it originally appeared in the AMC 10, this problem was stated incorrectly and had no answer; it has been modified here to be solvable.<br /> <br /> [[2015 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Tetrahedron &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;AC=3&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;BD=4&lt;/math&gt;, &lt;math&gt;AD=3&lt;/math&gt;, and &lt;math&gt;CD=\tfrac{12}5\sqrt2&lt;/math&gt;. What is the volume of the tetrahedron?<br /> <br /> &lt;math&gt;\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt;\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} &lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> The zeroes of the function &lt;math&gt;f(x)=x^2-ax+2a&lt;/math&gt; are integers. What is the sum of the possible values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> For some positive integers &lt;math&gt;p&lt;/math&gt;, there is a quadrilateral &lt;math&gt;ABCD&lt;/math&gt; with positive integer side lengths, perimeter &lt;math&gt;p&lt;/math&gt;, right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;AB=2&lt;/math&gt;, and &lt;math&gt;CD=AD&lt;/math&gt;. How many different values of &lt;math&gt;p&lt;2015&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Let &lt;math&gt;S&lt;/math&gt; be a square of side length &lt;math&gt;1&lt;/math&gt;. Two points are chosen independently at random on the sides of &lt;math&gt;S&lt;/math&gt;. The probability that the straight-line distance between the points is at least &lt;math&gt;\tfrac12&lt;/math&gt; is &lt;math&gt;\tfrac{a-b\pi}c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers with &lt;math&gt;\gcd(a,b,c)=1&lt;/math&gt;. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63&lt;/math&gt;<br /> <br /> [[2015 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2015|ab=A|before=[[2014 AMC 10B Problems]]|after=[[2015 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2015 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_13&diff=71629 1988 AIME Problems/Problem 13 2015-08-14T22:53:53Z <p>Npip99: /* Solution 4 */</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 ==<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 /> First, 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_{15}&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} + 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> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems&diff=71521 2007 AMC 12B Problems 2015-08-11T21:07:43Z <p>Npip99: </p> <hr /> <div>==Problem 1==<br /> Isabella's house has 3 bedrooms. Each bedroom is 12 feet long, 10 feet wide, and 8 feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy 60 square feet in each bedroom. How many square feet of walls must be painted? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 678\qquad \mathrm{(B) \ } 768\qquad \mathrm{(C) \ } 786\qquad \mathrm{(D) \ } 867\qquad \mathrm{(E) \ } 876&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> A college student drove his compact car 120 miles home for the weekend and averaged 30 miles per gallon. On the return trip the student drove his parents' SUV and averaged only 20 miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?<br /> <br /> &lt;math&gt;\mathrm {(A)} 22\qquad \mathrm {(B)} 24\qquad \mathrm {(C)} 25\qquad \mathrm {(D)} 26\qquad \mathrm {(E)} 28&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> The point &lt;math&gt;O&lt;/math&gt; is the center of the circle circumscribed about triangle &lt;math&gt;ABC&lt;/math&gt;, with &lt;math&gt;\angle BOC = 120^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AOB = 140^{\circ}&lt;/math&gt;, as shown. What is the degree measure of &lt;math&gt;\angle ABC&lt;/math&gt;?<br /> <br /> &lt;center&gt;[[Image:2007_12B_AMC-3.png]]&lt;/center&gt;<br /> <br /> &lt;math&gt;\mathrm {(A)} 35\qquad \mathrm {(B)} 40\qquad \mathrm {(C)} 45\qquad \mathrm {(D)} 50\qquad \mathrm {(E)} 60&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> At Frank's Fruit Market, 3 bananas cost as much as 2 apples, and 6 apples cost as much as 4 oranges. How many oranges cost as much as 18 bananas? <br /> <br /> &lt;math&gt;\mathrm {(A)} 6\qquad \mathrm {(B)} 8\qquad \mathrm {(C)} 9\qquad \mathrm {(D)} 12\qquad \mathrm {(E)} 18&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> The 2007 AMC 12 contests will be scored by awarding 6 points for each correct response, 0 points for each incorrect response, and 1.5 points for each problem left unanswered. After looking over the 25 problems, Sarah has decided to attempt the first 22 and leave the last 3 unanswered. How many of the first 22 problems must she solve correctly in order to score at least 100 points? <br /> <br /> &lt;math&gt;\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB = 5&lt;/math&gt;, &lt;math&gt;BC = 6&lt;/math&gt;, and &lt;math&gt;AC = 7&lt;/math&gt;. Two bugs start simultaneously from &lt;math&gt;A&lt;/math&gt; and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point &lt;math&gt;D&lt;/math&gt;. What is &lt;math&gt;BD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\mathrm {(A)} 1\qquad \mathrm {(B)} 2\qquad \mathrm {(C)} 3\qquad \mathrm {(D)} 4\qquad \mathrm {(E)} 5&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> All sides of the convex pentagon &lt;math&gt;ABCDE&lt;/math&gt; are of equal length, and &lt;math&gt;\angle A = \angle B = 90^{\circ}&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle E&lt;/math&gt;? <br /> <br /> &lt;math&gt;\mathrm {(A)} 90\qquad \mathrm {(B)} 108\qquad \mathrm {(C)} 120\qquad \mathrm {(D)} 144\qquad \mathrm {(E)} 150&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Tom's age is &lt;math&gt;T&lt;/math&gt; years, which is also the sum of the ages of his three children. His age &lt;math&gt;N&lt;/math&gt; years ago was twice the sum of their ages then. What is &lt;math&gt;T/N&lt;/math&gt; ? <br /> <br /> &lt;math&gt;\mathrm {(A)} 2\qquad \mathrm {(B)} 3\qquad \mathrm {(C)} 4\qquad \mathrm {(D)} 5\qquad \mathrm {(E)} 6&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> A function &lt;math&gt;f&lt;/math&gt; has the property that &lt;math&gt;f(3x-1)=x^2+x+1&lt;/math&gt; for all real numbers &lt;math&gt;x&lt;/math&gt;. What is &lt;math&gt;f(5)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} 7\qquad \mathrm {(B)} 13\qquad \mathrm {(C)} 31\qquad \mathrm {(D)} 111\qquad \mathrm {(E)} 211&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> Some boys and girls are having a car wash to raise money for a class trip to China. Initially &lt;math&gt;40&lt;/math&gt;% of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then &lt;math&gt;30&lt;/math&gt;% of the group are girls. How many girls were initially in the group?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\qquad \mathrm {(B)} 6\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> The angles of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfy &lt;math&gt;\angle A=2\angle B=3\angle C=4\angle D&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle A&lt;/math&gt;, rounded to the nearest whole number?<br /> <br /> &lt;math&gt;\mathrm {(A)} 125\qquad \mathrm {(B)} 144\qquad \mathrm {(C)} 153\qquad \mathrm {(D)} 173\qquad \mathrm {(E)} 180&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> A teacher gave a test to a class in which &lt;math&gt;10\%&lt;/math&gt; of the students are juniors and &lt;math&gt;90\%&lt;/math&gt; are seniors. The average score on the test was &lt;math&gt;84&lt;/math&gt;. The juniors all received the same score, and the average score of the seniors was &lt;math&gt;83&lt;/math&gt;. What score did each of the juniors receive on the test?<br /> <br /> &lt;math&gt;\mathrm {(A)} 85\qquad \mathrm {(B)} 88\qquad \mathrm {(C)} 93\qquad \mathrm {(D)} 94\qquad \mathrm {(E)} 98&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> A traffic light runs repeatedly through the following cycle: green for &lt;math&gt;30&lt;/math&gt; seconds, then yellow for &lt;math&gt;3&lt;/math&gt; seconds, and then red for &lt;math&gt;30&lt;/math&gt; seconds. Leah picks a random three-second time interval to watch the light. What is the probability that the color changes while she is watching?<br /> <br /> &lt;math&gt;\mathrm {(A)} \frac{1}{63}\qquad \mathrm {(B)} \frac{1}{21}\qquad \mathrm {(C)} \frac{1}{10}\qquad \mathrm {(D)} \frac{1}{7}\qquad \mathrm {(E)} \frac{1}{3}&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Point &lt;math&gt;P&lt;/math&gt; is inside equilateral &lt;math&gt;\triangle ABC&lt;/math&gt;. Points &lt;math&gt;Q&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt;, and &lt;math&gt;S&lt;/math&gt; are the feet of the perpendiculars from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{CA}&lt;/math&gt;, respectively. Given that &lt;math&gt;PQ=1&lt;/math&gt;, &lt;math&gt;PR=2&lt;/math&gt;, and &lt;math&gt;PS=3&lt;/math&gt;, what is &lt;math&gt;AB&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\qquad \mathrm {(B)} 3\sqrt{3}\qquad \mathrm {(C)} 6\qquad \mathrm {(D)} 4\sqrt{3}\qquad \mathrm {(E)} 9&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> The geometric series &lt;math&gt;a+ar+ar^2\cdots&lt;/math&gt; has a sum of &lt;math&gt;7&lt;/math&gt;, and the terms involving odd powers of &lt;math&gt;r&lt;/math&gt; have a sum of &lt;math&gt;3&lt;/math&gt;. What is &lt;math&gt;a+r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} \frac{4}{3}\qquad \mathrm {(B)} \frac{12}{7}\qquad \mathrm {(C)} \frac{3}{2}\qquad \mathrm {(D)} \frac{7}{3}\qquad \mathrm {(E)} \frac{5}{2}&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 15 | Solution]]<br /> <br /> ==Problem 16==<br /> Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?<br /> <br /> &lt;math&gt;\mathrm {(A)} 15\qquad \mathrm {(B)} 18\qquad \mathrm {(C)} 27\qquad \mathrm {(D)} 54\qquad \mathrm {(E)} 81&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 16 | Solution]]<br /> <br /> ==Problem 17==<br /> If &lt;math&gt;a&lt;/math&gt; is a nonzero integer and &lt;math&gt;b&lt;/math&gt; is a positive number such that &lt;math&gt;ab^2=\log_{10}b&lt;/math&gt;, what is the median of the set &lt;math&gt;\{0,1,a,b,1/b\}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} 0\qquad \mathrm {(B)} 1\qquad \mathrm {(C)} a\qquad \mathrm {(D)} b\qquad \mathrm {(E)} \frac{1}{b}&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 17 | Solution]]<br /> <br /> ==Problem 18==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; be digits with &lt;math&gt;a\ne 0&lt;/math&gt;. The three-digit integer &lt;math&gt;abc&lt;/math&gt; lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer &lt;math&gt;acb&lt;/math&gt; lies two thirds of the way between the same two squares. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} 10\qquad \mathrm {(B)} 13\qquad \mathrm {(C)} 16\qquad \mathrm {(D)} 18\qquad \mathrm {(E)} 21&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 18 | Solution]]<br /> <br /> ==Problem 19==<br /> Rhombus &lt;math&gt;ABCD&lt;/math&gt;, with side length &lt;math&gt;6&lt;/math&gt;, is rolled to form a cylinder of volume &lt;math&gt;6&lt;/math&gt; by taping &lt;math&gt;\overline{AB}&lt;/math&gt; to &lt;math&gt;\overline{DC}&lt;/math&gt;. What is &lt;math&gt;\sin(\angle ABC)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} \frac{\pi}{9}\qquad \mathrm {(B)} \frac{1}{2}\qquad \mathrm {(C)} \frac{\pi}{6}\qquad \mathrm {(D)} \frac{\pi}{4}\qquad \mathrm {(E)} \frac{\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 19 | Solution]]<br /> <br /> ==Problem 20==<br /> The parallelogram bounded by the lines &lt;math&gt;y=ax+c&lt;/math&gt;, &lt;math&gt;y=ax+d&lt;/math&gt;, &lt;math&gt;y=bx+c&lt;/math&gt;, and &lt;math&gt;y=bx+d&lt;/math&gt; has area &lt;math&gt;18&lt;/math&gt;. The parallelogram bounded by the lines &lt;math&gt;y=ax+c&lt;/math&gt;, &lt;math&gt;y=ax-d&lt;/math&gt;, &lt;math&gt;y=bx+c&lt;/math&gt;, and &lt;math&gt;y=bx-d&lt;/math&gt; has area &lt;math&gt;72&lt;/math&gt;. Given that &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; are positive integers, what is the smallest possible value of &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 20 | Solution]]<br /> <br /> ==Problem 21==<br /> The first &lt;math&gt;2007&lt;/math&gt; positive integers are each written in base &lt;math&gt;3&lt;/math&gt;. How many of these base-&lt;math&gt;3&lt;/math&gt; representations are palindromes? (A palindrome is a number that reads the same forward and backward.)<br /> <br /> &lt;math&gt;\mathrm {(A)} 100\qquad \mathrm {(B)} 101\qquad \mathrm {(C)} 102\qquad \mathrm {(D)} 103\qquad \mathrm {(E)} 104&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 21 | Solution]]<br /> <br /> ==Problem 22==<br /> Two particles move along the edges of equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; in the direction &lt;cmath&gt;A\Rightarrow B\Rightarrow C\Rightarrow A,&lt;/cmath&gt; starting simultaneously and moving at the same speed. One starts at &lt;math&gt;A&lt;/math&gt;, and the other starts at the midpoint of &lt;math&gt;\overline{BC}&lt;/math&gt;. The midpoint of the line segment joining the two particles traces out a path that encloses a region &lt;math&gt;R&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;R&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} \frac{1}{16}\qquad \mathrm {(B)} \frac{1}{12}\qquad \mathrm {(C)} \frac{1}{9}\qquad \mathrm {(D)} \frac{1}{6}\qquad \mathrm {(E)} \frac{1}{4}&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 22 | Solution]]<br /> <br /> ==Problem 23==<br /> How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to &lt;math&gt;3&lt;/math&gt; times their perimeters?<br /> <br /> &lt;math&gt;\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 23 | Solution]]<br /> <br /> ==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;gcd(a,b)=1&lt;/math&gt; and &lt;cmath&gt;\frac{a}{b}+\frac{14b}{9a}&lt;/cmath&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\qquad \mathrm {(B)} 6\qquad \mathrm {(C)} 9\qquad \mathrm {(D)} 12\qquad \mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 24 | Solution]]<br /> <br /> ==Problem 25==<br /> Points &lt;math&gt;A,B,C,D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are located in 3-dimensional space with &lt;math&gt;AB=BC=CD=DE=EA=2&lt;/math&gt; and &lt;math&gt;\angle ABC=\angle CDE=\angle DEA=90^o&lt;/math&gt;. The plane of &lt;math&gt;\triangle ABC&lt;/math&gt; is parallel to &lt;math&gt;\overline{DE}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle BDE&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}&lt;/math&gt;<br /> <br /> [[2007 AMC 12B Problems/Problem 25 | Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2007|ab=B|before=[[2007 AMC 12A Problems|2007 AMC 12A]]|after=[[2008 AMC 12A Problems|2008 AMC 12A]]}}<br /> * [[AMC 12]]<br /> * [[AMC 12 Problems and Solutions]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=142 2007 AMC B Math Jam Transcript]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems&diff=71520 2007 AMC 12A Problems 2015-08-11T21:05:12Z <p>Npip99: </p> <hr /> <div>==Problem 1==<br /> One ticket to a show costs &lt;math&gt;20&lt;/math&gt; at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> <br /> An aquarium has a [[rectangular prism|rectangular base]] that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0.5\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 1.5\qquad \mathrm{(D)}\ 2\qquad \mathrm{(E)}\ 2.5&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> <br /> <br /> The larger of two consecutive odd integers is three times the smaller. What is their sum?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> <br /> <br /> Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> Last year Mr. Jon Q. Public received an inheritance. He paid &lt;math&gt;20\%&lt;/math&gt; in federal taxes on the inheritance, and paid &lt;math&gt;10\%&lt;/math&gt; of what he had left in state taxes. He paid a total of &lt;math&gt;\textdollar10\,500&lt;/math&gt; for both taxes. How many dollars was his inheritance?<br /> <br /> &lt;math&gt;(\mathrm {A})\ 30\,000 \qquad (\mathrm {B})\ 32\,500 \qquad (\mathrm {C})\ 35\,000 \qquad (\mathrm {D})\ 37\,500 \qquad (\mathrm {E})\ 40\,000&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;ADC&lt;/math&gt; are [[isosceles]] with &lt;math&gt;AB=BC&lt;/math&gt; and &lt;math&gt;AD=DC&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; is inside triangle &lt;math&gt;ABC&lt;/math&gt;, angle &lt;math&gt;ABC&lt;/math&gt; measures 40 degrees, and angle &lt;math&gt;ADC&lt;/math&gt; measures 140 degrees. What is the degree measure of angle &lt;math&gt;BAD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> Let &lt;math&gt;a, b, c, d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; be five consecutive terms in an arithmetic sequence, and suppose that &lt;math&gt;a+b+c+d+e=30&lt;/math&gt;. Which of &lt;math&gt;a, b, c, d,&lt;/math&gt; or &lt;math&gt;e&lt;/math&gt; can be found?<br /> <br /> &lt;math&gt;\textrm{(A)} \ a\qquad \textrm{(B)}\ b\qquad \textrm{(C)}\ c\qquad \textrm{(D)}\ d\qquad \textrm{(E)}\ e&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> <br /> A star-[[polygon]] is drawn on a clock face by drawing a [[chord]] from each number to the fifth number counted clockwise from that number. That is, chords are drawn from 12 to 5, from 5 to 10, from 10 to 3, and so on, ending back at 12. What is the degree measure of the [[angle]] at each [[vertex]] in the star polygon?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 24\qquad \mathrm{(C)}\ 30\qquad \mathrm{(D)}\ 36\qquad \mathrm{(E)}\ 60&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> <br /> Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the [[ratio]] of Yan's distance from his home to his distance from the stadium?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> A [[triangle]] with side lengths in the [[ratio]] &lt;math&gt;3 : 4 : 5&lt;/math&gt; is inscribed in a [[circle]] with [[radius]] 3. What is the area of the triangle?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> A finite [[sequence]] of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let &lt;math&gt;S&lt;/math&gt; be the sum of all the terms in the sequence. What is the largest [[prime]] [[factor]] that always divides &lt;math&gt;S&lt;/math&gt;? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Integers &lt;math&gt;a, b, c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the [[probability]] that &lt;math&gt;ad-bc&lt;/math&gt; is [[even]]?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> <br /> A piece of cheese is located at &lt;math&gt;(12,10)&lt;/math&gt; in a [[coordinate plane]]. A mouse is at &lt;math&gt;(4,-2)&lt;/math&gt; and is running up the [[line]] &lt;math&gt;y=-5x+18&lt;/math&gt;. At the point &lt;math&gt;(a,b)&lt;/math&gt; the mouse starts getting farther from the cheese rather than closer to it. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Let a, b, c, d, and e be distinct [[integer]]s such that<br /> <br /> &lt;center&gt;&lt;math&gt;(6-a)(6-b)(6-c)(6-d)(6-e)=45&lt;/math&gt;&lt;/center&gt;<br /> <br /> What is &lt;math&gt;a+b+c+d+e&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 5\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 25\qquad \mathrm{(D)}\ 27\qquad \mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> <br /> The [[set]] &lt;math&gt;\{3,6,9,10\}&lt;/math&gt; is augmented by a fifth element &lt;math&gt;n&lt;/math&gt;, not equal to any of the other four. The [[median]] of the resulting set is equal to its [[mean]]. What is the sum of all possible values of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 19\qquad \mathrm{(D)}\ 24\qquad \mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 15 | Solution]]<br /> <br /> ==Problem 16==<br /> <br /> How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 16 | Solution]]<br /> <br /> ==Problem 17==<br /> Suppose that &lt;math&gt;\sin a + \sin b = \sqrt{\frac{5}{3}}&lt;/math&gt; and &lt;math&gt;\cos a + \cos b = 1&lt;/math&gt;. What is &lt;math&gt;\cos (a - b)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \sqrt{\frac{5}{3}} - 1\qquad \mathrm{(B)}\ \frac 13\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 23\qquad \mathrm{(E)}\ 1&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 17 | Solution]]<br /> <br /> ==Problem 18==<br /> <br /> The [[polynomial]] &lt;math&gt;f(x) = x^{4} + ax^{3} + bx^{2} + cx + d&lt;/math&gt; has real [[coefficient]]s, and &lt;math&gt;f(2i) = f(2 + i) = 0.&lt;/math&gt; What is &lt;math&gt;a + b + c + d?