https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Icewolf10&feedformat=atomAoPS Wiki - User contributions [en]2021-04-13T02:33:48ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_24&diff=1487811989 AHSME Problems/Problem 242021-03-07T18:14:42Z<p>Icewolf10: corrected number in one pair</p>
<hr />
<div>== Problem ==<br />
<br />
Five people are sitting at a round table. Let <math>f\geq 0</math> be the number of people sitting next to at least 1 female and <math>m\geq0</math> be the number of people sitting next to at least one male. The number of possible ordered pairs <math>(f,m)</math> is<br />
<br />
<math> \mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 } </math><br />
<br />
== Solution ==<br />
Suppose there are more men than women; then there are between zero and two women.<br />
<br />
If there are no women, the pair is <math>(0,5)</math>. If there is one woman, the pair is <math>(2,5)</math>.<br />
<br />
If there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs <math>(4,5)</math> and <math>(3,4)</math>.<br />
<br />
All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so <math>\rm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1989|num-b=23|num-a=25}} <br />
<br />
[[Category: Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=1449392021 AMC 12A Problems/Problem 232021-02-11T19:29:17Z<p>Icewolf10: </p>
<hr />
<div>==Problem==<br />
Frieda the frog begins a sequence of hops on a <math>3 \times 3</math> grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br />
<br />
<math>\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}</math><br />
<br />
==Solution==<br />
We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have <math>1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}</math>, so the answer is <math>\boxed{D}</math>. ~IceWolf10<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=1449382021 AMC 12A Problems/Problem 232021-02-11T19:28:29Z<p>Icewolf10: added sol</p>
<hr />
<div>==Problem==<br />
Frieda the frog begins a sequence of hops on a <math>3 \times 3</math> grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br />
<br />
<math>\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}</math><br />
<br />
==Solution==<br />
We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have <math>1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&diff=1449272021 AMC 12A Problems/Problem 232021-02-11T19:22:23Z<p>Icewolf10: added problem</p>
<hr />
<div>==Problem==<br />
Frieda the frog begins a sequence of hops on a <math>3 \times 3</math> grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br />
<br />
<math>\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}</math><br />
<br />
==Solution==<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems&diff=1449232021 AMC 10A Problems2021-02-11T19:21:04Z<p>Icewolf10: /* Problem 23 */</p>
<hr />
<div>{{AMC10 Problems|year=2021|ab=A}}<br />
<br />
<br />
<br />
February 4, 2021, is when the AMC 10A starts.<br />
<br />
==Problem 1==<br />
What is the value of<cmath>(2^2-2)-(3^2-3)+(4^2-4)?</cmath><math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~12 </math><br />
<br />
[[2021 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have?<br />
<br />
<math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math><br />
<br />
[[2021 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br />
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math><br />
<br />
[[2021 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
A cart rolls down a hill, travelling <math>5</math> inches the first second and accelerating so that during each successive <math>1</math>-second time interval, it travels <math>7</math> inches more than during the previous <math>1</math>-second interval. The cart takes <math>30</math> seconds to reach the bottom of the hill. How far, in inches, does it travel?<br />
<br />
<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math><br />
<br />
[[2021 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores of terms of <math>k</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at <math>4</math> miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to <math>2</math> miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at <math>3</math> miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?<br />
<br />
<math>\textbf{(A)} ~\frac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\frac{13}{12} \qquad\textbf{(D)} <br />
~\frac{24}{13} \qquad\textbf{(E)} ~2</math><br />
<br />
[[2021 AMC 10A Problems/Problem 6|Solution]]<br />
==Problem 7==<br />
Tom has a collection of <math>13</math> snakes, <math>4</math> of which are purple and <math>5</math> of which are happy. He knows that:<br />
All of his happy snakes can add<br />
None of his purple snakes can subtract<br />
All of his snakes that can't subtract also can't add<br />
<br />
Which of these conclusions can be drawn about Tom's snakes?<br />
<br />
<math>\textbf{(A)}</math> Purple snakes can add.<br />
<math>\textbf{(B)}</math> Purple snakes are happy.<br />
<math>\textbf{(C)}</math> Snakes that can add are purple.<br />
<math>\textbf{(D)}</math> Happy snakes are not purple.<br />
<math>\textbf{(E)}</math> Happy snakes can't subtract.<br />
<br />
[[2021 AMC 10A Problems/Problem 7|Solution]]<br />
==Problem 8==<br />
When a student multiplied the number <math>66</math> by the repeating decimal<br />
<cmath>\underline{1}.\underline{a} \underline{b} \underline{a} \underline{b} \cdots = <br />
\underline{1}.\overline{\underline{ab}}</cmath>Where <math>a</math> and <math>b</math> are digits. He did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit integer <math>\underline{ab}</math>?<br />
<br />
<math>\textbf{(A)} ~15\qquad\textbf{(B)} ~30\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~60\qquad\textbf{(E)} ~75</math><br />
<br />
[[2021 AMC 10A Problems/Problem 8|Solution]]<br />
==Problem 9==<br />
What is the least possible value of <math>(xy-1)^2 + (x+y)^2</math> for real numbers <math>x</math> and <math>y</math>?<br />
<br />
==Problem 10==<br />
Which of the following is equivalent to<br />
<cmath>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</cmath><br />
<math>\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}</math><br />
<br />
==Problem 11==<br />
For which of the following integers <math>b</math> is the base-<math>b</math> number <math>2021_b - 221_b</math> not divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8</math><br />
<br />
==Problem 12==<br />
==Problem 13==<br />
What is the volume of tetrahedron <math>ABCD</math> with edge lengths <math>AB = 2</math>, <math>AC = 3</math>, <math>AD = 4</math>, <math>BC = \sqrt{13}</math>, <math>BD = 2\sqrt{5}</math>, and <math>CD = 5</math> ?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math><br />
<br />
[[2021 AMC 10A Problems/Problem 13|Solution]]<br />
==Problem 14==<br />
All the roots of polynomial <math>z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16</math> are positive integers, possibly repeated. What is the value of <math>B</math>?<br />
<br />
<math>\textbf{(A)} ~-88\qquad\textbf{(B)} ~-80\qquad\textbf{(C)} ~-64\qquad\textbf{(D)} ~-41\qquad\textbf{(E)} ~-40</math><br />
<br />
==Problem 15==<br />
Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)<br />
<br />
<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math><br />
<br />
[[2021 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
In the following list of numbers, the integer <math>n</math> appears <math>n</math> times in the list for <math>1 \leq n \leq 200</math>.<br />
<cmath>1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \cdot, 200, 200, \cdot , 200</cmath>What is the median of the numbers in this list?<br />
<br />
<math>\textbf{(A)} ~100.5\qquad\textbf{(B)} ~134\qquad\textbf{(C)} ~142\qquad\textbf{(D)} ~150.5\qquad\textbf{(E)} ~167</math><br />
<br />
==Problem 17==<br />
Trapezoid <math>ABCD</math> has <math>\overline{AB} \parallel \overline{CD}</math>, <math>BC = CD = 43</math>, and <math>\overline{AD} \perp \overline{BD}</math>. Let <math>O</math> be the intersection of the diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>, and let <math>P</math> be the midpoint of <math>\overline{BD}</math>. GIven that <math>OP = 11</math>, the length <math>AD</math> can be written in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. What is <math>m + n</math>?<br />
<br />
<math>\textbf{(A)} ~65\qquad\textbf{(B)} ~132\qquad\textbf{(C)} ~157\qquad\textbf{(D)} ~194\qquad\textbf{(E)} ~215</math><br />
<br />
==Problem 18==<br />
Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a \cdot b) = f(a) + f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Suppose that <math>f</math> also have the property that <math>f(p) = p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x) < 0</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{17}{32}\qquad\textbf{(B)} ~\frac{11}{16}\qquad\textbf{(C)} ~\frac{7}{9}\qquad\textbf{(D)} ~\frac{7}{6} \qquad\textbf{(E)} ~\frac{25}{11}</math><br />
<br />
==Problem 19==<br />
The area of the region bounded by the graph of <cmath>x^2+y^2 = 3|x-y| + 3|x+y|</cmath>is <math>m+n\pi</math>, where <math>m</math> and <math>n</math> are integers. What is <math>m + n</math>?<br />
<br />
<math>\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54</math><br />
<br />
[[2021 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
In how many ways can the sequence <math>1, 2, 3, 4, 5</math> be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?<br />
<br />
<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~32\qquad\textbf{(E)} ~44</math><br />
<br />
==Problem 21==<br />
Let <math>ABCDEF</math> be an equiangular hexagon. The lines <math>AB, CD,</math> and <math>EF</math> determine a triangle with area <math>192\sqrt{3}</math>, and the lines <math>BC, DE,</math> and <math>FA</math> determine a triangle with area <math>324\sqrt{3}</math>. The perimeter of hexagon <math>ABCDEF</math> can be expressed as <math>m +n\sqrt{p}</math>, where <math>m, n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>?<br />
<br />
<math>\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63</math><br />
<br />
==Problem 22==<br />
Hiram's algebra notes are <math>50</math> pages long and are printed on <math>25</math> sheets of paper; the first sheet contains pages <math>1</math> and <math>2</math>, the second sheet contains pages <math>3</math> and <math>4</math>, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly <math>19</math>. How many sheets were borrowed?<br />
<br />
<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20</math><br />
<br />
==Problem 23==<br />
Frieda the frog begins a sequence of hops on a <math>3 \times 3</math> grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br />
<br />
<math>\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}</math><br />
<br />
==Problem 25==<br />
How many ways are there to place <math>3</math> indistinguishable red chips, <math>3</math> indistinguishable blue chips, and <math>3</math> indistinguishable green chips in the squares of a <math>3 \times 3</math> grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally.<br />
<br />
<math>\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36</math></div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_13&diff=1449142021 AMC 10A Problems/Problem 132021-02-11T19:18:20Z<p>Icewolf10: revised solution</p>
<hr />
<div>==Problem==<br />
What is the volume of tetrahedron <math>ABCD</math> with edge lengths <math>AB = 2</math>, <math>AC = 3</math>, <math>AD = 4</math>, <math>BC = \sqrt{13}</math>, <math>BD = 2\sqrt{5}</math>, and <math>CD = 5</math> ?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math><br />
<br />
==Solution==<br />
