https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Icepenguin&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T20:47:57ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Aime&diff=146823Aime2021-02-15T17:35:11Z<p>Icepenguin: Redirected page to American Invitational Mathematics Examination</p>
<hr />
<div>#REDIRECT [[American Invitational Mathematics Examination]]</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120196User:Icepenguin2020-03-27T19:09:32Z<p>Icepenguin: </p>
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<div>Content Deleted (page was becoming too active)<br />
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But still, please friend me. I accept all friend requests.</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120195User:Icepenguin2020-03-27T19:08:34Z<p>Icepenguin: </p>
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<div>Deleted (page was becoming too active)<br />
<br />
But still, please friend me. I accept all friend requests.</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120112User:Icepenguin2020-03-25T21:29:31Z<p>Icepenguin: </p>
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<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 6 \text{ different users.} \text{Please do not troll.}</cmath><br />
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Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}=33</cmath><br />
<br />
Thank you for who ever figured it out. Choosing new number every day from now on. Will post new anagram. Here you go:<br />
<br />
<cmath>\text{Peoe wolves minus west tent}</cmath><br />
<br />
>>>>> Oh, and I forgot. Please friend me. <<<<<<br />
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Wall of shame (users who trolled):<br />
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1. Bibear</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120111User:Icepenguin2020-03-25T20:02:26Z<p>Icepenguin: </p>
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<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 6 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}=33</cmath><br />
<br />
Thank you for who ever figured it out. Choosing new number every day from now on. Will post new anagram. Here you go:<br />
<br />
<cmath>\text{Peoe wolves minus west tent}</cmath><br />
<br />
Wall of shame (users who trolled):<br />
<br />
1. Bibear</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120110User:Icepenguin2020-03-25T19:49:26Z<p>Icepenguin: </p>
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<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 6 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}=33</cmath><br />
<br />
Thank you for who ever figured it out. Choosing new number every day from now on. Will post new anagram.<br />
<br />
Wall of shame (users who trolled):<br />
<br />
1. Bibear</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120109User:Icepenguin2020-03-25T19:48:44Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 6 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}=33</cmath><br />
<br />
Thank you for who ever figured it out. Choosing new number every day from now on. Will post new anagram.<br />
Wall of shame (users who trolled):<br />
<br />
1. Bibear</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120073User:Icepenguin2020-03-25T02:51:11Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 2 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath><br />
<br />
Wall of shame (users who trolled):<br />
<br />
1. Bibear</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120072User:Icepenguin2020-03-25T02:49:32Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 2 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath><br />
<br />
Wall of shame (users who trolled):<br />
<br />
1. bibear</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120071User:Icepenguin2020-03-25T02:49:03Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 2 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath><br />
<br />
Wall of shame (people who trolled):<br />
<br />
1. bibear</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120070User:Icepenguin2020-03-25T02:48:13Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 2 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120069User:Icepenguin2020-03-25T02:45:25Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 2 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120068User:Icepenguin2020-03-25T02:44:49Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 2 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120067User:Icepenguin2020-03-25T02:44:24Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 1 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120063User:Icepenguin2020-03-24T23:29:27Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 1 \text{ different users.} </cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120058User:Icepenguin2020-03-24T22:43:04Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 1 \text{ different users.} </cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>42</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120057User:Icepenguin2020-03-24T22:42:33Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 1 \text{ different users.} </cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>42</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram:<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120056User:Icepenguin2020-03-24T22:35:44Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 1 \text{ different users.} </cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>42</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram:<br />
<br />
<cmath>\text{Burn cinematic foe moths}</cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120052User:Icepenguin2020-03-24T22:28:29Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 1 \text{ different users.} </cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>42</math></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120051User:Icepenguin2020-03-24T22:26:27Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 1 \text{ different users.} </cmath></div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120050User:Icepenguin2020-03-24T22:26:14Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, please add one to the number below.<br />