&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 9 \qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 18 | Solution]]<br /> <br /> ==Problem 19==<br /> [[Triangle]]s &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;ADE&lt;/math&gt; have [[area]]s &lt;math&gt;2007&lt;/math&gt; and &lt;math&gt;7002,&lt;/math&gt; respectively, with &lt;math&gt;B = (0,0),&lt;/math&gt; &lt;math&gt;C = (223,0),&lt;/math&gt; &lt;math&gt;D = (680,380),&lt;/math&gt; and &lt;math&gt;E = (689,389).&lt;/math&gt; What is the sum of all possible x-coordinates of &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 282 \qquad \mathrm{(B)}\ 300 \qquad \mathrm{(C)}\ 600 \qquad \mathrm{(D)}\ 900 \qquad \mathrm{(E)}\ 1200&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 19 | Solution]]<br /> <br /> ==Problem 20==<br /> Corners are sliced off a [[unit cube]] so that the six [[face]]s each become regular [[octagon]]s. What is the total [[volume]] of the removed [[tetrahedra]]?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac{5\sqrt{2}-7}{3}\qquad \mathrm{(B)}\ \frac{10-7\sqrt{2}}{3}\qquad \mathrm{(C)}\ \frac{3-2\sqrt{2}}{3}\qquad \mathrm{(D)}\ \frac{8\sqrt{2}-11}{3}\qquad \mathrm{(E)}\ \frac{6-4\sqrt{2}}{3}&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 20 | Solution]]<br /> <br /> ==Problem 21==<br /> The sum of the [[root|zeros]], the product of the zeros, and the sum of the [[coefficient]]s of the [[function]] &lt;math&gt;f(x)=ax^{2}+bx+c&lt;/math&gt; are equal. Their common value must also be which of the following? <br /> <br /> &lt;math&gt;\textrm{(A)}\ \textrm{the\ coefficient\ of\ }x^{2}~~~ \textrm{(B)}\ \textrm{the\ coefficient\ of\ }x&lt;/math&gt;<br /> &lt;math&gt;\textrm{(C)}\ \textrm{the\ y-intercept\ of\ the\ graph\ of\ }y=f(x)&lt;/math&gt;<br /> &lt;math&gt;\textrm{(D)}\ \textrm{one\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)&lt;/math&gt;<br /> &lt;math&gt;\textrm{(E)}\ \textrm{the\ mean\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 21 | Solution]]<br /> <br /> ==Problem 22==<br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;S(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n.&lt;/math&gt; For how many values of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;n + S(n) + S(S(n)) = 2007?&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 22 | Solution]]<br /> <br /> ==Problem 23==<br /> [[Square]] &lt;math&gt;ABCD&lt;/math&gt; has area &lt;math&gt;36,&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt; is [[parallel]] to the [[x-axis]]. Vertices &lt;math&gt;A,&lt;/math&gt; &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are on the graphs of &lt;math&gt;y = \log_{a}x,&lt;/math&gt; &lt;math&gt;y = 2\log_{a}x,&lt;/math&gt; and &lt;math&gt;y = 3\log_{a}x,&lt;/math&gt; respectively. What is &lt;math&gt;a?&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \sqrt {3}\qquad \mathrm{(B)}\ \sqrt {3}\qquad \mathrm{(C)}\ \sqrt {6}\qquad \mathrm{(D)}\ \sqrt {6}\qquad \mathrm{(E)}\ 6&lt;/math&gt;<br /> <br /> <br /> [[2007 AMC 12A Problems/Problem 23 | Solution]]<br /> <br /> ==Problem 24==<br /> For each integer &lt;math&gt;n&gt;1&lt;/math&gt;, let &lt;math&gt;F(n)&lt;/math&gt; be the number of solutions to the equation &lt;math&gt;\sin{x}=\sin{(nx)}&lt;/math&gt; on the interval &lt;math&gt;[0,\pi]&lt;/math&gt;. What is &lt;math&gt;\sum_{n=2}^{2007} F(n)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2014524&lt;/math&gt; &lt;math&gt;\mathrm{(B)}\ 2015028&lt;/math&gt; &lt;math&gt;\mathrm{(C)}\ 2015033&lt;/math&gt; &lt;math&gt;\mathrm{(D)}\ 2016532&lt;/math&gt; &lt;math&gt;\mathrm{(E)}\ 2017033&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 24 | Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Call a set of integers ''spacy'' if it contains no more than one out of any three consecutive integers. How many [[subset]]s of &lt;math&gt;\{1,2,3,\ldots,12\},&lt;/math&gt; including the [[empty set]], are spacy?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129&lt;/math&gt;<br /> <br /> [[2007 AMC 12A Problems/Problem 25 | Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2007|ab=A|before=[[2006 AMC 12B Problems|2006 AMC 12B]]|after=[[2007 AMC 12B Problems|2007 AMC 12B]]}}<br /> * [[AMC 12]]<br /> * [[AMC 12 Problems and Solutions]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=142 2007 AMC A Math Jam Transcript]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_24&diff=71514 2006 AMC 12A Problems/Problem 24 2015-08-11T01:29:02Z <p>Npip99: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> The expression <br /> <br /> &lt;math&gt;(x+y+z)^{2006}+(x-y-z)^{2006}&lt;/math&gt;<br /> <br /> is simplified by expanding it and combining like terms. How many terms are in the simplified expression?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028&lt;/math&gt;<br /> <br /> == Solution 1==<br /> By the [[Multinomial Theorem]], the summands can be written as<br /> <br /> &lt;cmath&gt;\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},&lt;/cmath&gt;<br /> <br /> respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:<br /> <br /> &lt;cmath&gt;{2006+2\choose 2} = 2015028&lt;/cmath&gt;<br /> <br /> terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; must be opposite. Now we find a pattern:<br /> <br /> if the exponent of &lt;math&gt;y&lt;/math&gt; is 1, the exponent of &lt;math&gt;z&lt;/math&gt; can be all even integers up to 2004, so there are 1003 terms.<br /> <br /> if the exponent of &lt;math&gt;y&lt;/math&gt; is 3, the exponent of &lt;math&gt;z&lt;/math&gt; can go up to 2002, so there are 1002 terms.<br /> <br /> &lt;math&gt;\vdots&lt;/math&gt;<br /> <br /> if the exponent of &lt;math&gt;y&lt;/math&gt; is 2005, then &lt;math&gt;z&lt;/math&gt; can only be 0, so there is 1 term.<br /> <br /> If we add them up, we get &lt;math&gt;\frac{1003\cdot1004}{2}&lt;/math&gt; terms. However, we can switch the exponents of &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; and these terms will still have a negative sign. So there are a total of &lt;math&gt;1003\cdot1004&lt;/math&gt; negative terms.<br /> <br /> By subtracting this number from 2015028, we obtain &lt;math&gt;\boxed{D}&lt;/math&gt; or &lt;math&gt;1008016&lt;/math&gt; as our answer.<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we can use a generating function to solve this problem. <br /> The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of &lt;math&gt;k&lt;/math&gt;). In other words, we want to find &lt;math&gt;f(x)&lt;/math&gt; where the coefficient of &lt;math&gt;x^k&lt;/math&gt; equals the number of unique terms in &lt;math&gt;(x+y+z)^k + (x-y-z)^k&lt;/math&gt;.<br /> <br /> <br /> First, we note that all unique terms in the expression have the form, &lt;math&gt;Cx^ay^bz^c&lt;/math&gt;, where &lt;math&gt;a+b+c=k&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of &lt;math&gt;(x+y+z)^k + (x-y-z)^k&lt;/math&gt; is<br /> &lt;cmath&gt;(1+x+x^2+x^3\cdots)^3 = \frac{1}{(1-x)^3}&lt;/cmath&gt;<br /> <br /> <br /> Secondly, we note that a certain number of terms of the form, &lt;math&gt;Cx^ay^bz^c&lt;/math&gt;, do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when &lt;math&gt;1 \equiv b+c\text{ (mod }2\text{)}&lt;/math&gt; because every unique term is of the form:<br /> &lt;cmath&gt;\binom{k}{a,b,c}x^ay^bz^c+(-1)^{b+c}\binom{k}{a,b,c}x^ay^bz^c&lt;/cmath&gt;<br /> for all possible &lt;math&gt;a,b,c&lt;/math&gt;.<br /> <br /> <br /> Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of &lt;math&gt;k&lt;/math&gt;. With some thought, we see that this desired generating function is the following:<br /> &lt;cmath&gt;2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}&lt;/cmath&gt;<br /> <br /> <br /> Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in &lt;math&gt;(x+y+z)^k + (x-y-z)^k&lt;/math&gt;, our initial goal:<br /> &lt;cmath&gt;\frac{1}{(1-x)^3}-\frac{2x}{(1-x)^3(1+x)^2} = \frac{x^2+1}{(1-x)^3(1+x)^2}&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;(x^2+1)(1+x+x^2\cdots)^3(1-x+x^2-x^3\cdots)^2&lt;/cmath&gt;<br /> <br /> <br /> The coefficient of &lt;math&gt;x^{2006}&lt;/math&gt; of the above expression equals<br /> &lt;cmath&gt;\sum_{a=0}^{2006}\binom{2+a}{2}\binom{1+2006-a}{1}(-1)^a + \sum_{a=0}^{2004}\binom{2+a}{2}\binom{1+2004-a}{1}(-1)^a&lt;/cmath&gt;<br /> <br /> <br /> Evaluating the expression, we get &lt;math&gt;1008016&lt;/math&gt;, as expected.<br /> <br /> == Solution 3 ==<br /> <br /> Define &lt;math&gt;P&lt;/math&gt; such that &lt;math&gt;P=y+z&lt;/math&gt;. Then the expression in the problem becomes:<br /> &lt;math&gt;(x+P)^{2006}+(x-P)^{2006}&lt;/math&gt;. <br /> <br /> Expanding this using binomial theorem gives &lt;math&gt;(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)&lt;/math&gt;, where &lt;math&gt;n=2006&lt;/math&gt; (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves). <br /> <br /> Simplifying gives: &lt;math&gt;2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)&lt;/math&gt;. Note that two terms that come out of different powers of &lt;math&gt;P&lt;/math&gt; cannot combine and simplify, as their exponent of &lt;math&gt;x&lt;/math&gt; will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, &lt;math&gt;P^k=(y+z)^k&lt;/math&gt; will have &lt;math&gt;k+1&lt;/math&gt; terms, so the answer is &lt;math&gt;1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{D}&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> [[Category:Introductory Combinatorics Problems]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems&diff=71506 2005 AMC 12B Problems 2015-08-10T12:54:25Z <p>Npip99: /* Problem 23 */</p> <hr /> <div>== Problem 1 ==<br /> A scout troop buys &lt;math&gt;1000&lt;/math&gt; candy bars at a price of five for &lt;math&gt;2&lt;/math&gt; dollars. They sell all the candy bars at the price of two for &lt;math&gt;1&lt;/math&gt; dollar. What was their profit, in dollars?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 100 \qquad<br /> \mathrm{(B)}\ 200 \qquad<br /> \mathrm{(C)}\ 300 \qquad<br /> \mathrm{(D)}\ 400 \qquad<br /> \mathrm{(E)}\ 500<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> A positive number &lt;math&gt;x&lt;/math&gt; has the property that &lt;math&gt;x\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 2 \qquad<br /> \mathrm{(B)}\ 4 \qquad<br /> \mathrm{(C)}\ 10 \qquad<br /> \mathrm{(D)}\ 20 \qquad<br /> \mathrm{(E)}\ 40<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one ﬁfth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs? <br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ \frac15 \qquad<br /> \mathrm{(B)}\ \frac13 \qquad<br /> \mathrm{(C)}\ \frac25 \qquad<br /> \mathrm{(D)}\ \frac23 \qquad<br /> \mathrm{(E)}\ \frac45<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> At the beginning of the school year, Lisa's goal was to earn an A on at least &lt;math&gt;80\%&lt;/math&gt; of her &lt;math&gt;50&lt;/math&gt; quizzes for the year. She earned an A on &lt;math&gt;22&lt;/math&gt; of the first &lt;math&gt;30&lt;/math&gt; quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 1 \qquad<br /> \mathrm{(B)}\ 2 \qquad<br /> \mathrm{(C)}\ 3 \qquad<br /> \mathrm{(D)}\ 4 \qquad<br /> \mathrm{(E)}\ 5<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> An &lt;math&gt;8&lt;/math&gt;-foot by &lt;math&gt;10&lt;/math&gt;-foot floor is tiled with square tiles of size &lt;math&gt;1&lt;/math&gt; foot by &lt;math&gt;1&lt;/math&gt; foot. Each tile has a pattern consisting of four white quarter circles of radius &lt;math&gt;1/2&lt;/math&gt; foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(linewidth(.8pt));<br /> fill(unitsquare,gray);<br /> filldraw(Arc((0,0),.5,0,90)--(0,0)--cycle,white,black);<br /> filldraw(Arc((1,0),.5,90,180)--(1,0)--cycle,white,black);<br /> filldraw(Arc((1,1),.5,180,270)--(1,1)--cycle,white,black);<br /> filldraw(Arc((0,1),.5,270,360)--(0,1)--cycle,white,black);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 80-20\pi \qquad<br /> \mathrm{(B)}\ 60-10\pi \qquad<br /> \mathrm{(C)}\ 80-10\pi \qquad<br /> \mathrm{(D)}\ 60+10\pi \qquad<br /> \mathrm{(E)}\ 80+10\pi<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, we have &lt;math&gt;AC=BC=7&lt;/math&gt; and &lt;math&gt;AB=2&lt;/math&gt;. Suppose that &lt;math&gt;D&lt;/math&gt; is a point on line &lt;math&gt;AB&lt;/math&gt; such that &lt;math&gt;B&lt;/math&gt; lies between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;CD=8&lt;/math&gt;. What is &lt;math&gt;BD&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 3 \qquad<br /> \mathrm{(B)}\ 2\sqrt{3} \qquad<br /> \mathrm{(C)}\ 4 \qquad<br /> \mathrm{(D)}\ 5 \qquad<br /> \mathrm{(E)}\ 4\sqrt{2}<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> What is the area enclosed by the graph of &lt;math&gt;|3x|+|4y|=12&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 6 \qquad<br /> \mathrm{(B)}\ 12 \qquad<br /> \mathrm{(C)}\ 16 \qquad<br /> \mathrm{(D)}\ 24 \qquad<br /> \mathrm{(E)}\ 25<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> For how many values of &lt;math&gt;a&lt;/math&gt; is it true that the line &lt;math&gt;y = x + a&lt;/math&gt; passes through the<br /> vertex of the parabola &lt;math&gt;y = x^2 + a^2&lt;/math&gt; ?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 0 \qquad<br /> \mathrm{(B)}\ 1 \qquad<br /> \mathrm{(C)}\ 2 \qquad<br /> \mathrm{(D)}\ 10 \qquad<br /> \mathrm{(E)}\ \text{infinitely many}<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> On a certain math exam, &lt;math&gt;10\%&lt;/math&gt; of the students got &lt;math&gt;70&lt;/math&gt; points, &lt;math&gt;25\%&lt;/math&gt; got &lt;math&gt;80&lt;/math&gt; points, &lt;math&gt;20\%&lt;/math&gt; got &lt;math&gt;85&lt;/math&gt; points, &lt;math&gt;15\%&lt;/math&gt; got &lt;math&gt;90&lt;/math&gt; points, and the rest got &lt;math&gt;95&lt;/math&gt; points. What is the difference between the mean and the median score on this exam?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{4}}} \qquad \mathrm{(E)}\ {{{5}}}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The first term of a sequence is &lt;math&gt;2005&lt;/math&gt;. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the &lt;math&gt;2005^{\text{th}}&lt;/math&gt; term of the sequence?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{29}}} \qquad \mathrm{(B)}\ {{{55}}} \qquad \mathrm{(C)}\ {{{85}}} \qquad \mathrm{(D)}\ {{{133}}} \qquad \mathrm{(E)}\ {{{250}}}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> An envelope contains eight bills: &lt;math&gt;2&lt;/math&gt; ones, &lt;math&gt;2&lt;/math&gt; fives, &lt;math&gt;2&lt;/math&gt; tens, and &lt;math&gt;2&lt;/math&gt; twenties. Two bills are drawn at random without replacement. What is the probability that their sum is &amp;#36;&lt;math&gt;20&lt;/math&gt; or more?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{7}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The [[quadratic equation]] &lt;math&gt;x^2+mx+n&lt;/math&gt; has roots twice those of &lt;math&gt;x^2+px+m&lt;/math&gt;, and none of &lt;math&gt;m,n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; is zero. What is the value of &lt;math&gt;n/p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{1}}} \qquad \mathrm{(B)}\ {{{2}}} \qquad \mathrm{(C)}\ {{{4}}} \qquad \mathrm{(D)}\ {{{8}}} \qquad \mathrm{(E)}\ {{{16}}}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Suppose that &lt;math&gt;4^{x_1}=5&lt;/math&gt;, &lt;math&gt;5^{x_2}=6&lt;/math&gt;, &lt;math&gt;6^{x_3}=7&lt;/math&gt;, ... , &lt;math&gt;127^{x_{124}}=128&lt;/math&gt;. What is &lt;math&gt;x_1x_2...x_{124}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{2}}} \qquad \mathrm{(B)}\ {{{\frac{5}{2}}}} \qquad \mathrm{(C)}\ {{{3}}} \qquad \mathrm{(D)}\ {{{\frac{7}{2}}}} \qquad \mathrm{(E)}\ {{{4}}}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> A circle having center &lt;math&gt;(0,k)&lt;/math&gt;, with &lt;math&gt;k&gt;6&lt;/math&gt;,is tangent to the lines &lt;math&gt;y=x&lt;/math&gt;, &lt;math&gt;y=-x&lt;/math&gt; and &lt;math&gt;y=6&lt;/math&gt;. What is the radius of this circle?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 6\sqrt{2}-6 \qquad<br /> \mathrm{(B)}\ 6 \qquad<br /> \mathrm{(C)}\ 6\sqrt{2} \qquad<br /> \mathrm{(D)}\ 12 \qquad<br /> \mathrm{(E)}\ 6+6\sqrt{2}<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> The sum of four two-digit numbers is &lt;math&gt;221&lt;/math&gt;. None of the eight digits is &lt;math&gt;0&lt;/math&gt; and no two of them are the same. Which of the following is '''not''' included among the eight digits?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 1 \qquad<br /> \mathrm{(B)}\ 2 \qquad<br /> \mathrm{(C)}\ 3 \qquad<br /> \mathrm{(D)}\ 4 \qquad<br /> \mathrm{(E)}\ 5<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Eight spheres of radius 1, one per octant, are each tangent to the coordinate planes. What is the radius of the smallest sphere, centered at the origin, that contains these eight spheres?<br /> <br /> &lt;math&gt;<br /> \mathrm (A)\ \sqrt{2} \qquad<br /> \mathrm (B)\ \sqrt{3} \qquad<br /> \mathrm (C)\ 1+\sqrt{2}\qquad<br /> \mathrm (D)\ 1+\sqrt{3}\qquad<br /> \mathrm (E)\ 3<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> How many distinct four-tuples &lt;math&gt;(a, b, c, d)&lt;/math&gt; of rational numbers are there with<br /> <br /> &lt;math&gt;a \cdot \log_{10} 2+b \cdot \log_{10} 3 +c \cdot \log_{10} 5 + d \cdot \log_{10} 7 = 2005&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 0 \qquad<br /> \mathrm{(B)}\ 1 \qquad<br /> \mathrm{(C)}\ 17 \qquad<br /> \mathrm{(D)}\ 2004 \qquad<br /> \mathrm{(E)}\ \text{infinitely many}<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Let &lt;math&gt;A(2,2)&lt;/math&gt; and &lt;math&gt;B(7,7)&lt;/math&gt; be points in the plane. Define &lt;math&gt;R&lt;/math&gt; as the region in the first quadrant consisting of those points &lt;math&gt;C&lt;/math&gt; such that &lt;math&gt;\triangle ABC&lt;/math&gt; is an acute triangle. What is the closest integer to the area of the region &lt;math&gt;R&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 25 \qquad<br /> \mathrm{(B)}\ 39 \qquad<br /> \mathrm{(C)}\ 51 \qquad<br /> \mathrm{(D)}\ 60 \qquad<br /> \mathrm{(E)}\ 80 \qquad<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be two-digit integers such that &lt;math&gt;y&lt;/math&gt; is obtained by reversing the digits of &lt;math&gt;x&lt;/math&gt;. The integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x^{2}-y^{2}=m^{2}&lt;/math&gt; for some positive integer &lt;math&gt;m&lt;/math&gt;. What is &lt;math&gt;x+y+m&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 88 \qquad<br /> \mathrm{(B)}\ 112 \qquad<br /> \mathrm{(C)}\ 116 \qquad<br /> \mathrm{(D)}\ 144 \qquad<br /> \mathrm{(E)}\ 154 \qquad<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> Let &lt;math&gt;a,b,c,d,e,f,g&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; be distinct elements in the set<br /> <br /> &lt;cmath&gt;\{-7,-5,-3,-2,2,4,6,13\}.&lt;/cmath&gt;<br /> <br /> What is the minimum possible value of<br /> <br /> &lt;cmath&gt;(a+b+c+d)^{2}+(e+f+g+h)^{2}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 30 \qquad<br /> \mathrm{(B)}\ 32 \qquad<br /> \mathrm{(C)}\ 34 \qquad<br /> \mathrm{(D)}\ 40 \qquad<br /> \mathrm{(E)}\ 50<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> A positive integer &lt;math&gt;n&lt;/math&gt; has &lt;math&gt;60&lt;/math&gt; divisors and &lt;math&gt;7n&lt;/math&gt; has &lt;math&gt;80&lt;/math&gt; divisors. What is the greatest integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;7^k&lt;/math&gt; divides &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> A sequence of complex numbers &lt;math&gt;z_{0}, z_{1}, z_{2}, ...&lt;/math&gt; is defined by the rule<br /> <br /> &lt;cmath&gt;z_{n+1} = \frac {iz_{n}}{\overline {z_{n}}},&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;\overline {z_{n}}&lt;/math&gt; is the [[complex conjugate]] of &lt;math&gt;z_{n}&lt;/math&gt; and &lt;math&gt;i^{2}=-1&lt;/math&gt;. Suppose that &lt;math&gt;|z_{0}|=1&lt;/math&gt; and &lt;math&gt;z_{2005}=1&lt;/math&gt;. How many possible values are there for &lt;math&gt;z_{0}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 1 \qquad <br /> \textbf{(B)}\ 2 \qquad <br /> \textbf{(C)}\ 4 \qquad <br /> \textbf{(D)}\ 2005 \qquad <br /> \textbf{(E)}\ 2^{2005}<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of real numbers for which<br /> <br /> &lt;cmath&gt;\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.&lt;/cmath&gt;<br /> There are real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that for all ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; in &lt;math&gt;S&lt;/math&gt; we have &lt;math&gt;x^{3}+y^{3}=a \cdot 10^{3z} + b \cdot 10^{2z}.&lt;/math&gt; What is the value of &lt;math&gt;a+b?