https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Michaelchang1&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-27T19:31:50Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=141973 2017 AMC 10A Problems/Problem 19 2021-01-12T21:38:40Z <p>Michaelchang1: /* Solution 3: PIE */</p> <hr /> <div>==Problem==<br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of &lt;math&gt;5&lt;/math&gt; chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution 1: Casework==<br /> <br /> For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br /> We can split this problem up into two cases:<br /> <br /> &lt;math&gt;\textbf{Case 1: }&lt;/math&gt; A sits on an edge seat.<br /> <br /> Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of &lt;math&gt;2 \cdot 2 \cdot 2 \cdot 2 = 16&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2: }&lt;/math&gt; A does not sit in an edge seat.<br /> <br /> In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are &lt;math&gt;3 \cdot 2 \cdot 2 = 12&lt;/math&gt; seatings in this case.<br /> <br /> Adding up all the cases, we have &lt;math&gt;16+12 = \boxed{\textbf{(C) } 28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Label the seats (from left to right) &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;5&lt;/math&gt;. The number of ways to seat Derek and Eric in the five seats with no restrictions is &lt;math&gt;5 \cdot 4=20&lt;/math&gt;. The number of ways to seat Derek and Eric such that they sit next to each other is &lt;math&gt;8&lt;/math&gt; (we can treat Derek and Eric as a &quot;block&quot;. There are four ways to seat this &quot;block&quot;, and two ways to permute Derek and Eric, for a total of &lt;math&gt;4\cdot 2=8&lt;/math&gt;), so the number of ways such that Derek and Eric don't sit next to each other is &lt;math&gt;20-8=12&lt;/math&gt;. Note that once Derek and Eric are seated, there are three cases. <br /> <br /> The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us &lt;math&gt;0&lt;/math&gt; ways.<br /> <br /> Another possible case is if Derek and Eric seat in seats &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; in some order. There are &lt;math&gt;2&lt;/math&gt; possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are &lt;math&gt;3!=6&lt;/math&gt; ways to do this. So the second case gives us &lt;math&gt;2 \cdot 6=12&lt;/math&gt; total ways for the second case.<br /> <br /> The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are &lt;math&gt;12-2-2=8&lt;/math&gt; ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us &lt;math&gt;8 \cdot 2=16&lt;/math&gt; ways.<br /> <br /> So in total there are &lt;math&gt;12+16=28&lt;/math&gt;. So our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 28}&lt;/math&gt;.<br /> &lt;br&gt;<br /> Minor LaTeX edits by fasterthanlight<br /> <br /> Minor clarity edits by Carrot_Karen<br /> <br /> ==Solution 3: PIE==<br /> <br /> We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, then when they aren't.<br /> <br /> PIE:<br /> Let's count the total number of cases where one of these is true:<br /> <br /> When Alice is with Bob:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> <br /> When Alice is with Carla:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> <br /> When Carla is with Eric:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> <br /> Then, we count the cases where two of these are true.<br /> <br /> Alice is next to Bob Carla, and Alice is also next to Bob.<br /> There are two ways to rearrange Alice, Bob, and Carla so this is true:<br /> BAC and CAB.<br /> &lt;math&gt;2\cdot3!=12&lt;/math&gt;<br /> <br /> Alice is next to Carla, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> <br /> Alice is next to Bob, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> <br /> Finally, we count the cases where all three of these are true:<br /> &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt;<br /> <br /> We add the cases where one of these are true up:<br /> &lt;math&gt;48\cdot3=144&lt;/math&gt;<br /> <br /> Subtract the cases where two of these are true:<br /> &lt;math&gt;144-60=84&lt;/math&gt;<br /> <br /> And finally add back the cases where three of these are true:<br /> &lt;math&gt;84+8=92&lt;/math&gt;<br /> <br /> Thus, our answer is &lt;math&gt;5!-92=28&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> ~michaelchang1<br /> <br /> ==Video Solution==<br /> https://youtu.be/umr2Aj9ViOA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=141972 2017 AMC 10A Problems/Problem 19 2021-01-12T21:38:23Z <p>Michaelchang1: /* Solution 3: PIE */</p> <hr /> <div>==Problem==<br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of &lt;math&gt;5&lt;/math&gt; chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution 1: Casework==<br /> <br /> For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br /> We can split this problem up into two cases:<br /> <br /> &lt;math&gt;\textbf{Case 1: }&lt;/math&gt; A sits on an edge seat.