https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Justin6688&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-23T09:08:20Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=164555 2016 AMC 10B Problems/Problem 17 2021-11-04T00:41:17Z <p>Justin6688: /* Solution 3 (Cheap Solution) */</p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad<br /> \textbf{(B)}\ 343 \qquad<br /> \textbf{(C)}\ 625 \qquad<br /> \textbf{(D)}\ 729 \qquad<br /> \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let us call the six sides of our cube &lt;math&gt;a,b,c,d,e,&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt; (where &lt;math&gt;a&lt;/math&gt; is opposite &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; is opposite &lt;math&gt;e&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; is opposite &lt;math&gt;f&lt;/math&gt;.<br /> Thus, for the eight vertices, we have the following products: &lt;math&gt;abc,abe,bcd,bde,acf,cdf,aef,&lt;/math&gt; and &lt;math&gt;def&lt;/math&gt;.<br /> Let us find the sum of these products:<br /> &lt;cmath&gt;abc+abe+bcd+bde+acf+cdf+aef+def&lt;/cmath&gt;<br /> We notice &lt;math&gt;b&lt;/math&gt; is a factor of the first four terms, and &lt;math&gt;f&lt;/math&gt; is a factor of the last four terms.<br /> &lt;cmath&gt;b(ac+ae+cd+de)+f(ac+ae+cd+de)&lt;/cmath&gt;<br /> Now, we can factor even more:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (b+f)(ac+ae+cd+de)<br /> \\<br /> = &amp;(b+f)(a(c+e)+d(c+e))<br /> \\<br /> = &amp;(b+f)(a+d)(c+e)<br /> \end{align*}&lt;/cmath&gt;<br /> We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is &lt;math&gt;(7+2)&lt;/math&gt;,&lt;math&gt;(6+3)&lt;/math&gt;, and &lt;math&gt;(5+4)&lt;/math&gt; all factors.<br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (7+2)(6+3)(5+4)<br /> \\<br /> = &amp; 9 \cdot 9 \cdot 9<br /> \\<br /> = &amp; 729.<br /> \end{align*}&lt;/cmath&gt;<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> ==Solution 2(cheap parity)==<br /> We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We first find the factorization &lt;math&gt;(b+f)(a+d)(c+e)&lt;/math&gt; using the method in Solution 1. By using AM-GM, we get, &lt;math&gt;(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3&lt;/math&gt;. To maximize the factorization, we get the answer is &lt;math&gt;\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}&lt;/math&gt;<br /> <br /> ==Solution 4 (Cheap Solution)==<br /> <br /> <br /> ===Solution===<br /> Create a pairing that seems to intuitively be the optimal value, or, in other words, put a number and it's complement (the number that's the difference of 9 and this number) on opposite sides. &lt;math&gt;1680&lt;/math&gt; is way too high using reasonability after you do this so you put &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> https://youtu.be/mgEZOXgIZXs?t=117<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=164554 2016 AMC 10B Problems/Problem 17 2021-11-04T00:41:07Z <p>Justin6688: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad<br /> \textbf{(B)}\ 343 \qquad<br /> \textbf{(C)}\ 625 \qquad<br /> \textbf{(D)}\ 729 \qquad<br /> \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let us call the six sides of our cube &lt;math&gt;a,b,c,d,e,&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt; (where &lt;math&gt;a&lt;/math&gt; is opposite &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; is opposite &lt;math&gt;e&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; is opposite &lt;math&gt;f&lt;/math&gt;.<br /> Thus, for the eight vertices, we have the following products: &lt;math&gt;abc,abe,bcd,bde,acf,cdf,aef,&lt;/math&gt; and &lt;math&gt;def&lt;/math&gt;.<br /> Let us find the sum of these products:<br /> &lt;cmath&gt;abc+abe+bcd+bde+acf+cdf+aef+def&lt;/cmath&gt;<br /> We notice &lt;math&gt;b&lt;/math&gt; is a factor of the first four terms, and &lt;math&gt;f&lt;/math&gt; is a factor of the last four terms.<br /> &lt;cmath&gt;b(ac+ae+cd+de)+f(ac+ae+cd+de)&lt;/cmath&gt;<br /> Now, we can factor even more:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (b+f)(ac+ae+cd+de)<br /> \\<br /> = &amp;(b+f)(a(c+e)+d(c+e))<br /> \\<br /> = &amp;(b+f)(a+d)(c+e)<br /> \end{align*}&lt;/cmath&gt;<br /> We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is &lt;math&gt;(7+2)&lt;/math&gt;,&lt;math&gt;(6+3)&lt;/math&gt;, and &lt;math&gt;(5+4)&lt;/math&gt; all factors.<br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (7+2)(6+3)(5+4)<br /> \\<br /> = &amp; 9 \cdot 9 \cdot 9<br /> \\<br /> = &amp; 729.<br /> \end{align*}&lt;/cmath&gt;<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> ==Solution 2(cheap parity)==<br /> We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We first find the factorization &lt;math&gt;(b+f)(a+d)(c+e)&lt;/math&gt; using the method in Solution 1. By using AM-GM, we get, &lt;math&gt;(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3&lt;/math&gt;. To maximize the factorization, we get the answer is &lt;math&gt;\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}&lt;/math&gt;<br /> <br /> ==Solution 3 (Cheap Solution)==<br /> <br /> <br /> ===Solution===<br /> Create a pairing that seems to intuitively be the optimal value, or, in other words, put a number and it's complement (the number that's the difference of 9 and this number) on opposite sides. &lt;math&gt;1680&lt;/math&gt; is way too high using reasonability after you do this so you put &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> https://youtu.be/mgEZOXgIZXs?t=117<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=164553 2016 AMC 10B Problems/Problem 17 2021-11-04T00:40:57Z <p>Justin6688: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad<br /> \textbf{(B)}\ 343 \qquad<br /> \textbf{(C)}\ 625 \qquad<br /> \textbf{(D)}\ 729 \qquad<br /> \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let us call the six sides of our cube &lt;math&gt;a,b,c,d,e,&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt; (where &lt;math&gt;a&lt;/math&gt; is opposite &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; is opposite &lt;math&gt;e&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; is opposite &lt;math&gt;f&lt;/math&gt;.<br /> Thus, for the eight vertices, we have the following products: &lt;math&gt;abc,abe,bcd,bde,acf,cdf,aef,&lt;/math&gt; and &lt;math&gt;def&lt;/math&gt;.<br /> Let us find the sum of these products:<br /> &lt;cmath&gt;abc+abe+bcd+bde+acf+cdf+aef+def&lt;/cmath&gt;<br /> We notice &lt;math&gt;b&lt;/math&gt; is a factor of the first four terms, and &lt;math&gt;f&lt;/math&gt; is a factor of the last four terms.<br /> &lt;cmath&gt;b(ac+ae+cd+de)+f(ac+ae+cd+de)&lt;/cmath&gt;<br /> Now, we can factor even more:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (b+f)(ac+ae+cd+de)<br /> \\<br /> = &amp;(b+f)(a(c+e)+d(c+e))<br /> \\<br /> = &amp;(b+f)(a+d)(c+e)<br /> \end{align*}&lt;/cmath&gt;<br /> We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is &lt;math&gt;(7+2)&lt;/math&gt;,&lt;math&gt;(6+3)&lt;/math&gt;, and &lt;math&gt;(5+4)&lt;/math&gt; all factors.<br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (7+2)(6+3)(5+4)<br /> \\<br /> = &amp; 9 \cdot 9 \cdot 9<br /> \\<br /> = &amp; 729.<br /> \end{align*}&lt;/cmath&gt;<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> ==Solution 2(cheap parity)==<br /> We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We first find the factorization &lt;math&gt;(b+f)(a+d)(c+e)&lt;/math&gt; using the method in Solution 1. By using AM-GM, we get, &lt;math&gt;(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3&lt;/math&gt;. To maximize the factorization, we get the answer is &lt;math&gt;\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}&lt;/math&gt;<br /> <br /> ==Solution 3 (Cheap Solution)==<br /> <br /> <br /> ===Solution===<br /> Create a pairing that seems to intuitively be the optimal value, or, in other words, put a number and it's complement (the number that's the difference of 9 and this number) on opposite sides. &lt;math&gt;1680&lt;/math&gt; is way too high using reasonability after you do this so you put &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> https://youtu.be/mgEZOXgIZXs?t=117<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=164552 2016 AMC 10B Problems/Problem 17 2021-11-04T00:40:38Z <p>Justin6688: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad<br /> \textbf{(B)}\ 343 \qquad<br /> \textbf{(C)}\ 625 \qquad<br /> \textbf{(D)}\ 729 \qquad<br /> \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let us call the six sides of our cube &lt;math&gt;a,b,c,d,e,&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt; (where &lt;math&gt;a&lt;/math&gt; is opposite &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; is opposite &lt;math&gt;e&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; is opposite &lt;math&gt;f&lt;/math&gt;.<br /> Thus, for the eight vertices, we have the following products: &lt;math&gt;abc,abe,bcd,bde,acf,cdf,aef,&lt;/math&gt; and &lt;math&gt;def&lt;/math&gt;.<br /> Let us find the sum of these products:<br /> &lt;cmath&gt;abc+abe+bcd+bde+acf+cdf+aef+def&lt;/cmath&gt;<br /> We notice &lt;math&gt;b&lt;/math&gt; is a factor of the first four terms, and &lt;math&gt;f&lt;/math&gt; is a factor of the last four terms.<br /> &lt;cmath&gt;b(ac+ae+cd+de)+f(ac+ae+cd+de)&lt;/cmath&gt;<br /> Now, we can factor even more:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (b+f)(ac+ae+cd+de)<br /> \\<br /> = &amp;(b+f)(a(c+e)+d(c+e))<br /> \\<br /> = &amp;(b+f)(a+d)(c+e)<br /> \end{align*}&lt;/cmath&gt;<br /> We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is &lt;math&gt;(7+2)&lt;/math&gt;,&lt;math&gt;(6+3)&lt;/math&gt;, and &lt;math&gt;(5+4)&lt;/math&gt; all factors.<br /> &lt;cmath&gt;\begin{align*}<br /> &amp; (7+2)(6+3)(5+4)<br /> \\<br /> = &amp; 9 \cdot 9 \cdot 9<br /> \\<br /> = &amp; 729.<br /> \end{align*}&lt;/cmath&gt;<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> ==Solution 2==<br /> We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.