&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ \frac {15}{2} \qquad <br /> \textbf{(B)}\ \frac {29}{2} \qquad <br /> \textbf{(C)}\ 15 \qquad <br /> \textbf{(D)}\ \frac {39}{2} \qquad <br /> \textbf{(E)}\ 24<br /> &lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> All three vertices of an equilateral triangle are on the parabola &lt;math&gt;y=x^2&lt;/math&gt;, and one of its sides has a slope of &lt;math&gt;2&lt;/math&gt;. The &lt;math&gt;x&lt;/math&gt;-coordinates of the three vertices have a sum of &lt;math&gt;m/n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is the value of &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{14}}} \qquad \mathrm{(B)}\ {{{15}}} \qquad \mathrm{(C)}\ {{{16}}} \qquad \mathrm{(D)}\ {{{17}}} \qquad \mathrm{(E)}\ {{{18}}}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> Six ants simultaneously stand on the six [[vertex|vertices]] of a regular [[octahedron]], with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal [[probability]]. What is the probability that no two ants arrive at the same vertex?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac {5}{256}<br /> \qquad\mathrm{(B)}\ \frac {21}{1024}<br /> \qquad\mathrm{(C)}\ \frac {11}{512}<br /> \qquad\mathrm{(D)}\ \frac {23}{1024}<br /> \qquad\mathrm{(E)}\ \frac {3}{128}&lt;/math&gt;<br /> <br /> [[2005 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> * [[AMC 12]]<br /> * [[AMC 12 Problems and Solutions]]<br /> * [[2005 AMC 12B]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=49 2005 AMC B Math Jam Transcript]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1988_USAMO_Problems/Problem_4&diff=71497 1988 USAMO Problems/Problem 4 2015-08-09T16:43:46Z <p>Npip99: /* Solution */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with incenter &lt;math&gt;I&lt;/math&gt;. Show that the circumcenters of &lt;math&gt;\Delta IAB&lt;/math&gt;, &lt;math&gt;\Delta IBC&lt;/math&gt;, and &lt;math&gt;\Delta ICA&lt;/math&gt; lie on a circle whose center is the circumcenter of &lt;math&gt;\Delta ABC&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let the circumcenters of &lt;math&gt;\Delta IAB&lt;/math&gt;, &lt;math&gt;\Delta IBC&lt;/math&gt;, and &lt;math&gt;\Delta ICA&lt;/math&gt; be &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;O_a&lt;/math&gt;, and &lt;math&gt;O_b&lt;/math&gt;, respectively. It then suffices to show that &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, and &lt;math&gt;O_c&lt;/math&gt; are concyclic.<br /> <br /> We shall prove that quadrilateral &lt;math&gt;ABO_aC&lt;/math&gt; is cyclic first. Let &lt;math&gt;\angle BAC=\alpha&lt;/math&gt;, &lt;math&gt;\angle CBA=\beta&lt;/math&gt;, and &lt;math&gt;\angle ACB=\gamma&lt;/math&gt;. Then &lt;math&gt;\angle ICB=\gamma/2&lt;/math&gt; and &lt;math&gt;\angle IBC=\beta/2&lt;/math&gt;. Therefore minor arc &lt;math&gt;\overarc{BIC}&lt;/math&gt; in the circumcircle of &lt;math&gt;IBC&lt;/math&gt; has a degree measure of &lt;math&gt;\beta+\gamma&lt;/math&gt;. This shows that &lt;math&gt;\angle CO_aB=\beta+\gamma&lt;/math&gt;, implying that &lt;math&gt;\angle BAC+\angle BO_aC=\alpha+\beta+\gamma=180^{\circ}&lt;/math&gt;. Therefore quadrilateral &lt;math&gt;ABO_aC&lt;/math&gt; is cyclic.<br /> <br /> This shows that point &lt;math&gt;O_a&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta ABC&lt;/math&gt;. Analagous proofs show that &lt;math&gt;O_b&lt;/math&gt; and &lt;math&gt;O_c&lt;/math&gt; are also on the circumcircle of &lt;math&gt;ABC&lt;/math&gt;, which completes the proof. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;M&lt;/math&gt; denote the midpoint of arc &lt;math&gt;AC&lt;/math&gt;. It is well known that &lt;math&gt;M&lt;/math&gt; is equidistant from &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;I&lt;/math&gt; (to check, prove &lt;math&gt;&lt;IAM = &lt;AIM = \frac{&lt;BAC + &lt;ABC}{2}&lt;/math&gt;), so that &lt;math&gt;M&lt;/math&gt; is the circumcenter of &lt;math&gt;AIC&lt;/math&gt;. Similar results hold for &lt;math&gt;BIC&lt;/math&gt; and &lt;math&gt;CIA&lt;/math&gt;, and hence &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;O_a&lt;/math&gt;, and &lt;math&gt;O_b&lt;/math&gt; all lie on the circumcircle of &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{USAMO box|year=1988|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&diff=71491 1983 AIME Problems/Problem 15 2015-08-08T17:33:48Z <p>Npip99: /* Problem */</p> <hr /> <div>== Problem ==<br /> The adjoining figure shows two intersecting [[chord]]s in a [[circle]], with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is [[bisect]]ed by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the [[sine]] of the minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this fraction is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> &lt;asy&gt;size(100);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png|200px]] --&gt;<br /> <br /> == Solution ==<br /> Let &lt;math&gt;A&lt;/math&gt; be any fixed point on [[circle]] &lt;math&gt;O&lt;/math&gt; and let &lt;math&gt;AD&lt;/math&gt; be a [[chord]] of circle &lt;math&gt;O&lt;/math&gt;. The [[locus]] of [[midpoint]]s &lt;math&gt;N&lt;/math&gt; of the chord &lt;math&gt;AD&lt;/math&gt; is a circle &lt;math&gt;P&lt;/math&gt;, with diameter &lt;math&gt;AO&lt;/math&gt;. Generally, the circle &lt;math&gt;P&lt;/math&gt; can intersect the chord &lt;math&gt;BC&lt;/math&gt; at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle &lt;math&gt;P&lt;/math&gt; is tangent to BC at point N. <br /> <br /> Let M be the midpoint of the chord &lt;math&gt;BC&lt;/math&gt;. From [[right triangle]] &lt;math&gt;OMB&lt;/math&gt;, &lt;math&gt;OM = \sqrt{OB^2 - BM^2} =4&lt;/math&gt;. Thus, &lt;math&gt;\tan \angle BOM = \frac{BM}{OM} = \frac 3 4&lt;/math&gt;.<br /> <br /> Notice that the distance &lt;math&gt;OM&lt;/math&gt; equals &lt;math&gt;PN + PO \cos AOM = r(1 + \cos AOM)&lt;/math&gt; (Where &lt;math&gt;r&lt;/math&gt; is the radius of circle P). Evaluating this, &lt;math&gt;\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5&lt;/math&gt;. From &lt;math&gt;\cos \angle AOM&lt;/math&gt;, we see that &lt;math&gt;\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3&lt;/math&gt;<br /> <br /> Next, notice that &lt;math&gt;\angle AOB = \angle AOM - \angle BOM&lt;/math&gt;. We can therefore apply the tangent subtraction formula to obtain , &lt;math&gt;\tan AOB =\frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}&lt;/math&gt;. It follows that &lt;math&gt;\sin AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}&lt;/math&gt;, resulting in an answer of &lt;math&gt;7 \cdot 25=\boxed{175}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> &lt;asy&gt;<br /> size(10cm);<br /> import olympiad;<br /> pair O = (0,0);dot(O);label(&quot;$O$&quot;,O,SW);<br /> pair M = (4,0);dot(M);label(&quot;$M$&quot;,M,SE);<br /> pair N = (4,2);dot(N);label(&quot;$N$&quot;,N,NE);<br /> draw(circle(O,5));<br /> pair B = (4,3);dot(B);label(&quot;$B$&quot;,B,NE);<br /> pair C = (4,-3);dot(C);label(&quot;$C$&quot;,C,SE);<br /> draw(B--C);draw(O--M);<br /> pair P = (1.5,2);dot(P);label(&quot;$P$&quot;,P,W);<br /> draw(circle(P,2.5));<br /> pair A=(3,4);dot(A);label(&quot;$A$&quot;,A,NE);<br /> draw(O--A);<br /> draw(O--B);<br /> pair Q = (1.5,0); dot(Q); label(&quot;$Q$&quot;,Q,S);<br /> pair R = (3,0); dot(R); label(&quot;$R$&quot;,R,S);<br /> draw(P--Q,dotted); draw(A--R,dotted);<br /> pair D=(5,0); dot(D); label(&quot;$D$&quot;,D,E);<br /> draw(A--D);<br /> &lt;/asy&gt;<br /> <br /> The above solution works, but is quite messy and somewhat difficult to follow. This solution provides a diagram to scale, and the motivation behind the solution.<br /> <br /> First of all, where did the statement &quot;&lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; and bisected by &lt;math&gt;BC&lt;/math&gt; &quot; come from? What is its significance in this problem? What is the criterion for this statement to be true?<br /> <br /> We consider the locus of midpoints of the chords from &lt;math&gt;A&lt;/math&gt;. It is well known that this is the circle with diameter &lt;math&gt;AO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with ratio &lt;math&gt;1/2&lt;/math&gt; with center &lt;math&gt;A&lt;/math&gt;. Thus, the locus is the result of the dilation with ratio &lt;math&gt;1/2&lt;/math&gt; of circle &lt;math&gt;O&lt;/math&gt; with center &lt;math&gt;A&lt;/math&gt;. Let the center of this circle be &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Aha! Now we see. &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt; if they cross at some point &lt;math&gt;N&lt;/math&gt; on the circle. Moreover, since &lt;math&gt;AD&lt;/math&gt; is the only chord, &lt;math&gt;BC&lt;/math&gt; must be tangent to the circle &lt;math&gt;P&lt;/math&gt;.<br /> <br /> The rest of this problem is straight forward.<br /> <br /> Our goal is to find &lt;math&gt;\sin AOB = \sin (AOM - BOM)&lt;/math&gt; where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;. Then we have &lt;math&gt;BM=3&lt;/math&gt; and &lt;math&gt;OM=4&lt;/math&gt;.<br /> Let &lt;math&gt;R&lt;/math&gt; be the projection of &lt;math&gt;A&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;, and similarly &lt;math&gt;Q&lt;/math&gt; be the projection of &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;. Then it remains to find &lt;math&gt;AR&lt;/math&gt; so we can use the sine addition formula.<br /> <br /> As &lt;math&gt;PN&lt;/math&gt; is a radius of circle &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;PN=2.5&lt;/math&gt;, and similarly, &lt;math&gt;PO=2.5&lt;/math&gt;. Since &lt;math&gt;OM=4&lt;/math&gt;, &lt;math&gt;OQ=OM-QM=OM-PN=4-2.5=1.5&lt;/math&gt;. Thus, &lt;math&gt;PQ=\sqrt{(2.5)^2-1.5^2}=2&lt;/math&gt;.<br /> <br /> From here, we see that &lt;math&gt;\triangle OAR&lt;/math&gt; is a dilation of &lt;math&gt;\triangle OPQ&lt;/math&gt; about center &lt;math&gt;O&lt;/math&gt; with ratio &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;AR=2PQ=4&lt;/math&gt;.<br /> <br /> Lastly, we apply the formula:<br /> &lt;cmath&gt; \sin (AOM - BOM) = \sin AOM \cos BOM - \sin BOM \cos AOM = (4/5)(4/5)-(3/5)(3/5)=7/25.&lt;/cmath&gt;<br /> <br /> Thus, our answer is &lt;math&gt;7*25=\boxed{175}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems&diff=71490 1983 AIME Problems 2015-08-08T17:31:33Z <p>Npip99: /* Problem 15 */</p> <hr /> <div>{{AIME Problems|year=1983}}<br /> <br /> == Problem 1 ==<br /> Let &lt;math&gt;x&lt;/math&gt;,&lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt;, and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_xw=24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt;, and &lt;math&gt;\log_{xyz}w=12&lt;/math&gt;. Find &lt;math&gt;\log_zw&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;f(x)=|x-p|+|x-15|+|x-p-15|&lt;/math&gt;, where &lt;math&gt;p \leq x \leq 15&lt;/math&gt;. Determine the minimum value taken by &lt;math&gt;f(x)&lt;/math&gt; by &lt;math&gt;x&lt;/math&gt; in the interval &lt;math&gt;0 &lt; x \leq 15&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> What is the product of the real roots of the equation &lt;math&gt;x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is &lt;math&gt;\sqrt{50}&lt;/math&gt; cm, the length of &lt;math&gt;AB&lt;/math&gt; is 6 cm, and that of &lt;math&gt;BC&lt;/math&gt; is 2 cm. The angle &lt;math&gt;ABC&lt;/math&gt; is a right angle. Find the square of the distance (in centimeters) from &lt;math&gt;B&lt;/math&gt; to the center of the circle.<br /> <br /> &lt;asy&gt;<br /> size(150); defaultpen(linewidth(0.6)+fontsize(11));<br /> real r=10;<br /> pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br /> path P=circle(O,r);<br /> C=intersectionpoint(B--(B.x+r,B.y),P);<br /> draw(P);<br /> draw(C--B--A);<br /> dot(O); dot(A); dot(B); dot(C);<br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$A$&quot;,A,NE);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,SE);&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Suppose that the sum of the squares of two complex numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;7&lt;/math&gt; and the sum of the cubes is &lt;math&gt;10&lt;/math&gt;. What is the largest real value that &lt;math&gt;x + y&lt;/math&gt; can have?<br /> <br /> [[1983 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;a_n&lt;/math&gt; equal &lt;math&gt;6^{n}+8^{n}&lt;/math&gt;. Determine the remainder upon dividing &lt;math&gt;a_ {83}&lt;/math&gt; by &lt;math&gt;49&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let &lt;math&gt;P&lt;/math&gt; be the probability that at least two of the three had been sitting next to each other. If &lt;math&gt;P&lt;/math&gt; is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?<br /> <br /> [[1983 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> What is the largest 2-digit prime factor of the integer &lt;math&gt;{200\choose 100}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Find the minimum value of &lt;math&gt;\frac{9x^2\sin^2 x + 4}{x\sin x}&lt;/math&gt; for &lt;math&gt;0 &lt; x &lt; \pi&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The numbers &lt;math&gt;1337&lt;/math&gt;, &lt;math&gt;1005&lt;/math&gt;, and &lt;math&gt;1231&lt;/math&gt; have something in common. Each is a four-digit number beginning with &lt;math&gt;1&lt;/math&gt; that has exactly two identical digits. How many such numbers are there?<br /> <br /> [[1983 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The solid shown has a square base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is parallel to the base and has length &lt;math&gt;2s&lt;/math&gt;. All edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> size(170);<br /> pathpen = black+linewidth(0.65);<br /> pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C); <br /> draw(E--F);<br /> label(&quot;A&quot;,A, S);<br /> label(&quot;B&quot;,B, S);<br /> label(&quot;C&quot;,C, S);<br /> label(&quot;D&quot;,D, S);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The length of diameter &lt;math&gt;AB&lt;/math&gt; is a two digit integer. Reversing the digits gives the length of a perpendicular chord &lt;math&gt;CD&lt;/math&gt;. The distance from their intersection point &lt;math&gt;H&lt;/math&gt; to the center &lt;math&gt;O&lt;/math&gt; is a positive rational number. Determine the length of &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> &lt;asy&gt;pointpen=black; pathpen=black+linewidth(0.65);<br /> pair O=(0,0),A=(-65/2,0),B=(65/2,0);<br /> pair H=(-((65/2)^2-28^2)^.5,0),C=(H.x,28),D=(H.x,-28);<br /> D(CP(O,A));D(MP(&quot;A&quot;,A,W)--MP(&quot;B&quot;,B,E));D(MP(&quot;C&quot;,C,N)--MP(&quot;D&quot;,D));<br /> dot(MP(&quot;H&quot;,H,SE));dot(MP(&quot;O&quot;,O,SE));&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> For &lt;math&gt;\{1, 2, 3, \ldots, n\}&lt;/math&gt; and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for &lt;math&gt;\{1, 2, 3, 6,9\}&lt;/math&gt; is &lt;math&gt;9-6+3-2+1=6&lt;/math&gt; and for &lt;math&gt;\{5\}&lt;/math&gt; it is simply &lt;math&gt;5&lt;/math&gt;. Find the sum of all such alternating sums for &lt;math&gt;n=7&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> In the adjoining figure, two circles with radii &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are drawn with their centers &lt;math&gt;12&lt;/math&gt; units apart. At &lt;math&gt;P&lt;/math&gt;, one of the points of intersection, a line is drawn in such a way that the chords &lt;math&gt;QP&lt;/math&gt; and &lt;math&gt;PR&lt;/math&gt; have equal length. (&lt;math&gt;P&lt;/math&gt; is the midpoint of &lt;math&gt;QR&lt;/math&gt;) Find the square of the length of &lt;math&gt;QP&lt;/math&gt;. <br /> <br /> &lt;!-- [[Image:1983_AIME-14.png]] --&gt;<br /> &lt;asy&gt;size(160);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=3;<br /> pair O1=(0,0), O2=(12,0);<br /> path C1=Circle(O1,8), C2=Circle(O2,6);<br /> pair P=intersectionpoints(C1,C2);<br /> path C3=Circle(P,sqrt(130));<br /> pair Q=intersectionpoints(C3,C1);<br /> pair R=intersectionpoints(C3,C2);<br /> draw(C1);<br /> draw(C2);<br /> draw(O2--O1);<br /> dot(O1);<br /> dot(O2);<br /> draw(Q--R);<br /> label(&quot;$Q$&quot;,Q,NW);<br /> label(&quot;$P$&quot;,P,1.5*dir(80));<br /> label(&quot;$R$&quot;,R,NE);<br /> label(&quot;12&quot;,waypoint(O1--O2,0.4),S);&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this fraction is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> &lt;asy&gt;size(140);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png]] --&gt;<br /> <br /> [[1983 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 /> {{MAA Notice}}<br /> [[Category:AIME Problems]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems&diff=71489 1983 AIME Problems 2015-08-08T17:30:41Z <p>Npip99: /* Problem 15 */</p> <hr /> <div>{{AIME Problems|year=1983}}<br /> <br /> == Problem 1 ==<br /> Let &lt;math&gt;x&lt;/math&gt;,&lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt;, and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_xw=24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt;, and &lt;math&gt;\log_{xyz}w=12&lt;/math&gt;. Find &lt;math&gt;\log_zw&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;f(x)=|x-p|+|x-15|+|x-p-15|&lt;/math&gt;, where &lt;math&gt;p \leq x \leq 15&lt;/math&gt;. Determine the minimum value taken by &lt;math&gt;f(x)&lt;/math&gt; by &lt;math&gt;x&lt;/math&gt; in the interval &lt;math&gt;0 &lt; x \leq 15&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> What is the product of the real roots of the equation &lt;math&gt;x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is &lt;math&gt;\sqrt{50}&lt;/math&gt; cm, the length of &lt;math&gt;AB&lt;/math&gt; is 6 cm, and that of &lt;math&gt;BC&lt;/math&gt; is 2 cm. The angle &lt;math&gt;ABC&lt;/math&gt; is a right angle. Find the square of the distance (in centimeters) from &lt;math&gt;B&lt;/math&gt; to the center of the circle.<br /> <br /> &lt;asy&gt;<br /> size(150); defaultpen(linewidth(0.6)+fontsize(11));<br /> real r=10;<br /> pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br /> path P=circle(O,r);<br /> C=intersectionpoint(B--(B.x+r,B.y),P);<br /> draw(P);<br /> draw(C--B--A);<br /> dot(O); dot(A); dot(B); dot(C);<br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$A$&quot;,A,NE);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,SE);&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Suppose that the sum of the squares of two complex numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;7&lt;/math&gt; and the sum of the cubes is &lt;math&gt;10&lt;/math&gt;. What is the largest real value that &lt;math&gt;x + y&lt;/math&gt; can have?<br /> <br /> [[1983 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;a_n&lt;/math&gt; equal &lt;math&gt;6^{n}+8^{n}&lt;/math&gt;. Determine the remainder upon dividing &lt;math&gt;a_ {83}&lt;/math&gt; by &lt;math&gt;49&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let &lt;math&gt;P&lt;/math&gt; be the probability that at least two of the three had been sitting next to each other. If &lt;math&gt;P&lt;/math&gt; is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?<br /> <br /> [[1983 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> What is the largest 2-digit prime factor of the integer &lt;math&gt;{200\choose 100}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Find the minimum value of &lt;math&gt;\frac{9x^2\sin^2 x + 4}{x\sin x}&lt;/math&gt; for &lt;math&gt;0 &lt; x &lt; \pi&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The numbers &lt;math&gt;1337&lt;/math&gt;, &lt;math&gt;1005&lt;/math&gt;, and &lt;math&gt;1231&lt;/math&gt; have something in common. Each is a four-digit number beginning with &lt;math&gt;1&lt;/math&gt; that has exactly two identical digits. How many such numbers are there?