https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Somebodyyouusedtoknow&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T18:38:20ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=The_Devil%27s_Triangle&diff=159650The Devil's Triangle2021-08-06T04:47:55Z<p>Somebodyyouusedtoknow: /* Generalized Wooga Looga Theorem (The Devil's Triangle) */</p>
<hr />
<div>=Definition=<br />
==Generalized Wooga Looga Theorem (The Devil's Triangle)==<br />
For any triangle <math>\triangle ABC</math>, let <math>D, E</math> and <math>F</math> be points on <math>BC, AC</math> and <math>AB</math> respectively. The Generalized Wooga Looga Theorem (Gwoologth) or the Devil's Triangle Theorem states that if <math>\frac{BD}{CD}=r, \frac{CE}{AE}=s</math> and <math>\frac{AF}{BF}=t</math>, then <math>\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\frac{rst+1}{(r+1)(s+1)(t+1)}</math>.<br />
<br />
(*Simplification found by @Gogobao)<br />
<br />
=Proofs=<br />
==Proof 1==<br />
Proof by CoolJupiter:<br />
<br />
We have the following ratios:<br />
<math>\frac{BD}{BC}=\frac{r}{r+1}, \frac{CD}{BC}=\frac{1}{r+1},\frac{CE}{AC}=\frac{s}{s+1}, \frac{AE}{AC}=\frac{1}{s+1},\frac{AF}{AB}=\frac{t}{t+1}, \frac{BF}{AB}=\frac{1}{t+1}</math>.<br />
<br />
Now notice that <math>[DEF]=[ABC]-([BDF]+[CDE]+[AEF])</math>.<br />
<br />
We attempt to find the area of each of the smaller triangles. <br />
<br />
<br />
Notice that <math>\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}</math> using the ratios derived earlier.<br />
<br />
<br />
Similarly, <math>\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}</math> and <math>\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}</math>.<br />
<br />
<br />
Thus, <math>\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}</math>.<br />
<br />
Finally, we have <math>\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\boxed{\frac{rst+1}{(r+1)(s+1)(t+1)}}</math>.<br />
<br />
~@CoolJupiter<br />
==Proof 2==<br />
Proof by math_comb01<br />
Apply Barycentrics <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. also <math>D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {s}{s+1},0,\tfrac {1}{s+1}\right),F=\left(\tfrac {1}{t+1},\tfrac {t}{t+1},0\right)</math><br />
In the barycentrics, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we <cmath>\frac{[DEF]}{[ABC]}</cmath>=<math> \begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {s}{s+1}&0&\tfrac {1}{s+1}\\ \tfrac {1}{t+1}&\tfrac {t}{t+1}&0 \end{vmatrix}</math>=<math>\frac{1}{[s+1][r+1][t+1]}</math><math>+\frac{rst}{([s+1][r+1][t+1]}</math>=<math>\frac{rst+1}{([s+1][r+1][t+1]}</math><br />
~@Math_comb01<br />
<br />
=Other Remarks=<br />
This theorem is a generalization of the Wooga Looga Theorem, which @RedFireTruck claims to have "rediscovered". The link to the theorem can be found here:<br />
https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem<br />
<br />
Essentially, Wooga Looga is a special case of this, specifically when <math>r=s=t</math>.<br />
<br />
=Testimonials=<br />
<br />
This is Routh's theorem isn't it~ Ilovepizza2020<br />
<br />
Wow this generalization of my theorem is amazing. good job. - Foogle and Hoogle, Members of the Ooga Booga Tribe of The Caveman Society<br />
<br />
trivial by <math>\frac{1}{2}ab\sin(C)</math> but ok ~ bissue<br />
<br />
"Very nice theorem" - [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:12, 1 February 2021 (EST)</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=1586402021 AMC 10B Problems/Problem 182021-07-17T11:22:33Z<p>Somebodyyouusedtoknow: /* Solution 7 */</p>
<hr />
<div>==Problem==<br />
<br />
A fair <math>6</math>-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br />
<br />
<math>\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}</math><br />
<br />
<br />
==Solution 1==<br />
<br />
There is a <math>\frac36</math> chance that the first number we choose is even.<br />
<br />
There is a <math>\frac{2}{5}</math> chance that the next number that is distinct from the first is even.<br />
<br />
There is a <math>\frac{1}{4}</math> chance that the next number distinct from the first two is even.<br />
<br />
<math>\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{1}{20}}.</math><br />
<br />
~Tucker<br />
<br />
==Solution 2==<br />
<br />
Every set of three numbers chosen from <math>\{1,2,3,4,5,6\}</math> has an equal chance of being the first 3 distinct numbers rolled.<br />
<br />
Therefore, the probability that the first 3 distinct numbers are <math>\{2,4,6\}</math> is <math>\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}</math><br />
<br />
~kingofpineapplz<br />
<br />
==Solution 3 ==<br />
<br />
Note that the problem is basically asking us to find the probability that in some permutation of <math>1,2,3,4,5,6</math> that we get the three even numbers in the first three spots.<br />
<br />
There are <math>6!</math> ways to order the <math>6</math> numbers and <math>3!(3!)</math> ways to order the evens in the first three spots and the odds in the next three spots.<br />
<br />
Therefore the probability is <math>\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}</math>.<br />
<br />
<br />
--abhinavg0627<br />
<br />
==Solution 4==<br />
Let <math>P_n</math> denote the probability that the first odd number appears on roll <math>n</math> and all our conditions are met. We now proceed with complementary counting. <br />
<br />
For <math>n \le 3</math>, it's impossible to have all <math>3</math> evens appear before an odd. Note that for <math>n \ge 4,</math> <cmath>P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{1}}{3^{n-1}}\right)</cmath> since there's a <math>\frac {1}{2^{n}}</math> chance that the first odd appears on roll <math>n</math> (disregarding the other conditions) and the other term is subtracting the probability that less than <math>3</math> of the evens show up before the first odd roll. Simplifying, we arrive at <cmath>P_n= \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \frac {1}{2 \cdot 3^{n-2}} + \frac{1}{2^{n} \cdot 3^{n-2}}.</cmath><br />
<br />
Summing for all <math>n</math>, we get our answer of <cmath>\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \frac {1}{8} - \frac {1}{12} + \frac{1}{120} = \boxed{\textbf{(C) }\frac{1}{20}.}</cmath><br />
<br />
~ike.chen<br />
<br />
==Solution 5==<br />
Let <math> E_n </math> be that probability that the condition in the problem is satisfied given that we need <math> n </math> more distinct even numbers. Then, <br />
<cmath> E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0, </cmath><br />
since there is a <math> \frac{1}{3} </math> probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that <math> E_1=\frac{1}{4} </math>. <br />
<br />