<br />
<math></math> \text{This page has been visited by } 1 \text{ different users.}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120049User:Icepenguin2020-03-24T22:24:36Z<p>Icepenguin: </p>
<hr />
<div>It's fun to have a user page all to yourself<br />
<br />
Anyway, welcome! If you are a registered user and are reading this, <br />
TBA</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120048User:Icepenguin2020-03-24T22:23:11Z<p>Icepenguin: Created page with "It's fun to have a user page all to yourself"</p>
<hr />
<div>It's fun to have a user page all to yourself</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_6&diff=932172018 AMC 10B Problems/Problem 62018-03-13T22:37:26Z<p>Icepenguin: /* Solution 1 */</p>
<hr />
<div>A box contains <math>5</math> chips, numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds <math>4</math>. What is the probability that <math>3</math> draws are required?<br />
<br />
<math>\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}</math><br />
<br />
== Solution 1 ==<br />
<br />
Notice that the only four ways such that no more than <math>2</math> draws are required are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>3,1</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
Notice that only the first two draws are important, it doesn't matter what number we get third because no matter what combination of <math>3</math> numbers is picked, the sum will always be greater than 5. Also, note that it is necessary to draw a <math>1</math> in order to have 3 draws, otherwise <math>5</math> will be attainable in two or less draws. So the probability of getting a <math>1</math> is <math>\frac{1}{5}</math>. It is necessary to pull either a <math>2</math> or <math>3</math> on the next draw and the probability of that is <math>\frac{1}{2}</math>. But, the order of the draws can be switched so we get:<br />
<br />
<math>\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math><br />
<br />
By: Soccer_JAMS<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=931812018 AMC 10B Problems/Problem 242018-03-12T17:22:51Z<p>Icepenguin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote by <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(3/4,3sqrt(3)/4);<br />
O=(3/4,sqrt(3)/4);<br />
P=(3/8,7sqrt(3)/8);<br />
Q=(9/8, 3sqrt(3)/8);<br />
R=(0,sqrt(3)/4);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, NW);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
label("$P$", P, NNW);<br />
label("$Q$", Q, ESE);<br />
label("$R$", R, SW);<br />
fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br />
draw(A--B--C--D--E--F--cycle);<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O--cycle);<br />
<br />
</asy><br />
<br />
The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN</math>, <math>\triangle NQO</math>, and <math>\triangle ORM</math>).<br />
<br />
Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio. Thus, it must also divide transversal <math>\overline {AC}</math> and transversal <math>\overline {CO}</math> into a <math>1:1</math> ratio. By symmetry, the same applies for <math>\overline {CE}</math> and <math>\overline {EA}</math> as well as <math>\overline {EM}</math> and <math>\overline {AN}</math><br />
<br />
<br />
In <math>\triangle ACE</math>, we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}</math>. Our desired area becomes <br />
<br />
<cmath>(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}</cmath><br />
<br />
==Solution 2 ==<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (<math>AXFZ</math>, <math>XBCY</math>, and <math>ZYED</math>), and 3 right triangles, with one right angle on each of <math>X</math>, <math>Y</math>, and <math>Z</math>. <br />
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal, which has length 2) with a height of <math>\frac{\sqrt{3}}{4}</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>\frac{5\sqrt{3}}{16}</math> for a total area of <math>\frac{15\sqrt{3}}{16}.</math> (Alternatively, we could have calculated the area of hexagon <math>ABCDEF</math> and subtracted the area of <math>\triangle XYZ</math>, which, as we showed before, had a side length of 3/2). <br />
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of <math>YC = 1/2.</math> Using similar triangles we calculate the base to be 1/4 and the height to be <math>\frac{\sqrt{3}}{4}</math> giving us an area of <math>\frac{\sqrt{3}}{32}</math> per triangle, and a total area of <math>3\frac{\sqrt{3}}{32}</math>. Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}</math>. Finding the total area, we get <math>6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}</math>. Taking the complement, we get <math>\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \frac{15\sqrt{3}}{32} = (C)\frac{15}{32}\sqrt{3}</math><br />
<br />
==Solution 3 (Trig)==<br />
Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br />
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^{\textrm{o}}</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^{\textrm{o}}</math>, and thus the side length of this triangle is <math>1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}</math>. So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}</math><br />
Now let's find the area of the smaller triangles. Notice, triangle <math>ACE</math> cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then <math>\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}</math> and the sum of the areas is <math>3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}</math><br />
Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=931802018 AMC 10B Problems/Problem 242018-03-12T17:20:39Z<p>Icepenguin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote by <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(3/4,3sqrt(3)/4);<br />
O=(3/4,sqrt(3)/4);<br />
P=(3/8,7sqrt(3)/8);<br />
Q=(9/8, 3sqrt(3)/8);<br />
R=(0,sqrt(3)/4);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, NW);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
label("$P$", P, NNW);<br />
label("$Q$", Q, ESE);<br />
