https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Donutvan&feedformat=atomAoPS Wiki - User contributions [en]2022-01-25T18:13:26ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Talk:2010_AMC_12B_Problems/Problem_25&diff=161167Talk:2010 AMC 12B Problems/Problem 252021-08-29T15:54:19Z<p>Donutvan: </p>
<hr />
<div>How do we know that 67 will yield the smallest result of 77?<br />
<br />
I created a Mathematica function to check all primes <= 67, and yes indeed 67 gave the smallest result of 77 but 2 gives a problematically close 78. And the trend is not always decreasing:<br />
x = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67 <br />
f(x) = 78, 140, 162, 189, 164, 177, 162, 162, 149, 130, 128, 113, 108, 105, 99, 92, 83, 82, 77<br />
<br />
So if the target problem was not <math>2010^m</math> but <math>2001^m</math>, then 13 as the largest prime in 2001=7*11*13 will give an incorrect answer of <math>177</math>. The correct answer will be <math>164</math>.<br />
<br />
<math>\phantom{7*11*13=1001??????????}</math><br />
<br />
Do we just have to cross our fingers and guess that it is f(67)=77 is indeed the smallest?<br />
<br />
--- buhiroshi0205<br />
<br />
Because the AMC has a short time limit, yes, you "just have to cross our fingers and guess".<br />
<br />
Looking at the results of the Mathematica function mentioned above, you would likely want to guess both <math>2</math> and <math>67</math>, as the function appears to increase then decrease. That's probably the best way to have a chance of solving the problem quickly, and hope that they didn't throw in an outlier like 11.<br />
<br />
--- donutvan</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_8&diff=1567672003 AIME II Problems/Problem 82021-06-24T19:31:27Z<p>Donutvan: Continued to add solution</p>
<hr />
<div>==Problem==<br />
Find the eighth term of the sequence <math>1440,</math> <math>1716,</math> <math>1848,\ldots,</math> whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.<br />
<br />
==Solution 1==<br />
If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic <math>ax^2+bx+c</math> such that <math>f(1)=1440</math>, <math>f(2)=1716</math>, and <math>f(3)=1848</math>. Plugging in the values for x gives us a system of three equations:<br />
<br />
<math>a+b+c=1440</math><br />
<br />
<math>4a+2b+c=1716</math><br />
<br />
<math>9a+3b+c=1848</math><br />
<br />
Solving gives <math>a=-72, b=492,</math> and <math>c=1020</math>. Thus, the answer is <math>-72(8)^2+492\cdot8+1020= \boxed{348}.</math><br />
<br />
== Solution 2 ==<br />
Setting one of the sequences as <math>a+nr_1</math> and the other as <math>b+nr_2</math>, we can set up the following equalities<br />
<br />
<math>ab = 1440</math><br />
<br />
<math>(a+r_1)(b+r_2)=1716</math><br />
<br />
<math>(a+2r_1)(b+2r_2)=1848</math><br />
<br />
We want to find <math>(a+7r_1)(b+7r_2)</math><br />
<br />
Foiling out the two above, we have<br />
<br />
<math>ab + ar_2 + br_1 + r_1r_2 = 1716</math> and <math>ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848</math><br />
<br />
Plugging in <math>ab=1440</math> and bringing the constant over yields<br />
<br />
<math>ar_2 + br_1 + r_1r_2 = 276</math><br />
<br />
<math>ar_2 + br_1 + 2r_1r_2 = 204</math><br />
<br />
Subtracting the two yields <math>r_1r_2 = -72</math> and plugging that back in yields <math>ar_2 + br_1 = 348</math><br />
<br />
Now we find<br />
<br />
<math>(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}</math>.<br />
<br />
==Solution 3==<br />
Let the first sequence be <br />
<br />
<math>A={a+d_1, a + 2d_1, a + 3d_1, \cdots}</math><br />
<br />
and the second be <br />
<br />
<math>B={b+d_2, b + 2d_2, b + 3d_2, \cdots}</math>,<br />
<br />
with <math>(a+d_1)(b+d_2)=1440</math>. Now, note that the <math>n^{\text{th}}</math> term of sequence <math>A</math> is <math>a+d_1 n</math> and the <math>n^{\text{th}}</math> term of <math>B</math> is <math>b + d_2 <br />
n</math>. Thus, the <math>n^{\text{th}}</math> term of the given sequence is <br />
<br />
<math>n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab</math>,<br />
<br />
a quadratic in <math>n</math>. Now, letting the given sequence be <math>C</math>, we see that <br />
<br />
<math>C_n - C_{n-1} = n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab - (n-1)^2(d_1 + d_2) - (n-1)(ad_2 + bd_1) - ab = n(2d_1 + 2d_2) + ad_2 + bd_1 - d_1 - d_2</math>,<br />
<br />
a linear equation in <math>n</math>! Since <math>C_2 - C_1 = 276</math> and <math>C_3 - C_2 = 132</math>, we can see that, in general, we have <br />
<br />
<math>C_n - C_{n-1} = 420 - 144n</math>.<br />
<br />
Thus, we can easily find <br />
<br />
<math>C_4 - C_3 = -12 \rightarrow C_4 = 1836</math>,<br />
<br />
<math>C_5 - C_4 = -156 \rightarrow C_5 = 1680</math>,<br />
<br />
<math>C_6 - C_5 = -300 \rightarrow C_6 = 1380</math>, <br />
<br />
<math>C_7 - C_6 = -444 \rightarrow C_7 = 936</math>, and finally <br />
<br />
<math>C_8 - C_7 = -588 \rightarrow \boxed{C_8 = 348}</math>.<br />
<br />
==Solution 4(Tedious) ==<br />
Start by labeling the two sequences:<br />
<br />
Sequence 1:<math>a,a+d_1,a+2d_1,\dots a+(n-1)d_1</math>,<br />
<br />
Sequence 2:<math>b,b+d_2,b+2d_2,\dots b+(n-1)d_2</math>.<br />
<br />
Additionally, label the sequence given in the problem the function <math>f</math>, such that <br />
<br />
<math>f(1)=1440,f(2)=1716,f(3)=1848</math>.<br />
<br />
Then, <br />
<br />
<math>f(1)=ab,</math><br />
<br />
<math>f(2)=(a+d_1)(b+d_2)=a+ad_2+d_1b+d_1d_2,</math><br />
<br />
<math>f(3)=(a+2d_1)(b+2d_2)=ab+2ad_2+2bd_1+4d_1d_2,</math><br />
<br />
and <math>f(8)=(a+7d_1)(b+7d_2)=ab+7ad_2+7bd_1+49d_1d_2</math>.<br />
<br />
In order to find <math>f(8)</math> add <math>f(3)</math> enough times to get the difference between the <math>d_1d_2</math> and <math>ad_2+bd_1</math> terms, then add <math>f(2)</math> and <math>f(1)</math> to get the other terms:<br />
<br />
<math>21f(3)=21ab+42ad_2+42bd_1+84d_1d_2</math><br />
<br />
<math>21f(3)-35f(2)=21ab+42ad_2+42bd_1+84d_1d_2-35a-35ad_2-35d_1b-35d_1d_2=-14ab+7ad_2+7bd_1+49d_1d_2</math><br />
<br />
<math>21f(3)-35f(2)+15f(1)=-14ab+7ad_2+7bd_1+49d_1d_2+15ab=ab+7ad_2+7bd_1+49d_1d_2</math><br />
<br />
Now that the expression is in terms of the given values, insert values and solve:<br />
<br />
<math>21*1848-35*1716+15*1440=1848+20*132+(20*1716-35*1716+15*1716)+15*(-276)</math> <br />
<br />
<math>=1848+5*132+15(132-276)</math><br />
<br />
<math>=1848+660+15(-144)</math><br />
<br />
<math>=348</math><br />
<br />
==See also ==<br />
{{AIME box|year=2003|n=II|num-b=7|num-a=9}}<br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_8&diff=1566282003 AIME II Problems/Problem 82021-06-22T17:45:22Z<p>Donutvan: Added an additional, though tedious, solution.</p>