We can apply the same concept for <math> E_2 </math> and <math> E_3 </math>. We find that <cmath> E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0, </cmath> and so <math> E_2=\frac{1}{10} </math>. Also, <cmath> E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0, </cmath> so <math> E_3=\frac{1}{20} </math>. Since the problem is asking for <math> E_3 </math>, our answer is <math> \boxed{\textbf{(C) }\frac{1}{20}} </math>. -BorealBear<br />
<br />
==Solution 1a (same as solution 1 but with a little more of explanation)==<br />
The probability of choosing an even number on the first turn is <math>\frac{1}{2}</math>, now since you already chose that number, it is irrelevant to the problem now, so, if you chose the number again, it doesn't really matter to our problem anymore. Now, with our remaining <math>5</math> numbers, the probability of choosing another even number is <math>\frac{2}{5}</math>, and again, after you have chosen that number, it is out of our problem. Now, you just have <math>4</math> numbers left and the probability of choosing the last even number is <math>\frac{1}{4}</math>, so the answer is <math>\frac{1}{2} \times \frac{2}{5} \times \frac{1}{4}</math> <math>=</math> <math>\frac{1}{20}</math>.<br />
<br />
~math31415926535<br />
<br />
==Solution 6 (Infinite Geometric Sequence Method)==<br />
Let's say that our even integers are found in the first <math>n</math> numbers where n must be greater than or equal to <math>3</math>. Then, we can form an argument based on this. There are <math>3^n</math> total ways to assign even numbers being <math>(</math>2<math>, </math>4<math>, </math>6<math>)</math> to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total <math>n</math> spaces and the case where there are only <math>2</math> distinct even integers present. There is <math>1</math> way we can have a single even integer in the entire <math>n</math> spaces, therefore giving us just <math>1</math> option for each of the three integers, so we have <math>3</math> total cases for this. Moreover, the amount of cases with just <math>2</math> distinct even integers is <math>2^n</math> but subtract the cases where all of the n spaces is either a single integer giving us <math>2^n-2</math>, but we multiply by <math>3C2</math> because of the ways to choose <math>2</math> distinct even integers that are used in the sequence of <math>n</math>. Finally, we have <math>\sum_{n=3}^{\infty} \frac{(3^n-3-3(2^n-2))}{6^n}</math> note: we must divide all of this by <math>6^n</math> for probability. Additionally, over the entire summation, we multiply by <math>1/2</math> because of the <math>1/2</math> probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get <math>1/20</math>. <math>\boxed{\textbf{(C) }\frac{1}{20}.}</math><br />
<br />
~Jske25<br />
<br />
https://www.youtube.com/watch?v=T-kdDFeHzns&t=17s<br />
<br />
==Solution 2a (same as 2 but with more explanation)==<br />
The question basically asks for the probability that <math>2</math>, <math>4</math>, and <math>6</math> are all rolled before <math>1</math>, <math>3</math> or <math>5</math>. This probability is equal to the probability that any 3 number combination is rolled before the remaining numbers are rolled (for example, rolling <math>1, 2</math> and <math>3</math> before <math>4</math>, <math>5</math> and <math>6</math>). There are <math>\binom{6}{3}</math> combinations possible and summing them all up must result in 1 so the probability of the specific combination <math>2</math> <math>4</math> <math>6</math> being chosen is<math>\boxed{\textbf{(C) }\frac{1}{20}.}</math><br />
<br />
~AwesomeK<br />
<br />
== Video Solution by hurdler (complementary probability) ==<br />
https://www.youtube.com/watch?v=k2Jy4ni9tK8<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/FV9AnyERgJQ?t=480<br />
<br />
~IceMatrix<br />
==Video Solution by Interstigation (Simple Bash With PIE)==<br />
(which stands for Principle of Inclusion and Exclusion)<br />
https://youtu.be/2SGmSYZ5bqU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=1586392021 AMC 10B Problems/Problem 182021-07-17T11:21:59Z<p>Somebodyyouusedtoknow: /* Solution 6 (same as solution 1 but with a little more of explanation) */</p>
<hr />
<div>==Problem==<br />
<br />
A fair <math>6</math>-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br />
<br />
<math>\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}</math><br />
<br />
<br />
==Solution 1==<br />
<br />
There is a <math>\frac36</math> chance that the first number we choose is even.<br />
<br />
There is a <math>\frac{2}{5}</math> chance that the next number that is distinct from the first is even.<br />
<br />
There is a <math>\frac{1}{4}</math> chance that the next number distinct from the first two is even.<br />
<br />
<math>\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}</math>, so the answer is <math>\boxed{\textbf{(C) }\frac{1}{20}}.</math><br />
<br />
~Tucker<br />
<br />
==Solution 2==<br />
<br />
Every set of three numbers chosen from <math>\{1,2,3,4,5,6\}</math> has an equal chance of being the first 3 distinct numbers rolled.<br />
<br />
Therefore, the probability that the first 3 distinct numbers are <math>\{2,4,6\}</math> is <math>\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}</math><br />
<br />
~kingofpineapplz<br />
<br />
==Solution 3 ==<br />
<br />
Note that the problem is basically asking us to find the probability that in some permutation of <math>1,2,3,4,5,6</math> that we get the three even numbers in the first three spots.<br />
<br />
There are <math>6!</math> ways to order the <math>6</math> numbers and <math>3!(3!)</math> ways to order the evens in the first three spots and the odds in the next three spots.<br />
<br />
Therefore the probability is <math>\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}</math>.<br />
<br />
<br />
--abhinavg0627<br />
<br />
==Solution 4==<br />
Let <math>P_n</math> denote the probability that the first odd number appears on roll <math>n</math> and all our conditions are met. We now proceed with complementary counting. <br />
<br />
For <math>n \le 3</math>, it's impossible to have all <math>3</math> evens appear before an odd. Note that for <math>n \ge 4,</math> <cmath>P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{1}}{3^{n-1}}\right)</cmath> since there's a <math>\frac {1}{2^{n}}</math> chance that the first odd appears on roll <math>n</math> (disregarding the other conditions) and the other term is subtracting the probability that less than <math>3</math> of the evens show up before the first odd roll. Simplifying, we arrive at <cmath>P_n= \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \frac {1}{2 \cdot 3^{n-2}} + \frac{1}{2^{n} \cdot 3^{n-2}}.</cmath><br />
<br />
Summing for all <math>n</math>, we get our answer of <cmath>\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \frac {1}{8} - \frac {1}{12} + \frac{1}{120} = \boxed{\textbf{(C) }\frac{1}{20}.}</cmath><br />
<br />
~ike.chen<br />
<br />
==Solution 5==<br />