<br /> <br /> [[1983 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The solid shown has a square base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is parallel to the base and has length &lt;math&gt;2s&lt;/math&gt;. All edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> size(170);<br /> pathpen = black+linewidth(0.65);<br /> pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C); <br /> draw(E--F);<br /> label(&quot;A&quot;,A, S);<br /> label(&quot;B&quot;,B, S);<br /> label(&quot;C&quot;,C, S);<br /> label(&quot;D&quot;,D, S);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The length of diameter &lt;math&gt;AB&lt;/math&gt; is a two digit integer. Reversing the digits gives the length of a perpendicular chord &lt;math&gt;CD&lt;/math&gt;. The distance from their intersection point &lt;math&gt;H&lt;/math&gt; to the center &lt;math&gt;O&lt;/math&gt; is a positive rational number. Determine the length of &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> &lt;asy&gt;pointpen=black; pathpen=black+linewidth(0.65);<br /> pair O=(0,0),A=(-65/2,0),B=(65/2,0);<br /> pair H=(-((65/2)^2-28^2)^.5,0),C=(H.x,28),D=(H.x,-28);<br /> D(CP(O,A));D(MP(&quot;A&quot;,A,W)--MP(&quot;B&quot;,B,E));D(MP(&quot;C&quot;,C,N)--MP(&quot;D&quot;,D));<br /> dot(MP(&quot;H&quot;,H,SE));dot(MP(&quot;O&quot;,O,SE));&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> For &lt;math&gt;\{1, 2, 3, \ldots, n\}&lt;/math&gt; and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for &lt;math&gt;\{1, 2, 3, 6,9\}&lt;/math&gt; is &lt;math&gt;9-6+3-2+1=6&lt;/math&gt; and for &lt;math&gt;\{5\}&lt;/math&gt; it is simply &lt;math&gt;5&lt;/math&gt;. Find the sum of all such alternating sums for &lt;math&gt;n=7&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> In the adjoining figure, two circles with radii &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are drawn with their centers &lt;math&gt;12&lt;/math&gt; units apart. At &lt;math&gt;P&lt;/math&gt;, one of the points of intersection, a line is drawn in such a way that the chords &lt;math&gt;QP&lt;/math&gt; and &lt;math&gt;PR&lt;/math&gt; have equal length. (&lt;math&gt;P&lt;/math&gt; is the midpoint of &lt;math&gt;QR&lt;/math&gt;) Find the square of the length of &lt;math&gt;QP&lt;/math&gt;. <br /> <br /> &lt;!-- [[Image:1983_AIME-14.png]] --&gt;<br /> &lt;asy&gt;size(160);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=3;<br /> pair O1=(0,0), O2=(12,0);<br /> path C1=Circle(O1,8), C2=Circle(O2,6);<br /> pair P=intersectionpoints(C1,C2);<br /> path C3=Circle(P,sqrt(130));<br /> pair Q=intersectionpoints(C3,C1);<br /> pair R=intersectionpoints(C3,C2);<br /> draw(C1);<br /> draw(C2);<br /> draw(O2--O1);<br /> dot(O1);<br /> dot(O2);<br /> draw(Q--R);<br /> label(&quot;$Q$&quot;,Q,NW);<br /> label(&quot;$P$&quot;,P,1.5*dir(80));<br /> label(&quot;$R$&quot;,R,NE);<br /> label(&quot;12&quot;,waypoint(O1--O2,0.4),S);&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this fraction is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> &lt;!-- [[Image:1983_AIME-15.png]] --&gt;<br /> <br /> &lt;asy&gt;size(160);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> <br /> [[1983 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 /> {{MAA Notice}}<br /> [[Category:AIME Problems]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_24&diff=71486 2003 AMC 12B Problems/Problem 24 2015-08-08T04:30:36Z <p>Npip99: /* Solution */</p> <hr /> <div>== Problem ==<br /> Positive integers &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are chosen so that &lt;math&gt;a&lt;b&lt;c&lt;/math&gt;, and the system of [[equation]]s<br /> &lt;center&gt;&lt;math&gt;2x + y = 2003 \quad&lt;/math&gt; and &lt;math&gt;\quad y = |x-a| + |x-b| + |x-c|&lt;/math&gt;&lt;/center&gt;<br /> has exactly one solution. What is the minimum value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 668<br /> \qquad\mathrm{(B)}\ 669<br /> \qquad\mathrm{(C)}\ 1002<br /> \qquad\mathrm{(D)}\ 2003<br /> \qquad\mathrm{(E)}\ 2004&lt;/math&gt;<br /> == Solution 1 ==<br /> <br /> Consider the graph of &lt;math&gt;f(x)=|x-a|+|x-b|+|x-c|&lt;/math&gt;.<br /> <br /> When &lt;math&gt;x&lt;a&lt;/math&gt;, the slope is &lt;math&gt;-3&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a&lt;x&lt;b&lt;/math&gt;, the slope is &lt;math&gt;-1&lt;/math&gt;.<br /> <br /> When &lt;math&gt;b&lt;x&lt;c&lt;/math&gt;, the slope is &lt;math&gt;1&lt;/math&gt;.<br /> <br /> When &lt;math&gt;c&lt;x&lt;/math&gt;, the slope is &lt;math&gt;3&lt;/math&gt;.<br /> <br /> Setting &lt;math&gt;x=b&lt;/math&gt; gives &lt;math&gt;y=|b-a|+|b-b|+|b-c|=c-a&lt;/math&gt;, so &lt;math&gt;(b,c-a)&lt;/math&gt; is a point on &lt;math&gt;f(x)&lt;/math&gt;. In fact, it is the minimum of &lt;math&gt;f(x)&lt;/math&gt; considering the slope of lines to the left and right of &lt;math&gt;(b,c-a)&lt;/math&gt;. Thus, graphing this will produce a figure that looks like a cup:<br /> &lt;asy&gt;<br /> import graph;<br /> size(100);<br /> draw((0,6)--(3,0));<br /> xaxis(0,8.5);<br /> yaxis(0,10);<br /> real f(real x)<br /> {<br /> return -3(x-2)+5;<br /> }<br /> real f2(real x)<br /> {<br /> return -1(x-2)+5;<br /> }<br /> real f3(real x)<br /> {<br /> return 1(x-4)+3;<br /> }<br /> real f4(real x)<br /> {<br /> return 3(x-7)+6;<br /> }<br /> draw(graph(f,0,2));<br /> draw(graph(f2,2,4));<br /> draw(graph(f3,4,7));<br /> draw(graph(f4,7,8.5));<br /> draw((2,-0.25)--(2,0.25));<br /> label(&quot;a&quot;,(2,0),N);<br /> draw((4,-0.25)--(4,0.25));<br /> label(&quot;b&quot;,(4,0),N);<br /> draw((7,-0.25)--(7,0.25));<br /> label(&quot;c&quot;,(7,0),N);<br /> dot((2,5));<br /> label(&quot;P&quot;,(1.9,5.2),E);<br /> &lt;/asy&gt;<br /> From the graph, it is clear that &lt;math&gt;f(x)&lt;/math&gt; and &lt;math&gt;2x+y=2003&lt;/math&gt; have one intersection point if and only if they intersect at &lt;math&gt;x=a&lt;/math&gt;. Since the line where &lt;math&gt;a&lt;x&lt;b&lt;/math&gt; has slope &lt;math&gt;-1&lt;/math&gt;, the positive difference in &lt;math&gt;y&lt;/math&gt;-coordinates from &lt;math&gt;x=a&lt;/math&gt; to &lt;math&gt;x=b&lt;/math&gt; must be &lt;math&gt;b-a&lt;/math&gt;. Together with the fact that &lt;math&gt;(b,c-a)&lt;/math&gt; is on &lt;math&gt;f(x)&lt;/math&gt;, we see that &lt;math&gt;P=(a,c-a+b-a)&lt;/math&gt;. Since this point is on &lt;math&gt;x=a&lt;/math&gt;, the only intersection point with &lt;math&gt;2x+y=2003&lt;/math&gt;, we have &lt;math&gt;2 \cdot a+(b+c-2a)=2003 \implies a+b=2003&lt;/math&gt;. As &lt;math&gt;c&gt;b&lt;/math&gt;, the smallest possible value of &lt;math&gt;c&lt;/math&gt; occurs when &lt;math&gt;b=1001&lt;/math&gt; and &lt;math&gt;c=1002&lt;/math&gt;. This is indeed a solution as &lt;math&gt;a=1000&lt;/math&gt; puts &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;y=2003-2x&lt;/math&gt;, and thus the answer is &lt;math&gt;\boxed{\mathrm{(C)}\ 1002}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> === Step 1: Finding some promising bound ===<br /> <br /> Does the system have a solution where &lt;math&gt;x\leq a&lt;/math&gt;? <br /> <br /> For such a solution we would have &lt;math&gt;y=(a-x)+(b-x)+(c-x)&lt;/math&gt;, hence &lt;math&gt;2x+(a+b+c-3x)=2003&lt;/math&gt;, which solves to &lt;math&gt;x=a+b+c-2003&lt;/math&gt;.<br /> If we want to avoid this solution, we need to have &lt;math&gt;a+b+c-2003&gt;a&lt;/math&gt;, hence &lt;math&gt;b+c&gt;2003&lt;/math&gt;, hence &lt;math&gt;c\geq 1003&lt;/math&gt;. <br /> In other words, if &lt;math&gt;c&lt;1003&lt;/math&gt;, there will always be one solution &lt;math&gt;(x,y)&lt;/math&gt; such that &lt;math&gt;x\leq a&lt;/math&gt;. <br /> <br /> === Step 2: Showing one solution ===<br /> <br /> We will now find out whether there is a &lt;math&gt;c&lt;1003&lt;/math&gt; for which (and some &lt;math&gt;a,b&lt;/math&gt;) the system has only one solution. We already know of one such solution, so we need to make sure that no other solution appears.<br /> <br /> Obviously, there are three more theoretically possible solutions: one &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\left(a,b\right]&lt;/math&gt;, one in &lt;math&gt;\left(b,c\right]&lt;/math&gt;, and one in &lt;math&gt;\left(c,\infty\right)&lt;/math&gt;.<br /> The first case solves to &lt;math&gt;x=2003+a-b-c&lt;/math&gt;, the second to &lt;math&gt;3x=2003+a+b-c&lt;/math&gt;, and the third to &lt;math&gt;5x=2003+a+b+c&lt;/math&gt;.<br /> We need to make sure that the following three conditions hold:<br /> # &lt;math&gt;2003+a-b-c\not\in\left(a,b\right]&lt;/math&gt;<br /> # &lt;math&gt;\frac{2003+a+b-c}3\not\in \left(b,c\right]&lt;/math&gt;<br /> # &lt;math&gt;\frac{2003+a+b+c}5\not\in\left(c,\infty\right)&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;c=1002&lt;/math&gt; and &lt;math&gt;b=1001&lt;/math&gt;. We then have:<br /> # &lt;math&gt;2003+a-b-c=a&lt;/math&gt;<br /> # &lt;math&gt;\frac{2003+a+b-c}3 = \frac{2002+a}3 \leq \frac{2002+1000}3 &lt; 1001 = b&lt;/math&gt;<br /> # &lt;math&gt;\frac{2003+a+b+c}5 = \frac{4006+a}5 \leq \frac{4006+1006}5 &lt; 1002 = a&lt;/math&gt;<br /> <br /> Hence for &lt;math&gt;c=1002&lt;/math&gt;, &lt;math&gt;b=1001&lt;/math&gt; and any valid &lt;math&gt;a&lt;/math&gt; the system has exactly one solution &lt;math&gt;(x,y)=(a,2003-2a)&lt;/math&gt;.<br /> <br /> === Step 3: Proving the optimality of our solution ===<br /> <br /> We will now show that for &lt;math&gt;c&lt;1002&lt;/math&gt; the system always has a solution such that &lt;math&gt;x&gt;a&lt;/math&gt;. This will mean that the system has at least two solutions, and thus the solution with &lt;math&gt;c=1002&lt;/math&gt; is optimal.<br /> <br /> # As we are looking for a &lt;math&gt;c&lt;1002&lt;/math&gt;, we have &lt;math&gt;b+c\leq 2001&lt;/math&gt;, hence &lt;math&gt;2003+a-b-c &gt; a&lt;/math&gt;. To make sure that the value falls outside &lt;math&gt;\left(a,b\right]&lt;/math&gt;, we need to make it larger than &lt;math&gt;b&lt;/math&gt;, thus &lt;math&gt;2003+a-b-c &gt; b&lt;/math&gt;, or equivalently &lt;math&gt;2003+a &gt; 2b+c&lt;/math&gt;.<br /> # The condition we just derived, &lt;math&gt;2003+a &gt; 2b+c&lt;/math&gt;, can be rewritten as &lt;math&gt;2003+a+b &gt; 3b+c&lt;/math&gt;, then as &lt;math&gt;2003+a+b-c &gt; 3b&lt;/math&gt;, which becomes &lt;math&gt;\frac{2003+a+b-c}3 &gt; b&lt;/math&gt;. Thus to make sure that the second value falls outside &lt;math&gt;\left(b,c\right]&lt;/math&gt;, we need to make it larger than &lt;math&gt;c&lt;/math&gt;. The inequality &lt;math&gt;\frac{2003+a+b-c}3 &gt; c&lt;/math&gt; simplifies to &lt;math&gt;2003+a+b &gt; 4c&lt;/math&gt;.<br /> # To avoid the last solution, we must have &lt;math&gt;\frac{2003+a+b+c}5\leq c&lt;/math&gt;, which simplifies to &lt;math&gt;2003+a+b \leq 4c&lt;/math&gt;. <br /> <br /> The last two inequalities contradict each other, thus there are no &lt;math&gt;a,b,c&lt;/math&gt; that would satisfy both of them.<br /> <br /> === Conclusion ===<br /> <br /> We just showed that whenever &lt;math&gt;c&lt;1002&lt;/math&gt;, the system has at least two different solutions: one with &lt;math&gt;x\leq a&lt;/math&gt; and one with &lt;math&gt;x&gt;a&lt;/math&gt;. <br /> <br /> We also showed that for &lt;math&gt;c=1002&lt;/math&gt; there are some &lt;math&gt;a,b&lt;/math&gt; for which the system has exactly one solution. <br /> <br /> Hence the optimal value of &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;\boxed{\mathrm{(C)}\ 1002}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2003|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=1972_AHSME_Problems/Problem_24&diff=71437 1972 AHSME Problems/Problem 24 2015-08-03T01:06:27Z <p>Npip99: /* Solution */</p> <hr /> <div>A man walked a certain distance at a constant rate. If he had gone &lt;math&gt;\textstyle\frac{1}{2}&lt;/math&gt; mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone &lt;math&gt;\textstyle\frac{1}{2}&lt;/math&gt; mile per hour slower, he would have been &lt;math&gt;2\textstyle\frac{1}{2}&lt;/math&gt; hours longer on the road. The distance in miles he walked was<br /> <br /> &lt;math&gt;\textbf{(A) }13\textstyle\frac{1}{2}\qquad \textbf{(B) }15\qquad \textbf{(C) }17\frac{1}{2}\qquad \textbf{(D) }20\qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can make three equations out of the information, and since the distances are the same, we can equate these equations.<br /> <br /> &lt;math&gt;\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{1}{2})(x-\frac{1}{2})&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; is the man's rate and &lt;math&gt;t&lt;/math&gt; is the time it takes him. <br /> <br /> Looking at the first two parts of the equations, <br /> <br /> &lt;math&gt;\frac{4t}{5}(x+\frac{1}{2})=xt&lt;/math&gt;<br /> <br /> we note that we can solve for &lt;math&gt;x&lt;/math&gt;. <br /> <br /> Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x=2&lt;/math&gt;<br /> <br /> Now we look at the last two parts of the equation:<br /> <br /> &lt;math&gt;xt=(t+\frac{1}{2})(x-\frac{1}{2})&lt;/math&gt;<br /> <br /> we note that we can solve for &lt;math&gt;t&lt;/math&gt; and we get &lt;math&gt;t=\frac{15}{2}&lt;/math&gt; <br /> <br /> We want the find the distance so we just need to find &lt;math&gt;xt&lt;/math&gt; which is &lt;math&gt;\boxed{\mathrm{(B) \ } 15}&lt;/math&gt;</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_6&diff=71432 2000 AMC 12 Problems/Problem 6 2015-08-02T03:23:05Z <p>Npip99: /* Solution */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #6]] and [[2000 AMC 10 Problems|2000 AMC 10 #11]]}}<br /> <br /> == Problem ==<br /> Two different [[prime number]]s between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is &lt;math&gt;(13)(17)-(13+17) = 221 - 30 = 191&lt;/math&gt;. Thus, we can eliminate E. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is &lt;math&gt;(5)(7)-(5+7) = 23&lt;/math&gt;. Therefore, A cannot be an answer. So, the answer must be &lt;math&gt;\mathrm{(C)}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2000|num-b=10|num-a=12}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems/Problem_25&diff=71395 2007 AMC 10B Problems/Problem 25 2015-07-30T14:47:57Z <p>Npip99: /* Solution 1 */</p> <hr /> <div>How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and:<br /> <br /> &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; <br /> <br /> is an integer?<br /> <br /> &lt;math&gt; \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} &lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Getting common denominators, we have to find coprime &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;9ab|9a^2+14b^2&lt;/math&gt;. Clearly, &lt;math&gt;3|b&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are coprime, &lt;math&gt;a|9a^2+14b^2 \implies a|14&lt;/math&gt;. Similarly, &lt;math&gt;b|9&lt;/math&gt;. However, &lt;math&gt;b&lt;/math&gt; cannot be &lt;math&gt;9&lt;/math&gt; as &lt;math&gt;81a|81 \cdot 14 + 9a^2&lt;/math&gt; only has solutions when &lt;math&gt;3|a&lt;/math&gt;. Therefore, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;a \in \{1,2,7,14\}&lt;/math&gt;. Checking them all (Or noting that &lt;math&gt;4&lt;/math&gt; is the smallest answer choice), we see that they work and the answer is &lt;math&gt;\boxed{\mathrm{(A) \ } 4}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;x = \frac{a}{b}&lt;/math&gt;. We can then write the given expression as &lt;math&gt;x+\frac{14}{9x} = k&lt;/math&gt; where &lt;math&gt;k&lt;/math&gt; is an integer. We can rewrite this as a quadratic, &lt;math&gt;9x^2 - 9kx + 14 = 0&lt;/math&gt;. By the Quadratic Formula, &lt;math&gt;x = \frac{9k\pm\sqrt{81k^2-504}}{18} = \frac{k}{2}\pm\frac{\sqrt{9k^2-56}}{6}&lt;/math&gt;. We know that &lt;math&gt;x&lt;/math&gt; must be rational, so &lt;math&gt;9k^2-56&lt;/math&gt; must be a perfect square. Let &lt;math&gt;9k^2-56 = n^2&lt;/math&gt;. Then, &lt;math&gt;56 = 9k^2-n^2 = (3k - n)(3k + n)&lt;/math&gt;. The factors pairs of &lt;math&gt;56&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;56&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;28&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;14&lt;/math&gt;, and &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. Only &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;28&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;14&lt;/math&gt; give integer solutions, &lt;math&gt;k = 5&lt;/math&gt; and &lt;math&gt;n = 13&lt;/math&gt; and &lt;math&gt;k = 3&lt;/math&gt; and &lt;math&gt;n = 5&lt;/math&gt;, respectively. Plugging these back into the original equation, we get &lt;math&gt;\boxed{\mathrm{(A) \ } 4}&lt;/math&gt; possibilities for &lt;math&gt;x&lt;/math&gt;, namely &lt;math&gt;\frac{1}{3}, \frac{14}{3}, \frac{2}{3},&lt;/math&gt; and &lt;math&gt;\frac{7}{3}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2007|ab=B|num-b=24|after=Last question}}<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems/Problem_25&diff=71394 2007 AMC 10B Problems/Problem 25 2015-07-30T14:28:24Z <p>Npip99: /* Solution */</p> <hr /> <div>How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and:<br /> <br /> &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; <br /> <br /> is an integer?<br /> <br /> &lt;math&gt; \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} &lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Getting common denominators, we have to find coprime &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;9ab|9a^2+14b^2&lt;/math&gt;. Clearly, &lt;math&gt;3|b&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are coprime, &lt;math&gt;a|9a^2+14b^2 \implies a|14&lt;/math&gt;. Similarly, &lt;math&gt;b|9&lt;/math&gt;. However, &lt;math&gt;b&lt;/math&gt; cannot be &lt;math&gt;9&lt;/math&gt; as &lt;math&gt;81a|81 \cdot 14 + 9a^2&lt;/math&gt; only has solutions when &lt;math&gt;3|a&lt;/math&gt;. Therefore, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;a \in \{1,2,7,14\}&lt;/math&gt;. Checking them all, we see that they work and the answer is &lt;math&gt;\boxed{\mathrm{(A) \ } 4}&lt;/math&gt;.<br /> === Solution 2 ===<br /> Let &lt;math&gt;x = \frac{a}{b}&lt;/math&gt;. We can then write the given expression as &lt;math&gt;x+\frac{14}{9x} = k&lt;/math&gt; where &lt;math&gt;k&lt;/math&gt; is an integer. We can rewrite this as a quadratic, &lt;math&gt;9x^2 - 9kx + 14 = 0&lt;/math&gt;. By the Quadratic Formula, &lt;math&gt;x = \frac{9k\pm\sqrt{81k^2-504}}{18} = \frac{k}{2}\pm\frac{\sqrt{9k^2-56}}{6}&lt;/math&gt;. We know that &lt;math&gt;x&lt;/math&gt; must be rational, so &lt;math&gt;9k^2-56&lt;/math&gt; must be a perfect square. Let &lt;math&gt;9k^2-56 = n^2&lt;/math&gt;. Then, &lt;math&gt;56 = 9k^2-n^2 = (3k - n)(3k + n)&lt;/math&gt;. The factors pairs of &lt;math&gt;56&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;56&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;28&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;14&lt;/math&gt;, and &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. Only &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;28&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;14&lt;/math&gt; give integer solutions, &lt;math&gt;k = 5&lt;/math&gt; and &lt;math&gt;n = 13&lt;/math&gt; and &lt;math&gt;k = 3&lt;/math&gt; and &lt;math&gt;n = 5&lt;/math&gt;, respectively. Plugging these back into the original equation, we get &lt;math&gt;\boxed{\mathrm{(A) \ } 4}&lt;/math&gt; possibilities for &lt;math&gt;x&lt;/math&gt;, namely &lt;math&gt;\frac{1}{3}, \frac{14}{3}, \frac{2}{3},&lt;/math&gt; and &lt;math&gt;\frac{7}{3}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2007|ab=B|num-b=24|after=Last question}}<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems&diff=71391 2006 AMC 10B Problems 2015-07-30T12:55:13Z <p>Npip99: /* Problem 22 */</p> <hr /> <div>== Problem 1 ==<br /> What is &lt;math&gt; (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} &lt;/math&gt; ?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -2006\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } 2006 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> For real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, define &lt;math&gt; x \mathop{\spadesuit} y = (x+y)(x-y) &lt;/math&gt;. What is &lt;math&gt; 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 17\qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 24 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A &lt;math&gt; 2 \times 3 &lt;/math&gt; rectangle and a &lt;math&gt; 3 \times 4 &lt;/math&gt; rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 49\qquad \mathrm{(E) \ } 64 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> A region is bounded by semicircular arcs constructed on the side of a square whose sides measure &lt;math&gt; \frac{2}{\pi} &lt;/math&gt;, as shown. What is the perimeter of this region? <br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(.8);<br /> <br /> filldraw( circle( (0,1), 1 ), lightgray, black );<br /> filldraw( circle( (0,-1), 1 ), lightgray, black );<br /> filldraw( circle( (1,0), 1 ), lightgray, black );<br /> filldraw( circle( (-1,0), 1 ), lightgray, black );<br /> filldraw( (-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle, lightgray, black );<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{4}{\pi}\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } \frac{8}{\pi}\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } \frac{16}{\pi} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Which of the folowing is equivalent to &lt;math&gt; \sqrt{\frac{x}{1-\frac{x-1}{x}}} &lt;/math&gt; when &lt;math&gt; x &lt; 0 &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? <br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(.8);<br /> <br /> draw( (-sqrt(5),0) -- (sqrt(5),0), dashed );<br /> draw( (-1,0)--(-1,2)--(1,2)--(1,0)--cycle );<br /> draw( arc( (0,0), sqrt(5), 0, 180 ) );<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 129\qquad \mathrm{(B) \ } 137\qquad \mathrm{(C) \ } 174\qquad \mathrm{(D) \ } 233\qquad \mathrm{(E) \ } 411 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 43\qquad \mathrm{(B) \ } 44\qquad \mathrm{(C) \ } 45\qquad \mathrm{(D) \ } 46\qquad \mathrm{(E) \ } 47 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> What is the tens digit in the sum &lt;math&gt; 7!