https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Will3145&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-04T07:30:06Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Logarithm&diff=118788 Logarithm 2020-03-05T04:36:32Z <p>Will3145: /* Problems */</p> <hr /> <div>'''Logarithms''' and [[exponents]] are very closely related. In fact, they are [[Function/Introduction#The_Inverse_of_a_Function|inverse]] [[function]]s. This means that logarithms can be used to reverse the result of exponentiation and vice versa, just as addition can be used to reverse the result of subtraction. Thus, if we have &lt;math&gt; a^x = b &lt;/math&gt;, then taking the logarithm with base &lt;math&gt; a&lt;/math&gt; on both sides will give us &lt;math&gt;x=\log_a{b}&lt;/math&gt;.<br /> <br /> We would read this as &quot;the logarithm of b, base a, is x&quot;. For example, we know that &lt;math&gt;3^4=81&lt;/math&gt;. To express the same fact in logarithmic notation we would write &lt;math&gt;\log_3 81=4&lt;/math&gt;.<br /> <br /> <br /> ==Conventions==<br /> Depending on the field, the symbol &lt;math&gt;\log&lt;/math&gt; without a base can have different meanings. Sometimes in high schools, the symbol is used to refer to a base 10 logarithm. Thus, &lt;math&gt;\log(100)&lt;/math&gt; can mean &lt;math&gt;\log_{10}(100)=2&lt;/math&gt;. In these contexts, the symbol &lt;math&gt;\ln&lt;/math&gt; (an abbreviation of the French &quot;logarithme normal,&quot; meaning &quot;natural logarithm&quot;) is introduced to refer to the logarithm base [[e]], or '''natural logarithm'''. However, the choice of base 10 is arbitrary, and convenient only for computations in a base-10 number system. The natural logarithm, however, has many convenient mathematical properties, so practicing mathematicians often take the symbol &lt;math&gt;\log&lt;/math&gt; to mean the natural logarithm and do not use the symbol &lt;math&gt;\ln&lt;/math&gt;. (This is an example of conflicting [[mathematical convention]]s.) In addition, the notation &lt;math&gt;\lg&lt;/math&gt; is often used by [[combinatorics | combinatorists]] and computer scientists to refer to the logarithm base &lt;math&gt;2&lt;/math&gt;. Occasionally, the base of the logarithms is irrelevant.<br /> <br /> ==Logarithmic Properties==<br /> We can use the properties of exponents to build a set of properties for logarithms.<br /> <br /> We know that &lt;math&gt;a^x\cdot a^y=a^{x+y}&lt;/math&gt;. We let &lt;math&gt; a^x=b&lt;/math&gt; and &lt;math&gt; a^y=c &lt;/math&gt;. This also makes &lt;math&gt;a^{x+y}=bc &lt;/math&gt;. From &lt;math&gt; a^x = b&lt;/math&gt;, we have &lt;math&gt; x = \log_a{b}&lt;/math&gt;, and from &lt;math&gt; a^y=c &lt;/math&gt;, we have &lt;math&gt; y=\log_a{c} &lt;/math&gt;. So, &lt;math&gt; x+y = \log_a{b}+\log_a{c}&lt;/math&gt;. But we also have from &lt;math&gt;a^{x+y} = bc&lt;/math&gt; that &lt;math&gt; x+y = \log_a{bc}&lt;/math&gt;. Thus, we have found two expressions for &lt;math&gt; x+y&lt;/math&gt; establishing the identity:<br /> <br /> &lt;center&gt;&lt;math&gt; \log_a{b} + \log_a{c} = \log_a{bc}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> Using the laws of exponents, one can derive and prove the following identities:<br /> <br /> *&lt;math&gt;\log_a b^n=n\log_a b&lt;/math&gt;<br /> *&lt;math&gt;\log_a b+ \log_a c=\log_a bc&lt;/math&gt;<br /> *&lt;math&gt;\log_a b-\log_a c=\log_a \frac{b}{c}&lt;/math&gt;<br /> *&lt;math&gt;(\log_a b)(\log_c d)= (\log_a d)(\log_c b)&lt;/math&gt;<br /> *&lt;math&gt;\frac{\log_a b}{\log_a c}=\log_c b&lt;/math&gt; (the [[change of base formula]])<br /> <br /> These formulas also have a number of common special cases:<br /> *&lt;math&gt;\log_{a}b=\frac{1}{\log_{b}a}&lt;/math&gt; (sometimes known as the ''inverse property'' of logarithms)<br /> *&lt;math&gt;\log_{a^n} b^n=\log_a b&lt;/math&gt;<br /> *&lt;math&gt;\log_{1/a} b=-\log_a b&lt;/math&gt;<br /> <br /> == Problems ==<br /> <br /> # Evaluate &lt;math&gt;(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})&lt;/math&gt;.<br /> # Evaluate &lt;math&gt;(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)&lt;/math&gt;.<br /> # Simplify &lt;math&gt;\frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}N} &lt;/math&gt; where &lt;math&gt; N=(100!)^3&lt;/math&gt;.<br /> <br /> == Natural Logarithm ==<br /> The '''natural logarithm''' is the logarithm with base [[e]]. It is usually denoted &lt;math&gt;\ln&lt;/math&gt;, an abbreviation of the French ''logarithme normal'', so that &lt;math&gt; \ln(x) = \log_e(x).&lt;/math&gt; However, in higher mathematics such as [[complex analysis]], the base 10 logarithm is typically disposed with entirely, the symbol &lt;math&gt;\log&lt;/math&gt; is taken to mean the logarithm base e and the symbol &lt;math&gt;\ln&lt;/math&gt; is not used at all. (This is an example of conflicting [[mathematical convention]]s.)<br /> <br /> &lt;math&gt;\ln a&lt;/math&gt; can also be defined as the area under the curve &lt;math&gt;y=\frac{1}{x}&lt;/math&gt; between 1 and a, or &lt;math&gt;\int^a_1 \frac{1}{x}\, dx&lt;/math&gt;.<br /> <br /> All logarithms are undefined in nonpositive reals, as they are complex. From the identity &lt;math&gt;e^{i\pi}=-1&lt;/math&gt;, we have &lt;math&gt;\ln (-1)=i\pi&lt;/math&gt;. Additionally, &lt;math&gt;\ln (-n)=\ln n+i\pi&lt;/math&gt; for positive real &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Problems ==<br /> === Introductory ===<br /> * What is the value of &lt;math&gt;a&lt;/math&gt; for which &lt;math&gt;\frac1{\log_2a}+\frac1{\log_3a}+\frac1{\log_4a}=1&lt;/math&gt;?<br /> [[2015_AMC_12A_Problems/Problem_14 | Source]]<br /> * Positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; satisfy the condition &lt;math&gt;\log_2(\log_{2^a}(\log_{2^b}(2^{1000})))=0.&lt;/math&gt; Find the sum of all possible values of &lt;math&gt;a+b&lt;/math&gt;. <br /> [[2013_AIME_II_Problems/Problem_2 | Source]]<br /> === Intermediate ===<br /> * The [[sequence]] &lt;math&gt; a_1, a_2, \ldots &lt;/math&gt; is [[geometric sequence|geometric]] with &lt;math&gt; a_1=a &lt;/math&gt; and common [[ratio]] &lt;math&gt; r, &lt;/math&gt; where &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; r &lt;/math&gt; are positive integers. Given that &lt;math&gt; \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, &lt;/math&gt; find the number of possible ordered pairs &lt;math&gt; (a,r). &lt;/math&gt;<br /> [[2006_AIME_I_Problems/Problem_9 | Source]]<br /> === Olympiad ===<br /> <br /> ==External Links==<br /> Two-minute Intro to Logarithms [http://www.youtube.com/watch?v=ey7ttABX9SM]<br /> <br /> [[Category:Definition]]<br /> [[Category:Functions]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&diff=118731 2017 AIME I Problems/Problem 2 2020-03-04T03:04:14Z <p>Will3145: replace ans &quot;62&quot; with &quot;062&quot;</p> <hr /> <div>==Problem 2==<br /> When each of &lt;math&gt;702&lt;/math&gt;, &lt;math&gt;787&lt;/math&gt;, and &lt;math&gt;855&lt;/math&gt; is divided by the positive integer &lt;math&gt;m&lt;/math&gt;, the remainder is always the positive integer &lt;math&gt;r&lt;/math&gt;. When each of &lt;math&gt;412&lt;/math&gt;, &lt;math&gt;722&lt;/math&gt;, and &lt;math&gt;815&lt;/math&gt; is divided by the positive integer &lt;math&gt;n&lt;/math&gt;, the remainder is always the positive integer &lt;math&gt;s \neq r&lt;/math&gt;. Find &lt;math&gt;m+n+r+s&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let's work on both parts of the problem separately. First, &lt;cmath&gt;855 \equiv 787 \equiv 702 \equiv r \pmod{m}.&lt;/cmath&gt; We take the difference of &lt;math&gt;855&lt;/math&gt; and &lt;math&gt;787&lt;/math&gt;, and also of &lt;math&gt;787&lt;/math&gt; and &lt;math&gt;702&lt;/math&gt;. We find that they are &lt;math&gt;85&lt;/math&gt; and &lt;math&gt;68&lt;/math&gt;, respectively. Since the greatest common divisor of the two differences is &lt;math&gt;17&lt;/math&gt; (and the only one besides one), it's safe to assume that &lt;math&gt;m = 17&lt;/math&gt;.<br /> <br /> Then, we divide &lt;math&gt;855&lt;/math&gt; by &lt;math&gt;17&lt;/math&gt;, and it's easy to see that &lt;math&gt;r = 5&lt;/math&gt;. Dividing &lt;math&gt;787&lt;/math&gt; and &lt;math&gt;702&lt;/math&gt; by &lt;math&gt;17&lt;/math&gt; also yields remainders of &lt;math&gt;5&lt;/math&gt;, which means our work up to here is correct.<br /> <br /> Doing the same thing with &lt;math&gt;815&lt;/math&gt;, &lt;math&gt;722&lt;/math&gt;, and &lt;math&gt;412&lt;/math&gt;, the differences between &lt;math&gt;815&lt;/math&gt; and &lt;math&gt;722&lt;/math&gt; and &lt;math&gt;412&lt;/math&gt; are &lt;math&gt;310&lt;/math&gt; and &lt;math&gt;93&lt;/math&gt;, respectively. Since the only common divisor (besides &lt;math&gt;1&lt;/math&gt;, of course) is &lt;math&gt;31&lt;/math&gt;, &lt;math&gt;n = 31&lt;/math&gt;. Dividing all &lt;math&gt;3&lt;/math&gt; numbers by &lt;math&gt;31&lt;/math&gt; yields a remainder of &lt;math&gt;9&lt;/math&gt; for each, so &lt;math&gt;s = 9&lt;/math&gt;. Thus, &lt;math&gt;m + n + r + s = 17 + 5 + 31 + 9 = \boxed{062}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_8&diff=114356 2010 AIME II Problems/Problem 8 2020-01-06T18:39:13Z <p>Will3145: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of [[ordered pair]]s of nonempty sets &lt;math&gt;\mathcal{A}&lt;/math&gt; and &lt;math&gt;\mathcal{B}&lt;/math&gt; that have the following properties:<br /> <br /> &lt;UL&gt;<br /> &lt;LI&gt; &lt;math&gt;\mathcal{A} \cup \mathcal{B} = \{1,2,3,4,5,6,7,8,9,10,11,12\}&lt;/math&gt;,&lt;/LI&gt;<br /> &lt;LI&gt; &lt;math&gt;\mathcal{A} \cap \mathcal{B} = \emptyset&lt;/math&gt;, &lt;/LI&gt;<br /> &lt;LI&gt; The number of elements of &lt;math&gt;\mathcal{A}&lt;/math&gt; is not an element of &lt;math&gt;\mathcal{A}&lt;/math&gt;,&lt;/LI&gt;<br /> &lt;LI&gt; The number of elements of &lt;math&gt;\mathcal{B}&lt;/math&gt; is not an element of &lt;math&gt;\mathcal{B}&lt;/math&gt;.<br /> &lt;/UL&gt;<br /> <br /> Find &lt;math&gt;N&lt;/math&gt;.<br /> <br /> == Solution==<br /> Let us [[partition]] the set &lt;math&gt;\{1,2,\cdots,12\}&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; numbers in &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;12-n&lt;/math&gt; numbers in &lt;math&gt;B&lt;/math&gt;, <br /> <br /> Since &lt;math&gt;n&lt;/math&gt; must be in &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;12-n&lt;/math&gt; must be in &lt;math&gt;A&lt;/math&gt; (&lt;math&gt;n\ne6&lt;/math&gt;, we cannot partition into two sets of 6 because &lt;math&gt;6&lt;/math&gt; needs to end up somewhere, &lt;math&gt;n\ne 0&lt;/math&gt; or &lt;math&gt;12&lt;/math&gt; either).<br /> <br /> We have &lt;math&gt;\dbinom{10}{n-1}&lt;/math&gt; ways of picking the numbers to be in &lt;math&gt;A&lt;/math&gt;. <br /> <br /> So the answer is &lt;math&gt;\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Regardless of the size &lt;math&gt;n&lt;/math&gt; of &lt;math&gt;A&lt;/math&gt; (ignoring the case when &lt;math&gt;n = 6&lt;/math&gt;), &lt;math&gt;n&lt;/math&gt; must not be in &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;12 - n&lt;/math&gt; must be in &lt;math&gt;A&lt;/math&gt;. <br /> <br /> There are &lt;math&gt;10&lt;/math&gt; remaining elements who’s placements have yet to be determined. Note that the actual value of &lt;math&gt;n&lt;/math&gt; does not matter; there is always &lt;math&gt;1&lt;/math&gt; necessary element, &lt;math&gt;1&lt;/math&gt; forbidden element, and &lt;math&gt;10&lt;/math&gt; other elements that need to be distributed. There are &lt;math&gt;2&lt;/math&gt; places to put each of these elements, for &lt;math&gt;2^{10}&lt;/math&gt; possibilities.<br /> <br /> However, there is the edge case of &lt;math&gt;n = 6; 6&lt;/math&gt; is forced not the be in either set, so we must subtract the &lt;math&gt;\dbinom{10}{5}&lt;/math&gt; cases where &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; have size &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2010|num-b=7|num-a=9|n=II}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&diff=113966 1990 AIME Problems/Problem 15 2020-01-01T19:00:48Z <p>Will3145: latex</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a_{}^{}x^5 + b_{}y^5&lt;/math&gt; if the [[real number]]s &lt;math&gt;a_{}^{}&lt;/math&gt;, &lt;math&gt;b_{}^{}&lt;/math&gt;, &lt;math&gt;x_{}^{}&lt;/math&gt;, and &lt;math&gt;y_{}^{}&lt;/math&gt; satisfy the [[equation]]s<br /> &lt;cmath&gt;ax + by = 3^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^2 + by^2 = 7^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^3 + by^3 = 16^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^4 + by^4 = 42^{}_{}.&lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> Set &lt;math&gt;S = (x + y)&lt;/math&gt; and &lt;math&gt;P = xy&lt;/math&gt;. Then the relationship<br /> <br /> &lt;cmath&gt;(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})&lt;/cmath&gt;<br /> <br /> can be exploited:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^2 + by^2)(x + y) &amp; = &amp; (ax^3 + by^3) + (xy)(ax + by) \\<br /> (ax^3 + by^3)(x + y) &amp; = &amp; (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}7S &amp; = &amp; 16 + 3P \\<br /> 16S &amp; = &amp; 42 + 7P\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Consequently, &lt;math&gt;S = - 14&lt;/math&gt; and &lt;math&gt;P = - 38&lt;/math&gt;. Finally:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^4 + by^4)(x + y) &amp; = &amp; (ax^5 + by^5) + (xy)(ax^3 + by^3) \\<br /> (42)(S) &amp; = &amp; (ax^5 + by^5) + (P)(16) \\<br /> (42)( - 14) &amp; = &amp; (ax^5 + by^5) + ( - 38)(16) \\<br /> ax^5 + by^5 &amp; = &amp; \boxed{020}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> <br /> A [[linear recurrence | recurrence]] of the form &lt;math&gt;T_n=AT_{n-1}+BT_{n-2}&lt;/math&gt; will have the closed form &lt;math&gt;T_n=ax^n+by^n&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are the values of the starting term that make the sequence geometric, and &lt;math&gt;a,b&lt;/math&gt; are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.<br /> <br /> Suppose we have such a recurrence with &lt;math&gt;T_1=3&lt;/math&gt; and &lt;math&gt;T_2=7&lt;/math&gt;. Then &lt;math&gt;T_3=ax^3+by^3=16=7A+3B&lt;/math&gt;, and &lt;math&gt;T_4=ax^4+by^4=42=16A+7B&lt;/math&gt;. <br /> <br /> Solving these simultaneous equations for &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;A=-14&lt;/math&gt; and &lt;math&gt;B=38&lt;/math&gt;. So, &lt;math&gt;ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:<br /> <br /> &lt;math&gt;a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16&lt;/math&gt;.<br /> <br /> Similarly take the first two terms, yielding:<br /> <br /> &lt;math&gt;ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10&lt;/math&gt;.<br /> <br /> Lastly take an alternating three-term sum,<br /> <br /> &lt;math&gt;a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12&lt;/math&gt;.<br /> <br /> Now to get the solution, let the answer be &lt;math&gt;k&lt;/math&gt;, so <br /> <br /> &lt;math&gt;ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68&lt;/math&gt;.<br /> <br /> Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated<br /> as done in the first solution.<br /> <br /> == Solution 4 ==<br /> We first let the answer to this problem be k. Multiplying the first equation by &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;ax^2 + bxy=3x&lt;/math&gt; Subtracting this equation from the second equation gives &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;. Similarily, doing the same for the other equations, we obtain:<br /> &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;, &lt;math&gt;by^3-bxy^2=16-7x&lt;/math&gt;, &lt;math&gt;by^4-bxy^3=42-16x&lt;/math&gt;, and &lt;math&gt;by^5-bxy^4=k-42x&lt;/math&gt;<br /> Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us &lt;math&gt;by^3-bxy^2=(7-3x)y&lt;/math&gt;. We can also obtain &lt;math&gt;by^4-bxy^3=(16-7x)y&lt;/math&gt;. <br /> Now we can solve for x and y! &lt;math&gt;(7-3x)y = 16-7x&lt;/math&gt; and &lt;math&gt;(16-7x)y=42-16x&lt;/math&gt;. Solving for x and y gives us &lt;math&gt;(-7+\sqrt{87},-7-\sqrt{87})&lt;/math&gt;.(It can be switched, but since the given equations are symmetric, it doesn't matter). &lt;math&gt;k-42x= (42-16x)y&lt;/math&gt;, and solving for k gives us &lt;math&gt;k= \boxed{020}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&diff=113965 1990 AIME Problems/Problem 15 2020-01-01T19:00:05Z <p>Will3145: latex</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a_{}^{}x^5 + b_{}y^5&lt;/math&gt; if the [[real number]]s &lt;math&gt;a_{}^{}&lt;/math&gt;, &lt;math&gt;b_{}^{}&lt;/math&gt;, &lt;math&gt;x_{}^{}&lt;/math&gt;, and &lt;math&gt;y_{}^{}&lt;/math&gt; satisfy the [[equation]]s<br /> &lt;cmath&gt;ax + by = 3^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^2 + by^2 = 7^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^3 + by^3 = 16^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^4 + by^4 = 42^{}_{}.&lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> Set &lt;math&gt;S = (x + y)&lt;/math&gt; and &lt;math&gt;P = xy&lt;/math&gt;. Then the relationship<br /> <br /> &lt;cmath&gt;(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})&lt;/cmath&gt;<br /> <br /> can be exploited:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^2 + by^2)(x + y) &amp; = &amp; (ax^3 + by^3) + (xy)(ax + by) \\<br /> (ax^3 + by^3)(x + y) &amp; = &amp; (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}7S &amp; = &amp; 16 + 3P \\<br /> 16S &amp; = &amp; 42 + 7P\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Consequently, &lt;math&gt;S = - 14&lt;/math&gt; and &lt;math&gt;P = - 38&lt;/math&gt;. Finally:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^4 + by^4)(x + y) &amp; = &amp; (ax^5 + by^5) + (xy)(ax^3 + by^3) \\<br /> (42)(S) &amp; = &amp; (ax^5 + by^5) + (P)(16) \\<br /> (42)( - 14) &amp; = &amp; (ax^5 + by^5) + ( - 38)(16) \\<br /> ax^5 + by^5 &amp; = &amp; \boxed{020}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> <br /> A [[linear recurrence | recurrence]] of the form &lt;math&gt;T_n=AT_{n-1}+BT_{n-2}&lt;/math&gt; will have the closed form &lt;math&gt;T_n=ax^n+by^n&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are the values of the starting term that make the sequence geometric, and &lt;math&gt;a,b&lt;/math&gt; are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.