Let <math> E_n </math> be that probability that the condition in the problem is satisfied given that we need <math> n </math> more distinct even numbers. Then, <br />
<cmath> E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0, </cmath><br />
since there is a <math> \frac{1}{3} </math> probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that <math> E_1=\frac{1}{4} </math>. <br />
<br />
We can apply the same concept for <math> E_2 </math> and <math> E_3 </math>. We find that <cmath> E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0, </cmath> and so <math> E_2=\frac{1}{10} </math>. Also, <cmath> E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0, </cmath> so <math> E_3=\frac{1}{20} </math>. Since the problem is asking for <math> E_3 </math>, our answer is <math> \boxed{\textbf{(C) }\frac{1}{20}} </math>. -BorealBear<br />
<br />
==Solution 1a (same as solution 1 but with a little more of explanation)==<br />
The probability of choosing an even number on the first turn is <math>\frac{1}{2}</math>, now since you already chose that number, it is irrelevant to the problem now, so, if you chose the number again, it doesn't really matter to our problem anymore. Now, with our remaining <math>5</math> numbers, the probability of choosing another even number is <math>\frac{2}{5}</math>, and again, after you have chosen that number, it is out of our problem. Now, you just have <math>4</math> numbers left and the probability of choosing the last even number is <math>\frac{1}{4}</math>, so the answer is <math>\frac{1}{2} \times \frac{2}{5} \times \frac{1}{4}</math> <math>=</math> <math>\frac{1}{20}</math>.<br />
<br />
~math31415926535<br />
<br />
==Solution 6 (Infinite Geometric Sequence Method)==<br />
Let's say that our even integers are found in the first <math>n</math> numbers where n must be greater than or equal to <math>3</math>. Then, we can form an argument based on this. There are <math>3^n</math> total ways to assign even numbers being <math>(</math>2<math>, </math>4<math>, </math>6<math>)</math> to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total <math>n</math> spaces and the case where there are only <math>2</math> distinct even integers present. There is <math>1</math> way we can have a single even integer in the entire <math>n</math> spaces, therefore giving us just <math>1</math> option for each of the three integers, so we have <math>3</math> total cases for this. Moreover, the amount of cases with just <math>2</math> distinct even integers is <math>2^n</math> but subtract the cases where all of the n spaces is either a single integer giving us <math>2^n-2</math>, but we multiply by <math>3C2</math> because of the ways to choose <math>2</math> distinct even integers that are used in the sequence of <math>n</math>. Finally, we have <math>\sum_{n=3}^{\infty} \frac{(3^n-3-3(2^n-2))}{6^n}</math> note: we must divide all of this by <math>6^n</math> for probability. Additionally, over the entire summation, we multiply by <math>1/2</math> because of the <math>1/2</math> probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get <math>1/20</math>. <math>\boxed{\textbf{(C) }\frac{1}{20}.}</math><br />
<br />
~Jske25<br />
<br />
https://www.youtube.com/watch?v=T-kdDFeHzns&t=17s<br />
<br />
==Solution 7==<br />
The question basically asks for the probability that <math>2</math>, <math>4</math>, and <math>6</math> are all rolled before <math>1</math>, <math>3</math> or <math>5</math>. This probability is equal to the probability that any 3 number combination is rolled before the remaining numbers are rolled (for example, rolling <math>1, 2</math> and <math>3</math> before <math>4</math>, <math>5</math> and <math>6</math>). There are <math>\binom{6}{3}</math> combinations possible and summing them all up must result in 1 so the probability of the specific combination <math>2</math> <math>4</math> <math>6</math> being chosen is<math>\boxed{\textbf{(C) }\frac{1}{20}.}</math><br />
<br />
~AwesomeK<br />
<br />
== Video Solution by hurdler (complementary probability) ==<br />
https://www.youtube.com/watch?v=k2Jy4ni9tK8<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/FV9AnyERgJQ?t=480<br />
<br />
~IceMatrix<br />
==Video Solution by Interstigation (Simple Bash With PIE)==<br />
(which stands for Principle of Inclusion and Exclusion)<br />
https://youtu.be/2SGmSYZ5bqU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2004_JBMO_Problems/Problem_1&diff=1561892004 JBMO Problems/Problem 12021-06-17T09:51:34Z<p>Somebodyyouusedtoknow: Fix to /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Prove that the inequality <cmath> \frac{ x+y}{x^2-xy+y^2 } \leq \frac{ 2\sqrt 2 }{\sqrt{ x^2 +y^2 } } </cmath> holds for all real numbers <math>x</math> and <math>y</math>, not both equal to 0.<br />
<br />
<br />
== Solution ==<br />
<br />
Since the inequality is homogeneous, we can assume WLOG that xy = 1.<br />
<br />
<br />
Now, substituting <math>m = (x+y)^2</math>, we have:<br />
<br />
<math>m = x^2 + y^2 + 2xy = x^2 + y^2 + 2 </math> <math> \geq 2\sqrt {xy} + 2 = 4</math>, thus we have <math>m \geq 4 </math><br />
<br />
<br />
Now squaring both sides of the inequality, we get:<br />
<cmath> \frac{m}{(m-3)^2 } \leq \frac{8}{m-2} </cmath><br />
after cross multiplication and simplification we get:<br />
<math>7m^2 -46m + 72 \geq 0</math><br />
or, <math>7(m-4)^2 +10(m-4) \geq 0</math> which is always true since <math>m \geq 4</math>.<br />
<br />
<br />
<math>Kris17</math><br />
<br />
==Solution 2==<br />
<br />
By Trivial Inequality,<br />
<cmath> (x - y)^2 \geq 0 \iff 2(x^2 - xy + y^2) \geq x^2 + y^2 \iff \frac{2}{x^2 + y^2} \geq \frac{1}{x^2 - xy + y^2}.</cmath><br />
<br />
Then by multiplying by <math>x + y</math> on both sides, we use the Trivial Inequality again to obtain <math>2(x^2 + y^2) \geq (x + y)^2</math> which means <cmath>\frac{x+y}{x^2 - xy + y^2} \leq \frac{2(x + y)}{x^2 + y^2} \leq \frac{2\sqrt{2(x^2+y^2)}}{x^2 + y^2}</cmath> which after simplifying, proves the problem.</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2004_JBMO_Problems/Problem_1&diff=1561882004 JBMO Problems/Problem 12021-06-17T09:50:23Z<p>Somebodyyouusedtoknow: A new /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Prove that the inequality <cmath> \frac{ x+y}{x^2-xy+y^2 } \leq \frac{ 2\sqrt 2 }{\sqrt{ x^2 +y^2 } } </cmath> holds for all real numbers <math>x</math> and <math>y</math>, not both equal to 0.<br />
<br />
<br />
== Solution ==<br />
<br />
Since the inequality is homogeneous, we can assume WLOG that xy = 1.<br />
<br />
<br />
Now, substituting <math>m = (x+y)^2</math>, we have:<br />
<br />
<math>m = x^2 + y^2 + 2xy = x^2 + y^2 + 2 </math> <math> \geq 2\sqrt {xy} + 2 = 4</math>, thus we have <math>m \geq 4 </math><br />
<br />
<br />
Now squaring both sides of the inequality, we get:<br />
<cmath> \frac{m}{(m-3)^2 } \leq \frac{8}{m-2} </cmath><br />