label("$R$", R, SW);<br />
fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br />
draw(A--B--C--D--E--F--cycle);<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O--cycle);<br />
<br />
</asy><br />
<br />
The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN</math>, <math>\triangle NQO</math>, and <math>\triangle ORM</math>).<br />
<br />
Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio. Thus, it must also divide transversal <math>\overline {AC}</math> and transversal <math>\overline {CO}</math> into a <math>1:1</math> ratio. By symmetry, the same applies for <math>\overline {CE}</math> and <math>\overline {EA}</math> as well as <math>\overline {EM}</math> and <math>\overline {AN}</math><br />
<br />
<br />
In <math>\triangle ACE</math>, we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}</math>. Our desired area becomes <br />
<br />
<cmath>(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}</cmath><br />
<br />
==Solution 2 ==<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (<math>AXFZ</math>, <math>XBCY</math>, and <math>ZYED</math>), and 3 right triangles, with one right angle on each of <math>X</math>, <math>Y</math>, and <math>Z</math>. <br />
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal, which has length 2) with a height of <math>\sqrt{3}/4</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>\frac{5\sqrt{3}}{16}</math> for a total area of <math>\frac{15\sqrt{3}}{16}.</math> (Alternatively, we could have calculated the area of hexagon <math>ABCDEF</math> and subtracted the area of <math>\triangle XYZ</math>, which, as we showed before, had a side length of 3/2). <br />
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of <math>YC = 1/2.</math> Using similar triangles we calculate the base to be 1/4 and the height to be <math>\frac{\sqrt{3}}{4}</math> giving us an area of <math>\frac{\sqrt{3}}{32}</math> per triangle, and a total area of <math>3\frac{\sqrt{3}}{32}</math>. Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}</math>. Finding the total area, we get <math>6*1^2*\sqrt{3}/4=3\sqrt{3}/2</math>. Taking the complement, we get <math>3\sqrt{3}/2-33\sqrt{3}/32=15\sqrt{3}/32 = (C)\frac{15}{32}\sqrt{3}</math><br />
<br />
==Solution 3 (Trig)==<br />
Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br />
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^{\textrm{o}}</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^{\textrm{o}}</math>, and thus the side length of this triangle is <math>1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}</math>. So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}</math><br />
Now let's find the area of the smaller triangles. Notice, triangle <math>ACE</math> cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then <math>\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}</math> and the sum of the areas is <math>3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}</math><br />
Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=931792018 AMC 10B Problems/Problem 242018-03-12T17:19:26Z<p>Icepenguin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote by <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(3/4,3sqrt(3)/4);<br />
O=(3/4,sqrt(3)/4);<br />
P=(3/8,7sqrt(3)/8);<br />
Q=(9/8, 3sqrt(3)/8);<br />
R=(0,sqrt(3)/4);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, NW);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
label("$P$", P, NNW);<br />
label("$Q$", Q, ESE);<br />
label("$R$", R, SW);<br />
fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br />
draw(A--B--C--D--E--F--cycle);<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O--cycle);<br />
<br />
</asy><br />
<br />
The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN</math>, <math>\triangle NQO</math>, and <math>\triangle ORM</math>).<br />
<br />
Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio. Thus, it must also divide transversal <math>\overline {AC}</math> and transversal <math>\overline {CO}</math> into a <math>1:1</math> ratio. By symmetry, the same applies for <math>\overline {CE}</math> and <math>\overline {EA}</math> as well as <math>\overline {EM}</math> and <math>\overline {AN}</math><br />
<br />
<br />
In <math>\triangle ACE</math>, we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}</math>. Our desired area becomes <br />
<br />
<cmath>(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}</cmath><br />
<br />
==Solution 2 ==<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (<math>AXFZ</math>, <math>XBCY</math>, and <math>ZYED</math>), and 3 right triangles, with one right angle on each of <math>X</math>, <math>Y</math>, and <math>Z</math>. <br />
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal, which has length 2) with a height of <math>\sqrt{3}/4</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>\frac{5\sqrt{3}}{16}</math> for a total area of <math>\frac{15\sqrt{3}}{16}.</math> (Alternatively, we could have calculated the area of hexagon <math>ABCDEF</math> and subtracted the area of <math>\triangle XYZ</math>, which, as we showed before, had a side length of 3/2). <br />
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of <math>YC = 1/2.</math> Using similar triangles we calculate the base to be 1/4 and the height to be <math>\sqrt{3}/4</math> giving us an area of <math>\sqrt{3}/32</math> per triangle, and a total area of <math>3\frac{\sqrt{3}}{32}</math>. Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}</math>. Finding the total area, we get <math>6*1^2*\sqrt{3}/4=3\sqrt{3}/2</math>. Taking the complement, we get <math>3\sqrt{3}/2-33\sqrt{3}/32=15\sqrt{3}/32 = \boxed {C}\frac{15}{32}\sqrt{3}</math><br />
<br />
==Solution 3 (Trig)==<br />
Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br />
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^{\textrm{o}}</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^{\textrm{o}}</math>, and thus the side length of this triangle is <math>1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}</math>. So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}</math><br />