<br /> <br /> Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of &lt;math&gt;2 \cdot 2 \cdot 2 \cdot 2 = 16&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2: }&lt;/math&gt; A does not sit in an edge seat.<br /> <br /> In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are &lt;math&gt;3 \cdot 2 \cdot 2 = 12&lt;/math&gt; seatings in this case.<br /> <br /> Adding up all the cases, we have &lt;math&gt;16+12 = \boxed{\textbf{(C) } 28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Label the seats (from left to right) &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;5&lt;/math&gt;. The number of ways to seat Derek and Eric in the five seats with no restrictions is &lt;math&gt;5 \cdot 4=20&lt;/math&gt;. The number of ways to seat Derek and Eric such that they sit next to each other is &lt;math&gt;8&lt;/math&gt; (we can treat Derek and Eric as a &quot;block&quot;. There are four ways to seat this &quot;block&quot;, and two ways to permute Derek and Eric, for a total of &lt;math&gt;4\cdot 2=8&lt;/math&gt;), so the number of ways such that Derek and Eric don't sit next to each other is &lt;math&gt;20-8=12&lt;/math&gt;. Note that once Derek and Eric are seated, there are three cases. <br /> <br /> The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us &lt;math&gt;0&lt;/math&gt; ways.<br /> <br /> Another possible case is if Derek and Eric seat in seats &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; in some order. There are &lt;math&gt;2&lt;/math&gt; possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are &lt;math&gt;3!=6&lt;/math&gt; ways to do this. So the second case gives us &lt;math&gt;2 \cdot 6=12&lt;/math&gt; total ways for the second case.<br /> <br /> The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are &lt;math&gt;12-2-2=8&lt;/math&gt; ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us &lt;math&gt;8 \cdot 2=16&lt;/math&gt; ways.<br /> <br /> So in total there are &lt;math&gt;12+16=28&lt;/math&gt;. So our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 28}&lt;/math&gt;.<br /> &lt;br&gt;<br /> Minor LaTeX edits by fasterthanlight<br /> <br /> Minor clarity edits by Carrot_Karen<br /> <br /> ==Solution 3: PIE==<br /> <br /> We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, then when they aren't.<br /> <br /> PIE:<br /> Let's count the total number of cases where one of these is true:<br /> <br /> When Alice is with Bob:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> <br /> When Alice is with Carla:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> <br /> When Carla is with Eric:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> <br /> Then, we count the cases where two of these are true.<br /> <br /> Alice is next to Bob Carla, and Alice is also next to Bob.<br /> There are two ways to rearrange Alice, Bob, and Carla so this is true:<br /> BAC and CAB.<br /> &lt;math&gt;2\cdot3!=12&lt;/math&gt;<br /> <br /> Alice is next to Carla, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> <br /> Alice is next to Bob, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> <br /> Finally, we count the cases where all three of these are true:<br /> &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt;<br /> <br /> We add the cases where one of these are true up:<br /> &lt;math&gt;48\cdot3=144&lt;/math&gt;<br /> <br /> Subtract the cases where two of these are true:<br /> &lt;math&gt;144-60=84&lt;/math&gt;<br /> <br /> And finally add back the cases where three of these are true:<br /> &lt;math&gt;84+8=92&lt;/math&gt;<br /> Thus, our answer is &lt;math&gt;5!-92=28&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/umr2Aj9ViOA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=141971 2017 AMC 10A Problems/Problem 19 2021-01-12T21:37:31Z <p>Michaelchang1: /* Solution 3: PIE */</p> <hr /> <div>==Problem==<br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of &lt;math&gt;5&lt;/math&gt; chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution 1: Casework==<br /> <br /> For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br /> We can split this problem up into two cases:<br /> <br /> &lt;math&gt;\textbf{Case 1: }&lt;/math&gt; A sits on an edge seat.<br /> <br /> Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of &lt;math&gt;2 \cdot 2 \cdot 2 \cdot 2 = 16&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2: }&lt;/math&gt; A does not sit in an edge seat.<br /> <br /> In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are &lt;math&gt;3 \cdot 2 \cdot 2 = 12&lt;/math&gt; seatings in this case.