Drawing the tetrahedron out and testing side lengths, we realize that the triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid: <math>\frac{3\cdot4\cdot2}{3\cdot2}=4</math>, so we have an answer of <math>\boxed{C}</math>. ~IceWolf10<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_13&diff=1449082021 AMC 10A Problems/Problem 132021-02-11T19:16:09Z<p>Icewolf10: added solution</p>
<hr />
<div>==Problem==<br />
What is the volume of tetrahedron <math>ABCD</math> with edge lengths <math>AB = 2</math>, <math>AC = 3</math>, <math>AD = 4</math>, <math>BC = \sqrt{13}</math>, <math>BD = 2\sqrt{5}</math>, and <math>CD = 5</math> ?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math><br />
<br />
==Solution==<br />
Drawing the tetrahedron out and testing side lengths, we realize that the triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron, obtaining an answer of <math>\boxed{C}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems&diff=1448932021 AMC 10A Problems2021-02-11T19:11:36Z<p>Icewolf10: added 13</p>
<hr />
<div>{{AMC10 Problems|year=2021|ab=A}}<br />
<br />
<br />
<br />
February 4, 2021, is when the AMC 10A starts.<br />
<br />
==Problem 1==<br />
What is the value of<cmath>(2^2-2)-(3^2-3)+(4^2-4)?</cmath><math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~12 </math><br />
<br />
[[2021 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have?<br />
<br />
<math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math><br />
<br />
[[2021 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br />
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math><br />
<br />
[[2021 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
A cart rolls down a hill, travelling <math>5</math> inches the first second and accelerating so that during each successive <math>1</math>-second time interval, it travels <math>7</math> inches more than during the previous <math>1</math>-second interval. The cart takes <math>30</math> seconds to reach the bottom of the hill. How far, in inches, does it travel?<br />
<br />
<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math><br />
<br />
[[2021 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores of terms of <math>k</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at <math>4</math> miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to <math>2</math> miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at <math>3</math> miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?<br />
<br />
<math>\textbf{(A)} ~\frac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\frac{13}{12} \qquad\textbf{(D)} <br />
~\frac{24}{13} \qquad\textbf{(E)} ~2</math><br />
<br />
[[2021 AMC 10A Problems/Problem 6|Solution]]<br />
==Problem 7==<br />
Tom has a collection of <math>13</math> snakes, <math>4</math> of which are purple and <math>5</math> of which are happy. He knows that:<br />
All of his happy snakes can add<br />
None of his purple snakes can subtract<br />
All of his snakes that can't subtract also can't add<br />
<br />
Which of these conclusions can be drawn about Tom's snakes?<br />
<br />
<math>\textbf{(A)}</math> Purple snakes can add.<br />
<math>\textbf{(B)}</math> Purple snakes are happy.<br />
<math>\textbf{(C)}</math> Snakes that can add are purple.<br />
<math>\textbf{(D)}</math> Happy snakes are not purple.<br />
<math>\textbf{(E)}</math> Happy snakes can't subtract.<br />
<br />
[[2021 AMC 10A Problems/Problem 7|Solution]]<br />
==Problem 8==<br />
When a student multiplied the number <math>66</math> by the repeating decimal<br />
<cmath>\underline{1}.\underline{a} \underline{b} \underline{a} \underline{b} \cdots = <br />
\underline{1}.\overline{\underline{ab}}</cmath>Where <math>a</math> and <math>b</math> are digits. He did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit integer <math>\underline{ab}</math>?<br />
<br />
<math>\textbf{(A)} ~15\qquad\textbf{(B)} ~30\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~60\qquad\textbf{(E)} ~75</math><br />
<br />
[[2021 AMC 10A Problems/Problem 8|Solution]]<br />
==Problem 9==<br />
What is the least possible value of <math>(xy-1)^2 + (x+y)^2</math> for real numbers <math>x</math> and <math>y</math>?<br />
<br />
==Problem 10==<br />
Which of the following is equivalent to<br />
<cmath>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</cmath><br />
<math>\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}</math><br />
<br />
==Problem 11==<br />
==Problem 12==<br />
==Problem 13==<br />
What is the volume of tetrahedron <math>ABCD</math> with edge lengths <math>AB = 2</math>, <math>AC = 3</math>, <math>AD = 4</math>, <math>BC = \sqrt{13}</math>, <math>BD = 2\sqrt{5}</math>, and <math>CD = 5</math> ?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math><br />
<br />
[[2021 AMC 10A Problems/Problem 13|Solution]]<br />
==Problem 14==<br />
==Problem 15==<br />
Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)<br />
<br />
<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math><br />
<br />
[[2021 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
==Problem 17==<br />
==Problem 18==<br />
==Problem 19==<br />
The area of the region bounded by the graph of <cmath>x^2+y^2 = 3|x-y| + 3|x+y|</cmath>is <math>m+n\pi</math>, where <math>m</math> and <math>n</math> are integers. What is <math>m + n</math>?<br />
<br />
<math>\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54</math><br />
<br />
[[2021 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
In how many ways can the sequence <math>1, 2, 3, 4, 5</math> be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?<br />
<br />
<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~32\qquad\textbf{(E)} ~44</math><br />
<br />
==Problem 21==<br />
Let <math>ABCDEF</math> be an equiangular hexagon. The lines <math>AB, CD,</math> and <math>EF</math> determine a triangle with area <math>192\sqrt{3}</math>, and the lines <math>BC, DE,</math> and <math>FA</math> determine a triangle with area <math>324\sqrt{3}</math>. The perimeter of hexagon <math>ABCDEF</math> can be expressed as <math>m +n\sqrt{p}</math>, where <math>m, n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>?<br />
<br />
<math>\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63</math><br />
<br />
==Problem 22==<br />
Hiram's algebra notes are <math>50</math> pages long and are printed on <math>25</math> sheets of paper; the first sheet contains pages <math>1</math> and <math>2</math>, the second sheet contains pages <math>3</math> and <math>4</math>, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly <math>19</math>. How many sheets were borrowed?<br />
<br />
<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20</math><br />
<br />
==Problem 23==<br />
==Problem 24==<br />
==Problem 25==<br />
How many ways are there to place <math>3</math> indistinguishable red chips, <math>3</math> indistinguishable blue chips, and <math>3</math> indistinguishable green chips in the squares of a <math>3 \times 3</math> grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally.<br />
<br />
<math>\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36</math></div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_15&diff=1448722021 AMC 10A Problems/Problem 152021-02-11T19:04:14Z<p>Icewolf10: </p>
<hr />
<div>==Problem==<br />
Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)<br />
<br />
<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math><br />
<br />
==Solution==<br />
Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>, and the answer is <math>\boxed{C}</math>. ~IceWolf10<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_15&diff=1448692021 AMC 10A Problems/Problem 152021-02-11T19:02:09Z<p>Icewolf10: added formatting</p>
<hr />
<div>==Problem==<br />
Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)<br />
<br />
<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math><br />
<br />
==Solution==<br />
Assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>, and the answer is <math>\boxed{C}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_15&diff=1448652021 AMC 10A Problems/Problem 152021-02-11T19:00:21Z<p>Icewolf10: added solution</p>
<hr />
<div>Assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>.</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems&diff=1448602021 AMC 10A Problems2021-02-11T18:59:27Z<p>Icewolf10: </p>
<hr />
<div>{{AMC10 Problems|year=2021|ab=A}}<br />
<br />
<br />
<br />
February 4, 2021, is when the AMC 10A starts.<br />
<br />
==Problem 1==<br />
What is the value of<cmath>(2^2-2)-(3^2-3)+(4^2-4)?</cmath><math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~12 </math><br />
<br />
[[2021 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have?<br />
<br />
<math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math><br />
<br />
[[2021 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br />
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math><br />
<br />
[[2021 AMC 10A Problems/Problem 3|Solution]]<br />
==Problem 4==<br />
A cart rolls down a hill, travelling <math>5</math> inches the first second and accelerating so that during each successive <math>1</math>-second time interval, it travels <math>7</math> inches more than during the previous <math>1</math>-second interval. The cart takes <math>30</math> seconds to reach the bottom of the hill. How far, in inches, does it travel?<br />
<br />
<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math><br />
<br />
[[2021 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores of terms of <math>k</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at <math>4</math> miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to <math>2</math> miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at <math>3</math> miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?<br />
<br />
<math>\textbf{(A)} ~\frac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\frac{13}{12} \qquad\textbf{(D)} <br />
~\frac{24}{13} \qquad\textbf{(E)} ~2</math><br />
<br />
[[2021 AMC 10A Problems/Problem 6|Solution]]<br />
==Problem 7==<br />
Tom has a collection of <math>13</math> snakes, <math>4</math> of which are purple and <math>5</math> of which are happy. He knows that:<br />
All of his happy snakes can add<br />
None of his purple snakes can subtract<br />
All of his snakes that can't subtract also can't add<br />
<br />
Which of these conclusions can be drawn about Tom's snakes?<br />
<br />
<math>\textbf{(A)}</math> Purple snakes can add.<br />
<math>\textbf{(B)}</math> Purple snakes are happy.<br />
<math>\textbf{(C)}</math> Snakes that can add are purple.<br />
<math>\textbf{(D)}</math> Happy snakes are not purple.<br />
<math>\textbf{(E)}</math> Happy snakes can't subtract.<br />
<br />
[[2021 AMC 10A Problems/Problem 7|Solution]]<br />
==Problem 8==<br />
When a student multiplied the number <math>66</math> by the repeating decimal<br />
<cmath>\underline{1}.\underline{a} \underline{b} \underline{a} \underline{b} \cdots = <br />
\underline{1}.\overline{\underline{ab}}</cmath>Where <math>a</math> and <math>b</math> are digits. He did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit integer <math>\underline{ab}</math>?<br />
<br />
<math>\textbf{(A)} ~15\qquad\textbf{(B)} ~30\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~60\qquad\textbf{(E)} ~75</math><br />
<br />
[[2021 AMC 10A Problems/Problem 8|Solution]]<br />
==Problem 9==<br />
==Problem 10==<br />
==Problem 11==<br />
==Problem 12==<br />
==Problem 13==<br />
==Problem 14==<br />
==Problem 15==<br />
Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)<br />
<br />
<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math><br />
<br />
[[2021 AMC 10A Problems/Problem 15|Solution]]<br />
==Problem 16==<br />
==Problem 17==<br />
==Problem 18==<br />
==Problem 19==<br />
==Problem 20==<br />
==Problem 21==<br />
==Problem 22==<br />
==Problem 23==<br />