Now let's find the area of the smaller triangles. Notice, triangle <math>ACE</math> cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then <math>\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}</math> and the sum of the areas is <math>3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}</math><br />
Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=931782018 AMC 10B Problems/Problem 242018-03-12T17:18:32Z<p>Icepenguin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote by <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(3/4,3sqrt(3)/4);<br />
O=(3/4,sqrt(3)/4);<br />
P=(3/8,7sqrt(3)/8);<br />
Q=(9/8, 3sqrt(3)/8);<br />
R=(0,sqrt(3)/4);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, NW);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
label("$P$", P, NNW);<br />
label("$Q$", Q, ESE);<br />
label("$R$", R, SW);<br />
fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br />
draw(A--B--C--D--E--F--cycle);<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O--cycle);<br />
<br />
</asy><br />
<br />
The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN</math>, <math>\triangle NQO</math>, and <math>\triangle ORM</math>).<br />
<br />
Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio. Thus, it must also divide transversal <math>\overline {AC}</math> and transversal <math>\overline {CO}</math> into a <math>1:1</math> ratio. By symmetry, the same applies for <math>\overline {CE}</math> and <math>\overline {EA}</math> as well as <math>\overline {EM}</math> and <math>\overline {AN}</math><br />
<br />
<br />
In <math>\triangle ACE</math>, we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}</math>. Our desired area becomes <br />
<br />
<cmath>(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}</cmath><br />
<br />
==Solution 2 ==<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (<math>AXFZ</math>, <math>XBCY</math>, and <math>ZYED</math>), and 3 right triangles, with one right angle on each of <math>X</math>, <math>Y</math>, and <math>Z</math>. <br />
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal, which has length 2) with a height of <math>\sqrt{3}/4</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>\frac{5\sqrt{3}}{16}</math> for a total area of <math>{15\sqrt{3}}{16}.</math> (Alternatively, we could have calculated the area of hexagon <math>ABCDEF</math> and subtracted the area of <math>\triangle XYZ</math>, which, as we showed before, had a side length of 3/2). <br />
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of <math>YC = 1/2.</math> Using similar triangles we calculate the base to be 1/4 and the height to be <math>\sqrt{3}/4</math> giving us an area of <math>\sqrt{3}/32</math> per triangle, and a total area of <math>3\frac{\sqrt{3}}{32}</math>. Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}</math>. Finding the total area, we get <math>6*1^2*\sqrt{3}/4=3\sqrt{3}/2</math>. Taking the complement, we get <math>3\sqrt{3}/2-33\sqrt{3}/32=15\sqrt{3}/32 = \boxed {C}\frac{15}{32}\sqrt{3}</math><br />
<br />
==Solution 3 (Trig)==<br />
Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br />
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^{\textrm{o}}</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^{\textrm{o}}</math>, and thus the side length of this triangle is <math>1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}</math>. So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}</math><br />
Now let's find the area of the smaller triangles. Notice, triangle <math>ACE</math> cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then <math>\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}</math> and the sum of the areas is <math>3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}</math><br />
Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=931772018 AMC 10B Problems/Problem 242018-03-12T17:17:43Z<p>Icepenguin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote by <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(3/4,3sqrt(3)/4);<br />
O=(3/4,sqrt(3)/4);<br />
P=(3/8,7sqrt(3)/8);<br />
Q=(9/8, 3sqrt(3)/8);<br />
R=(0,sqrt(3)/4);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, NW);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
label("$P$", P, NNW);<br />
label("$Q$", Q, ESE);<br />
label("$R$", R, SW);<br />
fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br />
draw(A--B--C--D--E--F--cycle);<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O--cycle);<br />
<br />
</asy><br />
<br />
The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN</math>, <math>\triangle NQO</math>, and <math>\triangle ORM</math>).<br />
<br />
Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio. Thus, it must also divide transversal <math>\overline {AC}</math> and transversal <math>\overline {CO}</math> into a <math>1:1</math> ratio. By symmetry, the same applies for <math>\overline {CE}</math> and <math>\overline {EA}</math> as well as <math>\overline {EM}</math> and <math>\overline {AN}</math><br />
<br />
<br />
In <math>\triangle ACE</math>, we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}</math>. Our desired area becomes <br />
<br />
<cmath>(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}</cmath><br />
<br />
==Solution 2 ==<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (<math>AXFZ</math>, <math>XBCY</math>, and <math>ZYED</math>), and 3 right triangles, with one right angle on each of <math>X</math>, <math>Y</math>, and <math>Z</math>. <br />
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal, which has length 2) with a height of <math>\sqrt{3}/4</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>\frac{5\sqrt{3}{16}</math> for a total area of <math>{15\sqrt{3}}{16}.</math> (Alternatively, we could have calculated the area of hexagon <math>ABCDEF</math> and subtracted the area of <math>\triangle XYZ</math>, which, as we showed before, had a side length of 3/2). <br />
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of <math>YC = 1/2.</math> Using similar triangles we calculate the base to be 1/4 and the height to be <math>\sqrt{3}/4</math> giving us an area of <math>\sqrt{3}/32</math> per triangle, and a total area of <math>3\frac{\sqrt{3}}{32}</math>. Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}</math>. Finding the total area, we get <math>6*1^2*\sqrt{3}/4=3\sqrt{3}/2</math>. Taking the complement, we get <math>3\sqrt{3}/2-33\sqrt{3}/32=15\sqrt{3}/32 = \boxed {C}\frac{15}{32}\sqrt{3}</math><br />