<br /> Thus our answer is &lt;math&gt;\textbf{\boxed{(D)729}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We first find the factorization &lt;math&gt;(b+f)(a+d)(c+e)&lt;/math&gt; using the method in Solution 1. By using AM-GM, we get, &lt;math&gt;(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3&lt;/math&gt;. To maximize the factorization, we get the answer is &lt;math&gt;\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}&lt;/math&gt;<br /> <br /> ==Solution 3 (Cheap Solution)==<br /> <br /> <br /> ===Solution===<br /> Create a pairing that seems to intuitively be the optimal value, or, in other words, put a number and it's complement (the number that's the difference of 9 and this number) on opposite sides. &lt;math&gt;1680&lt;/math&gt; is way too high using reasonability after you do this so you put &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> https://youtu.be/mgEZOXgIZXs?t=117<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2020_USOMO_Problems/Problem_2&diff=148725 2020 USOMO Problems/Problem 2 2021-03-06T21:39:47Z <p>Justin6688: Created blank page</p> <hr /> <div></div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=148210 AMC historical results 2021-03-02T17:55:39Z <p>Justin6688: /* 2021 */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> ==2021==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: 103.5(official)<br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AMC 12A===<br /> *Average score:<br /> *AIME floor: 93(official)<br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AMC 12B===<br /> <br /> *Average score: <br /> *AIME floor: <br /> *Distinction: <br /> *Distinguished Honor Roll:<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> ===AMC 8===<br /> *Average score:<br /> *Honor Roll: <br /> *DHR:<br /> <br /> ==2020==<br /> ===AMC 10A===<br /> *Average score: 64.29<br /> *AIME floor: 103.5<br /> *Distinction: 105<br /> *Distinguished Honor Roll: 124.5<br /> <br /> ===AMC 10B===<br /> *Average score: 61.22<br /> *AIME floor: 102<br /> *Distinction: 103.5 <br /> *Distinguished Honor Roll: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 61.42<br /> *AIME floor: 87<br /> *Distinction: 100.5<br /> *Distinguished Honor Roll: 123<br /> <br /> ===AMC 12B===<br /> *Average score: 60.47<br /> *AIME floor: 87<br /> *Distinction: 97.5<br /> *Distinguished Honor Roll: 120<br /> <br /> ===AIME I===<br /> *Average score: 5.69<br /> *Median score: 6<br /> *USAMO cutoff: 233.5 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 229.5 (AMC 10A), 230 (AMC 10B)<br /> <br /> ===AIME II===<br /> Due to COVID-19, the 2020 AIME II was administered online and referred to as the AOIME.<br /> *Average score: 6.13<br /> *Median score: 6<br /> *USAMO cutoff: 234 (AMC 12A), 234.5 (AMC 12B)<br /> *USAJMO cutoff: 233.5 (AMC 10A), 229.5 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 10.00<br /> *Honor Roll: 18<br /> *DHR: 21<br /> <br /> ==2019==<br /> ===AMC 10A===<br /> *Average score: 51.66<br /> *Honor roll: 96<br /> *AIME floor: 103.5<br /> *DHR: 123<br /> <br /> ===AMC 10B===<br /> *Average score: 58.42<br /> *Honor roll: 102<br /> *AIME floor: 108<br /> *Distinguished Honor Roll: 121.5<br /> <br /> ===AMC 12A===<br /> *Average score: 49.22<br /> *AIME floor: 84<br /> *DHR: 121.5<br /> <br /> ===AMC 12B===<br /> *Average score: 56.73<br /> *AIME floor: 94.5 <br /> *DHR: 123<br /> <br /> ===AIME I===<br /> *Average score: 5.88<br /> *Median score: 6<br /> *USAMO cutoff: 220 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 209.5 (AMC 10A), 216 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 6.47<br /> *Median score: 6<br /> *USAMO cutoff: 230.5 (AMC 12A), 236 (AMC 12B)<br /> *USAJMO cutoff: 216.5 (AMC 10A), 220.5 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 9.43<br /> *Honor roll: 19<br /> *DHR: 23<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score: 53.84<br /> *Honor roll: 100.5<br /> *AIME floor: 111<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 57.81<br /> *Honor roll: 97.5<br /> *AIME floor: 108<br /> *DHR: 123<br /> <br /> ===AMC 12A===<br /> *Average score: 56.36<br /> *AIME floor: 93<br /> *DHR: 120<br /> <br /> ===AMC 12B===<br /> *Average score: 57.85<br /> *AIME floor: 99<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.09<br /> *Median score: 5<br /> *USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.48<br /> *Median score: 5<br /> *USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 8.51<br /> *Honor roll: 15<br /> *DHR: 19<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.66<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score: 5.69<br /> *Median score: 5<br /> *USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 5<br /> *USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 8.96<br /> *Honor roll: 17<br /> *DHR: 20<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.31<br /> *AIME floor: 110<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.40<br /> *AIME floor: 110<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 93<br /> *DHR: 111<br /> <br /> ===AMC 12B===<br /> *Average score: 68.65<br /> *AIME floor: 100.5<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.43<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ===AMC 8===<br /> *Average score: 9.36<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.09<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.88<br /> *AIME floor: 100.5<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ===AMC 8===<br /> *Average score: 8.55<br /> *Honor roll: 16<br /> *DHR: 21<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.34<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.42<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 63.60<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.12<br /> *AIME floor: 100.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AMC 8===<br /> *Average score: 11.43<br /> *Honor roll: 19<br /> *DHR: 23<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AMC 8===<br /> *Average score: 10.69<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> *DHR: 121.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> *DHR: 114<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: 5<br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: 5<br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AMC 8===<br /> *Average score: 10.67<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 64.24<br /> *AIME floor: 117<br /> *DHR: 129<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 65.38<br /> *AIME floor: 93<br /> *DHR: 112.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 2.23<br /> *Median score: 2<br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 5.47<br /> *Median score: 5<br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AMC 8===<br /> *Average score: 10.75<br /> *Honor roll: 17<br /> *DHR: 22<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> *DHR cutoff: 108<br /> <br /> ===AMC 12B===<br /> *Average score: 59.98<br /> *AIME floor: 88.5<br /> *DHR cutoff: 109.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: 6<br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: 3<br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AMC 8===<br /> *Average score: 9.59<br /> *Honor roll: 17<br /> *DHR: 22<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Average score: 10.28<br /> *Honor roll: 17<br /> *DHR: 20<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: 60.25<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 4.77<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 5.27<br /> *Median score: 5<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Average score: 11.45<br /> *Honor roll: 19<br /> *DHR: 22<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B=== <br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Average score: 9.87<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 16<br /> *DHR: 20<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 115<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: &lt;math&gt;64.2&lt;/math&gt;<br /> *AIME floor: &lt;math&gt;110&lt;/math&gt;<br /> <br /> ===AMC 12===<br /> *Average score: &lt;math&gt;64.9&lt;/math&gt;<br /> *AIME floor: &lt;math&gt;92&lt;/math&gt;<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: &lt;math&gt;68.8&lt;/math&gt;<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=User:Justin6688&diff=148063 User:Justin6688 2021-03-01T01:02:18Z <p>Justin6688: Created page with &quot;hi friend me&quot;</p> <hr /> <div>hi<br /> friend me</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A&diff=148062 2022 AMC 10A 2021-03-01T01:01:33Z <p>Justin6688: </p> <hr /> <div>Hasn't happened yet. Please don't troll.</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_24&diff=144730 2018 AMC 10A Problems/Problem 24 2021-02-03T01:07:38Z <p>Justin6688: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #18]] and [[2018 AMC 10A Problems|2018 AMC 10A #24]]}}<br /> <br /> == Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB=50&lt;/math&gt; and &lt;math&gt;AC=10&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;, and let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt; intersects &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;, respectively. What is the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }60 \qquad<br /> \textbf{(B) }65 \qquad<br /> \textbf{(C) }70 \qquad<br /> \textbf{(D) }75 \qquad<br /> \textbf{(E) }80 \qquad<br /> &lt;/math&gt;<br /> <br /> ==hardness of problem==<br /> <br /> The hardness of this problem is average medium or on a scale of 10, a 5. The problem needs no extension or addition of lines, as all you need to know is the angle bisector theorem and how to calculate the area of divided portions.<br /> <br /> ~justin6688<br /> <br /> == Solution 1 ==<br /> <br /> Let &lt;math&gt;BC = a&lt;/math&gt;, &lt;math&gt;BG = x&lt;/math&gt;, &lt;math&gt;GC = y&lt;/math&gt;, and the length of the perpendicular to &lt;math&gt;BC&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;h&lt;/math&gt;. By angle bisector theorem, we have that &lt;cmath&gt;\frac{50}{x} = \frac{10}{y},&lt;/cmath&gt; where &lt;math&gt;y = -x+a&lt;/math&gt;. Therefore substituting we have that &lt;math&gt;BG=\frac{5a}{6}&lt;/math&gt;. By similar triangles, we have that &lt;math&gt;DF=\frac{5a}{12}&lt;/math&gt;, and the height of this trapezoid is &lt;math&gt;\frac{h}{2}&lt;/math&gt;. Then, we have that &lt;math&gt;\frac{ah}{2}=120&lt;/math&gt;. We wish to compute &lt;math&gt;\frac{5a}{8}\cdot\frac{h}{2}&lt;/math&gt;, and we have that it is &lt;math&gt;\boxed{75}&lt;/math&gt; by substituting.