<br /> <br /> Adding up all the cases, we have &lt;math&gt;16+12 = \boxed{\textbf{(C) } 28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Label the seats (from left to right) &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;5&lt;/math&gt;. The number of ways to seat Derek and Eric in the five seats with no restrictions is &lt;math&gt;5 \cdot 4=20&lt;/math&gt;. The number of ways to seat Derek and Eric such that they sit next to each other is &lt;math&gt;8&lt;/math&gt; (we can treat Derek and Eric as a &quot;block&quot;. There are four ways to seat this &quot;block&quot;, and two ways to permute Derek and Eric, for a total of &lt;math&gt;4\cdot 2=8&lt;/math&gt;), so the number of ways such that Derek and Eric don't sit next to each other is &lt;math&gt;20-8=12&lt;/math&gt;. Note that once Derek and Eric are seated, there are three cases. <br /> <br /> The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us &lt;math&gt;0&lt;/math&gt; ways.<br /> <br /> Another possible case is if Derek and Eric seat in seats &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; in some order. There are &lt;math&gt;2&lt;/math&gt; possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are &lt;math&gt;3!=6&lt;/math&gt; ways to do this. So the second case gives us &lt;math&gt;2 \cdot 6=12&lt;/math&gt; total ways for the second case.<br /> <br /> The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are &lt;math&gt;12-2-2=8&lt;/math&gt; ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us &lt;math&gt;8 \cdot 2=16&lt;/math&gt; ways.<br /> <br /> So in total there are &lt;math&gt;12+16=28&lt;/math&gt;. So our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 28}&lt;/math&gt;.<br /> &lt;br&gt;<br /> Minor LaTeX edits by fasterthanlight<br /> <br /> Minor clarity edits by Carrot_Karen<br /> <br /> ==Solution 3: PIE==<br /> <br /> We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, then when they aren't.<br /> <br /> PIE:<br /> Let's count the total number of cases where one of these is true:<br /> When Alice is with Bob:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Alice is with Carla:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Carla is with Eric:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> Then, we count the cases where two of these are true.<br /> Alice is next to Bob Carla, and Alice is also next to Bob.<br /> There are two ways to rearrange Alice, Bob, and Carla so this is true:<br /> BAC and CAB.<br /> &lt;math&gt;2\cdot3!=12&lt;/math&gt;<br /> Alice is next to Carla, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Alice is with Bob, and Derek is also with Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Finally, we count the cases where all three of these are true:<br /> &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt;<br /> We add the cases where one of these are true up:<br /> &lt;math&gt;48\cdot3=144&lt;/math&gt;<br /> Subtract the cases where two of these are true:<br /> &lt;math&gt;144-60=84&lt;/math&gt;<br /> And finally add back the cases where three of these are true:<br /> &lt;math&gt;84+8=92&lt;/math&gt;<br /> Thus, our answer is &lt;math&gt;5!-92=28&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/umr2Aj9ViOA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=141970 2017 AMC 10A Problems/Problem 19 2021-01-12T21:37:12Z <p>Michaelchang1: /* Solution 3: PIE */</p> <hr /> <div>==Problem==<br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of &lt;math&gt;5&lt;/math&gt; chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution 1: Casework==<br /> <br /> For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br /> We can split this problem up into two cases:<br /> <br /> &lt;math&gt;\textbf{Case 1: }&lt;/math&gt; A sits on an edge seat.<br /> <br /> Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of &lt;math&gt;2 \cdot 2 \cdot 2 \cdot 2 = 16&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2: }&lt;/math&gt; A does not sit in an edge seat.<br /> <br /> In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are &lt;math&gt;3 \cdot 2 \cdot 2 = 12&lt;/math&gt; seatings in this case.<br /> <br /> Adding up all the cases, we have &lt;math&gt;16+12 = \boxed{\textbf{(C) } 28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Label the seats (from left to right) &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;5&lt;/math&gt;. The number of ways to seat Derek and Eric in the five seats with no restrictions is &lt;math&gt;5 \cdot 4=20&lt;/math&gt;. The number of ways to seat Derek and Eric such that they sit next to each other is &lt;math&gt;8&lt;/math&gt; (we can treat Derek and Eric as a &quot;block&quot;. There are four ways to seat this &quot;block&quot;, and two ways to permute Derek and Eric, for a total of &lt;math&gt;4\cdot 2=8&lt;/math&gt;), so the number of ways such that Derek and Eric don't sit next to each other is &lt;math&gt;20-8=12&lt;/math&gt;. Note that once Derek and Eric are seated, there are three cases. <br /> <br /> The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us &lt;math&gt;0&lt;/math&gt; ways.