after cross multiplication and simplification we get:<br />
<math>7m^2 -46m + 72 \geq 0</math><br />
or, <math>7(m-4)^2 +10(m-4) \geq 0</math> which is always true since <math>m \geq 4</math>.<br />
<br />
<br />
<math>Kris17</math><br />
<br />
==Solution 2==<br />
<br />
By Trivial inequality,<br />
<cmath> (x - y)^2 \geq 0 \iff 2(x^2 - xy + y^2) \geq 0 \iff \frac{2}{x^2 + y^2} \geq \frac{1}{x^2 - xy + y^2}.</cmath><br />
<br />
Then by multiplying by <math>x + y</math> on both sides, we use the Trivial Inequality again to obtain <math>2(x^2 + y^2) \geq (x + y)^2</math> which means <cmath>\frac{x+y}{x^2 - xy + y^2} \leq \frac{2(x + y)}{x^2 + y^2} \leq \frac{2\sqrt{2(x^2+y^2)}}{x^2 + y^2}</cmath> which after simplifying, proves the problem.</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_3&diff=1465662021 AMC 10B Problems/Problem 32021-02-14T06:07:03Z<p>Somebodyyouusedtoknow: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the <math>28</math> students in the program, <math>25\%</math> of the juniors and <math>10\%</math> of the seniors are on the debate team. How many juniors are in the program?<br />
<br />
<math>\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20</math><br />
<br />
==Solution 1==<br />
Say there are <math>j</math> juniors and <math>s</math> seniors in the program. Converting percentages to fractions, <math>\frac{j}{4}</math> and <math>\frac{s}{10}</math> are on the debate team, and since an equal number of juniors and seniors are on the debate team, <math>\frac{j}{4} = \frac{s}{10}.</math><br />
<br />
Cross-multiplying and simplifying we get <math>5j=2s.</math> Additionally, since there are <math>28</math> students in the program, <math>j+s = 28.</math> It is now a matter of solving the system of equations <cmath>5j=2s</cmath><cmath>j+s=28,</cmath> and the solution is <math>j = 8, s = 20.</math> Since we want the number of juniors, the answer is <cmath>\boxed{(C) \text{ } 8}.</cmath><br />
<br />
-PureSwag<br />
<br />
==Solution 2 (Fast and not rigorous)==<br />
We immediately see that <math>E</math> is the only possible amount of seniors, as <math>10\%</math> can only correspond with an answer choice ending with <math>0</math>. Thus the number of seniors is <math>20</math> and the number of juniors is <math>28-20=8\rightarrow \boxed{C}</math>. ~samrocksnature<br />
<br />
== Video Solution by OmegaLearn (System of Equations) ==<br />
https://youtu.be/BtEF-hJBGV8<br />
<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=2|num-a=4}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_16&diff=1298602007 AMC 12B Problems/Problem 162020-07-30T19:40:49Z<p>Somebodyyouusedtoknow: /* Solution 2 */</p>
<hr />
<div>==Problem 16==<br />
Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?<br />
<br />
<math>\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ 18 \qquad \mathrm{(C)}\ 27 \qquad \mathrm{(D)}\ 54 \qquad \mathrm{(E)}\ 81</math><br />
<br />
==Solution==<br />
A tetrahedron has 4 sides.<br />
The ratio of the number of faces with each color must be one of the following:<br />
<br />
<math>4:0:0</math>, <math>3:1:0</math>, <math>2:2:0</math>, or <math>2:1:1</math><br />
<br />
<br />
<br />
The first ratio yields <math>3</math> appearances, one of each color.<br />
<br />
The second ratio yields <math>3\cdot 2 = 6</math> appearances, three choices for the first color, and two choices for the second.<br />
<br />
The third ratio yields <math>\binom{3}{2} = 3</math> appearances since the two colors are interchangeable.<br />
<br />
The fourth ratio yields <math>3</math> appearances. There are three choices for the first color, and since the second two colors are interchangeable, there is only one distinguishable pair that fits them.<br />
<br />
The total is <math>3 + 6 + 3 + 3 = 15</math> appearances <math>\Rightarrow \mathrm{(A)}</math><br />
<br />
==Solution 2==<br />
Every colouring can be represented in the form <math>(w,r,b)</math>, where <math>w</math> is the number of white faces, <math>r</math> is the number of red faces, and <math>b</math> is the number of blue faces. Every distinguishable colouring pattern can be represented like this in exactly one way, and every ordered whole number triple with a total sum of 4 represents exactly one colouring pattern (if two tetrahedra have rearranged colours on their faces, it is always possible to rotate one so that it matches the other).<br />
<br />
Therefore, the number of colourings is equal to the number of ways 3 distinguishable nonnegative integers can add to 4. If you have 6 cockroaches in a row, this number is equal to the number of ways to pick two of the cockroaches to eat for dinner (because the remaining cockroaches in between are separated in to three sections with a non-negative number of cockroaches each), which is <math>\binom{6}{2} = 15</math><br />
<br />
Alternative explanation to solution 2:<br />
A regular tetrahedron is the only platonic solid in which any of the faces is adjacent to all the other 3 faces. Hence we only need to think about the number of faces we can colour for each face. Let the number of faces coloured with red, blue and white be <math>r, b, w</math> respectively. So we are solving for the number of solutions of the equation:<br />
<math>r + b + w = 4</math><br />
for nonnegative integers <math>r, b, w</math>. By Stars and Bars, we obtain the final answer which is <math>\binom{6}{2} = 15/(A)</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2007|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_16&diff=1298592007 AMC 12B Problems/Problem 162020-07-30T19:38:19Z<p>Somebodyyouusedtoknow: /* Solution 3 */</p>
<hr />
<div>==Problem 16==<br />
Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?<br />
<br />
<math>\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ 18 \qquad \mathrm{(C)}\ 27 \qquad \mathrm{(D)}\ 54 \qquad \mathrm{(E)}\ 81</math><br />
<br />
==Solution==<br />
A tetrahedron has 4 sides.<br />
The ratio of the number of faces with each color must be one of the following:<br />
<br />
<math>4:0:0</math>, <math>3:1:0</math>, <math>2:2:0</math>, or <math>2:1:1</math><br />
<br />
<br />
<br />
The first ratio yields <math>3</math> appearances, one of each color.<br />
<br />
The second ratio yields <math>3\cdot 2 = 6</math> appearances, three choices for the first color, and two choices for the second.<br />
<br />
The third ratio yields <math>\binom{3}{2} = 3</math> appearances since the two colors are interchangeable.<br />
<br />
The fourth ratio yields <math>3</math> appearances. There are three choices for the first color, and since the second two colors are interchangeable, there is only one distinguishable pair that fits them.<br />
<br />
The total is <math>3 + 6 + 3 + 3 = 15</math> appearances <math>\Rightarrow \mathrm{(A)}</math><br />
<br />
==Solution 2==<br />