==Problem 24==<br />
==Problem 25==</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems&diff=1448592021 AMC 10A Problems2021-02-11T18:58:33Z<p>Icewolf10: </p>
<hr />
<div>{{AMC10 Problems|year=2021|ab=A}}<br />
<br />
<br />
<br />
February 4, 2021, is when the AMC 10A starts.<br />
<br />
==Problem 1==<br />
What is the value of<cmath>(2^2-2)-(3^2-3)+(4^2-4)?</cmath><math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~12 </math><br />
<br />
[[2021 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have?<br />
<br />
<math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math><br />
<br />
[[2021 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br />
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math><br />
<br />
[[2021 AMC 10A Problems/Problem 3|Solution]]<br />
==Problem 4==<br />
A cart rolls down a hill, travelling <math>5</math> inches the first second and accelerating so that during each successive <math>1</math>-second time interval, it travels <math>7</math> inches more than during the previous <math>1</math>-second interval. The cart takes <math>30</math> seconds to reach the bottom of the hill. How far, in inches, does it travel?<br />
<br />
<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math><br />
<br />
[[2021 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores of terms of <math>k</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at <math>4</math> miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to <math>2</math> miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at <math>3</math> miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?<br />
<br />
<math>\textbf{(A)} ~\frac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\frac{13}{12} \qquad\textbf{(D)} <br />
~\frac{24}{13} \qquad\textbf{(E)} ~2</math><br />
<br />
[[2021 AMC 10A Problems/Problem 6|Solution]]<br />
==Problem 7==<br />
Tom has a collection of <math>13</math> snakes, <math>4</math> of which are purple and <math>5</math> of which are happy. He knows that:<br />
All of his happy snakes can add<br />
None of his purple snakes can subtract<br />
All of his snakes that can't subtract also can't add<br />
<br />
Which of these conclusions can be drawn about Tom's snakes?<br />
<br />
<math>\textbf{(A)}</math> Purple snakes can add.<br />
<math>\textbf{(B)}</math> Purple snakes are happy.<br />
<math>\textbf{(C)}</math> Snakes that can add are purple.<br />
<math>\textbf{(D)}</math> Happy snakes are not purple.<br />
<math>\textbf{(E)}</math> Happy snakes can't subtract.<br />
<br />
[[2021 AMC 10A Problems/Problem 7|Solution]]<br />
==Problem 8==<br />
When a student multiplied the number <math>66</math> by the repeating decimal<br />
<cmath>\underline{1}.\underline{a} \underline{b} \underline{a} \underline{b} \cdots = <br />
\underline{1}.\overline{\underline{ab}}</cmath>Where <math>a</math> and <math>b</math> are digits. He did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit integer <math>\underline{ab}</math>?<br />
<br />
<math>\textbf{(A)} ~15\qquad\textbf{(B)} ~30\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~60\qquad\textbf{(E)} ~75</math><br />
<br />
[[2021 AMC 10A Problems/Problem 8|Solution]]<br />
==Problem 9==<br />
==Problem 10==<br />
==Problem 11==<br />
==Problem 12==<br />
==Problem 13==<br />
==Problem 14==<br />
==Problem 15==<br />
Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)<br />
<br />
<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math><br />
[[2021 AMC 10A Problems/Problem 15|Solution]]<br />
==Problem 16==<br />
==Problem 17==<br />
==Problem 18==<br />
==Problem 19==<br />
==Problem 20==<br />
==Problem 21==<br />
==Problem 22==<br />
==Problem 23==<br />
==Problem 24==<br />
==Problem 25==</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=User:Icewolf10&diff=142890User:Icewolf102021-01-21T01:45:26Z<p>Icewolf10: Created page with "an icy wolf"</p>
<hr />
<div>an icy wolf</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=142889IceWolf102021-01-21T01:44:01Z<p>Icewolf10: Blanked the page</p>
<hr />
<div></div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Answer_Key&diff=1428882020 AMC 8 Answer Key2021-01-21T01:41:57Z<p>Icewolf10: </p>
<hr />
<div>#E<br />
#C<br />
#D<br />
#B<br />
#C<br />
#A<br />
#C<br />
#C<br />
#D<br />
#C<br />
#E<br />
#A<br />
#B<br />
#D<br />
#C<br />
#E<br />
#B<br />
#A<br />
#B<br />
#B<br />
#A<br />
#E<br />
#B<br />
#A<br />
#A</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Answer_Key&diff=1428872020 AMC 8 Answer Key2021-01-21T01:39:52Z<p>Icewolf10: added answer key</p>
<hr />
<div>1. E<br />
2. C<br />
3. D<br />
4. B<br />
5. C<br />
6. A<br />
7. C<br />
8. C<br />
9. D<br />
10. C<br />
11. E<br />
12. A<br />
13. B<br />
14. D<br />
15. C<br />
16. E<br />
17. B<br />
18. A<br />
19. B<br />
20. B<br />
21. A<br />
22. E<br />
23. B<br />
24. A<br />
25. A</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_23&diff=1399552002 AMC 12A Problems/Problem 232020-12-19T02:34:47Z<p>Icewolf10: Undo revision 138116 by Mathdragon3.14 (talk)</p>
<hr />
<div>==Problem==<br />
In triangle <math>ABC</math>, side <math>AC</math> and the [[perpendicular bisector]] of <math>BC</math> meet in point <math>D</math>, and <math>BD</math> bisects <math>\angle ABC</math>. If <math>AD=9</math> and <math>DC=7</math>, what is the area of triangle ABD? <br />
<br />
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math><br />
<br />
==Solution==<br />
'''Solution 1'''<br />
<asy><br />
unitsize(0.25 cm);<br />
pair A, B, C, D, M;<br />
A = (0,0);<br />
B = (88/9, 28*sqrt(5)/9);<br />
C = (16,0);<br />
D = 9/16*C;<br />
M = (B + C)/2;<br />
draw(A--B--C--cycle);<br />
draw(B--D--M);<br />
label("$A$", A, SW);<br />
label("$B$", B, N);<br />
label("$C$", C, SE);<br />
label("$D$", D, S);<br />
</asy><br />
Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let <math>x = \angle C</math>, so that <math>B=2x</math> from given and the previous deducted. Then <math>\angle ABD=x, \angle ADB=2x</math> because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means <math> \triangle ABD</math> and <math>\triangle ACB</math> are [[Similarity (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>.<br />
<br />
Then by using [[Heron's Formula]] on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>.<br />
<br />
'''Solution 2'''<br />
<br />
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, <math>BD = DC = 7</math> and <math>BM = MC</math>. Also, by the angle bisector theorem, <math>\frac {AB}{BC} = \frac{9}{7}</math>. Thus, let <math>AB = 9x</math> and <math>BC = 7x</math>. In addition, <math>BM = 3.5x</math>. <br />
<br />
Thus, <math>\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}</math>. Additionally, using the Law of Cosines and the fact that <math>\angle CBD = \angle ABD</math>, <math>81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD</math><br />
<br />
Substituting and simplifying, we get <math>x = 4/3</math><br />
<br />
Thus, <math>AB = 12</math>. We now know all sides of <math> \triangle ABD</math>. Using [[Heron's Formula]] on <math>\triangle ABD</math>, <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math><br />
<br />
'''Solution 3'''<br />
<br />
Note that because the perpendicular bisector and angle bisector meet at side <math>AC</math> and <math>CD = BD</math> as triangle <math>BDC</math> is isosceles, so <math>BD = 7</math>. By the angle bisector theorem, we can express <math>AB</math> and <math>BC</math> as <math>9x</math> and <math>7x</math> respectively. We try to find <math>x</math> through Stewart's Theorem. So<br />
<br />
<math>16(7^2+9\cdot7) = (7x)^2 \cdot 9 + (9x)^2 \cdot 7</math><br />
<br />
<math>16(49+63) = (49 \cdot 9 + 81 \cdot 7)x^2</math><br />
<br />
<math>16(49+63) = 9(49+63) \cdot x^2</math><br />
<br />
<math>x^2 = \frac{16}{9}</math><br />
<br />
<math>x=\frac{4}{3}</math><br />
<br />
We plug this to find that the sides of <math>\triangle ABD</math> are <math>12,7,9</math>. By Heron's formula, the area is <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. ~skyscraper<br />
<br />
==See Also==<br />
{{AMC12 box|year=2002|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=138571IceWolf102020-11-26T22:46:55Z<p>Icewolf10: </p>
<hr />
<div>hello join the pink penguinaphants army lmao</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=138570IceWolf102020-11-26T22:46:30Z<p>Icewolf10: Replaced content with "hello join the pp army lmfao"</p>
<hr />
<div>hello join the pp army lmfao</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_4&diff=1296572020 AMC 10A Problems/Problem 42020-07-28T19:59:07Z<p>Icewolf10: Undo revision 129653 by Icewolf10 (talk)</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #3]] and [[2020 AMC 10A Problems|2020 AMC 10A #4]]}}<br />
<br />
==Problem==<br />
<br />
A driver travels for <math>2</math> hours at <math>60</math> miles per hour, during which her car gets <math>30</math> miles per gallon of gasoline. She is paid <math>\$0.50</math> per mile, and her only expense is gasoline at <math>\$2.00</math> per gallon. What is her net rate of pay, in dollars per hour, after this expense?<br />
<br />
<math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26 </math><br />
<br />
==Solution==<br />
Since the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends \$4 per hour on gas. If she gets \$0.50 per mile, then she gets \$30 per hour of driving. Subtracting the gas cost, her net rate of pay per hour is <math>\boxed{\textbf{(E)}\ 26}</math>.<br />
~mathsmiley<br />
<br />
==Video Solution==<br />
https://youtu.be/WUcbVNy2uv0<br />
<br />
~IceMatrix<br />
<br />
https://www.youtube.com/watch?v=7-3sl1pSojc<br />
<br />
~bobthefam<br />
<br />
https://youtu.be/Dj_DFoZO-xw<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=3|num-a=5}}<br />
{{AMC12 box|year=2020|ab=A|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_4&diff=1296532020 AMC 10A Problems/Problem 42020-07-28T18:36:21Z<p>Icewolf10: </p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #3]] and [[2020 AMC 10A Problems|2020 AMC 10A #4]]}}<br />
<br />
==Problem==<br />
<br />
A driver travels for <math>2</math> hours at <math>60</math> miles per hour, during which her car gets <math>30</math> miles per gallon of gasoline. She is paid <math>\$0.50</math> per mile, and her only expense is gasoline at <math>\$2.00</math> per gallon. What is her net rate of pay, in dollars per hour, after this expense?<br />
<br />
<math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26 </math><br />
<br />
==Solution==<br />
Since the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends \$4 per hour on gas. If she gets \$0.50 per mile, then she gets \$30 per hour of driving. Subtracting the gas cost, her net rate of pay per hour is <math>\boxed{\textbf{(E)}\ 26}</math>.<br />
~mathsmiley<br />
<br />
==Video Solution==<br />
https://youtu.be/WUcbVNy2uv0<br />
<br />
~IceMatrix<br />
<br />
https://www.youtube.com/watch?v=7-3sl1pSojc<br />
<br />
~bobthefam<br />
<br />
https://youtu.be/Dj_DFoZO-xw<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Random_module&diff=127109Random module2020-06-30T16:55:32Z<p>Icewolf10: typo in spelling "them" lol</p>
<hr />
<div>The '''random [[module (Python)|module]]''' is used to generate pseudo-random numbers in [[Python]].<br />
<br />
==Useful Functions==<br />
The following are functions in the random module, see the page on [[module (Python)|modules]] for directions on how to import them into your program.<br />
*<tt>random.'''random'''()</tt> returns a random [[floating point number]] between 0.0 and 1.0.<br />
*<tt>random.'''randint'''(a,b)</tt> returns a random [[integer]] between a and b, inclusive.<br />
*<tt>random.'''uniform'''(a,b)</tt> returns a random floating point number between a and b. <br />