<hr />
<div>==Problem==<br />
Find the eighth term of the sequence <math>1440,</math> <math>1716,</math> <math>1848,\ldots,</math> whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.<br />
<br />
==Solution 1==<br />
If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic <math>ax^2+bx+c</math> such that <math>f(1)=1440</math>, <math>f(2)=1716</math>, and <math>f(3)=1848</math>. Plugging in the values for x gives us a system of three equations:<br />
<br />
<math>a+b+c=1440</math><br />
<br />
<math>4a+2b+c=1716</math><br />
<br />
<math>9a+3b+c=1848</math><br />
<br />
Solving gives <math>a=-72, b=492,</math> and <math>c=1020</math>. Thus, the answer is <math>-72(8)^2+492\cdot8+1020= \boxed{348}.</math><br />
<br />
== Solution 2 ==<br />
Setting one of the sequences as <math>a+nr_1</math> and the other as <math>b+nr_2</math>, we can set up the following equalities<br />
<br />
<math>ab = 1440</math><br />
<br />
<math>(a+r_1)(b+r_2)=1716</math><br />
<br />
<math>(a+2r_1)(b+2r_2)=1848</math><br />
<br />
We want to find <math>(a+7r_1)(b+7r_2)</math><br />
<br />
Foiling out the two above, we have<br />
<br />
<math>ab + ar_2 + br_1 + r_1r_2 = 1716</math> and <math>ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848</math><br />
<br />
Plugging in <math>ab=1440</math> and bringing the constant over yields<br />
<br />
<math>ar_2 + br_1 + r_1r_2 = 276</math><br />
<br />
<math>ar_2 + br_1 + 2r_1r_2 = 204</math><br />
<br />
Subtracting the two yields <math>r_1r_2 = -72</math> and plugging that back in yields <math>ar_2 + br_1 = 348</math><br />
<br />
Now we find<br />
<br />
<math>(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}</math>.<br />
<br />
==Solution 3==<br />
Let the first sequence be <br />
<br />
<math>A={a+d_1, a + 2d_1, a + 3d_1, \cdots}</math><br />
<br />
and the second be <br />
<br />
<math>B={b+d_2, b + 2d_2, b + 3d_2, \cdots}</math>,<br />
<br />
with <math>(a+d_1)(b+d_2)=1440</math>. Now, note that the <math>n^{\text{th}}</math> term of sequence <math>A</math> is <math>a+d_1 n</math> and the <math>n^{\text{th}}</math> term of <math>B</math> is <math>b + d_2 <br />
n</math>. Thus, the <math>n^{\text{th}}</math> term of the given sequence is <br />
<br />
<math>n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab</math>,<br />
<br />
a quadratic in <math>n</math>. Now, letting the given sequence be <math>C</math>, we see that <br />
<br />
<math>C_n - C_{n-1} = n^2(d_1 + d_2) + n(ad_2 + bd_1) + ab - (n-1)^2(d_1 + d_2) - (n-1)(ad_2 + bd_1) - ab = n(2d_1 + 2d_2) + ad_2 + bd_1 - d_1 - d_2</math>,<br />
<br />
a linear equation in <math>n</math>! Since <math>C_2 - C_1 = 276</math> and <math>C_3 - C_2 = 132</math>, we can see that, in general, we have <br />
<br />
<math>C_n - C_{n-1} = 420 - 144n</math>.<br />
<br />
Thus, we can easily find <br />
<br />
<math>C_4 - C_3 = -12 \rightarrow C_4 = 1836</math>,<br />
<br />
<math>C_5 - C_4 = -156 \rightarrow C_5 = 1680</math>,<br />
<br />
<math>C_6 - C_5 = -300 \rightarrow C_6 = 1380</math>, <br />
<br />
<math>C_7 - C_6 = -444 \rightarrow C_7 = 936</math>, and finally <br />
<br />
<math>C_8 - C_7 = -588 \rightarrow \boxed{C_8 = 348}</math>.<br />
<br />
==Solution 4(Tedious) ==<br />
Start by labeling the two sequences:<br />
<br />
Sequence 1:<math>a,a+d_1,a+2d_1,\dots a+(n-1)d_1</math>,<br />
<br />
Sequence 2:<math>b,b+d_2,b+2d_2,\dots b+(n-1)d_2</math>.<br />
<br />
Additionally, we'll label the sequence given in the problem the function <math>f</math>, such that <br />
<br />
<math>f(1)=1440,f(2)=1716,f(3)=1848</math>.<br />
<br />
Then, <math>f(1)=ab,</math> <math>f(2)=(a+d_1)(b+d_2),</math> and <math>f(3)=(a+2d_1)(b+2d_2)</math><br />
<br />
==See also ==<br />
{{AIME box|year=2003|n=II|num-b=7|num-a=9}}<br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_3&diff=1566182003 AIME II Problems/Problem 32021-06-22T16:30:47Z<p>Donutvan: Text formatting fix</p>
<hr />
<div>== Problem ==<br />
Define a <math>\text{good~word}</math> as a sequence of letters that consists only of the letters <math>A</math>, <math>B</math>, and <math>C</math> - some of these letters may not appear in the sequence - and in which <math>A</math> is never immediately followed by <math>B</math>, <math>B</math> is never immediately followed by <math>C</math>, and <math>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there?<br />
<br />
== Solution 1 == <br />
<br />
There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is <math>3*2^6=192</math><br />
<br />
Therefore, there are <math>\boxed{192}</math> seven-letter good words.<br />
<br />
== Solution 2 ==<br />
<br />
There are three choices for the first letter and two choices for each subsequent letter, so there are <math>3\cdot2^{n-1}\ n</math>-letter good words. Substitute <math>n=7</math> to find there are <math>3\cdot2^6=\boxed{192}</math> seven-letter good words. ~ aopsav (Credit to AoPS Alcumus)<br />
<br />
== Solution 3 (Recursion) ==<br />
<br />
We solve this problem using recursion. Let <math>f(x)</math> be the number of <math>x</math>-letter good words. Thus <math>f(1) = 3</math> (A, B or C) and the answer is just <math>f(7)</math>. The recurrence relation can be found by considering the last letter of one of the valid strings of length <math>x - 1</math>. There are <math>2</math> possibilities for the next letter and thus <math>f(x) = 2 \cdot f(x-1)</math>. Now we can find a closed form as <math>f(x) = 3 \cdot 2 ^{x-1}</math> (easy to prove by induction) and thus <math>f(7) = 64 * 3 = \boxed{192}</math> seven-letter good words. ~AK2006<br />
<br />
== See also ==<br />
{{AIME box|year=2003|n=II|num-b=2|num-a=4}}<br />
<br />
[[Category: Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems&diff=1566172003 AIME II Problems2021-06-22T16:27:43Z<p>Donutvan: Minor text formatting fix</p>
<hr />
<div>{{AIME Problems|year=2003|n=II}}<br />
<br />
== Problem 1 ==<br />
The product <math>N</math> of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of <math>N</math>.<br />
<br />
[[2003 AIME II Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Let <math>N</math> be the greatest integer multiple of 8, whose digits are all different. What is the remainder when <math>N</math> is divided by 1000?<br />
<br />
[[2003 AIME II Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Define a <math>\text{good word}</math> as a sequence of letters that consists only of the letters <math>A</math>, <math>B</math>, and <math>C</math> - some of these letters may not appear in the sequence - and in which <math>A</math> is never immediately followed by <math>B</math>, <math>B</math> is never immediately followed by <math>C</math>, and <math>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there?<br />