<br /> <br /> == Solution 2 ==<br /> <br /> For this problem, we have &lt;math&gt;\triangle{ADE}\sim\triangle{ABC}&lt;/math&gt; because of SAS and &lt;math&gt;DE = \frac{BC}{2}&lt;/math&gt;. Therefore, &lt;math&gt;\bigtriangleup ADE&lt;/math&gt; is a quarter of the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, which is &lt;math&gt;30&lt;/math&gt;. Subsequently, we can compute the area of quadrilateral &lt;math&gt;BDEC&lt;/math&gt; to be &lt;math&gt;120 - 30 = 90&lt;/math&gt;. Using the angle bisector theorem in the same fashion as the previous problem, we get that &lt;math&gt;\overline{BG}&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt; times the length of &lt;math&gt;\overline{GC}&lt;/math&gt;. We want the larger piece, as described by the problem. Because the heights are identical, one area is &lt;math&gt;5&lt;/math&gt; times the other, and &lt;math&gt;\frac{5}{6} \cdot 90 = \boxed{75}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;5:1&lt;/math&gt; by Law of Sines. So the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt;. Since &lt;math&gt;\overline{DE}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, so &lt;math&gt;\overline{DF}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; . So the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;1:4&lt;/math&gt; , so the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, by similar triangles. Therefore the area of quad &lt;math&gt;FDBG&lt;/math&gt; is &lt;math&gt;100-25=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 4 ==<br /> The area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; is the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; minus the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt;. Notice, &lt;math&gt;\overline{DE} || \overline{BC}&lt;/math&gt;, so &lt;math&gt;\bigtriangleup ABG \sim \bigtriangleup ADF&lt;/math&gt;, and since &lt;math&gt;\overline{AD}:\overline{AB}=1:2&lt;/math&gt;, the area of &lt;math&gt;\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4&lt;/math&gt;. Given that the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt;, using &lt;math&gt;\frac{bh}{2}&lt;/math&gt; on side &lt;math&gt;AB&lt;/math&gt; yields &lt;math&gt;\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}&lt;/math&gt;. Using the Angle Bisector Theorem, &lt;math&gt;\overline{BG}:\overline{BC}=50:(10+50)=5:6&lt;/math&gt;, so the height of &lt;math&gt;\bigtriangleup ABG: \bigtriangleup ACB=5:6&lt;/math&gt;. Therefore our answer is &lt;math&gt;\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 5: Trig ==<br /> We try to find the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; by subtracting the area outside the quadrilateral but inside triangle &lt;math&gt;ABC&lt;/math&gt;. Note that the area of &lt;math&gt;\triangle ADE&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A&lt;/math&gt;. The ratio &lt;math&gt;\frac{[ADE]}{[ABC]}&lt;/math&gt; is thus equal to &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the area of triangle &lt;math&gt;ADE&lt;/math&gt; is &lt;math&gt;\frac{1}{4} \cdot 120 = 30&lt;/math&gt;. Let side &lt;math&gt;BC&lt;/math&gt; be equal to &lt;math&gt;6x&lt;/math&gt;, then &lt;math&gt;BG = 5x, GC = x&lt;/math&gt; by the angle bisector theorem. Similarly, we find the area of triangle &lt;math&gt;AGC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 10 \cdot x \cdot \sin C&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C&lt;/math&gt;. A ratio between these two triangles yields &lt;math&gt;\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}&lt;/math&gt;, so &lt;math&gt;[AGC] = 20&lt;/math&gt;. Now we just need to find the area of triangle &lt;math&gt;AFE&lt;/math&gt; and subtract it from the combined areas of &lt;math&gt;[ADE]&lt;/math&gt; and &lt;math&gt;[ACG]&lt;/math&gt;, since we count it twice. Note that the angle bisector theorem also applies for &lt;math&gt;\triangle ADE&lt;/math&gt; and &lt;math&gt;\frac{AE}{AD} = \frac{1}{5}&lt;/math&gt;, so thus &lt;math&gt;\frac{EF}{ED} = \frac{1}{6}&lt;/math&gt; and we find &lt;math&gt;[AFE] = \frac{1}{6} \cdot 30 = 5&lt;/math&gt;, and the area outside &lt;math&gt;FDBG&lt;/math&gt; must be &lt;math&gt; [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45&lt;/math&gt;, and we finally find &lt;math&gt;[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}&lt;/math&gt;, and we are done. <br /> <br /> ==Solution 6: Areas ==<br /> &lt;asy&gt;<br /> draw((0,0)--(1,3)--(5,0)--cycle);<br /> draw((0,0)--(2,2.25));<br /> draw((0.5,1.5)--(2.5,0));<br /> label(&quot;A&quot;,(0,0),SW);<br /> label(&quot;B&quot;,(5,0),SE);<br /> label(&quot;C&quot;,(1,3),N);<br /> label(&quot;G&quot;,(2,2.25),NE);<br /> label(&quot;D&quot;,(2.5,0),S);<br /> label(&quot;E&quot;,(0.5,1.5),NW);<br /> label(&quot;3Y&quot;,(2.5,0.75),N);<br /> label(&quot;Y&quot;,(1,0.2),N);<br /> label(&quot;X&quot;,(0.5,0.5),N);<br /> label(&quot;3X&quot;,(1.25,1.75),N);<br /> &lt;/asy&gt;<br /> Give triangle &lt;math&gt;AEF&lt;/math&gt; area X. Then, by similarity, since &lt;math&gt;\frac{AC}{AE} = \frac{2}{1}&lt;/math&gt;, &lt;math&gt;ACG&lt;/math&gt; has area 4X. Thus, &lt;math&gt;FGCE&lt;/math&gt; has area 3X.<br /> Doing the same for triangle &lt;math&gt;AGB&lt;/math&gt;, we get that triangle &lt;math&gt;AFD&lt;/math&gt; has area Y and quadrilateral &lt;math&gt;GFDB&lt;/math&gt; has area 3Y. Since &lt;math&gt;AEF&lt;/math&gt; has the same height as &lt;math&gt;AFD&lt;/math&gt;, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, &lt;math&gt;\frac{CG}{GB} = \frac{1}{5}&lt;/math&gt;. So, &lt;math&gt;\frac{[AEF]}{[AFD]} = \frac{1}{5}&lt;/math&gt;. Since &lt;math&gt;AEF&lt;/math&gt; has area X, we can write the equation 5X = Y and substitute 5X for Y.<br /> &lt;asy&gt;<br /> draw((0,0)--(1,3)--(5,0)--cycle);<br /> draw((0,0)--(2,2.25));<br /> draw((0.5,1.5)--(2.5,0));<br /> label(&quot;A&quot;,(0,0),SW);<br /> label(&quot;B&quot;,(5,0),SE);<br /> label(&quot;C&quot;,(1,3),N);<br /> label(&quot;G&quot;,(2,2.25),NE);<br /> label(&quot;D&quot;,(2.5,0),S);<br /> label(&quot;E&quot;,(0.5,1.5),NW);<br /> label(&quot;&quot;,(2.5,0.75),N);<br /> label(&quot;&quot;,(1,0.2),N);<br /> label(&quot;F&quot;, (1, 1.5), N);<br /> label(&quot;&quot;,(0.5,1.5),N);<br /> label(&quot;&quot;,(1.25,1.75),N);<br /> &lt;/asy&gt;<br /> Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find &lt;math&gt;GFDB&lt;/math&gt;, we substitute 5 for 15X to get &lt;math&gt;\boxed{75}&lt;/math&gt;.<br /> &lt;math&gt;\sim&lt;/math&gt;krishkhushi09<br /> <br /> == Video Solution by Richard Rusczyk ==<br /> <br /> https://artofproblemsolving.com/videos/amc/2018amc10a/469<br /> <br /> ~ dolphin7<br /> <br /> == Video Solution ==<br /> https://youtu.be/4_x1sgcQCp4?t=4898<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_23&diff=144729 2018 AMC 10A Problems/Problem 23 2021-02-03T01:02:03Z <p>Justin6688: /* Solution 8(Similar but cleaner than first solution): */</p> <hr /> <div>{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #17]] and [[2018 AMC 10A Problems|2018 AMC 10A #23]]}}<br /> <br /> == Problem ==<br /> <br /> Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square &lt;math&gt;S&lt;/math&gt; so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from &lt;math&gt;S&lt;/math&gt; to the hypotenuse is 2 units. What fraction of the field is planted?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), N);<br /> label(&quot;$3$&quot;, (0,1.5), E);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the square have side length &lt;math&gt;x&lt;/math&gt;. Connect the upper-right vertex of square &lt;math&gt;S&lt;/math&gt; with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is &lt;math&gt;6&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> draw((0.3,0.3)--(4,0), dashed);<br /> draw((0.3,0.3)--(0,3), dashed);<br /> label(&quot;$\small{x}$&quot;, (0.15,0.3), N);<br /> label(&quot;$\small{x}$&quot;, (0.3,0.15), E);<br /> &lt;/asy&gt;<br /> <br /> Square &lt;math&gt;S&lt;/math&gt; has area &lt;math&gt;x^2&lt;/math&gt;, and the two thin triangle regions have area &lt;math&gt;\dfrac{x(3-x)}{2}&lt;/math&gt; and &lt;math&gt;\dfrac{x(4-x)}{2}&lt;/math&gt;. The final triangular region with the hypotenuse as its base and height &lt;math&gt;2&lt;/math&gt; has area &lt;math&gt;5&lt;/math&gt;. Thus, we have &lt;cmath&gt;x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6&lt;/cmath&gt;<br /> <br /> Solving gives &lt;math&gt;x=\dfrac{2}{7}&lt;/math&gt;. The area of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\dfrac{4}{49}&lt;/math&gt; and the desired ratio is &lt;math&gt;\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}&lt;/math&gt;.<br /> <br /> Alternatively, once you get &lt;math&gt;x=\frac{2}{7}&lt;/math&gt;, you can avoid computation by noticing that there is a denominator of &lt;math&gt;7&lt;/math&gt;, so the answer must have a factor of &lt;math&gt;7&lt;/math&gt; in the denominator, which only &lt;math&gt;\boxed{\dfrac{145}{147}}&lt;/math&gt; does.<br /> <br /> ==Solution 2==<br /> Let the square have side length &lt;math&gt;s&lt;/math&gt;. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is &lt;math&gt;\frac{5}{3}(2)=\frac{10}{3}&lt;/math&gt;. Now, let's extend this larger similar right triangle to the left until it hits the side of length 3. Now, the length is &lt;math&gt;\frac{10}{3}+s&lt;/math&gt;, and using the ratios of the side lengths, the height is &lt;math&gt;\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}&lt;/math&gt;. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so &lt;cmath&gt;\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}&lt;/cmath&gt;<br /> <br /> Now comes the easy part: finding the ratio of the areas: &lt;math&gt;\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We use coordinate geometry. Let the right angle be at &lt;math&gt;(0,0)&lt;/math&gt; and the hypotenuse be the line &lt;math&gt;3x+4y = 12&lt;/math&gt; for &lt;math&gt;0\le x\le 3&lt;/math&gt;. Denote the position of &lt;math&gt;S&lt;/math&gt; as &lt;math&gt;(s,s)&lt;/math&gt;, and by the point to line distance formula, we know that &lt;cmath&gt;\frac{|3s+4s-12|}{5} = 2&lt;/cmath&gt; &lt;cmath&gt;\Rightarrow |7s-12| = 10&lt;/cmath&gt; Obviously &lt;math&gt;s&lt;\frac{22}{7}&lt;/math&gt;, so &lt;math&gt;s = \frac{2}{7}&lt;/math&gt;, and from here the rest of the solution follows to get &lt;math&gt;\boxed{\frac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let the side length of the square be &lt;math&gt;x&lt;/math&gt;. First off, let us make a similar triangle with the segment of length &lt;math&gt;2&lt;/math&gt; and the top-right corner of &lt;math&gt;S&lt;/math&gt;. Therefore, the longest side of the smaller triangle must be &lt;math&gt;2 \cdot \frac54 = \frac52&lt;/math&gt;. We then do operations with that side in terms of &lt;math&gt;x&lt;/math&gt;. We subtract &lt;math&gt;x&lt;/math&gt; from the bottom, and &lt;math&gt;\frac{3x}{4}&lt;/math&gt; from the top. That gives us the equation of &lt;math&gt;3-\frac{7x}{4} = \frac{5}{2}&lt;/math&gt;. Solving, &lt;cmath&gt;12-7x = 10 \implies x = \frac{2}{7}.