<br />
==Solution 3 (Trig)==<br />
Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br />
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^{\textrm{o}}</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^{\textrm{o}}</math>, and thus the side length of this triangle is <math>1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}</math>. So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}</math><br />
Now let's find the area of the smaller triangles. Notice, triangle <math>ACE</math> cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then <math>\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}</math> and the sum of the areas is <math>3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}</math><br />
Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_25&diff=931762018 AMC 10B Problems/Problem 252018-03-12T17:04:32Z<p>Icepenguin: /* Solution 5 */</p>
<hr />
<div>== Problem ==<br />
Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many real numbers <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>?<br />
<br />
<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math><br />
<br />
==Solution 1==<br />
This rewrites itself to <math>x^2=10,000\{x\}</math>.<br />
<br />
Graphing <math>y=10,000\{x\}</math> and <math>y=x^2</math> we see that the former is a set of line segments with slope <math>10,000</math> from <math>0</math> to <math>1</math> with a hole at <math>x=1</math>, then <math>1</math> to <math>2</math> with a hole at <math>x=2</math> etc.<br />
<br />
Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization.<br />
<br />
<asy><br />
import graph;<br />
size(400);<br />
xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5}));<br />
yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}));<br />
real y(real x) {return x^2;}<br />
draw(circle((-4,16), 0.1));<br />
draw(circle((-3,16), 0.1));<br />
draw(circle((-2,16), 0.1));<br />
draw(circle((-1,16), 0.1));<br />
draw(circle((0,16), 0.1));<br />
draw(circle((1,16), 0.1));<br />
draw(circle((2,16), 0.1));<br />
draw(circle((3,16), 0.1));<br />
draw(circle((4,16), 0.1));<br />
draw((-5,0)--(-4,16), black);<br />
draw((-4,0)--(-3,16), black);<br />
draw((-3,0)--(-2,16), black);<br />
draw((-2,0)--(-1,16), black);<br />
draw((-1,0)--(-0,16), black);<br />
draw((0,0)--(1,16), black);<br />
draw((1,0)--(2,16), black);<br />
draw((2,0)--(3,16), black);<br />
draw((3,0)--(4,16), black);<br />
draw(graph(y,-4.2,4.2),green);<br />
</asy><br />
<br />
Now notice that when <math>x=\pm 100</math> then graph has a hole at <math>(\pm 100,10,000)</math> which the equation <math>y=x^2</math> passes through and then continues upwards. Thus our set of possible solutions is bounded by <math>(-100,100)</math>. We can see that <math>y=x^2</math> intersects each of the lines once and there are <math>99-(-99)+1=199</math> lines for an answer of <math>\boxed{\text{(C)}~199}</math>.<br />
<br />
==Solution 2==<br />
<br />
Same as the first solution, <math>x^2=10,000\{x\} </math>.<br />
<br />
<br />
We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>{x}</math> in terms of <math>\lfloor x \rfloor</math>:<br />
<math>\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0</math><br />
<br />
<br />
We use the quadratic formula to solve for {x}:<br />
<math>\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{( -2 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 ) }}{2} </math><br />
<br />
<br />
Since <math> 0 \leq \{x\} < 1 </math>, we get an inequality which we can then solve. After simplifying a lot, we get that <math>\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0</math>.<br />
<br />
<br />
Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of<math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.<br />
<br />
==Solution 3==<br />
<br />
Let <math>x = a+k</math> where <math>a</math> is the integer portion of <math>x</math> and <math>k</math> is the decimal portion.<br />
We can then rewrite the problem below:<br />
<br />
<math>(a+k)^2 + 10000a = 10000(a+k)</math><br />
<br />
From here, we get<br />
<br />
<math>(a+k)^2 + 10000a = 10000a + 10000k</math><br />
<br />
Solving for <math>a+k</math>...<br />
<br />
<math>(a+k)^2 = 10000k</math><br />
<br />
<math>a+k = \pm100\sqrt{k}</math><br />
<br />
Because <math>0 \leq k < 1</math>, we know that <math>a+k</math> cannot be less than or equal to <math>-100</math> nor greater than or equal to <math>100</math>. Therefore:<br />
<br />
<math>-99 \leq a+k = x \leq 99</math><br />
<br />
There are 199 elements in this range, so the answer is <math>\boxed{\textbf{(C)} \text{ 199}}</math>.<br />
<br />
== Solution 4 ==<br />
<br />
The following C++ code confirms that there are <math>\boxed{\textbf{(C)} \text{ 199}}</math> solutions in total.<br />
<br />
<source lang="cpp"><br />
typedef long double ld;<br />
<br />
int main() {<br />
// x^2=10000{x}<br />
// (t+b)^2=10000b <br />
// b^2+(2t-10000)b+t^2=0 <br />
<br />
FOR(t,-200,201) {<br />
ld b = 2*t-10000, c = t*t;<br />
ld s1 = (-b+sqrt(b*b-4*c))/2;<br />
ld s2 = (-b-sqrt(b*b-4*c))/2;<br />
if (0 <= s1 && s1 < 1) cout << t+s1 << "\n";<br />
if (0 <= s2 && s2 < 1) cout << t+s2 << "\n";<br />
}<br />
}<br />
</source><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=1996_AHSME_Problems/Problem_30&diff=860971996 AHSME Problems/Problem 302017-06-18T18:31:05Z<p>Icepenguin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
<math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409 </math><br />
<br />
==Solution 1==<br />
In hexagon <math>ABCDEF</math>, let <math>AB=BC=CD=3</math> and let <math>DE=EF=FA=5</math>. Since arc <math>BAF</math> is one third of the circumference of the circle, it follows that <math>\angle BCF = \angle BEF=60^{\circ}</math>. Similarly, <math>\angle CBE =\angle CFE=60^{\circ}</math>. Let <math>P</math> be the intersection of <math>\overline{BE}</math> and <math>\overline{CF}</math>, <math>Q</math> that of <math>\overline{BE}</math> and <math>\overline{AD}</math>, and <math>R</math> that of <math>\overline{CF}</math> and <math>\overline{AD}</math>. Triangles <math>EFP</math> and <math>BCP</math> are equilateral, and by symmetry, triangle <math>PQR</math> is isosceles and thus also equilateral. <br />