<br /> <br /> Another possible case is if Derek and Eric seat in seats &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; in some order. There are &lt;math&gt;2&lt;/math&gt; possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are &lt;math&gt;3!=6&lt;/math&gt; ways to do this. So the second case gives us &lt;math&gt;2 \cdot 6=12&lt;/math&gt; total ways for the second case.<br /> <br /> The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are &lt;math&gt;12-2-2=8&lt;/math&gt; ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us &lt;math&gt;8 \cdot 2=16&lt;/math&gt; ways.<br /> <br /> So in total there are &lt;math&gt;12+16=28&lt;/math&gt;. So our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 28}&lt;/math&gt;.<br /> &lt;br&gt;<br /> Minor LaTeX edits by fasterthanlight<br /> <br /> Minor clarity edits by Carrot_Karen<br /> <br /> ==Solution 3: PIE==<br /> <br /> We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, then when they aren't.<br /> <br /> PIE:<br /> Let's count the total number of cases where one of these is true:<br /> When Alice is with Bob:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Alice is with Carla:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Carla is with Eric:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> Then, we count the cases where two of these are true.<br /> Alice is next to Bob Carla, and Alice is also next to Bob.<br /> There are two ways to rearrange Alice, Bob, and Carla so this is true:<br /> BAC and CAB.<br /> &lt;math&gt;2\cdot3!=12&lt;/math&gt;<br /> Alice is next to Carla, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Alice is with Bob, and Derek is also with Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Finally, we count the cases where all three of these are true:<br /> &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt;<br /> We add the cases where one of these are true up:<br /> &lt;math&gt;48\cdot3=144&lt;/math&gt;<br /> Subtract the cases where two of these are true:<br /> &lt;math&gt;144-60=84&lt;/math&gt;<br /> And finally add back the cases where three of these are true:<br /> &lt;math&gt;84+8=92&lt;/math&gt;<br /> Thus, our answer is &lt;math&gt;5!-92=28&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C) }&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/umr2Aj9ViOA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=141969 2017 AMC 10A Problems/Problem 19 2021-01-12T21:36:31Z <p>Michaelchang1: /* Solution 3: PIE */</p> <hr /> <div>==Problem==<br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of &lt;math&gt;5&lt;/math&gt; chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution 1: Casework==<br /> <br /> For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br /> We can split this problem up into two cases:<br /> <br /> &lt;math&gt;\textbf{Case 1: }&lt;/math&gt; A sits on an edge seat.<br /> <br /> Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of &lt;math&gt;2 \cdot 2 \cdot 2 \cdot 2 = 16&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2: }&lt;/math&gt; A does not sit in an edge seat.<br /> <br /> In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are &lt;math&gt;3 \cdot 2 \cdot 2 = 12&lt;/math&gt; seatings in this case.<br /> <br /> Adding up all the cases, we have &lt;math&gt;16+12 = \boxed{\textbf{(C) } 28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Label the seats (from left to right) &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;5&lt;/math&gt;. The number of ways to seat Derek and Eric in the five seats with no restrictions is &lt;math&gt;5 \cdot 4=20&lt;/math&gt;. The number of ways to seat Derek and Eric such that they sit next to each other is &lt;math&gt;8&lt;/math&gt; (we can treat Derek and Eric as a &quot;block&quot;. There are four ways to seat this &quot;block&quot;, and two ways to permute Derek and Eric, for a total of &lt;math&gt;4\cdot 2=8&lt;/math&gt;), so the number of ways such that Derek and Eric don't sit next to each other is &lt;math&gt;20-8=12&lt;/math&gt;. Note that once Derek and Eric are seated, there are three cases. <br /> <br /> The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us &lt;math&gt;0&lt;/math&gt; ways.<br /> <br /> Another possible case is if Derek and Eric seat in seats &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; in some order. There are &lt;math&gt;2&lt;/math&gt; possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are &lt;math&gt;3!=6&lt;/math&gt; ways to do this. So the second case gives us &lt;math&gt;2 \cdot 6=12&lt;/math&gt; total ways for the second case.<br /> <br /> The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are &lt;math&gt;12-2-2=8&lt;/math&gt; ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us &lt;math&gt;8 \cdot 2=16&lt;/math&gt; ways.