+8!+9!+...+2006!&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The lines &lt;math&gt; x=\frac{1}{4}y+a &lt;/math&gt; and &lt;math&gt; y=\frac{1}{4}x+b &lt;/math&gt; intersect at the point &lt;math&gt; (1,2) &lt;/math&gt;. What is &lt;math&gt; a+b &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Joe and JoAnn each bought 12 ounces of coffee in a 16 ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{6}{7}\qquad \mathrm{(B) \ } \frac{13}{14}\qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ } \frac{14}{13}\qquad \mathrm{(E) \ } \frac{7}{6} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the roots of the equation &lt;math&gt; x^2-mx+2=0 &lt;/math&gt;. Suppose that &lt;math&gt; a+(1/b) &lt;/math&gt; and &lt;math&gt; b+(1/a) &lt;/math&gt; are the roots of the equation &lt;math&gt; x^2-px+q=0 &lt;/math&gt;. What is &lt;math&gt;q&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{5}{2}\qquad \mathrm{(B) \ } \frac{7}{2}\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } \frac{9}{2}\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Rhombus &lt;math&gt;ABCD&lt;/math&gt; is similar to rhombus &lt;math&gt;BFDE&lt;/math&gt;. The area of rhombus &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; and &lt;math&gt; \angle BAD = 60^\circ &lt;/math&gt;. What is the area of rhombus &lt;math&gt;BFDE&lt;/math&gt;? <br /> <br /> &lt;asy&gt;<br /> unitsize(3cm);<br /> defaultpen(.8);<br /> <br /> pair A=(0,0), B=(1,0), D=dir(60), C=B+D;<br /> <br /> draw(A--B--C--D--cycle);<br /> pair Ep = intersectionpoint( B -- (B+10*dir(150)), D -- (D+10*dir(270)) );<br /> pair F = intersectionpoint( B -- (B+10*dir(90)), D -- (D+10*dir(330)) );<br /> <br /> draw(B--Ep--D--F--cycle);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,NW);<br /> label(&quot;$E$&quot;,Ep,SW);<br /> label(&quot;$F$&quot;,F,NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 4\sqrt{3}\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 6\sqrt{3} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Leap Day, February 29, 2004, occured on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \textrm{Tuesday} \qquad \mathrm{(B) \ } \textrm{Wednesday} \qquad \mathrm{(C) \ } \textrm{Thursday} \qquad \mathrm{(D) \ } \textrm{Friday} \qquad \mathrm{(E) \ } \textrm{Saturday} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> Let &lt;math&gt; a_1 , a_2 , ... &lt;/math&gt; be a sequence for which<br /> <br /> &lt;math&gt; a_1=2 &lt;/math&gt; , &lt;math&gt; a_2=3 &lt;/math&gt;, and &lt;math&gt;a_n=\frac{a_{n-1}}{a_{n-2}} &lt;/math&gt; for each positive integer &lt;math&gt; n \ge 3 &lt;/math&gt;. <br /> <br /> What is &lt;math&gt; a_{2006} &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> A circle of radius &lt;math&gt;2&lt;/math&gt; is centered at &lt;math&gt;O&lt;/math&gt;. Square &lt;math&gt;OABC&lt;/math&gt; has side length &lt;math&gt;1&lt;/math&gt;. Sides &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CB&lt;/math&gt; are extended past &lt;math&gt;B&lt;/math&gt; to meet the circle at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;, respectively. What is the area of the shaded region in the figure, which is bounded by &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;BE&lt;/math&gt;, and the minor arc connecting &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(1.5cm);<br /> defaultpen(.8);<br /> <br /> draw( circle( (0,0), 2 ) );<br /> draw( (-2,0) -- (2,0) );<br /> draw( (0,-2) -- (0,2) );<br /> <br /> pair D = intersectionpoint( circle( (0,0), 2 ), (1,0) -- (1,2) );<br /> pair Ep = intersectionpoint( circle( (0,0), 2 ), (0,1) -- (2,1) );<br /> draw( (1,0) -- D );<br /> draw( (0,1) -- Ep );<br /> <br /> filldraw( (1,1) -- arc( (0,0),Ep,D ) -- cycle, mediumgray, black );<br /> <br /> label(&quot;$O$&quot;,(0,0),SW);<br /> label(&quot;$A$&quot;,(1,0),S);<br /> label(&quot;$C$&quot;,(0,1),W);<br /> label(&quot;$B$&quot;,(1,1),SW);<br /> label(&quot;$D$&quot;,D,N);<br /> label(&quot;$E$&quot;,Ep,E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})\qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, we have &lt;math&gt;A=(6,-22)&lt;/math&gt;, &lt;math&gt;B=(2006,178)&lt;/math&gt;, &lt;math&gt;D=(8,y)&lt;/math&gt;, for some integer &lt;math&gt;y&lt;/math&gt;. What is the area of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> For a particular peculiar pair of dice, the probabilities of rolling &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt;, on each die are in the ratio &lt;math&gt;1:2:3:4:5:6&lt;/math&gt;. What is the probability of rolling a total of &lt;math&gt;7&lt;/math&gt; on the two dice? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> Elmo makes &lt;math&gt;N&lt;/math&gt; sandwiches for a fundraiser. For each sandwich he uses &lt;math&gt;B&lt;/math&gt; globs of peanut butter at &lt;math&gt;4&lt;/math&gt;¢ per glob and &lt;math&gt;J&lt;/math&gt; blobs of jam at &lt;math&gt;5&lt;/math&gt;¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is \$&lt;math&gt;2.53&lt;/math&gt;. Assume that &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;J&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; are positive integers with &lt;math&gt;N&gt;1&lt;/math&gt;. What is the cost of the jam Elmo uses to make the sandwiches?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1.05\qquad \mathrm{(B) \ } 1.25\qquad \mathrm{(C) \ } 1.45\qquad \mathrm{(D) \ } 1.65\qquad \mathrm{(E) \ } 1.85 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br /> <br /> &lt;asy&gt;<br /> unitsize(1.5cm);<br /> defaultpen(.8);<br /> <br /> pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br /> pair F = intersectionpoint( A--D, B--Ep );<br /> <br /> draw( A -- B -- C -- cycle );<br /> draw( A -- D );<br /> draw( B -- Ep );<br /> filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br /> <br /> label(&quot;$7$&quot;,(1.25,0.2));<br /> label(&quot;$7$&quot;,(2.2,0.45));<br /> label(&quot;$3$&quot;,(0.45,0.35));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> Circles with centers &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; have radii &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, respectively, and are externally tangent. Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; on the circle with center &lt;math&gt;O&lt;/math&gt; and points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; on the circle with center &lt;math&gt;P&lt;/math&gt; are such that &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; are common external tangents to the circles. What is the area of the concave hexagon &lt;math&gt;AOBCPD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(.7cm);<br /> defaultpen(.8);<br /> <br /> pair O = (0,0), P = (6,0), Q = (-6,0);<br /> pair A = intersectionpoint( arc( (-3,0), (0,0), (-6,0) ), circle( O, 2 ) );<br /> pair B = (A.x, -A.y );<br /> pair D = Q + 2*(A-Q);<br /> pair C = Q + 2*(B-Q);<br /> <br /> draw( circle(O,2) );<br /> draw( circle(P,4) );<br /> draw( (Q + 0.8*(A-Q)) -- ( Q + 2.3*(A-Q) ) );<br /> draw( (Q + 0.8*(B-Q)) -- ( Q + 2.3*(B-Q) ) );<br /> draw( A -- O -- B );<br /> draw( C -- P -- D );<br /> draw( O -- P );<br /> <br /> label(&quot;$O$&quot;,O,W);<br /> label(&quot;$P$&quot;,P,E);<br /> <br /> label(&quot;$A$&quot;,A,NNW);<br /> label(&quot;$B$&quot;,B,SSW);<br /> <br /> label(&quot;$D$&quot;,D,NNW);<br /> label(&quot;$C\$&quot;,C,SSW);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2} &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. &quot;Look, daddy!&quot; she exclaims. &quot;That number is evenly divisible by the age of each of us kids!&quot; &quot;That's right,&quot; replies Mr. Jones, &quot;and the last two digits just happen to be my age.&quot; Which of the following is &lt;b&gt;&lt;i&gt;not&lt;/i&gt;&lt;/b&gt; the age of one of Mr. Jones's children? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> [[2006 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|before=[[2006 AMC 10A Problems]]|after=[[2007 AMC 10A Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2006 AMC 10B]]<br /> * [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=143 2006 AMC B Math Jam Transcript]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=Resources_for_mathematics_competitions&diff=71063 Resources for mathematics competitions 2015-07-10T02:32:15Z <p>Npip99: /* Olympiad Problem Solvers */</p> <hr /> <div>The [[Art of Problem Solving]] hosts this [[AoPSWiki]] as well as many other online resources for students interested in [[mathematics competitions]]. Look around the AoPSWiki. Individual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources relevant to the topic.<br /> <br /> * [[Math books]]<br /> * [[Mathematics forums]]<br /> * [[Mathematics websites]]<br /> <br /> <br /> == Math competition classes ==<br /> * [[Art of Problem Solving]] hosts classes that are popular among many of the highest performing students in the United States. [http://www.artofproblemsolving.com/Classes/AoPS_C_PSeries.php AoPS Problem Series].<br /> * [[EPGY]] hosts classes for [[AMC]] students.<br /> <br /> <br /> == Math competition problems ==<br /> === Problem books ===<br /> Many mathematics competitions sell books of past competitions and solutions. These books can be great supplementary material for avid students of mathematics.<br /> * [[ARML]] has four problem books covering most ARML as well as some [[NYSML]] competitions. However, they are generally difficult to find. Some can be ordered [http://www.arml.com/books.htm here].<br /> * [[MOEMS]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MOEMS.php here] at [[AoPS]].<br /> * [[MathCounts]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php here] at [[AoPS]].<br /> * [[AMC]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php here] at [[AoPS]].<br /> * [[Mandelbrot Competition]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_Mand.php here] at [[AoPS]].<br /> * [[William Lowell Putnam Mathematical Competition | Putnam Competition]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php here] at [[AoPS]].<br /> <br /> <br /> === Problems online ===<br /> [[Art of Problem Solving]] maintains a very large database of [http://www.artofproblemsolving.com/Forum/resources.php math contest problems]. Many math contest websites include archives of past problems. The [[List of mathematics competitions]] leads to links for many of these competition homepages. Here are a few exmaples:<br /> ==== Introductory Problem Solvers ====<br /> * [[Mu Alpha Theta]].org hosts past [http://www.mualphatheta.org/National_Convention/PastTests.aspx contest problems].<br /> * Noetic Learning [http://www.noetic-learning.com/gifted/index.jsp Challenge Math] - Problem Solving for the Gifted Elementary Students .<br /> * Elias Saab's [[MathCounts]] [http://mathcounts.saab.org/ Drills page].<br /> * [[Alabama Statewide High School Mathematics Contest]] [http://mcis.jsu.edu/mathcontest/ homepage].<br /> * The [[South African Mathematics Olympiad]] [http://www.samf.ac.za/QuestionPapers.aspx here] includes many years of past problems with solutions.<br /> * [http://www.beestar.org/index.jsp?adid=106 Beestar.org] - Beestar weekly problem solving tests for grade 1 - 8<br /> <br /> ==== Intermediate Problem Solvers ====<br /> * [[AoPS]] [http://www.artofproblemsolving.com/Forum/resources.php math contest problems and solutions]<br /> * Past [[United States of America Mathematical Talent Search | USAMTS]] problems can be found at the [http://usamts.org USAMTS homepage].<br /> * The [http://www.kalva.demon.co.uk/ Kalva site] is one of the best resources for math problems on the planet.<br /> * Past [[Colorado Mathematical Olympiad]] (CMO) problems can be found at the [http://www.uccs.edu/%7Easoifer/olympiad.html CMO homepage].<br /> * Past [[International Mathematical Talent Search]] (IMTS) problems can be found [http://www.cms.math.ca/Competitions/IMTS/ here]<br /> * [https://brilliant.org/ Brilliant] is a website where one can solve problems to gain points and go to higher levels.<br /> *[https://www.clevermath.org/] Is similar to above<br /> <br /> ==== Olympiad Problem Solvers ====<br /> * [[AoPS]] [http://www.artofproblemsolving.com/Forum/resources.php math contest problems and solutions]<br /> * Past [[United States of America Mathematical Talent Search | USAMTS]] problems can be found at the [http://usamts.org USAMTS homepage].<br /> * The [http://www.kalva.demon.co.uk/ Kalva site] is one of the best resources for math problems on the planet. (Currently offline - but a few mirrors are available, e.g [https://webspace.utexas.edu/ag6823/www/www.kalva.demon.co.uk/index.html here].)<br /> * [http://www.qbyte.org/puzzles/ Nick's Mathematical Puzzles] -- Challenging problems with hints and solutions.<br /> * [[Canadian Mathematical Olympiad]] are hosted [http://www.cms.math.ca/Competitions/CMO/ here by the Canadian Mathematical Society].<br /> * [http://web.archive.org/web/20120825124642/http://pertselv.tripod.com/RusMath.html Problems of the All-Soviet-Union math competitions 1961-1986] - Many problems, no solutions. [Site no longer exists. Site has been replaced by a web capture]<br /> * Past [[International Mathematical Talent Search]] (IMTS) problems can be found [http://www.cms.math.ca/Competitions/IMTS/ here]<br /> * [http://web.archive.org/web/20091027032345/http://geocities.com/CapeCanaveral/Lab/4661/ Olympiad Math Madness] - Stacks of challenging problems, no solutions. [Site no longer exists. Site has been replaced by a web capture]<br /> <br /> == Articles ==<br /> <br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Contests.php Pros and Cons and Math Competitions] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_CultureofExpectation.php Establishing a Positive Culture of Expectation in Math Education] by [[Sister Scholastica Award]] winner Darryl Hill.<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php Stop Making Stupid Mistakes] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Questions.php Stupid Questions] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Teaching.php Learning Through Teaching]<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php How to Write a Solution] by [[Richard Rusczyk]] and [[user:MCrawford | Mathew Crawford]].<br /> <br /> == A Huge List of Links ==<br /> === AoPS Course Recommendations ===<br /> [http://www.artofproblemsolving.com/School/recommendations.php Art of Problem Solving Course Recommendations]<br /> [http://www.artofproblemsolving.com/Store/personalrec.php Still have trouble deciding which course? Ask for personal recommendations.]<br /> <br /> ===AMC 8 Preparation===<br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=42 AMC 8 Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AMC_8_Problems_and_Solutions AMC 8 Problems in the AoPS wiki]<br /> <br /> ===AMC 10/12 Preparation===<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=388108&amp;hilit=preparation How preparing for the AIME will help AMC 10/12 Score] <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396741&amp;hilit=preparation What class to take?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=387918&amp;hilit=preparation AMC 10 for AMC 12 practice]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385418&amp;hilit=preparation AMC prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384828&amp;hilit=preparation AMC 10/12 Preparation]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384747&amp;hilit=preparation AIME/AMC 10 Overlap and Preparation]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=378851&amp;hilit=preparation How to prepare for amc10 and aime?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=369849&amp;hilit=preparation Preparation for AMC 10?]<br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=43 AMC 10 Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AMC_10_Problems_and_Solutions AMC 10 Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=44 AMC 12 Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AHSME_Problems_and_Solutions AHSME (Old AMC 12) Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AMC_12_Problems_and_Solutions AMC 12 Problems in the AoPS Wiki]<br /> <br /> ===AIME Preparation===<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=397954&amp;hilit=preparation Studying to qualify for USAMO]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=400442&amp;hilit=preparation How to prepare for the AIME]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399160&amp;hilit=preparation Preparation for the AIME]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=357602&amp;hilit=preparation Using non-AIME questions to prepare for AIME]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=355918&amp;hilit=preparation Best books to prepare for AIME?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344816&amp;hilit=preparation How to improve AIME score to make JMO?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=341827&amp;hilit=preparation Preparation for AIME and USAMO]<br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=45 AIME Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AIME_Problems_and_Solutions AIME Problems in the AoPS Wiki]<br /> <br /> ===Beginning Olympiad Preparation===<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=480253 General]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=481746&amp;p=2698978 General]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401061&amp;hilit=preparation How to Prepare for USAJMO?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399023&amp;hilit=preparation USAMO preparation/doing problems]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396736&amp;hilit=preparation Easier Olympiads for USAJMO practice?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=366383&amp;hilit=preparation For the USAMO: ACoPS or Engel?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=360619&amp;hilit=preparation Olympiad problems- how to prepare]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=354103&amp;hilit=preparation USAMO/Olympiads Preparation: Where to start?