<br />
[[2003 AIME II Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
[[2003 AIME II Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A cylindrical log has diameter <math>12</math> inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a <math>45^\circ</math> angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as <math>n\pi</math>, where n is a positive integer. Find <math>n</math>.<br />
<br />
[[2003 AIME II Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
In triangle <math>ABC,</math> <math>AB = 13,</math> <math>BC = 14,</math> <math>AC = 15,</math> and point <math>G</math> is the intersection of the medians. Points <math>A',</math> <math>B',</math> and <math>C',</math> are the images of <math>A,</math> <math>B,</math> and <math>C,</math> respectively, after a <math>180^\circ</math> rotation about <math>G.</math> What is the area of the union of the two regions enclosed by the triangles <math>ABC</math> and <math>A'B'C'?</math><br />
<br />
[[2003 AIME II Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Find the area of rhombus <math>ABCD</math> given that the radii of the circles circumscribed around triangles <math>ABD</math> and <math>ACD</math> are <math>12.5</math> and <math>25</math>, respectively.<br />
<br />
[[2003 AIME II Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Find the eighth term of the sequence <math>1440,</math> <math>1716,</math> <math>1848,\ldots,</math> whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.<br />
<br />
[[2003 AIME II Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Consider the polynomials <math>P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x</math> and <math>Q(x) = x^{4} - x^{3} - x^{2} - 1.</math> Given that <math>z_{1},z_{2},z_{3},</math> and <math>z_{4}</math> are the roots of <math>Q(x) = 0,</math> find <math>P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).</math><br />
<br />
[[2003 AIME II Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Two positive integers differ by <math>60.</math> The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?<br />
<br />
[[2003 AIME II Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Triangle <math>ABC</math> is a right triangle with <math>AC = 7,</math> <math>BC = 24,</math> and right angle at <math>C.</math> Point <math>M</math> is the midpoint of <math>AB,</math> and <math>D</math> is on the same side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math><br />
<br />
[[2003 AIME II Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The members of a distinguished committee were choosing a president, and each member gave one vote to one of the <math>27</math> candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least <math>1</math> than the number of votes for that candidate. What is the smallest possible number of members of the committee?<br />
<br />
[[2003 AIME II Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m + n.</math><br />
<br />
[[2003 AIME II Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Let <math>A = (0,0)</math> and <math>B = (b,2)</math> be points on the coordinate plane. Let <math>ABCDEF</math> be a convex equilateral hexagon such that <math>\angle FAB = 120^\circ,</math> <math>\overline{AB}\parallel \overline{DE},</math> <math>\overline{BC}\parallel \overline{EF,}</math> <math>\overline{CD}\parallel \overline{FA},</math> and the y-coordinates of its vertices are distinct elements of the set <math>\{0,2,4,6,8,10\}.</math> The area of the hexagon can be written in the form <math>m\sqrt {n},</math> where <math>m</math> and <math>n</math> are positive integers and n is not divisible by the square of any prime. Find <math>m + n.</math><br />
<br />
[[2003 AIME II Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Let <math>P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}</math>. Let <math>z_{1},z_{2},\ldots,z_{r}</math> be the distinct zeros of <math>P(x),</math> and let <math>z_{k}^{2} = a_{k} + b_{k}i</math> for <math>k = 1,2,\ldots,r,</math> where <math>a_{k}</math> and <math>b_{k}</math> are real numbers. Let<br />
<center><math>\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},</math></center><br />
where <math>m, n,</math> and <math>p</math> are integers and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p.</math><br />
<br />
[[2003 AIME II Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
<br />
{{AIME box|year = 2003|n=II|before=[[2003 AIME I Problems]]|after=[[2004 AIME I Problems]]}}<br />
<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems/Problem_6&diff=1526531999 AIME Problems/Problem 62021-04-25T19:40:47Z<p>Donutvan: /* Added note explaining that the solution is not accurate */</p>
<hr />
<div>== Problem ==<br />
A transformation of the first [[quadrant]] of the [[coordinate plane]] maps each point <math>(x,y)</math> to the point <math>(\sqrt{x},\sqrt{y}).</math> The [[vertex|vertices]] of [[quadrilateral]] <math>ABCD</math> are <math>A=(900,300), B=(1800,600), C=(600,1800),</math> and <math>D=(300,900).</math> Let <math>k_{}</math> be the area of the region enclosed by the image of quadrilateral <math>ABCD.</math> Find the greatest integer that does not exceed <math>k_{}.</math><br />
<br />
The below solution may not be accurate, see talk page for detail and to discuss.<br />
== Solution ==<br />
<cmath>\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\<br />
B' = & (\sqrt {1800}, \sqrt {600})\\<br />
C' = & (\sqrt {600}, \sqrt {1800})\\<br />
D' = & (\sqrt {300}, \sqrt {900}) \end{eqnarray*}</cmath><br />
<br />
First we see that lines passing through <math>AB</math> and <math>CD</math> have [[equation]]s <math>y = \frac {1}{3}x</math> and <math>y = 3x</math>, respectively. Looking at the points above, we see the equations for <math>A'B'</math> and <math>C'D'</math> are <math>y^2 = \frac {1}{3}x^2</math> and <math>y^2 = 3x^2</math>, or, after manipulation <math>y = \frac {x}{\sqrt {3}}</math> and <math>y = \sqrt {3}x</math>, respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines. <br />
<br />
Now take a look at <math>BC</math> and <math>AD</math>, which have the equations <math>y = - x + 2400</math> and <math>y = - x + 1200</math>. The image equations hence are <math>x^2 + y^2 = 2400</math> and <math>x^2 + y^2 = 1200</math>, respectively, which are the equations for [[circle]]s. <br />
<br />
[[Image:1999_AIME-6.png]]<br />
<br />