<br /> <br /> Suppose we have such a recurrence with &lt;math&gt;T_1=3&lt;/math&gt; and &lt;math&gt;T_2=7&lt;/math&gt;. Then &lt;math&gt;T_3=ax^3+by^3=16=7A+3B&lt;/math&gt;, and &lt;math&gt;T_4=ax^4+by^4=42=16A+7B&lt;/math&gt;. <br /> <br /> Solving these simultaneous equations for &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;A=-14&lt;/math&gt; and &lt;math&gt;B=38&lt;/math&gt;. So, &lt;math&gt;ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:<br /> <br /> &lt;math&gt;a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16&lt;/math&gt;.<br /> <br /> Similarly take the first two terms, yielding:<br /> <br /> &lt;math&gt;ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10&lt;/math&gt;.<br /> <br /> Lastly take an alternating three-term sum,<br /> <br /> &lt;math&gt;a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12&lt;/math&gt;.<br /> <br /> Now to get the solution, let the answer be &lt;math&gt;k&lt;/math&gt;, so <br /> <br /> &lt;math&gt;ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68&lt;/math&gt;.<br /> <br /> Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated<br /> as done in the first solution.<br /> <br /> == Solution 4 ==<br /> We first let the answer to this problem be k. Multiplying the first equation by &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;ax^2 + bxy=3x&lt;/math&gt; Subtracting this equation from the second equation gives &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;. Similarily, doing the same for the other equations, we obtain:<br /> &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;, &lt;math&gt;by^3-bxy^2=16-7x&lt;/math&gt;, &lt;math&gt;by^4-bxy^3=42-16x&lt;/math&gt;, and &lt;math&gt;by^5-bxy^4=k-42x&lt;/math&gt;<br /> Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us &lt;math&gt;by^3-bxy^2=(7-3x)y&lt;/math&gt;. We can also obtain &lt;math&gt;by^4-bxy^3=(16-7x)y&lt;/math&gt;. <br /> Now we can solve for x and y! &lt;math&gt;(7-3x)y = 16-7x&lt;/math&gt; and &lt;math&gt;(16-7x)y=42-16x&lt;/math&gt;. Solving for x and y gives us &lt;math&gt;(-7+\sqrt(87),-7-\sqrt(87))&lt;/math&gt;.(It can be switched, but since the given equations are symmetric, it doesn't matter). &lt;math&gt;k-42x= (42-16x)y&lt;/math&gt;, and solving for k gives us &lt;math&gt;k= \boxed{020}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_13&diff=113707 2008 AIME II Problems/Problem 13 2019-12-29T19:32:32Z <p>Will3145: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A [[regular polygon|regular]] [[hexagon]] with center at the [[origin]] in the [[complex plane]] has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let &lt;math&gt;R&lt;/math&gt; be the region outside the hexagon, and let &lt;math&gt;S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace&lt;/math&gt;. Then the area of &lt;math&gt;S&lt;/math&gt; has the form &lt;math&gt;a\pi + \sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with it's conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution.<br /> ~First~<br /> <br /> == Solution 2==<br /> If a point &lt;math&gt;z = r\text{cis}\,\theta&lt;/math&gt; is in &lt;math&gt;R&lt;/math&gt;, then the point &lt;math&gt;\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; (where [[cis]] denotes &lt;math&gt;\text{cis}\, \theta = \cos \theta + i \sin \theta&lt;/math&gt;). Since &lt;math&gt;R&lt;/math&gt; is symmetric every &lt;math&gt;60^{\circ}&lt;/math&gt; about the origin, it suffices to consider the area of the result of the transformation when &lt;math&gt;-30 \le \theta \le 30&lt;/math&gt;, and then to multiply by &lt;math&gt;6&lt;/math&gt; to account for the entire area.<br /> <br /> We note that if the region &lt;math&gt;S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace&lt;/math&gt;, where &lt;math&gt;R_2&lt;/math&gt; is the region (in green below) outside the circle of radius &lt;math&gt;1/\sqrt{3}&lt;/math&gt; centered at the origin, then &lt;math&gt;S_2&lt;/math&gt; is simply the region inside a circle of radius &lt;math&gt;\sqrt{3}&lt;/math&gt; centered at the origin. It now suffices to find what happens to the mapping of the region &lt;math&gt;R_2 - R&lt;/math&gt; (in blue below). <br /> <br /> The equation of the hexagon side in that region is &lt;math&gt;x = r \cos \theta = \frac{1}{2}&lt;/math&gt;, which is transformed to &lt;math&gt;\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta = &lt;/math&gt;2 . Let &lt;math&gt;r\text{cis}\,\theta = a+bi&lt;/math&gt; where &lt;math&gt;a,b \in \mathbb{R}&lt;/math&gt;; then &lt;math&gt;r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}&lt;/math&gt;, so the equation becomes &lt;math&gt;a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1&lt;/math&gt;. Hence the side is sent to an arc of the unit circle centered at &lt;math&gt;(1,0)&lt;/math&gt;, after considering the restriction that the side of the hexagon is a segment of length &lt;math&gt;1/\sqrt{3}&lt;/math&gt;. <br /> <br /> Including &lt;math&gt;S_2&lt;/math&gt;, we find that &lt;math&gt;S&lt;/math&gt; is the union of six unit circles centered at &lt;math&gt;\text{cis}\, \frac{k\pi}{6}&lt;/math&gt;, &lt;math&gt;k = 0,1,2,3,4,5&lt;/math&gt;, as shown below. <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype(&quot;4 4&quot;); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1));<br /> draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i &lt; 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label(&quot;$1/\sqrt{3}$&quot;,(0,-0.5),W,fontsize(8)); <br /> &lt;/asy&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;math&gt;\Longrightarrow&lt;/math&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;asy&gt;<br /> defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1));<br /> draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype(&quot;4 4&quot;)); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype(&quot;4 4&quot;)); draw(Circle((0,0),1),linetype(&quot;4 4&quot;)); label(&quot;$\sqrt{3}$&quot;,(0,-0.5),W,fontsize(8));<br /> add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The area of the regular hexagon is &lt;math&gt;6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}&lt;/math&gt;. The total area of the six &lt;math&gt;120^{\circ}&lt;/math&gt; sectors is &lt;math&gt;6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}&lt;/math&gt;. Their sum is &lt;math&gt;2\pi + \sqrt{27}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{029}&lt;/math&gt;.<br /> - Th3Numb3rThr33<br /> <br /> == Solution 3 (Calculus) ==<br /> One can describe the line parallel to the imaginary axis &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; using polar coordinates as &lt;math&gt;r(\theta)=\dfrac{1}{2\cos{\theta}}&lt;/math&gt;<br /> <br /> so &lt;math&gt;z&lt;/math&gt; is equal to &lt;math&gt;z=(\dfrac{1}{2\cos{\theta}})(cis{\theta})<br /> \rightarrow \frac{1}{z}=2\cos{\theta}cis(-\theta)&lt;/math&gt;<br /> <br /> Dividing the hexagon to 12 equal parts we get that <br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta&lt;/math&gt;<br /> <br /> which is a routine computation:<br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta&lt;/math&gt;<br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12[\frac{1}{2}\sin{2\theta}+\theta]_0^{\frac{\pi}{6}}=12(\frac{\sqrt{3}}{4}+\frac{\pi}{6})=2\pi+3\sqrt{3}=2\pi + \sqrt{27}&lt;/math&gt;<br /> <br /> &lt;math&gt;a+b = \boxed{029}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=113209 2018 AMC 10B Problems/Problem 24 2019-12-23T05:10:28Z <p>Will3145: /* Solution 2 (Alternate Geometrical Approach to 1) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be a regular hexagon with side length &lt;math&gt;1&lt;/math&gt;. Denote by &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; the midpoints of sides &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{EF}&lt;/math&gt;, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of &lt;math&gt;\triangle ACE&lt;/math&gt; and &lt;math&gt;\triangle XYZ&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> M=(0,sqrt(3)/2);<br /> N=(3/4,3sqrt(3)/4);<br /> O=(3/4,sqrt(3)/4);<br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$M$&quot;, M, NW);<br /> label(&quot;$N$&quot;, N, NE);<br /> label(&quot;$O$&quot;, O, SE);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> The desired area (hexagon &lt;math&gt;MPNQOR&lt;/math&gt;) consists of an equilateral triangle (&lt;math&gt;\triangle MNO&lt;/math&gt;) and three right triangles (&lt;math&gt;\triangle MPN&lt;/math&gt;, &lt;math&gt;\triangle NQO&lt;/math&gt;, and &lt;math&gt;\triangle ORM&lt;/math&gt;).<br /> <br /> Notice that &lt;math&gt;\overline {AD}&lt;/math&gt; (not shown) and &lt;math&gt;\overline {BC}&lt;/math&gt; are parallel. &lt;math&gt;\overline {XY}&lt;/math&gt; divides transversals &lt;math&gt;\overline {AB}&lt;/math&gt; and &lt;math&gt;\overline {CD}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. Thus, it must also divide transversal &lt;math&gt;\overline {AC}&lt;/math&gt; and transversal &lt;math&gt;\overline {CO}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. By symmetry, the same applies for &lt;math&gt;\overline {CE}&lt;/math&gt; and &lt;math&gt;\overline {EA}&lt;/math&gt; as well as &lt;math&gt;\overline {EM}&lt;/math&gt; and &lt;math&gt;\overline {AN}&lt;/math&gt;.<br /> <br /> <br /> In &lt;math&gt;\triangle ACE&lt;/math&gt;, we see that &lt;math&gt;\frac{[MNO]}{[ACE]} = \frac{1}{4}&lt;/math&gt; and &lt;math&gt;\frac{[MPN]}{[ACE]} = \frac{1}{8}&lt;/math&gt;. Our desired area becomes <br /> <br /> &lt;cmath&gt;(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Alternate Geometrical Approach to 1)==<br /> Instead of directly finding the desired hexagonal area, &lt;math&gt;\triangle XYZ&lt;/math&gt; can be found. It consists of three triangles and the desired hexagon. Given triangle rotational symmetry, the three triangles are congruent. See that &lt;math&gt;\triangle XYZ&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equilateral, so &lt;math&gt;m\angle PXN=60&lt;/math&gt;, so &lt;math&gt;m\angle AXP = \frac{180-60}{2}=60&lt;/math&gt;. As &lt;math&gt;\overline {AC}&lt;/math&gt; is a transversal running through &lt;math&gt;\overline {FC}&lt;/math&gt; (use your imagination) and &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;m\angle BAC=m\angle FCA = \frac{m\angle ACE}{2}=30&lt;/math&gt;. <br /> <br /> <br /> Then, &lt;math&gt;\triangle APX&lt;/math&gt; is a &lt;math&gt;30&lt;/math&gt;-&lt;math&gt;60&lt;/math&gt;-&lt;math&gt;90&lt;/math&gt; triangle. By HL congruence, &lt;math&gt;\triangle APX \cong \triangle NPX&lt;/math&gt;. &lt;math&gt;AX=\frac{1}{2}&lt;/math&gt;. Then, the area of &lt;math&gt;\triangle PXN&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt;. There are three such triangles for a total area of &lt;math&gt;\triangle XYZ&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt{3}}{32}&lt;/math&gt;. Find the side of &lt;math&gt;\triangle XYZ&lt;/math&gt; to be &lt;math&gt;\frac{3}{2}&lt;/math&gt;, so the area is &lt;math&gt;\frac{9\sqrt{3}}{16}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{15\sqrt{3}}{32} \implies \boxed{C}&lt;/math&gt;<br /> <br /> <br /> ~BJHHar<br /> <br /> ==Solution 3 ==<br /> <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> <br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, SE);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of &lt;math&gt;3&lt;/math&gt; isosceles trapezoids (&lt;math&gt;AXFZ&lt;/math&gt;, &lt;math&gt;XBCY&lt;/math&gt;, and &lt;math&gt;ZYED&lt;/math&gt;), and &lt;math&gt;3&lt;/math&gt; right triangles, with one right angle on each of &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt;. <br /> Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is &lt;math&gt;1&lt;/math&gt;, and the other base is &lt;math&gt;\frac{3}{2}&lt;/math&gt; (it is halfway in between the side and the longest diagonal, which has length &lt;math&gt;2&lt;/math&gt;) with a height of &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; (by using the Pythagorean Theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of &lt;math&gt;\frac{5\sqrt{3}}{16}&lt;/math&gt; for a total area of &lt;math&gt;\frac{15\sqrt{3}}{16}.&lt;/math&gt; (Alternatively, we could have calculated the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt; and subtracted the area of &lt;math&gt;\triangle XYZ&lt;/math&gt;, which, as we showed before, had a side length of &lt;math&gt;\frac{3}{2}&lt;/math&gt;). <br /> Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on &lt;math&gt;X&lt;/math&gt;, is similar to the triangle with a base of &lt;math&gt;YC = 1/2.&lt;/math&gt; Using similar triangles, we calculate the base to be &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the height to be &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; giving us an area of &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt; per triangle, and a total area of &lt;math&gt;3\frac{\sqrt{3}}{32}&lt;/math&gt;. Adding the two areas together, we get &lt;math&gt;\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}&lt;/math&gt;. Finding the total area, we get &lt;math&gt;6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}&lt;/math&gt;. Taking the complement, we get &lt;math&gt;\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \frac{15\sqrt{3}}{32} = (C)\frac{15}{32}\sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution 4 (Trig)==<br /> Notice, the area of the convex hexagon formed through the intersection of the &lt;math&gt;2&lt;/math&gt; triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br /> To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is &lt;math&gt;120^{\textrm{o}}&lt;/math&gt; and the trapezoid is isosceles, we know that the angle opposite is &lt;math&gt;60^{\textrm{o}}&lt;/math&gt;, and thus the side length of this triangle is &lt;math&gt;1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}&lt;/math&gt;. So the area of this triangle is &lt;math&gt;\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}&lt;/math&gt;<br /> Now let's find the area of the smaller triangles. Notice, triangle &lt;math&gt;ACE&lt;/math&gt; cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then &lt;math&gt;\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}&lt;/math&gt; and the sum of the areas is &lt;math&gt;3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}&lt;/math&gt;<br /> Therefore, the area of the convex hexagon is &lt;math&gt;\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> Dividing &lt;math&gt;\triangle MNO&lt;/math&gt; into two right triangles congruent to &lt;math&gt;\triangle PMN&lt;/math&gt;, we see that &lt;math&gt;[MPNQOR]=\dfrac{5}{8}[ACE]&lt;/math&gt;. Because &lt;math&gt;[ACE] = \dfrac{1}{2}[ABCDEF]&lt;/math&gt;, we have &lt;math&gt;[MPNQOR]=\dfrac{5}{16}[ABCDEF]&lt;/math&gt;. From here, you should be able to tell that the answer will have a factor of &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;\boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt; is the only answer that has a factor of &lt;math&gt;5&lt;/math&gt;. However, if you want to actually calculate the area, you would calculate &lt;math&gt;[ABCDEF]&lt;/math&gt; to be &lt;math&gt;6 \cdot \dfrac{\sqrt{3}}{2 \cdot 2} = \dfrac{3\sqrt{3}}{2}&lt;/math&gt;, so &lt;math&gt;[MPNQOR] = \dfrac{5}{16} \cdot \dfrac{3\sqrt{3}}{2} = \boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=yDbn9Mx2myw<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=113208 2018 AMC 10B Problems/Problem 24 2019-12-23T05:08:45Z <p>Will3145: /* Solution 2 (Alternate Geometrical Approach to 1) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be a regular hexagon with side length &lt;math&gt;1&lt;/math&gt;. Denote by &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; the midpoints of sides &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{EF}&lt;/math&gt;, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of &lt;math&gt;\triangle ACE&lt;/math&gt; and &lt;math&gt;\triangle XYZ&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> M=(0,sqrt(3)/2);<br /> N=(3/4,3sqrt(3)/4);<br /> O=(3/4,sqrt(3)/4);<br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$M$&quot;, M, NW);<br /> label(&quot;$N$&quot;, N, NE);<br /> label(&quot;$O$&quot;, O, SE);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> The desired area (hexagon &lt;math&gt;MPNQOR&lt;/math&gt;) consists of an equilateral triangle (&lt;math&gt;\triangle MNO&lt;/math&gt;) and three right triangles (&lt;math&gt;\triangle MPN&lt;/math&gt;, &lt;math&gt;\triangle NQO&lt;/math&gt;, and &lt;math&gt;\triangle ORM&lt;/math&gt;).<br /> <br /> Notice that &lt;math&gt;\overline {AD}&lt;/math&gt; (not shown) and &lt;math&gt;\overline {BC}&lt;/math&gt; are parallel. &lt;math&gt;\overline {XY}&lt;/math&gt; divides transversals &lt;math&gt;\overline {AB}&lt;/math&gt; and &lt;math&gt;\overline {CD}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. Thus, it must also divide transversal &lt;math&gt;\overline {AC}&lt;/math&gt; and transversal &lt;math&gt;\overline {CO}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. By symmetry, the same applies for &lt;math&gt;\overline {CE}&lt;/math&gt; and &lt;math&gt;\overline {EA}&lt;/math&gt; as well as &lt;math&gt;\overline {EM}&lt;/math&gt; and &lt;math&gt;\overline {AN}&lt;/math&gt;.