<br /> <br /> So in total there are &lt;math&gt;12+16=28&lt;/math&gt;. So our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 28}&lt;/math&gt;.<br /> &lt;br&gt;<br /> Minor LaTeX edits by fasterthanlight<br /> <br /> Minor clarity edits by Carrot_Karen<br /> <br /> ==Solution 3: PIE==<br /> <br /> We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, then when they aren't.<br /> <br /> PIE:<br /> Let's count the total number of cases where one of these is true:<br /> When Alice is with Bob:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Alice is with Carla:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Carla is with Eric:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> Then, we count the cases where two of these are true.<br /> Alice is next to Bob Carla, and Alice is also next to Bob.<br /> There are two ways to rearrange Alice, Bob, and Carla so this is true:<br /> BAC and CAB.<br /> &lt;math&gt;2\cdot3!=12&lt;/math&gt;<br /> Alice is next to Carla, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Alice is with Bob, and Derek is also with Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Finally, we count the cases where all three of these are true:<br /> &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt;<br /> We add the cases where one of these are true up:<br /> &lt;math&gt;48\cdot3=144&lt;/math&gt;<br /> Subtract the cases where two of these are true:<br /> &lt;math&gt;144-60=84&lt;/math&gt;<br /> And finally add back the cases where three of these are true:<br /> &lt;math&gt;84+8=92&lt;/math&gt;<br /> Thus, our answer is &lt;math&gt;5!-92=\boxed{28}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C) } 28}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/umr2Aj9ViOA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=141968 2017 AMC 10A Problems/Problem 19 2021-01-12T21:36:15Z <p>Michaelchang1: </p> <hr /> <div>==Problem==<br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of &lt;math&gt;5&lt;/math&gt; chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution 1: Casework==<br /> <br /> For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br /> We can split this problem up into two cases:<br /> <br /> &lt;math&gt;\textbf{Case 1: }&lt;/math&gt; A sits on an edge seat.<br /> <br /> Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of &lt;math&gt;2 \cdot 2 \cdot 2 \cdot 2 = 16&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2: }&lt;/math&gt; A does not sit in an edge seat.<br /> <br /> In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are &lt;math&gt;3 \cdot 2 \cdot 2 = 12&lt;/math&gt; seatings in this case.<br /> <br /> Adding up all the cases, we have &lt;math&gt;16+12 = \boxed{\textbf{(C) } 28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Label the seats (from left to right) &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;5&lt;/math&gt;. The number of ways to seat Derek and Eric in the five seats with no restrictions is &lt;math&gt;5 \cdot 4=20&lt;/math&gt;. The number of ways to seat Derek and Eric such that they sit next to each other is &lt;math&gt;8&lt;/math&gt; (we can treat Derek and Eric as a &quot;block&quot;. There are four ways to seat this &quot;block&quot;, and two ways to permute Derek and Eric, for a total of &lt;math&gt;4\cdot 2=8&lt;/math&gt;), so the number of ways such that Derek and Eric don't sit next to each other is &lt;math&gt;20-8=12&lt;/math&gt;. Note that once Derek and Eric are seated, there are three cases. <br /> <br /> The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us &lt;math&gt;0&lt;/math&gt; ways.<br /> <br /> Another possible case is if Derek and Eric seat in seats &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; in some order. There are &lt;math&gt;2&lt;/math&gt; possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are &lt;math&gt;3!=6&lt;/math&gt; ways to do this. So the second case gives us &lt;math&gt;2 \cdot 6=12&lt;/math&gt; total ways for the second case.<br /> <br /> The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are &lt;math&gt;12-2-2=8&lt;/math&gt; ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us &lt;math&gt;8 \cdot 2=16&lt;/math&gt; ways.<br /> <br /> So in total there are &lt;math&gt;12+16=28&lt;/math&gt;. So our answer is &lt;math&gt;\boxed{\textbf{(C)}\ 28}&lt;/math&gt;.<br /> &lt;br&gt;<br /> Minor LaTeX edits by fasterthanlight<br /> <br /> Minor clarity edits by Carrot_Karen<br /> <br /> ==Solution 3: PIE==<br /> <br /> We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, then when they aren't.<br /> <br /> PIE:<br /> Let's count the total number of cases where one of these is true:<br /> When Alice is with Bob:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Alice is with Carla:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> When Carla is with Eric:<br /> &lt;math&gt;2\cdot4!=48&lt;/math&gt;<br /> Then, we count the cases where two of these are true.