&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;x^2 = \frac{4}{49}&lt;/math&gt;, so the fraction of the triangle (area &lt;math&gt;6&lt;/math&gt;) covered by the square is &lt;math&gt;\frac{2}{147}&lt;/math&gt;. The answer is then &lt;math&gt;\boxed{\dfrac{145}{147}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> draw((0.3,0.3)--(3.6,0.3), dashed);<br /> draw((0.3,2.7)--(0.3,0.3), dashed);<br /> label(&quot;$S$&quot;, (-0.05,-0.05), NE);<br /> draw((0.3,0.3)--(1.41,1.91));<br /> draw((1.63,1.78)--(1.48,1.56));<br /> draw((1.28,1.70)--(1.48,1.56));<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$\frac{10}{3}$&quot;, (2,0.3), N);<br /> label(&quot;$\frac{5}{2}$&quot;, (0.3,1.5), E);<br /> label(&quot;$2$&quot;, (1,1.2), E);<br /> draw((3.6,0)--(3.6,0.3), dashed);<br /> draw((0,2.7)--(0.3,2.7), dashed);<br /> label(&quot;$\small{l}$&quot;, (3.6,0.15), W);<br /> label(&quot;$\small{l}$&quot;, (0.15,2.7), S);<br /> label(&quot;$\small{l}$&quot;, (0.3,0.15), E);<br /> label(&quot;$\small{l}$&quot;, (0.15,0.3), N);<br /> &lt;/asy&gt;<br /> <br /> On the diagram above, find two smaller triangles similar to the large one with side lengths &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt;; consequently, the segments with length &lt;math&gt;\frac{5}{2}&lt;/math&gt; and &lt;math&gt;\frac{10}{3}&lt;/math&gt;.<br /> <br /> With &lt;math&gt;l&lt;/math&gt; being the side length of the square, we need to find an expression for &lt;math&gt;l&lt;/math&gt;. Using the hypotenuse, we can see that &lt;math&gt;\frac{3}{2}+\frac{8}{3}+\frac{5}{4}l+\frac{5}{3}l=5&lt;/math&gt;. Simplifying, &lt;math&gt;\frac{35}{12}l=\frac{5}{6}&lt;/math&gt;, or &lt;math&gt;l=2/7&lt;/math&gt;.<br /> <br /> A different calculation would yield &lt;math&gt;l+\frac{3}{4}l+\frac{5}{2}=3&lt;/math&gt;, so &lt;math&gt;\frac{7}{4}l=\frac{1}{2}&lt;/math&gt;. In other words, &lt;math&gt;l=\frac{2}{7}&lt;/math&gt;, while to check, &lt;math&gt;l+\frac{4}{3}l+\frac{10}{3}=4&lt;/math&gt;. As such, &lt;math&gt;\frac{7}{3}l=\frac{2}{3}&lt;/math&gt;, and &lt;math&gt;l=\frac{2}{7}&lt;/math&gt;.<br /> <br /> Finally, we get &lt;math&gt;A(\Square S)=l^2=\frac{4}{49}&lt;/math&gt;, to finish. As a proportion of the triangle with area &lt;math&gt;6&lt;/math&gt;, the answer would be &lt;math&gt;1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}&lt;/math&gt;, so &lt;math&gt;\boxed{\textit D}&lt;/math&gt; is correct.<br /> <br /> == Solution 6: Also Coordinate Geo ==<br /> Let the right angle be at &lt;math&gt;(0,0)&lt;/math&gt;, the point &lt;math&gt;(x,x)&lt;/math&gt; be the far edge of the unplanted square and the hypotenuse be the line &lt;math&gt;y=-\frac{3}{4}x+3&lt;/math&gt;. Since the line from &lt;math&gt;(x,x)&lt;/math&gt; to the hypotenuse is the shortest possible distance, we know this line, call it line &lt;math&gt;\l&lt;/math&gt;, is perpendicular to the hypotenuse and therefore has a slope of &lt;math&gt;\frac{4}{3}&lt;/math&gt;. <br /> <br /> Since we know &lt;math&gt;m=\frac{4}{3}&lt;/math&gt; , we can see that the line rises by &lt;math&gt;\frac{8}{5}&lt;/math&gt; and moves to the right by &lt;math&gt;\frac{6}{5}&lt;/math&gt; to meet the hypotenuse. (Let &lt;math&gt;2 = 5x&lt;/math&gt; and the rise be &lt;math&gt;4x&lt;/math&gt; and the run be &lt;math&gt;3x&lt;/math&gt; and then solve.) Therefore, line &lt;math&gt;\l&lt;/math&gt; intersects the hypotenuse at the point &lt;math&gt;(x+\frac{6}{5}, x+\frac{8}{5})&lt;/math&gt;. Plugging into the equation for the hypotenuse we have &lt;math&gt;x=\frac{2}{7}&lt;/math&gt; , and after a bit of computation we get &lt;math&gt;\boxed{\textbf{(D) } \frac{145}{147}}&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution 7(Slightly different from first solution):==<br /> Same drawing as before:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (0,1.5), W);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> draw((0.3,0.3)--(4,0), dashed);<br /> draw((0.3,0.3)--(0,3), dashed);<br /> label(&quot;$\small{a}$&quot;, (0.15,0.3), N);<br /> label(&quot;$\small{a}$&quot;, (0.3,0.15), E);<br /> &lt;/asy&gt;<br /> <br /> Let's assign &lt;math&gt;a&lt;/math&gt; as the side length of box S. We then get each of the smaller triangle areas.<br /> The sum of all the triangular areas(not including the box) is equal to<br /> &lt;math&gt;\frac{(3-a) \cdot a}{2} + \frac{(4-a) \cdot a}{2} + \frac{5 \cdot 2}{2} = \frac{3 \cdot 4}{2} - a^2&lt;/math&gt;<br /> <br /> You can solve for &lt;math&gt;a=\frac{2}{7}&lt;/math&gt;<br /> <br /> Then, the ratio would be &lt;math&gt;1-\dfrac{{\frac{2}{7}}^2}{6}&lt;/math&gt; which is equal to &lt;math&gt;\boxed{\textbf{(D) } \frac{145}{147}}&lt;/math&gt;<br /> <br /> ~Starshooter11<br /> <br /> ==Solution 8(Similar but cleaner than first solution):==<br /> <br /> Instead of dividing the large triangle into three right triangles plus a square, simply draw one diagonal of the square (can someone asymptote this for me I don't know how) divide the large triangle into three triangles with the sides &lt;math&gt;3,4,&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; as their bases. The total area is &lt;math&gt;6&lt;/math&gt; as found above, and the square's side length is &lt;math&gt;x&lt;/math&gt;. The area equation is:<br /> &lt;cmath&gt;\dfrac{3x}{2}+\dfrac{4x}{2}+\dfrac{5\cdot2}{2}=6,&lt;/cmath&gt; which solves to &lt;math&gt;x=\dfrac{2}{7},&lt;/math&gt; the same as the first solution (but easier to calculate). The final answer is &lt;math&gt;\boxed{\dfrac{145}{147}}&lt;/math&gt;.<br /> <br /> ==hardness of problem==<br /> <br /> This problem is medium hard or on a scale of 10, a 6.5. The problem itself requires the drawing of a few obvious lines and algebra, although the image deceives the solver.<br /> <br /> == Video Solution by Richard Rusczyk ==<br /> <br /> https://www.youtube.com/watch?v=p9npzq4FY_Y<br /> <br /> ~ dolphin7<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2018|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_22&diff=144728 2018 AMC 10A Problems/Problem 22 2021-02-03T00:55:27Z <p>Justin6688: /* Solution 4 (Fastest) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a, b, c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; be positive integers such that &lt;math&gt;\gcd(a, b)=24&lt;/math&gt;, &lt;math&gt;\gcd(b, c)=36&lt;/math&gt;, &lt;math&gt;\gcd(c, d)=54&lt;/math&gt;, and &lt;math&gt;70&lt;\gcd(d, a)&lt;100&lt;/math&gt;. Which of the following must be a divisor of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> The GCD information tells us that &lt;math&gt;24&lt;/math&gt; divides &lt;math&gt;a&lt;/math&gt;, both &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;36&lt;/math&gt; divide &lt;math&gt;b&lt;/math&gt;, both &lt;math&gt;36&lt;/math&gt; and &lt;math&gt;54&lt;/math&gt; divide &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;54&lt;/math&gt; divides &lt;math&gt;d&lt;/math&gt;. Note that we have the prime factorizations:<br /> &lt;cmath&gt;\begin{align*}<br /> 24 &amp;= 2^3\cdot 3,\\<br /> 36 &amp;= 2^2\cdot 3^2,\\<br /> 54 &amp;= 2\cdot 3^3.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Hence we have<br /> &lt;cmath&gt;\begin{align*}<br /> a &amp;= 2^3\cdot 3\cdot w\\<br /> b &amp;= 2^3\cdot 3^2\cdot x\\<br /> c &amp;= 2^2\cdot 3^3\cdot y\\<br /> d &amp;= 2\cdot 3^3\cdot z<br /> \end{align*}&lt;/cmath&gt;<br /> for some positive integers &lt;math&gt;w,x,y,z&lt;/math&gt;. Now if &lt;math&gt;3&lt;/math&gt; divides &lt;math&gt;w&lt;/math&gt;, then &lt;math&gt;\gcd(a,b)&lt;/math&gt; would be at least &lt;math&gt;2^3\cdot 3^2&lt;/math&gt; which is too large, hence &lt;math&gt;3&lt;/math&gt; does not divide &lt;math&gt;w&lt;/math&gt;. Similarly, if &lt;math&gt;2&lt;/math&gt; divides &lt;math&gt;z&lt;/math&gt;, then &lt;math&gt;\gcd(c,d)&lt;/math&gt; would be at least &lt;math&gt;2^2\cdot 3^3&lt;/math&gt; which is too large, so &lt;math&gt;2&lt;/math&gt; does not divide &lt;math&gt;z&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)&lt;/cmath&gt;<br /> where neither &lt;math&gt;2&lt;/math&gt; nor &lt;math&gt;3&lt;/math&gt; divide &lt;math&gt;\gcd(w,z)&lt;/math&gt;. In other words, &lt;math&gt;\gcd(w,z)&lt;/math&gt; is divisible only by primes that are at least &lt;math&gt;5&lt;/math&gt;. The only possible value of &lt;math&gt;\gcd(a,d)&lt;/math&gt; between &lt;math&gt;70&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt; and which fits this criterion is &lt;math&gt;78=2\cdot3\cdot13&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(D) }13}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> We can say that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; 'have' &lt;math&gt;2^3 * 3&lt;/math&gt;, that &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; have &lt;math&gt;2^2 * 3^2&lt;/math&gt;, and that &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; have &lt;math&gt;3^3 * 2&lt;/math&gt;. Combining &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; yields &lt;math&gt;b&lt;/math&gt; has (at a minimum) &lt;math&gt;2^3 * 3^2&lt;/math&gt;, and thus &lt;math&gt;a&lt;/math&gt; has &lt;math&gt;2^3 * 3&lt;/math&gt; (and no more powers of &lt;math&gt;3&lt;/math&gt; because otherwise &lt;math&gt;gcd(a,b)&lt;/math&gt; would be different). In addition, &lt;math&gt;c&lt;/math&gt; has &lt;math&gt;3^3 * 2^2&lt;/math&gt;, and thus &lt;math&gt;d&lt;/math&gt; has &lt;math&gt;3^3 * 2&lt;/math&gt; (similar to &lt;math&gt;a&lt;/math&gt;, we see that &lt;math&gt;d&lt;/math&gt; cannot have any other powers of &lt;math&gt;2&lt;/math&gt;). We now assume the simplest scenario, where &lt;math&gt;a = 2^3 * 3&lt;/math&gt; and &lt;math&gt;d = 3^3 * 2&lt;/math&gt;. According to this base case, we have &lt;math&gt;gcd(a, d) = 2 * 3 = 6&lt;/math&gt;. We want an extra factor between the two such that this number is between &lt;math&gt;70&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt;, and this new factor cannot be divisible by &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt;. Checking through, we see that &lt;math&gt;6 * 13&lt;/math&gt; is the only one that works. Therefore the answer is &lt;math&gt;\boxed{\textbf{(D) } 13}&lt;/math&gt;<br /> <br /> Solution by JohnHankock<br /> <br /> == Solution 2.1 (updated with better notation)==<br /> Do casework on &lt;math&gt;v_2&lt;/math&gt; and &lt;math&gt;v_3.