<br /> <br /> <br /> In &lt;math&gt;\triangle ACE&lt;/math&gt;, we see that &lt;math&gt;\frac{[MNO]}{[ACE]} = \frac{1}{4}&lt;/math&gt; and &lt;math&gt;\frac{[MPN]}{[ACE]} = \frac{1}{8}&lt;/math&gt;. Our desired area becomes <br /> <br /> &lt;cmath&gt;(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Alternate Geometrical Approach to 1)==<br /> Instead of directly finding the desired hexagonal area, &lt;math&gt;\triangle XYZ&lt;/math&gt; can be found. It consists of three triangles and the desired hexagon. Given triangle rotational symmetry, the three triangles are congruent. See that &lt;math&gt;\triangle XYZ&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equilateral, so &lt;math&gt;m\angle PXN=60&lt;/math&gt;, so &lt;math&gt;m\angle AXP = \frac{180-60}{2}=60&lt;/math&gt;. As &lt;math&gt;\overline {AC}&lt;/math&gt; is a transversal running through &lt;math&gt;\overline {FC}&lt;/math&gt; (use your imagination) and &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;m\angle BAC=m\angle FCA = \frac{m\angle ACE}{2}=30&lt;/math&gt;. <br /> <br /> <br /> Then, &lt;math&gt;\triangle APX&lt;/math&gt; is a &lt;math&gt;30&lt;/math&gt;-&lt;math&gt;60&lt;/math&gt;-&lt;math&gt;90&lt;/math&gt; triangle. By HL congruence, &lt;math&gt;\triangle APX \cong \triangle PXN&lt;/math&gt;. &lt;math&gt;AX=\frac{1}{2}&lt;/math&gt;. Then, the area of &lt;math&gt;\triangle PXN&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt;. There are three such triangles for a total area of &lt;math&gt;\triangle XYZ&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt{3}}{32}&lt;/math&gt;. Find the side of &lt;math&gt;\triangle XYZ&lt;/math&gt; to be &lt;math&gt;\frac{3}{2}&lt;/math&gt;, so the area is &lt;math&gt;\frac{9\sqrt{3}}{16}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{15\sqrt{3}}{32} \implies \boxed{C}&lt;/math&gt;<br /> <br /> <br /> ~BJHHar<br /> <br /> ==Solution 3 ==<br /> <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> <br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, SE);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of &lt;math&gt;3&lt;/math&gt; isosceles trapezoids (&lt;math&gt;AXFZ&lt;/math&gt;, &lt;math&gt;XBCY&lt;/math&gt;, and &lt;math&gt;ZYED&lt;/math&gt;), and &lt;math&gt;3&lt;/math&gt; right triangles, with one right angle on each of &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt;. <br /> Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is &lt;math&gt;1&lt;/math&gt;, and the other base is &lt;math&gt;\frac{3}{2}&lt;/math&gt; (it is halfway in between the side and the longest diagonal, which has length &lt;math&gt;2&lt;/math&gt;) with a height of &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; (by using the Pythagorean Theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of &lt;math&gt;\frac{5\sqrt{3}}{16}&lt;/math&gt; for a total area of &lt;math&gt;\frac{15\sqrt{3}}{16}.&lt;/math&gt; (Alternatively, we could have calculated the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt; and subtracted the area of &lt;math&gt;\triangle XYZ&lt;/math&gt;, which, as we showed before, had a side length of &lt;math&gt;\frac{3}{2}&lt;/math&gt;). <br /> Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on &lt;math&gt;X&lt;/math&gt;, is similar to the triangle with a base of &lt;math&gt;YC = 1/2.&lt;/math&gt; Using similar triangles, we calculate the base to be &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the height to be &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; giving us an area of &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt; per triangle, and a total area of &lt;math&gt;3\frac{\sqrt{3}}{32}&lt;/math&gt;. Adding the two areas together, we get &lt;math&gt;\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}&lt;/math&gt;. Finding the total area, we get &lt;math&gt;6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}&lt;/math&gt;. Taking the complement, we get &lt;math&gt;\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \frac{15\sqrt{3}}{32} = (C)\frac{15}{32}\sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution 4 (Trig)==<br /> Notice, the area of the convex hexagon formed through the intersection of the &lt;math&gt;2&lt;/math&gt; triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br /> To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is &lt;math&gt;120^{\textrm{o}}&lt;/math&gt; and the trapezoid is isosceles, we know that the angle opposite is &lt;math&gt;60^{\textrm{o}}&lt;/math&gt;, and thus the side length of this triangle is &lt;math&gt;1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}&lt;/math&gt;. So the area of this triangle is &lt;math&gt;\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}&lt;/math&gt;<br /> Now let's find the area of the smaller triangles. Notice, triangle &lt;math&gt;ACE&lt;/math&gt; cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then &lt;math&gt;\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}&lt;/math&gt; and the sum of the areas is &lt;math&gt;3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}&lt;/math&gt;<br /> Therefore, the area of the convex hexagon is &lt;math&gt;\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> Dividing &lt;math&gt;\triangle MNO&lt;/math&gt; into two right triangles congruent to &lt;math&gt;\triangle PMN&lt;/math&gt;, we see that &lt;math&gt;[MPNQOR]=\dfrac{5}{8}[ACE]&lt;/math&gt;. Because &lt;math&gt;[ACE] = \dfrac{1}{2}[ABCDEF]&lt;/math&gt;, we have &lt;math&gt;[MPNQOR]=\dfrac{5}{16}[ABCDEF]&lt;/math&gt;. From here, you should be able to tell that the answer will have a factor of &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;\boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt; is the only answer that has a factor of &lt;math&gt;5&lt;/math&gt;. However, if you want to actually calculate the area, you would calculate &lt;math&gt;[ABCDEF]&lt;/math&gt; to be &lt;math&gt;6 \cdot \dfrac{\sqrt{3}}{2 \cdot 2} = \dfrac{3\sqrt{3}}{2}&lt;/math&gt;, so &lt;math&gt;[MPNQOR] = \dfrac{5}{16} \cdot \dfrac{3\sqrt{3}}{2} = \boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=yDbn9Mx2myw<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_2&diff=112429 2006 AIME II Problems/Problem 2 2019-12-01T00:58:16Z <p>Will3145: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The lengths of the sides of a [[triangle]] with positive area are &lt;math&gt;\log_{10} 12&lt;/math&gt;, &lt;math&gt;\log_{10} 75&lt;/math&gt;, and &lt;math&gt;\log_{10} n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer. Find the number of possible values for &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> By the [[Triangle Inequality]] and applying the well-known logarithmic property &lt;math&gt;\log_{c} a \cdot \log_{c} b = \log_{c} ab&lt;/math&gt;, we have that<br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt; 12n &gt; 75 &lt;/math&gt; <br /> <br /> &lt;math&gt; n &gt; \frac{75}{12} = \frac{25}{4} = 6.25 &lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Also, <br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} 75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12\cdot75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt; n &lt; 900 &lt;/math&gt; <br /> &lt;/div&gt;<br /> Combining these two inequalities: <br /> <br /> &lt;cmath&gt; 6.25 &lt; n &lt; 900 &lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;n&lt;/math&gt; is in the set &lt;math&gt;(6.25 , 900)&lt;/math&gt;; the number of positive integer &lt;math&gt;n&lt;/math&gt; which satisfies this requirement is &lt;math&gt;\boxed{893}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_2&diff=112428 2006 AIME II Problems/Problem 2 2019-12-01T00:48:33Z <p>Will3145: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The lengths of the sides of a [[triangle]] with positive area are &lt;math&gt;\log_{10} 12&lt;/math&gt;, &lt;math&gt;\log_{10} 75&lt;/math&gt;, and &lt;math&gt;\log_{10} n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer. Find the number of possible values for &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> By the [[Triangle Inequality]] and applying the well-known logarithmic property &lt;math&gt;\log_{c} a \cdot \log_{c} b = \log_{c} ab&lt;/math&gt;, we have that<br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt; 12n &gt; 75 &lt;/math&gt; <br /> <br /> &lt;math&gt; n &gt; \frac{75}{12} = \frac{25}{4} = 6.25 &lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Also, <br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} 75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12\cdot75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt; n &lt; 900 &lt;/math&gt; <br /> &lt;/div&gt;<br /> Combining these two inequalities: <br /> <br /> &lt;cmath&gt; 6.25 &lt; n &lt; 900 &lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;n \subseteq (6.25 , 900)&lt;/math&gt;, and the number of positive integer &lt;math&gt;n&lt;/math&gt; which satisfies this requirement is &lt;math&gt;\boxed{893}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_2&diff=112427 2006 AIME II Problems/Problem 2 2019-12-01T00:46:49Z <p>Will3145: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The lengths of the sides of a [[triangle]] with positive area are &lt;math&gt;\log_{10} 12&lt;/math&gt;, &lt;math&gt;\log_{10} 75&lt;/math&gt;, and &lt;math&gt;\log_{10} n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer. Find the number of possible values for &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> By the [[Triangle Inequality]] and applying the well-known logarithmic property &lt;math&gt;\log_{c} a \cdot \log_{c} b = \log_{c} ab&lt;/math&gt;, we have that<br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt; 12n &gt; 75 &lt;/math&gt; <br /> <br /> &lt;math&gt; n &gt; \frac{75}{12} = \frac{25}{4} = 6.25 &lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Also, <br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} 75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12\cdot75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt; n &lt; 900 &lt;/math&gt; <br /> &lt;/div&gt;<br /> Combining these two inequalities: <br /> <br /> &lt;cmath&gt; 6.25 &lt; n &lt; 900 &lt;/cmath&gt;<br /> <br /> The number of possible integer values for &lt;math&gt;n&lt;/math&gt; is the number of integers over the interval &lt;math&gt;(6.25 , 900)&lt;/math&gt;, which is &lt;math&gt;\boxed{893}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_2&diff=112426 2006 AIME II Problems/Problem 2 2019-12-01T00:45:52Z <p>Will3145: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The lengths of the sides of a [[triangle]] with positive area are &lt;math&gt;\log_{10} 12&lt;/math&gt;, &lt;math&gt;\log_{10} 75&lt;/math&gt;, and &lt;math&gt;\log_{10} n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer. Find the number of possible values for &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> By the [[Triangle Inequality]]:<br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12n &gt; \log_{10} 75 &lt;/math&gt;<br /> <br /> &lt;math&gt; 12n &gt; 75 &lt;/math&gt; <br /> <br /> &lt;math&gt; n &gt; \frac{75}{12} = \frac{25}{4} = 6.25 &lt;/math&gt;<br /> &lt;/div&gt;<br /> Also, applying the well-known logarithmic property &lt;math&gt;\log_{c} a \cdot \log_{c} b = \log_{c} ab&lt;/math&gt;, we have<br /> &lt;div style=&quot;text-align:center;&quot;&gt;<br /> &lt;math&gt;\log_{10} 12 + \log_{10} 75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_{10} 12\cdot75 &gt; \log_{10} n &lt;/math&gt;<br /> <br /> &lt;math&gt; n &lt; 900 &lt;/math&gt; <br /> &lt;/div&gt;<br /> Combining these two inequalities: <br /> <br /> &lt;cmath&gt; 6.25 &lt; n &lt; 900 &lt;/cmath&gt;<br /> <br /> The number of possible integer values for &lt;math&gt;n&lt;/math&gt; is the number of integers over the interval &lt;math&gt;(6.25 , 900)&lt;/math&gt;, which is &lt;math&gt;\boxed{893}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_22&diff=112350 2019 AMC 8 Problems/Problem 22 2019-11-29T04:31:59Z <p>Will3145: /* Solution 1 */</p> <hr /> <div>==Problem 22==<br /> A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose the fraction of discount is &lt;math&gt;x&lt;/math&gt;. That means &lt;math&gt;(1-x)(1+x)=0.84&lt;/math&gt;; so &lt;math&gt;1-x^{2}=0.84&lt;/math&gt;, and &lt;math&gt;(x^{2})=0.16&lt;/math&gt;, obtaining &lt;math&gt;x=0.4&lt;/math&gt;. Therefore, the price was increased and decreased by &lt;math&gt;40&lt;/math&gt;%, or &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;.<br /> <br /> ==Solution 2(Answer options)==<br /> Let the price be &lt;math&gt;100&lt;/math&gt; and then trying for each option leads to &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> ==Solution 3==<br /> Let x be the discount. We can also work in reverse such as (&lt;math&gt;84&lt;/math&gt;)&lt;math&gt;(\frac{100}{100-x})&lt;/math&gt;&lt;math&gt;(\frac{100}{100+x})&lt;/math&gt; = &lt;math&gt;100&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;8400&lt;/math&gt; = &lt;math&gt;(100+x)(100-x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; gives us &lt;math&gt;x = 40, -40&lt;/math&gt;. But &lt;math&gt;x&lt;/math&gt; has to be positive. Thus &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;40&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=21|num-a=23}}<br /> <br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_5&diff=112232 2005 AIME II Problems/Problem 5 2019-11-27T18:27:50Z <p>Will3145: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Determine the number of [[ordered pair]]s &lt;math&gt; (a,b) &lt;/math&gt; of [[integer]]s such that &lt;math&gt; \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, &lt;/math&gt; and &lt;math&gt; 2 \leq b \leq 2005. &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> The equation can be rewritten as &lt;math&gt; \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5 &lt;/math&gt; Multiplying through by &lt;math&gt;\log a \log b &lt;/math&gt; and factoring yields &lt;math&gt;(\log b - 3\log a)(\log b - 2\log a)=0 &lt;/math&gt;. Therefore, &lt;math&gt;\log b=3\log a &lt;/math&gt; or &lt;math&gt;\log b=2\log a &lt;/math&gt;, so either &lt;math&gt; b=a^3 &lt;/math&gt; or &lt;math&gt; b=a^2 &lt;/math&gt;. <br /> *For the case &lt;math&gt; b=a^2 &lt;/math&gt;, note that &lt;math&gt; 44^2=1936 &lt;/math&gt; and &lt;math&gt; 45^2=2025 &lt;/math&gt;. Thus, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; will work. <br /> *For the case &lt;math&gt; b=a^3 &lt;/math&gt;, note that &lt;math&gt; 12^3=1728 &lt;/math&gt; while &lt;math&gt; 13^3=2197 &lt;/math&gt;. Therefore, for this case, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt; work. <br /> There are &lt;math&gt; 44-2+1=43 &lt;/math&gt; possibilities for the square case and &lt;math&gt; 12-2+1=11 &lt;/math&gt; possibilities for the cube case. Thus, the answer is &lt;math&gt; 43+11= \boxed{054}&lt;/math&gt;.<br /> <br /> Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs &lt;math&gt;(a,b)&lt;/math&gt;, and not for the number of possible values of &lt;math&gt;b&lt;/math&gt;. Were the problem to ask for the number of possible values of &lt;math&gt;b&lt;/math&gt;, the values of &lt;math&gt;b^6&lt;/math&gt; under &lt;math&gt;2005&lt;/math&gt; would have to be subtracted, which would just be &lt;math&gt;2&lt;/math&gt; values: &lt;math&gt;2^6&lt;/math&gt; and &lt;math&gt;3^6&lt;/math&gt;. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)<br /> <br /> ==Solution 2 ==<br /> Let &lt;math&gt;k=\log_a b&lt;/math&gt;. Then our equation becomes &lt;math&gt;k+\frac{6}{k}=5&lt;/math&gt;. Multiplying through by &lt;math&gt;k&lt;/math&gt; and solving the quadratic gives us &lt;math&gt;k=2&lt;/math&gt; or &lt;math&gt;k=3&lt;/math&gt;. Hence &lt;math&gt;a^2=b&lt;/math&gt; or &lt;math&gt;a^3=b&lt;/math&gt;. <br /> <br /> For the first case &lt;math&gt;a^2=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 44, a total of 43 values.<br /> For the second case &lt;math&gt;a^3=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 12, a total of 11 values.<br /> <br /> <br /> Thus the total number of possible values is &lt;math&gt;43+11=\boxed{54}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_5&diff=112231 2005 AIME II Problems/Problem 5 2019-11-27T18:27:36Z <p>Will3145: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Determine the number of [[ordered pair]]s &lt;math&gt; (a,b) &lt;/math&gt; of [[integer]]s such that &lt;math&gt; \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, &lt;/math&gt; and &lt;math&gt; 2 \leq b \leq 2005. &lt;/math&gt;<br /> <br /> == Solution I ==<br /> The equation can be rewritten as &lt;math&gt; \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5 &lt;/math&gt; Multiplying through by &lt;math&gt;\log a \log b &lt;/math&gt; and factoring yields &lt;math&gt;(\log b - 3\log a)(\log b - 2\log a)=0 &lt;/math&gt;. Therefore, &lt;math&gt;\log b=3\log a &lt;/math&gt; or &lt;math&gt;\log b=2\log a &lt;/math&gt;, so either &lt;math&gt; b=a^3 &lt;/math&gt; or &lt;math&gt; b=a^2 &lt;/math&gt;. <br /> *For the case &lt;math&gt; b=a^2 &lt;/math&gt;, note that &lt;math&gt; 44^2=1936 &lt;/math&gt; and &lt;math&gt; 45^2=2025 &lt;/math&gt;. Thus, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; will work. <br /> *For the case &lt;math&gt; b=a^3 &lt;/math&gt;, note that &lt;math&gt; 12^3=1728 &lt;/math&gt; while &lt;math&gt; 13^3=2197 &lt;/math&gt;. Therefore, for this case, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt; work. <br /> There are &lt;math&gt; 44-2+1=43 &lt;/math&gt; possibilities for the square case and &lt;math&gt; 12-2+1=11 &lt;/math&gt; possibilities for the cube case. Thus, the answer is &lt;math&gt; 43+11= \boxed{054}&lt;/math&gt;.<br /> <br /> Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs &lt;math&gt;(a,b)&lt;/math&gt;, and not for the number of possible values of &lt;math&gt;b&lt;/math&gt;. Were the problem to ask for the number of possible values of &lt;math&gt;b&lt;/math&gt;, the values of &lt;math&gt;b^6&lt;/math&gt; under &lt;math&gt;2005&lt;/math&gt; would have to be subtracted, which would just be &lt;math&gt;2&lt;/math&gt; values: &lt;math&gt;2^6&lt;/math&gt; and &lt;math&gt;3^6&lt;/math&gt;. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)<br /> <br /> ==Solution 2 ==<br /> Let &lt;math&gt;k=\log_a b&lt;/math&gt;. Then our equation becomes &lt;math&gt;k+\frac{6}{k}=5&lt;/math&gt;. Multiplying through by &lt;math&gt;k&lt;/math&gt; and solving the quadratic gives us &lt;math&gt;k=2&lt;/math&gt; or &lt;math&gt;k=3&lt;/math&gt;. Hence &lt;math&gt;a^2=b&lt;/math&gt; or &lt;math&gt;a^3=b&lt;/math&gt;. <br /> <br /> For the first case &lt;math&gt;a^2=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 44, a total of 43 values.<br /> For the second case &lt;math&gt;a^3=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 12, a total of 11 values.<br /> <br /> <br /> Thus the total number of possible values is &lt;math&gt;43+11=\boxed{54}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_4&diff=112207 1991 AIME Problems/Problem 4 2019-11-26T20:01:33Z <p>Will3145: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> How many [[real number]]s &lt;math&gt;x^{}_{}&lt;/math&gt; satisfy the [[equation]] &lt;math&gt;\frac{1}{5}\log_2 x = \sin (5\pi x)&lt;/math&gt;?