https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Alexwin0806&feedformat=atom AoPS Wiki - User contributions [en] 2021-11-29T04:01:08Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_USOJMO_Problems/Problem_1&diff=141614 2020 USOJMO Problems/Problem 1 2021-01-06T23:41:43Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;n \geq 2&lt;/math&gt; be an integer. Carl has &lt;math&gt;n&lt;/math&gt; books arranged on a bookshelf.<br /> Each book has a height and a width. No two books have the same height, and no two<br /> books have the same width.<br /> Initially, the books are arranged in increasing order of height from left to right. In a<br /> move, Carl picks any two adjacent books where the left book is wider and shorter than<br /> the right book, and swaps their locations. Carl does this repeatedly until no further<br /> moves are possible.<br /> Prove that regardless of how Carl makes his moves, he must stop after a finite number<br /> of moves, and when he does stop, the books are sorted in increasing order of width<br /> from left to right.<br /> <br /> ==Solution==<br /> <br /> Without loss of generality, let the books have widths and heights of 1, 2, 3,..., and n. First we will show that Carl can only make finitely many moves.<br /> <br /> Note that two books can never be swapped more than once with each other. This is because after a first swap, the wider book will be to the left of the other, so they can never be swapped back. Thus there can be at most &lt;math&gt;{n\choose 2}&lt;/math&gt; moves.<br /> <br /> Now we will show that the books are in increasing order of width when Carl is done making moves. Assume otherwise for the sake of contradiction. Now let &lt;math&gt;f(k)&lt;/math&gt; be the width of the book in the kth spot from the left. Since the books are not in increasing order of width, there must be i such that &lt;math&gt;f(i)&gt;f(i+1)&lt;/math&gt;, for &lt;math&gt;1\leq i\leq n-1&lt;/math&gt;. <br /> <br /> &lt;b&gt;Claim:&lt;/b&gt; The ith book must be shorter than the &lt;math&gt;(i+1)&lt;/math&gt;th book. <br /> <br /> &lt;b&gt;Proof:&lt;/b&gt; Assume otherwise for contradiction. Then the ith book must have started to the right of the &lt;math&gt;(i+1)&lt;/math&gt;th book, since the ith book is taller. However, the ith book is now to the left of the &lt;math&gt;(i+1)&lt;/math&gt;th book, so they must have been swapped at some point. But since &lt;math&gt;f(i)&gt;f(i+1)&lt;/math&gt;, the ith book is also wider. But the ith book started to the right, so since it was wider it cannot have been swapped with the &lt;math&gt;(i+1)&lt;/math&gt;th book, so we arrive at a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> From the claim, the ith book must be shorter than the &lt;math&gt;(i+1)&lt;/math&gt;th book. But since the ith book is also wider by definition (as &lt;math&gt;f(i)&gt;f(i+1)&lt;/math&gt;), Carl can swap those two books, so he still has legal moves and we arrive at a contradiction, so we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ~MortemEtInteritum</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_24&diff=121899 1981 AHSME Problems/Problem 24 2020-05-02T02:27:40Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem ==<br /> If &lt;math&gt; \theta&lt;/math&gt; is a constant such that &lt;math&gt; 0 &lt; \theta &lt; \pi&lt;/math&gt; and &lt;math&gt; x + \dfrac{1}{x} = 2\cos{\theta}&lt;/math&gt;, then for each positive integer &lt;math&gt; n&lt;/math&gt;, &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; equals<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta&lt;/math&gt;<br /> <br /> == Solution ==<br /> Multiply both sides by &lt;math&gt;x&lt;/math&gt; and rearrange to &lt;math&gt;x^2-2x\cos(\theta)+1=0&lt;/math&gt;. Using the quadratic equation, we can solve for &lt;math&gt;x&lt;/math&gt;. After some simplifying:<br /> <br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + i\sin(\theta)&lt;/cmath&gt;<br /> <br /> Substituting this expression in to the desired &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; gives:<br /> <br /> &lt;cmath&gt;(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}&lt;/cmath&gt;<br /> <br /> Using DeMoivre's Theorem:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;\cos&lt;/math&gt; is even and &lt;math&gt;\sin&lt;/math&gt; is odd:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)&lt;/cmath&gt;<br /> &lt;cmath&gt;=\boxed{\textbf{2\cos(n\theta)}}&lt;/cmath&gt;<br /> <br /> Which gives the answer &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_24&diff=121898 1981 AHSME Problems/Problem 24 2020-05-02T02:20:17Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem ==<br /> If &lt;math&gt; \theta&lt;/math&gt; is a constant such that &lt;math&gt; 0 &lt; \theta &lt; \pi&lt;/math&gt; and &lt;math&gt; x + \dfrac{1}{x} = 2\cos{\theta}&lt;/math&gt;, then for each positive integer &lt;math&gt; n&lt;/math&gt;, &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; equals<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta&lt;/math&gt;<br /> <br /> == Solution ==<br /> Multiply both sides by &lt;math&gt;x&lt;/math&gt; and rearrange to &lt;math&gt;x^2-2x\cos(\theta)+1=0&lt;/math&gt;. Using the quadratic equation, we can solve for &lt;math&gt;x&lt;/math&gt;. After some simplifying:<br /> <br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + i\sin(\theta)&lt;/cmath&gt;<br /> <br /> Substituting this expression in to the desired &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; gives:<br /> <br /> &lt;cmath&gt;(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}&lt;/cmath&gt;<br /> <br /> Using DeMoivre's Theorem:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;\cos&lt;/math&gt; is even and &lt;math&gt;\sin&lt;/math&gt; is odd:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)&lt;/cmath&gt;<br /> &lt;cmath&gt;=2\cos(n\theta)&lt;/cmath&gt;<br /> <br /> Which gives the answer &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_24&diff=121897 1981 AHSME Problems/Problem 24 2020-05-02T02:19:52Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem ==<br /> If &lt;math&gt; \theta&lt;/math&gt; is a constant such that &lt;math&gt; 0 &lt; \theta &lt; \pi&lt;/math&gt; and &lt;math&gt; x + \dfrac{1}{x} = 2\cos{\theta}&lt;/math&gt;, then for each positive integer &lt;math&gt; n&lt;/math&gt;, &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; equals<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta&lt;/math&gt;<br /> <br /> == Solution ==<br /> Multiply both sides by &lt;math&gt;x&lt;/math&gt; and rearrange to &lt;math&gt;x^2-2x\cos(\theta)+1=0&lt;/math&gt;. Using the quadratic equation, we can solve for &lt;math&gt;x&lt;/math&gt;. After some simplifying:<br /> <br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta)}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + i\sin(\theta)&lt;/cmath&gt;<br /> <br /> Substituting this expression in to the desired &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; gives:<br /> <br /> &lt;cmath&gt;(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}&lt;/cmath&gt;<br /> <br /> Using DeMoivre's Theorem:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;\cos&lt;/math&gt; is even and &lt;math&gt;\sin&lt;/math&gt; is odd:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)&lt;/cmath&gt;<br /> &lt;cmath&gt;=2\cos(n\theta)&lt;/cmath&gt;<br /> <br /> Which gives the answer &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_24&diff=121896 1981 AHSME Problems/Problem 24 2020-05-02T02:17:46Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem ==<br /> If &lt;math&gt; \theta&lt;/math&gt; is a constant such that &lt;math&gt; 0 &lt; \theta &lt; \pi&lt;/math&gt; and &lt;math&gt; x + \dfrac{1}{x} = 2\cos{\theta}&lt;/math&gt;, then for each positive integer &lt;math&gt; n&lt;/math&gt;, &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; equals<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta&lt;/math&gt;<br /> <br /> == Solution ==<br /> Multiply both sides by &lt;math&gt;x&lt;/math&gt; and rearrange to &lt;math&gt;x^2-2x\cos(\theta)+1=0&lt;/math&gt;. Using the quadratic equation, we can solve for &lt;math&gt;x&lt;/math&gt;. After some simplifying:<br /> <br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\cos(\theta) + i\sin(\theta)&lt;/cmath&gt;<br /> <br /> Substituting this expression in to the desired &lt;math&gt; x^n + \dfrac{1}{x^n}&lt;/math&gt; gives:<br /> <br /> &lt;cmath&gt;(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}&lt;/cmath&gt;<br /> <br /> Using DeMoivre's Theorem:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;\cos&lt;/math&gt; is even and &lt;math&gt;\sin&lt;/math&gt; is odd:<br /> <br /> &lt;cmath&gt;=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)&lt;/cmath&gt;<br /> &lt;cmath&gt;=2\cos(n\theta)&lt;/cmath&gt;<br /> <br /> Which gives the answer &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_20&diff=120009 1990 AHSME Problems/Problem 20 2020-03-23T21:05:29Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>== Problem ==<br /> &lt;asy&gt;<br /> pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0);<br /> draw(A--B--C--D--cycle,dot);<br /> draw(A--E--F--C,dot);<br /> draw(D--E--F--B,dot);<br /> markscalefactor = 0.1;<br /> draw(rightanglemark(B, A, D));<br /> draw(rightanglemark(D, E, C));<br /> draw(rightanglemark(B, F, A));<br /> draw(rightanglemark(D, C, B));<br /> MP(&quot;A&quot;,(0,0),W);<br /> MP(&quot;B&quot;,(7,4.2),N);<br /> MP(&quot;C&quot;,(10,0),E);<br /> MP(&quot;D&quot;,(3,-5),S);<br /> MP(&quot;E&quot;,(3,0),N);<br /> MP(&quot;F&quot;,(7,0),S);<br /> &lt;/asy&gt;<br /> <br /> In the figure &lt;math&gt;ABCD&lt;/math&gt; is a quadrilateral with right angles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BF}&lt;/math&gt; are perpendicual to &lt;math&gt;\overline{AC}&lt;/math&gt;. If &lt;math&gt;AE=3, DE=5, &lt;/math&gt; and &lt;math&gt;CE=7&lt;/math&gt;, then &lt;math&gt;BF=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 3.6\quad<br /> \text{(B) } 4\quad<br /> \text{(C) } 4.2\quad<br /> \text{(D) } 4.5\quad<br /> \text{(E) } 5&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Label the angles as shown in the diagram. Since &lt;math&gt;\angle DEC&lt;/math&gt; forms a linear pair with &lt;math&gt;\angle DEA&lt;/math&gt;, &lt;math&gt;\angle DEA&lt;/math&gt; is a right angle.<br /> <br /> &lt;asy&gt;<br /> pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0);<br /> draw(A--B--C--D--cycle,dot);<br /> draw(A--E--F--C,dot);<br /> draw(D--E--F--B,dot);<br /> <br /> markscalefactor = 0.075;<br /> <br /> draw(rightanglemark(B, A, D));<br /> draw(rightanglemark(D, E, A));<br /> draw(rightanglemark(B, F, A));<br /> draw(rightanglemark(D, C, B));<br /> draw(rightanglemark(D, E, C));<br /> draw(rightanglemark(B, F, C));<br /> MP(&quot;A&quot;,(0,0),W);<br /> MP(&quot;B&quot;,(7,4.2),N);<br /> MP(&quot;C&quot;,(10,0),E);<br /> MP(&quot;D&quot;,(3,-5),S);<br /> MP(&quot;E&quot;,(3,0),N);<br /> MP(&quot;F&quot;,(7,0),S);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;\angle DAE = \alpha&lt;/math&gt; and &lt;math&gt;\angle ADE = \beta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\alpha + \beta = 90^\circ&lt;/math&gt;, and &lt;math&gt;\alpha + \angle BAF = 90^\circ&lt;/math&gt;, then &lt;math&gt;\beta = \angle BAF&lt;/math&gt;. By the same logic, &lt;math&gt;\angle ABF = \alpha&lt;/math&gt;.<br /> <br /> <br /> As a result, &lt;math&gt;\triangle AED \sim \triangle BFA&lt;/math&gt;. By the same logic, &lt;math&gt;\triangle CFB \sim \triangle DEC&lt;/math&gt;.<br /> <br /> Then, &lt;math&gt;\frac{BF}{AF} = \frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\frac{CF}{BF} = \frac{5}{7}&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;7CF = 5BF&lt;/math&gt;, and &lt;math&gt;5BF = 3AF&lt;/math&gt;.<br /> <br /> By the transitive property, &lt;math&gt;7CF = 3AF&lt;/math&gt;. &lt;math&gt;AC = AF + CF = 10&lt;/math&gt;, and plugging in, we get &lt;math&gt;CF = 3&lt;/math&gt;.<br /> <br /> Finally, plugging in to &lt;math&gt;\frac{CF}{BF} = \frac{5}{7}&lt;/math&gt;, we get &lt;math&gt;BF = 4.2&lt;/math&gt; &lt;math&gt;\fbox{C}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1990|num-b=19|num-a=21}} <br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_20&diff=120008 1990 AHSME Problems/Problem 20 2020-03-23T21:03:38Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>== Problem ==<br /> &lt;asy&gt;<br /> pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0);<br /> draw(A--B--C--D--cycle,dot);<br /> draw(A--E--F--C,dot);<br /> draw(D--E--F--B,dot);<br /> markscalefactor = 0.1;<br /> draw(rightanglemark(B, A, D));<br /> draw(rightanglemark(D, E, C));<br /> draw(rightanglemark(B, F, A));<br /> draw(rightanglemark(D, C, B));<br /> MP(&quot;A&quot;,(0,0),W);<br /> MP(&quot;B&quot;,(7,4.2),N);<br /> MP(&quot;C&quot;,(10,0),E);<br /> MP(&quot;D&quot;,(3,-5),S);<br /> MP(&quot;E&quot;,(3,0),N);<br /> MP(&quot;F&quot;,(7,0),S);<br /> &lt;/asy&gt;<br /> <br /> In the figure &lt;math&gt;ABCD&lt;/math&gt; is a quadrilateral with right angles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BF}&lt;/math&gt; are perpendicual to &lt;math&gt;\overline{AC}&lt;/math&gt;. If &lt;math&gt;AE=3, DE=5, &lt;/math&gt; and &lt;math&gt;CE=7&lt;/math&gt;, then &lt;math&gt;BF=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 3.6\quad<br /> \text{(B) } 4\quad<br /> \text{(C) } 4.2\quad<br /> \text{(D) } 4.5\quad<br /> \text{(E) } 5&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Label the angles as shown in the diagram. Since &lt;math&gt;\angle DEC&lt;/math&gt; forms a linear pair with &lt;math&gt;\angle DEA&lt;/math&gt;, &lt;math&gt;\angle DEA&lt;/math&gt; is a right angle.<br /> <br /> &lt;asy&gt;<br /> pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0);<br /> draw(A--B--C--D--cycle,dot);<br /> draw(A--E--F--C,dot);<br /> draw(D--E--F--B,dot);<br /> <br /> markscalefactor = 0.075;<br /> <br /> draw(rightanglemark(B, A, D));<br /> draw(rightanglemark(D, E, A));<br /> draw(rightanglemark(B, F, A));<br /> draw(rightanglemark(D, C, B));<br /> draw(rightanglemark(D, E, C));<br /> draw(rightanglemark(B, F, C));<br /> MP(&quot;A&quot;,(0,0),W);<br /> MP(&quot;B&quot;,(7,4.2),N);<br /> MP(&quot;C&quot;,(10,0),E);<br /> MP(&quot;D&quot;,(3,-5),S);<br /> MP(&quot;E&quot;,(3,0),N);<br /> MP(&quot;F&quot;,(7,0),S);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;\angle DAE = \alpha&lt;/math&gt; and &lt;math&gt;\angle BAF = \beta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\alpha + \beta = 90^\circ&lt;/math&gt;, and &lt;math&gt;\alpha + \angle BAF = 90^\circ&lt;/math&gt;, then &lt;math&gt;\beta = \angle BAF&lt;/math&gt;. By the same logic, &lt;math&gt;\angle ABF = \alpha&lt;/math&gt;.<br /> <br /> <br /> As a result, &lt;math&gt;\triangle AED \sim \triangle BFA&lt;/math&gt;. By the same logic, &lt;math&gt;\triangle CFB \sim \triangle DEC&lt;/math&gt;.<br /> <br /> Then, &lt;math&gt;\frac{BF}{AF} = \frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\frac{CF}{BF} = \frac{5}{7}&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;7CF = 5BF&lt;/math&gt;, and &lt;math&gt;5BF = 3AF&lt;/math&gt;.<br /> <br /> By the transitive property, &lt;math&gt;7CF = 3AF&lt;/math&gt;. &lt;math&gt;AC = AF + CF = 10&lt;/math&gt;, and plugging in, we get &lt;math&gt;CF = 3&lt;/math&gt;.<br /> <br /> Finally, plugging in to &lt;math&gt;\frac{CF}{BF} = \frac{5}{7}&lt;/math&gt;, we get &lt;math&gt;BF = 4.2&lt;/math&gt; &lt;math&gt;\fbox{C}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1990|num-b=19|num-a=21}} <br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1987_AJHSME_Problems/Problem_23&diff=119963 1987 AJHSME Problems/Problem 23 2020-03-22T20:22:04Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Assume the adjoining chart shows the &lt;math&gt;1980&lt;/math&gt; U.S. population, in millions, for each region by ethnic group. To the nearest [[percent]], what percent of the U.S. Black population lived in the South?<br /> <br /> &lt;cmath&gt;\begin{tabular}[t]{c|cccc}<br /> &amp; NE &amp; MW &amp; South &amp; West \\ \hline<br /> White &amp; 42 &amp; 52 &amp; 57 &amp; 35 \\<br /> Black &amp; 5 &amp; 5 &amp; 15 &amp; 2 \\<br /> Asian &amp; 1 &amp; 1 &amp; 1 &amp; 3 \\<br /> Other &amp; 1 &amp; 1 &amp; 2 &amp; 4 <br /> \end{tabular}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 20\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 56\% \qquad \text{(E)}\ 80\% &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> There are &lt;math&gt;5+5+15+2=27&lt;/math&gt; million Blacks living in the U.S. Out of these, &lt;math&gt;15&lt;/math&gt; of them live in the South, so the percentage is &lt;math&gt;\frac{15}{27}\approx \frac{60}{108}\approx 56\% &lt;/math&gt;.