&lt;/math&gt; Notice that we must have &lt;math&gt;v_3(a) = 1&lt;/math&gt; and &lt;math&gt;v_2(d)=1&lt;/math&gt; and the values of &lt;math&gt;b,d&lt;/math&gt; does not matter. Therefore, &lt;math&gt;\gcd(d,a) = 6k,&lt;/math&gt; where &lt;math&gt;k&lt;/math&gt; is not divisible by &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3.&lt;/math&gt; We see that &lt;math&gt;13&lt;/math&gt; is the only possible answer.<br /> <br /> -Williamgolly<br /> <br /> ==Solution 3 (Better notation)==<br /> <br /> First off, note that &lt;math&gt;24&lt;/math&gt;, &lt;math&gt;36&lt;/math&gt;, and &lt;math&gt;54&lt;/math&gt; are all of the form &lt;math&gt;2^x\times3^y&lt;/math&gt;. The prime factorizations are &lt;math&gt;2^3\times 3^1&lt;/math&gt;, &lt;math&gt;2^2\times 3^2&lt;/math&gt; and &lt;math&gt;2^1\times 3^3&lt;/math&gt;, respectively. Now, let &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; be the number of times &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; go into &lt;math&gt;a&lt;/math&gt;,respectively. Define &lt;math&gt;b_2&lt;/math&gt;, &lt;math&gt;b_3&lt;/math&gt;, &lt;math&gt;c_2&lt;/math&gt;, and &lt;math&gt;c_3&lt;/math&gt; similiarly. Now, translate the &lt;math&gt;lcm&lt;/math&gt;s into the following: <br /> &lt;cmath&gt;\min(a_2,b_2)=3&lt;/cmath&gt; &lt;cmath&gt;\min(a_3,b_3)=1&lt;/cmath&gt; &lt;cmath&gt;\min(b_2,c_2)=2&lt;/cmath&gt; &lt;cmath&gt;\min(b_3,c_3)=2&lt;/cmath&gt; &lt;cmath&gt;\min(a_2,c_2)=1&lt;/cmath&gt; &lt;cmath&gt;\min(a_3,c_3)=3&lt;/cmath&gt; .<br /> <br /> (Unfinished)<br /> ~Rowechen Zhong<br /> <br /> ==Solution 4 (Fastest)==<br /> Notice that &lt;math&gt;gcd (a,b,c,d)=gcd(gcd(a,b),gcd(b,c),gcd(c,d))=gcd(24,36,54)=6&lt;/math&gt;, so &lt;math&gt;gcd(d,a)&lt;/math&gt; must be a multiple of &lt;math&gt;6&lt;/math&gt;. The only answer choice that gives a value between &lt;math&gt;70&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt; when multiplied by 6 is &lt;math&gt;\boxed{\textbf{(D) } 13}&lt;/math&gt;. - mathleticguyyy + einstein<br /> <br /> In the case where there can be 2 possible answers, we can do casework on gcd(d,a)<br /> ~Williamgolly<br /> <br /> ==Hardness of problem==<br /> <br /> The ranking of this number theory problem is easy hard(just starting to be hard) or a 7 out of 10. This is because you've got to think through out although it is more of which and easy solution<br /> <br /> ~justin6688<br /> <br /> == Video Solution by Richard Rusczyk ==<br /> <br /> https://artofproblemsolving.com/videos/amc/2018amc10a/467<br /> <br /> ~ dolphin7<br /> <br /> ==Video Solution==<br /> https://youtu.be/yjrqINsQP5c<br /> <br /> ~savannahsolver<br /> <br /> == Video Solution (Meta-Solving Technique) ==<br /> https://youtu.be/GmUWIXXf_uk?t=1003<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=21|num-a=23}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_18&diff=144517 2018 AMC 10A Problems/Problem 18 2021-02-02T01:41:59Z <p>Justin6688: /* Video Solution &amp; difficulty */</p> <hr /> <div>{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #13]] and [[2018 AMC 10A Problems|2018 AMC 10A #18]]}}<br /> <br /> ==Problem==<br /> How many nonnegative integers can be written in the form &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 512 \qquad <br /> \textbf{(B) } 729 \qquad <br /> \textbf{(C) } 1094 \qquad <br /> \textbf{(D) } 3281 \qquad <br /> \textbf{(E) } 59,048 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> This looks like balanced ternary, in which all the integers with absolute values less than &lt;math&gt;\frac{3^n}{2}&lt;/math&gt; are represented in &lt;math&gt;n&lt;/math&gt; digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of &lt;math&gt;|x|=3280.5&lt;/math&gt;, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are &lt;math&gt;3280+1=\boxed{3281}&lt;/math&gt; integers or &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all &lt;math&gt;a_i=0&lt;/math&gt;. The total number of ways to pick &lt;math&gt;a_i&lt;/math&gt; from &lt;math&gt;i=0, 1, 2, 3, ... 7&lt;/math&gt; is &lt;math&gt;3^8=6561&lt;/math&gt;. &lt;math&gt;\frac{6561-1}{2}=3280&lt;/math&gt; gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives &lt;math&gt;6561-3280=\boxed{\textbf{(D) } 3281}&lt;/math&gt;. (RegularHexagon, KLBBC minor changes)<br /> <br /> ==Solution ==<br /> Note that the number of total possibilities (ignoring the conditions set by the problem) is &lt;math&gt;3^8=6561&lt;/math&gt;. So, E is clearly unrealistic. <br /> <br /> Note that if &lt;math&gt;a_7&lt;/math&gt; is 1, then it's impossible for &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt; to be negative. Therefore, if &lt;math&gt;a_7&lt;/math&gt; is 1, there are &lt;math&gt;3^7=2187&lt;/math&gt; possibilities. (We also must convince ourselves that these &lt;math&gt;2187&lt;/math&gt; different sets of coefficients must necessarily yield &lt;math&gt;2187&lt;/math&gt; different integer results.)<br /> <br /> As A, B, and C are all less than 2187, the answer must be &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;<br /> <br /> ==Solution ==<br /> Note that we can do some simple casework:<br /> If &lt;math&gt;a_7=1&lt;/math&gt;, then we can choose anything for the other 7 variables, so this give us &lt;math&gt;3^7&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt; and &lt;math&gt;a_6=1&lt;/math&gt;, then we can choose anything for the other 6 variables, giving us &lt;math&gt;3^6&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt;, &lt;math&gt;a_6=0&lt;/math&gt;, and &lt;math&gt;a_5=1&lt;/math&gt;, then we have &lt;math&gt;3^5&lt;/math&gt;.<br /> Continuing in this vein, we have &lt;math&gt;3^7+3^6+\cdots+3^1+3^0&lt;/math&gt; ways to choose the variables' values, except we have to add 1<br /> because we haven't counted the case where all variables are 0. So our total sum is &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;.<br /> Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.<br /> <br /> ==Casework and Recursion Solution==<br /> This solution has a similar idea to that of above.<br /> We will start off with casework on &lt;math&gt;a_{7}&lt;/math&gt;:<br /> If &lt;math&gt;a_7 = -1&lt;/math&gt;, we can show that there is no way to assign the values of &lt;math&gt;-1, 0, 1&lt;/math&gt; to the other &lt;math&gt;a_{i}&lt;/math&gt;'s for &lt;math&gt;0 \leq i \leq 6&lt;/math&gt;. This is because even if we make all of the other &lt;math&gt;a_{i}&lt;/math&gt; equal to &lt;math&gt;1&lt;/math&gt;, we will have that our number is &lt;math&gt;(-1)(3^{7}) + (3^{6}+3^{5}+...+3^{0}) = -3^{7} + \frac{3^{7}-1}{3-1}&lt;/math&gt;. However, note that since &lt;math&gt;\frac{3^{7}-1}{3-1} &lt; 3^{7}&lt;/math&gt;, &lt;math&gt;-3^{7} + \frac{3^{7}-1}{3-1} &lt; 0&lt;/math&gt;.<br /> If &lt;math&gt;a_{7}=0&lt;/math&gt;, we can ignore &lt;math&gt;a_{7}&lt;/math&gt; and pretend like it never even existed. This then simplifies the problem: the question remains the same, except that instead of having &lt;math&gt;a_{0}...a_{7}&lt;/math&gt;, we only have &lt;math&gt;a_{0}...a_{6}&lt;/math&gt;. Now, we introduce some notation: let the number of nonnegative integers that can be written in the form of &lt;math&gt;a_{n} \cdot 3^{n} + ... + a_{0} \cdot 3^{0}&lt;/math&gt; be denoted as &lt;math&gt;S_{n}&lt;/math&gt;. Then, the problem is asking us for &lt;math&gt;S_{7}&lt;/math&gt;, and the total number of nonnegative integers we get from this case is &lt;math&gt;S_{6}&lt;/math&gt;. Let's keep this in mind and revisit this idea later.<br /> If &lt;math&gt;a_{7}=1&lt;/math&gt;, we can show that no matter what values we assign to the rest of the &lt;math&gt;a_{i}&lt;/math&gt;, we will always get a nonnegative integer (we can make this even stricter and say positive integer). This is because of something we figured out in the first case: &lt;math&gt;\frac{3^{7}-1}{3-1} &lt; 3^{7}&lt;/math&gt;. So, even if we let &lt;math&gt;a_{i}=1&lt;/math&gt; for &lt;math&gt;0 \leq i \leq 6&lt;/math&gt;, we will still have &lt;math&gt;3^{7} - (3^{6}+...+3^{0}) &gt; 0&lt;/math&gt;. So, for this case, we have &lt;math&gt;3^7&lt;/math&gt; possible nonnegative integers.<br /> Totalling up our values from the three cases, we see that &lt;math&gt;S_{7} = 0 + S_{6} + 3^{7}&lt;/math&gt;. In fact, using the reasoning from the three cases above, we can deduce that, for all positive integers &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S_{n} = 3^{n} + S_{n-1}&lt;/math&gt;. We can now start building up our values for &lt;math&gt;S&lt;/math&gt;: &lt;math&gt;S_{0} = 2&lt;/math&gt; (&lt;math&gt;a_{0} = 0, 1&lt;/math&gt;), &lt;math&gt;S_{1} = 3^{1}+S_{0} = 5, S_{2} = 14, ..., S_{6}=1094&lt;/math&gt;. So, we have &lt;math&gt;S_{7} = 2187 + 1094 = \boxed{3281}&lt;/math&gt;, which is answer choice &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;.<br /> ~advanture<br /> <br /> ==Solution ==<br /> The key is to realize that this question is basically taking place in &lt;math&gt;a\in\{0,1,2\}&lt;/math&gt; if each value of &lt;math&gt;a&lt;/math&gt; was increased by &lt;math&gt;1&lt;/math&gt;, essentially making it into base &lt;math&gt;3&lt;/math&gt;. Then the range would be from &lt;math&gt;0\cdot3^7+&lt;/math&gt; &lt;math&gt;0\cdot3^6+&lt;/math&gt; &lt;math&gt;0\cdot3^5+&lt;/math&gt; &lt;math&gt;0\cdot3^4+&lt;/math&gt; &lt;math&gt;0\cdot3^3+&lt;/math&gt; &lt;math&gt;0\cdot3^2+&lt;/math&gt; &lt;math&gt;0\cdot3^1+&lt;/math&gt; &lt;math&gt;0\cdot3^0=&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;2\cdot3^7+&lt;/math&gt; &lt;math&gt;2\cdot3^6+&lt;/math&gt; &lt;math&gt;2\cdot3^5+&lt;/math&gt; &lt;math&gt;2\cdot3^4+&lt;/math&gt; &lt;math&gt;2\cdot3^3+&lt;/math&gt; &lt;math&gt;2\cdot3^2+&lt;/math&gt; &lt;math&gt;2\cdot3^1+&lt;/math&gt; &lt;math&gt;2\cdot3^0=&lt;/math&gt; &lt;math&gt;3^8-1=&lt;/math&gt; &lt;math&gt;6561-1=&lt;/math&gt; &lt;math&gt;6560&lt;/math&gt;, yielding &lt;math&gt;6561&lt;/math&gt; different values. Since the distribution for all &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; the question originally gave is symmetrical, we retain the &lt;math&gt;3280&lt;/math&gt; positive integers and one &lt;math&gt;0&lt;/math&gt; but discard the &lt;math&gt;3280&lt;/math&gt; negative integers. Thus, we are left with the answer, &lt;math&gt;\boxed{\textbf{(D)} 3281}\qquad&lt;/math&gt;. --anna0kear<br /> <br /> ==Solution ==<br /> First, set &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq1&lt;/math&gt;. The range would be the integers for which &lt;math&gt;[-1,1]&lt;/math&gt;. If &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq2&lt;/math&gt;, our set expands to include all integers such that &lt;math&gt;-4\leq\mathbb{Z}\leq4&lt;/math&gt;. Similarly, when &lt;math&gt;i\geq3&lt;/math&gt; we get &lt;math&gt;-13\leq\mathbb{Z}\leq13&lt;/math&gt;, and when &lt;math&gt;i\geq4&lt;/math&gt; the range is &lt;math&gt;-40\leq\mathbb{Z}\leq40&lt;/math&gt;. The pattern continues until we reach &lt;math&gt;i=7&lt;/math&gt;, where &lt;math&gt;-3280\leq\mathbb{Z}\leq3280&lt;/math&gt;. Because we are only looking for positive integers, we filter out all &lt;math&gt;\mathbb{Z}&lt;0&lt;/math&gt;, leaving us with all integers between &lt;math&gt;0\leq\mathbb{Z}\leq3280&lt;/math&gt;, inclusive. The answer becomes &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;. --anna0kear<br /> <br /> ==Solution ==<br /> To get the number of integers, we can get the highest positive integer that can be represented using &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;.