<asy><br />
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));<br />
real angleUnit = 15;<br />
draw(Circle(origin,1));<br />
pair D = dir(22.5);<br />
pair C = dir(3*angleUnit + degrees(D));<br />
pair B = dir(3*angleUnit + degrees(C));<br />
pair A = dir(3*angleUnit + degrees(B));<br />
pair F = dir(5*angleUnit + degrees(A));<br />
pair E = dir(5*angleUnit + degrees(F));<br />
draw(A--B--C--D--E--F--cycle);<br />
dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F);<br />
draw(A--D^^B--E^^C--F);<br />
label("$3$",D--C,SW); label("$3$",B--C,S); label("$3$",A--B,SE); label("$5$",A--F,NE); label("$5$",F--E,N); label("$5$",D--E,NW);<br />
</asy><br />
<br />
Furthermore, <math>\angle BAD</math> and <math>\angle BED</math> subtend the same arc, as do <math>\angle ABE</math> and <math>\angle ADE</math>. Hence triangles <math>ABQ</math> and <math>EDQ</math> are similar. Therefore, <cmath>\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.</cmath> It follows that <cmath>\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad<br />
\mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.</cmath> Solving the two equations simultaneously yields <math>AD=360/49,</math> so <math>m+n=\boxed{409}. \blacksquare</math><br />
<br />
==Solution 2==<br />
All angle measures are in degrees.<br />
Let the first trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=3</math>. Then the second trapezoid is <math>AFED</math>, where <math>AF=FE=ED=5</math>. We look for <math>AD</math>.<br />
<br />
Since <math>ABCD</math> is an isosceles trapezoid, we know that <math>\angle BAD=\angle CDA</math> and, since <math>AB=BC</math>, if we drew <math>AC</math>, we would see <math>\angle BCA=\angle BAC</math>. Anyway, <math>\widehat{AB}=\widehat{BC}=\widehat{CD}</math> (<math>\widehat{AB}</math> means arc AB). Using similar reasoning, <math>\widehat{AF}=\widehat{FE}=\widehat{ED}</math>.<br />
<br />
Let <math>\widehat{AB}=2\phi</math> and <math>\widehat{AF}=2\theta</math>. Since <math>6\theta+6\phi=360</math> (add up the angles), <math>2\theta+2\phi=120</math> and thus <math>\widehat{AB}+\widehat{AF}=\widehat{BF}=120</math>. Therefore, <math>\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120</math>. <math>\angle CDE=120</math> as well.<br />
<br />
Now I focus on triangle <math>FAB</math>. By the Law of Cosines, <math>BF^2=3^2+5^2-30\cos{120}=9+25+15=49</math>, so <math>BF=7</math>. Seeing <math>\angle ABF=\theta</math> and <math>\angle AFB=\phi</math>, we can now use the Law of Sines to get:<br />
<cmath>\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.</cmath><br />
<br />
Now I focus on triangle <math>AFD</math>. <math>\angle AFD=3\phi</math> and <math>\angle ADF=\theta</math>, and we are given that <math>AF=5</math>, so<br />
<cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.</cmath><br />
We know <math>\sin{\theta}=\frac{5\sqrt{3}}{14}</math>, but we need to find <math>\sin{3\phi}</math>. Using various identities, we see<br />
<cmath>\begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\<br />
&=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\<br />
&=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\<br />
&=\sin{\phi}(3-4\sin^2{\phi})\\<br />
&=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343}<br />
\end{align*}</cmath><br />
Returning to finding <math>AD</math>, we remember <cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.</cmath><br />
Plugging in and solving, we see <math>AD=\frac{360}{49}</math>. Thus, the answer is <math>360 + 49 = 409</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>.<br />
<br />
==Solution 3==<br />
<br />
Let <math>x</math> be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides <math>a, b, c, d</math> the circumradius <math>R</math> satisfies <cmath>R^2=\frac{1}{16}\cdot\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)},</cmath> where <math>s</math> is the semiperimeter. Applying this to the trapezoid with sides <math>3, 3, 3, x</math>, we see that many terms cancel and we are left with <cmath>R^2=\frac{27}{9-x}</cmath> Similar canceling occurs for the trapezoid with sides <math>5, 5, 5, x</math>, and since the two quadrilaterals share the same circumradius, we can equate: <cmath>\frac{27}{9-x}=\frac{125}{15-x}</cmath> Solving for <math>x</math> gives <math>x=\frac{360}{49}</math>, so the answer is <math>\fbox{(E) 409}</math>.<br />
<br />
== See also ==<br />
{{AHSME box | year = 1996 | num-b = 29 | after = Last Problem}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=84245Gmaas2017-02-27T23:11:15Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas === <br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br />
<br />
- themoocow is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^{267794} (k+1)(k+2)+GMAAS</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1164. (Year: 2015) (gmaas code: 6666)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Number Theory 1229 (AGAIN)<br />
~Calculus 1226<br />
~Gallifrey Meeting #123456, 1888.<br />
~On his account of course.<br />
~On FEFF once<br />
~On Mt. Everest at 13:37 in the year 1337 on 1/3 37<br />
~Everywhere in outer space<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Epecially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens (also known as Vsaace XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=84244Gmaas2017-02-27T23:10:15Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas === <br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard.<br />
<br />
- themoocow is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^{267794} (k+1)(k+2)+GMAAS</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1164. (Year: 2015) (gmaas code: 6666)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Number Theory 1229 (AGAIN)<br />
~Calculus 1226<br />
~Gallifrey Meeting #123456, 1888.<br />
~On his account of course.<br />
~On FEFF once<br />
~On Mt. Everest at 13:37 in the year 1337 on 1/3 37<br />