Every colouring can be represented in the form <math>(w,r,b)</math>, where <math>w</math> is the number of white faces, <math>r</math> is the number of red faces, and <math>b</math> is the number of blue faces. Every distinguishable colouring pattern can be represented like this in exactly one way, and every ordered whole number triple with a total sum of 4 represents exactly one colouring pattern (if two tetrahedra have rearranged colours on their faces, it is always possible to rotate one so that it matches the other).<br />
<br />
Therefore, the number of colourings is equal to the number of ways 3 distinguishable nonnegative integers can add to 4. If you have 6 cockroaches in a row, this number is equal to the number of ways to pick two of the cockroaches to eat for dinner (because the remaining cockroaches in between are separated in to three sections with a non-negative number of cockroaches each), which is <math>\binom{6}{2} = 15</math><br />
<br />
Alternative explanation to solution 2:<br />
A regular tetrahedron is the only platonic solid in which any of the faces is adjacent to all the other 3 faces. Hence we onlg need to think about the number of faces we can colour for each face. Let the number of faces coloured with red, blue and white be <math>r, b, w</math> respectively. So we are solving for the number of solutions of the equation:<br />
<math>r + b + w = 4</math><br />
for nonnegative integers <math>r, b, w</math>. By Stars and Bars, we obtain the final answer which is <math>\binom{6}{2} = 15/(A)</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2007|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=1998_JBMO_Problems/Problem_2&diff=1272691998 JBMO Problems/Problem 22020-07-02T12:31:45Z<p>Somebodyyouusedtoknow: /* Solution 3 */</p>
<hr />
<div>==Problem 2==<br />
<br />
Let <math>ABCDE</math> be a convex pentagon such that <math>AB=AE=CD=1</math>, <math>\angle ABC=\angle DEA=90^\circ</math> and <math>BC+DE=1</math>. Compute the area of the pentagon.<br />
<br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
<br />
Let <math>BC = a, ED = 1 - a</math><br />
<br />
Let <math>\angle DAC = X</math><br />
<br />
Applying cosine rule to <math>\triangle DAC</math> we get:<br />
<br />
<math>\cos X = \frac{AC ^ {2} + AD ^ {2} - DC ^ {2}}{ 2 \cdot AC \cdot AD }</math><br />
<br />
Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get:<br />
<br />
<math>\cos^{2} X = \frac{(1 - a - a ^ {2}) ^ {2}}{(1 + a^{2})(2 - 2a + a^{2})}</math><br />
<br />
From above, <math>\sin^{2} X = 1 - \cos^{2} X = \frac{1}{(1 + a^{2})(2 - 2a + a^{2})} = \frac{1}{AC^{2} \cdot AD^{2}}</math><br />
<br />
Thus, <math>\sin X \cdot AC \cdot AD = 1</math><br />
<br />
So, area of <math>\triangle DAC</math> = <math>\frac{1}{2}\cdot \sin X \cdot AC \cdot AD = \frac{1}{2}</math><br />
<br />
Let <math>AF</math> be the altitude of <math>\triangle DAC</math> from <math>A</math>. <br />
<br />
So <math>\frac{1}{2}\cdot DC\cdot AF = \frac{1}{2}</math><br />
<br />
This implies <math>AF = 1</math>. <br />
<br />
Since <math>AFCB</math> is a cyclic quadrilateral with <math>AB = AF</math>, <math>\triangle ABC</math> is congruent to <math>\triangle AFC</math>. <br />
Similarly <math>AEDF</math> is a cyclic quadrilateral and <math>\triangle AED</math> is congruent to <math>\triangle AFD</math>. <br />
<br />
So area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> = area of <math>\triangle ADC</math>.<br />
Thus area of pentagon <math>ABCD</math> = area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> + area of <math>\triangle DAC</math> = <math>\frac{1}{2}+\frac{1}{2} = 1</math><br />
<br />
<br />
<br />
By <math>Kris17</math><br />
<br />
=== Solution 2 ===<br />
<br />
Let <math>BC = x, DE = y</math>. Denote the area of <math>\triangle XYZ</math> by <math>[XYZ]</math>.<br />
<br />
<math>[ABC]+[AED]=\frac{1}{2}(x+y)=\frac{1}{2}</math><br />
<br />
<math>[ACD]</math> can be found by [[Heron's formula]].<br />
<br />
<math>AC=\sqrt{x^2+1}</math><br />
<br />
<math>AD=\sqrt{y^2+1}</math><br />
<br />
Let <math>AC=b, AD=c</math>.<br />
<br />
<cmath><br />
\begin{align*}<br />
[ACD]&=\frac{1}{4}\sqrt{(1+b+c)(-1+b+c)(1-b+c)(1+b-c)}\\<br />
&=\frac{1}{4}\sqrt{((b+c)^2-1)(1-(b-c)^2)}\\<br />
&=\frac{1}{4}\sqrt{(b+c)^2+(b-c)^2-(b^2-c^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2(b^2+c^2)-(b^2-c^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2(x^2+y^2+2)-(x^2-y^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2((x+y)^2-2xy+2)-(x+y)^2(x-y)^2-1}\\<br />
&=\frac{1}{4}\sqrt{5-4xy-(x-y)^2}\\<br />
&=\frac{1}{4}\sqrt{5-(x+y)^2}\\<br />
&=\frac{1}{2}<br />
\end{align*}<br />
</cmath><br />
<br />
Total area <math>=[ABC]+[AED]+[ACD]=\frac{1}{2}+\frac{1}{2}=1</math>.<br />
<br />
By durianice<br />
<br />
=== Solution 3 ===<br />
Construct <math>AD</math> and <math>AC</math> to partition the figure into <math>ABC</math>, <math>ACD</math> and <math>ADE</math>.<br />
<br />
Rotate <math>ADE</math> with centre <math>A</math> such that <math>AE</math> coincides with <math>AB</math> and <math>AD</math> is mapped to <math>AD'</math>. Hence the area of the pentagon is still preserved and it suffices to find the area of the quadrilateral <math>AD'CD</math>.<br />
<br />
Hence <math>[AD'C]</math> = <math>\frac{1}{2}</math> (<math>D'E + BC</math>)<math>AB</math>= <math>\frac{1}{2}</math><br />
<br />
Since <math>CD</math> = <math>CD'</math>, <math>AC</math> = <math>AC</math> and <math>AD</math> = <math>AD'</math>, by SSS Congruence, <math>ACD</math> and <math>ACD'</math> are congruent, so <math>[ACD]</math> = <math>\frac{1}{2}</math><br />
<br />
So the area of pentagon <math>ABCDE = \frac{1}{2} + \frac{1}{2} = 1</math>.<br />
<br />
- SomebodyYouUsedToKnow<br />
<br />
==See Also==<br />
{{JBMO box|year=1998|num-b=1|num-a=3|five=}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=1998_JBMO_Problems/Problem_2&diff=1272681998 JBMO Problems/Problem 22020-07-02T12:30:58Z<p>Somebodyyouusedtoknow: /* Solution 3 */</p>
<hr />
<div>==Problem 2==<br />
<br />
Let <math>ABCDE</math> be a convex pentagon such that <math>AB=AE=CD=1</math>, <math>\angle ABC=\angle DEA=90^\circ</math> and <math>BC+DE=1</math>. Compute the area of the pentagon.<br />
<br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
<br />
Let <math>BC = a, ED = 1 - a</math><br />
<br />
Let <math>\angle DAC = X</math><br />
<br />
Applying cosine rule to <math>\triangle DAC</math> we get:<br />
<br />
<math>\cos X = \frac{AC ^ {2} + AD ^ {2} - DC ^ {2}}{ 2 \cdot AC \cdot AD }</math><br />
<br />
Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get:<br />
<br />
<math>\cos^{2} X = \frac{(1 - a - a ^ {2}) ^ {2}}{(1 + a^{2})(2 - 2a + a^{2})}</math><br />
<br />
From above, <math>\sin^{2} X = 1 - \cos^{2} X = \frac{1}{(1 + a^{2})(2 - 2a + a^{2})} = \frac{1}{AC^{2} \cdot AD^{2}}</math><br />
<br />
Thus, <math>\sin X \cdot AC \cdot AD = 1</math><br />