<br /> <br /> &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1987|num-b=22|num-a=24}}<br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=119774 2020 AMC 10A Problems/Problem 16 2020-03-19T14:37:05Z <p>Alexwin0806: /* Solution 3 (Estimating) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br /> <br /> == Problem ==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> === Diagram ===<br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray);<br /> draw(arc((1,0), 0.3989, 90, 180));<br /> filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray);<br /> draw(arc((1,1), 0.3989, 180, 270));<br /> filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray);<br /> draw(arc((0,1), 0.3989, 270, 360));<br /> filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> Diagram by [[User:Mathandski|MathandSki]] Using Asymptote<br /> <br /> Note: The diagram represents each unit square of the given &lt;math&gt;2020 * 2020&lt;/math&gt; square.<br /> <br /> ===Solution===<br /> <br /> We consider an individual one-by-one block.<br /> <br /> If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius &lt;math&gt;d&lt;/math&gt;, the area covered by the circles should be &lt;math&gt;0.5&lt;/math&gt;. Because of this, and the fact that there are four circles, we write<br /> <br /> &lt;cmath&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;d&lt;/math&gt;, we obtain &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;, where with &lt;math&gt;\pi \approx 3&lt;/math&gt;, we get &lt;math&gt;d = \frac{1}{\sqrt{6}}&lt;/math&gt;, and from here, we simplify and see that &lt;math&gt;d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}&lt;/math&gt; ~Crypthes<br /> <br /> &lt;math&gt;\textbf{Note:}&lt;/math&gt; To be more rigorous, note that &lt;math&gt;d&lt;0.5&lt;/math&gt; since if &lt;math&gt;d\geq0.5&lt;/math&gt; then clearly the probability is greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This would make sure the above solution works, as if &lt;math&gt;d\geq0.5&lt;/math&gt; there is overlap with the quartercircles. &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> <br /> == Solution 2 ==<br /> As in the previous solution, we obtain the equation &lt;math&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/math&gt;, which simplifies to &lt;math&gt;\pi d^2 = \frac{1}{2} = 0.5&lt;/math&gt;. Since &lt;math&gt;\pi&lt;/math&gt; is slightly more than &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;d^2&lt;/math&gt; is slightly less than &lt;math&gt;\frac{0.5}{3} = 0.1\bar{6}&lt;/math&gt;. We notice that &lt;math&gt;0.1\bar{6}&lt;/math&gt; is slightly more than &lt;math&gt;0.4^2 = 0.16&lt;/math&gt;, so &lt;math&gt;d&lt;/math&gt; is roughly &lt;math&gt;\boxed{\textbf{(B) } 0.4}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> == Solution 3 (Estimating) ==<br /> <br /> As above, we find that we need to estimate &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;. <br /> <br /> Note that we can approximate &lt;math&gt;2\pi \approx 6.28318 \approx 6.25&lt;/math&gt; and so &lt;math&gt;\frac{1}{\sqrt{2\pi}}&lt;/math&gt; &lt;math&gt;\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4&lt;/math&gt;.<br /> <br /> And so our answer is &lt;math&gt;\boxed{\textbf{(B) } 0.4}&lt;/math&gt;.<br /> <br /> ~Silverdragon<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> {{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_10&diff=119432 2020 AMC 10A Problems/Problem 10 2020-03-15T14:19:41Z <p>Alexwin0806: /* Solution 3 (a bit more tedious than others) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #7]] and [[2020 AMC 10A Problems|2020 AMC 10A #10]]}}<br /> <br /> ==Problem==<br /> <br /> Seven cubes, whose volumes are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;343&lt;/math&gt; cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The volume of each cube follows the pattern of &lt;math&gt;n^3&lt;/math&gt; ascending, for &lt;math&gt;n&lt;/math&gt; is between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;.<br /> <br /> We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the &lt;math&gt;7\times 7\times 7&lt;/math&gt; cube (which is just &lt;math&gt;7 \times 7 = 49&lt;/math&gt;). The sides areas can be measured as the sum &lt;math&gt;4\sum_{n=0}^{7} n^2&lt;/math&gt;, giving us &lt;math&gt;560&lt;/math&gt;. Structurally, if we examine the tower from the top, we see that it really just forms a &lt;math&gt;7\times 7&lt;/math&gt; square of area &lt;math&gt;49&lt;/math&gt;. Therefore, we can say that the total surface area is &lt;math&gt;560 + 49 + 49 = \boxed{\textbf{(B) }658}&lt;/math&gt;.<br /> Alternatively, for the area of the tops, we could have found the sum &lt;math&gt;\sum_{n=0}^{6}((n+1)^{2}-n^{2})&lt;/math&gt;, giving us &lt;math&gt;49&lt;/math&gt; as well.<br /> <br /> ~ciceronii<br /> <br /> ==Solution 2==<br /> <br /> It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.<br /> <br /> First, we will calculate the total surface area of the cubes, ignoring overlap. This value is &lt;math&gt;6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840&lt;/math&gt;. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to &lt;math&gt;2\sum_{n=1}^{6} n^2 = 182&lt;/math&gt;. Subtracting the overlapped surface area from the total surface area, we get &lt;math&gt;840 - 182 = \boxed{\textbf{(B) }658}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3 (a bit more tedious than other solutions)==<br /> It can be seen that the side lengths of the cubes using cube roots are all integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;7&lt;/math&gt;, inclusive.<br /> <br /> Only the cubes with side length &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; have &lt;math&gt;5&lt;/math&gt; faces in the surface area and the rest have &lt;math&gt;3&lt;/math&gt;. Also, since the <br /> <br /> cubes are stacked, we have to find the difference between each &lt;math&gt;n^2&lt;/math&gt; and &lt;math&gt;(n-1)^2&lt;/math&gt; side length as &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;7&lt;/math&gt; to <br /> <br /> &lt;math&gt;2&lt;/math&gt;. <br /> <br /> We then come up with this: &lt;math&gt;5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)&lt;/math&gt;. <br /> <br /> We then add all of this and get &lt;math&gt;\boxed{\textbf{(B) }658}&lt;/math&gt;.<br /> <br /> ~aryam<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=9|num-a=11}}<br /> {{AMC12 box|year=2020|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_9&diff=119431 2020 AMC 10A Problems/Problem 9 2020-03-15T14:18:15Z <p>Alexwin0806: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A single bench section at a school event can hold either &lt;math&gt;7&lt;/math&gt; adults or &lt;math&gt;11&lt;/math&gt; children. When &lt;math&gt;N&lt;/math&gt; bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77&lt;/math&gt;<br /> <br /> == Solution == <br /> <br /> The least common multiple of &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt; is &lt;math&gt;77&lt;/math&gt;. Therefore, there must be &lt;math&gt;77&lt;/math&gt; adults and &lt;math&gt;77&lt;/math&gt; children. The total number of benches is &lt;math&gt;\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\text{(B) }18}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> <br /> This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt; are relatively prime, their LCM must be their product. So the answer would be &lt;math&gt;7 + 11 = \boxed{\text{(B) } 18}&lt;/math&gt;. ~Baolan<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_5&diff=119420 2016 AMC 10A Problems/Problem 5 2020-03-15T01:56:39Z <p>Alexwin0806: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the smallest side length be &lt;math&gt;x&lt;/math&gt;. Then the volume is &lt;math&gt;x \cdot 3x \cdot 4x =12x^3&lt;/math&gt;. If &lt;math&gt;x=2&lt;/math&gt;, then &lt;math&gt;12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> As seen in the first solution, we end up with &lt;math&gt;12x^3&lt;/math&gt;. Taking the answer choices and dividing by &lt;math&gt;12&lt;/math&gt;, we get &lt;math&gt;(A) 4&lt;/math&gt;,<br /> &lt;math&gt;(B) 4 \frac{2}{3}&lt;/math&gt;, &lt;math&gt;(C) 5 \frac{1}{3}&lt;/math&gt;, &lt;math&gt;(D) 8&lt;/math&gt;, &lt;math&gt;(E) 12&lt;/math&gt; and the final answer has to equal &lt;math&gt;x^3&lt;/math&gt;. The only answer choice that works is &lt;math&gt;(D)&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_5&diff=119419 2016 AMC 10A Problems/Problem 5 2020-03-15T01:56:16Z <p>Alexwin0806: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the smallest side length be &lt;math&gt;x&lt;/math&gt;. Then the volume is &lt;math&gt;x \cdot 3x \cdot 4x =12x^3&lt;/math&gt;. If &lt;math&gt;x=2&lt;/math&gt;, then &lt;math&gt;12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> As seen in the first solution, we end up with &lt;math&gt;12x^3&lt;/math&gt;. Taking the answer choices and dividing by &lt;math&gt;12&lt;/math&gt;, we get &lt;math&gt;(A) 4&lt;/math&gt;,<br /> &lt;math&gt;(B) 4 \frac{2}{3}&lt;/math&gt;, &lt;math&gt;(C) 5 \frac{1}{3}&lt;/math&gt;, &lt;math&gt;(D) 8&lt;/math&gt;, &lt;math&gt;(E) 12&lt;/math&gt; and the final answer has to equal &lt;math&gt;x^3&lt;/math&gt;. The only answer choice that works is &lt;math&gt;(D)&lt;/math&gt;.