<br /> <br /> == Solution ==<br /> &lt;center&gt;[[Image:AIME_1991_Solution_04.png]]&lt;/center&gt;<br /> The [[range]] of the [[sine]] function is &lt;math&gt;-1 \le y \le 1&lt;/math&gt;. It is [[periodic function|periodic]] (in this problem) with a period of &lt;math&gt;\frac{2}{5}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;-1 \le \frac{1}{5} \log_2 x \le 1&lt;/math&gt;, and &lt;math&gt;-5 \le \log_2 x \le 5&lt;/math&gt;. The solutions for &lt;math&gt;x&lt;/math&gt; occur in the domain of &lt;math&gt;\frac{1}{32} \le x \le 32&lt;/math&gt;. When &lt;math&gt;x &gt; 1&lt;/math&gt; the [[logarithm]] function returns a [[positive]] value; up to &lt;math&gt;x = 32&lt;/math&gt; it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of &lt;math&gt;x&lt;/math&gt;) of the sine curve and another curve that is &lt;math&gt;&lt; 1&lt;/math&gt;, so there are &lt;math&gt;\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154&lt;/math&gt; values (the subtraction of 6 since all the “intersections” when &lt;math&gt;x &lt; 1&lt;/math&gt; must be disregarded). When &lt;math&gt;y = 0&lt;/math&gt;, there is exactly &lt;math&gt;1&lt;/math&gt; touching point between the two functions: &lt;math&gt;\left(\frac{1}{5},0\right)&lt;/math&gt;. When &lt;math&gt;y &lt; 0&lt;/math&gt; or &lt;math&gt;x &lt; 1&lt;/math&gt;, we can count &lt;math&gt;4&lt;/math&gt; more solutions. The solution is &lt;math&gt;154 + 1 + 4 = \boxed{159}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Notice that the equation is satisfied twice for every sine period (which is &lt;math&gt;\frac{2}{5}&lt;/math&gt;), except in the sole case when the two equations equate to &lt;math&gt;0&lt;/math&gt;. In that case, the equation is satisfied twice but only at the one instance when &lt;math&gt;y=0&lt;/math&gt;. Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute:<br /> &lt;math&gt;32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed {159}&lt;/math&gt;<br /> <br /> == See also ==<br /> *[[Trigonometry]]<br /> *[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70180 Aops Topic]<br /> {{AIME box|year=1991|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_7&diff=112204 2004 AIME I Problems/Problem 7 2019-11-26T19:09:04Z <p>Will3145: /* Solutions */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; C &lt;/math&gt; be the [[coefficient]] of &lt;math&gt; x^2 &lt;/math&gt; in the expansion of the product &lt;math&gt; (1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x). &lt;/math&gt; Find &lt;math&gt; |C|. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Let our [[polynomial]] be &lt;math&gt;P(x)&lt;/math&gt;.<br /> <br /> It is clear that the coefficient of &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;P(x)&lt;/math&gt; is &lt;math&gt;-1 + 2 - 3 + \ldots + 14 - 15 = -8&lt;/math&gt;, so &lt;math&gt;P(x) = 1 -8x + Cx^2 + Q(x)&lt;/math&gt;, where &lt;math&gt;Q(x)&lt;/math&gt; is some polynomial [[divisibility | divisible]] by &lt;math&gt;x^3&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;P(-x) = 1 + 8x + Cx^2 + Q(-x)&lt;/math&gt; and so &lt;math&gt;P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)&lt;/math&gt;, where &lt;math&gt;R(x)&lt;/math&gt; is some polynomial divisible by &lt;math&gt;x^3&lt;/math&gt;.<br /> <br /> However, we also know &lt;math&gt;P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)&lt;/math&gt; &lt;math&gt;= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)&lt;/math&gt; &lt;math&gt;= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)&lt;/math&gt;.<br /> <br /> Equating coefficients, we have &lt;math&gt;2C - 64 = -(1 + 4 + \ldots + 225) = -1240&lt;/math&gt;, so &lt;math&gt;-2C = 1176&lt;/math&gt; and &lt;math&gt;|C| = \boxed{588}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of integers &lt;math&gt;\{-1,2,-3,\ldots,14,-15\}&lt;/math&gt;. The coefficient of &lt;math&gt;x^2&lt;/math&gt; in the expansion is equal to the sum of the product of each pair of distinct terms, or &lt;math&gt;C = \sum_{1 \le i \neq j}^{15} S_iS_j&lt;/math&gt;. Also, we know that <br /> &lt;cmath&gt;\begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &amp;= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &amp;= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}&lt;/cmath&gt;<br /> where the left-hand sum can be computed from:<br /> &lt;center&gt;&lt;math&gt;\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8&lt;/math&gt;&lt;/center&gt;<br /> and the right-hand sum comes from the formula for the sum of the first &lt;math&gt;n&lt;/math&gt; perfect squares. Therefore, &lt;math&gt;|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}&lt;/math&gt;.<br /> <br /> === Solution 3 (Bash)===<br /> <br /> Consider the set &lt;math&gt;[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]&lt;/math&gt;. Denote by &lt;math&gt;S&lt;/math&gt; all size 2 subsets of this set. Replace each element of &lt;math&gt;S&lt;/math&gt; by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;, we can simplify this to &lt;math&gt;|-1\cdot(-7)+2\cdot(-9)-3\cdot(-6)+4\cdot(-10)-5\cdot(-5)+\ldots+12\cdot(-14)-13\cdot(-1)+14\cdot(-15)|=|-588|=\boxed{588}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let set &lt;math&gt;N&lt;/math&gt; be &lt;math&gt;\{-1, -3, \ldots -15\}&lt;/math&gt; and set &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;\{2, 4, \ldots 14\}&lt;/math&gt;. The sum of the negative &lt;math&gt;x^2&lt;/math&gt; coefficients is the sum of the products of the elements in all two element sets such that one element is from &lt;math&gt;N&lt;/math&gt; and the other is from &lt;math&gt;P&lt;/math&gt;. Each summand is a term in the expansion of<br /> &lt;cmath&gt;(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)&lt;/cmath&gt;<br /> which equals &lt;math&gt;-56 * 64 = -(60^2 - 4^2) = -3584&lt;/math&gt;. The sum of the positive &lt;math&gt;x^2&lt;/math&gt; coefficients is the sum of the products of all two element sets such that the two elements are either both in &lt;math&gt;N&lt;/math&gt; or both in &lt;math&gt;P&lt;/math&gt;. By counting, the sum is &lt;math&gt;2992&lt;/math&gt;, so the sum of all &lt;math&gt;x^2&lt;/math&gt; coefficients is &lt;math&gt;-588&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{588}&lt;/math&gt;.<br /> <br /> <br /> == Solution 5==<br /> <br /> We can find out the coefficient of &lt;math&gt;x^2&lt;/math&gt; by multiplying every pair of two coefficients for &lt;math&gt;x&lt;/math&gt;. This means that we multiply &lt;math&gt;-1&lt;/math&gt; by &lt;math&gt;2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15&lt;/math&gt;. and etc. This sum can be easily simplified and is equal to &lt;math&gt;(-1)(-7)+(-3)(-6)+(-5)(-5)+(-7)(-4)+(-9)(-3)+(-11)(-2)+(-13)(-1)+2(-9)+4(-10)+6(-11)+8(-12)+10(-13)+12(-14)+14(-15)&lt;/math&gt; or &lt;math&gt;588&lt;/math&gt;.<br /> <br /> -David Camacho<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=112170 2019 AIME I Problems/Problem 14 2019-11-25T18:41:18Z <p>Will3145: /* Note to solution 1 */</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> The problem tells us that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we get &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, &lt;math&gt;ord_p(2019)&lt;/math&gt; = &lt;math&gt;1, 2, 4, 8,&lt;/math&gt; or &lt;math&gt;16&lt;/math&gt;<br /> <br /> However, if &lt;math&gt;ord_p(2019)&lt;/math&gt; = &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; clearly will be &lt;math&gt;1 \pmod{p} &lt;/math&gt; instead of &lt;math&gt;-1 \pmod{p}&lt;/math&gt;, causing a contradiction.<br /> <br /> Therefore, &lt;math&gt;ord_p(2019) = 16&lt;/math&gt;. Because &lt;math&gt;ord_p(2019) \vert \phi(p)&lt;/math&gt;, &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of 16. Since we know &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p(1 - \frac{1}{p})&lt;/math&gt; or &lt;math&gt;p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p&lt;/math&gt; must be &lt;math&gt;1 \pmod{16}&lt;/math&gt;. The two smallest primes that are &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt;, but &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{97}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(p)&lt;/math&gt; is called the &quot;Euler Function&quot; of integer &lt;math&gt;p&lt;/math&gt;.<br /> Euler theorem: define &lt;math&gt;\phi(p)&lt;/math&gt; as the number of positive integers less than &lt;math&gt;n&lt;/math&gt; but relatively prime to &lt;math&gt;n&lt;/math&gt;, then we have &lt;cmath&gt;\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})&lt;/cmath&gt; where &lt;math&gt;p_1,p_2,...,p_n&lt;/math&gt; are the prime factors of &lt;math&gt;p&lt;/math&gt;. Then, we have &lt;cmath&gt;a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)&lt;/cmath&gt; if &lt;math&gt;(a,p)=1&lt;/math&gt;.<br /> <br /> Furthermore, &lt;math&gt;ord_n(a)&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\ (\mathrm{mod}\ n)&lt;/math&gt;. An important property of the order is that &lt;math&gt;ord_n(a)|\phi(n)&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=112169 2019 AIME I Problems/Problem 14 2019-11-25T18:41:03Z <p>Will3145: /* Solution*/</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> The problem tells us that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we get &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, &lt;math&gt;ord_p(2019)&lt;/math&gt; = &lt;math&gt;1, 2, 4, 8,&lt;/math&gt; or &lt;math&gt;16&lt;/math&gt;<br /> <br /> However, if &lt;math&gt;ord_p(2019)&lt;/math&gt; = &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; clearly will be &lt;math&gt;1 \pmod{p} &lt;/math&gt; instead of &lt;math&gt;-1 \pmod{p}&lt;/math&gt;, causing a contradiction.<br /> <br /> Therefore, &lt;math&gt;ord_p(2019) = 16&lt;/math&gt;. Because &lt;math&gt;ord_p(2019) \vert \phi(p)&lt;/math&gt;, &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of 16. Since we know &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p(1 - \frac{1}{p})&lt;/math&gt; or &lt;math&gt;p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p&lt;/math&gt; must be &lt;math&gt;1 \pmod{16}&lt;/math&gt;. The two smallest primes that are &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt;, but &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{97}&lt;/math&gt;.<br /> <br /> ===Note to solution 1===<br /> &lt;math&gt;\phi(p)&lt;/math&gt; is called the &quot;Euler Function&quot; of integer &lt;math&gt;p&lt;/math&gt;.<br /> Euler theorem: define &lt;math&gt;\phi(p)&lt;/math&gt; as the number of positive integers less than &lt;math&gt;n&lt;/math&gt; but relatively prime to &lt;math&gt;n&lt;/math&gt;, then we have &lt;cmath&gt;\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})&lt;/cmath&gt; where &lt;math&gt;p_1,p_2,...,p_n&lt;/math&gt; are the prime factors of &lt;math&gt;p&lt;/math&gt;. Then, we have &lt;cmath&gt;a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)&lt;/cmath&gt; if &lt;math&gt;(a,p)=1&lt;/math&gt;.<br /> <br /> Furthermore, &lt;math&gt;ord_n(a)&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\ (\mathrm{mod}\ n)&lt;/math&gt;. An important property of the order is that &lt;math&gt;ord_n(a)|\phi(n)&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_6&diff=112140 2012 AIME I Problems/Problem 6 2019-11-25T02:16:15Z <p>Will3145: /* Solution */</p> <hr /> <div>==Problem 6==<br /> The complex numbers &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; satisfy &lt;math&gt;z^{13} = w,&lt;/math&gt; &lt;math&gt;w^{11} = z,&lt;/math&gt; and the imaginary part of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;\sin{\frac{m\pi}{n}}&lt;/math&gt;, for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;m&lt;n.&lt;/math&gt; Find &lt;math&gt;n.&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Substituting the first equation into the second, we find that &lt;math&gt;(z^{13})^{11} = z&lt;/math&gt; and thus &lt;math&gt;z^{143} = z.&lt;/math&gt; We know that &lt;math&gt;z \neq 0,&lt;/math&gt; because we are given the imaginary part of &lt;math&gt;z,&lt;/math&gt; so we can divide by &lt;math&gt;z&lt;/math&gt; to get &lt;math&gt;z^{142} = 1.&lt;/math&gt; So, &lt;math&gt;z&lt;/math&gt; must be a &lt;math&gt;142&lt;/math&gt;nd root of unity, and thus, by De Moivre's formula, the imaginary part of &lt;math&gt;z&lt;/math&gt; will be of the form &lt;math&gt;\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},&lt;/math&gt; where &lt;math&gt;k \in \{1, 2, \ldots, 70\}.&lt;/math&gt; Note that &lt;math&gt;71&lt;/math&gt; is prime and &lt;math&gt;k&lt;71&lt;/math&gt; by the conditions of the problem, so the denominator in the argument of this value will always be &lt;math&gt;71.&lt;/math&gt; Thus, &lt;math&gt;n = \boxed{071}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2012|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=111984 2019 AMC 8 Problems/Problem 5 2019-11-22T01:21:48Z <p>Will3145: shouldnt exist since it is the same thing as sol 1</p> <hr /> <div>== Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> <br /> [[File:2019_AMC_8_-4_Image_1.png|900px]]<br /> <br /> [[File:2019_AMC_8_-4_Image_2.png|600px]]<br /> <br /> ==Solution 1 (Using the answer choices)==<br /> First, the tortoise walks at a constant rate, ruling out &lt;math&gt;(D)&lt;/math&gt;<br /> Second, when the hare is resting, the distance will stay the same, ruling out &lt;math&gt;(E)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt;.<br /> Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> Therefore, the answer is the only one left.<br /> -Lcz<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=111785 2019 AMC 8 Problems/Problem 3 2019-11-20T22:42:28Z <p>Will3145: /* Problem 3 */</p> <hr /> <div>==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider subtracting 1 from each of the fractions. Our new fractions would then be &lt;math&gt;\frac{4}{11}, \frac{4}{15}, and \frac{4}{13}&lt;/math&gt;. Since &lt;math&gt;\frac{4}{15}&lt;\frac{4}{13}&lt;\frac{4}{11}&lt;/math&gt;, it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;<br /> <br /> -will3145<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=111783 2019 AMC 8 Problems/Problem 3 2019-11-20T22:42:12Z <p>Will3145: /* Solution*/</p> <hr /> <div>===Problem 3===<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Consider subtracting 1 from each of the fractions. Our new fractions would then be &lt;math&gt;\frac{4}{11}, \frac{4}{15}, and \frac{4}{13}&lt;/math&gt;. Since &lt;math&gt;\frac{4}{15}&lt;\frac{4}{13}&lt;\frac{4}{11}&lt;/math&gt;, it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;<br /> <br /> -will3145<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=111781 2019 AMC 8 Problems/Problem 3 2019-11-20T22:41:45Z <p>Will3145: /* Solution 1 */</p> <hr /> <div>===Problem 3===<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> Consider subtracting 1 from each of the fractions. Our new fractions would then be &lt;math&gt;\frac{4}{11}, \frac{4}{15}, and \frac{4}{13}&lt;/math&gt;. Since &lt;math&gt;\frac{4}{15}&lt;\frac{4}{13}&lt;\frac{4}{11}&lt;/math&gt;, it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=111780 2019 AMC 8 Problems/Problem 3 2019-11-20T22:40:53Z <p>Will3145: /* Solution 1 */</p> <hr /> <div>===Problem 3===<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> Consider subtracting 1 from each of the fractions. Our new fractions would then be &lt;math&gt;\frac{4}{11}, \frac{4}{15}, and \frac{4}{13}&lt;/math&gt;. Since &lt;math&gt;\frac{4}{15}&lt;\frac{4}{13}&lt;\frac{4}{11}&lt;/math&gt;, it follows that the answer is &lt;math&gt;\boxed{\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=111600 2019 AMC 8 Problems/Problem 3 2019-11-20T04:59:39Z <p>Will3145: Problem 3</p> <hr /> <div>===Problem===<br /> 3. Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ===Solution===<br /> Consider subtracting 1 from each of the fractions. Our new fractions would then be &lt;math&gt;\frac{4}{11}, \frac{4}{15}, and \frac{4}{13}&lt;/math&gt;. Since &lt;math&gt;\frac{4}{15}&lt;\frac{4}{13}&lt;\frac{4}{11}&lt;/math&gt;, it follows that the answer is &lt;math&gt;\boxed{\qquad\textbf{(E) }} \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_14&diff=111599 2013 AIME II Problems/Problem 14 2019-11-20T04:54:49Z <p>Will3145: /* Solution */</p> <hr /> <div>==Problem 14==<br /> For positive integers &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, let &lt;math&gt;f(n, k)&lt;/math&gt; be the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k&lt;/math&gt;, and for &lt;math&gt;n &gt; 1&lt;/math&gt; let &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;. Find the remainder when &lt;math&gt;\sum\limits_{n=20}^{100} F(n)&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> ===The Pattern===<br /> <br /> We can find that <br /> <br /> &lt;math&gt;20\equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;21\equiv 5 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;22\equiv 6 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;23\equiv 7 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;24\equiv 6 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;25\equiv 7 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;26\equiv 8 \pmod{9}&lt;/math&gt;<br /> <br /> Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get<br /> <br /> &lt;math&gt;99\equiv 31 \pmod{34}&lt;/math&gt;<br /> <br /> &lt;math&gt;100\equiv 32 \pmod{34}&lt;/math&gt;<br /> <br /> So the sum is &lt;math&gt;6+3\times(6+...+31)+31+32=1512&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;.<br /> By: Kris17<br /> <br /> ===The Intuition===<br /> First, let's see what happens if we remove a restriction. Let's define &lt;math&gt;G(x)&lt;/math&gt; as<br /> <br /> &lt;math&gt;G(x):=\max_{\substack{1\le k}} f(n, k)&lt;/math&gt;<br /> <br /> Now, if you set &lt;math&gt;k&lt;/math&gt; as any number greater than &lt;math&gt;n&lt;/math&gt;, you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence &lt;math&gt;H(x)&lt;/math&gt; is defined as<br /> <br /> &lt;math&gt;H(x):=\max_{\substack{1\le k\le n}} f(n, k)&lt;/math&gt;<br /> <br /> Now, after some thought, we find that if we set &lt;math&gt;k=\lfloor \frac{n}{2} \rfloor+1&lt;/math&gt; we get a remainder of &lt;math&gt;\lfloor \frac{n-1}{2} \rfloor&lt;/math&gt;, the max possible. Once we have gotten this far, it is easy to see that the original equation, <br /> <br /> &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;<br /> <br /> has a solution with &lt;math&gt;k=\lfloor \frac{n}{3} \rfloor+1&lt;/math&gt;.