<br /> Alice is next to Bob Carla, and Alice is also next to Bob.<br /> There are two ways to rearrange Alice, Bob, and Carla so this is true:<br /> BAC and CAB.<br /> &lt;math&gt;2\cdot3!=12&lt;/math&gt;<br /> Alice is next to Carla, and Derek is also next to Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Alice is with Bob, and Derek is also with Eric.<br /> &lt;math&gt;2\cdot2\cdot3!=24&lt;/math&gt;<br /> Finally, we count the cases where all three of these are true:<br /> &lt;math&gt;2\cdot2\cdot2=8&lt;/math&gt;<br /> We add the cases where one of these are true up:<br /> &lt;math&gt;48\cdot3=144&lt;/math&gt;<br /> Subtract the cases where two of these are true:<br /> &lt;math&gt;144-60=84&lt;/math&gt;<br /> And finally add back the cases where three of these are true:<br /> &lt;math&gt;84+8=92&lt;/math&gt;<br /> Thus, our answer is &lt;math&gt;5!-92=\boxed{28}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(C) } 28}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/umr2Aj9ViOA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=User:Michaelchang1&diff=140644 User:Michaelchang1 2020-12-26T14:05:28Z <p>Michaelchang1: Created page with &quot;Is a human&quot;</p> <hr /> <div>Is a human</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=139474 User:Piphi 2020-12-12T03:00:05Z <p>Michaelchang1: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;405&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#B1B2B3;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.&lt;br&gt;<br /> <br /> Piphi started the signature trend at around May 2020.&lt;br&gt;<br /> <br /> Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].&lt;br&gt;<br /> <br /> Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].&lt;br&gt;<br /> <br /> Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.&lt;br&gt;<br /> <br /> Piphi has also added a lot of the info that is in the [[Reaper Archives]].&lt;br&gt;<br /> <br /> Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].&lt;br&gt;<br /> <br /> Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br /> <br /> Piphi has a post that was made an announcement in a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br /> <br /> Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br /> <br /> Piphi has won 2 gold trophies at the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] and is now part of the staff.<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#727373;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;#f0f2f3&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;#f0f2f3&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 500<br /> {{User:Piphi/Template:Progress_Bar|80.2|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|64.5|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|47.5|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|20.36|width=100%}}&lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=137902 Wooga Looga Theorem 2020-11-20T01:27:23Z <p>Michaelchang1: /* Testimonials */</p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;D,E,F&lt;/math&gt; on the sides &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively such that &lt;math&gt;\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> <br /> Created by Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Proof by Gogobao:<br /> <br /> We have: &lt;math&gt;\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1} &lt;/math&gt;<br /> <br /> We have: &lt;math&gt;[DEF] = [ABC] - [DCE] - [FAE] - [FBD]&lt;/math&gt;<br /> <br /> &lt;math&gt;[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;<br /> <br /> ==Proof 2==<br /> Proof by franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&amp;\tfrac {1}{r+1}&amp;\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&amp;0&amp;\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&amp;\tfrac {r}{r+1}&amp;0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.&lt;/cmath&gt;<br /> <br /> ==Proof 3==<br /> Proof by RedFireTruck:<br /> <br /> WLOG we let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y\in\mathbb{R}&lt;/math&gt;. We then use Shoelace Forumla to get &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt;. We then figure out that &lt;math&gt;D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)&lt;/math&gt;, &lt;math&gt;E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)&lt;/math&gt;, and &lt;math&gt;F=\left(\frac{r}{r+1}, 0\right)&lt;/math&gt; so we know that by Shoelace Formula &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|&lt;/math&gt;. We know that &lt;math&gt;\frac{r^2-r+1}{(r+1)^2}\ge0&lt;/math&gt; for all &lt;math&gt;r\in\mathbb{R}&lt;/math&gt; so &lt;math&gt;\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> ==Proof 4==<br /> Proof by jasperE3:<br /> <br /> The Wooga Looga theorem is a direct application of Routh's Theorem when a=b=c.