<br /> <br /> Note that the least nonnegative integer that can be represented is &lt;math&gt;0&lt;/math&gt;, when all &lt;math&gt;a_i=0&lt;/math&gt;. The highest number will be the number when all &lt;math&gt;a_i=1&lt;/math&gt;. That will be &lt;cmath&gt;3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}&lt;/cmath&gt; &lt;cmath&gt;=3280&lt;/cmath&gt;<br /> <br /> Therefore, there are &lt;math&gt;3280&lt;/math&gt; positive integers and &lt;math&gt;(3280+1)&lt;/math&gt; nonnegative integers (while including &lt;math&gt;0&lt;/math&gt;) that can be represented. Our answer is &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==Solution ==<br /> Notice that there are &lt;math&gt;3^8&lt;/math&gt; options for &lt;math&gt;a_7, a_6, \cdots a_0&lt;/math&gt; since each &lt;math&gt;a_i&lt;/math&gt; can take the value of &lt;math&gt;-1&lt;/math&gt; or &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;. Now we want to find how many of them are positive and then we can add one in the end to account for &lt;math&gt;0&lt;/math&gt;(they are asking for non-negative). <br /> <br /> By symmetry(look out for these on the contest), we see that exactly half of them are positive. So &lt;math&gt;\lfloor{\tfrac{3^8}{2}}\rfloor = 3280.&lt;/math&gt; So now we will add &lt;math&gt;1&lt;/math&gt; because of the &lt;math&gt;0&lt;/math&gt; to account for the non-negative solutions. <br /> <br /> So our final answer is &lt;math&gt;3280 + 1 = 3281&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/0xmbHDcUI2w<br /> <br /> ~IceMatrix<br /> <br /> ==Difficulty==<br /> Out of a scale of 10, we rate this as 7 due to its complexion, however the solution is both traditional and easy. <br /> ~ Justin6688 and friend me<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=144193 Georgeooga-Harryooga Theorem 2021-01-31T23:09:43Z <p>Justin6688: </p> <hr /> <div>=Definition=<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; objects are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects in a line.<br /> <br /> <br /> Created by George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> ==Proof 2==<br /> Let us call the &lt;math&gt;b&lt;/math&gt; people &lt;math&gt;1, 2, ... b&lt;/math&gt;<br /> <br /> Let the number of people before &lt;math&gt;1&lt;/math&gt; in line be &lt;math&gt;y_1&lt;/math&gt;, between &lt;math&gt;1, 2&lt;/math&gt; be &lt;math&gt;y_2&lt;/math&gt;, ... after &lt;math&gt;b&lt;/math&gt; be &lt;math&gt;y_{b+1}&lt;/math&gt;.<br /> We have &lt;cmath&gt;y_1 + y_2 + y_3 + \dots y_{b+1} = a-b&lt;/cmath&gt;<br /> <br /> The number of ways to determine &lt;math&gt;y_1, y_2, \dots&lt;/math&gt; is equivalent to the number of positive integer solutions to:<br /> &lt;cmath&gt;x_1 + x_2 + .. + x_{b+1} = a-b + 2&lt;/cmath&gt; where &lt;math&gt;(x_2, ... x_b) = (y_2, ..., y_b) &lt;/math&gt; and &lt;math&gt;(x_1, x_{b+1}) = (y_1 +1, y_{b+1} + 1)&lt;/math&gt;.<br /> <br /> So, by stars and bars, the number of ways to determine &lt;math&gt;(y_2, ..., y_b) &lt;/math&gt; is &lt;cmath&gt;F(a,b) = \dbinom{a-b+1}{b} = \frac {(a-b+1)!}{b!(a-2b+1)!}&lt;/cmath&gt;<br /> <br /> Furthermore, after picking positions for the people, we have &lt;math&gt;(a-b)!&lt;/math&gt; ways to order the &lt;math&gt;(a-b)&lt;/math&gt; people who can be together, and &lt;math&gt;b!&lt;/math&gt; ways to order the &lt;math&gt;b&lt;/math&gt; people who cannot be together. So for each &lt;math&gt;(y_1, y_2, ... y_{b+1}&lt;/math&gt;, we have &lt;math&gt;b! (a-b)!&lt;/math&gt; orderings.<br /> <br /> Therefore, the final answer is &lt;cmath&gt;b! (a-b)! F(a,b) = \frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/cmath&gt;<br /> <br /> <br /> Proof by Aryabhata000<br /> <br /> =Applications=<br /> ==Application 1==<br /> ===Problem===<br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream.<br /> Fred and George are identical twins, so they are indistinguishable.<br /> Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> <br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> <br /> Problem by [https://artofproblemsolving.com/community/c4h2342517p18900597 Math4Life2020]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the [[Georgeooga-Harryooga Theorem]]. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ==Application 2==<br /> ===Problem===<br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> <br /> Problem by [[2020 AMC 8 Problems/Problem 10|The Mathematical Association of America's American Mathematics Competitions]]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> By the [[Georgeooga-Harryooga Theorem]] there are &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}&lt;/math&gt; way to arrange the marbles.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ====Solution 2====<br /> We can arrange our marbles like so &lt;math&gt;\square A\square B\square&lt;/math&gt;.<br /> <br /> To arrange the &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; we have &lt;math&gt;2!=2&lt;/math&gt; ways.<br /> <br /> To place the &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the blanks we have &lt;math&gt;_3P_2=6&lt;/math&gt; ways.<br /> <br /> By fundamental counting principle our final answer is &lt;math&gt;2\cdot6=\boxed{\textbf{(C) }12}&lt;/math&gt;<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> ====Solution 3====<br /> Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by &lt;math&gt;A,B,S,&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt;, respectively. If we ignore the constraint that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; cannot be next to each other, we get a total of &lt;math&gt;4!=24&lt;/math&gt; ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; can be next to each other. If we place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. &lt;math&gt;ST\square\square, \square ST\square, \square\square ST&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;TS\square\square, \square TS\square, \square\square TS&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; directly next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other. Subtracting this from the total number of arrangements gives us &lt;math&gt;24-12=12&lt;/math&gt; total arrangements &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> <br /> We can also solve this problem directly by looking at the number of ways that we can place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not directly next to each other. Observe that there are three ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (in that order) into the four slots so they are not next to each other (i.e. &lt;math&gt;S\square T\square, \square S\square T, S\square\square T&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;T\square S\square, \square T\square S, T\square\square S&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not next to each other &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ====Solution 4====<br /> Let's try complementary counting. There &lt;math&gt;4!&lt;/math&gt; ways to arrange the 4 marbles. However, there are &lt;math&gt;2\cdot3!&lt;/math&gt; arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, &lt;cmath&gt;4!-2\cdot3!=\boxed{12 \textbf{(C)}}&lt;/cmath&gt;<br /> <br /> ====Solution 5====<br /> <br /> We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a &quot;super marble.&quot; There are &lt;math&gt;2!&lt;/math&gt; ways to arrange Steelie and Tiger within this &quot;super marble.&quot; Then there are &lt;math&gt;3!&lt;/math&gt; ways to arrange the &quot;super marble&quot; and Zara's two other marbles in a row. Since there are &lt;math&gt;4!&lt;/math&gt; ways to arrange the marbles without any restrictions, the answer is given by &lt;math&gt;4!-2!\cdot 3!=\textbf{(C) }12&lt;/math&gt;<br /> <br /> -franzliszt<br /> <br /> ====Solution 6====<br /> <br /> We will use the following<br /> <br /> &lt;math&gt;\textbf{Georgeooga-Harryooga Theorem:}&lt;/math&gt; The [[Georgeooga-Harryooga Theorem]] states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; of them cannot be together, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;math&gt;\textit{Proof. (Created by AoPS user RedFireTruck)}&lt;/math&gt;<br /> <br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> <br /> Back to the problem. By the [[Georgeooga-Harryooga Theorem]], our answer is &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\textbf{(C) }12&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ====Solution 7====<br /> https://youtu.be/pB46JzBNM6g<br /> <br /> ~savannahsolver<br /> <br /> =Testimonials=<br /> &quot;Thanks for rediscovering our theorem [[User:Redfiretruck|RedFireTruck]]&quot; - George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> &quot;Wow! George and Harry are alive???&quot; ~ samrocksnature<br /> <br /> &quot;Hi&quot; ~ jasperE3<br /> <br /> &quot;I used this theorem on the AMC 8 and got a 25. Very useful!&quot; - [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &quot;This is very complicated, but great.&quot; - Jiseop55406<br /> <br /> I used this theorem on the AMC 8 too! ~ ilp<br /> <br /> &quot;Very nice theorem&quot; ~ IdkHowToAddNumbers<br /> <br /> &quot;This theorem is really worth of being used in the AMC 8! Very useful!&quot; ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 20:48, 21 December 2020 (EST)<br /> <br /> &quot;I have realized the way&quot; ~ hi..