<br /> ---AlexWin0806<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_5&diff=119418 2016 AMC 10A Problems/Problem 5 2020-03-15T01:55:45Z <p>Alexwin0806: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the smallest side length be &lt;math&gt;x&lt;/math&gt;. Then the volume is &lt;math&gt;x \cdot 3x \cdot 4x =12x^3&lt;/math&gt;. If &lt;math&gt;x=2&lt;/math&gt;, then &lt;math&gt;12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> As seen in the first solution, we end up with &lt;math&gt;12x^3&lt;/math&gt;. Taking the answer choices and dividing by &lt;math&gt;12&lt;/math&gt;, we get &lt;math&gt;(A) 4&lt;/math&gt;,<br /> &lt;math&gt;(B) 4 \frac{2}{3}&lt;/math&gt;, &lt;math&gt;(C) 5 \frac{1}{3}&lt;/math&gt;, &lt;math&gt;(D) 8&lt;/math&gt;, &lt;math&gt;(E) 12&lt;/math&gt; and the final answer has to equal &lt;math&gt;x^3&lt;/math&gt;. The only answer choice that works is &lt;math&gt;(D)&lt;/math&gt;.<br /> [[User:Alexwin0806|Alexwin0806]] ([[User talk:Alexwin0806|talk]])AlexWin0806<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_5&diff=119417 2016 AMC 10A Problems/Problem 5 2020-03-15T01:54:06Z <p>Alexwin0806: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the smallest side length be &lt;math&gt;x&lt;/math&gt;. Then the volume is &lt;math&gt;x \cdot 3x \cdot 4x =12x^3&lt;/math&gt;. If &lt;math&gt;x=2&lt;/math&gt;, then &lt;math&gt;12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> As seen in the first solution, we end up with &lt;math&gt;12x^3&lt;/math&gt;. Taking the answer choices and dividing by &lt;math&gt;12&lt;/math&gt;, we get &lt;math&gt;(A) 4&lt;/math&gt;,<br /> &lt;math&gt;(B) 4 \frac{2}{3}&lt;/math&gt;, &lt;math&gt;(C) 5 \frac{1}{3}&lt;/math&gt;, &lt;math&gt;(D) 8&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_5&diff=119416 2016 AMC 10A Problems/Problem 5 2020-03-15T01:51:07Z <p>Alexwin0806: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the smallest side length be &lt;math&gt;x&lt;/math&gt;. Then the volume is &lt;math&gt;x \cdot 3x \cdot 4x =12x^3&lt;/math&gt;. If &lt;math&gt;x=2&lt;/math&gt;, then &lt;math&gt;12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> As seen in the first solution, we end up with &lt;math&gt;12x^3&lt;/math&gt;. Taking the answer choices and dividing by &lt;math&gt;12&lt;/math&gt;, we get &lt;math&gt;(A) 4&lt;/math&gt;,<br /> &lt;math&gt;(B) 4 2/3&lt;/math&gt;, &lt;math&gt;(C) 5 1/3&lt;/math&gt;, &lt;math&gt;(D) 8&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_5&diff=119415 2016 AMC 10A Problems/Problem 5 2020-03-15T01:50:33Z <p>Alexwin0806: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the smallest side length be &lt;math&gt;x&lt;/math&gt;. Then the volume is &lt;math&gt;x \cdot 3x \cdot 4x =12x^3&lt;/math&gt;. If &lt;math&gt;x=2&lt;/math&gt;, then &lt;math&gt;12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> As seen in the first solution, we end up with &lt;math&gt;12x^3&lt;/math&gt;. Taking the answer choices and dividing by &lt;math&gt;12&lt;/math&gt;, we get &lt;math&gt;(A) 4&lt;/math&gt;,<br /> &lt;math&gt;(B) 4 2/3&lt;/math&gt;, &lt;math&gt;(C) 5 1/3&lt;/math&gt;, &lt;math&gt;(D)&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_5&diff=119413 2016 AMC 10A Problems/Problem 5 2020-03-15T01:47:34Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the smallest side length be &lt;math&gt;x&lt;/math&gt;. Then the volume is &lt;math&gt;x \cdot 3x \cdot 4x =12x^3&lt;/math&gt;. If &lt;math&gt;x=2&lt;/math&gt;, then &lt;math&gt;12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> As seen in the first solution, we end up with &lt;math&gt;12x^3&lt;/math&gt;. Taking the answer choices and dividing by &lt;math&gt;12&lt;/math&gt;, we get<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_6&diff=119073 1993 AIME Problems/Problem 6 2020-03-11T22:16:34Z <p>Alexwin0806: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> What is the smallest [[positive]] [[integer]] that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Denote the first of each of the series of consecutive integers as &lt;math&gt;a,\ b,\ c&lt;/math&gt;. Therefore, &lt;math&gt;n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;. Simplifying, &lt;math&gt;9a = 10b + 9 = 11c + 19&lt;/math&gt;. The relationship between &lt;math&gt;a,\ b&lt;/math&gt; suggests that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Also, &lt;math&gt;10b -10 = 10(b-1) = 11c&lt;/math&gt;, so &lt;math&gt;b-1&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;. We find that the least possible value of &lt;math&gt;b = 45&lt;/math&gt;, so the answer is &lt;math&gt;10(45) + 45 = 495&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let the desired integer be &lt;math&gt;n&lt;/math&gt;. From the information given, it can be determined that, for positive integers &lt;math&gt;a, \ b, \ c&lt;/math&gt;:<br /> <br /> &lt;math&gt;n = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;<br /> <br /> This can be rewritten as the following congruences:<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{9}&lt;/math&gt; <br /> <br /> &lt;math&gt;n \equiv 5 \pmod{10}&lt;/math&gt;<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{11}&lt;/math&gt;<br /> <br /> Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is &lt;math&gt;\boxed{495}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the desired integer. From the given information, we have<br /> &lt;cmath&gt; \begin{align*}9x &amp;= a \\ 11y &amp;= a \\ 10z + 5 &amp;= a, \end{align*}&lt;/cmath&gt; here, &lt;math&gt;x,&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have &lt;math&gt;z&lt;/math&gt; as the 4th term of the sequence. Since, &lt;math&gt;a&lt;/math&gt; is a multiple of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11,&lt;/math&gt; it is also a multiple of &lt;math&gt;\text{lcm}[9,11]=99.&lt;/math&gt; Hence, &lt;math&gt;a=99m,&lt;/math&gt; for some &lt;math&gt;m.&lt;/math&gt; So, we have &lt;math&gt;10z + 5 = 99m.&lt;/math&gt; It follows that &lt;math&gt;99(5) = \boxed{495}&lt;/math&gt; is the smallest integer that can be represented in such a way.<br /> <br /> === Solution 4 ===<br /> By the method in Solution 1, we find that the number &lt;math&gt;n&lt;/math&gt; can be written as &lt;math&gt;9a+36=10b+45=11c+55&lt;/math&gt; for some integers &lt;math&gt;a,b,c&lt;/math&gt;. From this, we can see that &lt;math&gt;n&lt;/math&gt; must be divisible by 9, 5, and 11. This means &lt;math&gt;n&lt;/math&gt; must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that &lt;math&gt;n&lt;/math&gt; cannot be divisible by 10, so &lt;math&gt;n&lt;/math&gt; must equal &lt;math&gt;\boxed{495}&lt;/math&gt;. Solution by Zeroman.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_16&diff=119046 2015 AMC 10A Problems/Problem 16 2020-03-11T01:39:52Z <p>Alexwin0806: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> If &lt;math&gt;y+4 = (x-2)^2, x+4 = (y-2)^2&lt;/math&gt;, and &lt;math&gt;x \neq y&lt;/math&gt;, what is the value of &lt;math&gt;x^2+y^2&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }\text{30} &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> Note that we can add the two equations to yield the equation<br /> <br /> &lt;math&gt;x^2 + y^2 - 4x - 4y + 8 = x + y + 8.&lt;/math&gt;<br /> <br /> Moving terms gives the equation<br /> <br /> &lt;math&gt;x^2+y^2=5 \left( x + y \right).&lt;/math&gt;<br /> <br /> We can also subtract the two equations to yield the equation<br /> <br /> &lt;math&gt;x^2 - y^2 - 4x +4y = y - x.&lt;/math&gt;<br /> <br /> Moving terms gives the equation<br /> <br /> &lt;math&gt;x^2 - y^2 = (x + y)(x - y) = 3x - 3y = 3(x - y).&lt;/math&gt;<br /> <br /> Because &lt;math&gt;x \neq y,&lt;/math&gt; we can divide both sides of the equation by &lt;math&gt;x - y&lt;/math&gt; to yield the equation<br /> <br /> &lt;math&gt;x + y = 3.