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=preparation USAJMO prep]<br /> ====Bunch of General links====<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=31888&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=71008&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=79077&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=81296&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=143168&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=273572&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=294132&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=olympiad+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385092&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=397424&amp;hilit=olympiad+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401201&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401640&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=406402&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=411476&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=411476&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=419800&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=447454]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=453638&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=474960&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385654]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=420845]<br /> '''[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2379622#p2379622]'''<br /> <br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=176 USAJMO Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/USAJMO_Problems_and_Solutions USAJMO Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27 USAMO Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/USAMO_Problems_and_Solutions USAMO Problems in the AoPS Wiki]<br /> <br /> ===Middle/Advanced Olympiad Preparation===<br /> <br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38803 Practice Olympiad 1]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38804 Practice Olympiad 2]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38805 Practice Olympiad 3]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38806 Practice Olympiad Solutions]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27 USAMO Problems in the Resources Section] <br /> [http://www.artofproblemsolving.com/Wiki/index.php/USAMO_Problems_and_Solutions USAMO Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=1&amp;cid=16 IMO Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions IMO Problems in the AoPS Wiki]<br /> <br /> <br /> ===Book Links:===<br /> ====Olympiad Level====<br /> =====Free=====<br /> [http://students.imsa.edu/~tliu/Math/planegeo.pdf Plane Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38802 Hidden Discoveries -- How To]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38801 Infinity]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38816 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38817 Diophantine Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38818 More Diophantine Number Theory]<br /> <br /> =====Not Free=====<br /> [http://www.amazon.com/Plane-Euclidean-Geometry-Theory-Problems/dp/0953682366/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742080&amp;sr=1-1 Plane Euclidean Geometry: Theory and Problems]<br /> [http://www.amazon.com/Complex-Geometry-Mathematical-Association-Textbooks/dp/0883855100/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742131&amp;sr=1-1 Complex Numbers and Geometry]<br /> [http://www.amazon.com/Geometry-Complex-Numbers-Dover-Mathematics/dp/0486638308/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742156&amp;sr=1-1 Geometry of Complex Numbers]<br /> [http://www.amazon.com/Complex-Numbers-Z-Titu-Andreescu/dp/0817643265/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741912&amp;sr=1-1 Complex Numbers from A to …Z]<br /> [http://www.amazon.com/103-Trigonometry-Problems-Training-Team/dp/0817643346/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742048&amp;sr=1-1 103 Trigonometry Problems: From the Training of the USA IMO Team]<br /> [http://www.amazon.com/An-Introduction-Diophantine-Equations-Problem-Based/dp/0817645489/ref=sr_1_1?ie=UTF8&amp;qid=1338741533&amp;sr=8-1 An Introduction to Diophantine Equations: A Problem-Based Approach]<br /> [http://www.amazon.com/Introductions-Number-Theory-Inequalities-Bradley/dp/0953682382/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741653&amp;sr=1-1 Introductions to Number Theory and Inequalities]<br /> [http://www.amazon.com/104-Number-Theory-Problems-Training/dp/0817645276/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741697&amp;sr=1-1 104 Number Theory Problems: From the Training of the USA IMO Team]<br /> [http://www.amazon.com/102-Combinatorial-Problems-Titu-Andreescu/dp/0817643176/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741741&amp;sr=1-1 102 Combinatorial Problems]<br /> [http://www.amazon.com/Path-Combinatorics-Undergraduates-Counting-Strategies/dp/8181283368/ref=sr_1_2?s=books&amp;ie=UTF8&amp;qid=1338741874&amp;sr=1-2 A Path to Combinatorics for Undergraduates: Counting Strategies]<br /> [http://www.amazon.com/Mathematical-Olympiads-1972-1986-Problems-Solutions/dp/0883856344/ref=sr_1_fkmr1_1?s=books&amp;ie=UTF8&amp;qid=1338742228&amp;sr=1-1 -fkmr1 USA Mathematical Olympiads 1972-1986 Problems and Solutions]<br /> [http://www.amazon.com/s/ref=nb_sb_noss_1?url=search-alias%3Daps&amp;field-keywords=art+and+craft+of+problem+solving Art and Craft of Problem Solving]<br /> [http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191/ref=sr_1_1?ie=UTF8&amp;qid=1338865322&amp;sr=8-1 Problem Solving Strategies]<br /> <br /> ===Problem Sets===<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=1068 31 Olympiad problems about Probabilistic Method]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=30721 567 Nice and Hard Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32201 Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32093 100 Polynomial Problems]<br /> [http://http://www.artofproblemsolving.com/Forum/download/file.php?id=32270 161 Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=31329 Trigonometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32212 General all levels]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32310 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32228 Olympiad Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33993 33 Functional Equations]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33874 Induction Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33873 Induction Solutions]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32128 260 Geometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=30649 150 Geometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35398 50 Diophantine Equation Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35716 60 Geometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33026 116 Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32361 Algebraic Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33543 100 Combinatorics Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=30597 100 Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32007 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37234 Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37233 General]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33551 100 Number Theory Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33486 100 Functional Equation Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37457 Beginning/Intermediate Counting and Probability]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37628 40 Functional Equations]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35164 100 Geometric Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38538 10 Fun Unconventional Problems :)]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35831 169 Functional Equations]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38916 Triangle Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38915 Probability]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38914 Algebra]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38920 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38918 Circle Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38919 Other Geometry]<br /> <br /> '''[http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Competition_ratings Ranking of all Olympiads (Difficulty Level)]'''<br /> <br /> == See also ==<br /> <br /> * [[List of mathematics competitions]]<br /> * [[Mathematics scholarships]]<br /> * [[Science competitions]]<br /> * [[Informatics competitions]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=Resources_for_mathematics_competitions&diff=71062 Resources for mathematics competitions 2015-07-10T02:30:45Z <p>Npip99: /* Olympiad Problem Solvers */ The link is valid.</p> <hr /> <div>The [[Art of Problem Solving]] hosts this [[AoPSWiki]] as well as many other online resources for students interested in [[mathematics competitions]]. Look around the AoPSWiki. Individual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources relevant to the topic.<br /> <br /> * [[Math books]]<br /> * [[Mathematics forums]]<br /> * [[Mathematics websites]]<br /> <br /> <br /> == Math competition classes ==<br /> * [[Art of Problem Solving]] hosts classes that are popular among many of the highest performing students in the United States. [http://www.artofproblemsolving.com/Classes/AoPS_C_PSeries.php AoPS Problem Series].<br /> * [[EPGY]] hosts classes for [[AMC]] students.<br /> <br /> <br /> == Math competition problems ==<br /> === Problem books ===<br /> Many mathematics competitions sell books of past competitions and solutions. These books can be great supplementary material for avid students of mathematics.<br /> * [[ARML]] has four problem books covering most ARML as well as some [[NYSML]] competitions. However, they are generally difficult to find. Some can be ordered [http://www.arml.com/books.htm here].<br /> * [[MOEMS]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MOEMS.php here] at [[AoPS]].<br /> * [[MathCounts]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php here] at [[AoPS]].<br /> * [[AMC]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php here] at [[AoPS]].<br /> * [[Mandelbrot Competition]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_Mand.php here] at [[AoPS]].<br /> * [[William Lowell Putnam Mathematical Competition | Putnam Competition]] books are available [http://www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php here] at [[AoPS]].<br /> <br /> <br /> === Problems online ===<br /> [[Art of Problem Solving]] maintains a very large database of [http://www.artofproblemsolving.com/Forum/resources.php math contest problems]. Many math contest websites include archives of past problems. The [[List of mathematics competitions]] leads to links for many of these competition homepages. Here are a few exmaples:<br /> ==== Introductory Problem Solvers ====<br /> * [[Mu Alpha Theta]].org hosts past [http://www.mualphatheta.org/National_Convention/PastTests.aspx contest problems].<br /> * Noetic Learning [http://www.noetic-learning.com/gifted/index.jsp Challenge Math] - Problem Solving for the Gifted Elementary Students .<br /> * Elias Saab's [[MathCounts]] [http://mathcounts.saab.org/ Drills page].<br /> * [[Alabama Statewide High School Mathematics Contest]] [http://mcis.jsu.edu/mathcontest/ homepage].<br /> * The [[South African Mathematics Olympiad]] [http://www.samf.ac.za/QuestionPapers.aspx here] includes many years of past problems with solutions.<br /> * [http://www.beestar.org/index.jsp?adid=106 Beestar.org] - Beestar weekly problem solving tests for grade 1 - 8<br /> <br /> ==== Intermediate Problem Solvers ====<br /> * [[AoPS]] [http://www.artofproblemsolving.com/Forum/resources.php math contest problems and solutions]<br /> * Past [[United States of America Mathematical Talent Search | USAMTS]] problems can be found at the [http://usamts.org USAMTS homepage].<br /> * The [http://www.kalva.demon.co.uk/ Kalva site] is one of the best resources for math problems on the planet.<br /> * Past [[Colorado Mathematical Olympiad]] (CMO) problems can be found at the [http://www.uccs.edu/%7Easoifer/olympiad.html CMO homepage].<br /> * Past [[International Mathematical Talent Search]] (IMTS) problems can be found [http://www.cms.math.ca/Competitions/IMTS/ here]<br /> * [https://brilliant.org/ Brilliant] is a website where one can solve problems to gain points and go to higher levels.<br /> *[https://www.clevermath.org/] Is similar to above<br /> <br /> ==== Olympiad Problem Solvers ====<br /> * [[AoPS]] [http://www.artofproblemsolving.com/Forum/resources.php math contest problems and solutions]<br /> * Past [[United States of America Mathematical Talent Search | USAMTS]] problems can be found at the [http://usamts.org USAMTS homepage].<br /> * The [http://www.kalva.demon.co.uk/ Kalva site] is one of the best resources for math problems on the planet. (Currently offline - but a few mirrors are available, e.g [https://webspace.utexas.edu/ag6823/www/www.kalva.demon.co.uk/index.html here].)<br /> * [http://www.qbyte.org/puzzles/ Nick's Mathematical Puzzles] -- Challenging problems with hints and solutions.<br /> * [[Canadian Mathematical Olympiad]] are hosted [http://www.cms.math.ca/Competitions/CMO/ here by the Canadian Mathematical Society].<br /> * [http://web.archive.org/web/20120825124642/http://pertselv.tripod.com/RusMath.html Problems of the All-Soviet-Union math competitions 1961-1986] - Many problems, no solutions. [Site no longer exists. Site has been replaced by a web capture ]<br /> * Past [[International Mathematical Talent Search]] (IMTS) problems can be found [http://www.cms.math.ca/Competitions/IMTS/ here]<br /> * [http://web.archive.org/web/20091027032345/http://geocities.com/CapeCanaveral/Lab/4661/ Olympiad Math Madness] - Stacks of challenging problems, no solutions.<br /> <br /> == Articles ==<br /> <br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Contests.php Pros and Cons and Math Competitions] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_CultureofExpectation.php Establishing a Positive Culture of Expectation in Math Education] by [[Sister Scholastica Award]] winner Darryl Hill.<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php Stop Making Stupid Mistakes] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Questions.php Stupid Questions] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_Teaching.php Learning Through Teaching]<br /> * [http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php How to Write a Solution] by [[Richard Rusczyk]] and [[user:MCrawford | Mathew Crawford]].<br /> <br /> == A Huge List of Links ==<br /> === AoPS Course Recommendations ===<br /> [http://www.artofproblemsolving.com/School/recommendations.php Art of Problem Solving Course Recommendations]<br /> [http://www.artofproblemsolving.com/Store/personalrec.php Still have trouble deciding which course? Ask for personal recommendations.]<br /> <br /> ===AMC 8 Preparation===<br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=42 AMC 8 Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AMC_8_Problems_and_Solutions AMC 8 Problems in the AoPS wiki]<br /> <br /> ===AMC 10/12 Preparation===<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=388108&amp;hilit=preparation How preparing for the AIME will help AMC 10/12 Score] <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396741&amp;hilit=preparation What class to take?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=387918&amp;hilit=preparation AMC 10 for AMC 12 practice]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385418&amp;hilit=preparation AMC prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384828&amp;hilit=preparation AMC 10/12 Preparation]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384747&amp;hilit=preparation AIME/AMC 10 Overlap and Preparation]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=378851&amp;hilit=preparation How to prepare for amc10 and aime?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=369849&amp;hilit=preparation Preparation for AMC 10?]<br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=43 AMC 10 Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AMC_10_Problems_and_Solutions AMC 10 Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=44 AMC 12 Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AHSME_Problems_and_Solutions AHSME (Old AMC 12) Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AMC_12_Problems_and_Solutions AMC 12 Problems in the AoPS Wiki]<br /> <br /> ===AIME Preparation===<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=397954&amp;hilit=preparation Studying to qualify for USAMO]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=400442&amp;hilit=preparation How to prepare for the AIME]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399160&amp;hilit=preparation Preparation for the AIME]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=357602&amp;hilit=preparation Using non-AIME questions to prepare for AIME]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=355918&amp;hilit=preparation Best books to prepare for AIME?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344816&amp;hilit=preparation How to improve AIME score to make JMO?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=341827&amp;hilit=preparation Preparation for AIME and USAMO]<br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=45 AIME Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/AIME_Problems_and_Solutions AIME Problems in the AoPS Wiki]<br /> <br /> ===Beginning Olympiad Preparation===<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=480253 General]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=481746&amp;p=2698978 General]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401061&amp;hilit=preparation How to Prepare for USAJMO?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399023&amp;hilit=preparation USAMO preparation/doing problems]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396736&amp;hilit=preparation Easier Olympiads for USAJMO practice?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=366383&amp;hilit=preparation For the USAMO: ACoPS or Engel?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=360619&amp;hilit=preparation Olympiad problems- how to prepare]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=354103&amp;hilit=preparation USAMO/Olympiads Preparation: Where to start?]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=preparation USAJMO prep]<br /> ====Bunch of General links====<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=31888&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=71008&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=79077&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=81296&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=143168&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=273572&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=294132&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=olympiad+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385092&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=397424&amp;hilit=olympiad+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401201&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401640&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=406402&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=411476&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=411476&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=419800&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=447454]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=453638&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=474960&amp;hilit=USAMO+prep]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385654]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=420845]<br /> '''[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2379622#p2379622]'''<br /> <br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=176 USAJMO Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/USAJMO_Problems_and_Solutions USAJMO Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27 USAMO Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/USAMO_Problems_and_Solutions USAMO Problems in the AoPS Wiki]<br /> <br /> ===Middle/Advanced Olympiad Preparation===<br /> <br /> ====Problems====<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38803 Practice Olympiad 1]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38804 Practice Olympiad 2]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38805 Practice Olympiad 3]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38806 Practice Olympiad Solutions]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27 USAMO Problems in the Resources Section] <br /> [http://www.artofproblemsolving.com/Wiki/index.php/USAMO_Problems_and_Solutions USAMO Problems in the AoPS Wiki]<br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=1&amp;cid=16 IMO Problems in the Resources Section]<br /> [http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions IMO Problems in the AoPS Wiki]<br /> <br /> <br /> ===Book Links:===<br /> ====Olympiad Level====<br /> =====Free=====<br /> [http://students.imsa.edu/~tliu/Math/planegeo.pdf Plane Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38802 Hidden Discoveries -- How To]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38801 Infinity]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38816 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38817 Diophantine Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38818 More Diophantine Number Theory]<br /> <br /> =====Not Free=====<br /> [http://www.amazon.com/Plane-Euclidean-Geometry-Theory-Problems/dp/0953682366/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742080&amp;sr=1-1 Plane Euclidean Geometry: Theory and Problems]<br /> [http://www.amazon.com/Complex-Geometry-Mathematical-Association-Textbooks/dp/0883855100/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742131&amp;sr=1-1 Complex Numbers and Geometry]<br /> [http://www.amazon.com/Geometry-Complex-Numbers-Dover-Mathematics/dp/0486638308/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742156&amp;sr=1-1 Geometry of Complex Numbers]<br /> [http://www.amazon.com/Complex-Numbers-Z-Titu-Andreescu/dp/0817643265/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741912&amp;sr=1-1 Complex Numbers from A to …Z]<br /> [http://www.amazon.com/103-Trigonometry-Problems-Training-Team/dp/0817643346/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742048&amp;sr=1-1 