To find the area between the circles (actually, parts of the circles), we need to figure out the [[angle]] of the [[arc]]. This could be done by <math>\arctan \sqrt {3} - \arctan \frac {1}{\sqrt {3}} = 60^\circ - 30^\circ = 30^\circ</math>. So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = <math>\frac {30^\circ}{360^\circ}(R^2\pi - r^2\pi) = \frac {1}{12}(2400\pi - 1200\pi) = 100\pi</math>. Hence the answer is <math>\boxed{314}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1999|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=Talk:1999_AIME_Problems/Problem_6&diff=152650Talk:1999 AIME Problems/Problem 62021-04-25T19:35:58Z<p>Donutvan: The solution given on this page is incorrect</p>
<hr />
<div>This solution is not accurate, as checking through a computer graphing program reveals the actual area of the parellelogram to be 300. A quick solution may involve finding the area of the circular sector and multiplying by <math>\frac{3}{\pi}</math>, though I am unsure how to find that reasoning without first knowing the answer. <br />
Another approach is coordinate bashing and attempting to find the altitude of the trapezoid, though this solution is excessively long with plenty of room for errors.</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_3&diff=1526492012 AIME I Problems/Problem 32021-04-25T19:29:11Z<p>Donutvan: Reworded note to avoid a potential ad hominem attack on the author</p>
<hr />
<div>==Problem==<br />
Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.<br />
<br />
==Solution 1==<br />
Call a beef meal <math>B,</math> a chicken meal <math>C,</math> and a fish meal <math>F.</math> Now say the nine people order meals <math>\text{BBBCCCFFF}</math> respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by <math>9</math> to account for the <math>9</math> different ways in which the person to receive the correct meal could be picked. Note, this implies that the dishes are indistinguishable, though the people aren't. For example, two people who order chicken are separate, though if they receive fish, there is only 1 way to order them.<br />
<br />
The problem we must solve is to distribute meals <math>\text{BBCCCFFF}</math> to orders <math>\text{BBCCCFFF}</math> with no matches. The two people who ordered <math>B</math>'s can either both get <math>C</math>'s, both get <math>F</math>'s, or get one <math>C</math> and one <math>F.</math> We proceed with casework.<br />
<br />
<UL><br />
<LI> If the two <math>B</math> people both get <math>C</math>'s, then the three <math>F</math> meals left to distribute must all go to the <math>C</math> people. The <math>F</math> people then get <math>BBC</math> in some order, which gives three possibilities. The indistinguishability is easier to see here, as we distribute the <math>F</math> meals to the <math>C</math> people, and there is only 1 way to order this, as all three meals are the same.</LI><br />
<br />
<LI> If the two <math>B</math> people both get <math>F</math>'s, the situation is identical to the above and three possibilities arise. </LI><br />
<br />
<LI> If the two <math>B</math> people get <math>CF</math> in some order, then the <math>C</math> people must get <math>FFB</math> and the <math>F</math> people must get <math>CCB.</math> This gives <math>2 \cdot 3 \cdot 3 = 18</math> possibilities. </LI><br />
</UL><br />
<br />
Summing across the cases we see there are <math>24</math> possibilities, so the answer is <math>9 \cdot 24 = \boxed{216.}</math><br />
<br />
==Solution 2==<br />
<br />
We only need to figure out the number of ways to order the string <math>BBBCCCFFF</math>, where exactly one <math>B</math> is in the first three positions, one <math>C</math> is in the <math>4^{th}</math> to <math>6^{th}</math> positions, and one <math>F</math> is in the last three positions. There are <math>3^3=27</math> ways to place the first <math>3</math> meals. Then for the other two people, there are <math>2</math> ways to serve their meals. Thus, there are <math>(3\cdot2)^3=\boxed{216}</math> ways to serve their meals.<br />
<br />
Note: This solution gets the correct answer through coincidence and should not be used.<br />
<br />
== Video Solution by Richard Rusczyk ===<br />
<br />
https://artofproblemsolving.com/videos/amc/2012aimei/331<br />
<br />
~ dolphin7<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=T8Ox412AkZc<br />
~Shreyas S<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Donutvanhttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=98003Gmaas2018-10-02T00:58:57Z<p>Donutvan: donutvan is not eaten by Gmaas?</p>
<hr />
<div> :) :O :P {{{=== Gmaas Facts ===}}} P: O: (: :) :)<br />
<br />
- Gmass would like to go to Taco Bell, but Gmaas goes to Wendy's instead. No one knows why.<br />
<br />
- Wasp5 is Gmaas' Priest<br />
<br />
- Gmaas has the following powers: trout, carp, earthworm, and catfish. Gmaas never uses any of them, because Gmaas has an infinite number of powers.<br />
<br />
- Gmaas likes to suprise unsuspecting people.<br />
<br />
- Gmaas loves sparkly mechanical pencils. <br />
<br />
- The bible of Gmaas is this page, and people go to worship Gmaas in the Maas.<br />
<br />
- Gmaas is said to taste like a furry meatball (this was said by BigSams)<br />
<br />
- Leon2000 is Gmaas' Rabbi.<br />
<br />
- Gmaas is both singular and plural.<br />
<br />
- Gmaas eats disbelievers like donutvan for breakfast.(Yet I'm somehow still alive. Do you think Gmaas actually ate my soul?)<br />
<br />
- Gmaas knows Jon Snow's parents.<br />
<br />
- Gmaas is Aegon VI Targaryen.<br />
<br />
- Gmaas created the Marvel Universe.<br />
<br />
- All editors will be escorted to Gmaas heaven after they die.<br />
<br />
- Gmaas won all the wars.<br />
<br />
- Gmail was named after Gmaas.<br />
<br />
- Google was named after Gmaas.<br />
<br />
- Gmaas disapproves of the DC universe.<br />
<br />
- Gmaas won Battle For Dream Island, and also Total Drama Island.<br />
<br />
- Gmaas can lift with Gmaas' will.<br />
<br />
- Gmaas is also the rightful heir to the Iron Throne.<br />
<br />
- Gmaas started the Game of Thrones .<br />
<br />
- Gmaas will also end the Game of Thrones.<br />
<br />
- Gmaas killed Joffrey.<br />
<br />
- Gmaas is Azor Ahai.<br />
<br />
- Gmaas is the prince who was promised.<br />
<br />
- Gmaas created everything after puking.<br />
<br />
- Gmaas is Gordon Maas in disguise (5Space).<br />
<br />
- Gordan's last name was named after Gmaas.<br />
<br />
- Gmaas is more powerful than Gohan.<br />
<br />
- Gmaas is over 90000 years old.<br />
<br />
- Gmaas has 100000000000000000000000000000000000000000000000 cat lives, maybe even more. Gmaas has 0 dog lives.<br />
<br />
- Who wrote ''Harry Potter''? None other than Gmaas himself.<br />
<br />
- Gmaas created the catfish.<br />