<br /> <br /> &lt;math&gt;W^5&lt;/math&gt;~Rowechen<br /> <br /> ===The Proof===<br /> The solution presented above does not prove why &lt;math&gt;F(x)&lt;/math&gt; is found by dividing &lt;math&gt;x&lt;/math&gt; by &lt;math&gt;3&lt;/math&gt;. Indeed, that is the case, as rigorously shown below.<br /> <br /> Consider the case where &lt;math&gt;x = 3k&lt;/math&gt;. We shall prove that &lt;math&gt;F(x) = f(x, k+1)&lt;/math&gt;.<br /> For all &lt;math&gt;x/2\geq n &gt; k+1, x = 2n + q&lt;/math&gt;, where &lt;math&gt;0\leq q\leq n&lt;/math&gt;. This is because &lt;math&gt;x &gt; 3k + 3 = 3n&lt;/math&gt; and &lt;math&gt;x &lt; n&lt;/math&gt;. Also, as n increases, &lt;math&gt;q&lt;/math&gt; decreases. Thus, &lt;math&gt;q = f(x, n) &lt; f(x, k+1) = k - 2&lt;/math&gt; for all &lt;math&gt;n &gt; k+1&lt;/math&gt;.<br /> Consider all &lt;math&gt;n &lt; k+1. f(x, k) = 0&lt;/math&gt; and &lt;math&gt;f(x, k-1) = 3&lt;/math&gt;. Also, &lt;math&gt;0 &lt; f(x, k-2) &lt; k-2&lt;/math&gt;. Thus, for &lt;math&gt;k &gt; 5, f(x, k+1) &gt; f(x, n)&lt;/math&gt; for &lt;math&gt;n &lt; k+1&lt;/math&gt;.<br /> <br /> Similar proofs apply for &lt;math&gt;x = 3k + 1&lt;/math&gt; and &lt;math&gt;x = 3k + 2&lt;/math&gt;. The reader should feel free to derive these proofs themself.<br /> <br /> ===Generalized Solution===<br /> <br /> &lt;math&gt;Lemma:&lt;/math&gt; Highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n/2&lt;/math&gt; is obtained for &lt;math&gt;k_0 = (n + (3 - n \pmod{3}))/3&lt;/math&gt; and the remainder thus obtained is &lt;math&gt;(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]&lt;/math&gt;. <br /> <br /> &lt;math&gt;Note:&lt;/math&gt; This is the second highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n&lt;/math&gt; and the highest remainder occurs when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+1)/2&lt;/math&gt; for odd &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+2)/2&lt;/math&gt; for even &lt;math&gt;n&lt;/math&gt;.<br /> <br /> Using the lemma above:<br /> <br /> &lt;math&gt;\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}] &lt;/math&gt;<br /> &lt;math&gt;= [(120*81/2)/3 - 2*81 + (2/3)*81]<br /> = 1512&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;<br /> <br /> Proof of Lemma: It is similar to &lt;math&gt;The Proof&lt;/math&gt; stated above. <br /> <br /> Kris17<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_14&diff=111598 2013 AIME II Problems/Problem 14 2019-11-20T04:54:30Z <p>Will3145: /* The Pattern */</p> <hr /> <div>==Problem 14==<br /> For positive integers &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, let &lt;math&gt;f(n, k)&lt;/math&gt; be the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k&lt;/math&gt;, and for &lt;math&gt;n &gt; 1&lt;/math&gt; let &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;. Find the remainder when &lt;math&gt;\sum\limits_{n=20}^{100} F(n)&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solutions==<br /> <br /> ===The Pattern===<br /> <br /> We can find that <br /> <br /> &lt;math&gt;20\equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;21\equiv 5 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;22\equiv 6 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;23\equiv 7 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;24\equiv 6 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;25\equiv 7 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;26\equiv 8 \pmod{9}&lt;/math&gt;<br /> <br /> Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get<br /> <br /> &lt;math&gt;99\equiv 31 \pmod{34}&lt;/math&gt;<br /> <br /> &lt;math&gt;100\equiv 32 \pmod{34}&lt;/math&gt;<br /> <br /> So the sum is &lt;math&gt;6+3\times(6+...+31)+31+32=1512&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;.<br /> By: Kris17<br /> <br /> ===The Intuition===<br /> First, let's see what happens if we remove a restriction. Let's define &lt;math&gt;G(x)&lt;/math&gt; as<br /> <br /> &lt;math&gt;G(x):=\max_{\substack{1\le k}} f(n, k)&lt;/math&gt;<br /> <br /> Now, if you set &lt;math&gt;k&lt;/math&gt; as any number greater than &lt;math&gt;n&lt;/math&gt;, you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence &lt;math&gt;H(x)&lt;/math&gt; is defined as<br /> <br /> &lt;math&gt;H(x):=\max_{\substack{1\le k\le n}} f(n, k)&lt;/math&gt;<br /> <br /> Now, after some thought, we find that if we set &lt;math&gt;k=\lfloor \frac{n}{2} \rfloor+1&lt;/math&gt; we get a remainder of &lt;math&gt;\lfloor \frac{n-1}{2} \rfloor&lt;/math&gt;, the max possible. Once we have gotten this far, it is easy to see that the original equation, <br /> <br /> &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;<br /> <br /> has a solution with &lt;math&gt;k=\lfloor \frac{n}{3} \rfloor+1&lt;/math&gt;.<br /> <br /> &lt;math&gt;W^5&lt;/math&gt;~Rowechen<br /> <br /> ===The Proof===<br /> The solution presented above does not prove why &lt;math&gt;F(x)&lt;/math&gt; is found by dividing &lt;math&gt;x&lt;/math&gt; by &lt;math&gt;3&lt;/math&gt;. Indeed, that is the case, as rigorously shown below.<br /> <br /> Consider the case where &lt;math&gt;x = 3k&lt;/math&gt;. We shall prove that &lt;math&gt;F(x) = f(x, k+1)&lt;/math&gt;.<br /> For all &lt;math&gt;x/2\geq n &gt; k+1, x = 2n + q&lt;/math&gt;, where &lt;math&gt;0\leq q\leq n&lt;/math&gt;. This is because &lt;math&gt;x &gt; 3k + 3 = 3n&lt;/math&gt; and &lt;math&gt;x &lt; n&lt;/math&gt;. Also, as n increases, &lt;math&gt;q&lt;/math&gt; decreases. Thus, &lt;math&gt;q = f(x, n) &lt; f(x, k+1) = k - 2&lt;/math&gt; for all &lt;math&gt;n &gt; k+1&lt;/math&gt;.<br /> Consider all &lt;math&gt;n &lt; k+1. f(x, k) = 0&lt;/math&gt; and &lt;math&gt;f(x, k-1) = 3&lt;/math&gt;. Also, &lt;math&gt;0 &lt; f(x, k-2) &lt; k-2&lt;/math&gt;. Thus, for &lt;math&gt;k &gt; 5, f(x, k+1) &gt; f(x, n)&lt;/math&gt; for &lt;math&gt;n &lt; k+1&lt;/math&gt;.<br /> <br /> Similar proofs apply for &lt;math&gt;x = 3k + 1&lt;/math&gt; and &lt;math&gt;x = 3k + 2&lt;/math&gt;. The reader should feel free to derive these proofs themself.<br /> <br /> ===Generalized Solution===<br /> <br /> &lt;math&gt;Lemma:&lt;/math&gt; Highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n/2&lt;/math&gt; is obtained for &lt;math&gt;k_0 = (n + (3 - n \pmod{3}))/3&lt;/math&gt; and the remainder thus obtained is &lt;math&gt;(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]&lt;/math&gt;. <br /> <br /> &lt;math&gt;Note:&lt;/math&gt; This is the second highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n&lt;/math&gt; and the highest remainder occurs when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+1)/2&lt;/math&gt; for odd &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+2)/2&lt;/math&gt; for even &lt;math&gt;n&lt;/math&gt;.<br /> <br /> Using the lemma above:<br /> <br /> &lt;math&gt;\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}] &lt;/math&gt;<br /> &lt;math&gt;= [(120*81/2)/3 - 2*81 + (2/3)*81]<br /> = 1512&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;<br /> <br /> Proof of Lemma: It is similar to &lt;math&gt;The Proof&lt;/math&gt; stated above. <br /> <br /> Kris17<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_6&diff=110236 2004 AIME I Problems/Problem 6 2019-10-11T21:17:33Z <p>Will3145: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> An integer is called snakelike if its decimal representation &lt;math&gt; a_1a_2a_3\cdots a_k &lt;/math&gt; satisfies &lt;math&gt; a_i&lt;a_{i+1} &lt;/math&gt; if &lt;math&gt; i &lt;/math&gt; is [[odd integer | odd]] and &lt;math&gt; a_i&gt;a_{i+1} &lt;/math&gt; if &lt;math&gt; i &lt;/math&gt; is [[even integer | even]]. How many snakelike integers between 1000 and 9999 have four distinct digits?<br /> <br /> == Solution 1 ==<br /> <br /> We divide the problem into two cases: one in which zero is one of the digits and one in which it is not. In the latter case, suppose we pick digits <br /> &lt;math&gt;x_1,x_2,x_3,x_4&lt;/math&gt; such that &lt;math&gt;x_1&lt;x_2&lt;x_3&lt;x_4&lt;/math&gt;. There are five arrangements of these digits that satisfy the condition of being snakelike: &lt;math&gt;x_1x_3x_2x_4&lt;/math&gt;, &lt;math&gt;x_1x_4x_2x_3&lt;/math&gt;, &lt;math&gt;x_2x_3x_1x_4&lt;/math&gt;, &lt;math&gt;x_2x_4x_1x_3&lt;/math&gt;, &lt;math&gt;x_3x_4x_1x_2&lt;/math&gt;. Thus there are &lt;math&gt;5\cdot {9\choose 4}=630&lt;/math&gt; snakelike numbers which do not contain the digit zero.<br /> <br /> In the second case we choose zero and three other digits such that &lt;math&gt;0&lt;x_2&lt;x_3&lt;x_4&lt;/math&gt;. There are three arrangements of these digits that satisfy the condition of being snakelike: &lt;math&gt;x_2x_30x_4&lt;/math&gt;, &lt;math&gt;x_2x_40x_3&lt;/math&gt;, &lt;math&gt;x_3x_40x_2&lt;/math&gt;. Because we know that zero is a digit, there are &lt;math&gt;3\cdot{9\choose 3}=252&lt;/math&gt; snakelike numbers which contain the digit zero. Thus there are &lt;math&gt;630+252=\boxed{882}&lt;/math&gt; snakelike numbers.<br /> <br /> == Solution 2 ==<br /> Let's create the snakelike number from digits &lt;math&gt;a &lt; b &lt; c &lt; d&lt;/math&gt;, and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of &lt;math&gt;5\cdot{10 \choose 4}&lt;/math&gt; But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits &lt;math&gt;a &lt; b &lt; c&lt;/math&gt;. There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is &lt;math&gt;2\cdot{9 \choose 3}&lt;/math&gt;. Thus our answer is &lt;math&gt;5\cdot{10 \choose 4} - 2\cdot{9 \choose 3} = \boxed{882}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_11&diff=107551 1993 AIME Problems/Problem 11 2019-07-11T00:26:05Z <p>Will3145: LaTeX</p> <hr /> <div>== Problem ==<br /> Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is &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 are the last three digits of &lt;math&gt;m+n\,&lt;/math&gt;? <br /> <br /> == Solution ==<br /> The probability that the &lt;math&gt;n&lt;/math&gt;th flip in each game occurs and is a head is &lt;math&gt;\frac{1}{2^n}&lt;/math&gt;. The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is &lt;math&gt;\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots = \frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}&lt;/math&gt;, and the probability of the second person winning is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the probability that Alfred wins the &lt;math&gt;n&lt;/math&gt;th game, and let &lt;math&gt;b_n&lt;/math&gt; be the probability that Bonnie wins the &lt;math&gt;n&lt;/math&gt;th game. <br /> <br /> If Alfred wins the &lt;math&gt;n&lt;/math&gt;th game, then the probability that Alfred wins the &lt;math&gt;n+1&lt;/math&gt;th game is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. If Bonnie wins the &lt;math&gt;n&lt;/math&gt;th game, then the probability that Alfred wins the &lt;math&gt;n+1&lt;/math&gt;th game is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;a_{n+1}=\frac{1}{3}a_n+\frac{2}{3}b_n&lt;/math&gt;. <br /> <br /> Similarly, &lt;math&gt;b_{n+1}=\frac{2}{3}a_n+\frac{1}{3}b_n&lt;/math&gt;. <br /> <br /> Since Alfred goes first in the &lt;math&gt;1&lt;/math&gt;st game, &lt;math&gt;(a_1,b_1)=\left(\frac{2}{3}, \frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Using these [[recursion|recursive]] equations: <br /> <br /> &lt;math&gt;(a_2,b_2)=\left(\frac{4}{9}, \frac{5}{9}\right)&lt;/math&gt; <br /> <br /> &lt;math&gt;(a_3,b_3)=\left(\frac{14}{27}, \frac{13}{27}\right)&lt;/math&gt;<br /> <br /> &lt;math&gt;(a_4,b_4)=\left(\frac{40}{81}, \frac{41}{81}\right)&lt;/math&gt; <br /> <br /> &lt;math&gt;(a_5,b_5)=\left(\frac{122}{243}, \frac{121}{243}\right)&lt;/math&gt; <br /> <br /> &lt;math&gt;(a_6,b_6)=\left(\frac{364}{729}, \frac{365}{729}\right)&lt;/math&gt; <br /> <br /> Since &lt;math&gt;a_6=\frac{364}{729}&lt;/math&gt;, &lt;math&gt;m+n = 1093 \equiv \boxed{093} \pmod{1000}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Kind of Bashy) ==<br /> In order to begin this problem, we need to calculate the probability that Alfred will win on the first round. Because he goes first, Alfred has a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of winning (getting heads) on his first flip. Then, Bonnie, who goes second, has a &lt;math&gt;\frac{1}{2}&lt;/math&gt; * &lt;math&gt;\frac{1}{2}&lt;/math&gt; = &lt;math&gt;\frac{1}{4}&lt;/math&gt;, chance of winning on her first coin toss. Therefore Alfred’s chance of winning on his second flip is &lt;math&gt;\frac{1}{4}&lt;/math&gt; * &lt;math&gt;\frac{1}{2}&lt;/math&gt; = &lt;math&gt;\frac{1}{8}&lt;/math&gt;<br /> <br /> From this, we can see that Alfred’s (who goes first) chance of winning the first round is:<br /> <br /> &lt;cmath&gt;\frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots = \frac{2}{3}&lt;/cmath&gt;<br /> <br /> Bonnie’s (who goes second) chance of winning the first round is then 1 - &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> This means that the person who goes first has a &lt;math&gt;\frac{2}{3}&lt;/math&gt; chance of winning the round, while the person who goes second has a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance of winning.<br /> <br /> Now, through casework, we can calculate Alfred’s chance of winning the second round.<br /> <br /> Case 1: Alfred wins twice; &lt;math&gt;\frac{2}{3}&lt;/math&gt; * &lt;math&gt;\frac{1}{3}&lt;/math&gt; (Bonnie goes first this round) = &lt;math&gt;\frac{2}{9}&lt;/math&gt;.<br /> <br /> Case 2: Alfred loses the first round, but wins the second; &lt;math&gt;\frac{1}{3}&lt;/math&gt; * &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\frac{2}{9}&lt;/math&gt;.<br /> <br /> Adding up the cases, we get &lt;math&gt;\frac{2}{9}&lt;/math&gt; + &lt;math&gt;\frac{2}{9}&lt;/math&gt; = &lt;math&gt;\frac{4}{9}&lt;/math&gt;.<br /> <br /> Alfred, therefore, has a &lt;math&gt;\frac{4}{9}&lt;/math&gt; of winnning the second round, and Bonnie has a 1- &lt;math&gt;\frac{4}{9}&lt;/math&gt; = &lt;math&gt;\frac{5}{9}&lt;/math&gt; chance of winning this round.<br /> <br /> From here, it is not difficult to see that the probabilities alternate in a pattern. Make A the probability that Alfred wins a round.<br /> <br /> The chances of Alfred and Bonnie, respectively, winning the first round are A and A - &lt;math&gt;\frac{1}{3}&lt;/math&gt;, which can be written as 2A - &lt;math&gt;\frac{1}{3}&lt;/math&gt; = 1<br /> <br /> The second round’s for chances are A and A + &lt;math&gt;\frac{1}{9}&lt;/math&gt;, which can also be written as 2A + &lt;math&gt;\frac{1}{9}&lt;/math&gt;<br /> <br /> From this, we can conclude that for the &lt;math&gt;n&lt;/math&gt;th even round, the probability that Alfred (A) wins can be calculated through the equation 2A + &lt;math&gt;\frac{1}{3^n} = 1&lt;/math&gt;.<br /> <br /> Solving the equation for &lt;math&gt;n&lt;/math&gt; = 6, we get A = &lt;math&gt;\frac{364}{729}&lt;/math&gt;.<br /> <br /> 364 + 729 = 1093, so our answer is &lt;math&gt;\boxed{093}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_11&diff=107550 1993 AIME Problems/Problem 11 2019-07-11T00:24:28Z <p>Will3145: LaTeX</p> <hr /> <div>== Problem ==<br /> Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is &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 are the last three digits of &lt;math&gt;m+n\,&lt;/math&gt;? <br /> <br /> == Solution ==<br /> The probability that the &lt;math&gt;n&lt;/math&gt;th flip in each game occurs and is a head is &lt;math&gt;\frac{1}{2^n}&lt;/math&gt;. The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is &lt;math&gt;\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots = \frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}&lt;/math&gt;, and the probability of the second person winning is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the probability that Alfred wins the &lt;math&gt;n&lt;/math&gt;th game, and let &lt;math&gt;b_n&lt;/math&gt; be the probability that Bonnie wins the &lt;math&gt;n&lt;/math&gt;th game. <br /> <br /> If Alfred wins the &lt;math&gt;n&lt;/math&gt;th game, then the probability that Alfred wins the &lt;math&gt;n+1&lt;/math&gt;th game is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. If Bonnie wins the &lt;math&gt;n&lt;/math&gt;th game, then the probability that Alfred wins the &lt;math&gt;n+1&lt;/math&gt;th game is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;a_{n+1}=\frac{1}{3}a_n+\frac{2}{3}b_n&lt;/math&gt;. <br /> <br /> Similarly, &lt;math&gt;b_{n+1}=\frac{2}{3}a_n+\frac{1}{3}b_n&lt;/math&gt;. <br /> <br /> Since Alfred goes first in the &lt;math&gt;1&lt;/math&gt;st game, &lt;math&gt;(a_1,b_1)=\left(\frac{2}{3}, \frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Using these [[recursion|recursive]] equations: <br /> <br /> &lt;math&gt;(a_2,b_2)=\left(\frac{4}{9}, \frac{5}{9}\right)&lt;/math&gt; <br /> <br /> &lt;math&gt;(a_3,b_3)=\left(\frac{14}{27}, \frac{13}{27}\right)&lt;/math&gt;<br /> <br /> &lt;math&gt;(a_4,b_4)=\left(\frac{40}{81}, \frac{41}{81}\right)&lt;/math&gt; <br /> <br /> &lt;math&gt;(a_5,b_5)=\left(\frac{122}{243}, \frac{121}{243}\right)&lt;/math&gt; <br /> <br /> &lt;math&gt;(a_6,b_6)=\left(\frac{364}{729}, \frac{365}{729}\right)&lt;/math&gt; <br /> <br /> Since &lt;math&gt;a_6=\frac{364}{729}&lt;/math&gt;, &lt;math&gt;m+n = 1093 \equiv \boxed{093} \pmod{1000}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Kind of Bashy) ==<br /> In order to begin this problem, we need to calculate the probability that Alfred will win on the first round. Because he goes first, Alfred has a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of winning (getting heads) on his first flip. Then, Bonnie, who goes second, has a &lt;math&gt;\frac{1}{2}&lt;/math&gt; * &lt;math&gt;\frac{1}{2}&lt;/math&gt; = &lt;math&gt;\frac{1}{4}&lt;/math&gt;, chance of winning on her first coin toss. Therefore Alfred’s chance of winning on his second flip is &lt;math&gt;\frac{1}{4}&lt;/math&gt; * &lt;math&gt;\frac{1}{2}&lt;/math&gt; = &lt;math&gt;\frac{1}{8}&lt;/math&gt;<br /> <br /> From this, we can see that Alfred’s (who goes first) chance of winning the first round is:<br /> <br /> &lt;cmath&gt;\frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots = \frac{2}{3}&lt;/cmath&gt;<br /> <br /> Bonnie’s (who goes second) chance of winning the first round is then 1 - &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\frac{1}{3}&lt;/math&gt;.