*<tt>random.'''choice'''(s)</tt> returns a random element of [[sequence (Python)|sequence]] <tt>s</tt>. For example, <tt>random.choice([1,3,7])</tt> will return either 1, 3, or 7.<br />
*<tt>random.'''shuffle'''(s)</tt> shuffles sequence <tt>s</tt> in place. Hence, the original s is changed and nothing is returned. If you want to preserve the original order also you need to copy the list first (<tt>myListCopy = myList[:]</tt> would work).<br />
<br />
==See Also==<br />
*[http://docs.python.org/py3k/library/random.html Python 3.2 Documentation]<br />
*[http://en.wikipedia.org/wiki/Pseudorandom_number_generator Pseudorandom Number Generators on Wikipedia]<br />
<br />
[[Category: Introduction to Programming]]<br />
[[Category:Python Modules]]</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=126093IceWolf102020-06-20T23:32:46Z<p>Icewolf10: </p>
<hr />
<div>Hello!<br />
Please join the [[Pink Penguinaphants Army]], if you haven't already.<br />
<br />
I am the second member to join the PP Army, more than 2 years old. Sixth place on the leaderboard. I am also the founder of the Syntagma RPG Series, a successful RPG series, as its name implies. The first installment, Syntagma, got more than 20k posts in 81 days.<br />
<br />
Bragging aside...<br />
Uhh check out my blog if you want lol</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Talk:Gmaas&diff=126092Talk:Gmaas2020-06-20T23:28:48Z<p>Icewolf10: </p>
<hr />
<div>== Caution GMAAS followers ==<br />
Take caution GMAAS followers as some evil anti-GMAAS people will try to lure you and make you believe that GMAAS is fake. I say, however, GMAAS is real! And will always be real!<br />
<br />
Now let's get some things straight. GMAAS never changes his oh so powerful name. Neither is he Grumpy Cat. Grumpy Cat is a mere mortal, GMAAS on the other hand lives forever. Another thing, the new GMAAS page is [[GMAAS | here]], it was restored by me as the old one was locked.<br />
<br />
Let me remind all of you that times like these will be hard as the number of GMAAS followers has decreased.<br />
But remain faithful!<br />
<br />
--[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 18:49, 2 April 2020 (EDT)<br />
<br />
It seems that the only Gmaas follower is Piphi, however, it shows that this tab is in use and being updated daily, so the page is still in use. This page WILL NOT be deleted, since it shows that at least one user is watching this page.<br />
<br />
== Summoning GMAAS ==<br />
Follow the following steps to summon [[Gmaas]]:<br />
<br />
1. Draw a circle and circumscribe it with a regular hexagon and an equilateral triangle.<br />
<br />
2. Write the numerical value of <math>17^{36}</math> along the edge of the circle.<br />
<br />
3. Write the numerical value of <math>33^{29}</math> along the edge of the hexagon.<br />
<br />
4. Write the numerical value of <math>\cot(0)</math> 5 centimeters above the hexagon.<br />
<br />
5. Write the numerical value of <math>\tan(\frac{\pi}{2} rad)</math> 5 centimeters below the hexagon.<br />
<br />
6. Write the numerical value of <math>\ln(0)</math> inside the circle.<br />
<br />
7. Write the numerical value of the melting point of water inside the circle. Include units and write 15 significant digits.<br />
<br />
8. Write the numerical value of the gravitational constant inside the circle. Include units and write 25 significant digits.<br />
<br />
9. Carry the paper with the circle and hexagon in your left hand.<br />
<br />
10. Recite the value of <math>\pi^{e^2}</math> and include <math>\pi^{e^2}</math> (rounded to the nearest septillionth) digits.<br />
<br />
11. Travel at the speed of light with that paper and Gmaas will be summoned.<br />
<div><br />
<br />
<div><br />
EDIT: The author congratulates the previous editor for his research on the cutting edge of Gmaasology. He/she has been awarded the Nobel Prize in Gmaasology for his discovery. However, the aforementioned summoning has several problems: it does not specify the dimensions of the hexagon or the circle; the fact that it has to be drawn on a paper; the size of the paper; the pressure of the water being melted; whether <math>\frac{\pi}{2}</math> radians, degrees, or gradians; the exact location if the value of <math>17^{36}</math> and <math>33^{29}</math> in the diagram and the base those numbers should be written in; the units of the melting point of water or the units of the gravitational constant; or the location where Gmaas will be summoned. He might be summoned on the other side of the universe.<br />
<div><br />
<br />
<div><br />
The person who wrote the steps to summon Gmaas says:<br />
<div><br />
<br />
<div><br />
When I followed this procedure to summon Gmaas, I used a paper that is 15 inches long and 11 inches wide. It is currently unknown whether paper of other dimensions can be used to summon Gmaas. For step 7, the pressure of the water being melted is at 611.66 Pascals, which matches the triple point of water. Also, the units of the melting points of water and the gravitational constant should be written in terms of SI base units. The circle used should have a radius of 5 inches. If you follow this procedure, Gmaas will be summoned for about 3.53 nanoseconds at a random position in the planet you are on. The exact time Gmaas will be summoned depends on how accurately you follow the procedure. For example, if the melting point of water and the gravitational constant are accurate to 200 significant digits, then Gmaas will be summoned for 77.8 microseconds. If they are accurate to 400 significant digits, Gmaas will be summoned for 1.7 seconds. The function that relates the number of significant digits and the time Gmaas will be summoned is still unknown.<br />
<br />
EDIT: Based on the two data points, the function that relates the number of significant digits and the time Gmaas will be summoned is <math>t=.0081111s+.5444</math>.<br />
<br />
Edit: Gmaas no longer appears with a hexagon. You must now draw a heptagon.<br />
<br />
Edit: Gmaas never appears for more than 10 seconds, and burns the paper with him, forcing you do do it all over again to summon him again<br />
<br />
Edit: If you are able to edit these steps, then you have been partially (<math>\frac{1}{1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000}</math>) enlightened by Gmaas<br />
<br />
Edit Names For Challenge:<br />
juliankuang<br />
<br />
== Gmaas ==<br />
<br />
NOOOOO. THEY DELETED THE PAGE!!! https://artofproblemsolving.com/wiki/index.php?title=Gmaas#Known_Facts_About_gmaas<br />
<br />
<br />
HOW DARE THEY! THIS IS HORRIBLE!!!!!! THIS IS NOT GOOD AT ALL!!!! - asdf334 ;( currently mourning this loss<br />
<br />
<br />
NOPE. IT'S THERE NOW LAST TIME I CHECKED. -MATHGUY49<br />
<br />
<br />
Collball: Talk: Gmaas is not the place to post your article.<br />
<br />
<br />
<br />
(Mistyketchum28: EDIT: ALL HAIL GMAAS<br />
btw ONLY THE PART IN PARENTHESIS IS WRITTEN BY ME <br />
IM NOT CoPYING)<br />
<br />
NOOOOOOOOOOOOOOOOOOOOOO GMAAS PaGE IS NOW PROTECTED NOOOOOOOOOOOOOOOOoo - juliankuang, crying in bed right now<br />
<br />
There's even a mistake in the first line :(<br />
<br />
It's been protected since 16:49, December 18, 2019 someone should make a petition to bring it back --[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 14:43, 2 March 2020 (EST)<br />
<br />
== No more sseraj ==<br />
<br />
By the way, Samer Seraj (GMAAS's slave) doesn't use the account sseraj anymore.<br />
<br />
== Locking the GMAAS page ==<br />
<br />
It's a sad thing the 5space locked the GMAAS page. But at least we still have this Talk page<br />
<br />
--[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 18:47, 26 December 2019 (EST)<br />
<br />
It might not be 5space. Let us just put hope, even if absurd the idea is, that the Gmaas page will be unprotected soon. juliankuang 14:15, 31 December 2019 (EST) juliankuang<br />
<br />
It is 5space, that's what the history page says. I think he locked it because there was a user that blanked the page like 3 times.<br />
<br />
--[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 18:47, 30 December 2019 (EST)<br />
<br />
--[[User:ARay10|ARay10]] ([[User talk:Aray10|talk]]) 02:08 2 January 2020 (EST)<br />
Even if it is locked, honor Gmass! He is really important.<br />
<br />
Respect Gmaas. Spell His name correctly. juliankuang 10:17, 2 January 2020 (EST)<br />
<br />
Actually it's spelled GMAAS, that's what sseraj called him.<br />
--[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 14:15, 2 January 2020 (EST)<br />
<br />
But there is no more sseraj ARay 10 20:34, 3 January 2020 (EST)<br />
<br />
- I respectfully disapprove of 5space's decision to lock this article. 5space is probably trying to stop vandalism and spam from ruining this page. The Gmaas article has been a subject to much vandalism and spam over the years, but other AoPSers have always been able to repair the damage. The Gmaas article will keep living as long as we let it live. But once the article is locked, it cannot keep living. This article is now a mummy, preserved indefinitely. But Gmaas should not die. Please, 5space, I respect your decision, but would you please unlock this article? -Idefix<br />
<br />
For now you can edit this GMAAS page that's still open, [[GMAAS]]. --[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 13:55, 11 June 2020 (EDT)<br />
<br />
==Some of Gmass's facts==<br />
<br />
Notice, this is not all, as the real Gmass page has to be the biggest.<br />
<br />
<br />
<br />
<br />
<br />
gmaaas is our leader. To prove yourself worthy of GMAAS, you have to memorize facts: _________________________________________________<br />
<br />
-1. GMAAS looks like this when he is mad: https://cdn.artofproblemsolving.com/images/1/d/4/1d4ec3462f2c90f5a6c6b194e0a40b8418f554ec.jpg GMAAS permits to post this picture.<br />
<br />
0. THE FACTS ABOUT THE GREAT EPIC AWESOME GMAAS: Words fail to describe the epic nature of GMAAS, for he is too almighty and powerful to be bound by the plebeian and trifling words. But even the Great Epic Awesome Plenipotentiary GMAAS cannot get rid of a language that's been around for a <math>999999999999999999999999999999999999999999^{9999999999999999999999999999999999999}</math> years and counting. That's older than he is. EDIT: It is not older than he is because Gmaas is infinity years old. EDIT EDIT: He has now<br />
<br />
1. GMAAS's theorem states that for any math problem, GMAAS knows the answer to it. This theorem was proved by GMAAS. But then GMAAS forgot about the theorem so later, the mathematician named ARay10 proved it again: [b]Proof of the MAAS theorem[/b]: The GMAAS theorem states that for every math problem, GMAAS knows the answer. Using the 3.141592653589793238462643383... GMAAS theorem, stating that "26. GMAAS's Theorem states that GMAAS knows the answer to any math problem. EDIT: That theorem was proved by GMAAS. EDIT: GMAAS's Theorem has real-world applications: because GMAAS knows the answer to any math problem, you can use GMAAS to solve math problems. GMAAS is busy, so he charges a fee of one dollar for 1,000,000 math problems. EDIT: The fee has gone up. It is now 1,000,000 dollars for one math problem. Gmaas's technician made a mistake and reciprocated the fee," you must pay <math>1,000,000</math> to solve a problem using the Games theorem. You wouldn't waste that much money to solve one problem. Therefore, I proved that you cannot use the Games theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Games can turn things into anything. Using that fact, you can use the Games theorem to solve any problem.<br />
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2. The GMATS were supposed to be called the GMAAS, but the manager, games, wanted to eat some gnats. Games have stopped eating gnats because they taste dreadful. EDIT: He now eats gnats again. He thinks that they taste like pi(e). EDIT: He ate 3,141,592,653,589,793,238,462,643,383 so far.<br />