<br />
- Gmaas founded Target, but then Gmaas sued them for making the mascot look like a dog when it was supposed to look like Gmaas. <br />
<br />
- Everyone has a bit of Gmaas inside them. EDIT: Except you. You are pure Gmaas. <br />
<br />
- Gmaas likes to eat popsicles. Especially the fudge ones that get him hyper. <br />
<br />
- When Gmaas is hyper Gmaas runs across Washington D.C. grabbing unsuspecting pedestrians and steals their phone, hacks into them and downloads Pub G onto their poor phone.<br />
<br />
- Gmaas' favorite cereal is fruit loops. Gmaas thinks it tastes like unicorns jumping on rainbows.<br />
<br />
- Gmaas thinks that the McChicken has way too much mayonnaise.<br />
<br />
- Gmaas is a champion pillow-fighter.<br />
<br />
- Gmaas colonized Mars.<br />
<br />
- Gmaas also colonized Jupiter, Pluto, and several other galaxies. Gmaas cloned some little Gmaas robots (with Gmaas' amazingly robotic skill of coding) and put them all over a galaxy called Gmaasalaxy. <br />
<br />
- Gmaas has the ability to make every device play "The Duck Song" at will.<br />
<br />
- "The Duck Song" was copied off of the "Gmaas song", but the animators though Gmaas wasn't catchy enough.<br />
<br />
- Gmaas once caught the red dot and ate it.<br />
<br />
- Gmaas' favorite color is neon klfhsadkhfd.<br />
<br />
- Gmaas is a champion pvp Minecraft player.<br />
<br />
- Gmaas is the coach of Tfue, Ninja, Muselk, and Myth.<br />
<br />
- Gmass caught a CP 6000 Mewtwo with a normal Pokeball in Pokemon Go.<br />
<br />
- Gmaas founded Costco.<br />
<br />
- Gmaas does not need to attend the FIFA World Cup. If Gmaas did, Gmaas would automatically beat any team.<br />
<br />
- Gmaas can solve any puzzle instantly besides the 3x3 Rubik's Cube.<br />
<br />
- Gmaas caught a CP 20,000 Mewtwo with a normal Pokeball and no berries blindfolded first try in Pokemon Go. <br />
<br />
- When Gmaas flips coins, they always land tails. Except once when Gmaas was making a bet with Zeus.<br />
<br />
- On Gmaas's math tests, Gmaas always gets <math>\infty</math>%.<br />
<br />
- Gmaas' favorite number is pi. It's also one of Gmaas' favorite foods. <br />
<br />
- Gmaas' burps created all gaseous planets.<br />
<br />
- Gmass beat Luke Robatille in an epic showdown of catnip consumption.<br />
<br />
- Gmaas' wealth is unknown, but it is estimated to be way more than Scrooge's.<br />
<br />
- Gmaas has a summer house on Mars. <br />
<br />
- Gmaas has a winter house on Jupiter.<br />
<br />
- The Earth and all known planets are simply Gmaas' hairballs.<br />
<br />
- Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time using a time-turner.<br />
<br />
- Gmaas also attended Hogwarts and was a prefect. Edit: Gmaas was headmaster.<br />
<br />
- Mrs.Norris is Gmaas's archenemy.<br />
<br />
- Gmaas is a demigod and attends Camp Half-Blood over summer. Gmaas is the counselor for the Apollo cabin. Because cats are demigod counselors too.<br />
<br />
- Gmaas has completed over 2,000 quests, and is very popular throughout Camp Half-Blood. Gmaas has also been to Camp Jupiter.<br />
<br />
- Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br />
<br />
- Gmass knows that their real names are Gmassa Lisa, The Last Domestic Meal, and Far-away Light.<br />
<br />
- Gmaas actually attended all the Ivy Leagues.<br />
<br />
- I am Gmaas.<br />
<br />
- I too am Gmaas.<br />
<br />
- In 2018, Gmaas once challenged Magnus Carlsen to a chess match. Gmaas won every game. <br />
<br />
- But it is I who is Gmaas.<br />
<br />
- Gmaas is us all.<br />
<br />
- Gmaas is all of us yet none of us.<br />
<br />
- Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain.<br />
<br />
- Gmaas's fur is White, Black, Grey, Yellow, Red, Blue, Green, Brown, Pink, Orange, and Purple all at the same time. <br />
<br />
- Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br />
<br />
- Gmaas crossed the Delaware River with Washington.<br />
<br />
- Gmaas also crossed the Atlantic with the pilgrims.<br />
<br />
- If you are able to capture a Gmaas hair, Gmaas will give you some of his Gmaas power.<br />
<br />
- Chuck Norris makes Gmaas jokes.<br />
<br />
- Gmaas is also the ruler of Oceania, Eastasia, and Eurasia.<br />
<br />
- Gmaas killed Big Brother by farting on him. Though Gmaas was caught by the Ministry of Love, Gmaas escaped easily. Edit: Gmaas destroyed the Ministry of Love.<br />
<br />
- Gmaas was not affected by Thano's snap, in fact Gmaas is the creator of the Infinity Stones. <br />
<br />
- Everyone knows that Gmaas is a god.<br />
<br />
- Gmaas also owns Animal Farm. Napoleon was Gmaas servant.<br />
<br />
- Gmaas is the only one who knows where Amelia Earhart is.<br />
<br />
- Gmaas is the only cat that has been proven transcendental.<br />
<br />
- Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br />
<br />
- Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmaas isn't very happy about that, either.<br />
<br />
- The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br />
<br />
- Gmaas dueled Grumpy Cat and won. Gmaas wasn't trying.<br />
<br />
- Gmaas sits on the statue of Pallas and says forevermore.<br />
<br />
- Gmaas is a big fan of Edgar Allan Poe, because he is actually Poe. <br />
<br />
- Gmaas does merely not use USD. He owns it.<br />
<br />
- Gmaas really knows that Roblox is awful and does not play it seriously, thank Gmaas our lord is sane<br />
<br />
- The only god is Gmaas.<br />
<br />
- In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br />
<br />
- "Actually, my name is spelled "GMAAS".<br />
<br />
- Gmaas is the smartest living being in the universe.<br />
<br />
- It was Gmaas who helped Sun Wukong on the Journey to the West.<br />
<br />
- Gmaas is the real creator of Wikipedia.<br />
<br />
- It is said Gmaas could hack any website he desires.<br />
<br />
- Gmaas is the basis of Greek mythology and also Egyptian mythology.<br />
<br />
- Gmaas once sold Google to a man for around <math>12</math> dollars!<br />
<br />
- Gmaas uses a HP printer. It is specifically a HP 21414144124124142141414412412414214141441241241421414144124124142141414412412414 printer.<br />
<br />
- Gmaas owns all AoPS staff including Richard Rusczyk.<br />
<br />
- Richard Rusczyk is one of Gmaas' many code names.<br />
<br />
- Gmaas was there when Yoda was born.<br />
EDIT: Gmaas is Yoda's father<br />
<br />
- Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas. Edit: This is all not true.<br />
<br />
- sseraj once spelled Gmaas as gmASS on accident in Introduction to Geometry (1532).<br />
<br />
- Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br />