<br />
So, area of <math>\triangle DAC</math> = <math>\frac{1}{2}\cdot \sin X \cdot AC \cdot AD = \frac{1}{2}</math><br />
<br />
Let <math>AF</math> be the altitude of <math>\triangle DAC</math> from <math>A</math>. <br />
<br />
So <math>\frac{1}{2}\cdot DC\cdot AF = \frac{1}{2}</math><br />
<br />
This implies <math>AF = 1</math>. <br />
<br />
Since <math>AFCB</math> is a cyclic quadrilateral with <math>AB = AF</math>, <math>\triangle ABC</math> is congruent to <math>\triangle AFC</math>. <br />
Similarly <math>AEDF</math> is a cyclic quadrilateral and <math>\triangle AED</math> is congruent to <math>\triangle AFD</math>. <br />
<br />
So area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> = area of <math>\triangle ADC</math>.<br />
Thus area of pentagon <math>ABCD</math> = area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> + area of <math>\triangle DAC</math> = <math>\frac{1}{2}+\frac{1}{2} = 1</math><br />
<br />
<br />
<br />
By <math>Kris17</math><br />
<br />
=== Solution 2 ===<br />
<br />
Let <math>BC = x, DE = y</math>. Denote the area of <math>\triangle XYZ</math> by <math>[XYZ]</math>.<br />
<br />
<math>[ABC]+[AED]=\frac{1}{2}(x+y)=\frac{1}{2}</math><br />
<br />
<math>[ACD]</math> can be found by [[Heron's formula]].<br />
<br />
<math>AC=\sqrt{x^2+1}</math><br />
<br />
<math>AD=\sqrt{y^2+1}</math><br />
<br />
Let <math>AC=b, AD=c</math>.<br />
<br />
<cmath><br />
\begin{align*}<br />
[ACD]&=\frac{1}{4}\sqrt{(1+b+c)(-1+b+c)(1-b+c)(1+b-c)}\\<br />
&=\frac{1}{4}\sqrt{((b+c)^2-1)(1-(b-c)^2)}\\<br />
&=\frac{1}{4}\sqrt{(b+c)^2+(b-c)^2-(b^2-c^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2(b^2+c^2)-(b^2-c^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2(x^2+y^2+2)-(x^2-y^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2((x+y)^2-2xy+2)-(x+y)^2(x-y)^2-1}\\<br />
&=\frac{1}{4}\sqrt{5-4xy-(x-y)^2}\\<br />
&=\frac{1}{4}\sqrt{5-(x+y)^2}\\<br />
&=\frac{1}{2}<br />
\end{align*}<br />
</cmath><br />
<br />
Total area <math>=[ABC]+[AED]+[ACD]=\frac{1}{2}+\frac{1}{2}=1</math>.<br />
<br />
By durianice<br />
<br />
=== Solution 3 ===<br />
Construct <math>AD</math> and <math>AC</math> to partition the figure into <math>ABC</math>, <math>ACD</math> and <math>ADE</math>.<br />
<br />
Rotate <math>ADE</math> with centre <math>A</math> such that <math>AE</math> coincides with <math>AB</math> and <math>AD</math> is mapped to <math>AD'</math>. Hence the area of the pentagon is still preserved and it suffices to find the area of the pentagon <math>AD'BCD</math>.<br />
<br />
Hence <math>[AD'C]</math> = <math>\frac{1}{2}</math> (<math>D'E + BC</math>)<math>AB</math>= <math>\frac{1}{2}</math><br />
<br />
Since <math>CD</math> = <math>CD'</math>, <math>AC</math> = <math>AC</math> and <math>AD</math> = <math>AD'</math>, by SSS Congruence, <math>ACD</math> and <math>ACD'</math> are congruent, so <math>[ACD]</math> = <math>\frac{1}{2}</math><br />
<br />
So the area of pentagon <math>ABCDE = \frac{1}{2} + \frac{1}{2} = 1</math>.<br />
<br />
- SomebodyYouUsedToKnow<br />
<br />
==See Also==<br />
{{JBMO box|year=1998|num-b=1|num-a=3|five=}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=1998_JBMO_Problems/Problem_2&diff=1272671998 JBMO Problems/Problem 22020-07-02T12:29:43Z<p>Somebodyyouusedtoknow: Added a new solution</p>
<hr />
<div>==Problem 2==<br />
<br />
Let <math>ABCDE</math> be a convex pentagon such that <math>AB=AE=CD=1</math>, <math>\angle ABC=\angle DEA=90^\circ</math> and <math>BC+DE=1</math>. Compute the area of the pentagon.<br />
<br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
<br />
Let <math>BC = a, ED = 1 - a</math><br />
<br />
Let <math>\angle DAC = X</math><br />
<br />
Applying cosine rule to <math>\triangle DAC</math> we get:<br />
<br />
<math>\cos X = \frac{AC ^ {2} + AD ^ {2} - DC ^ {2}}{ 2 \cdot AC \cdot AD }</math><br />
<br />
Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get:<br />
<br />
<math>\cos^{2} X = \frac{(1 - a - a ^ {2}) ^ {2}}{(1 + a^{2})(2 - 2a + a^{2})}</math><br />
<br />
From above, <math>\sin^{2} X = 1 - \cos^{2} X = \frac{1}{(1 + a^{2})(2 - 2a + a^{2})} = \frac{1}{AC^{2} \cdot AD^{2}}</math><br />
<br />
Thus, <math>\sin X \cdot AC \cdot AD = 1</math><br />
<br />
So, area of <math>\triangle DAC</math> = <math>\frac{1}{2}\cdot \sin X \cdot AC \cdot AD = \frac{1}{2}</math><br />
<br />
Let <math>AF</math> be the altitude of <math>\triangle DAC</math> from <math>A</math>. <br />
<br />
So <math>\frac{1}{2}\cdot DC\cdot AF = \frac{1}{2}</math><br />
<br />
This implies <math>AF = 1</math>. <br />
<br />
Since <math>AFCB</math> is a cyclic quadrilateral with <math>AB = AF</math>, <math>\triangle ABC</math> is congruent to <math>\triangle AFC</math>. <br />
Similarly <math>AEDF</math> is a cyclic quadrilateral and <math>\triangle AED</math> is congruent to <math>\triangle AFD</math>. <br />
<br />
So area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> = area of <math>\triangle ADC</math>.<br />
Thus area of pentagon <math>ABCD</math> = area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> + area of <math>\triangle DAC</math> = <math>\frac{1}{2}+\frac{1}{2} = 1</math><br />
<br />
<br />
<br />
By <math>Kris17</math><br />
<br />
=== Solution 2 ===<br />
<br />
Let <math>BC = x, DE = y</math>. Denote the area of <math>\triangle XYZ</math> by <math>[XYZ]</math>.<br />
<br />
<math>[ABC]+[AED]=\frac{1}{2}(x+y)=\frac{1}{2}</math><br />
<br />
<math>[ACD]</math> can be found by [[Heron's formula]].<br />
<br />
<math>AC=\sqrt{x^2+1}</math><br />
<br />
<math>AD=\sqrt{y^2+1}</math><br />
<br />
Let <math>AC=b, AD=c</math>.<br />
<br />
<cmath><br />
\begin{align*}<br />
[ACD]&=\frac{1}{4}\sqrt{(1+b+c)(-1+b+c)(1-b+c)(1+b-c)}\\<br />
&=\frac{1}{4}\sqrt{((b+c)^2-1)(1-(b-c)^2)}\\<br />
&=\frac{1}{4}\sqrt{(b+c)^2+(b-c)^2-(b^2-c^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2(b^2+c^2)-(b^2-c^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2(x^2+y^2+2)-(x^2-y^2)^2-1}\\<br />
&=\frac{1}{4}\sqrt{2((x+y)^2-2xy+2)-(x+y)^2(x-y)^2-1}\\<br />
&=\frac{1}{4}\sqrt{5-4xy-(x-y)^2}\\<br />
&=\frac{1}{4}\sqrt{5-(x+y)^2}\\<br />
&=\frac{1}{2}<br />
\end{align*}<br />
</cmath><br />
<br />
Total area <math>=[ABC]+[AED]+[ACD]=\frac{1}{2}+\frac{1}{2}=1</math>.<br />
<br />
By durianice<br />
<br />
=== Solution 3 ===<br />
Construct <math>AD</math> and <math>AC</math> to partition the figure into <math>ABC</math>, <math>ACD</math> and <math>ADE</math>.<br />
<br />
Rotate <math>ADE</math> with centre <math>A</math> such that <math>AE</math> coincides with <math>AB</math> and <math>AD</math> is mapped to <math>AD'</math>. Hence the area of the pentagon is still preserved and it suffices to find the area of the pentagon <math>AD'BCD</math>.<br />
<br />
Hence <math>[AD'C]</math> = <math>\frac{1}{2}</math> (<math>D'E + BC</math>)<math>AB</math>= <math>\frac{1}{2}</math><br />