<br /> <br /> &quot;This theorem even works on AMC10 and 12&quot; - TryhardMathlete<br /> <br /> &quot;This is very cool&quot;~Physicsisfun123<br /> <br /> &quot;Interesting!&quot; ~justin6688<br /> <br /> I used this on AMC 8 problem 10 and I got it right! THANK YOU! -Onafets {Satire}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_17&diff=144152 2015 AMC 10A Problems/Problem 17 2021-01-31T19:44:39Z <p>Justin6688: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> A line that passes through the origin intersects both the line &lt;math&gt; x = 1&lt;/math&gt; and the line &lt;math&gt;y=1+ \frac{\sqrt{3}}{3} x&lt;/math&gt;. The three lines create an equilateral triangle. What is the perimeter of the triangle?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)} \ 2+2\sqrt{3} \qquad\textbf{(C)} \ 6 \qquad\textbf{(D)} \ 3 + 2\sqrt{3} \qquad\textbf{(E)} \ 6 + \frac{\sqrt{3}}{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since the triangle is equilateral and one of the sides is a vertical line, the triangle must have a horizontal line of symmetry, and therefore the other two sides will have opposite slopes. The slope of the other given line is &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; so the third must be &lt;math&gt;-\frac{\sqrt{3}}{3}&lt;/math&gt;. Since this third line passes through the origin, its equation is simply &lt;math&gt;y = -\frac{\sqrt{3}}{3}x&lt;/math&gt;. To find two vertices of the triangle, plug in &lt;math&gt;x=1&lt;/math&gt; to both the other equations. <br /> <br /> &lt;math&gt;y = -\frac{\sqrt{3}}{3}&lt;/math&gt;<br /> <br /> &lt;math&gt;y = 1 + \frac{\sqrt{3}}{3}&lt;/math&gt;<br /> <br /> We now have the coordinates of two vertices, &lt;math&gt;\left(1, -\frac{\sqrt{3}}{3}\right)&lt;/math&gt; and &lt;math&gt;\left(1, 1 + \frac{\sqrt{3}}{3}\right)&lt;/math&gt;. The length of one side is the distance between the y-coordinates, or &lt;math&gt;1 + \frac{2\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> The perimeter of the triangle is thus &lt;math&gt;3\left(1 + \frac{2\sqrt{3}}{3}\right)&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(D) }3 + 2\sqrt{3}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Draw a line from the y-intercept of the equation &lt;math&gt;y=1+ \frac{\sqrt{3}}{3} x&lt;/math&gt; perpendicular to the line &lt;math&gt;x=1&lt;/math&gt;. There is a square of side length 1 inscribed in the equilateral triangle. The problem becomes reduced to finding the perimeter of an equilateral triangle with a square of side length 1 inscribed in it. The side length is &lt;math&gt;2\left(\frac{1}{\sqrt{3}}\right) + 1&lt;/math&gt;. After multiplying the side length by 3 and rationalizing, you get &lt;math&gt;\boxed{\textbf{(D) }3 + 2\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/-l1Kawq_hds<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> Video Solution:<br /> <br /> https://www.youtube.com/watch?v=2kvSRL8KMac<br /> <br /> <br /> {{AMC10 box|year=2015|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}a</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=137067 2016 AMC 8 Problems/Problem 25 2020-11-09T00:45:43Z <p>Justin6688: </p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> ==Note==<br /> There are many solutions here, and all of them are equally good. For your own benefit, look at all of the solutions, as they employ many unique techniques to get to the final answer.<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> pair A, B, C, D; <br /> A=(0,0); <br /> B=(16,0); <br /> C=(8,15); <br /> D=B/2; <br /> draw(A--B--C--cycle); <br /> draw(C--D); <br /> draw(arc(D,120/17,0,180)); <br /> draw(rightanglemark(B,D,C,25)); <br /> label(&quot;$A$&quot;,A,SW); <br /> label(&quot;$B$&quot;,B,SE); <br /> label(&quot;$C$&quot;,C,N); <br /> label(&quot;$D$&quot;,D,S); <br /> label(&quot;$8$&quot;,(A+D)/2,S);<br /> label(&quot;$15$&quot;,(C+D)/2,NE);<br /> label(&quot;$17$&quot;,(A+C)/2,W);<br /> &lt;/asy&gt;<br /> <br /> First, we drop a perpendicular, shown above, to the base of the triangle, cutting the triangle into two congruent right triangles. This triangle is isosceles, which means perpendiculars are medians and vice versa. The base of the resulting right triangle is &lt;math&gt;8&lt;/math&gt; for both sides, and the height is &lt;math&gt;15,&lt;/math&gt; as given. Using the Pythagorean theorem, we can find the length of the hypotenuse, or &lt;math&gt;17.&lt;/math&gt; Using the two legs of the right triangle, we find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; results in the radius, which is the height of the right triangle when using the hypotenuse as the base. Hence, the answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> *Note — There are several other following solutions below that use similar methods of finding the area two different ways shown in this solution. Try to eliminate any words (such as 'can' or 'would be') that make your solution less concise. Try to write 'sure of yourself.'<br /> <br /> ==Solution 2: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> Notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AC}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> ==Solution 3: Inscribed Circle==<br /> <br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB = BC = CD = DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area &lt;math&gt;[ABCD]&lt;/math&gt; of the rhombus is twice that, which is &lt;math&gt;bh = (16)(15) = 240&lt;/math&gt;.<br /> <br /> The [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34r = 240&lt;/math&gt;. Solving this, &lt;math&gt;r = \frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> Noting that we have a 8-15-17 triangle, we can find &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CE.&lt;/math&gt; Let &lt;math&gt;AE=x&lt;/math&gt;, &lt;math&gt;CE=17-x.&lt;/math&gt; Then by similar triangles (or &quot;Altitude on Hypotenuse&quot;) we have &lt;math&gt;15^2=x*17.&lt;/math&gt; Thus, &lt;math&gt;AE=x=225/17, CE=64/17.&lt;/math&gt; Now again by &quot;Altitude on Hypotenuse”, &lt;math&gt;r=\sqrt{AE*CE}.&lt;/math&gt; Therefore &lt;math&gt;r=\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 5: Simple Trigonometry==<br /> Note: This solution uses [http://artofproblemsolving.com/wiki/index.php?title=Trigonometry#Word Trigonometric Concepts]<br /> &lt;asy&gt;<br /> pair A, B, C, D; <br /> A=(0,0); <br /> B=(16,0); <br /> C=(8,15); <br /> D=B/2; <br /> draw(A--B--C--cycle); <br /> draw(C--D); <br /> draw(arc(D,120/17,0,180)); <br /> draw(rightanglemark(B,D,C,25)); <br /> label(&quot;$A$&quot;,A,SW); <br /> label(&quot;$B$&quot;,B,SE); <br /> label(&quot;$C$&quot;,C,N); <br /> label(&quot;$D$&quot;,D,S); <br /> &lt;/asy&gt;<br /> <br /> Denote the bottom left vertex of the isosceles triangle to be &lt;math&gt;8&lt;/math&gt;<br /> <br /> Denote the bottom right vertex of the isosceles triangle to be &lt;math&gt;8&lt;/math&gt; <br /> <br /> Denote the top vertex of the isosceles triangle to be &lt;math&gt;x&lt;/math&gt;<br /> <br /> Drop an altitude from &lt;math&gt;x&lt;/math&gt; to side &lt;math&gt;AB&lt;/math&gt;. Denote the foot of the intersection to be &lt;math&gt;D&lt;/math&gt;.<br /> <br /> By the Pythagorean Theorem, &lt;math&gt;AC=17&lt;/math&gt;.<br /> <br /> Now, we see that by sin(x), &lt;math&gt;\sin{A}=\frac{15}{17}&lt;/math&gt;.<br /> <br /> This implies that &lt;math&gt;\sin{A}=\frac{r}{8}&lt;/math&gt; (r=radius of semicircle).<br /> <br /> Hence, &lt;math&gt;r=\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 6: Area==<br /> <br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Credits for Asymptote go to whoever wrote this diagram up in Solution 2. <br /> <br /> There are two ways to find the area of &lt;math&gt;\triangle ABC&lt;/math&gt;. The first way is the most obvious, and that is to multiply the base times the height (&lt;math&gt;16\cdot15&lt;/math&gt;) and then divide it by two. The second way is, in a way, a little more complex. Note that &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt; are congruent. This means that if we find the area of one triangle, we can just multiply it's area by 2 and we've found the area of the larger triangle &lt;math&gt;ABC&lt;/math&gt;. But since we always divide by two, as it is in the formula for finding the area of a triangle, the multiply by two and divide by two cancel out, giving us that if we just multiply base times height of &lt;math&gt;\triangle ACD&lt;/math&gt;, we will get the area of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> Now you might think: &quot;But what is the other way of finding the area&quot;. Well that is &lt;math&gt;AC&lt;/math&gt;, which would be the base, times &lt;math&gt;DE&lt;/math&gt;, the radius, which would be the height. <br /> <br /> The first way to find the area gives us the area of the &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\frac{15\cdot16}{2}=120&lt;/math&gt;. This gives us &lt;math&gt;120=AC\cdot r&lt;/math&gt; (&lt;math&gt;r&lt;/math&gt; signifies radius). We can find &lt;math&gt;AC&lt;/math&gt; using the Pythagorean Theorem on &lt;math&gt;\triangle ACD&lt;/math&gt;. The two legs are &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;, which gives us that the hypotenuse, &lt;math&gt;AC&lt;/math&gt;, is equal to &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now that we have an equation with only one variable for the radius, &lt;math&gt;120=17r&lt;/math&gt;, we can just solve for the radius. We get &lt;math&gt;r=\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;. <br /> <br /> This may seem like a lengthy explanation, but doing it yourself when you know what to do, it actually takes very little time. Try it yourself!<br /> <br /> ==Solution 7==<br /> Let us draw altitude &lt;math&gt;\overline{CD}.&lt;/math&gt; This cuts our base into line segments with length &lt;math&gt;\dfrac{16}{2}=8.&lt;/math&gt; Finding the area of the resulting triangles gives &lt;math&gt;[\triangle ADC] = [\triangle BDC] = \dfrac{8 \cdot 15}{2} = 60.&lt;/math&gt; Since &lt;math&gt;m\angle CDB = m\angle CDA = 90^\circ,&lt;/math&gt; we use the Pythagorean Theorem to find length &lt;math&gt;\overline{AC}&lt;/math&gt;:<br /> &lt;cmath&gt;\begin{align*}<br /> (\overline{AD})^2+(\overline{CD})^2 &amp;= (\overline{AC})^2 \\<br /> 8^2+15^2 &amp;= (\overline{AC})^2 \\<br /> 64+225 &amp;= (\overline{AC})^2 \\<br /> 289 &amp;= (\overline{AC})^2 \\<br /> \sqrt{289} &amp;= \sqrt{(\overline{AC})^2} \\<br /> 17 &amp;= AC.