~Everywhere in outer space<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Epecially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens (also known as Vsaace XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=84243Gmaas2017-02-27T23:09:39Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas === <br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard.<br />
<br />
- themoocow is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^{10483} (k+1)(k+2)+GMAAS</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1164. (Year: 2015) (gmaas code: 6666)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Number Theory 1229 (AGAIN)<br />
~Calculus 1226<br />
~Gallifrey Meeting #123456, 1888.<br />
~On his account of course.<br />
~On FEFF once<br />
~On Mt. Everest at 13:37 in the year 1337 on 1/3 37<br />
~Everywhere in outer space<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Epecially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens (also known as Vsaace XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=84242Gmaas2017-02-27T23:09:19Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas === <br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard.<br />
<br />
- themoocow is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^10483 (k+1)(k+2)+GMAAS</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1164. (Year: 2015) (gmaas code: 6666)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Number Theory 1229 (AGAIN)<br />
~Calculus 1226<br />
~Gallifrey Meeting #123456, 1888.<br />
~On his account of course.<br />
~On FEFF once<br />
~On Mt. Everest at 13:37 in the year 1337 on 1/3 37<br />
~Everywhere in outer space<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Epecially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens (also known as Vsaace XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=84241Gmaas2017-02-27T23:08:40Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas === <br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard.<br />
<br />
- themoocow is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^1 (k+1)(k+2)+5</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1164. (Year: 2015) (gmaas code: 6666)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Number Theory 1229 (AGAIN)<br />
~Calculus 1226<br />
~Gallifrey Meeting #123456, 1888.<br />
~On his account of course.<br />
~On FEFF once<br />
~On Mt. Everest at 13:37 in the year 1337 on 1/3 37<br />
~Everywhere in outer space<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Epecially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens (also known as Vsaace XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=81915Gmaas2016-12-15T22:13:06Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas === <br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard.<br />
<br />
- themoocow is gmass. gmass is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^1 (k+1)(k+2)+5</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so. <br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1343 (Year: 2015)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Number Theory 1229 (AGAIN)<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (except the USF castle walls), but deforestation threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity. (Only when everyone in the world blinks at the same time.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens (also known as Vsauce XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
- Gmaas has two tails, one for everyday life, and one for special occasions<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_21&diff=815752016 AMC 8 Problems/Problem 212016-11-24T17:41:00Z<p>Icepenguin: /* Solution 1 */</p>
<hr />
<div>A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?<br />
<br />
<math>\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}</math><br />
<br />
==Solution 1==<br />
We put five chips randomly in order, and then pick the chips from the left to the right. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. Because a green chip will be last 4 out of 10 times and a red chip will be last 6 out of 10 times, our answer is <math>\boxed{\text{(B) }\dfrac{2}{5}}</math>.<br />
<br />
==Solution 2==<br />
There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out 3 red chips, 3 red chips and a green chip, 2 green chips, 2 green chips and a red chip, and 2 green chip and 2 red chips. Because order is important in this problem, there are <math>1+4+1+3+6=15</math> ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which <math>15-5=10</math>. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is <math>\frac4{10}</math>, or <math>\boxed{\text{(B) }\dfrac{2}{5}}</math>.<br />
<br />
{{AMC8 box|year=2016|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_21&diff=815722016 AMC 8 Problems/Problem 212016-11-24T17:34:08Z<p>Icepenguin: /* Solution 2 */</p>
<hr />
<div>A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?<br />
<br />
<math>\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}</math><br />
<br />
==Solution 1==<br />
We put five chips randomly in order, and then pick the chips from the left to the right. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. Because a green chip will be last 6 out of 10 times and a red chip will be last 4 out of 10 times, our answer is <math>\boxed{\text{(B) }\dfrac{2}{5}}</math>.<br />
<br />
==Solution 2==<br />
There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out 3 red chips, 3 red chips and a green chip, 2 green chips, 2 green chips and a red chip, and 2 green chip and 2 red chips. Because order is important in this problem, there are <math>1+4+1+3+6=15</math> ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which <math>15-5=10</math>. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is <math>\frac4{10}</math>, or <math>\boxed{\text{(B) }\dfrac{2}{5}}</math>.<br />
<br />
{{AMC8 box|year=2016|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=80754Gmaas2016-10-24T23:36:28Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas ===<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats fish fingers and custard and plays Minecraft and Agario(totally not cancerous games, says the Doctor.)<br />
<br />