<br />
Since <math>CD</math> = <math>CD'</math>, <math>AC</math> = <math>AC</math> and <math>AD</math> = <math>AD'</math>, by SSS Congruence, <math>ACD</math> and <math>ACD'</math> are congruent, so [ACD]&=\frac{1}{2}\<br />
<br />
So the area of pentagon <math>ABCDE = \frac{1}{2} + \frac{1}{2} = 1</math>.<br />
<br />
- SomebodyYouUsedToKnow<br />
<br />
==See Also==<br />
{{JBMO box|year=1998|num-b=1|num-a=3|five=}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_25&diff=1257482019 AMC 12B Problems/Problem 252020-06-17T21:35:12Z<p>Somebodyyouusedtoknow: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCD</math> be a convex quadrilateral with <math>BC=2</math> and <math>CD=6.</math> Suppose that the centroids of <math>\triangle ABC,\triangle BCD,</math> and <math>\triangle ACD</math> form the vertices of an equilateral triangle. What is the maximum possible value of the area of <math>ABCD</math>?<br />
<br />
<math>\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30</math><br />
<br />
==Solution 1 (vectors)==<br />
Place an origin at <math>A</math>, and assign position vectors of <math>B = \vec{p}</math> and <math>D = \vec{q}</math>. Since <math>AB</math> is not parallel to <math>AD</math>, vectors <math>\vec{p}</math> and <math>\vec{q}</math> are linearly independent, so we can write <math>C = m\vec{p} + n\vec{q}</math> for some constants <math>m</math> and <math>n</math>. Now, recall that the centroid of a triangle <math>\triangle XYZ</math> has position vector <math>\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)</math>. <br />
<br />
Thus the centroid of <math>\triangle ABC</math> is <math>g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}</math>; the centroid of <math>\triangle BCD</math> is <math>g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>; and the centroid of <math>\triangle ACD</math> is <math>g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>. <br />
<br />
Hence <math>\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}</math>, <math>\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}</math>, and <math>\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}</math>. For <math>\triangle G_{1}G_{2}G_{3}</math> to be equilateral, we need <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD</math>. Further, <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD</math>. Hence we have <math>AB = AD = BD</math>, so <math>\triangle ABD</math> is equilateral.<br />
<br />
Now let the side length of <math>\triangle ABD</math> be <math>k</math>, and let <math>\angle BCD = \theta</math>. By the Law of Cosines in <math>\triangle BCD</math>, we have <math>k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}</math>. Since <math>\triangle ABD</math> is equilateral, its area is <math>\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}</math>, while the area of <math>\triangle BCD</math> is <math>\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}</math>. Thus the total area of <math>ABCD</math> is <math>10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\sin{\theta} \frac{1}{2} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}</math>, where in the last step we used the subtraction formula for <math>\sin</math>. Alternatively, we can use calculus to find the local maximum. Observe that <math>\sin{\left(\theta-60^{\circ}\right)}</math> has maximum value <math>1</math> when e.g. <math>\theta = 150^{\circ}</math>, which is a valid configuration, so the maximum area is <math>10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}</math>.<br />
<br />
==Solution 2==<br />
<br />
Let <math>G_1</math>, <math>G_2</math>, <math>G_3</math> be the centroids of <math>ABC</math>, <math>BCD</math>, and <math>CDA</math> respectively, and let <math>M</math> be the midpoint of <math>BC</math>. <math>A</math>, <math>G_1</math>, and <math>M</math> are collinear due to well-known properties of the centroid. Likewise, <math>D</math>, <math>G_2</math>, and <math>M</math> are collinear as well. Because (as is also well-known) <math>3MG_1 = AM</math> and <math>3MG_2 = DM</math>, we have <math>\triangle MG_1G_2\sim\triangle MAD</math>. This implies that <math>AD</math> is parallel to <math>G_1G_2</math>, and in terms of lengths, <math>AD = 3G_1G_2</math>. (SAS Similarity)<br />
<br />
We can apply the same argument to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>, concluding that <math>AB</math> is parallel to <math>G_2G_3</math> and <math>AB = 3G_2G_3</math>. Because <math>3G_1G_2 = 3G_2G_3</math> (due to the triangle being equilateral), <math>AB = AD</math>, and the pair of parallel lines preserve the <math>60^{\circ}</math> angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore <math>\triangle BAD</math> is equilateral.<br />
<br />
At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:<br />
<br />
Let <math>BD = 2x</math>, where <math>2 < x < 4</math> due to the Triangle Inequality in <math>\triangle BCD</math>. By breaking the quadrilateral into <math>\triangle ABD</math> and <math>\triangle BCD</math>, we can create an expression for the area of <math>ABCD</math>. We use the formula for the area of an equilateral triangle given its side length to find the area of <math>\triangle ABD</math> and Heron's formula to find the area of <math>\triangle BCD</math>.<br />
<br />
After simplifying,<br />
<br />
<cmath>[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}</cmath><br />
<br />
Substituting <math>k = x^2 - 10</math>, the expression becomes<br />
<br />
<cmath>[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}</cmath><br />
<br />
We can ignore the <math>10\sqrt{3}</math> for now and focus on <math>k\sqrt{3} + \sqrt{36 - k^2}</math>.<br />
<br />
By the Cauchy-Schwarz inequality,<br />
<br />
<cmath>\left(k\sqrt 3 + \sqrt{36-k^2}\right)^2 \leq \left(\left(\sqrt{3}\right)^2+1^2\right)\left(\left(k\right)^2 + \left(\sqrt{36-k^2}\right)^2\right)</cmath><br />
<br />
The RHS simplifies to <math>12^2</math>, meaning the maximum value of <math>k\sqrt{3} + \sqrt{36 - k^2}</math> is <math>12</math>.<br />
<br />
Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=1257472019 AMC 10B Problems/Problem 252020-06-17T21:09:43Z<p>Somebodyyouusedtoknow: /* Solution 1 (recursion) */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br />
<br />
==Problem==<br />
<br />
How many sequences of <math>0</math>s and <math>1</math>s of length <math>19</math> are there that begin with a <math>0</math>, end with a <math>0</math>, contain no two consecutive <math>0</math>s, and contain no three consecutive <math>1</math>s?<br />
<br />
<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math><br />
<br />
==Solution 1 (recursion)==<br />
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.<br />
Thus we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, where <math>f(n)</math> is the number of valid sequences of length <math>n</math>.<br />