<br /> <br /> This means that the person who goes first has a &lt;math&gt;\frac{2}{3}&lt;/math&gt; chance of winning the round, while the person who goes second has a &lt;math&gt;\frac{1}{3}&lt;/math&gt; chance of winning.<br /> <br /> Now, through casework, we can calculate Alfred’s chance of winning the second round.<br /> <br /> Case 1: Alfred wins twice; &lt;math&gt;\frac{2}{3}&lt;/math&gt; * &lt;math&gt;\frac{1}{3}&lt;/math&gt; (Bonnie goes first this round) = &lt;math&gt;\frac{2}{9}&lt;/math&gt;.<br /> <br /> Case 2: Alfred loses the first round, but wins the second; &lt;math&gt;\frac{1}{3}&lt;/math&gt; * &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\frac{2}{9}&lt;/math&gt;.<br /> <br /> Adding up the cases, we get &lt;math&gt;\frac{2}{9}&lt;/math&gt; + &lt;math&gt;\frac{2}{9}&lt;/math&gt; = &lt;math&gt;\frac{4}{9}&lt;/math&gt;.<br /> <br /> Alfred, therefore, has a &lt;math&gt;\frac{4}{9}&lt;/math&gt; of winnning the second round, and Bonnie has a 1- &lt;math&gt;\frac{4}{9}&lt;/math&gt; = &lt;math&gt;\frac{5}{9}&lt;/math&gt; chance of winning this round.<br /> <br /> From here, it is not difficult to see that the probabilities alternate in a pattern. Make A the probability that Alfred wins a round.<br /> <br /> The chances of Alfred and Bonnie, respectively, winning the first round are A and A - &lt;math&gt;\frac{1}{3}&lt;/math&gt;, which can be written as 2A - &lt;math&gt;\frac{1}{3}&lt;/math&gt; = 1<br /> <br /> The second round’s for chances are A and A + &lt;math&gt;\frac{1}{9}&lt;/math&gt;, which can also be written as 2A + &lt;math&gt;\frac{1}{9}&lt;/math&gt;<br /> <br /> From this, we can conclude that for the &lt;math&gt;n&lt;/math&gt;th even round, the probability that Alfred (A) wins can be calculated through the equation 2A + &lt;math&gt;\frac{1}{3^n} = 1&lt;/math&gt;.<br /> <br /> Solving the equation for &lt;math&gt;n&lt;/math&gt; = 6, we get A = &lt;math&gt;\frac{364}{729}&lt;/math&gt;.<br /> <br /> 364 + 729 = 1093, so our answer is &lt;math&gt;\boxed{093}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1996_AHSME_Problems/Problem_25&diff=107498 1996 AHSME Problems/Problem 25 2019-07-08T18:05:34Z <p>Will3145: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Given that &lt;math&gt;x^2 + y^2 = 14x + 6y + 6&lt;/math&gt;, what is the largest possible value that &lt;math&gt;3x + 4y&lt;/math&gt; can have? <br /> <br /> &lt;math&gt; \text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Complete the square to get <br /> &lt;cmath&gt;(x-7)^2 + (y-3)^2 = 64.&lt;/cmath&gt;<br /> Applying Cauchy-Schwarz directly,<br /> &lt;cmath&gt;64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.&lt;/cmath&gt;<br /> &lt;cmath&gt; 40 \ge 3x+4y-33 &lt;/cmath&gt;<br /> &lt;cmath&gt;3x+4y \le 73.&lt;/cmath&gt;<br /> Thus our answer is &lt;math&gt;\boxed{(B)}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Geometric)==<br /> <br /> The first equation is a [[circle]], so we find its center and [[radius]] by [[completing the square]]: <br /> &lt;math&gt;x^2 - 14x + y^2 - 6y = 6&lt;/math&gt;, so &lt;cmath&gt;(x-7)^2 + (y-3)^2 = (x^2- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.&lt;/cmath&gt;<br /> <br /> So we have a circle centered at &lt;math&gt;(7,3)&lt;/math&gt; with radius &lt;math&gt;8&lt;/math&gt;, and we want to find the max of &lt;math&gt;3x + 4y&lt;/math&gt;.<br /> <br /> The set of lines &lt;math&gt;3x + 4y = A&lt;/math&gt; are all [[parallel]], with slope &lt;math&gt;-\frac{3}{4}&lt;/math&gt;. Increasing &lt;math&gt;A&lt;/math&gt; shifts the lines up and/or to the right.<br /> <br /> We want to shift this line up high enough that it's [[tangent (geometry)|tangent]] to the circle, but not so high that it misses the circle altogether. This means &lt;math&gt;3x + 4y = A&lt;/math&gt; will be tangent to the circle.<br /> <br /> Imagine that this line hits the circle at point &lt;math&gt;(a,b)&lt;/math&gt;. The [[slope]] of the radius connecting the center of the circle, &lt;math&gt;(7,3)&lt;/math&gt;, to tangent point &lt;math&gt;(a,b)&lt;/math&gt; will be &lt;math&gt;\frac{4}{3}&lt;/math&gt;, since the radius is perpendicular to the tangent line.<br /> <br /> So we have a point, &lt;math&gt;(7,3)&lt;/math&gt;, and a slope of &lt;math&gt;\frac{4}{3}&lt;/math&gt; that represents the slope of the radius to the tangent point. Let's start at the point &lt;math&gt;(7,3)&lt;/math&gt;. If we go &lt;math&gt;4k&lt;/math&gt; units up and &lt;math&gt;3k&lt;/math&gt; units right from &lt;math&gt;(7,3)&lt;/math&gt;, we would arrive at a point that's &lt;math&gt;5k&lt;/math&gt; units away. But in reality we want &lt;math&gt;5k = 8&lt;/math&gt; to reach the tangent point, since the radius of the circle is &lt;math&gt;8&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;k = \frac{8}{5}&lt;/math&gt;, and we want to travel &lt;math&gt;4\cdot \frac{8}{5}&lt;/math&gt; up and &lt;math&gt;3\cdot \frac{8}{5}&lt;/math&gt; over from the point &lt;math&gt;(7,3)&lt;/math&gt; to reach our maximum. This means the maximum value of &lt;math&gt;3x + 4y&lt;/math&gt; occurs at &lt;math&gt;\left(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5}\right)&lt;/math&gt;, which is &lt;math&gt;\left(\frac{59}{5}, \frac{47}{5}\right).&lt;/math&gt;<br /> <br /> Plug in those values for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, and you get the maximum value of &lt;math&gt;3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}&lt;/math&gt;, which is option &lt;math&gt;\boxed{(B)}&lt;/math&gt;.<br /> ==Solution 2B==<br /> Let the tangent point be &lt;math&gt;P&lt;/math&gt;, and the tangent line's x-intercept be &lt;math&gt;Q&lt;/math&gt;. Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle &lt;math&gt;OPK&lt;/math&gt; is 3-4-5, &lt;math&gt;OP=8&lt;/math&gt;, so &lt;math&gt;OK = \frac{5}{3}*8 = \frac{40}{3}&lt;/math&gt;. Note that the horizontal distance from &lt;math&gt;O&lt;/math&gt; to the origin is &lt;math&gt;7&lt;/math&gt;, and the horizontal distance from K to Q is 4, (&lt;math&gt;\frac{4}{3}&lt;/math&gt; of its y coordinate), so the x-intercept is &lt;math&gt;7+4+OK = 73/3&lt;/math&gt;. The value of &lt;math&gt;3x+4y&lt;/math&gt; is 73 at point &lt;math&gt;Q&lt;/math&gt;. Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of &lt;math&gt;P&lt;/math&gt;. &lt;math&gt;\boxed{(B)}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;z = 3x + 4y&lt;/math&gt;. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y = (z - 3x)/4&lt;/math&gt;. Substituting into the given equation, we get<br /> &lt;cmath&gt;x^2 + \left( \frac{z - 3x}{4} \right)^2 = 14x + 6 \cdot \frac{z - 3x}{4} + 6,&lt;/cmath&gt;<br /> which simplifies to<br /> &lt;cmath&gt;25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.&lt;/cmath&gt;<br /> <br /> This quadratic equation has real roots in &lt;math&gt;x&lt;/math&gt; if and only if its discriminant is nonnegative, so<br /> &lt;cmath&gt;(6z + 152)^2 - 4 \cdot 25 \cdot (z^2 - 24z - 96) \ge 0,&lt;/cmath&gt;<br /> which simplifies to<br /> &lt;cmath&gt;-64z^2 + 4224z + 32704 \ge 0,&lt;/cmath&gt;<br /> which can be factored as<br /> &lt;cmath&gt;-64(z + 7)(z - 73) \ge 0.&lt;/cmath&gt;<br /> The largest value of &lt;math&gt;z&lt;/math&gt; that satisfies this inequality is &lt;math&gt;\boxed{73}&lt;/math&gt;, which is &lt;math&gt;\boxed{(B)}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 4 (Using Answer Choice + Calculus)==<br /> Implicitly differentiating the given equation with respect to &lt;math&gt;x&lt;/math&gt; yields:<br /> <br /> &lt;math&gt;2x + 2y\frac{dy}{dx} = 14 + 6\frac{dy}{dx}&lt;/math&gt;<br /> <br /> Now solve for &lt;math&gt;\frac{dy}{dx}&lt;/math&gt; to obtain:<br /> <br /> &lt;math&gt;\frac{dy}{dx} = -\frac{x - 7}{y - 3}&lt;/math&gt;<br /> <br /> Set the equation equal to zero to find the maximum occurs at &lt;math&gt;x = 7&lt;/math&gt;<br /> <br /> Plug this back into the equation that we are trying to maximize and see that we are left with: <br /> &lt;math&gt;21 + 4y&lt;/math&gt;.<br /> <br /> The only answer choice that can be obtained from this equation is &lt;math&gt;\bf{73}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AHSME box|year=1996|num-b=24|num-a=26}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Introductory Algebra Problems]]<br /> <br /> [[Category:Circle Problems]]<br /> {{MAA Notice}}<br /> &lt;math&gt;aopsswag&lt;/math&gt;</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=Ball-and-urn&diff=107388 Ball-and-urn 2019-07-05T20:14:10Z <p>Will3145: LaTeX</p> <hr /> <div>The '''ball-and-urn''' technique, also known as '''stars-and-bars''', is a commonly used technique in [[combinatorics]].<br /> <br /> It is used to solve problems of the form: how many ways can one distribute &lt;math&gt;k&lt;/math&gt; indistinguishable objects into &lt;math&gt;n&lt;/math&gt; bins? We can imagine this as finding the number of ways to drop &lt;math&gt;k&lt;/math&gt; balls into &lt;math&gt;n&lt;/math&gt; urns, or equivalently to drop &lt;math&gt;k&lt;/math&gt; balls amongst &lt;math&gt;n-1&lt;/math&gt; dividers. The number of ways to do such is &lt;math&gt;{n+k-1 \choose k}&lt;/math&gt;.<br /> <br /> <br /> == Reasoning (One of Several) ==<br /> If you could only put one ball in each urn, then there would be &lt;math&gt;{n \choose k}&lt;/math&gt; possibilities; the problem is that you can repeat urns, so this does not work. You can, however, reframe the problem as so: imagine that you have the &lt;math&gt;n&lt;/math&gt; urns (numbered 1 through &lt;math&gt;n&lt;/math&gt;) and then you also have &lt;math&gt;k-1&lt;/math&gt; urns labeled &quot;repeat 1st&quot;, &quot;repeat 2nd&quot;, ..., and &quot;repeat &lt;math&gt;k-1&lt;/math&gt;-th&quot;. After the balls are in urns you can imagine that any balls in the &quot;repeat&quot; urns are moved on top of the correct balls in the first &lt;math&gt;n&lt;/math&gt; urns, moving from left to right. There is a one-to-one correspondence between the non-repeating arrangements in these new urns and the repeats-allowed arrangements in the original urns. <br /> <br /> For a simple example, consider &lt;math&gt;4&lt;/math&gt; balls and &lt;math&gt;5&lt;/math&gt; urns. The one to one correspondence between several of the possibilities and the &quot;repeated urns&quot; version is shown. Since there are 4 balls, these examples will have three possible &quot;repeat&quot; urns. For simplicity, I am listing the numbers of the urns with balls in them, so &quot;1,1,2,4&quot; means &lt;math&gt;2&lt;/math&gt; balls in urn &lt;math&gt;1, 1&lt;/math&gt; in urn &lt;math&gt;2,&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; in urn &lt;math&gt;4.&lt;/math&gt; The same is true for the &quot;repeat&quot; urns options but I use the notation &lt;math&gt;r_1&lt;/math&gt; etc.<br /> <br /> *&lt;math&gt;1,2,3,4 \leftrightarrow 1,2,3,4&lt;/math&gt; (no repeats).<br /> *&lt;math&gt;1,1,2,4 \leftrightarrow 1,2,4,&lt;/math&gt; &lt;math&gt;r_1&lt;/math&gt;.<br /> *&lt;math&gt;1,2,2,2 \leftrightarrow 1,2,&lt;/math&gt; &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt;.<br /> *&lt;math&gt;4,4,5,5 \leftrightarrow 4,5,&lt;/math&gt; &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;.<br /> <br /> Since the re-framed version of the problem has &lt;math&gt;n+k-1&lt;/math&gt; urns, and &lt;math&gt;k&lt;/math&gt; balls that can each only go in one urn, the number of possible scenarios is simply &lt;math&gt;{n+k-1 <br /> \choose k}.&lt;/math&gt;<br /> <br /> == Problems ==<br /> *[[2003 AMC 10A Problems/Problem 21]]<br /> *[[Mock AIME 3 Pre 2005 Problems/Problem 2]]<br /> *[[2007 AIME I Problems/Problem 10]]<br /> *[[1986 AIME Problems/Problem 13]]<br /> *[[2018 AMC 10A Problems/Problem 11]]<br /> [[Category:Combinatorics]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_4&diff=106620 1993 AIME Problems/Problem 4 2019-06-19T01:14:14Z <p>Will3145: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> How many ordered four-tuples of integers &lt;math&gt;(a,b,c,d)\,&lt;/math&gt; with &lt;math&gt;0 &lt; a &lt; b &lt; c &lt; d &lt; 500\,&lt;/math&gt; satisfy &lt;math&gt;a + d = b + c\,&lt;/math&gt; and &lt;math&gt;bc - ad = 93\,&lt;/math&gt;?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;k = a + d = b + c&lt;/math&gt; so &lt;math&gt;d = k-a, b=k-c&lt;/math&gt;. It follows that &lt;math&gt;(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93&lt;/math&gt;. Hence &lt;math&gt;(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)&lt;/math&gt;.<br /> <br /> Solve them in tems of &lt;math&gt;c&lt;/math&gt; to get <br /> &lt;math&gt;(a,b,c,d) = (c - 93,c - 92,c,c + 1),&lt;/math&gt; &lt;math&gt;(c - 31,c - 28,c,c + 3),&lt;/math&gt; &lt;math&gt;(c - 1,c + 92,c,c + 93),&lt;/math&gt; &lt;math&gt;(c - 3,c + 28,c,c + 31)&lt;/math&gt;. The last two solutions don't follow &lt;math&gt;a &lt; b &lt; c &lt; d&lt;/math&gt;, so we only need to consider the first two solutions.<br /> <br /> The first solution gives us &lt;math&gt;c - 93\geq 1&lt;/math&gt; and &lt;math&gt;c + 1\leq 499&lt;/math&gt; &lt;math&gt;\implies 94\leq c\leq 498&lt;/math&gt;, and the second one gives us &lt;math&gt;32\leq c\leq 496&lt;/math&gt;.<br /> <br /> So the total number of such four-tuples is &lt;math&gt;405 + 465 = \huge{\boxed{870}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;b = a + m&lt;/math&gt; and &lt;math&gt;c = a + m + n&lt;/math&gt;. From &lt;math&gt;a + d = b + c&lt;/math&gt;, &lt;math&gt;d = b + c - a = a + 2m + n&lt;/math&gt;. <br /> <br /> Substituting &lt;math&gt;b = a + m&lt;/math&gt;, &lt;math&gt;c = a + m + n&lt;/math&gt;, and &lt;math&gt;d = b + c - a = a + 2m + n&lt;/math&gt; into &lt;math&gt;bc - ad = 93&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> bc - ad = (1 + m)(1 + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)<br /> &lt;/cmath&gt;<br /> Hence, &lt;math&gt;(m,n) = (1,92)&lt;/math&gt; or &lt;math&gt;(3,28)&lt;/math&gt;. <br /> <br /> For &lt;math&gt;(m,n) = (1,92)&lt;/math&gt;, we know that &lt;math&gt;0 &lt; a &lt; a + 1 &lt; a + 93 &lt; a + 94 &lt; 500&lt;/math&gt;, so there are &lt;math&gt;405&lt;/math&gt; four-tuples. For &lt;math&gt;(m,n) = (3,28)&lt;/math&gt;, &lt;math&gt;0 &lt; a &lt; a + 3 &lt; a + 31 &lt; a + 34 &lt; 500&lt;/math&gt;, and there are &lt;math&gt;465&lt;/math&gt; four-tuples. In total, we have &lt;math&gt;405 + 465 = \boxed{870}&lt;/math&gt; four-tuples.<br /> <br /> === Solution 3 ===<br /> Square both sides of the first equation in order to get &lt;math&gt;bc&lt;/math&gt; and &lt;math&gt;ad&lt;/math&gt; terms, which we can plug &lt;math&gt;93&lt;/math&gt; in for.<br /> <br /> &lt;math&gt;(a+d)^2 = (b+c)^2 \implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \implies 2bc-2ad = a^2-b^2 + d^2-c^2 \implies 2(bc-ad) = (a-b)(a+b)<br /> +(d-c)(d+c)&lt;/math&gt;<br /> <br /> We can plug &lt;math&gt;93&lt;/math&gt; in for &lt;math&gt;bc - ad&lt;/math&gt; to get &lt;math&gt;186&lt;/math&gt; on the left side, and also observe that &lt;math&gt;a-b = c-d&lt;/math&gt; after rearranging the first equation. Plug in &lt;math&gt;c-d&lt;/math&gt; for &lt;math&gt;a-b&lt;/math&gt;.<br /> <br /> &lt;math&gt;186 = (c-d)(a+b) + (d-c)(d+c) \implies 186 = -(d-c)(a+b) + (d-c)(d+c) \implies 186 = (d-c)(d+c-a-b)&lt;/math&gt;<br /> <br /> Now observe the possible factors of &lt;math&gt;186&lt;/math&gt;, which are &lt;math&gt;1 \cdot 186, 2\cdot 93, 3 \cdot 62, 6\cdot 31&lt;/math&gt;. &lt;math&gt;(d-c)&lt;/math&gt; and &lt;math&gt;(d+c-a-b)&lt;/math&gt; must be factors of &lt;math&gt;186&lt;/math&gt;, and &lt;math&gt;(d+c-a-b)&lt;/math&gt; must be greater than &lt;math&gt;(d-c)&lt;/math&gt;.<br /> <br /> &lt;math&gt;1 \cdot 186&lt;/math&gt; work, and yields &lt;math&gt;405&lt;/math&gt; possible solutions.<br /> &lt;math&gt;2 \cdot 93&lt;/math&gt; does not work, because if &lt;math&gt;c-d = 2&lt;/math&gt;, then &lt;math&gt;a+b&lt;/math&gt; must differ by 2 as well, but an odd number &lt;math&gt;93&lt;/math&gt; can only result from two numbers of different parity. &lt;math&gt;c-d&lt;/math&gt; will be even, and &lt;math&gt;a+b&lt;/math&gt; will be even, so &lt;math&gt;c+d - (a+b)&lt;/math&gt; must be even.<br /> &lt;math&gt;3 \cdot 62&lt;/math&gt; works, and yields &lt;math&gt;465&lt;/math&gt; possible solutions, while &lt;math&gt;6 \cdot 31&lt;/math&gt; fails for the same reasoning above.<br /> <br /> Thus, the answer is &lt;math&gt;405 + 465 = \boxed{870}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> <br /> Add the two conditions together to get &lt;math&gt;a+d+ad+93=b+c+bc&lt;/math&gt;. Rearranging and factorising with SFFT, &lt;math&gt;(a+1)(d+1)+93=(b+1)(c+1)&lt;/math&gt;. This implies that for every quadruple &lt;math&gt;(a,b,c,d)&lt;/math&gt;, we can replace &lt;math&gt;a\longrightarrow a+1&lt;/math&gt;, &lt;math&gt;b\longrightarrow b+1&lt;/math&gt;, etc. and this will still produce a valid quadruple. This means, that we can fix &lt;math&gt;a=1&lt;/math&gt;, and then just repeatedly add &lt;math&gt;1&lt;/math&gt; to get the other quadruples. <br /> <br /> Now, our conditions are &lt;math&gt;b+c=d+1&lt;/math&gt; and &lt;math&gt;bc=d+93&lt;/math&gt;. Replacing &lt;math&gt;d&lt;/math&gt; in the first equation, we get &lt;math&gt;bc-b-c=92&lt;/math&gt;. Factorising again with SFFT gives &lt;math&gt;(b-1)(c-1)=93&lt;/math&gt;. Since &lt;math&gt;b&lt;c&lt;/math&gt;, we have two possible cases to consider.<br /> <br /> Case 1: &lt;math&gt;b=2&lt;/math&gt;, &lt;math&gt;c=94&lt;/math&gt;. This produces the quadruple &lt;math&gt;(1,2,94,95)&lt;/math&gt;, which indeed works.<br /> <br /> Case 2: &lt;math&gt;b=4&lt;/math&gt;, &lt;math&gt;c=32&lt;/math&gt;. This produces the quadruple &lt;math&gt;(1,4,32,35)&lt;/math&gt;, which indeed works. <br /> <br /> Now, for case 1, we can add &lt;math&gt;1&lt;/math&gt; to each term exactly &lt;math&gt;404&lt;/math&gt; times (until we get the quadruple &lt;math&gt;(405,406,498,499)&lt;/math&gt;), until we violate &lt;math&gt;d&lt;500&lt;/math&gt;. This gives &lt;math&gt;405&lt;/math&gt; quadruples for case 1.<br /> <br /> For case 2, we can add &lt;math&gt;1&lt;/math&gt; to each term exactly &lt;math&gt;464&lt;/math&gt; times (until we get the quadruple &lt;math&gt;(465,468,496,499)&lt;/math&gt;). this gives &lt;math&gt;465&lt;/math&gt; quadruples for case 2.<br /> <br /> In conclusion, having exhausted all cases, we can finish. There are hence &lt;math&gt;405+465=\boxed{870}&lt;/math&gt; possible quadruples.