&lt;/math&gt;<br /> <br /> Substituting this into the equation for &lt;math&gt;x^2 + y^2&lt;/math&gt; that we derived earlier gives<br /> <br /> &lt;math&gt;x^2 + y^2 = 5 \left( x + y \right) = 5 \left( 3 \right) = \boxed{\textbf{(B) } 15}&lt;/math&gt;<br /> <br /> ===Solution 2 (Algebraic) ===<br /> <br /> Subtract &lt;math&gt;4&lt;/math&gt; from the left hand side of both equations, and use difference of squares to yield the equations<br /> <br /> &lt;math&gt;x = y(y-4)&lt;/math&gt; and &lt;math&gt;y = x(x-4)&lt;/math&gt;.<br /> <br /> It may save some time to find two solutions, &lt;math&gt;(0, 0)&lt;/math&gt; and &lt;math&gt;(5, 5)&lt;/math&gt;, at this point. However, &lt;math&gt; x = y&lt;/math&gt; in these solutions.<br /> <br /> <br /> Substitute &lt;math&gt;y = x(x-4)&lt;/math&gt; into &lt;math&gt;x = y(y-4)&lt;/math&gt;.<br /> <br /> <br /> This gives the equation<br /> <br /> &lt;math&gt;x = x(x-4)(x^2-4x-4)&lt;/math&gt;<br /> <br /> which can be simplified to<br /> <br /> &lt;math&gt;x(x^3 - 8x^2 +12x + 15) = 0&lt;/math&gt;.<br /> <br /> Knowing &lt;math&gt;x = 0&lt;/math&gt; and &lt;math&gt; x = 5&lt;/math&gt; are solutions is now helpful, as you divide both sides by &lt;math&gt;x(x-5)&lt;/math&gt;. This can also be done using polynomial division to find &lt;math&gt;x = 5&lt;/math&gt; as a factor. This gives<br /> <br /> &lt;math&gt;x^2 - 3x -3 = 0&lt;/math&gt;.<br /> <br /> Because the two equations &lt;math&gt;x = y(y-4)&lt;/math&gt; and &lt;math&gt;y = x(x-4)&lt;/math&gt; are symmetric, the &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; values are the roots of the equation, which are &lt;math&gt;x = \frac{3 + \sqrt{21}}{2}&lt;/math&gt; and &lt;math&gt;x = \frac{3 - \sqrt{21}}{2}&lt;/math&gt;.<br /> <br /> Squaring these and adding them together gives<br /> <br /> &lt;math&gt;\frac{3^2 + 21 + 6\sqrt{21}}{4} + \frac{3^2 + 21 - 6\sqrt{21}}{4} = \frac{2(3^2 +21)}{4} = \boxed{\textbf{(B) } 15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> By graphing the two equations on a piece of graph paper, we can see that the point where they intersect that is not on the line &lt;math&gt;y=x&lt;/math&gt; is close to the point &lt;math&gt;(4,-1)&lt;/math&gt; (or &lt;math&gt;(-1, 4)&lt;/math&gt;). &lt;math&gt;(-1)^2+4^2=17&lt;/math&gt;, and the closest answer choice to &lt;math&gt;17&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) } 15}&lt;/math&gt;.<br /> <br /> ====Note for solution 3:====<br /> This is risky, as &lt;math&gt;20&lt;/math&gt; could be a viable answer too. Do not use this method unless you're sure about the answer. In other words, this solution is less reliable than the others, so only use it if you can't do the other methods.<br /> <br /> ===Solution 4 (When you can't algebraically manipulate)===<br /> This, by the way, is a really cheap way of solving the problem but as long as you get an answer, it doesn't matter.<br /> <br /> Looking at the first given equation, begin searching for solutions. Notice how &lt;math&gt;(4,0)&lt;/math&gt; works but when plugged into the<br /> <br /> second equation, you get &lt;math&gt;8=4&lt;/math&gt;. Now, if you decrease the &lt;math&gt;y&lt;/math&gt; value by &lt;math&gt;1&lt;/math&gt; to obtain &lt;math&gt;(2+\sqrt3,-1)&lt;/math&gt;, plugging it into <br /> <br /> the second equation will yield &lt;math&gt;6+\sqrt3=9&lt;/math&gt;. Now, notice how the LHS is now less than the RHS as opposed to what it had <br /> <br /> been when we plugged &lt;math&gt;(4,0)&lt;/math&gt; into the second given equation. We then conclude that there must be a solution between <br /> <br /> &lt;math&gt;y=0&lt;/math&gt; and &lt;math&gt;y=-1&lt;/math&gt;, so calculating &lt;math&gt;x^2+y^2&lt;/math&gt; of &lt;math&gt;(4,0)&lt;/math&gt; and &lt;math&gt;(2+\sqrt3,-1)&lt;/math&gt; we obtain<br /> <br /> &lt;cmath&gt;4^2+0^2=16&lt;/cmath&gt; and<br /> &lt;cmath&gt;(2+\sqrt3)^2+(-1)^2=8+4\sqrt3\approx14.9&lt;/cmath&gt;<br /> <br /> Thus we know the answer to be &lt;math&gt;\boxed{\textbf{(B) } 15}&lt;/math&gt;.<br /> <br /> This is a really bad solution, but it works. :/<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_7&diff=119037 2015 AMC 10A Problems/Problem 7 2020-03-11T01:20:05Z <p>Alexwin0806: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> How many terms are in the arithmetic sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;73-13 = 60&lt;/math&gt;, so the amount of terms in the sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt; is the same as in the sequence &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;57&lt;/math&gt;, &lt;math&gt;60&lt;/math&gt;. <br /> <br /> In this sequence, the terms are the multiples of &lt;math&gt;3&lt;/math&gt; going up to &lt;math&gt;60&lt;/math&gt;, and there are &lt;math&gt;20&lt;/math&gt; multiples of &lt;math&gt;3&lt;/math&gt; in &lt;math&gt;60&lt;/math&gt;. <br /> <br /> However, one more must be added to include the first term. So, the answer is &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Using the formula for arithmetic sequence's nth term, we see that &lt;math&gt;a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21&lt;/math&gt;<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Minus each of the terms by &lt;math&gt;12&lt;/math&gt; to make the the sequence &lt;math&gt;1 , 4 , 7,..., 61&lt;/math&gt;. &lt;math&gt;61-1/3=20, 20 + 1 = 21&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Subtract each of the terms by &lt;math&gt;10&lt;/math&gt; to make the sequence &lt;math&gt;3 , 6 , 9,..., 60, 63&lt;/math&gt;. Then divide the each term in the sequence by &lt;math&gt;3&lt;/math&gt; to get &lt;math&gt;1, 2, 3,..., 20, 21&lt;/math&gt;. Now it is clear to see that there are &lt;math&gt;21&lt;/math&gt; terms in the sequence.<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_7&diff=119036 2015 AMC 10A Problems/Problem 7 2020-03-11T01:19:35Z <p>Alexwin0806: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> <br /> How many terms are in the arithmetic sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;73-13 = 60&lt;/math&gt;, so the amount of terms in the sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt; is the same as in the sequence &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;57&lt;/math&gt;, &lt;math&gt;60&lt;/math&gt;. <br /> <br /> In this sequence, the terms are the multiples of &lt;math&gt;3&lt;/math&gt; going up to &lt;math&gt;60&lt;/math&gt;, and there are &lt;math&gt;20&lt;/math&gt; multiples of &lt;math&gt;3&lt;/math&gt; in &lt;math&gt;60&lt;/math&gt;. <br /> <br /> However, one more must be added to include the first term. So, the answer is &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Using the formula for arithmetic sequence's nth term, we see that &lt;math&gt;a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21&lt;/math&gt;<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Minus each of the terms by 12 to make the the sequence &lt;math&gt;1 , 4 , 7,..., 61&lt;/math&gt;. <br /> <br /> &lt;math&gt;61-1/3=20, 20 + 1 = 21&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Subtract each of the terms by &lt;math&gt;10&lt;/math&gt; to make the sequence &lt;math&gt;3 , 6 , 9,..., 60, 63&lt;/math&gt;. Then divide the each term in the sequence by &lt;math&gt;3&lt;/math&gt; to get &lt;math&gt;1, 2, 3,..., 20, 21&lt;/math&gt;. Now it is clear to see that there are &lt;math&gt;21&lt;/math&gt; terms in the sequence.<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_7&diff=119032 2015 AMC 10A Problems/Problem 7 2020-03-10T23:55:38Z <p>Alexwin0806: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> <br /> How many terms are in the arithmetic sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;73-13 = 60&lt;/math&gt;, so the amount of terms in the sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt; is the same as in the sequence &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;57&lt;/math&gt;, &lt;math&gt;60&lt;/math&gt;. <br /> <br /> In this sequence, the terms are the multiples of &lt;math&gt;3&lt;/math&gt; going up to &lt;math&gt;60&lt;/math&gt;, and there are &lt;math&gt;20&lt;/math&gt; multiples of &lt;math&gt;3&lt;/math&gt; in &lt;math&gt;60&lt;/math&gt;. <br /> <br /> However, one more must be added to include the first term. So, the answer is &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Using the formula for arithmetic sequence's nth term, we see that &lt;math&gt;a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21&lt;/math&gt;<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Minus each of the terms by 12 to make the the sequence &lt;math&gt;1 , 4 , 7,..., 61&lt;/math&gt;. <br /> <br /> &lt;math&gt;61-1/3=20, 20 + 1 = 21&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Subtract each of the terms by 10 to make the sequence &lt;math&gt;3 , 6 , 9,..., 60, 63&lt;/math&gt;. Then divide the each term in the sequence by &lt;math&gt;3&lt;/math&gt; to get &lt;math&gt;1, 2, 3,..., 20, 21&lt;/math&gt;. Now it is clear to see that there are &lt;math&gt;21&lt;/math&gt; terms in the sequence.