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3. GMAAS's archenemy is John Wick. John Wick once gave Gmaas 31,415,926,535,897,932,384 Oren Berries, but Gmaas thought they were Oran Berries. That is why Games hates John Wick. Germans knew they were Oren Berries because Gmaas knows everything, but he decided to play along.<br />
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4. Gmaas wants every fact to have a pi reference. Gmaas created pi. EDIT: He made it 3.1415926 times.<br />
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5. Gmaas owned many memes. Then they died. He left the meme business because others took it over. Gmaas decided to make some more memes after that. Since then, he has been working for the government. All of the politicians are his henchmen. He controls the government. He wrote all of the laws and documents including the U.S. constitution. However, Gmaas doesn't believe in constitutions. He simply writes them for fun (or not, only games knows!!!) EDIT: His favorite is memedog EDIT EDIT: Gmaas does not like meme dog. Games was the one who told the AoPSSheriff to give warnings for posting meme dog.<br />
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6. Gmaas owns Scratch. Gmaas sued them because Scratch cat was supposed to be a picture of Gmaas. But this was one lawsuit he lost. Gmaas won that lawsuit, but he ate food so he calmed down and dropped the case.<br />
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7. Thanos wishes he could be like Gmaas. It's why he got the Infinity Stones and snapped.<br />
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8. Gmaas dies in Endgame, but he somehow possesses Thanos and kills everyone. Gmaas never dies. Gmaas has the power to die whenever he wants and frequently does this to escape others' woes in life. Of course, he never faces such troubles.<br />
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9. He hates contemporary music. He only like Renaissance music, Baroque music, Classical music, and Romantic music. His only exception is Queen music(Especially Bohemian Rhapsody). Gmaas also likes Debussy. But he thinks rock'n'roll is awful.EDIT: Gmass doesn't like music that much; it hurts his ears. EDIT EDIT: GMAAS is now confirmed to like al video game OSTs, especially Undertale. His favorite song is Megalovania. EDIT EDIT EDIT: The previous edit is false.<br />
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10. Gmaas is superior to you. If you see Gmaas or a picture of the Great Gmaas, DON'T bow down. It is proper to have a painting or lifesize sculpture of Gmaas. If you do not, Gmaas will put one of them there himself. EDIT: That is exactly why you should not have a statue of Gmaas: Gmaas will give you one for free. A life-size sculpture of Gmaas will be infinitely large because that is how superior Gmaas is. Pictures also have power, so pictures require power to make. Making a sculpture of Gmaas will require almost an infinite amount of power. Only Gmaas can make a quality sculpture of himself, and if anyone makes a low-quality sculpture of Gmaas, it will be disrespectful to such a superior power. Therefore, it is best to let Gmaas make a proper sculpture of himself, so he is respected. If you want to respect him the most, you should NOT have a high-quality sculpture of Gmaas.<br />
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11. Gmaas is known to barf entire universes at will. EDIT: Every once in a while, he barfs a furball, which always turns into a black hole. The less impressive ones turn into neutron stars.<br />
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12. Gmaas farted and created a false vacuum, but then he burped, destroying the false vacuum.<br />
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13. Gmaas was the first person to use the Infinity Gauntlet. EDIT: Gmaas created the Infinity Gauntlet and the Infinity Stones. EDIT: But Thanos stole them after Gmaas died for 0.31415926535 seconds.<br />
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14. Gmaas killed Thanos with a swat of Gmaas's tail. Gmaas barfed and created a furball, which leads to a new dimension, where he put a newly revived Thanos to become a farmer when Thor aimed for the head. Gmaas got mad and made him fat. EDIT: The presence of Gmaas killed Thanos.<br />
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15. Gmaas won an infinite amount of games against AlphaGo and Gary Kasparov while eating dinner, chasing sseraj, and doing a handstand.<br />
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16. Gmaas owns a pet: sseraj. Gmaas likes to play chess with his pets. EDIT: Gmaas has long moved on from chess because he was too good for it. EDIT EDIT: Gmaas has now invented a new form of chess<br />
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17. Gmaas is so interesting that an entire science has been devoted to studying him: Gmaasology.<br />
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18. Gmaas is the only known living being who has a Ph.D. in Gmaasology.<br />
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19. Most universities, including Harvard, are beginning to offer MAs in Gmaasology.<br />
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20. Gmaasology is one of the most eminent fields of science, falling behind Physics, Biology, Chemistry, Economics, Geology, and Computer Programming.<br />
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21. Gmaas wants his AoPS Wiki page to be the longest ever. EDIT: Sadly, the Gmaas page is only the fourth-longest page on AoPS wiki. It has around 50,000 bytes. EDIT: The Gmaas page is getting closer and closer! Gmaas has beaten Primitive Pythagorean Triple and Proofs without words and is the second-longest AoPS Wiki page. It now has around 60,000 bytes. However, it will still be a challenger to overcome 2008 most iT Problems, which has around 73,000 bytes. EDIT: Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Gmaas's article has 89,119 bytes.<br />
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22. Gmaas has the longest lifespan of any cat. He has lived for 3,141,592,653,589,793,238,462,643,383,279 years. (He is older than the universe.)<br />
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23. Gmaas is the ancient Greek god of cats and catfish. EDIT: He is the god of everything<br />
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24. Gmaas can turn things into gold. EDIT: Gmaas can turn anything into anything.<br />
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25. Gmaas can eat lava. EDIT: Everyone can eat lava once. After you eat it once, you die. EDIT EDIT: Gmaas can eat anything.<br />
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26. Gmaas's Theorem states that Gmaas knows the answer to any math problem. EDIT: That theorem was proved by Gmaas EDIT: Gmaas's Theorem has real-world applications: because Gmaas knows the answer to any math problem, you can use Gmaas to solve math problems. Gmaas is busy, so he charges a fee of one dollar for 1,000,000 math problems. EDIT: The fee has gone up. It is now 1,000,000 dollars for one math problem. Gmaas's technician made a mistake and reciprocated the fee.<br />
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27. Gmaas can only be described as Gmaas. EDIT: Gmaas has an age, but his age changes all the time. Every second his age increases by one second.<br />
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28. Gmaas should be capitalized to show respect. EDIT: It is normal to capitalize on people's names. EDIT: Gmaas isn't a person, he is a divine entity that takes the form of a cat, and should, therefore, be worshipped.<br />
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29. The steps to summon Gmaas can be found at Talk: Gmaas.<br />
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30. Gmaas owns your AoPS account 31.415926% of the time. EDIT: There is an exception to this: Gmaas owns his own account 99.9999% of the time. The only time when Gmaas did not control his AoPS account was when he had slow internet. For Gmaas, "slow internet" happens when it takes more than a nanosecond to load a webpage. EDIT: Gmaas owns every AoPS account at some point. EDIT: You are controlled by Gmaas. So actually, Gmaas owns your AoPS account 100% of the time, and his AoPS account is owned by him 314% of the time.<br />
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31. Everyone, except for me, is Gmaas. EDIT: You are also Gmaas. Gmaas is not me nor you. Gmaas is in us all and also not in us all. Why? The power of Gmaas. Gmaas is an energy field created by all living things. It surrounds us and penetrates us. It binds the galaxy together.<br />
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32. Gmaas used to be a dog, but he didn't like to be a dog. So he became a cat.<br />
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33. Gmaas can be spotted in The Matrix at 31:41:59, metric time. EDIT: This fact is incorrect because there have only been 30 sightings, all of them inconsistent.<br />
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34. The Metropolitan Museum of Art and the Louvre are Gmaas's private art collections from 3:14 AM to 3:15 AM. EDIT: The only exception was on the 3141st day after its opening. EDIT: All the paintings in these museums are secretly portraits of Gmaas in his most glorious human/animal forms.<br />
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35. Gmaas is a Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is blah blah blah blah blah blah blah blah blah blah blah blah blah blah.<br />
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36. Gmaas says hello. EDIT: Gmaas does not speak; he only uses mind signals. Speaking is too primitive for Gmaas.<br />
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37. For one day, Gmaas was Richard Rusczyk, Gmaas, and David Patrick at the same time. It felt strange, so Gmaas stopped.<br />
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38. Gmaas, in addition to being the oldest cat in history, the most powerful cat in history, and the most confusing cat in history is also the largest cat in history.<br />
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39. Gmaas is dead--he was eaten for lunch. EDIT: The above sentence is incorrect. Gmaas has not sent a gamma-ray burst to destroy the planet, which has happened every time someone has eaten Gmaas.<br />
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40. Gmaas is the creator of the BCPI Ai project.<br />
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41. Gmaas exists in <math>2\pi^2</math> dimensions because he doesn't like string theory.<br />
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42. Christmas was supposed to be called Gmaasmas. Until it wasn't. EDIT: Christ is Gmaas. EDIT: Why? Because of the power of Gmaas.<br />
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43. <math>\pi</math> is a representation of how many pies Gmaas has eaten. EDIT: The number <math>\pi</math> was created by Gmaas. He took a ten-sided die and flipped it an infinite number of times. The numbers he rolled became the digits of <math>\pi</math>.<br />