<br /> <br /> ===Solution 5===<br /> <br /> Let &lt;math&gt;r = d-c&lt;/math&gt;. From the equation &lt;math&gt;a+d = b+c&lt;/math&gt;, we have &lt;cmath&gt; r = d-c = b-a , &lt;/cmath&gt; so &lt;math&gt;b = a+r&lt;/math&gt; and &lt;math&gt;c = d-r&lt;/math&gt;. We then have &lt;cmath&gt; 93 = (a+r)(d-r) - ad = rd - ra - r^2 = r(d-a-r) . &lt;/cmath&gt; Since &lt;math&gt;c &gt; b&lt;/math&gt;, &lt;math&gt;d-r &gt; a+r&lt;/math&gt;, or &lt;math&gt;d-a-r &gt; r&lt;/math&gt;. Since the prime factorization of 93 is &lt;math&gt;3 \cdot 31&lt;/math&gt;, we must either have &lt;math&gt;r=1&lt;/math&gt; and &lt;math&gt;d-a-r = 93&lt;/math&gt;, or &lt;math&gt;r=3&lt;/math&gt; and &lt;math&gt;d-a-r = 31&lt;/math&gt;. We consider these cases separately.<br /> <br /> If &lt;math&gt;r=1&lt;/math&gt;, then &lt;math&gt;d-a = 94&lt;/math&gt;, &lt;math&gt;b= a+1&lt;/math&gt;, and &lt;math&gt;c = d-1&lt;/math&gt;. Thus &lt;math&gt;d&lt;/math&gt; can be any integer between 95 and 499, inclusive, and our choice of &lt;math&gt;d&lt;/math&gt; determines the four-tuple &lt;math&gt;(a,b,c,d)&lt;/math&gt;. We therefore have &lt;math&gt;499-95+1 = 405&lt;/math&gt; possibilities in this case.<br /> <br /> If &lt;math&gt;r=3&lt;/math&gt;, then &lt;math&gt;d-a = 34&lt;/math&gt;, &lt;math&gt;b = a+3&lt;/math&gt;, and &lt;math&gt;c=d-3&lt;/math&gt;. Thus &lt;math&gt;d&lt;/math&gt; can be any integer between 35 and 499, inclusive, and our choice of &lt;math&gt;d&lt;/math&gt; determines the four-tuple &lt;math&gt;(a,b,c,d)&lt;/math&gt;, as before. We therefore have &lt;math&gt;499-35+1 = 465&lt;/math&gt; possibilities in this case.<br /> <br /> Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are &lt;math&gt;405 + 465 = \boxed{870}&lt;/math&gt; four-tuples.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_2&diff=105756 1981 IMO Problems/Problem 2 2019-05-11T18:02:51Z <p>Will3145: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;1 \le r \le n &lt;/math&gt; and consider all subsets of &lt;math&gt;r &lt;/math&gt; elements of the set &lt;math&gt; \{ 1, 2, \ldots , n \} &lt;/math&gt;. Each of these subsets has a smallest member. Let &lt;math&gt;F(n,r) &lt;/math&gt; denote the arithmetic mean of these smallest numbers; prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> F(n,r) = \frac{n+1}{r+1}.<br /> &lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> <br /> Clearly, the sum of the desired least elements is &lt;math&gt; \sum_{k=1}^{n} k {n-k \choose r-1} = \sum_{k=1}^{n} {k \choose 1}{n-k \choose r-1} &lt;/math&gt;, which we claim to be equal to &lt;math&gt;{n+1 \choose r+1} &lt;/math&gt; by virtue of the following argument.<br /> <br /> Consider a binary string of length &lt;math&gt;n+1 &lt;/math&gt; which contains &lt;math&gt;r+1 &lt;/math&gt; 1s. For some value of &lt;math&gt;k&lt;/math&gt; between 1 and &lt;math&gt;n&lt;/math&gt;, inclusive, we say that the second 1 will occur in the &lt;math&gt;(k+1) &lt;/math&gt;th place. Clearly, there are &lt;math&gt; {k \choose 1} &lt;/math&gt; ways to arrange the bits coming before the second 1, and &lt;math&gt;{n-k \choose r-1} &lt;/math&gt; ways to arrange the bits after the second 1. Our identity follows.<br /> <br /> Since the sum of the least elements of the sets is &lt;math&gt;{n+1 \choose r+1}&lt;/math&gt;, the mean of the least elements is &lt;math&gt;\frac{{n+1 \choose r+1}}{{n \choose r}} = \frac{n+1}{r+1}&lt;/math&gt;, Q.E.D.<br /> <br /> === Solution 2 ===<br /> <br /> We proceed as in the previous solution, but we prove our identity using the following manipulations:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \begin{matrix}\sum_{k=1}^{n}k{n-k \choose r-1 } &amp;=&amp;\sum_{k=1}^{n-(r-1)}{k}{n-k \choose r-1}\\<br /> &amp;=&amp; \sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}{n-k \choose r-1}\\<br /> &amp;=&amp;\sum_{i=1}^{n-(r-1)}{n-i+1 \choose r}\\<br /> &amp;=&amp;{n+1 \choose r+1}<br /> \end{matrix}<br /> &lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Q.E.D.<br /> <br /> === Solution 3 ===<br /> <br /> We proceed by strong induction.<br /> <br /> We define &lt;math&gt;F(k, k-1)&lt;/math&gt; to be zero (the empty sum).<br /> <br /> We consider &lt;math&gt;r&lt;/math&gt; to be fixed. The assertion obviously holds for &lt;math&gt;r = n&lt;/math&gt;. We now assume the problem to hold for values of &lt;math&gt;n&lt;/math&gt; less than or equal to &lt;math&gt;k&lt;/math&gt;. By considering subsets containing &lt;math&gt;k+1&lt;/math&gt; and not containing &lt;math&gt;k+1&lt;/math&gt;, respectively, we conclude that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> F(k+1, r) = \frac{{k \choose r-1}F(k,r-1) + {k \choose r}F(k,r)}{{k+1 \choose r}} = 1 + \frac{k-r+1}{r+1} = \frac{k+2}{r+1}<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> This completes our induction, Q.E.D.<br /> <br /> ===Solution 4===<br /> <br /> Consider a bipartite graph &lt;math&gt;G&lt;/math&gt; with bipartition &lt;math&gt;\{A,B\}&lt;/math&gt;. The vertices in &lt;math&gt;A&lt;/math&gt; are the &lt;math&gt;(r+1)&lt;/math&gt;-element subsets of &lt;math&gt;\{0, \dots , n\}&lt;/math&gt;, and the vertices in &lt;math&gt;B&lt;/math&gt; are the &lt;math&gt;r&lt;/math&gt;-element subsets of &lt;math&gt;\{1, \dots , n\}&lt;/math&gt;, and we draw an edge &lt;math&gt;\overline{ab}&lt;/math&gt; iff the subset &lt;math&gt;b \in B&lt;/math&gt; may be obtained from &lt;math&gt;a \in A&lt;/math&gt; by deleting the smallest element in &lt;math&gt;a&lt;/math&gt;. <br /> <br /> Note that <br /> <br /> &lt;center&gt;<br /> &lt;math&gt;|A|=\binom{n+1}{r+1}, |B|=\binom{n}{r}, |E(G)|=\binom{n+1}{r+1}=\frac{n+1}{r+1} \binom{n}{r}.&lt;/math&gt;<br /> &lt;/center&gt; <br /> <br /> The degree of a vertex in &lt;math&gt;B&lt;/math&gt; is the value of the least element of its corresponding subset. Hence &lt;cmath&gt;F(n,r)=\frac{1}{\binom{n}{r}} \sum_{v \in B} \deg (v)= \frac{n+1}{r+1}.&lt;/cmath&gt;<br /> {{alternate solutions}}<br /> <br /> {{IMO box|num-b=1|num-a=3|year=1981}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_2&diff=105755 1981 IMO Problems/Problem 2 2019-05-11T18:02:23Z <p>Will3145: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;1 \le r \le n &lt;/math&gt; and consider all subsets of &lt;math&gt;r &lt;/math&gt; elements of the set &lt;math&gt; \{ 1, 2, \ldots , n \} &lt;/math&gt;. Each of these subsets has a smallest member. Let &lt;math&gt;F(n,r) &lt;/math&gt; denote the arithmetic mean of these smallest numbers; prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> F(n,r) = \frac{n+1}{r+1}.<br /> &lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> <br /> Clearly, the sum of the desired least elements is &lt;math&gt; \sum_{k=1}^{n} k {n-k \choose r-1} = \sum_{k=1}^{n} {k \choose 1}{n-k \choose r-1} &lt;/math&gt;, which we claim to be equal to &lt;math&gt;{n+1 \choose r+1} &lt;/math&gt; by virtue of the following argument.<br /> <br /> Consider a binary string of length &lt;math&gt;n+1 &lt;/math&gt; which contains &lt;math&gt;r+1 &lt;/math&gt; 1s. For some value of &lt;math&gt;k&lt;/math&gt; between 1 and &lt;math&gt;n&lt;/math&gt;, inclusive, we say that the second 1 will occur in the &lt;math&gt;(k+1) &lt;/math&gt;th place. Clearly, there are &lt;math&gt; {k \choose 1} &lt;/math&gt; ways to arrange the bits coming before the second 1, and &lt;math&gt;{n-k \choose r-1} &lt;/math&gt; ways to arrange the bits after the second 1. Our identity follows.<br /> <br /> Since the sum of the least elements of the sets is &lt;math&gt;{n+1 \choose r+1}&lt;/math&gt;, the mean of the least elements is &lt;math&gt;\frac{{n+1 \choose r+1}}{{n \choose r}} = \frac{n+1}{r+1}&lt;/math&gt;, Q.E.D.<br /> <br /> === Solution 2 ===<br /> <br /> We proceed as in the previous solution, but we prove our identity using the following manipulations:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \begin{matrix}\sum_{k=1}^{n}k{n-k \choose r-1 } &amp;=&amp;\sum_{k=1}^{n-(r-1)}{k}{n-k \choose r-1}\\<br /> &amp;=&amp; \sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}{n-k \choose r-1}\\<br /> &amp;=&amp;\sum_{i=1}^{n-(r-1)}{n-i+1 \choose r}\\<br /> &amp;=&amp;{n+1 \choose r+1}<br /> \end{matrix}<br /> &lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Q.E.D.<br /> <br /> === Solution 3 ===<br /> <br /> We proceed by strong induction.<br /> <br /> We define &lt;math&gt;F(k, k-1)&lt;/math&gt; to be zero (the empty sum).<br /> <br /> We consider &lt;math&gt;r&lt;/math&gt; to be fixed. The assertion obviously holds for &lt;math&gt;r = n&lt;/math&gt;. We now assume the problem to hold for values of &lt;math&gt;n&lt;/math&gt; less than or equal to &lt;math&gt;k&lt;/math&gt;. By considering subsets containing &lt;math&gt;k+1&lt;/math&gt; and not containing &lt;math&gt;k+1&lt;/math&gt;, respectively, we conclude that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> F(k+1, r) = \frac{{k \choose r-1}F(k,r-1) + {k \choose r}F(k,r)}{{k+1 \choose r}} = 1 + \frac{k-r+1}{r+1} = \frac{k+2}{r+1}<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> This completes our induction, Q.E.D.<br /> <br /> ===Solution 4===<br /> <br /> Consider a bipartite graph &lt;math&gt;G&lt;/math&gt; with bipartition &lt;math&gt;\{A,B\}&lt;/math&gt;. The vertices in &lt;math&gt;A&lt;/math&gt; are the &lt;math&gt;(r+1)&lt;/math&gt;-element subsets of &lt;math&gt;\{0, \dots , n\}&lt;/math&gt;, and the vertices in &lt;math&gt;B&lt;/math&gt; are the &lt;math&gt;r&lt;/math&gt;-element subsets of &lt;math&gt;\{1, \dots , n\}&lt;/math&gt;, and we draw an edge &lt;math&gt;\overline{ab}&lt;/math&gt; iff the subset &lt;math&gt;b \in B&lt;/math&gt; may be obtained from &lt;math&gt;a \in A&lt;/math&gt; by deleting the smallest element in &lt;math&gt;a&lt;/math&gt;. <br /> <br /> Note that <br /> <br /> &lt;center&gt;<br /> &lt;math&gt;|A|=\binom{n+1}{r+1}, |B|=\binom{n}{r}, |E(G)|=\binom{n+1}{r+1}=\frac{n+1}{r+1} \binom{n}{r}.&lt;/math&gt;<br /> &lt;/center&gt; <br /> <br /> The degree of a vertex in &lt;math&gt;B&lt;/math&gt; is the value of the least element of its corresponding subset. Hence &lt;cmath&gt;F(n,r)=\frac{1}{\binom{n}{r}} \sum_{v \in B} \deg (v)= \frac{n+1}{r+1}.&lt;/cmath&gt;<br /> {{alternate solutions}}<br /> <br /> {{IMO box|num-b=1|num-a=3|year=1988}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_7&diff=103154 2019 AIME I Problems/Problem 7 2019-02-17T17:39:10Z <p>Will3145: Created page with &quot;Seriously!!! IP address sent... -AoPS division of cheating&quot;</p> <hr /> <div>Seriously!!! IP address sent...<br /> <br /> -AoPS division of cheating</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_4&diff=103057 2019 AIME II Problems/Problem 4 2019-02-16T06:02:16Z <p>Will3145: Created page with &quot;&lt;math&gt;1+1=x&lt;/math&gt; solve for y.&quot;</p> <hr /> <div>&lt;math&gt;1+1=x&lt;/math&gt;<br /> <br /> solve for y.</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_3&diff=103056 2019 AIME II Problems/Problem 3 2019-02-16T06:01:45Z <p>Will3145: Created page with &quot;I spy with my little eye... A cheater.&quot;</p> <hr /> <div>I spy with my little eye...<br /> <br /> <br /> A cheater.</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_2&diff=103055 2019 AIME II Problems/Problem 2 2019-02-16T05:53:32Z <p>Will3145: Created page with &quot;You do not belong here.&quot;</p> <hr /> <div>You do not belong here.</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_21&diff=102585 2019 AMC 10B Problems/Problem 21 2019-02-14T22:23:29Z <p>Will3145: LaTeX edits</p> <hr /> <div>==Problem==<br /> <br /> Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We first want to find out the sequences of coin flips that satisfy this equation. Since Debra sees two tails before two heads, her first flip can't be heads, as that would mean she would either end at tails or see two heads before she sees two tails. Therefore, her first flip must be tails. If you calculate the shortest way she can get two heads in a row and see two tails before she sees two heads, it would be T H T H H, which would be &lt;math&gt;\frac{1}{2^5}&lt;/math&gt;, or &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Following this, she can prolong her coin flipping by adding an extra (T H), which is an extra &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance. Since she can do this indefinitely, this is an infinite geometric sequence, which means the answer is &lt;math&gt;\frac{\frac{1}{32}}{1-\frac{1}{4}}&lt;/math&gt; or &lt;math&gt;\boxed{B) \frac{1}{24}}&lt;/math&gt;. -chen1046, LaTeX edit by will3145<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_21&diff=102583 2019 AMC 10B Problems/Problem 21 2019-02-14T22:23:03Z <p>Will3145: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We first want to find out the sequences of coin flips that satisfy this equation. Since Debra sees two tails before two heads, her first flip can't be heads, as that would mean she would either end at tails or see two heads before she sees two tails. Therefore, her first flip must be tails. If you calculate the shortest way she can get two heads in a row and see two tails before she sees two heads, it would be T H T H H, which would be &lt;math&gt;\frac{1}{2^5}&lt;/math&gt;, or &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Following this, she can prolong her coin flipping by adding an extra (T H), which is an extra &lt;math&gt;frac{1}{4}&lt;/math&gt; chance. Since she can do this indefinitely, this is an infinite geometric sequence, which means the answer is &lt;math&gt;\frac{\frac{1}{32}}{1-\frac{1}{4}}&lt;/math&gt; or &lt;math&gt;\boxed{B) \frac{1}{24}}&lt;/math&gt;. -chen1046, LaTeX edit by will3145<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_21&diff=102582 2019 AMC 10B Problems/Problem 21 2019-02-14T22:22:45Z <p>Will3145: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We first want to find out the sequences of coin flips that satisfy this equation. Since Debra sees two tails before two heads, her first flip can't be heads, as that would mean she would either end at tails or see two heads before she sees two tails. Therefore, her first flip must be tails. If you calculate the shortest way she can get two heads in a row and see two tails before she sees two heads, it would be T H T H H, which would be &lt;math&gt;\frac{1}{2^5}&lt;/math&gt;, or &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Following this, she can prolong her coin flipping by adding an extra (T H), which is an extra &lt;math&gt;frac{1}{4}&lt;/math&gt; chance. Since she can do this indefinitely, this is an infinite geometric sequence, which means the answer is &lt;math&gt;\frac{\frac{1}{32}}{1-\frac{1}{4}}or &lt;/math&gt;\boxed{B) \frac{1}{24}}$. -chen1046, LaTeX edit by will3145<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems&diff=102568 2019 AMC 10B Problems 2019-02-14T22:14:02Z <p>Will3145: minor formatting</p> <hr /> <div>==Problem 1==<br /> <br /> Alicia had two containers. The first was &lt;math&gt;\tfrac{5}{6}&lt;/math&gt; full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; full of water. What is the ratio of the volume of the first container to the volume of the second container?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Consider the statement, &quot;If &lt;math&gt;n&lt;/math&gt; is not prime, then &lt;math&gt;n-2&lt;/math&gt; is prime.&quot; Which of the following values of &lt;math&gt;n&lt;/math&gt; is a counterexample to this statement.<br /> <br /> &lt;math&gt;\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> In a high school with &lt;math&gt;500&lt;/math&gt; students, &lt;math&gt;40\%&lt;/math&gt; of the seniors play a musical instrument, while &lt;math&gt;30\%&lt;/math&gt; of the non-seniors do not play a musical instrument. In all, &lt;math&gt;46.8\%&lt;/math&gt; of the students do not play a musical instrument. How many non-seniors play a musical instrument?<br /> <br /> &lt;math&gt;\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> All lines with equation &lt;math&gt;ax+by=c&lt;/math&gt; such that &lt;math&gt;a,b,c&lt;/math&gt; form an arithmetic progression pass through a common point. What are the coordinates of that point?<br /> <br /> &lt;math&gt;\textbf{(A) } (-1,2)<br /> \qquad\textbf{(B) } (0,1)<br /> \qquad\textbf{(C) } (1,-2)<br /> \qquad\textbf{(D) } (1,0)<br /> \qquad\textbf{(E) } (1,2)&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are reflected across the line &lt;math&gt;y=x&lt;/math&gt; to points &lt;math&gt;A'&lt;/math&gt;, &lt;math&gt;B'&lt;/math&gt;, and &lt;math&gt;C'&lt;/math&gt;, respectively. Assume that none of the vertices of the triangle lie on the line &lt;math&gt;y=x&lt;/math&gt;. Which of the following statements is not always true?<br /> <br /> ==Problem 6==<br /> <br /> There is a real &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;(n+1)! + (n+2)! = n! \cdot 440&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or &lt;math&gt;n&lt;/math&gt; pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> [[2019 AMC 10B Problems/Problem 8|Solution]]<br /> //make asy diagram <br /> The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?<br /> &lt;math&gt;(A)&lt;/math&gt; &lt;math&gt;4&lt;/math&gt;<br /> &lt;math&gt;(B)&lt;/math&gt; &lt;math&gt;12 - 4\sqrt{3}&lt;/math&gt;<br /> &lt;math&gt;(C)&lt;/math&gt; &lt;math&gt;3\sqrt{3}&lt;/math&gt;<br /> &lt;math&gt;(D)&lt;/math&gt; &lt;math&gt;4\sqrt{3}&lt;/math&gt;<br /> &lt;math&gt;(E)&lt;/math&gt; &lt;math&gt;16 - \sqrt{3}&lt;/math&gt;<br /> <br /> ==Problem 9==<br /> <br /> The function &lt;math&gt;f&lt;/math&gt; is defined by &lt;cmath&gt;f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|&lt;/cmath&gt;for all real numbers &lt;math&gt;x&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to the real number &lt;math&gt;r&lt;/math&gt;. What is the range of &lt;math&gt;f&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\} \qquad\textbf{(E) } \text{The set of nonnegative integers} &lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> In a given plane, points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;10&lt;/math&gt; units apart. How many points &lt;math&gt;C&lt;/math&gt; are there in the plane such that the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; units and the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?