<br />
- Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br />
<br />
- Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., Soviet, Russian, and Chinese citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br />
<br />
- "I am sand" destroyed Gmaas in FTW.<br />
<br />
- sseraj posted a picture of Gmaas with a game controller in Introduction to Geometry (1532).<br />
<br />
- Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also <math>\boxed{\text{loves}}</math> <b>Cat</b>ch that fish.<br />
<br />
- Gmaas is Roy Moore's horse in the shape of a cat.<br />
<br />
- Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over <math>289547987693</math> robux and <math>190348</math> in CPR.<br />
<br />
- This is all hypothetical.<br />
<br />
- EDIT: This is all factual.<br />
<br />
- Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br />
(Warrior cats reference).<br />
<br />
- He is capable of salmon powers, according to PunSpark (ask him).<br />
<br />
- The Gmaas told Richard Rusczyk to make AoPS.<br />
<br />
- The Gmaas is everything. Yes, you are part of the Gmaas-Dw789.<br />
<br />
- The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999999999999999999999999999999999999th dimension.<br />
<br />
- He went into a black hole, entered the white hole, got into dimension 15 where people drink tea every day, and stole 154 buckets of tea.<br />
<br />
- Gmaas is "TIRED OF PEOPLE ADDING TO HIS PAGE!!" (Maas 45).<br />
<br />
- Gmaas is Gmaas who is actually Gmaas.<br />
<br />
- Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. His other penguin is called PotatoPenguin19. He is most definitely alive.<br />
<br />
- Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br />
<br />
- Gmaas knows how to hack into top secret aops community pages.<br />
<br />
- Gmaas was a river clan cat who crossed the event horizon of a black hole and came out the other end!<br />
<br />
- Gmaas is king of the first men, the anduls.<br />
<br />
- Gmaas is a well known professor at MEOWston Academy.<br />
<br />
- Gmaas is also the CEO of Caterpillar.<br />
<br />
- Gmaas drinks Starbucks everyday.<br />
<br />
- Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout.<br />
<br />
- Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat. (He also out-grumped grumpy cat!!!)<br />
<br />
- Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM.<br />
<br />
- The owner of sseraj, not pet.<br />
<br />
- The embodiment of life and universe and beyond.<br />
<br />
- Gmaas watches memes of Gmaas.<br />
<br />
- After Death became the GOD OF HYPERDEATH and obtained over 9000 souls.<br />
<br />
-Gmaas invented Rick Rolling.<br />
<br />
- Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)].<br />
<br />
- Gmaas is a certified Slytherin.<br />
<br />
- Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom. <br />
<br />
- Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom).<br />
<br />
- Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br />
<br />
- Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br />
<br />
- Gmaas is a supreme overlord who must be given <math>10^{1000000000000000000000^{1000000000000000000000}}</math> minecraft DIAMONDS.<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 />
- Gmaas is everyone's favorite animal. <br />
<br />
- He lives with sseraj. <br />
<br />
- Gmaas is my favorite pokemon.<br />
<br />
- Gmaas dislikes number theory but enjoys geometry.<br />
<br />
- Gmaas is cool.<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=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS</cmath> supercars, excluding the Purrari and the 138838383 Teslas. <br />
<br />
- He employs AoPS.<br />
<br />
- He is a Gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He has the ability to divide by zero.<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 />
- His stare also turned Medusa into rock, King Midas into gold, and sseraj into sseraj.<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, Gmaas is a cat. Gmaas said so. And science also says so.<br />
<br />
- He is distant relative of Mathcat1234.<br />
<br />
- He cannot be Force choked. Darth Vader learned that the hard way...<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 />
- Then he ate all the food in sseraj's freezer.<br />
<br />
- Gmass once demanded Epic Games to give him 5,000,000 V-bucks for his 569823rd birthday.<br />
<br />
- This is why he does not have an Epic Games account anymore.<br />
<br />
- Gmaas created Epic games, though.<br />
<br />
- Gmaas sightings are not very common. There have only been 30 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 />
<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
<br />
-Prealgebra 2 (1440)<br />
<br />
~Alligator Swamp A 1072 <br />
<br />
~Alligator Swamp B 1073<br />
<br />
~Prealgebra A (1488)<br />
<br />
~Introduction to Algebra A (1170)<br />
<br />
~Introduction to Algebra B (1529)<br />
<br />
~Welcome to Panda Town Gate 1076<br />
<br />
~Welcome to Gmaas Town Gate 1221<br />
<br />
~Welcome to Gmaas Town Gate 1125<br />
<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
<br />
~The other side of the ice in Antarctica<br />
<br />
~Feisty Alligator Swamp 1115<br />
<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
<br />
~Introduction to Counting and Probability 1142 <br />
<br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
<br />
~Intermediate Counting and Probability 1137<br />
<br />
~Intermediate Counting and Probability 1207<br />
<br />
~Posting student surveys<br />
<br />
~USF Castle Walls - Elven Tribe 1203<br />
<br />
~Dark Lord's Hut 1210<br />
<br />
~AMC 10 Problem Series 1200<br />
<br />
~Intermediate Number Theory 1138<br />
<br />
~Intermediate Number Theory 1476<br />
<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
<br />
~Algebra B 1112<br />
<br />
~Intermediate Algebra 1561 7:17 PM 12/11/16<br />
<br />
~Nowhere Else, Tasmania<br />
<br />
~Earth Dimension C-137<br />
<br />
~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br />
<br />
~Intermediate Algebra 1710 9/24/2018<br />
<br />
<br />
- These have all been designated as the most glorious sections of AoPSland now (especially 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 once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<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 />
EDIT: Nobody has yet seen him atop a tribal base yet.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: Gmaas rarely disguises himself as a penguin.<br />
<br />
- Many know that leafy stole dream island. In truth, After leafy stole it, Gmaas stole it himself. (BFDI Reference)<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 />