<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_7&diff=119030 2015 AMC 10A Problems/Problem 7 2020-03-10T23:54:04Z <p>Alexwin0806: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> How many terms are in the arithmetic sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;73-13 = 60&lt;/math&gt;, so the amount of terms in the sequence &lt;math&gt;13&lt;/math&gt;, &lt;math&gt;16&lt;/math&gt;, &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;70&lt;/math&gt;, &lt;math&gt;73&lt;/math&gt; is the same as in the sequence &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;\dotsc&lt;/math&gt;, &lt;math&gt;57&lt;/math&gt;, &lt;math&gt;60&lt;/math&gt;. <br /> <br /> In this sequence, the terms are the multiples of &lt;math&gt;3&lt;/math&gt; going up to &lt;math&gt;60&lt;/math&gt;, and there are &lt;math&gt;20&lt;/math&gt; multiples of &lt;math&gt;3&lt;/math&gt; in &lt;math&gt;60&lt;/math&gt;. <br /> <br /> However, one more must be added to include the first term. So, the answer is &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Using the formula for arithmetic sequence's nth term, we see that &lt;math&gt;a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21&lt;/math&gt;<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Minus each of the terms by 12 to make the the sequence &lt;math&gt;1 , 4 , 7,..., 61&lt;/math&gt;. <br /> <br /> &lt;math&gt;61-1/3=20, 20 + 1 = 21&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Subtract each of the terms by 10 to make the sequence 3,6,9,....,60,63. Then divide the each term in the sequence by 3 to get 1,2,3,...,20,21. Now it is clear to see that there are 21 terms in the sequence<br /> &lt;math&gt;\boxed{\textbf{(B)}\ 21}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_6&diff=119029 2015 AMC 10A Problems/Problem 6 2020-03-10T23:41:55Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #4]] and [[2015 AMC 10A Problems|2015 AMC 10A #6]]}}<br /> ==Problem==<br /> <br /> The sum of two positive numbers is &lt;math&gt; 5 &lt;/math&gt; times their difference. What is the ratio of the larger number to the smaller number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the bigger number and &lt;math&gt;b&lt;/math&gt; be the smaller.<br /> <br /> &lt;math&gt;a + b = 5(a - b)&lt;/math&gt;.<br /> <br /> Multiplying out gives &lt;math&gt;a + b = 5a - 5b&lt;/math&gt; and rearranging gives &lt;math&gt;4a = 6b&lt;/math&gt; and then solving gives<br /> <br /> &lt;math&gt;\frac{a}{b} = \frac32&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(B) }\frac32}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2015|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_6&diff=119028 2015 AMC 10A Problems/Problem 6 2020-03-10T23:41:35Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #4]] and [[2015 AMC 10A Problems|2015 AMC 10A #6]]}}<br /> ==Problem==<br /> <br /> The sum of two positive numbers is &lt;math&gt; 5 &lt;/math&gt; times their difference. What is the ratio of the larger number to the smaller number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the bigger number and &lt;math&gt;b&lt;/math&gt; be the smaller.<br /> <br /> &lt;math&gt;a + b = 5(a - b)&lt;/math&gt;.<br /> <br /> Multiplying out gives &lt;math&gt;a + b = 5a - 5b&lt;/math&gt; and rearranging gives 4a = 6b and then solving gives<br /> <br /> &lt;math&gt;\frac{a}{b} = \frac32&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(B) }\frac32}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2015|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_2&diff=119027 2015 AMC 10A Problems/Problem 2 2020-03-10T23:27:53Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A box contains a collection of triangular and square tiles. There are &lt;math&gt;25&lt;/math&gt; tiles in the box, containing &lt;math&gt;84&lt;/math&gt; edges total. How many square tiles are there in the box?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the amount of triangular tiles and &lt;math&gt;b&lt;/math&gt; be the amount of square tiles.<br /> <br /> Triangles have &lt;math&gt;3&lt;/math&gt; edges and squares have &lt;math&gt;4&lt;/math&gt; edges, so we have a system of equations.<br /> <br /> We have &lt;math&gt;a + b&lt;/math&gt; tiles total, so &lt;math&gt;a + b = 25&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;3a + 4b&lt;/math&gt; edges total, so &lt;math&gt;3a + 4b = 84&lt;/math&gt;.<br /> <br /> Multiplying the first equation by &lt;math&gt;3&lt;/math&gt; on both sides gives &lt;math&gt;3a + 3b = 3(25) = 75&lt;/math&gt;.<br /> <br /> Second equation minus the first equation gives &lt;math&gt;b = 9&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(D) }9}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> If all of the tiles were triangles, there would be &lt;math&gt;75&lt;/math&gt; edges. This is not enough, so there need to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out &lt;math&gt;9&lt;/math&gt; triangles for squares. Answer: &lt;math&gt;\boxed{\textbf{(D) }9}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_2&diff=119026 2015 AMC 10A Problems/Problem 2 2020-03-10T23:27:21Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A box contains a collection of triangular and square tiles. There are &lt;math&gt;25&lt;/math&gt; tiles in the box, containing &lt;math&gt;84&lt;/math&gt; edges total. How many square tiles are there in the box?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the amount of triangular tiles and &lt;math&gt;b&lt;/math&gt; be the amount of square tiles.<br /> <br /> Triangles have &lt;math&gt;3&lt;/math&gt; edges and squares have &lt;math&gt;4&lt;/math&gt; edges, so we have a system of equations.<br /> <br /> We have &lt;math&gt;a + b&lt;/math&gt; tiles total, so &lt;math&gt;a + b = 25&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;3a + 4b&lt;/math&gt; edges total, so &lt;math&gt;3a + 4b = 84&lt;/math&gt;.<br /> <br /> Multiplying the first equation by &lt;math&gt;3&lt;/math&gt; on both sides gives &lt;math&gt;3a + 3b = 3(25) = 75&lt;/math&gt;.<br /> <br /> Second equation minus the first equation gives &lt;math&gt;b = 9&lt;/math&gt;.<br /> <br /> Solving gives, &lt;math&gt;a = 16&lt;/math&gt; and &lt;math&gt;b = 9&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(D) }9}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> If all of the tiles were triangles, there would be &lt;math&gt;75&lt;/math&gt; edges. This is not enough, so there need to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out &lt;math&gt;9&lt;/math&gt; triangles for squares. Answer: &lt;math&gt;\boxed{\textbf{(D) }9}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_24&diff=118973 2014 AMC 10A Problems/Problem 24 2020-03-09T23:41:16Z <p>Alexwin0806: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> A sequence of natural numbers is constructed by listing the first &lt;math&gt;4&lt;/math&gt;, then skipping one, listing the next &lt;math&gt;5&lt;/math&gt;, skipping &lt;math&gt;2&lt;/math&gt;, listing &lt;math&gt;6&lt;/math&gt;, skipping &lt;math&gt;3&lt;/math&gt;, and, on the &lt;math&gt;n&lt;/math&gt;th iteration, listing &lt;math&gt;n+3&lt;/math&gt; and skipping &lt;math&gt;n&lt;/math&gt;. The sequence begins &lt;math&gt;1,2,3,4,6,7,8,9,10,13&lt;/math&gt;. What is the &lt;math&gt;500,\!000&lt;/math&gt;th number in the sequence?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If we list the rows by iterations, then we get <br /> <br /> &lt;math&gt;1,2,3,4&lt;/math&gt;<br /> <br /> &lt;math&gt;6,7,8,9,10&lt;/math&gt;<br /> <br /> &lt;math&gt;13,14,15,16,17,18&lt;/math&gt; etc.<br /> <br /> so that the &lt;math&gt;500,000&lt;/math&gt;th number is the &lt;math&gt;506&lt;/math&gt;th number on the &lt;math&gt;997&lt;/math&gt;th row because &lt;math&gt;4+5+6+7......+999 = 499,494&lt;/math&gt;. The last number of the &lt;math&gt;996&lt;/math&gt;th row (when including the numbers skipped) is &lt;math&gt;499,494 + (1+2+3+4.....