- themoocow is gmass. gmass is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed(with probability <math>\frac{3972}{7891}</math>), or malnourished(with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^1 (k+1)(k+2)-2</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so. <br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
-Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
-Gmaas is a sage omniscient cat.<br />
<br />
-He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1343 (Year: 2015)<br />
~Intermediate Counting and Probability 1137<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (except the USF castle walls), but deforestation threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity. (Only when everyone in the world blinks at the same time.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens(also known as Vsauce XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
<br />
-Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
-Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
-Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
<br />
-Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=80681Gmaas2016-10-18T21:17:50Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas ===<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats fish fingers and custard and plays Minecraft and Agario(totally not cancerous games, says the Doctor.)<br />
<br />
- themoocow is gmass. gmass is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed or malnourished by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^1 (k+1)(k+2)-2</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so. <br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
-Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
-Gmaas is a sage omniscient cat.<br />
<br />
-He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1343 (Year: 2015)<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (except the USF castle walls), but deforestation threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity. (Only when everyone in the world blinks at the same time.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens(also known as Vsauce XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
<br />
-Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
-Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
-Gmaas is a Super Cat Time Lord. He has 57843504 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
<br />
-Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguinhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=80668Gmaas2016-10-18T00:54:06Z<p>Icepenguin: /* Known Facts About gmaas */</p>
<hr />
<div>=== Known Facts About gmaas ===<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats fish fingers and custard and plays Minecraft and Agario(totally not cancerous games, says the Doctor.)<br />
<br />
- themoocow is gmass. gmass is gmaas's cousin.<br />
<br />
- Gmass is <math>\bold{not}</math> Colonel Meow.<br />
<br />
- Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
- He is often overfed or malnourished by sseraj.<br />
<br />
- He has <cmath>\sum_{k=0}^1 (k+1)(k+2)-2</cmath> supercars, excluding the Purrari.<br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so. <br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
-Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
-Gmaas is a sage omniscient cat.<br />
<br />
-He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
~Alligator Swamp A 1072 <br />
~Alligator Swamp B 1073<br />
~Introduction to Algebra A (1170)<br />
~Welcome to Panda Town Gate 1076<br />
~Welcome to Gmaas Town Gate 1221<br />
~Welcome to Gmaas Town Gate 1125<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
~The other side of the ice in Antarctica<br />
~Feisty Alligator Swamp 1115<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
~Introduction to Counting and Probability 1142 <br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
~Intermediate Counting and Probability 1137<br />
~Intermediate Counting and Probability 1207<br />
~Posting student surveys<br />
~USF Castle Walls 1203<br />
~Dark Lord's Hut 1210<br />
~AMC 10 Problem Series 1200<br />
~Intermediate Number Theory 1138<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
~Algebra B 1112<br />
~33°81'199.4"N 167°05'45.1"W (Unknown location, please try again later)<br />
~Ocelot Rainforest 1111<br />
~Intermediate Counting and Probability 1137 <br />
~Intermediate Number Theory 1138 (AGAIN)<br />
~AMC 10 Problem Series 1200 (AGAIN)<br />
~Cat Gathering #339 1222<br />
~Cat Gathering #312 1535<br />
~Introduction to Geometry (1190)<br />
~Hogwarts Introduction to Transfiguration (...16168)<br />
~Cat Meow Meow Purr 1110<br />
~Bilbo's Hobbit Hole 1234<br />
~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br />
~Pre-Algebra 1 with sseraj 1343 (Year: 2015)<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (except the USF castle walls), but deforestation threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99.999}{100}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Vu.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- Gmaas's left eye contains the singularity. (Only when everyone in the world blinks at the same time.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens(also known as Vsauce XD), or his page name, Purrshanks.<br />
<br />
- Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
<br />
-Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
-Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, "We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing." The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br />
<br />
-Gmaas is a Super Cat Time Lord. He has 216 regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2=216</cmath>. <br />
<br />
<br />
-Gmaas loves to eat turnips. At <math>\frac{7}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath is also writing a book about Gmaas<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
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- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
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- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div>Icepenguin