- Aha!! An impostor!! <br />
http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br />
(look at the acronym).<br />
<br />
- EDIT. The above fact is slightly irrelevant.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99999.\overline{9}}{100000}</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 />
- Gmaas is neither he nor she, Gmaas is above gender.<br />
<br />
- Gmaas is love, Gmaas is life<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mew.<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 />
- EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br />
https://artofproblemsolving.com/community/c287916h1291232<br />
<br />
- EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<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 />
- EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure. <br />
<br />
- Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br />
<br />
- The original owner of Gmaas is Gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into a 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.) EDIT: he wasn't the fourth Peverell brother, but he was a cousin of theirs, and he was the one who advised Ignotus to give up his cloak.<br />
<br />
- It is suspected that Gmaas may be ordering his cyber hairballs 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 was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<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 />
- Edit: Gmaas is neither Anakin Skywalker or Obi-Wan Kenobi as he is trillions of years older.<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 gmewal 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. That was the 15 minutes after he tried to play Taylor Swift music on his 34,000 year old MP3 player in front of sseraj, who, at the time, was handling a bucket of dangerous, radioactive material.<br />
<br />
- Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks. EDIT: Crookshanks was his brother.<br />
<br />
- Gmaas 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 seen 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 has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br />
<br />
- The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br />
<br />
- EDIT: The above fact is somewhat irrelevant.<br />
<br />
- EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has <math>57843504</math> regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
- Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<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 a secret hidden garden full of turnips under sseraj's house.<br />
<br />
- sseraj is "gmaas's person."<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 with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows (except him), not even the leader of the Cult of Skaro.<br />
<br />
- Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br />
<br />
- Gmaas is an excellent driver. EDIT: he was to one who designed the driver's license test, although he didn't bother with the permit test.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
- Donald Trump once sang "All Hail the Chief" to Gmaas, 3 days after being sworn in as US President.<br />
<br />
- Gmaas likes to talk with rrusczyk from time to time.<br />
<br />
- Gmaas can shoot fire from his smelly butt.<br />
<br />
- Gmaas is the reason why the USF has the longest thread on AoPS.<br />
<br />
- Gmass is an avid watcher of the popular T.V. show "Bernie Sanders and the Gauntlet of DOOM".<br />
<br />
- sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying "Who wants to see 5space?" at around 5:16 PM Mountain Time, noting Gmaas was "also 5space".<br />
<br />
- EDIT: he also did it in Introduction to Algebra A once.<br />
<br />
- Gmaas is now my HD background on my Mac. <br />
<br />
- Gmaas is not retarded.<br />
<br />
- In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br />
<br />
- Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br />
<br />
- In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of <math>-GMAAS</math> seconds.<br />
<br />
- Gmaas beat Superman in a fight with ease.<br />
<br />
- Gmaas was an admin of Roblox.<br />
<br />
EDIT: He created Roblox.<br />
<br />
- Gmaas traveled around the world, paying so much <math>MONEY</math> just to eat :D<br />
<br />
- Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br />
Summary:<br />
<br />
- When Gmaas subtracts <math>0.\overline{99}</math> from <math>1</math>, the difference is greater than <math>0</math>.<br />
<br />
- Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br />
<br />
- Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br />
<br />
- The results of the revival are top secret, and nobody knows what happened.<br />
<br />
- sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br />
<br />
- sseraj likes to post a picture of gmaas in every class he passes by.<br />
<br />
- sseraj posted a picture of Gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Endor is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought Gmaas is now wandering space in search for a home.<br />
EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br />
EDIT EDIT: also, glass doesn't care. He can live there no matter what the climate is.<br />
<br />
- Gmaas is the lord of the pokemans.<br />
<br />
- Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will.<br />
<br />
- Picture of Gmaas http://i.imgur.com/PP9xi.png<br />
<br />
- Known by Mike Miller.<br />
<br />
- Gmaas got mad at sseraj once, so he locked him in his own freezer.<br />
<br />
- Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment.<br />
<br />
- Gmass ate slester.<br />
<br />
- A gmass bite is 7000 psi.<br />
<br />
- haha0201 met him.<br />
<br />
- haha0201 comfirms that gmass can talk.<br />
<br />
- gmass likes to eat fur.<br />
<br />
- gmass is bigger than an ant.<br />
<br />
- gmass lives somewhere over the rainbow.<br />
<br />
- Gmaas is an obviously omnipotent cat.<br />
<br />
- ehawk11 met him.<br />
<br />
- sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br />
<br />
- sseraj has posted pictures of gmaas in '"intro to algebra", before class started, with the title, "caption contest". anyone who posted a caption mysteriously vanished in the middle of the night. <br />
EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br />
<br />
- Gmaas has once slept in your bed and made it gry.<br />
<br />
- It is rumored that rrusczyk is actually Gmaas in disguise.<br />
<br />
- Gmaas is suspected to be a Mewtwo in disguise.<br />
<br />