+996)= 996,000&lt;/math&gt;, (we add the &lt;math&gt;1-996&lt;/math&gt; because of the numbers we skip) so our answer is &lt;math&gt;996,000 + 506 = \boxed{\textbf{(A)}996,506}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let's start with natural numbers, with no skips in between.<br /> <br /> &lt;math&gt;1,2,3,4,5,...,500,000&lt;/math&gt;<br /> <br /> All we need to do is count how many numbers are skipped, &lt;math&gt;n&lt;/math&gt;, and &quot;push&quot; (add on to) &lt;math&gt;500,000&lt;/math&gt; however many numbers are skipped.<br /> <br /> Clearly, &lt;math&gt;\frac{999(1000)}{2}&lt;500,000&lt;\frac{1000(1001)}{2}&lt;/math&gt;. This means that the number of skipped number &quot;blocks&quot; in the sequence is &lt;math&gt;999-3=996&lt;/math&gt; because we started counting from 4.<br /> <br /> Therefore &lt;math&gt;n=\frac{996(997)}{2}=496,506&lt;/math&gt;, and the answer is &lt;math&gt;496,506+500000=\boxed{\textbf{(A)}996,506}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_20&diff=118972 2014 AMC 10A Problems/Problem 20 2020-03-09T23:27:05Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #16]] and [[2014 AMC 10A Problems|2014 AMC 10A #20]]}}<br /> ==Problem==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;The product &lt;math&gt;(8)(888\dots8)&lt;/math&gt;, where the second factor has &lt;math&gt;k&lt;/math&gt; digits, is an integer whose digits have a sum of &lt;math&gt;1000&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999&lt;/math&gt;<br /> <br /> ==Solution==<br /> We can list the first few numbers in the form &lt;math&gt;8*(8....8)&lt;/math&gt;<br /> <br /> (Hard problem to do without the multiplication, but you can see the pattern early on)<br /> <br /> &lt;math&gt;8*8 = 64&lt;/math&gt;<br /> <br /> &lt;math&gt;8*88 = 704&lt;/math&gt;<br /> <br /> &lt;math&gt;8*888 = 7104&lt;/math&gt;<br /> <br /> &lt;math&gt;8*8888 = 71104&lt;/math&gt;<br /> <br /> &lt;math&gt;8*88888 = 711104&lt;/math&gt;<br /> <br /> By now it's clear that the numbers will be in the form &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;k-2&lt;/math&gt; &lt;math&gt;1&lt;/math&gt;s, and &lt;math&gt;04&lt;/math&gt;. We want to make the numbers sum to 1000, so &lt;math&gt;7+4+(k-2) = 1000&lt;/math&gt;. Solving, we get &lt;math&gt;k = 991&lt;/math&gt;, meaning the answer is &lt;math&gt;\fbox{(D)}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2014|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Number Theory Problems]]</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_13&diff=117876 2018 AMC 10A Problems/Problem 13 2020-02-17T22:10:45Z <p>Alexwin0806: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point &lt;math&gt;A&lt;/math&gt; falls on point &lt;math&gt;B&lt;/math&gt;. What is the length in inches of the crease?<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,3), NE);<br /> label(&quot;$C$&quot;, (4,0), SE);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt; draw((0,0)--(4,0)--(4,3)--(0,0)); label(&quot;$A$&quot;, (0,0), SW); label(&quot;$B$&quot;, (4,3), NE); label(&quot;$C$&quot;, (4,0), SE); label(&quot;$D$&quot;, (2,1.5), NW); label(&quot;$E$&quot;, (3.125,0), S);<br /> draw ((2,1.5)--(3.125,0),linewidth(1));<br /> draw(rightanglemark((0,0),(2,1.5),(3.125,0))); &lt;/asy&gt;<br /> <br /> <br /> First, we need to realize that the crease line is just the perpendicular bisector of side &lt;math&gt;AB&lt;/math&gt;, the hypotenuse of right triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;. Draw this line and call the intersection point with &lt;math&gt;AC&lt;/math&gt; as &lt;math&gt;E&lt;/math&gt;. Now, &lt;math&gt;\triangle ACB&lt;/math&gt; is similar to &lt;math&gt;\triangle ADE&lt;/math&gt; by &lt;math&gt;AA&lt;/math&gt; similarity. Setting up the ratios, we find that<br /> &lt;cmath&gt;\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{(D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> <br /> ===Note===<br /> In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of &lt;math&gt;AB&lt;/math&gt;, because &lt;math&gt;A&lt;/math&gt; must be reflected onto &lt;math&gt;B&lt;/math&gt;.<br /> <br /> ==Solution 2 (Draw it, not recommended)==<br /> <br /> Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than &lt;math&gt;\frac{7}{4}&lt;/math&gt; units and somewhat less than &lt;math&gt;2&lt;/math&gt; units. The only answer choice in range is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> This is pretty much a cop-out, but it's allowed in the rules technically. This is basically useless for problems without diagrams.<br /> <br /> ==Solution 3==<br /> <br /> Since &lt;math&gt;\triangle ABC&lt;/math&gt; is a right triangle, we can see that the slope of line &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;\frac{BC}{AC}&lt;/math&gt; = &lt;math&gt;\frac{3}{4}&lt;/math&gt;. We know that if we fold &lt;math&gt;\triangle ABC&lt;/math&gt; so that point &lt;math&gt;A&lt;/math&gt; meets point &lt;math&gt;B&lt;/math&gt; the crease line will be perpendicular to &lt;math&gt;AB&lt;/math&gt; and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is &lt;math&gt;-\frac{4}{3}&lt;/math&gt;.<br /> &lt;asy&gt;<br /> pen dotstyle = black;<br /> <br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> <br /> dot((0,0),dotstyle); <br /> label(&quot;$A$&quot;, (0.03153837092244126,0.07822624343603715), SW); <br /> dot((4,0),dotstyle); <br /> label(&quot;$C$&quot;, (4.028913881471271,0.07822624343603715), SE); <br /> dot((4,3),dotstyle); <br /> label(&quot;$B$&quot;, (4.028913881471271,3.078221223847919), NE); <br /> dot((2,1.5),dotstyle);<br /> label(&quot;$D$&quot;, (2.0341528211973956,1.578223733641978), SE); <br /> dot((2,0),dotstyle);<br /> label(&quot;$E$&quot;, (2.0341528211973956,0.07822624343603715), NE); <br /> dot((3.1249518689638927,0),dotstyle);<br /> label(&quot;$F$&quot;, (3.1571875913515854,0.07822624343603715), NE); <br /> <br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> &lt;/asy&gt;<br /> Let us call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;, the midpoint of &lt;math&gt;AC&lt;/math&gt; point &lt;math&gt;E&lt;/math&gt;, and the crease line &lt;math&gt;DF&lt;/math&gt;. We know that &lt;math&gt;DE&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt; and that &lt;math&gt;DE&lt;/math&gt;'s length is &lt;math&gt;\frac{BC}{2}=\frac{3}{2}&lt;/math&gt;. Using our slope calculation from earlier, we can see that&lt;math&gt;-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}&lt;/math&gt;. With this information, we can solve for &lt;math&gt;EF&lt;/math&gt;: <br /> &lt;cmath&gt;-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.&lt;/cmath&gt;<br /> We can then use the Pythagorean Theorem to find &lt;math&gt;DF&lt;/math&gt;.<br /> &lt;cmath&gt;\frac{3}{2}^2+\frac{9}{8}^2=DF^2 \Rightarrow \frac{9}{4}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8}{4\cdot2\cdot8}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8+9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9(2\cdot8+9)}{8\cdot8}=DF^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow DF=\sqrt{\frac{9(2\cdot8+9)}{8\cdot8}} \Rightarrow DF=\frac{3\cdot5}{8} \Rightarrow DF=\frac{15}{8}&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 4==<br /> Make use of the diagram in Solution 3. It can be deduced that &lt;math&gt;AF=BF&lt;/math&gt;. Let &lt;math&gt;DF=x&lt;/math&gt;. In &lt;math&gt;\triangle ADF&lt;/math&gt;, &lt;math&gt;AF^2=x^2+2.5^2 \Rightarrow AF=\sqrt{x^2+2.5^2}&lt;/math&gt;. Then &lt;math&gt;FC&lt;/math&gt; also would be &lt;math&gt;4-\sqrt{x^2+2.5^2}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle BCF&lt;/math&gt;, &lt;math&gt;BF^2=FC^2+BC^2 \Rightarrow (\sqrt{x^2+2.5^2})^2=(4-\sqrt{x^2+2.5^2})^2+3^2&lt;/math&gt;. After some quick math, we get &lt;math&gt;\sqrt{x^2+2.5^2}=\frac{25}{8}&lt;/math&gt;.<br /> Solving for &lt;math&gt;x&lt;/math&gt; will give &lt;math&gt;x=\frac{15}{8}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\therefore&lt;/math&gt; &lt;math&gt;DF=x=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 5==<br /> <br /> Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matter the positioning of the triangle, reorient the triangle so it has C at the origin and B and A at &lt;math&gt;(0,3)&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt; respectively.