<br /> \end{align*}&lt;/cmath&gt;<br /> Thus we have &lt;math&gt;\overline{AC}=\overline{BC}=17.&lt;/math&gt; <br /> <br /> &lt;asy&gt;<br /> pair A, B, C, D; <br /> A=(0,0); <br /> B=(16,0); <br /> C=(8,15); <br /> D=B/2; <br /> draw(A--B--C--cycle); <br /> draw(C--D); <br /> draw(arc(D,120/17,0,180)); <br /> draw(rightanglemark(B,D,C,25)); <br /> label(&quot;$A$&quot;,A,SW); <br /> label(&quot;$B$&quot;,B,SE); <br /> label(&quot;$C$&quot;,C,N); <br /> label(&quot;$D$&quot;,D,S); <br /> label(&quot;$8$&quot;,(A+D)/2,S);<br /> label(&quot;$8$&quot;,(B+D)/2,S);<br /> label(&quot;$15$&quot;,(C+D)/2,NE);<br /> label(&quot;$17$&quot;,(A+C)/2,W);<br /> label(&quot;$17$&quot;,(B+C)/2,E);<br /> &lt;/asy&gt;<br /> <br /> <br /> Letting &lt;math&gt;17&lt;/math&gt; be the base of the triangle makes our height the radius of the semicircle:<br /> &lt;asy&gt;<br /> pair A, B, C, D, EE; <br /> A=(0,0); <br /> B=(16,0); <br /> C=(8,15); <br /> D=B/2; <br /> EE=(64/17*8/17, 64/17*15/17);<br /> draw(A--B--C--cycle); <br /> draw(C--D); <br /> draw(D--EE);<br /> draw(arc(D,120/17,0,180)); <br /> draw(rightanglemark(B,D,C,25)); <br /> draw(rightanglemark(A,EE,D,25));<br /> label(&quot;$A$&quot;,A,SW); <br /> label(&quot;$B$&quot;,B,SE); <br /> label(&quot;$C$&quot;,C,N); <br /> label(&quot;$D$&quot;,D,S); <br /> label(&quot;$8$&quot;,(A+D)/2,S);<br /> label(&quot;$8$&quot;,(B+D)/2,S);<br /> label(&quot;$15$&quot;,(C+D)/2,NE);<br /> label(&quot;$17$&quot;,(A+C)/2,W);<br /> label(&quot;$17$&quot;,(B+C)/2,E);<br /> label(&quot;$r$&quot;,D--EE, N);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; be the radius of the semicircle. Then we have &lt;math&gt;\dfrac{17 \cdot r}{2}=60 \implies 17 \cdot r = 120 \implies \dfrac{17r}{17} = \dfrac{120}{17} \implies r = \boxed{\textbf{(B) }\dfrac{120}{17}}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Remember: Area can help in places where you least expect it!}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\definecolor{salmon}{rgb}{.918, .6, .6}\colorbox{salmon}{\color{white}{solution by \textbf{pog}}}&lt;/cmath&gt;<br /> <br /> ==Solution 8: equations==<br /> <br /> Draw tangent lines to connect the centre of the circle to the equal sides. Since the angle is 90 degrees and the height is perpendicular to the side not equal with any of the other side lengths, the two triangles(Not the big one but the two smaller ones) are similar. We've got one common side that gives us the ratio is &lt;math&gt;\frac{8}{15}&lt;/math&gt;. We've got a system of equations!<br /> <br /> 1. 8/15=x(the radius)/y(the side that is similar)<br /> 2. &lt;math&gt;x^2 + y^2 = 225&lt;/math&gt; (by the pythagorean theorem).<br /> <br /> Solving the equations we get &lt;math&gt;\boxed{\frac{120}{17}}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\definecolor{salmon}{rgb}{.789, .6, .6}\colorbox{salmon}{\color{black}{created by \textbf{justin6688}}}&lt;/cmath&gt;<br /> <br /> ==Solution 9: Process of Elimination==<br /> <br /> Not a true solution, but in the interest of creativity and alternative methods, here is a way to cheese the solution in seconds.<br /> <br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> <br /> First, notice that &lt;math&gt;AC=17&lt;/math&gt; (Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;). Since all the side lengths of &lt;math&gt;\triangle ADC&lt;/math&gt; are integers, and &lt;math&gt;\triangle AED \sim \triangle ADC \sim \triangle DEC&lt;/math&gt;, all of the side length ratios are rational, and therefore, &lt;math&gt;DE&lt;/math&gt; must be a rational value. We eliminate answer choices A, D, and E.<br /> <br /> Finally, notice that answer choice C is impossible because that would mean that in &lt;math&gt;\triangle AED&lt;/math&gt;, the leg &lt;math&gt;DE = 10&lt;/math&gt; would be bigger than the hypotenuse &lt;math&gt;AD = 8&lt;/math&gt;. The only answer choice remaining is &lt;math&gt;\boxed{\textbf{(B) }\dfrac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Video Solution==<br /> <br /> ===Video Solution 1===<br /> https://youtu.be/fMXtuMXfAcE - Happytwin<br /> <br /> ===Video Solution 2===<br /> https://www.youtube.com/watch?v=jfGc3hHPu2w&amp;feature=youtu.be<br /> <br /> ~IceMatrix2<br /> <br /> ==See Also==<br /> {{AMC8 box|year = 2016|num-b = 24|after = This is the Last Problem}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_21&diff=134988 2016 AMC 8 Problems/Problem 21 2020-10-13T00:40:44Z <p>Justin6688: /* Solution 3 */</p> <hr /> <div>A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/OOdK-nOzaII?t=1452<br /> <br /> ==Solution 1==<br /> We put five chips randomly in order, and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for &lt;math&gt;\binom{5}{2} = 10&lt;/math&gt;. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is &lt;math&gt;\binom{4}{3} = 4&lt;/math&gt;. Because a green chip will be last &lt;math&gt;4&lt;/math&gt; out of the &lt;math&gt;10&lt;/math&gt; situations, our answer is &lt;math&gt;\boxed{\textbf{(B) } \frac{2}{5}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out &lt;math&gt;3&lt;/math&gt; red chips, &lt;math&gt;3&lt;/math&gt; red chips and &lt;math&gt;1&lt;/math&gt; green chip, &lt;math&gt;2&lt;/math&gt; green chips, &lt;math&gt;2&lt;/math&gt; green chips and &lt;math&gt;1&lt;/math&gt; red chip, and &lt;math&gt;2&lt;/math&gt; green chips and &lt;math&gt;2&lt;/math&gt; red chips. Because order is important in this problem, there are &lt;math&gt;1+4+1+3+6=15&lt;/math&gt; ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which &lt;math&gt;15-5=10&lt;/math&gt;. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is &lt;math&gt;\frac{4}{10} = \frac{2}{5}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(B) } \frac{2}{5}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Assume that after you draw the three red chips in a row without drawing both green chips, you continue drawing for the next turn. The last/fifth chip that is drawn must be a green chip because if both green chips were drawn before, we would've already completed the game. So technically, the problem is asking for the probability that the &quot;fifth draw&quot; is a green chip. This probability is symmetric to the probability that the first chip drawn is green, which is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. Thus the probability is &lt;math&gt;\boxed{\textbf{(B) } \frac{2}{5}}&lt;/math&gt;. <br /> <br /> Note: This problem is almost identical to 2001 AMC 10 #23.<br /> <br /> ==Video Solution==<br /> Fast Solution:<br /> https://www.youtube.com/watch?v=w0y-JuRQvDc&amp;feature=youtu.be<br /> <br /> https://youtu.be/m834gDVyPJM<br /> <br /> https://youtu.be/fTtUAtfWKyQ - Happytwin<br /> <br /> {{AMC8 box|year=2016|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Justin6688 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_1&diff=128802 1991 AIME Problems/Problem 1 2020-07-21T13:15:49Z <p>Justin6688: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Find &lt;math&gt;x^2+y^2_{}&lt;/math&gt; if &lt;math&gt;x_{}^{}&lt;/math&gt; and &lt;math&gt;y_{}^{}&lt;/math&gt; are positive integers such that<br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;xy_{}^{}+x+y = 71&lt;/math&gt;&lt;/div&gt;<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;x^2y+xy^2 = 880^{}_{}.&lt;/math&gt;&lt;/div&gt;&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define &lt;math&gt;a = x + y&lt;/math&gt; and &lt;math&gt;b = xy&lt;/math&gt;. Then &lt;math&gt;a + b = 71&lt;/math&gt; and &lt;math&gt;ab = 880&lt;/math&gt;. Solving these two equations yields a [[quadratic equation|quadratic]]: &lt;math&gt;a^2 - 71a + 880 = 0&lt;/math&gt;, which [[factor]]s to &lt;math&gt;(a - 16)(a - 55) = 0&lt;/math&gt;. Either &lt;math&gt;a = 16&lt;/math&gt; and &lt;math&gt;b = 55&lt;/math&gt; or &lt;math&gt;a = 55&lt;/math&gt; and &lt;math&gt;b = 16&lt;/math&gt;. For the first case, it is easy to see that &lt;math&gt;(x,y)&lt;/math&gt; can be &lt;math&gt;(5,11)&lt;/math&gt; (or vice versa). In the second case, since all factors of &lt;math&gt;16&lt;/math&gt; must be &lt;math&gt;\le 16&lt;/math&gt;, no two factors of &lt;math&gt;16&lt;/math&gt; can sum greater than &lt;math&gt;32&lt;/math&gt;, and so there are no integral solutions for &lt;math&gt;(x,y)&lt;/math&gt;. The solution is &lt;math&gt;5^2 + 11^2 = \boxed{146}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Since &lt;math&gt;xy + x + y + 1 = 72&lt;/math&gt;, this can be factored to &lt;math&gt;(x + 1)(y + 1) = 72&lt;/math&gt;. As &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are [[integer]]s, the possible sets for &lt;math&gt;(x,y)&lt;/math&gt; (ignoring cases where &lt;math&gt;x &gt; y&lt;/math&gt; since it is symmetrical) are &lt;math&gt;(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)&lt;/math&gt;. The second equation factors to &lt;math&gt;(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11&lt;/math&gt;. The only set with a factor of &lt;math&gt;11&lt;/math&gt; is &lt;math&gt;(5,11)&lt;/math&gt;, and checking shows that it is correct.<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;a=x+y&lt;/math&gt;, &lt;math&gt;b=xy&lt;/math&gt; then we get the equations<br /> &lt;cmath&gt;\begin{align*}<br /> a+b&amp;=71\\<br /> ab&amp;=880<br /> \end{align*}&lt;/cmath&gt;<br /> After finding the [[prime factorization]] of &lt;math&gt;880=2^4\cdot5\cdot11&lt;/math&gt;, it's easy to obtain the solution &lt;math&gt;(a,b)=(16,55)&lt;/math&gt;. Thus<br /> &lt;cmath&gt;x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}&lt;/cmath&gt;<br /> Note that if &lt;math&gt;(a,b)=(55,16)&lt;/math&gt;, the answer would exceed &lt;math&gt;999&lt;/math&gt; which is invalid for an AIME answer.<br /> ~ Nafer<br /> === Solution 4 ===<br /> From the first equation, we know &lt;math&gt;x+y=71-xy&lt;/math&gt;. We factor the second equation as &lt;math&gt;xy(71-xy)=880&lt;/math&gt;. Let &lt;math&gt;a=xy&lt;/math&gt; and rearranging we get &lt;math&gt;a^2-71a+880=(a-16)(a-55)=0&lt;/math&gt;. We have two cases: (1) &lt;math&gt;x+y=16&lt;/math&gt; and &lt;math&gt;xy=55&lt;/math&gt; OR (2) &lt;math&gt;x+y=55&lt;/math&gt; and &lt;math&gt;xy=16&lt;/math&gt;. We find the former is true for &lt;math&gt;(x,y) = (5,11)&lt;/math&gt;. &lt;math&gt;x^2+y^2=121+25=146&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|before=First question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Justin6688