103 Trigonometry Problems: From the Training of the USA IMO Team]<br /> [http://www.amazon.com/An-Introduction-Diophantine-Equations-Problem-Based/dp/0817645489/ref=sr_1_1?ie=UTF8&amp;qid=1338741533&amp;sr=8-1 An Introduction to Diophantine Equations: A Problem-Based Approach]<br /> [http://www.amazon.com/Introductions-Number-Theory-Inequalities-Bradley/dp/0953682382/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741653&amp;sr=1-1 Introductions to Number Theory and Inequalities]<br /> [http://www.amazon.com/104-Number-Theory-Problems-Training/dp/0817645276/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741697&amp;sr=1-1 104 Number Theory Problems: From the Training of the USA IMO Team]<br /> [http://www.amazon.com/102-Combinatorial-Problems-Titu-Andreescu/dp/0817643176/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338741741&amp;sr=1-1 102 Combinatorial Problems]<br /> [http://www.amazon.com/Path-Combinatorics-Undergraduates-Counting-Strategies/dp/8181283368/ref=sr_1_2?s=books&amp;ie=UTF8&amp;qid=1338741874&amp;sr=1-2 A Path to Combinatorics for Undergraduates: Counting Strategies]<br /> [http://www.amazon.com/Mathematical-Olympiads-1972-1986-Problems-Solutions/dp/0883856344/ref=sr_1_fkmr1_1?s=books&amp;ie=UTF8&amp;qid=1338742228&amp;sr=1-1 -fkmr1 USA Mathematical Olympiads 1972-1986 Problems and Solutions]<br /> [http://www.amazon.com/s/ref=nb_sb_noss_1?url=search-alias%3Daps&amp;field-keywords=art+and+craft+of+problem+solving Art and Craft of Problem Solving]<br /> [http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191/ref=sr_1_1?ie=UTF8&amp;qid=1338865322&amp;sr=8-1 Problem Solving Strategies]<br /> <br /> ===Problem Sets===<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=1068 31 Olympiad problems about Probabilistic Method]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=30721 567 Nice and Hard Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32201 Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32093 100 Polynomial Problems]<br /> [http://http://www.artofproblemsolving.com/Forum/download/file.php?id=32270 161 Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=31329 Trigonometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32212 General all levels]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32310 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32228 Olympiad Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33993 33 Functional Equations]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33874 Induction Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33873 Induction Solutions]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32128 260 Geometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=30649 150 Geometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35398 50 Diophantine Equation Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35716 60 Geometry Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33026 116 Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32361 Algebraic Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33543 100 Combinatorics Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=30597 100 Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=32007 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37234 Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37233 General]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33551 100 Number Theory Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=33486 100 Functional Equation Problems]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37457 Beginning/Intermediate Counting and Probability]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=37628 40 Functional Equations]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35164 100 Geometric Inequalities]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38538 10 Fun Unconventional Problems :)]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=35831 169 Functional Equations]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38916 Triangle Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38915 Probability]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38914 Algebra]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38920 Number Theory]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38918 Circle Geometry]<br /> [http://www.artofproblemsolving.com/Forum/download/file.php?id=38919 Other Geometry]<br /> <br /> '''[http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Competition_ratings Ranking of all Olympiads (Difficulty Level)]'''<br /> <br /> == See also ==<br /> <br /> * [[List of mathematics competitions]]<br /> * [[Mathematics scholarships]]<br /> * [[Science competitions]]<br /> * [[Informatics competitions]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2&diff=69466 2014 USAMO Problems/Problem 2 2015-03-25T00:58:01Z <p>Npip99: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\mathbb{Z}&lt;/math&gt; be the set of integers. Find all functions &lt;math&gt;f : \mathbb{Z} \rightarrow \mathbb{Z}&lt;/math&gt; such that &lt;cmath&gt;xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))&lt;/cmath&gt; for all &lt;math&gt;x, y \in \mathbb{Z}&lt;/math&gt; with &lt;math&gt;x \neq 0&lt;/math&gt;.<br /> ==Solution==<br /> Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.<br /> <br /> Lemma 1: &lt;math&gt;f(0) = 0&lt;/math&gt;.<br /> Proof: Assume the opposite for a contradiction. Plug in &lt;math&gt;x = 2f(0)&lt;/math&gt; (because we assumed that &lt;math&gt;f(0) \neq 0&lt;/math&gt;), &lt;math&gt;y = 0&lt;/math&gt;. What you get eventually reduces to:<br /> &lt;cmath&gt;4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2&lt;/cmath&gt;<br /> which is a contradiction since the LHS is divisible by 2 but not 4.<br /> <br /> Then plug in &lt;math&gt;y = 0&lt;/math&gt; into the original equation and simplify by Lemma 1. We get:<br /> &lt;cmath&gt;x^2f(-x) = f(x)^2&lt;/cmath&gt;<br /> Then:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> x^6f(x) &amp;= x^4(-x)^2f(-(-x))\\<br /> &amp;= x^4f(-x)^2\\<br /> &amp;= f(x)^4<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;f(x)&lt;/math&gt; must be 0 or &lt;math&gt;x^2&lt;/math&gt;.<br /> <br /> Now either &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;x^2&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; or there exists &lt;math&gt;a \neq 0&lt;/math&gt; such that &lt;math&gt;f(a)=0&lt;/math&gt;. The first case gives a valid solution. In the second case, we let &lt;math&gt;y = a&lt;/math&gt; in the original equation and simplify to get:<br /> &lt;cmath&gt;xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}&lt;/cmath&gt;<br /> But we know that &lt;math&gt;xf(-x) = \frac{f(x)^2}{x}&lt;/math&gt;, so:<br /> &lt;cmath&gt;a^2f(2x) = 0&lt;/cmath&gt;<br /> Since &lt;math&gt;a&lt;/math&gt; is not 0, &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; (including 0). Now either &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, or there exists some &lt;math&gt;m \neq 0&lt;/math&gt; such that &lt;math&gt;f(m) = m^2&lt;/math&gt;. Then &lt;math&gt;m&lt;/math&gt; must be odd. We can let &lt;math&gt;x = 2k&lt;/math&gt; in the original equation, and since &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, stuff cancels and we get:<br /> &lt;cmath&gt;y^2f(4k - f(y)) = f(yf(y))&lt;/cmath&gt;<br /> [b]for &lt;math&gt;k \neq 0&lt;/math&gt;.[/b]<br /> Now, let &lt;math&gt;y = m&lt;/math&gt; and we get:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3)&lt;/cmath&gt;<br /> Now, either both sides are 0 or both are equal to &lt;math&gt;m^6&lt;/math&gt;. If both are &lt;math&gt;m^6&lt;/math&gt; then:<br /> &lt;cmath&gt;m^2(4k - m^2)^2 = m^6&lt;/cmath&gt;<br /> which simplifies to:<br /> &lt;cmath&gt;4k - m^2 = \pm m^2&lt;/cmath&gt;<br /> Since &lt;math&gt;k \neq 0&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; is odd, both cases are impossible, so we must have:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3) = 0&lt;/cmath&gt;<br /> Then we can let &lt;math&gt;k&lt;/math&gt; be anything except 0, and get &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 3 \pmod{4}&lt;/math&gt; except &lt;math&gt;-m^2&lt;/math&gt;. Also since &lt;math&gt;x^2f(-x) = f(x)^2&lt;/math&gt;, we have &lt;math&gt;f(x) = 0 \Rightarrow f(-x) = 0&lt;/math&gt;, so &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 1 \pmod{4}&lt;/math&gt; except &lt;math&gt;m^2&lt;/math&gt;. So &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; except &lt;math&gt;\pm m^2&lt;/math&gt;. Since &lt;math&gt;f(m) \neq 0&lt;/math&gt;, &lt;math&gt;m = \pm m^2&lt;/math&gt;. Squaring, &lt;math&gt;m^2 = m^4&lt;/math&gt; and dividing by &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;m = m^3&lt;/math&gt;. Since &lt;math&gt;f(m^3) = 0&lt;/math&gt;, &lt;math&gt;f(m) = 0&lt;/math&gt;, which is a contradiction, so our only solutions are &lt;math&gt;f(x) = 0&lt;/math&gt; and &lt;math&gt;f(x) = x^2&lt;/math&gt;.</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2&diff=69465 2014 USAMO Problems/Problem 2 2015-03-25T00:57:41Z <p>Npip99: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\mathbb{Z}&lt;/math&gt; be the set of integers. Find all functions &lt;math&gt;f : \mathbb{Z} \rightarrow \mathbb{Z}&lt;/math&gt; such that &lt;cmath&gt;xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))&lt;/cmath&gt; for all &lt;math&gt;x, y \in \mathbb{Z}&lt;/math&gt; with &lt;math&gt;x \neq 0&lt;/math&gt;.<br /> ==Solution==<br /> Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.<br /> <br /> Lemma 1: &lt;math&gt;f(0) = 0&lt;/math&gt;.<br /> Proof: Assume the opposite for a contradiction. Plug in &lt;math&gt;x = 2f(0)&lt;/math&gt; (because we assumed that &lt;math&gt;f(0) \neq 0&lt;/math&gt;), &lt;math&gt;y = 0&lt;/math&gt;. What you get eventually reduces to:<br /> &lt;cmath&gt;4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2&lt;/cmath&gt;<br /> which is a contradiction since the LHS is divisible by 2 but not 4.<br /> <br /> Then plug in &lt;math&gt;y = 0&lt;/math&gt; into the original equation and simplify by Lemma 1. We get:<br /> &lt;cmath&gt;x^2f(-x) = f(x)^2&lt;/cmath&gt;<br /> Then:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> x^6f(x) = x^4(-x)^2f(-(-x))\\<br /> &amp;= x^4f(-x)^2\\<br /> &amp;= f(x)^4<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;f(x)&lt;/math&gt; must be 0 or &lt;math&gt;x^2&lt;/math&gt;.<br /> <br /> Now either &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;x^2&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; or there exists &lt;math&gt;a \neq 0&lt;/math&gt; such that &lt;math&gt;f(a)=0&lt;/math&gt;. The first case gives a valid solution. In the second case, we let &lt;math&gt;y = a&lt;/math&gt; in the original equation and simplify to get:<br /> &lt;cmath&gt;xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}&lt;/cmath&gt;<br /> But we know that &lt;math&gt;xf(-x) = \frac{f(x)^2}{x}&lt;/math&gt;, so:<br /> &lt;cmath&gt;a^2f(2x) = 0&lt;/cmath&gt;<br /> Since &lt;math&gt;a&lt;/math&gt; is not 0, &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; (including 0). Now either &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, or there exists some &lt;math&gt;m \neq 0&lt;/math&gt; such that &lt;math&gt;f(m) = m^2&lt;/math&gt;. Then &lt;math&gt;m&lt;/math&gt; must be odd. We can let &lt;math&gt;x = 2k&lt;/math&gt; in the original equation, and since &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, stuff cancels and we get:<br /> &lt;cmath&gt;y^2f(4k - f(y)) = f(yf(y))&lt;/cmath&gt;<br /> [b]for &lt;math&gt;k \neq 0&lt;/math&gt;.[/b]<br /> Now, let &lt;math&gt;y = m&lt;/math&gt; and we get:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3)&lt;/cmath&gt;<br /> Now, either both sides are 0 or both are equal to &lt;math&gt;m^6&lt;/math&gt;. If both are &lt;math&gt;m^6&lt;/math&gt; then:<br /> &lt;cmath&gt;m^2(4k - m^2)^2 = m^6&lt;/cmath&gt;<br /> which simplifies to:<br /> &lt;cmath&gt;4k - m^2 = \pm m^2&lt;/cmath&gt;<br /> Since &lt;math&gt;k \neq 0&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; is odd, both cases are impossible, so we must have:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3) = 0&lt;/cmath&gt;<br /> Then we can let &lt;math&gt;k&lt;/math&gt; be anything except 0, and get &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 3 \pmod{4}&lt;/math&gt; except &lt;math&gt;-m^2&lt;/math&gt;. Also since &lt;math&gt;x^2f(-x) = f(x)^2&lt;/math&gt;, we have &lt;math&gt;f(x) = 0 \Rightarrow f(-x) = 0&lt;/math&gt;, so &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 1 \pmod{4}&lt;/math&gt; except &lt;math&gt;m^2&lt;/math&gt;. So &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; except &lt;math&gt;\pm m^2&lt;/math&gt;. Since &lt;math&gt;f(m) \neq 0&lt;/math&gt;, &lt;math&gt;m = \pm m^2&lt;/math&gt;. Squaring, &lt;math&gt;m^2 = m^4&lt;/math&gt; and dividing by &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;m = m^3&lt;/math&gt;. Since &lt;math&gt;f(m^3) = 0&lt;/math&gt;, &lt;math&gt;f(m) = 0&lt;/math&gt;, which is a contradiction, so our only solutions are &lt;math&gt;f(x) = 0&lt;/math&gt; and &lt;math&gt;f(x) = x^2&lt;/math&gt;.</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2006_Alabama_ARML_TST_Problems&diff=69464 2006 Alabama ARML TST Problems 2015-03-25T00:56:35Z <p>Npip99: /* Problem 6 */</p> <hr /> <div>==Problem 1==<br /> How many integers &lt;math&gt;x&lt;/math&gt; satisfy the inequality<br /> <br /> &lt;center&gt;&lt;math&gt;(x-2006)(x-2004)(x-2002)\cdots (x-4)&lt;0?&lt;/math&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 1 | Solution]]<br /> ==Problem 2==<br /> Compute &lt;math&gt;\sum_{k=1}^{5}\sum_{n=1}^{6}kn&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 2 | Solution]]<br /> ==Problem 3==<br /> River draws four cards from a standard 52 card deck of playing cards. Exactly 3 of them are 2’s. Find the probability River drew exactly one spade and one club from the deck.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 3 | Solution]]<br /> ==Problem 4==<br /> Find the number of six-digit positive integers for which the digits are in increasing order.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 4 | Solution]]<br /> ==Problem 5==<br /> There exist positive integers &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; with no common factor greater than 1, such that<br /> <br /> &lt;center&gt;&lt;math&gt;A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D.&lt;/math&gt;&lt;/center&gt;<br /> <br /> Find &lt;math&gt;A+B+C+D&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> Let &lt;math&gt;\lfloor a \rfloor&lt;/math&gt; be the greatest integer less than or equal to &lt;math&gt;a&lt;/math&gt; and let &lt;math&gt;\{a\}=a-\lfloor a \rfloor&lt;/math&gt;. Find &lt;math&gt;10(x+y+z)&lt;/math&gt; given that<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{align*}<br /> x+\lfloor y \rfloor +\{z\}=14.2,\\<br /> \lfloor x \rfloor+\{y\} +z=15.3,\\<br /> \{x\}+y +\lfloor z \rfloor=16.1.<br /> \end{align*}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> Four equilateral triangles are drawn such that each one shares a different side with a square of side length 10. None of the areas of the triangles overlap with the area of the square. The four vertices of the triangles that aren’t vertices of the square are connected to form a larger square. Find the area of this larger square.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> A bored mathematician has his computer calculate 1000 consecutive terms in the Fibonacci sequence. He notes that the smallest of the numbers is a multiple of 7. How many of the other 999 Fibonacci numbers are multiples of 7?<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 8 | Solution]]<br /> ==Problem 9==<br /> Amanda ordered a dozen donuts. She said she wanted only chocolate, glazed, and powdered donuts, and at least one of each kind. Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; be the number of chocolate, glazed, and powdered donuts she wound up with. Find the number of possible ordered triples &lt;math&gt;(a, b, c)&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> Let &lt;math&gt;p&lt;/math&gt; be the probability that Scooby Doo solves any given mystery. The probability that Scooby Doo solves 1800 out of 2006 given mysteries is the same as the probability that he solves 1801 of them. Find the probability that Scooby Doo solves the mystery of why Eddie Murphy decided to stop being funny.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 10 | Solution]]<br /> ==Problem 11==<br /> The integer &lt;math&gt;5^{2006}&lt;/math&gt; has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers &lt;math&gt;0\leq k \leq 2005&lt;/math&gt; does &lt;math&gt;5^k&lt;/math&gt; begin with the digit 1?<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> Yoda begins writing the positive integers starting from 1 and continuing consecutively as he writes. When he stops, he realizes that there is no set of 5 composite integers among the ones he wrote such that each pair of those 5 is relatively prime. What’s the largest possible number Yoda could have stopped on?<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 12 | Solution]]<br /> ==Problem 13==<br /> Find the sum of the solutions to the equation<br /> <br /> &lt;center&gt;&lt;math&gt;\sqrt{x+27}+\sqrt{55-x}=4.&lt;/math&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Find the real solution &lt;math&gt;(x, y)&lt;/math&gt; to the system of equations<br /> <br /> &lt;center&gt;&lt;math&gt;x^3-3xy^2=-610,&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;3x^2y-y^3=182.&lt;/math&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> Ying lives on Strangeland, a tiny planet with 4 little cities that are each 100 miles apart from each other. One day, Ying begins driving from her home city of Viavesta to the city of Havennew, which takes her about an hour. When she gets to Havennew, she decides she wants to go straight to another city on Strangeland, so she randomly chooses one of the other three cities (possibly Viavesta), and starts driving there. Ying drives like this for most of the day, making 8 total trips between cities on Strangeland, choosing randomly where to drive to next from each stop. She then stops at her final city of destination, digs a hole, and buries her car.<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; be the probability Ying buried her car in Viavesta and let q be the probability she buried it in Havennew. Find the value of &lt;math&gt;p + q&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 15 | Solution]]<br /> <br /> ==See also==<br /> * [[Alabama ARML TST]]<br /> * [[Alabama ARML]]<br /> * [[Alabama Math Contests]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2006_Alabama_ARML_TST_Problems&diff=69463 2006 Alabama ARML TST Problems 2015-03-25T00:56:08Z <p>Npip99: /* Problem 6 */</p> <hr /> <div>==Problem 1==<br /> How many integers &lt;math&gt;x&lt;/math&gt; satisfy the inequality<br /> <br /> &lt;center&gt;&lt;math&gt;(x-2006)(x-2004)(x-2002)\cdots (x-4)&lt;0?&lt;/math&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 1 | Solution]]<br /> ==Problem 2==<br /> Compute &lt;math&gt;\sum_{k=1}^{5}\sum_{n=1}^{6}kn&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 2 | Solution]]<br /> ==Problem 3==<br /> River draws four cards from a standard 52 card deck of playing cards. Exactly 3 of them are 2’s. Find the probability River drew exactly one spade and one club from the deck.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 3 | Solution]]<br /> ==Problem 4==<br /> Find the number of six-digit positive integers for which the digits are in increasing order.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 4 | Solution]]<br /> ==Problem 5==<br /> There exist positive integers &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; with no common factor greater than 1, such that<br /> <br /> &lt;center&gt;&lt;math&gt;A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D.&lt;/math&gt;&lt;/center&gt;<br /> <br /> Find &lt;math&gt;A+B+C+D&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> Let &lt;math&gt;\lfloor a \rfloor&lt;/math&gt; be the greatest integer less than or equal to &lt;math&gt;a&lt;/math&gt; and let &lt;math&gt;\{a\}=a-\lfloor a \rfloor&lt;/math&gt;. Find &lt;math&gt;10(x+y+z)&lt;/math&gt; given that<br /> <br /> &lt;center&gt;&lt;cmath&gt;\begin{align}<br /> x+\lfloor y \rfloor +\{z\}=14.2,\\<br /> \lfloor x \rfloor+\{y\} +z=15.3,\\<br /> \{x\}+y +\lfloor z \rfloor=16.1.