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44. <math>e</math> is a representation of how many days Gmaas forgot to eat. EDIT: The number <math>e</math> was created by Gmaas. He took a book that has infinite pages and flipped to a random page. The digits of the page number became digits of <math>e</math>. EDIT: That is impossible. The previous fact would mean that <math>e</math> has the last digit, which it does not. EDIT: That is why Gmaas is still flipping to this day. EDIT: Why would he still be doing that? That would be boring. EDIT: Gmaas is not flipping the book now because he can flip an infinite amount of pages in <math>\frac{1}{\infty}</math> seconds. EDIT: Then why isn't he done? EDIT: The power of Gmaas. EDIT: The editors of the Gmaas article like to make long chains of edits, don't we? : Yes, we do. : I do too. So do I. EDIT: The number of edits only verifies how high-quality this holy bible of Gmaas is, because, with each edit, this bible becomes better. That is exactly why we are editing this. I will make one more edit, just to respect Gmaas EDIT: Gmass rules!<br />
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45. Gmaas is the creator of everything including nothing.<br />
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46. According to recent DNA tests, famous historical people, such as Euclid, Julius Caesar, Omar Khayyam, George Washington, and Ramanujan are Gmaas in disguise. Gmaas has been thousands of people; only 99.9% of them were important historical figures. EDIT: They aren't dead. They're still alive. They are all dead reincarnations of Gmaas. Gmaas only has one reincarnation at a time. He is right now a cat living in sseraj's house.<br />
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47. Gmaas is a cat. EDIT: How many times do we say this? EDIT: Very many times. EDIT: Gmaas can be anything, but he chooses to be a cat. EDIT: He has been a human dozen of times. Gmaas painted the Lascaux cave paintings.<br />
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48. Gmaas has the following powers: trout, carp, earthworm, and catfish. Gmaas never uses any of them because Gmaas has an infinite number of powers. EDIT: Gmaas has used his catfish power several times. EDIT: Gmaas once used all his powers 3,141,592,653,589,793,238,462,643,383,279 times in 1 second.<br />
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49. Gmaas was once spotted in Minecraft chewing a tree. EDIT: The tree broke. EDIT: Gmaas once broke a Nokia. EDIT: Gmaas can break anything except for Gmaas's logic. It is too strong. EDIT: Gmaas is strong.<br />
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50. Gmaas was spotted in Roblox eating a taco cat. EDIT: Tacocat is just what happens when Gmaas sheds. Shedding is annoying for Gmaas, so he sheds no more. Therefore, there are only 271,828,182 taco cats in the world. EDIT: However, Tacocats reproduce, so they are not in danger of extinction.<br />
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51. Gmaas would like to go to Taco Bell, but Gmaas goes to Wendy's instead. No one knows why. EDIT: Because Taco Bell doesn't serve tacocats.<br />
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52. Gmaas real name is Grayson Maas. He is the CEO of AoPS. EDIT: Gmaas's real name is Gmaas. He is not the CEO of AoPS. EDIT: Gmaas has always been the master of AoPS as he is AoPS.<br />
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53. Gmaas has a pet pufferfish named Pafferfash. EDIT: He also has a goldfish named Sylar.<br />
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54. Gmaas has colonized the universe. EDIT: Gmaas created the universe, so he is allowed to claim control over it. EDIT EDIT: Gmaas created every universe.<br />
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55. Some people think that Gmaas is human. However, this has never been proven. Many AoPSers believe Gmaas is a cat. EDIT: Gmaas is a cat. EDIT: Yeah, we've said that already. EDIT: It does not matter how many times we say it; it will always be true. EDIT: Gmaas is a cat. EDIT: Right now he is a cat, but many of his lives have been other species. EDIT: He has been a dog, as said before, but it was too boring.<br />
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56. Gmaas started Pastafarianism. But then converted to Catholicism because Gmaas knows all EDIT: In that religion, one can only eat pasta.<br />
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57. Gmaas could eat your hand, but he would not because hands taste bad.<br />
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58. According to the Interuniversal Gmaas Society, 17.548 percent of the universe's population thinks that Gmaas is spelled "Gmass". EDIT: In case you don't know already, the name Gmass is spelled Gmaas. Alternative spellings include GMAAS, gmaas, gmaas, Gmaas, Gmail, Maas, G. Maas, G. Mass, Gabriel Maas, Genius of Maas, General Maas, Greek Mass, and the big fluffy kitty who lives in sseraj's house. These are no longer accepted spellings, and Gmaas is the current acceptable spelling.<br />
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59. The Interuniversal Gmaas Society was founded in 1314 on May 9th at 2:06:53 PM. EDIT: Ever since then, Gmaas day has been celebrated on May 9th.<br />
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60. The Interuniversal Gmaas Society has just reconstructed a lost book about Gmaas from Lucretius's De Rerum Natura. They researched this for three years. EDIT: A few years ago the Society compiled a biography of Gmaas's last twenty lives. EDIT: The only copies of these biographies are locked in the Gmaasian Library beneath the Library of Congress. EDIT: See the Gmaasology page for more information on these projects and others made by the Interuniversal Gmaas Society.<br />
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61. The Interuniversal Gmaas society has found out all of the information and more. For details on the history of the Interuniversal Gmaas Society, see the Gmaathamatics page.<br />
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62. GMAAS likes to surprise unsuspecting people. People cannot surprise GMAAS because GMAAS knows what everyone is doing and will do.<br />
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63. Gmaas loves sparkly mechanical pencils. EDIT: Gmaas has eaten several mechanical pencils. EDIT: He absorbed them and became colorful for 0.271828182846 seconds. EDIT: Then, he came colorful for 3.1415926535897932384626433832 more seconds.<br />
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64. This page is Gmaas's holy book where people go to worship Gmaas in Maas. EDIT: This is an unusual holy where everyone edits it. EDIT: That is the point. Gmaas is too lazy to write or to hire someone to write his holy book, so he lets people write it for free. EDIT: Gmaas does not like being called lazy. Our language is simply too unimportant for him to waste time on.<br />
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65. Gmaas tastes like a furry meatball. EDIT: No one has ever tasted Gmaas's sacred body before. EDIT: Once, he was a catfish, and someone ate him for dinner. Gmaas was angry and the entire planet exploded. (This was on the planet, Demeter. It blew up into so many pieces that the Asteroid Belt was formed. The biggest asteroid in the belt is called Ceres, the Roman version of Demeter, in honor of the lost planet.)<br />
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66. Gmaas's favorite food is pepperoni pizza. EDIT: Gmaas's favorite food is turnips. EDIT: Gmaas hates catnip and turnips, and pepperoni. He only likes alien alienish H2So4. EDIT: He does love turnips. He has a secret turnip garden under sseraj's house.<br />
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67. Gmaas is Johnny Johnny's Papa. EDIT: We'll never know. EDIT: Gmaas caught Johnny Johnny eating sugar and lying. He is Johnny Johnny's Papa.<br />
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68. Gmaas is in you, and Gmaas is in you, and Gmaas is in me. EDIT: Gmaas is in all cats. EDIT: He is not in every cat. The Guinness Book of World Records has a cat that does not have any Gmaas. EDIT: Gmaas invented The Guinness Book of World Records. EDIT: Gmaas invented the world. EDIT: Gmaas will also destroy the world.<br />
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69. Gmaas always remembers not to destroy the universe. EDIT: Once he forgot and had to travel back in time to stop it.<br />
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70. Gmaas created many user's accounts. EDIT: All accounts on AoPS are Gmaas in disguise since you are Gmaas in disguise. Except for maybe Geoflex and CrazyEyeMoody.EDIT: Everyone is Gmass, even Geoflex and CrazyEyeMoody EDIT: How? EDIT: The power of Gmass.<br />
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71. Gmaas is both living and nonliving. EDIT: He is living 90% of the time. Every once in a while he takes a break and dies.<br />
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72. Gmaas cannot comprehend the stupidity of humans.<br />
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73. All hail the Gmaas cloud! EDIT: Gmaas is a cloud: a cloud of electrons and nuclei.<br />
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74. Some things are beyond possible human comprehension. Nothing is beyond Gmaas. EDIT: The only thing beyond Gmaas is his tail; he has never managed to eat it. EDIT: Once he ate it. It tasted bad, so he didn't ever eat it again.<br />
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75. Gmaas eats food and resides at King Arthur's throne when he feels like it. EDIT: King Arthur is dead. However, Welsh folk stories say that he will come back if the Welsh people are in trouble. EDIT: King Arthur lives on paper flour bags. He came back when the Welsh people didn't have enough bread. EDIT: Because Gmaas ate 31415926535897932984626433 loaves of bread.<br />
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76. Gmaas owns a rabbit. EDIT: Gmaas ate it on March 19, 2019. It reincarnated on the other side of the planet. EDIT: Gmaas ate it again on April 31, 2019.<br />
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77. Gmaas is both singular and plural. EDIT: It is usually singular. EDIT: The plural of Gmaas in English and Latin is Gmaases. (Gmaas is a third declension noun: Gmaas, Gmaasis, m.) EDIT: But there is no use for the plural, because there is only one Gmaas.<br />
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78. Gmaas eats disbelievers as if they were donuts for breakfast. (Yet I'm somehow still alive. Do you think Gmaas ate my soul?) EDIT: Yes, I do. Gmaas is just typing through your account. EDIT: Gmaas typed through everyone's account. EDIT: Gmaas owns most AoPSers, including me. So that's why I'm typing. EDIT: That is why everyone here is typing.<br />
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79. Gmaas knows Jon Snow's parents. EDIT: Gmaas was Jon Snow's parents for 0.3141592653589793238 seconds, but it felt weird. So he became Gmaas again.<br />
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80. All Gmaas article editors will be escorted to Gmaas heaven after they die. EDIT: I hope so because I edited this article. EDIT: Me too. EDIT EDIT: Me three. EDIT EDIT EDIT: That are good!<br />
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81. Gmaas won all the wars. EDIT: He did not win the Intergmaasian War, which was between Gmaas's head and his tail. His tail won. Two flees died in the war. EDIT: Stefán Karl Stefansson died in the war, which made Gmaas very sad. Because of this, Gmaas started the world peace movement.<br />
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82. Gmail was named after Gmaas. EDIT: Google was named after Gmaas. EDIT: Almost every word starting with G is named after Gmaas. EDIT: Every word starting with G is named after Gmaas.<br />
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83. Gmaas ate cat-food today. EDIT: Gmaas also ate it yesterday. EDIT: Only because there was no alien alienish H2So4.<br />