- Gmaas is a cat but has characteristics of every other animal on Earth.<br />
<br />
- Pegasus was modeled off Gmaas.<br />
<br />
- Gmaas is the ruler of the universe and has been known to be the creator of the species "Gmaasians".<br />
<br />
- There is a rumor that Gmaas is starting a poll.<br />
<br />
- Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br />
<br />
- There is a rumored sport called "Gmaas Hunting" where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br />
<br />
- Gmaas burped and caused an earthquake.<br />
<br />
- Gmaas once drank from your pretty teacup.<br />
<br />
- GMAAS IS HERE.... PURRRRRRRRRRRRRRRRRRRR<br />
<br />
- Gmass made, and currently owns the Matrix.<br />
<br />
- The above fact is true. Therefore, this is an illusion.<br />
<br />
- Gmaas is the reason Salah will become better than Ronaldo.<br />
<br />
- Who is Gmaas, really?<br />
Gmass is a heavenly being.<br />
<br />
- Illuminati was a manifestation of Gmaas, but Gmass decided illuminati was not great enough for his godly self.<br />
<br />
- jlikemath has met Gmaas and Gmaas is his best friend.<br />
<br />
- Gmaas hates K-pop<br />
<br />
- Gmaas read Twilight EDIT: ...and SURVIVED.<br />
<br />
- there is a secret code when put into super smash, Gmass would be a playible character. Too bad he didn't say it.<br />
<br />
- Gmaas was a tribute to one of the Hunger Games and came out a Victor and now lives in District 4.<br />
<br />
- Gmaas is the only known creature to survive the destruction of earth in 99999999 years.<br />
<br />
- 5Space (side admin) is Gmaas. <br />
<br />
=== Gmaas photos ===<br />
http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br />
http://disneycreate.wikia.com/wiki/File:Troll_cat_gif_(1).gif <br />
<br />
He was also sighted here.<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- <s>Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".</s>Sorry, this was a rick roll troll.<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 and themoocow are also writing a book about Gmaas.<br />
<br />
- Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br />
<br />
- Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed.<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.<br />
<br />
- gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br />
<br />
- oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here.<br />
<br />
-Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br />
<br />
- No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br />
<br />
- In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br />
https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br />
https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br />
https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br />
https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br />
<br />
<br />
- Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br />
<br />
- <math>Another</math> sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br />
- Yet Another Gmaas sighting ? [https://supportforums.cisco.com/t5/user/viewprofilepage/user-id/45046]<br />
<br />
- Gmaas has been sighted several times on the Global Announcements forum.<br />
<br />
- Gmaas uses the following transportation: <img> http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg </img><br />
<br />
- When Gmaas was mad, he started world wars 1 & 2. It is only because of Gmaas that we have not had World War 3.<br />
<br />
- Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br />
<br />
- Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br />
<br />
- Gmass has beaten every demon in Geometry Dash along with their unnerfed versions and every upcoming demon too.<br />
<br />
- Gmaas likes to whiz on the wilzo.<br />
<br />
- Gmaas has been spotted in AMC 8 Basics.<br />
<br />
- Gmaas is cool.<br />
<br />
- Gmaas hemoon card that does over 9000000 dmg.<br />
<br />
- Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br />
<br />
- Kirby once swallowed Gmaas. Gmaas had to spit him out.<br />
<br />
- Gmaas was the creator of pokemon, and his pokemon card can OHKO anyone in one turn. He is invisible and he will always move first.<br />
<br />
- Gmaas beat Dongmin in The Genius Game Seasons 1, 2, 3, 4, 5, 6, and 7.<br />
<br />
- Gmaas has five letters. Pizza also has five letters. Pizzas are round. Eyes are round. There is an eye in the illuminati symbol. iLLuMiNaTii cOnFiRmEdd.<br />
<br />
- Gmaas knows both 'table' and 'tabular' in LaTeX, and can do them in his sleep. <br />
<br />
- Gmaas hates crawdads with the passion of a thousand burning stars. <br />
<br />
- Gmaas does not hate cheddar cheese. But he doesn't love it either. <br />
<br />
- Gmaas is a cat and not a cat.<br />
<br />
- Gmaas was born on the sun.<br />
<br />
- Gmaas eats tape.<br />
<br />
- Gmaas likes Bubble Gum.<br />
<br />
- Thomas Edison did not invent the lightbulb; Gmaas did.<br />
<br />
- Gmaas eats metal.<br />
<br />
- Gmaas is over 9000 years old!<br />
- EDIT: this is just a DBZ reference, and bears no reality to his true age.<br />
<br />
- Gmaas started the Iron Age.<br />
<br />
- Gmaas made the dinosaurs go extinct.<br />
<br />
- Gmaas created Life... <br />
<br />
- Gmaas created AoPS.<br />
- EDIT: AoPS was actually born out of a small fraction of Gmass's abstract reality, and on;y the sheer amount math can keep it here, (it is also rumored that when he reclaims it, the USF will be deleted, as that is where 83% of the factions of his abstract reality lives, and when people leave USF, more and more escapes.)<br />
<br />
- Gmaas does not like Roblox.<br />
<br />
- Gmaas told Steve Jobs to start a company.<br />
<br />
- Gmaas invented Geometry Dash.<br />
<br />
- Gmaas got to 1634 in Flappy Bird.<br />
<br />
- Gmaas invented Helix Jump.<br />
<br />
- Gmaas can play Happy Birthday on the Violin.<br />
<br />
- Gmaas has mastered Paganini<br />
<br />
- Gmaas discovered Atlantis after one dive underwater.<br />
<br />
- Gmaas made a piano with 89 keys.<br />
<br />
- Gmaas can see the future and change it.<br />
<br />
- Gmaas has every super power you can imagine.<br />
<br />
- Gmaas made a Violin with 9 strings.<br />
<br />
- Gmaas eats rubber bands.<br />
<br />
- Gmaas married Mrs. Norris.<br />
<br />
- Gmaas can fly faster than anything.<br />
<br />
- Grumpy cat is his son.<br />
<br />
- Gmaas eats paper.<br />
<br />
- Gmaas likes lollipops.<br />
<br />
- Gmaas nibbles on pencils.<br />
<br />
- Gmaas is alive.<br />
<br />
- Gmaas is Ninja in Fortnite.<br />
<br />
- Gmaas is love, Gmaas is life<br />
<br />
<math>\text{gmaas is watching you\dots}</math></div>Donutvan