<br /> \end{align}&lt;/cmath&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> Four equilateral triangles are drawn such that each one shares a different side with a square of side length 10. None of the areas of the triangles overlap with the area of the square. The four vertices of the triangles that aren’t vertices of the square are connected to form a larger square. Find the area of this larger square.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> A bored mathematician has his computer calculate 1000 consecutive terms in the Fibonacci sequence. He notes that the smallest of the numbers is a multiple of 7. How many of the other 999 Fibonacci numbers are multiples of 7?<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 8 | Solution]]<br /> ==Problem 9==<br /> Amanda ordered a dozen donuts. She said she wanted only chocolate, glazed, and powdered donuts, and at least one of each kind. Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; be the number of chocolate, glazed, and powdered donuts she wound up with. Find the number of possible ordered triples &lt;math&gt;(a, b, c)&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> Let &lt;math&gt;p&lt;/math&gt; be the probability that Scooby Doo solves any given mystery. The probability that Scooby Doo solves 1800 out of 2006 given mysteries is the same as the probability that he solves 1801 of them. Find the probability that Scooby Doo solves the mystery of why Eddie Murphy decided to stop being funny.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 10 | Solution]]<br /> ==Problem 11==<br /> The integer &lt;math&gt;5^{2006}&lt;/math&gt; has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers &lt;math&gt;0\leq k \leq 2005&lt;/math&gt; does &lt;math&gt;5^k&lt;/math&gt; begin with the digit 1?<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> Yoda begins writing the positive integers starting from 1 and continuing consecutively as he writes. When he stops, he realizes that there is no set of 5 composite integers among the ones he wrote such that each pair of those 5 is relatively prime. What’s the largest possible number Yoda could have stopped on?<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 12 | Solution]]<br /> ==Problem 13==<br /> Find the sum of the solutions to the equation<br /> <br /> &lt;center&gt;&lt;math&gt;\sqrt{x+27}+\sqrt{55-x}=4.&lt;/math&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Find the real solution &lt;math&gt;(x, y)&lt;/math&gt; to the system of equations<br /> <br /> &lt;center&gt;&lt;math&gt;x^3-3xy^2=-610,&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;3x^2y-y^3=182.&lt;/math&gt;&lt;/center&gt;<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> Ying lives on Strangeland, a tiny planet with 4 little cities that are each 100 miles apart from each other. One day, Ying begins driving from her home city of Viavesta to the city of Havennew, which takes her about an hour. When she gets to Havennew, she decides she wants to go straight to another city on Strangeland, so she randomly chooses one of the other three cities (possibly Viavesta), and starts driving there. Ying drives like this for most of the day, making 8 total trips between cities on Strangeland, choosing randomly where to drive to next from each stop. She then stops at her final city of destination, digs a hole, and buries her car.<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; be the probability Ying buried her car in Viavesta and let q be the probability she buried it in Havennew. Find the value of &lt;math&gt;p + q&lt;/math&gt;.<br /> <br /> [[2006 Alabama ARML TST Problems/Problem 15 | Solution]]<br /> <br /> ==See also==<br /> * [[Alabama ARML TST]]<br /> * [[Alabama ARML]]<br /> * [[Alabama Math Contests]]</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2&diff=69462 2014 USAMO Problems/Problem 2 2015-03-25T00:52:50Z <p>Npip99: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\mathbb{Z}&lt;/math&gt; be the set of integers. Find all functions &lt;math&gt;f : \mathbb{Z} \rightarrow \mathbb{Z}&lt;/math&gt; such that &lt;cmath&gt;xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))&lt;/cmath&gt; for all &lt;math&gt;x, y \in \mathbb{Z}&lt;/math&gt; with &lt;math&gt;x \neq 0&lt;/math&gt;.<br /> ==Solution==<br /> Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.<br /> <br /> Lemma 1: &lt;math&gt;f(0) = 0&lt;/math&gt;.<br /> Proof: Assume the opposite for a contradiction. Plug in &lt;math&gt;x = 2f(0)&lt;/math&gt; (because we assumed that &lt;math&gt;f(0) \neq 0&lt;/math&gt;), &lt;math&gt;y = 0&lt;/math&gt;. What you get eventually reduces to:<br /> &lt;cmath&gt;4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2&lt;/cmath&gt;<br /> which is a contradiction since the LHS is divisible by 2 but not 4.<br /> <br /> Then plug in &lt;math&gt;y = 0&lt;/math&gt; into the original equation and simplify by Lemma 1. We get:<br /> &lt;cmath&gt;x^2f(-x) = f(x)^2&lt;/cmath&gt;<br /> Then:<br /> <br /> &lt;cmath&gt;x^6f(x) = x^4(-x)^2f(-(-x))&lt;/cmath&gt;<br /> &lt;cmath&gt;= x^4f(-x)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;= f(x)^4&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;f(x)&lt;/math&gt; must be 0 or &lt;math&gt;x^2&lt;/math&gt;.<br /> <br /> Now either &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;x^2&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; or there exists &lt;math&gt;a \neq 0&lt;/math&gt; such that &lt;math&gt;f(a)=0&lt;/math&gt;. The first case gives a valid solution. In the second case, we let &lt;math&gt;y = a&lt;/math&gt; in the original equation and simplify to get:<br /> &lt;cmath&gt;xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}&lt;/cmath&gt;<br /> But we know that &lt;math&gt;xf(-x) = \frac{f(x)^2}{x}&lt;/math&gt;, so:<br /> &lt;cmath&gt;a^2f(2x) = 0&lt;/cmath&gt;<br /> Since &lt;math&gt;a&lt;/math&gt; is not 0, &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; (including 0). Now either &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, or there exists some &lt;math&gt;m \neq 0&lt;/math&gt; such that &lt;math&gt;f(m) = m^2&lt;/math&gt;. Then &lt;math&gt;m&lt;/math&gt; must be odd. We can let &lt;math&gt;x = 2k&lt;/math&gt; in the original equation, and since &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, stuff cancels and we get:<br /> &lt;cmath&gt;y^2f(4k - f(y)) = f(yf(y))&lt;/cmath&gt;<br /> [b]for &lt;math&gt;k \neq 0&lt;/math&gt;.[/b]<br /> Now, let &lt;math&gt;y = m&lt;/math&gt; and we get:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3)&lt;/cmath&gt;<br /> Now, either both sides are 0 or both are equal to &lt;math&gt;m^6&lt;/math&gt;. If both are &lt;math&gt;m^6&lt;/math&gt; then:<br /> &lt;cmath&gt;m^2(4k - m^2)^2 = m^6&lt;/cmath&gt;<br /> which simplifies to:<br /> &lt;cmath&gt;4k - m^2 = \pm m^2&lt;/cmath&gt;<br /> Since &lt;math&gt;k \neq 0&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; is odd, both cases are impossible, so we must have:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3) = 0&lt;/cmath&gt;<br /> Then we can let &lt;math&gt;k&lt;/math&gt; be anything except 0, and get &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 3 \pmod{4}&lt;/math&gt; except &lt;math&gt;-m^2&lt;/math&gt;. Also since &lt;math&gt;x^2f(-x) = f(x)^2&lt;/math&gt;, we have &lt;math&gt;f(x) = 0 \Rightarrow f(-x) = 0&lt;/math&gt;, so &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 1 \pmod{4}&lt;/math&gt; except &lt;math&gt;m^2&lt;/math&gt;. So &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; except &lt;math&gt;\pm m^2&lt;/math&gt;. Since &lt;math&gt;f(m) \neq 0&lt;/math&gt;, &lt;math&gt;m = \pm m^2&lt;/math&gt;. Squaring, &lt;math&gt;m^2 = m^4&lt;/math&gt; and dividing by &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;m = m^3&lt;/math&gt;. Since &lt;math&gt;f(m^3) = 0&lt;/math&gt;, &lt;math&gt;f(m) = 0&lt;/math&gt;, which is a contradiction, so our only solutions are &lt;math&gt;f(x) = 0&lt;/math&gt; and &lt;math&gt;f(x) = x^2&lt;/math&gt;.</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2&diff=69461 2014 USAMO Problems/Problem 2 2015-03-25T00:52:33Z <p>Npip99: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\mathbb{Z}&lt;/math&gt; be the set of integers. Find all functions &lt;math&gt;f : \mathbb{Z} \rightarrow \mathbb{Z}&lt;/math&gt; such that &lt;cmath&gt;xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))&lt;/cmath&gt; for all &lt;math&gt;x, y \in \mathbb{Z}&lt;/math&gt; with &lt;math&gt;x \neq 0&lt;/math&gt;.<br /> ==Solution==<br /> Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.<br /> <br /> Lemma 1: &lt;math&gt;f(0) = 0&lt;/math&gt;.<br /> Proof: Assume the opposite for a contradiction. Plug in &lt;math&gt;x = 2f(0)&lt;/math&gt; (because we assumed that &lt;math&gt;f(0) \neq 0&lt;/math&gt;), &lt;math&gt;y = 0&lt;/math&gt;. What you get eventually reduces to:<br /> &lt;cmath&gt;4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2&lt;/cmath&gt;<br /> which is a contradiction since the LHS is divisible by 2 but not 4.<br /> <br /> Then plug in &lt;math&gt;y = 0&lt;/math&gt; into the original equation and simplify by Lemma 1. We get:<br /> &lt;cmath&gt;x^2f(-x) = f(x)^2&lt;/cmath&gt;<br /> Then:<br /> <br /> &lt;cmath&gt;x^6f(x) &amp;= x^4(-x)^2f(-(-x))&lt;/cmath&gt;<br /> &lt;cmath&gt;= x^4f(-x)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;= f(x)^4&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;f(x)&lt;/math&gt; must be 0 or &lt;math&gt;x^2&lt;/math&gt;.<br /> <br /> Now either &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;x^2&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; or there exists &lt;math&gt;a \neq 0&lt;/math&gt; such that &lt;math&gt;f(a)=0&lt;/math&gt;. The first case gives a valid solution. In the second case, we let &lt;math&gt;y = a&lt;/math&gt; in the original equation and simplify to get:<br /> &lt;cmath&gt;xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}&lt;/cmath&gt;<br /> But we know that &lt;math&gt;xf(-x) = \frac{f(x)^2}{x}&lt;/math&gt;, so:<br /> &lt;cmath&gt;a^2f(2x) = 0&lt;/cmath&gt;<br /> Since &lt;math&gt;a&lt;/math&gt; is not 0, &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; (including 0). Now either &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, or there exists some &lt;math&gt;m \neq 0&lt;/math&gt; such that &lt;math&gt;f(m) = m^2&lt;/math&gt;. Then &lt;math&gt;m&lt;/math&gt; must be odd. We can let &lt;math&gt;x = 2k&lt;/math&gt; in the original equation, and since &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, stuff cancels and we get:<br /> &lt;cmath&gt;y^2f(4k - f(y)) = f(yf(y))&lt;/cmath&gt;<br /> [b]for &lt;math&gt;k \neq 0&lt;/math&gt;.[/b]<br /> Now, let &lt;math&gt;y = m&lt;/math&gt; and we get:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3)&lt;/cmath&gt;<br /> Now, either both sides are 0 or both are equal to &lt;math&gt;m^6&lt;/math&gt;. If both are &lt;math&gt;m^6&lt;/math&gt; then:<br /> &lt;cmath&gt;m^2(4k - m^2)^2 = m^6&lt;/cmath&gt;<br /> which simplifies to:<br /> &lt;cmath&gt;4k - m^2 = \pm m^2&lt;/cmath&gt;<br /> Since &lt;math&gt;k \neq 0&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; is odd, both cases are impossible, so we must have:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3) = 0&lt;/cmath&gt;<br /> Then we can let &lt;math&gt;k&lt;/math&gt; be anything except 0, and get &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 3 \pmod{4}&lt;/math&gt; except &lt;math&gt;-m^2&lt;/math&gt;. Also since &lt;math&gt;x^2f(-x) = f(x)^2&lt;/math&gt;, we have &lt;math&gt;f(x) = 0 \Rightarrow f(-x) = 0&lt;/math&gt;, so &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 1 \pmod{4}&lt;/math&gt; except &lt;math&gt;m^2&lt;/math&gt;. So &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; except &lt;math&gt;\pm m^2&lt;/math&gt;. Since &lt;math&gt;f(m) \neq 0&lt;/math&gt;, &lt;math&gt;m = \pm m^2&lt;/math&gt;. Squaring, &lt;math&gt;m^2 = m^4&lt;/math&gt; and dividing by &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;m = m^3&lt;/math&gt;. Since &lt;math&gt;f(m^3) = 0&lt;/math&gt;, &lt;math&gt;f(m) = 0&lt;/math&gt;, which is a contradiction, so our only solutions are &lt;math&gt;f(x) = 0&lt;/math&gt; and &lt;math&gt;f(x) = x^2&lt;/math&gt;.</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2&diff=69460 2014 USAMO Problems/Problem 2 2015-03-25T00:51:51Z <p>Npip99: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\mathbb{Z}&lt;/math&gt; be the set of integers. Find all functions &lt;math&gt;f : \mathbb{Z} \rightarrow \mathbb{Z}&lt;/math&gt; such that &lt;cmath&gt;xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))&lt;/cmath&gt; for all &lt;math&gt;x, y \in \mathbb{Z}&lt;/math&gt; with &lt;math&gt;x \neq 0&lt;/math&gt;.<br /> ==Solution==<br /> Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.<br /> <br /> Lemma 1: &lt;math&gt;f(0) = 0&lt;/math&gt;.<br /> Proof: Assume the opposite for a contradiction. Plug in &lt;math&gt;x = 2f(0)&lt;/math&gt; (because we assumed that &lt;math&gt;f(0) \neq 0&lt;/math&gt;), &lt;math&gt;y = 0&lt;/math&gt;. What you get eventually reduces to:<br /> &lt;cmath&gt;4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2&lt;/cmath&gt;<br /> which is a contradiction since the LHS is divisible by 2 but not 4.<br /> <br /> Then plug in &lt;math&gt;y = 0&lt;/math&gt; into the original equation and simplify by Lemma 1. We get:<br /> &lt;cmath&gt;x^2f(-x) = f(x)^2&lt;/cmath&gt;<br /> Then:<br /> <br /> &lt;cmath&gt;\x^6f(x) &amp;= x^4(-x)^2f(-(-x))&lt;/cmath&gt;<br /> &lt;cmath&gt;\= x^4f(-x)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\= f(x)^4&lt;/cmath&gt;<br /> <br /> Therefore, &lt;math&gt;f(x)&lt;/math&gt; must be 0 or &lt;math&gt;x^2&lt;/math&gt;.<br /> <br /> Now either &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;x^2&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; or there exists &lt;math&gt;a \neq 0&lt;/math&gt; such that &lt;math&gt;f(a)=0&lt;/math&gt;. The first case gives a valid solution. In the second case, we let &lt;math&gt;y = a&lt;/math&gt; in the original equation and simplify to get:<br /> &lt;cmath&gt;xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}&lt;/cmath&gt;<br /> But we know that &lt;math&gt;xf(-x) = \frac{f(x)^2}{x}&lt;/math&gt;, so:<br /> &lt;cmath&gt;a^2f(2x) = 0&lt;/cmath&gt;<br /> Since &lt;math&gt;a&lt;/math&gt; is not 0, &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; (including 0). Now either &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, or there exists some &lt;math&gt;m \neq 0&lt;/math&gt; such that &lt;math&gt;f(m) = m^2&lt;/math&gt;. Then &lt;math&gt;m&lt;/math&gt; must be odd. We can let &lt;math&gt;x = 2k&lt;/math&gt; in the original equation, and since &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, stuff cancels and we get:<br /> &lt;cmath&gt;y^2f(4k - f(y)) = f(yf(y))&lt;/cmath&gt;<br /> [b]for &lt;math&gt;k \neq 0&lt;/math&gt;.[/b]<br /> Now, let &lt;math&gt;y = m&lt;/math&gt; and we get:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3)&lt;/cmath&gt;<br /> Now, either both sides are 0 or both are equal to &lt;math&gt;m^6&lt;/math&gt;. If both are &lt;math&gt;m^6&lt;/math&gt; then:<br /> &lt;cmath&gt;m^2(4k - m^2)^2 = m^6&lt;/cmath&gt;<br /> which simplifies to:<br /> &lt;cmath&gt;4k - m^2 = \pm m^2&lt;/cmath&gt;<br /> Since &lt;math&gt;k \neq 0&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; is odd, both cases are impossible, so we must have:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3) = 0&lt;/cmath&gt;<br /> Then we can let &lt;math&gt;k&lt;/math&gt; be anything except 0, and get &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 3 \pmod{4}&lt;/math&gt; except &lt;math&gt;-m^2&lt;/math&gt;. Also since &lt;math&gt;x^2f(-x) = f(x)^2&lt;/math&gt;, we have &lt;math&gt;f(x) = 0 \Rightarrow f(-x) = 0&lt;/math&gt;, so &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 1 \pmod{4}&lt;/math&gt; except &lt;math&gt;m^2&lt;/math&gt;. So &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; except &lt;math&gt;\pm m^2&lt;/math&gt;. Since &lt;math&gt;f(m) \neq 0&lt;/math&gt;, &lt;math&gt;m = \pm m^2&lt;/math&gt;. Squaring, &lt;math&gt;m^2 = m^4&lt;/math&gt; and dividing by &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;m = m^3&lt;/math&gt;. Since &lt;math&gt;f(m^3) = 0&lt;/math&gt;, &lt;math&gt;f(m) = 0&lt;/math&gt;, which is a contradiction, so our only solutions are &lt;math&gt;f(x) = 0&lt;/math&gt; and &lt;math&gt;f(x) = x^2&lt;/math&gt;.</div> Npip99 https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2&diff=69459 2014 USAMO Problems/Problem 2 2015-03-25T00:49:03Z <p>Npip99: /* Solution */ RHS -&gt; LHS</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\mathbb{Z}&lt;/math&gt; be the set of integers. Find all functions &lt;math&gt;f : \mathbb{Z} \rightarrow \mathbb{Z}&lt;/math&gt; such that &lt;cmath&gt;xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))&lt;/cmath&gt; for all &lt;math&gt;x, y \in \mathbb{Z}&lt;/math&gt; with &lt;math&gt;x \neq 0&lt;/math&gt;.<br /> ==Solution==<br /> Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.<br /> <br /> Lemma 1: &lt;math&gt;f(0) = 0&lt;/math&gt;.<br /> Proof: Assume the opposite for a contradiction. Plug in &lt;math&gt;x = 2f(0)&lt;/math&gt; (because we assumed that &lt;math&gt;f(0) \neq 0&lt;/math&gt;), &lt;math&gt;y = 0&lt;/math&gt;. What you get eventually reduces to:<br /> &lt;cmath&gt;4f(0)-2 = \left( \frac{f(2f(0))}{f(0)} \right)^2&lt;/cmath&gt;<br /> which is a contradiction since the LHS is divisible by 2 but not 4.<br /> <br /> Then plug in &lt;math&gt;y = 0&lt;/math&gt; into the original equation and simplify by Lemma 1. We get:<br /> &lt;cmath&gt;x^2f(-x) = f(x)^2&lt;/cmath&gt;<br /> Then:<br /> <br /> \begin{align*}<br /> &lt;math&gt;x^6f(x) &amp;= x^4(-x)^2f(-(-x)) \\<br /> &amp;= x^4f(-x)^2 \\<br /> &amp;= f(x)^4&lt;/math&gt;<br /> \end{align*}<br /> <br /> Therefore, &lt;math&gt;f(x)&lt;/math&gt; must be 0 or &lt;math&gt;x^2&lt;/math&gt;.<br /> <br /> Now either &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;x^2&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; or there exists &lt;math&gt;a \neq 0&lt;/math&gt; such that &lt;math&gt;f(a)=0&lt;/math&gt;. The first case gives a valid solution. In the second case, we let &lt;math&gt;y = a&lt;/math&gt; in the original equation and simplify to get:<br /> &lt;cmath&gt;xf(-x) + a^2f(2x) = \frac{f(x)^2}{x}&lt;/cmath&gt;<br /> But we know that &lt;math&gt;xf(-x) = \frac{f(x)^2}{x}&lt;/math&gt;, so:<br /> &lt;cmath&gt;a^2f(2x) = 0&lt;/cmath&gt;<br /> Since &lt;math&gt;a&lt;/math&gt; is not 0, &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; (including 0). Now either &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, or there exists some &lt;math&gt;m \neq 0&lt;/math&gt; such that &lt;math&gt;f(m) = m^2&lt;/math&gt;. Then &lt;math&gt;m&lt;/math&gt; must be odd. We can let &lt;math&gt;x = 2k&lt;/math&gt; in the original equation, and since &lt;math&gt;f(2x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt;, stuff cancels and we get:<br /> &lt;cmath&gt;y^2f(4k - f(y)) = f(yf(y))&lt;/cmath&gt;<br /> [b]for &lt;math&gt;k \neq 0&lt;/math&gt;.[/b]<br /> Now, let &lt;math&gt;y = m&lt;/math&gt; and we get:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3)&lt;/cmath&gt;<br /> Now, either both sides are 0 or both are equal to &lt;math&gt;m^6&lt;/math&gt;. If both are &lt;math&gt;m^6&lt;/math&gt; then:<br /> &lt;cmath&gt;m^2(4k - m^2)^2 = m^6&lt;/cmath&gt;<br /> which simplifies to:<br /> &lt;cmath&gt;4k - m^2 = \pm m^2&lt;/cmath&gt;<br /> Since &lt;math&gt;k \neq 0&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; is odd, both cases are impossible, so we must have:<br /> &lt;cmath&gt;m^2f(4k - m^2) = f(m^3) = 0&lt;/cmath&gt;<br /> Then we can let &lt;math&gt;k&lt;/math&gt; be anything except 0, and get &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 3 \pmod{4}&lt;/math&gt; except &lt;math&gt;-m^2&lt;/math&gt;. Also since &lt;math&gt;x^2f(-x) = f(x)^2&lt;/math&gt;, we have &lt;math&gt;f(x) = 0 \Rightarrow f(-x) = 0&lt;/math&gt;, so &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x \equiv 1 \pmod{4}&lt;/math&gt; except &lt;math&gt;m^2&lt;/math&gt;. So &lt;math&gt;f(x)&lt;/math&gt; is 0 for all &lt;math&gt;x&lt;/math&gt; except &lt;math&gt;\pm m^2&lt;/math&gt;. Since &lt;math&gt;f(m) \neq 0&lt;/math&gt;, &lt;math&gt;m = \pm m^2&lt;/math&gt;. Squaring, &lt;math&gt;m^2 = m^4&lt;/math&gt; and dividing by &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;m = m^3&lt;/math&gt;. Since &lt;math&gt;f(m^3) = 0&lt;/math&gt;, &lt;math&gt;f(m) = 0&lt;/math&gt;, which is a contradiction, so our only solutions are &lt;math&gt;f(x) = 0&lt;/math&gt; and &lt;math&gt;f(x) = x^2&lt;/math&gt;.</div> Npip99