<br />
This is because for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> and the resulting sequence will still satisfy the given conditions.<br />
<br />
It is easy to find <math>f(5) = 1</math> with the only possible sequence being <math>01010</math> and <math>f(6) = 2</math> with the only two possible sequences being <math>011010</math> and <math>010110</math> by hand, and then by the recursive formula, we have <math>f(19) = \boxed{\textbf{(C) }65}</math>.<br />
<br />
*This solution is invalid. It requires knowing <math>f(7) = 1</math>. After doing that, following the recursion formula we obtain <math>f(19) = 53</math> which is not the correct answer. Instead, refer to the solutions below, in the meantime.<br />
<br />
==Solution 2 (casework)==<br />
After any particular <math>0</math>, the next <math>0</math> in the sequence must appear exactly <math>2</math> or <math>3</math> positions down the line. In this case, we start at position <math>1</math> and end at position <math>19</math>, i.e. we move a total of <math>18</math> positions down the line. Therefore, we must add a series of <math>2</math>s and <math>3</math>s to get <math>18</math>. There are a number of ways to do this:<br />
<br />
'''Case 1''': nine <math>2</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.<br />
<br />
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose4} = 35</math> ways to arrange them.<br />
<br />
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.<br />
<br />
==Solution 3 (casework and blocks)==<br />
We can simplify the original problem into a problem where there are <math>2^{17}</math> binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of <math>0</math>s, <math>1</math>s, and <math>11</math>s. Now, we use casework: <br />
<br />
'''Case 1''': Alternating 1s and 0s. There is simply 1 way to do this: <math>0101010101010101010</math>. <br />
Now, we note that there cannot be only one block of <math>11</math> in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of <math>11</math>s this cannot be satisfied. This is true for all odd numbers of <math>11</math> blocks. <br />
<br />
'''Case 2''': There are 2 <math>11</math> blocks. Using the zeroes in the sequence as dividers, we have a sample as <math>0110110101010101010</math>. We know there are 8 places for <math>11</math>s, which will be filled by <math>1</math>s if the <math>11</math>s don't fill them. This is <math>{8\choose2} = 28</math> ways. <br />
<br />
'''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math>11</math> blocks. <br />
<br />
'''Case 4''': No single <math>1</math> blocks, only <math>11</math> blocks. There is simply one case for this, which is <math>0110110110110110110</math>. <br />
<br />
Adding these four cases, we have <math>1+28+35+1=\boxed{\textbf{(C) }65}</math> as our final answer. <br />
<br />
~Equinox8<br />
<br />
==Video Solution==<br />
For those who want a video solution: https://youtu.be/VamT49PjmdI<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Somebodyyouusedtoknowhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=1257462019 AMC 10B Problems/Problem 252020-06-17T21:03:44Z<p>Somebodyyouusedtoknow: /* Solution 1 (recursion) */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br />
<br />
==Problem==<br />
<br />
How many sequences of <math>0</math>s and <math>1</math>s of length <math>19</math> are there that begin with a <math>0</math>, end with a <math>0</math>, contain no two consecutive <math>0</math>s, and contain no three consecutive <math>1</math>s?<br />
<br />
<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math><br />
<br />
==Solution 1 (recursion)==<br />
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.<br />
Thus we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, where <math>f(n)</math> is the number of valid sequences of length <math>n</math>.<br />
<br />
This is because for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> and the resulting sequence will still satisfy the given conditions.<br />
<br />
It is easy to find <math>f(5) = 1</math> with the only possible sequence being <math>01010</math> and <math>f(6) = 2</math> with the only two possible sequences being <math>011010</math> and <math>010110</math> by hand, and then by the recursive formula, we have <math>f(19) = \boxed{\textbf{(C) }65}</math>.<br />
<br />
*This solution is invalid. It requires knowing <math>f(7) = 1</math>. After doing that, following the recursion formula we obtain <math>f(19) = 53</math> which is not the correct answer.<br />
<br />
==Solution 2 (casework)==<br />
After any particular <math>0</math>, the next <math>0</math> in the sequence must appear exactly <math>2</math> or <math>3</math> positions down the line. In this case, we start at position <math>1</math> and end at position <math>19</math>, i.e. we move a total of <math>18</math> positions down the line. Therefore, we must add a series of <math>2</math>s and <math>3</math>s to get <math>18</math>. There are a number of ways to do this:<br />
<br />
'''Case 1''': nine <math>2</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.<br />
<br />
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose4} = 35</math> ways to arrange them.<br />
<br />
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.<br />
<br />
==Solution 3 (casework and blocks)==<br />
We can simplify the original problem into a problem where there are <math>2^{17}</math> binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of <math>0</math>s, <math>1</math>s, and <math>11</math>s. Now, we use casework: <br />
<br />
'''Case 1''': Alternating 1s and 0s. There is simply 1 way to do this: <math>0101010101010101010</math>. <br />
Now, we note that there cannot be only one block of <math>11</math> in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of <math>11</math>s this cannot be satisfied. This is true for all odd numbers of <math>11</math> blocks. <br />
<br />
'''Case 2''': There are 2 <math>11</math> blocks. Using the zeroes in the sequence as dividers, we have a sample as <math>0110110101010101010</math>. We know there are 8 places for <math>11</math>s, which will be filled by <math>1</math>s if the <math>11</math>s don't fill them. This is <math>{8\choose2} = 28</math> ways. <br />
<br />
'''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math>11</math> blocks. <br />
<br />
'''Case 4''': No single <math>1</math> blocks, only <math>11</math> blocks. There is simply one case for this, which is <math>0110110110110110110</math>. <br />
<br />
Adding these four cases, we have <math>1+28+35+1=\boxed{\textbf{(C) }65}</math> as our final answer. <br />
<br />
~Equinox8<br />
<br />
==Video Solution==<br />
For those who want a video solution: https://youtu.be/VamT49PjmdI<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Somebodyyouusedtoknow