<br /> <br /> =Application 1=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by franzliszt:<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;X,Y,Z&lt;/math&gt; are on sides &lt;math&gt;BC,CA,AB&lt;/math&gt; such that &lt;math&gt;\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71&lt;/math&gt;. Find the ratio of &lt;math&gt;[XYZ]&lt;/math&gt; to &lt;math&gt;[ABC]&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> is this solution by RedFireTruck:<br /> <br /> WLOG let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt;. Then &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt; by Shoelace Theorem and &lt;math&gt;X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)&lt;/math&gt;, &lt;math&gt;Y=\left(\frac{x}{8}, \frac{y}{8}\right)&lt;/math&gt;, &lt;math&gt;Z=\left(\frac78, 0\right)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12\left|\frac{43y}{64}\right|&lt;/math&gt; by Shoelace Theorem. Therefore the answer is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\<br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[XYZ]}{[ABC]}=\begin{vmatrix}<br /> 0&amp;\tfrac 18&amp;\tfrac 78\\<br /> \tfrac 78&amp;0&amp;\tfrac 18\\<br /> \tfrac18&amp;\tfrac78&amp;0<br /> \end{vmatrix}=\frac{43}{64}.&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> or this solution by aaja3427:<br /> <br /> According the the Wooga Looga Theorem, It is &lt;math&gt;\frac{49-7+1}{8^2}&lt;/math&gt;. This is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> or this solution by eduD_looC:<br /> <br /> This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. A very beautiful application, which leaves graders and readers speechless.<br /> <br /> ==Solution 6==<br /> or this solution by CoolJupiter:<br /> <br /> Wow. All of your solutions are slow, compared to my sol:<br /> <br /> By math, we have &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> ^<br /> |<br /> EVERYONE USE THIS SOLUTION IT'S BRILLIANT <br /> ~bsu1<br /> Yes, very BRILLIANT!<br /> ~ TheAoPSLebron<br /> <br /> ==The Best Solution==<br /> <br /> By the 1+1=2 principle, we get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. Definitely the best method. When asked, please say that OlympusHero taught you this method. Cuz he did.<br /> <br /> =Application 2=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by Matholic:<br /> <br /> The figure below shows a triangle ABC whose area is &lt;math&gt;72 \text{cm}^2&lt;/math&gt;. If &lt;math&gt;\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}&lt;/math&gt;, find &lt;math&gt;[DEF].&lt;/math&gt;<br /> <br /> ~LaTeX-ifyed by RP3.1415<br /> <br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\ <br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{}=\begin{vmatrix}<br /> \tfrac 56&amp;\tfrac 16&amp;0\\<br /> 0&amp;\tfrac 56&amp;\tfrac 16\\<br /> \tfrac16&amp;0&amp;\tfrac56<br /> \end{vmatrix}=\frac{7}{12}&lt;/cmath&gt; so &lt;math&gt;[DEF]=42&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 2==<br /> or this solution by RedFireTruck:<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}&lt;/math&gt;. We are given that &lt;math&gt;[ABC]=72&lt;/math&gt; so &lt;math&gt;[DEF]=\frac{7}{12}\cdot72=\boxed{42}&lt;/math&gt;<br /> <br /> =Application 3=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:<br /> <br /> Find the ratio &lt;math&gt;\frac{[GHI]}{[ABC]}&lt;/math&gt; if &lt;math&gt;\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12&lt;/math&gt; and &lt;math&gt;\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1&lt;/math&gt; in the diagram below.&lt;asy&gt;<br /> draw((0, 0)--(6, 0)--(4, 3)--cycle);<br /> draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle);<br /> draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle);<br /> label(&quot;A$&quot;, (0, 0), SW);<br /> label(&quot;$B$&quot;, (6, 0), SE);<br /> label(&quot;$C$&quot;, (4, 3), N);<br /> label(&quot;$D$&quot;, (2, 0), S);<br /> label(&quot;$E$&quot;, (16/3, 1), NE);<br /> label(&quot;$F$&quot;, (8/3, 2), NW);<br /> label(&quot;$G$&quot;, (11/3, 1/2), SE);<br /> label(&quot;$H$&quot;, (4, 3/2), NE);<br /> label(&quot;$I\$&quot;, (7/3, 1), W);<br /> &lt;/asy&gt;<br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13&lt;/math&gt;. Notice that &lt;math&gt;\triangle GHI&lt;/math&gt; is the medial triangle of '''Wooga Looga Triangle ''' of &lt;math&gt;\triangle ABC&lt;/math&gt;. So &lt;math&gt;\frac{[GHI]}{[DEF]}=\frac 14&lt;/math&gt; and &lt;math&gt;\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}&lt;/math&gt; by Chain Rule ideas.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt; so that &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then &lt;math&gt;D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)&lt;/math&gt;. And &lt;math&gt;G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)&lt;/math&gt;.