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,3)--(4,0)--(0,0));<br /> label(&quot;$C$&quot;, (0,0), SW); label(&quot;$B$&quot;, (0,3), NW); label(&quot;$A$&quot;, (4,0), SE);<br /> draw ((2,1.5)--(0.875,0),blue+linewidth(1));<br /> label(&quot;$(\frac{7}{8},0)$&quot;, (0.875,0), S); label(&quot;$(2,\frac{3}{2})$&quot;, (2,1.5), NE);<br /> draw(rightanglemark((4,0),(2,1.5),(0.875,0)));<br /> &lt;/asy&gt;<br /> <br /> We know that each point in the crease is equidistant from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, so the crease must pass through the midpoint of &lt;math&gt;AB&lt;/math&gt;, which is &lt;math&gt;({2,\frac{3}{2}})&lt;/math&gt; , and be perpendicular to hypotenuse &lt;math&gt;AB&lt;/math&gt;. The crease therefore has a slope of &lt;math&gt;\frac{4}{3}&lt;/math&gt;.<br /> <br /> Plugging into point-slope we find that the equation of the crease is, &lt;math&gt;y-y_1=m(x-x_1)\implies y-\frac{3}{2}=\frac{4}{3}(x-2)\implies y=\frac{4}{3}x-\frac{7}{6}&lt;/math&gt;<br /> <br /> Using 0 for y, we see that the crease intersects &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;(\frac{7}{8},0)&lt;/math&gt;. <br /> <br /> By the Distance Formula,<br /> <br /> &lt;math&gt;(2-\frac{7}{8})^2+({\frac{3}{2}})^2=d^2 \implies d^2=\frac{225}{64} \implies d=\boxed{\frac{15}{8}}&lt;/math&gt;<br /> <br /> <br /> ==Solution 6==<br /> Let &lt;math&gt;D&lt;/math&gt; be a point on AC such that &lt;math&gt;AD=DB&lt;/math&gt; and &lt;math&gt;\triangle{ADB}&lt;/math&gt; is isosceles. The altitude of this triangle is the crease of the paper triangle.<br /> <br /> Setting &lt;math&gt;DC=x&lt;/math&gt;, we have a right triangle &lt;math&gt;\triangle{BCD}&lt;/math&gt; with legs of x and 3, and, because &lt;math&gt;AD=DB,&lt;/math&gt; hypotenuse of &lt;math&gt;(4-x)&lt;/math&gt;.<br /> <br /> &lt;math&gt;x^2+3^2=(4-x)^2 \implies x^2+9=16-8x+x^2 \implies 8x=7 \implies x=\frac{7}{8}&lt;/math&gt;<br /> <br /> The area of &lt;math&gt;\triangle{BCD} \text{ is thus} = \frac{7}{8}*3*\frac{1}{2} \implies \frac{21}{16}.&lt;/math&gt; Because &lt;math&gt;[ADB]=[ABC]-[BCD],&lt;/math&gt; we can plug in and see that &lt;math&gt;[ADB]=6-\frac{21}{16}=\frac{75}{16}.&lt;/math&gt;<br /> <br /> Since &lt;math&gt;A=\frac{bh}{2}&lt;/math&gt;, we get &lt;math&gt;\frac{75}{16}=\frac{5h}{2} \implies 5h=\frac{75}{8} \implies h=\boxed{\frac{15}{8}}&lt;/math&gt;<br /> <br /> Our answer is &lt;math&gt;D&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_6&diff=117868 1993 AIME Problems/Problem 6 2020-02-17T17:43:15Z <p>Alexwin0806: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> What is the smallest [[positive]] [[integer]] that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Denote the first of each of the series of consecutive integers as &lt;math&gt;a,\ b,\ c&lt;/math&gt;. Therefore, &lt;math&gt;n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;. Simplifying, &lt;math&gt;9a = 10b + 9 = 11c + 19&lt;/math&gt;. The relationship between &lt;math&gt;a,\ b&lt;/math&gt; suggests that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Also, &lt;math&gt;10b -10 = 10(b-1) = 11c&lt;/math&gt;, so &lt;math&gt;b-1&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;. We find that the least possible value of &lt;math&gt;b = 45&lt;/math&gt;, so the answer is &lt;math&gt;10(45) + 45 = &lt;/math&gt;495.<br /> <br /> === Solution 2 ===<br /> <br /> Let the desired integer be &lt;math&gt;n&lt;/math&gt;. From the information given, it can be determined that, for positive integers &lt;math&gt;a, \ b, \ c&lt;/math&gt;:<br /> <br /> &lt;math&gt;n = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;<br /> <br /> This can be rewritten as the following congruences:<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{9}&lt;/math&gt; <br /> <br /> &lt;math&gt;n \equiv 5 \pmod{10}&lt;/math&gt;<br /> <br /> &lt;math&gt;n \equiv 0 \pmod{11}&lt;/math&gt;<br /> <br /> Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is &lt;math&gt;\boxed{495}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the desired integer. From the given information, we have<br /> &lt;cmath&gt; \begin{align*}9x &amp;= a \\ 11y &amp;= a \\ 10z + 5 &amp;= a, \end{align*}&lt;/cmath&gt; here, &lt;math&gt;x,&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have &lt;math&gt;z&lt;/math&gt; as the 4th term of the sequence. Since, &lt;math&gt;a&lt;/math&gt; is a multiple of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11,&lt;/math&gt; it is also a multiple of &lt;math&gt;\text{lcm}[9,11]=99.&lt;/math&gt; Hence, &lt;math&gt;a=99m,&lt;/math&gt; for some &lt;math&gt;m.&lt;/math&gt; So, we have &lt;math&gt;10z + 5 = 99m.&lt;/math&gt; It follows that &lt;math&gt;99(5) = \boxed{495}&lt;/math&gt; is the smallest integer that can be represented in such a way.<br /> <br /> === Solution 4 ===<br /> By the method in Solution 1, we find that the number &lt;math&gt;n&lt;/math&gt; can be written as &lt;math&gt;9a+36=10b+45=11c+55&lt;/math&gt; for some integers &lt;math&gt;a,b,c&lt;/math&gt;. From this, we can see that &lt;math&gt;n&lt;/math&gt; must be divisible by 9, 5, and 11. This means &lt;math&gt;n&lt;/math&gt; must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that &lt;math&gt;n&lt;/math&gt; cannot be divisible by 10, so &lt;math&gt;n&lt;/math&gt; must equal &lt;math&gt;\boxed{495}&lt;/math&gt;. Solution by Zeroman.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Alexwin0806 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_6&diff=117867 1993 AIME Problems/Problem 6 2020-02-17T17:42:28Z <p>Alexwin0806: /* Solution */</p> <hr /> <div>== Problem ==<br /> What is the smallest [[positive]] [[integer]] that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br /> <br /> ==Solution==<br /> === Solution 1 ===<br /> Denote the first of each of the series of consecutive integers as &lt;math&gt;a,\ b,\ c&lt;/math&gt;. Therefore, &lt;math&gt;n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55&lt;/math&gt;. Simplifying, &lt;math&gt;9a = 10b + 9 = 11c + 19&lt;/math&gt;. The relationship between &lt;math&gt;a,\ b&lt;/math&gt; suggests that &lt;math&gt;b&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Also, &lt;math&gt;10b -10 = 10(b-1) = 11c&lt;/math&gt;, so &lt;math&gt;b-1&lt;/math&gt; is divisible by &lt;math&gt;11&lt;/math&gt;. We find that the least possible value of &lt;math&gt;b = 45&lt;/math&gt;, so the answer is &lt;math&gt;10(45) + 45 = &lt;/math&gt;\boxed{495}&lt;math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let the desired integer be &lt;/math&gt;n&lt;math&gt;. From the information given, it can be determined that, for positive integers &lt;/math&gt;a, \ b, \ c&lt;math&gt;:<br /> <br /> &lt;/math&gt;n = 9a + 36 = 10b + 45 = 11c + 55&lt;math&gt;<br /> <br /> This can be rewritten as the following congruences:<br /> <br /> &lt;/math&gt;n \equiv 0 \pmod{9}&lt;math&gt; <br /> <br /> &lt;/math&gt;n \equiv 5 \pmod{10}&lt;math&gt;<br /> <br /> &lt;/math&gt;n \equiv 0 \pmod{11}&lt;math&gt;<br /> <br /> Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is &lt;/math&gt;\boxed{495}&lt;math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;/math&gt;n&lt;math&gt; be the desired integer. From the given information, we have<br /> &lt;cmath&gt; \begin{align*}9x &amp;= a \\ 11y &amp;= a \\ 10z + 5 &amp;= a, \end{align*}&lt;/cmath&gt; here, &lt;/math&gt;x,&lt;math&gt; and &lt;/math&gt;y&lt;math&gt; are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have &lt;/math&gt;z&lt;math&gt; as the 4th term of the sequence. Since, &lt;/math&gt;a&lt;math&gt; is a multiple of &lt;/math&gt;9&lt;math&gt; and &lt;/math&gt;11,&lt;math&gt; it is also a multiple of &lt;/math&gt;\text{lcm}[9,11]=99.&lt;math&gt; Hence, &lt;/math&gt;a=99m,&lt;math&gt; for some &lt;/math&gt;m.&lt;math&gt; So, we have &lt;/math&gt;10z + 5 = 99m.&lt;math&gt; It follows that &lt;/math&gt;99(5) = \boxed{495}&lt;math&gt; is the smallest integer that can be represented in such a way.<br /> <br /> === Solution 4 ===<br /> By the method in Solution 1, we find that the number &lt;/math&gt;n&lt;math&gt; can be written as &lt;/math&gt;9a+36=10b+45=11c+55&lt;math&gt; for some integers &lt;/math&gt;a,b,c&lt;math&gt;. From this, we can see that &lt;/math&gt;n&lt;math&gt; must be divisible by 9, 5, and 11. This means &lt;/math&gt;n&lt;math&gt; must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that &lt;/math&gt;n&lt;math&gt; cannot be divisible by 10, so &lt;/math&gt;n&lt;math&gt; must equal &lt;/math&gt;\boxed{495}\$. Solution by Zeroman.<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Alexwin0806