<br /> <br /> &lt;math&gt;\textbf{(A) } 5<br /> \qquad\textbf{(B) } 10<br /> \qquad\textbf{(C) } 25<br /> \qquad\textbf{(D) } 45<br /> \qquad\textbf{(E) } 50&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than &lt;math&gt;2019&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 11<br /> \qquad\textbf{(B) } 14<br /> \qquad\textbf{(C) } 22<br /> \qquad\textbf{(D) } 23<br /> \qquad\textbf{(E) } 27&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which the median of the numbers &lt;math&gt;4,6,8,17,&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; is equal to the mean of those five numbers?<br /> <br /> &lt;math&gt;\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> The base-ten representation for &lt;math&gt;19!&lt;/math&gt; is &lt;math&gt;121,6T5,100,40M,832,H00&lt;/math&gt;, where &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, and &lt;math&gt;H&lt;/math&gt; denote digits that are not given. What is &lt;math&gt;T+M+H&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17 &lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Two right triangles, &lt;math&gt;T_1&lt;/math&gt; and &lt;math&gt;T_2&lt;/math&gt;, have areas of 1 and 2, respectively. One side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other. What is the product of the third side lengths of &lt;math&gt;T_1&lt;/math&gt; and &lt;math&gt;T_2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{28}{3} \qquad\textbf{(B) }10\qquad\textbf{(C) } \frac{32}{3} \qquad\textbf{(D) } \frac{34}{3} \qquad\textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; with a right angle at &lt;math&gt;C,&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{AB}&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;AC=CD,&lt;/math&gt; &lt;math&gt;DE=EB,&lt;/math&gt; and the ratio &lt;math&gt;AC:DE=4:3.&lt;/math&gt; What is the ratio &lt;math&gt;AD:DB?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2:3<br /> \qquad\textbf{(B) } 2:\sqrt{5}<br /> \qquad\textbf{(C) } 1:1<br /> \qquad\textbf{(D) } 3:\sqrt{5}<br /> \qquad\textbf{(E) } 3:2&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;2^{-k}&lt;/math&gt; for &lt;math&gt;k=1,2,3,\ldots.&lt;/math&gt; What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Henry decides one morning to do a workout, and he walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the way from his home to his gym. The gym is &lt;math&gt;2&lt;/math&gt; kilometers away from Henry's home. At that point, he changes his mind and walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point &lt;math&gt;A&lt;/math&gt; kilometers from home and a point &lt;math&gt;B&lt;/math&gt; kilometers from home. What is &lt;math&gt;|A-B|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 1\frac{1}{5} \qquad \textbf{(D) } 1\frac{1}{4} \qquad \textbf{(E) } 1\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all positive integer divisors of &lt;math&gt;100,000.&lt;/math&gt; How many numbers are the product of two distinct elements of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> As shown in the figure, line segment &lt;math&gt;\overline{AD}&lt;/math&gt; is trisected by points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AB=BC=CD=2.&lt;/math&gt; Three semicircles of radius &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;\overarc{AEB},\overarc{BFC},&lt;/math&gt; and &lt;math&gt;\overarc{CGD},&lt;/math&gt; have their diameters on &lt;math&gt;\overline{AD},&lt;/math&gt; and are tangent to line &lt;math&gt;EG&lt;/math&gt; at &lt;math&gt;E,F,&lt;/math&gt; and &lt;math&gt;G,&lt;/math&gt; respectively. A circle of radius &lt;math&gt;2&lt;/math&gt; has its center on &lt;math&gt;F. &lt;/math&gt; The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form<br /> &lt;cmath&gt;\frac{a}{b}\cdot\pi-\sqrt{c}+d,&lt;/cmath&gt;where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), gray(0.7));<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D\$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> [[2019 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Define a sequence recursively by &lt;math&gt;x_0=5&lt;/math&gt; and<br /> &lt;cmath&gt;x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}&lt;/cmath&gt;for all nonnegative integers &lt;math&gt;n.&lt;/math&gt; Let &lt;math&gt;m&lt;/math&gt; be the least positive integer such that<br /> &lt;cmath&gt;x_m\leq 4+\frac{1}{2^{20}}.&lt;/cmath&gt;In which of the following intervals does &lt;math&gt;m&lt;/math&gt; lie?<br /> <br /> &lt;math&gt;\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> [[2019 AMC 10B Problems/Problem 25|Solution]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_14&diff=96877 2009 AMC 8 Problems/Problem 14 2018-08-08T23:16:43Z <p>Will3145: new solution</p> <hr /> <div>==Problem==<br /> <br /> Austin and Temple are &lt;math&gt; 50&lt;/math&gt; miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging &lt;math&gt; 60&lt;/math&gt; miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged &lt;math&gt; 40&lt;/math&gt; miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 46 \qquad<br /> \textbf{(B)}\ 48 \qquad<br /> \textbf{(C)}\ 50 \qquad<br /> \textbf{(D)}\ 52 \qquad<br /> \textbf{(E)}\ 54&lt;/math&gt;<br /> <br /> ==Solution==<br /> The way to Temple took &lt;math&gt;\frac{50}{60}=\frac56&lt;/math&gt; hours, and the way back took &lt;math&gt;\frac{50}{40}=\frac54&lt;/math&gt; for a total of &lt;math&gt;\frac56 + \frac54 = \frac{25}{12}&lt;/math&gt; hours. The trip is &lt;math&gt;50\cdot2=100&lt;/math&gt; miles. The average speed is &lt;math&gt;\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}&lt;/math&gt; miles per hour.<br /> <br /> ==Solution 2==<br /> This question simply asks for the harmonic mean of &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;40&lt;/math&gt;, regardless of how far Austin and Temple are. <br /> Plugging in, we have: &lt;math&gt;\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}&lt;/math&gt; miles per hour.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_13&diff=96876 2009 AMC 8 Problems/Problem 13 2018-08-08T23:09:54Z <p>Will3145: New solution</p> <hr /> <div>==Problem==<br /> <br /> A three-digit integer contains one of each of the digits &lt;math&gt; 1&lt;/math&gt;, &lt;math&gt; 3&lt;/math&gt;, and &lt;math&gt; 5&lt;/math&gt;. What is the probability that the integer is divisible by &lt;math&gt; 5&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{6} \qquad<br /> \textbf{(B)}\ \frac{1}{3} \qquad<br /> \textbf{(C)}\ \frac{1}{2} \qquad<br /> \textbf{(D)}\ \frac{2}{3} \qquad<br /> \textbf{(E)}\ \frac{5}{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The three digit numbers are &lt;math&gt;135,153,351,315,513,531&lt;/math&gt;. The numbers that end in &lt;math&gt;5&lt;/math&gt; are divisible are &lt;math&gt;5&lt;/math&gt;, and the probability of choosing those numbers is &lt;math&gt;\boxed{\textbf{(B)}\ \frac13}&lt;/math&gt;.<br /> <br /> ==Alternate Solution==<br /> The number is odd if and only if the number ends in &lt;math&gt;5&lt;/math&gt; (also &lt;math&gt;0&lt;/math&gt;, but that case can be ignored, as none of the digits are &lt;math&gt;0&lt;/math&gt;)<br /> If we randomly arrange the three digits, the probability of the last digit being &lt;math&gt;5&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B)}\ \frac13}&lt;/math&gt;.<br /> <br /> Note: The last sentence is true because there are &lt;math&gt;3&lt;/math&gt; randomly-arrangeable numbers)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_20&diff=96835 2009 AMC 10A Problems/Problem 20 2018-08-07T18:53:49Z <p>Will3145: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Andrea and Lauren are &lt;math&gt;20&lt;/math&gt; kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute. After &lt;math&gt;5&lt;/math&gt; minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 20<br /> \qquad<br /> \mathrm{(B)}\ 30<br /> \qquad<br /> \mathrm{(C)}\ 55<br /> \qquad<br /> \mathrm{(D)}\ 65<br /> \qquad<br /> \mathrm{(E)}\ 80<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Let their speeds in kilometers per hour be &lt;math&gt;v_A&lt;/math&gt; and &lt;math&gt;v_L&lt;/math&gt;. We know that &lt;math&gt;v_A=3v_L&lt;/math&gt; and that &lt;math&gt;v_A+v_L=60&lt;/math&gt;. (The second equation follows from the fact that &lt;math&gt;1\mathrm km/min = 60\mathrm km/h&lt;/math&gt;.) This solves to &lt;math&gt;v_A=45&lt;/math&gt; and &lt;math&gt;v_L=15&lt;/math&gt;. <br /> <br /> As the distance decreases at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute, after &lt;math&gt;5&lt;/math&gt; minutes the distance between them will be &lt;math&gt;20-5=15&lt;/math&gt; kilometers.<br /> <br /> From this point on, only Lauren will be riding her bike. As there are &lt;math&gt;15&lt;/math&gt; kilometers remaining and &lt;math&gt;v_L=15&lt;/math&gt;, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is &lt;math&gt;5+60 = \boxed{65}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is &lt;math&gt;\frac{3}{4} \textbf{km/min}&lt;/math&gt;, and Lauren's &lt;math&gt;\frac{1}{4} \textbf{km/min}&lt;/math&gt;. Therefore, after 5 minutes, Andrea will have biked &lt;math&gt;\frac{3}{4} \cdot 5 = \frac{15}{4}&lt;/math&gt;km.<br /> <br /> In all, Lauren will have to bike &lt;math&gt;20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}&lt;/math&gt;km. Because her speed is &lt;math&gt;\frac{1}{4} \textbf{km/min}&lt;/math&gt;, the time elapsed will be &lt;math&gt;\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_20&diff=96834 2009 AMC 10A Problems/Problem 20 2018-08-07T18:53:12Z <p>Will3145: new solution</p> <hr /> <div>== Problem ==<br /> <br /> Andrea and Lauren are &lt;math&gt;20&lt;/math&gt; kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute. After &lt;math&gt;5&lt;/math&gt; minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 20<br /> \qquad<br /> \mathrm{(B)}\ 30<br /> \qquad<br /> \mathrm{(C)}\ 55<br /> \qquad<br /> \mathrm{(D)}\ 65<br /> \qquad<br /> \mathrm{(E)}\ 80<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Let their speeds in kilometers per hour be &lt;math&gt;v_A&lt;/math&gt; and &lt;math&gt;v_L&lt;/math&gt;. We know that &lt;math&gt;v_A=3v_L&lt;/math&gt; and that &lt;math&gt;v_A+v_L=60&lt;/math&gt;. (The second equation follows from the fact that &lt;math&gt;1\mathrm km/min = 60\mathrm km/h&lt;/math&gt;.) This solves to &lt;math&gt;v_A=45&lt;/math&gt; and &lt;math&gt;v_L=15&lt;/math&gt;. <br /> <br /> As the distance decreases at a rate of &lt;math&gt;1&lt;/math&gt; kilometer per minute, after &lt;math&gt;5&lt;/math&gt; minutes the distance between them will be &lt;math&gt;20-5=15&lt;/math&gt; kilometers.<br /> <br /> From this point on, only Lauren will be riding her bike. As there are &lt;math&gt;15&lt;/math&gt; kilometers remaining and &lt;math&gt;v_L=15&lt;/math&gt;, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is &lt;math&gt;5+60 = \boxed{65}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is &lt;math&gt;\frac{3}{4} \textbf{km/min}&lt;/math&gt;, and Lauren's &lt;math&gt;\frac{1}{4} \textbf{km/min}&lt;/math&gt;. Therefore, after 5 minutes, Andrea will have biked &lt;math&gt;\frac{3}{4} \cdot 5 = \frac{15}{4}&lt;/math&gt;.<br /> <br /> In all, Lauren will have to bike &lt;math&gt;20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}&lt;/math&gt;km. Because her speed is &lt;math&gt;\frac{1}{4} \textbf{km/min}&lt;/math&gt;, the time elapsed will be &lt;math&gt;\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=Mean_Value_Theorem&diff=96829 Mean Value Theorem 2018-08-07T16:27:58Z <p>Will3145: </p> <hr /> <div>The '''Mean Value Theorem''' states that if &lt;math&gt;a &lt; b&lt;/math&gt; are [[real number]]s and the [[function]] &lt;math&gt;f:[a,b] \to \mathbb{R}&lt;/math&gt; is [[differentiable]] on the [[interval]] &lt;math&gt;(a,b)&lt;/math&gt;, then there exists a value &lt;math&gt;c&lt;/math&gt; in &lt;math&gt;(a,b)&lt;/math&gt; such that<br /> <br /> &lt;cmath&gt;f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx.&lt;/cmath&gt;<br /> <br /> In words, there is a number &lt;math&gt;c&lt;/math&gt; in &lt;math&gt;(a,b)&lt;/math&gt; such that &lt;math&gt;f(c)&lt;/math&gt; equals the average value of the function in the interval &lt;math&gt;[a,b]&lt;/math&gt;.<br /> <br /> {{stub}}<br /> <br /> ==Proof==<br /> <br /> <br /> <br /> ==Other==<br /> <br /> Rolle's Theorem is a sub-case of this theorem. It states that if &lt;math&gt;f(a)=f(b)=0&lt;/math&gt; for two real numbers a and b, then there is a real number c such that &lt;math&gt;a&lt;c&lt;b&lt;/math&gt; and &lt;math&gt;f'(c)=0&lt;/math&gt;.<br /> <br /> {{stub}}<br /> [[Category:Calculus]]<br /> [[Category:Definition]]<br /> [[Category:Theorems]]</div> Will3145 https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=96815 Ceva's Theorem 2018-08-06T22:15:18Z <p>Will3145: /* Proof by Barycentric coordinates */</p> <hr /> <div>'''Ceva's Theorem''' is a criterion for the [[concurrence]] of [[cevian]]s in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> <br /> [[Image:Ceva1.PNG|thumb|right]]<br /> Let &lt;math&gt;ABC &lt;/math&gt; be a triangle, and let &lt;math&gt;D, E, F &lt;/math&gt; be points on lines &lt;math&gt;BC, CA, AB &lt;/math&gt;, respectively. Lines &lt;math&gt;AD, BE, CF &lt;/math&gt; are [[concurrent]] if and only if<br /> &lt;br&gt;&lt;center&gt;<br /> &lt;math&gt;\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 &lt;/math&gt;,<br /> &lt;/center&gt;&lt;br&gt;<br /> where lengths are [[directed segments | directed]]. This also works for the [[reciprocal]] of each of the ratios, as the reciprocal of &lt;math&gt;1&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt;.<br /> <br /> <br /> (Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)<br /> <br /> <br /> The proof using [[Routh's Theorem]] is extremely trivial, so we will not include it.<br /> <br /> == Proof ==<br /> <br /> We will use the notation &lt;math&gt;[ABC] &lt;/math&gt; to denote the area of a triangle with vertices &lt;math&gt;A,B,C &lt;/math&gt;.<br /> <br /> First, suppose &lt;math&gt;AD, BE, CF &lt;/math&gt; meet at a point &lt;math&gt;X &lt;/math&gt;. We note that triangles &lt;math&gt;ABD, ADC &lt;/math&gt; have the same altitude to line &lt;math&gt;BC &lt;/math&gt;, but bases &lt;math&gt;BD &lt;/math&gt; and &lt;math&gt;DC &lt;/math&gt;. It follows that &lt;math&gt; \frac {BD}{DC} = \frac{[ABD]}{[ADC]} &lt;/math&gt;. The same is true for triangles &lt;math&gt;XBD, XDC &lt;/math&gt;, so <br /> <br /> &lt;center&gt;&lt;math&gt; \frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]} &lt;/math&gt;. &lt;/center&gt;<br /> Similarly, &lt;math&gt; \frac{CE}{EA} = \frac{[BCX]}{[BXA]} &lt;/math&gt; and &lt;math&gt; \frac{AF}{FB} = \frac{[CAX]}{[CXB]} &lt;/math&gt;,<br /> so<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Now, suppose &lt;math&gt;D, E,F &lt;/math&gt; satisfy Ceva's criterion, and suppose &lt;math&gt;AD, BE &lt;/math&gt; intersect at &lt;math&gt;X &lt;/math&gt;. Suppose the line &lt;math&gt;CX &lt;/math&gt; intersects line &lt;math&gt;AB &lt;/math&gt; at &lt;math&gt;F' &lt;/math&gt;. We have proven that &lt;math&gt;F' &lt;/math&gt; must satisfy Ceva's criterion. This means that &lt;center&gt;&lt;math&gt; \frac{AF'}{F'B} = \frac{AF}{FB} &lt;/math&gt;, &lt;/center&gt; so &lt;center&gt;&lt;math&gt;F' = F &lt;/math&gt;, &lt;/center&gt; and line &lt;math&gt;CF &lt;/math&gt; concurrs with &lt;math&gt;AD &lt;/math&gt; and &lt;math&gt;BE &lt;/math&gt;. {{Halmos}}<br /> <br /> ==Proof by [[Barycentric coordinates]]==<br /> <br /> Since &lt;math&gt;D\in BC&lt;/math&gt;, we can write its coordinates as &lt;math&gt;(0,d,1-d)&lt;/math&gt;. The equation of line &lt;math&gt;AD&lt;/math&gt; is then &lt;math&gt;z=\frac{1-d}{d}y&lt;/math&gt;. <br /> <br /> Similarly, since &lt;math&gt;E=(1-e,0,e)&lt;/math&gt;, and &lt;math&gt;F=(f,1-f,0)&lt;/math&gt;, we can see that the equations of &lt;math&gt;BE&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt; respectively are &lt;math&gt;x=\frac{1-e}{e}z&lt;/math&gt; and &lt;math&gt;y=\frac{1-f}{f}x&lt;/math&gt;<br /> <br /> [[Multiplying]] the three together yields the solution to the equation:<br /> <br /> &lt;math&gt;xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y&lt;/math&gt;<br /> <br /> Dividing by &lt;math&gt;xyz&lt;/math&gt; yields:<br /> <br /> <br /> &lt;math&gt;1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}&lt;/math&gt;, which is equivalent to Ceva's theorem<br /> <br /> QED<br /> <br /> == Trigonometric Form ==<br /> <br /> The [[trig | trigonometric]] form of Ceva's Theorem (Trig Ceva) states that cevians &lt;math&gt;AD,BE,CF&lt;/math&gt; concur if and only if<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> === Proof ===<br /> <br /> First, suppose &lt;math&gt;AD, BE, CF &lt;/math&gt; concur at a point &lt;math&gt;X &lt;/math&gt;. We note that<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} &lt;/math&gt;, &lt;/center&gt;<br /> and similarly,<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} &lt;/math&gt;. &lt;/center&gt;<br /> It follows that<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} &lt;/math&gt; &lt;br&gt; &lt;br&gt; &lt;math&gt; \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod &lt;math&gt;\pi &lt;/math&gt; to be either positive or negative.<br /> <br /> The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. {{Halmos}}<br /> <br /> == Problems ==<br /> ===Introductory===<br /> *Suppose &lt;math&gt;AB, AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; have lengths &lt;math&gt;13, 14&lt;/math&gt;, and &lt;math&gt;15&lt;/math&gt;, respectively. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;, find &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;DC&lt;/math&gt;. ([[Ceva's Theorem/Problems|Source]])<br /> <br /> ===Intermediate===<br /> *In &lt;math&gt;\Delta ABC, AD, BE, CF&lt;/math&gt; are concurrent lines. &lt;math&gt;P, Q, R&lt;/math&gt; are points on &lt;math&gt;EF, FD, DE&lt;/math&gt; such that &lt;math&gt;DP, EQ, FR&lt;/math&gt; are concurrent. Prove that (using ''plane geometry'') &lt;math&gt;AP, BQ, CR&lt;/math&gt; are concurrent. (&lt;url&gt;viewtopic.php?f=151&amp;t=543574 &lt;/url&gt;)<br /> <br /> == See also ==<br /> * [[Stewart's Theorem]]<br /> * [[Menelaus' Theorem]]<br /> <br /> [[Category:Geometry]]<br /> <br /> [[Category:Theorems]]</div> Will3145