<br /> <br /> In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&amp;\tfrac 12&amp;\tfrac 16\\ \tfrac 16&amp;\tfrac 13&amp;\tfrac 12\\ \tfrac 12&amp;\tfrac 16&amp;\tfrac 13 \end{vmatrix}=\frac{1}{12}.&lt;/cmath&gt;<br /> <br /> =Application 4=<br /> ==Problem==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle and &lt;math&gt;D,E,F&lt;/math&gt; be points on sides &lt;math&gt;BC,AC,&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; respectively. We have that &lt;math&gt;\frac{BD}{DC} = 3&lt;/math&gt; and similar for the other sides. If the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, then what is the area of triangle &lt;math&gt;DEF&lt;/math&gt;? (By ilovepizza2020)<br /> <br /> ==Solution 1==<br /> <br /> By Franzliszt<br /> <br /> By Wooga Looga, &lt;math&gt;\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}&lt;/math&gt; so the answer is &lt;math&gt;7&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> By franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&amp;\tfrac 14&amp;\tfrac 34\\ \tfrac 34&amp;0&amp;\tfrac 14\\ \tfrac 14&amp;\tfrac 34&amp;0 \end{vmatrix}=\frac{7}{16}.&lt;/cmath&gt;So the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> =Testimonials=<br /> Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> Franzlist is wooga looga howsopro - volkie boy<br /> <br /> The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm<br /> <br /> The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.<br /> ~ilp2020<br /> <br /> The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck<br /> <br /> The Wooga Looga Theorem is the best. -aaja3427<br /> <br /> The Wooga Looga Theorem is needed for everything and it is great-hi..<br /> <br /> The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click &quot;about&quot;. now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT<br /> <br /> This theorem has helped me with school and I am no longer failing my math class. -mchang<br /> <br /> &quot;I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman.&quot; ~CoolJupiter<br /> <br /> Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)<br /> <br /> Too powerful... ~franzliszt<br /> <br /> The Wooga Looga Theorem is so pro ~ ac142931<br /> <br /> It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)<br /> <br /> This theorem changed my life... ~ samrocksnature<br /> <br /> Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3<br /> <br /> It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3<br /> <br /> This is franzliszt and I endorse this theorem. ~franzliszt<br /> <br /> This theorem is too OP. ~bestzack66<br /> <br /> This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283<br /> <br /> Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415<br /> <br /> The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020<br /> <br /> It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321<br /> <br /> The Wooga Looga Theorem is a special case of [url=https://en.wikipedia.org/wiki/Routh%27s_theorem]Routh's Theorem.[/url] So this wiki article is DEFINITELY needed. -peace<br /> <br /> I actually thought this was a joke theorem until I read this page - HumanCalculator9<br /> <br /> I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825<br /> <br /> This is &lt;i&gt;almost&lt;/i&gt; as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun<br /> <br /> I ReAlLy don't get it - Senguamar<br /> <br /> The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.</div> Michaelchang1 https://artofproblemsolving.com/wiki/index.php?title=User:OlympusHero&diff=135831 User:OlympusHero 2020-10-26T03:02:16Z <p>Michaelchang1: /* User Count */</p> <hr /> <div>OlympusHero's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;107px&quot;&gt; 71th user MICHAELCHANG1<br /> &lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;OlympusHero is currently borderline AIME.&lt;br&gt;<br /> <br /> OlympusHero is 10 years old.&lt;br&gt;<br /> <br /> OlympusHero scored 47/46 (got an extra point for finishing the test under 10 minutes) when mocking the 2019 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State.&lt;br&gt;<br /> <br /> OlympusHero is a pro at math and chess<br /> <br /> OlympusHero for brown MOP 2021<br /> <br /> OlympusHero was #1 at the 2017 Chess World Cadets U8 (He was 7 years old)<br /> <br /> OlympusHero is pro at joking believe it.<br /> <br /> OlympusHero is much better than Rusczyk at math yes or no, no one knows.<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;A User Count of 300<br /> <br /> Make AIME 2021 (Currently borderline)<br /> <br /> Get 5 or more on AIME I 2021 (Mocked a 3 on the 2020 AIME I a few months ago)<br /> <br /> Get to 40 or more on the MATHCOUNTS Trainer Nationals Level (25/40)<br /> <br /> Get a perfect score and First Place in IMO 2021.<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Michaelchang1