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84. Gmaas won Battle For Dream Island and Total Drama Island. EDIT: Gmaas made Dream Island. And he ate it. It tasted like dirt.<br />
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85. Somehow Gmaas exists at all places at the same time. EDIT: Gmaas does not exist in my kitchen. EDIT: I'll just go check. Aaaaaaaaah! He is in my kitchen. He is eating everything! EDIT: Seriously? Oh ya, the power of Gmass.<br />
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86. Gmaas can lift with Gmaas's will. EDIT: Gmaas can lift with no one's will while he is sleepwalking.<br />
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87. Gmaas is the rightful heir to the Iron Throne. EDIT: Gmaas made the Iron Throne.<br />
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88. Gmaas is Teemo in League of Legends because Gmaas made LoL and they made an honorary Gmaas character. EDIT: LoL must be used in this article more than once. EDIT EDIT: It is. In fact, twice in this message.<br />
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89. Gmaas started the Game of Thrones. EDIT: Gmaas will end the Game of Thrones.<br />
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90. Gmaas has killed himself hundreds of times. He was reincarnated as a different species each time. EDIT: Gmaas has only died two times. Other times he was putting on a magic show. The kids were very impressed. EDIT: Gmaas has died and reincarnated thousands of times.<br />
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91. Gmaas created everything after puking. EDIT: He could not have puked because he is a god. Gods do not do that. EDIT: Gmaas does that. EDIT: Why? EDIT: The power of Gmaas. EDIT: He almost never does it, only if he wants to.. Also he did not puke and made everything. EDIT: No, because, GMAAS.<br />
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92. Gordan's last name was named after Gmaas. EDIT: But Gmaas did not want people disrupting his beauty sleep. So he changed it.<br />
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93. Gmaas is more powerful than Gohan.<br />
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94. Gmaas is over 90000 years old. EDIT: Gmaas is trillions of years old.<br />
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95. Gmaas has 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 cat lives, maybe even more. Gmaas has 0 dog lives. EDIT: He has 314,159,265 fish lives. EDIT: Out of Gmaas's 314,159,265 fish lives, 271,828,182 are catfish lives. He has used up 141,421,356 of them. EDIT: Gmaas has infinitely lives of all types. EDIT: How? EDIT: The power of Gmaas.<br />
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96. Who wrote Harry Potter? None other than Gmaas himself. EDIT: That is incorrect. EDIT: The above post is irrelevant. Gmaas created J. K. Rowling. Therefore, he created Harry Potter. EDIT: Gmaas has strong logic. No one can break it. Not even Gmaas himself. He is still trying to this day. EDIT: Why? EDIT: The power of Gmass.<br />
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97. Gmaas created the catfish. EDIT: See a few posts above for more information about Gmaas and catfish.<br />
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98. Gmaas has proven that the universe is infinite by traveling to the edge of the universe in a second. EDIT: He has never repeated this experiment because Gmaas is busy and has better things to do.<br />
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99. Gmaas founded Target, but then Gmaas sued them for making the mascot look like a dog when it was supposed to look like Gmaas. EDIT: They went broke because Gmaas sued them but then Gmaas ate a fudge popsicle that made him super hyper and he made Target not broke anymore in his hyperness.<br />
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100. Everyone has a bit of Gmaas inside them. EDIT: I don't. EDIT: Yes, you do. EDIT: Gmass: LoL<br />
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101. Gmaas likes to eat popsicles. Especially the fudge ones that get him hyper. EDIT: Gmaas is a popsicle. EDIT: Then how come Gmaas hasn't melted? EDIT: The power of Gmaas.<br />
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102. When Gmaas is hyper, he runs across Washington D.C. and grabs unsuspecting pedestrians, steals their phones, hacks into them, and downloads PubG onto their phone.<br />
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103. Gmaas's favorite cereal is fruit loops. Gmaas thinks it tastes like unicorns jumping on rainbows. EDIT: Gmaas eats unicorns jumping on rainbows like a toddler eats Goldfish.<br />
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104. Gmaas thinks that the McChicken has too much mayonnaise. EDIT: Gmaas thinks McDonald's is not good enough for him. He like KFC better.<br />
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105. Gmaas is a champion pillow-fighter. EDIT: Gmaas invented pillows.<br />
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106. Gmaas colonized Mars. EDIT: Gmaas also colonized Jupiter, Pluto, and several other galaxies. Gmaas cloned little Gmaas robots (with Gmaas's amazingly robotic skill of coding) and put them all over a galaxy called Gmaasalaxy. EDIT: Gmaas has colonized the whole universe.<br />
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This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam This article is spam 107. Gmaas can make every device play "The Duck Song" at will. EDIT: "The Duck Song" was copied off of the "Gmaas song," but the animators though Gmaas wasn't catchy enough.<br />
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108. Gmaas once caught the red dot and ate it. EDIT: Gmaas is a red dot.<br />
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109. Gmaas's favorite color is Gmaasian blue, a very rare blue that shines with the brightness of lightning. EDIT: it's <math>1000^{1000}</math> times brighter than lightning. Gmaasian blue occurs when a meteor hits the earth and usually is only seen for a few seconds. Only Gmaas has good enough eyesight to see it for that short a time.<br />
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110. Gmaas can create wormholes and false vacuums. EDIT: He is made out of exotic matter.<br />
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111. Gmaas is a champion PVP Minecraft player.<br />
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112. Gmaas doesn't like tacos. The last time he tried one he turned into a mouse and then caught himself.<br />
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113. Gmaas is the coach of True Ninja Music and Myth.<br />
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114. Gmaas caught a CP 6000 Mewtwo with a normal Pokeball in Pokemon Go.<br />
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115. Gmaas founded Costco.<br />
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116. Gmaas does not need to attend the FIFA World Cup. If Gmaas did, he would start the game with a goal and break the ankles of everyone watching the World Cup, including you, at the same time in a fraction of a second, even if you are watching from a device. EDIT: Gmaas created your device. EDIT: Gmaas is your device.<br />
<br />
117. Gmaas can solve any puzzle instantly except for the 3x3 Rubik's Cube. EDIT: When Gmaas is handed a 3x3 Rubik's Cube, it is always already solved. Why? The power of Gmaas.<br />
<br />
118. Gmaas caught a CP 20,000 Mewtwo with a normal Pokeball and no berries blindfolded first try in Pokemon Go.<br />
<br />
119. When Gmaas flips coins, they always land tails, except when Gmaas makes bets with Zeus.<br />
<br />
120. On Gmaas's math tests, Gmaas always gets <math>\infty</math>. EDIT: He gets a 26 on the AMC8 every year. The results never show him because Gmaas is no longer in middle school.<br />
<br />
121. Gmaas's favorite number is <math>\pi</math>. It is also one of Gmaas's favorite foods. EDIT: Gmaas created <math>\pi</math>, as well as most other constants, such as <math>e</math> and <math>\sqrt{2}</math>.<br />
<br />
122. Gmaas's burps created all gaseous planets.<br />
<br />
123. Gmaas beat Luke Robatille in an epic showdown of catnip consumption.<br />
<br />
124. Gmaas's wealth is unknown, but it is estimated to be more than Scrooge's. EDIT: It may be more than John D. Rockefeller. EDIT: More accurate estimates predict that Gmaas's wealth is infinite.<br />
<br />
125. Gmaas has a summer house on Mars. EDIT: Gmaas has a fall house on Venus. EDIT: Gmaas has a winter house on Jupiter. EDIT: Gmaas has a spring house on Earth. EDIT: Gmaas also experiences a fifth season called wintrautumn. It happens during November and December and is the cross between fall and winter. Everything is dreary, but there is no snow. Gmaas spends wintrautumn in his house on Venus, where he is incinerated daily. He reincarnates every night through the power of Gmaas. EDIT: Gmaas invented a new season on Feb. 31, 2019, called winter-summer. It is spring but transcends spring.<br />
<br />
126. The Earth and all known planets are Gmaas's hairballs.<br />
<br />
127. Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time using a time-turner.<br />
<br />
128. Gmaas attended Hogwarts and was a perfect. EDIT: Gmaas is headmaster. EDIT: Hogwarts was supposed to be called Hogmaas.<br />
<br />
129. Mrs. Norris is Gmaas's archenemy. Gmaas yawned, and Ms. Norris was petrified. EDIT: Gmaas is the basilisk, and he let Harry Potter pet him. Harry did not kill the basilisk, he only gave Gmaas a haircut. EDIT: Gmaas hates having a haircuts, so he reincarnated Voldemort to punish Harry.Edit: Not true at all.<br />
<br />
130. Gmaas is a demigod and attends Camp Half-Blood over summer. Gmaas is the counselor for the Apollo cabin because cats can be demigod counselors too. EDIT: Apollo was one of the many reincarnations of Gmaas.<br />
<br />
131. Gmaas has completed over 2,000 quests and is very popular throughout Camp Half-Blood. Gmaas has also been to Camp Jupiter. EDIT: Gmaas is Camp Jupiter, he is also Camp Half-Blood. EDIT: How? EDIT: The power of Gmaas.<br />
<br />
132. Percy Jackson was only able to complete his quests because Gmaas helped him. EDIT: Gmaas is Percy Jackson. You are Gmaas, but Gmaas is not you.<br />
<br />
133. Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night. EDIT: Gmaas knows that their real names are Gmaasa Lisa, The Last Domestic Meal, and Far-away Light.<br />
<br />
134. Gmaas knows that the Blood Moon is just the red dot. He has not caught it yet. EDIT: He caught it during the super blue blood moon.<br />
<br />
135. Gmaas attended all the Ivy Leagues.<br />
<br />
136. I am Gmaas. EDIT: No you are not. You only have part of Gmaas inside of you. EDIT EDIT: I am also Gmaas. EDIT EDIT EDIT: But it is I who am Gmaas. EDIT EDIT EDIT EDIT: Gmaas is in us all. EDIT EDIT EDIT EDIT EDIT: Gmaas is all of us yet none of us. EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas is a cat. EDIT EDIT EDIT EDIT EDIT EDIT EDIT: He is only in a cat form right now. He can be whatever he wants to be. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: This chain of edits has superseded another chain of edits on this page for the record of the longest chain of edits on AoPS Wiki. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: But the edits in this chain tend to be shorter than most. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: You're right. But we sure do love strings of edits, don't we? EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas makes the edits. He is in your head spinning edits in your brain. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas is made to be edited. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas pays people to edit. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Some people edit even without Gmaas's payment. They do it because they are believers. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas is Gmaas. You are not Gmaas. Gmaas does not pay people to edit; we do it because we want to.<br />
<br />
137. In 2018 Gmaas once challenged Magnus Carlsen to a chess game. Gmaas won every round. EDIT: Gmaas ate the chess pieces afterward. They tasted funny.<br />
<br />
138. Gmaas hates tiktok<br />
<br />
139. Gmaas got COVID-19. But He is so powerful he gave COVID-19 COVID-19 so all of the COVID-19 viruses died of COVID-19. He does not have COVID-19 anymore.<br />
<br />
All hail Gmaas!<br />
<br />
P.S. Gmaas won USAMO 5 times in a row. That's because he invented USAMO. When Gmaas took it nobody knew about it and he was the only one who took it. So of course, he won. How dare you spell Gmaas with a lowercase G? He is the supreme and awesome ruler of the universe!<br />
<br />
P.P.S. Gmaas invented the word Yeet.<br />
<br />
P.P.P.S. Gmass is the first person who played the game YECK (Moth poth 2019 reference)<br />
<br />
P.P.P.P.S. Gmaas is a non-human who can snap anyone out of existence without anything.<br />
<br />
P.P.P.P.P.S. Gmaas was the first person who did math.<br />
<br />
P.P.P.P.P.P.S. Gmaas invented proof by dictatorship.<br />
<br />
P.P.P.P.P.P.P.S Gmaas invented proof by procrastination.<br />
<br />
P.P.P.P.P.P.P.P.S Gmaas invented proofs!<br />
<br />
P.P.P.P.P.P.P.P.P.S Gmaas invented everything.<br />
<br />
P.P.P.P.P.P.P.P.P.P.S Gmaas did not invent roblox. Gmaas invented Minecraft.<br />
<br />
P.P.P.P.P.P.P.P.P.P.P.S Gmass likes cookies<br />
<br />
==Notice(Gmass's page)==<br />
<br />
Most of the starting facts are false. Plz don't post these false facts, 1 of the reasons 5space locked it(along with blanking)<br />
<br />
OK what happened here lmao ~IceWolf10</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=118605IceWolf102020-03-01T18:55:39Z<p>Icewolf10: </p>
<hr />
<div>Hello!<br />
Please join the [[Pink Penguinaphants Army]], if you haven't already.<br />
<br />
I am the second member to join the PP Army. Right now I am its leader, and it is thriving with the help of ryrychen, [[anc3]], Winter_Ryan, [[Firecobra100]], and many more users!<br />
<br />
We have games, marathons, spam battles, and a lot of fun with each other.<br />
<br />
An extension of the PP Army is the Pink Penguinaphants RPG. Join if you like imagination and gaming!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=118604IceWolf102020-03-01T18:55:00Z<p>Icewolf10: </p>
<hr />
<div>Hello!<br />
Please join the [[Pink Penguinaphants Army]], if you haven't already.<br />
<br />
I am the second member to join the PP Army. Right now I am its leader, and it is thriving with the help of ryrychen, [[anc3]], Winter_Ryan, [[FireCobra100]], and many more users!<br />
<br />
We have games, marathons, spam battles, and a lot of fun with each other.<br />
<br />
An extension of the PP Army is the Pink Penguinaphants RPG. Join if you like imagination and gaming!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Firecobra100&diff=118603Firecobra1002020-03-01T18:53:45Z<p>Icewolf10: </p>
<hr />
<div>FireCobra100 is a very smart person.<br />
Or is he? He actually is. Or maybe he isn't. He can't seem to decide.<br />
Now he can decide. He is a very smart person. He knows what 1+1=2!<br />
Do you know that? He knows that 80x80=6400! Do you know that? He also knows<br />
this:<br />
<math>\sqrt64</math> = <math>64.</math><br />
He's indeed very smart, if he does say so himself.</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Firecobra100&diff=118602Firecobra1002020-03-01T18:53:22Z<p>Icewolf10: </p>
<hr />
<div>FireCobra100 is a very smart person.<br />
Or is he? He actually is. Or maybe he isn't. He can't seem to decide.<br />
Now he can decide. He is a very smart person. He knows what 1+1=2!<br />
Do you know that? He knows that 80x80=6400! Do you know that? He also knows<br />
this:<br />
<math>\sqrt64</math> = <math>64</math><br />
He's indeed very smart, if he does say so himself.</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Firecobra100&diff=117249Firecobra1002020-02-08T03:26:21Z<p>Icewolf10: </p>
<hr />
<div>FireCobra100 is a very smart person.<br />
Or is he? He actually is. Or maybe he isn't. He can't seem to decide.<br />
Now he can decide. He is a very smart person. He knows what 1+1=2!<br />
Do you know that? He knows that 80x80=6400! Do you know that?</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Firecobra100&diff=117237Firecobra1002020-02-08T03:17:23Z<p>Icewolf10: </p>
<hr />
<div>FireCobra100 is a very smart person.<br />
Or is he? He actually is. Or maybe he isn't. He can't seem to decide.</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Firecobra100&diff=117235Firecobra1002020-02-08T03:15:43Z<p>Icewolf10: </p>
<hr />
<div>FireCobra100 is a very smart person.<br />
Or is he?</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Pink_Penguinaphants_Army&diff=117229Pink Penguinaphants Army2020-02-08T03:06:58Z<p>Icewolf10: </p>
<hr />
<div>The '''Pink Penguinaphants Army''', also called the PP Army by its members, is a forum founded on February 26th, 2018. It is still active as of February 7th, 2020.<br />
<br />
It was founded by [[natiecswim]] and [[sm24136]]. The first few members to join were em24136, [[IceWolf10]], and alligator1234. The cofounders stayed for one full year, then passed on leadership to IceWolf10 and Awanglnc about a month after the anniversary.<br />
alligator1234, [[IceWolf10]], and BZ1 were essential to the early development of the PP army. Later on, haha0201 gave the forum a huge boost in activity, creating games and contributing a lot.<br />
<br />
Currently, it is under the leadership of [[IceWolf10]], [[anc3]], and michaelwen. Join now!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Pink_Penguinaphants_Army&diff=117228Pink Penguinaphants Army2020-02-08T03:04:08Z<p>Icewolf10: </p>
<hr />
<div>The Pink Penguinaphants Army, also called the PP Army by its members, is a forum founded on February 26th, 2018. It is still active as of February 7th, 2020.<br />
<br />
It was founded by [[natiecswim]] and [[sm24136]]. The first few members to join were em24136, [[IceWolf10]], and alligator1234. The cofounders stayed for one full year, then passed on leadership to IceWolf10 and Awanglnc about a month after the anniversary.<br />
alligator1234, [[IceWolf10]], and BZ1 were essential to the early development of the PP army. Later on, haha0201 gave the forum a huge boost in activity, creating games and contributing a lot.<br />
<br />
Currently, it is under the leadership of [[IceWolf10]], [[anc3]], and michaelwen. Join now!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Pink_Penguinaphants_Army&diff=117227Pink Penguinaphants Army2020-02-08T03:03:24Z<p>Icewolf10: </p>
<hr />
<div>The Pink Penguinaphants Army, also called the PP Army by its members, is a forum founded on February 26th, 2018. It is still active as of February 7th, 2020.<br />
<br />
It was founded by [[natiecswim]] and [[sm24136]]. The first few members to join were em24136, [[IceWolf10]], and alligator1234. The cofounders stayed for one full year, then passed on leadership to IceWolf10 and Awanglnc about a month after the anniversary.<br />
alligator1234, [[IceWolf10]], and BZ1 were essential to the early development of the PP army. Later on, haha0201 gave the forum a huge boost in activity, creating games and contributing a lot.<br />
<br />
Currently, it is under the leadership of [[IceWolf10]], anc3, and michaelwen. Join now!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Sm24136&diff=117226Sm241362020-02-08T03:02:55Z<p>Icewolf10: Created page with "A cofounder of the Pink Penguinaphants Army; the other founder is natiecswim."</p>
<hr />
<div>A cofounder of the [[Pink Penguinaphants Army]]; the other founder is [[natiecswim]].</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Natiecswim&diff=117225Natiecswim2020-02-08T03:02:08Z<p>Icewolf10: Created page with "A legendary founder, who cofounded the Pink Penguinaphants Army with sm24136."</p>
<hr />
<div>A legendary founder, who cofounded the [[Pink Penguinaphants Army]] with [[sm24136]].</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=117223IceWolf102020-02-08T03:00:59Z<p>Icewolf10: </p>
<hr />
<div>Hello!<br />
Please join the [[Pink Penguinaphants Army]], if you haven't already.<br />
<br />
I am the second member to join the PP Army. Right now I am its leader, and it is thriving with the help of ryrychen, [[anc3]], Winter_Ryan, FireCobra100, and many more users!<br />
<br />
We have games, marathons, spam battles, and a lot of fun with each other.<br />
<br />
An extension of the PP Army is the Pink Penguinaphants RPG. Join if you like imagination and gaming!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=117221IceWolf102020-02-08T03:00:35Z<p>Icewolf10: Blanked the page</p>
<hr />
<div></div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Pink_Penguinaphants_Army&diff=117215Pink Penguinaphants Army2020-02-08T02:55:44Z<p>Icewolf10: </p>
<hr />
<div>The Pink Penguinaphants Army, also called the PP Army by its members, is a forum founded on February 26th, 2018. It is still active as of February 7th, 2020.<br />
<br />
It was founded by natiecswim and sm24136. The first few members to join were em24136, [[IceWolf10]], and alligator1234. The cofounders stayed for one full year, then passed on leadership to IceWolf10 and Awanglnc about a month after the anniversary.<br />
alligator1234, [[IceWolf10]], and BZ1 were essential to the early development of the PP army. Later on, haha0201 gave the forum a huge boost in activity, creating games and contributing a lot.<br />
<br />
Currently, it is under the leadership of [[IceWolf10]], anc3, and michaelwen. Join now!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=117214IceWolf102020-02-08T02:55:04Z<p>Icewolf10: </p>
<hr />
<div>Hello!<br />
Please join the [[Pink Penguinaphants Army]], if you haven't already.<br />
<br />
I am the second member to join the PP Army. Right now I am its leader, and it is thriving with the help of ryrychen, [[anc3]], Winter_Ryan, FireCobra100, and many more users!<br />
<br />
We have games, marathons, spam battles, and a lot of fun with each other.<br />
<br />
An extension of the PP Army is the Pink Penguinaphants RPG. Join if you like imagination and gaming!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Pink_Penguinaphants_Army&diff=117213Pink Penguinaphants Army2020-02-08T02:52:04Z<p>Icewolf10: </p>
<hr />
<div>The Pink Penguinaphants Army, also called the PP Army by its members, is a forum founded on February 26th, 2018. It is still active as of February 7th, 2020.<br />
It was founded by natiecswim and sm24136. The first few members to join were em24136, [[IceWolf10]], and alligator1234. The cofounders stayed for one full year, then passed on leadership to IceWolf10 and Awanglnc about a month after the anniversary.<br />
alligator1234, [[IceWolf10]], and BZ1 were essential to the early development of the PP army. Later on, haha0201 gave the forum a huge boost in activity, creating games and contributing a lot.<br />
Currently, it is under the leadership of [[IceWolf10]], anc3, and michaelwen. Join now!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=Pink_Penguinaphants_Army&diff=117210Pink Penguinaphants Army2020-02-08T02:50:14Z<p>Icewolf10: Created page with "The Pink Penguinaphants Army, also called the PP Army by its members, is a forum founded on February 26th, 2018. It is still active as of February 7th, 2020. It was founded by..."</p>
<hr />
<div>The Pink Penguinaphants Army, also called the PP Army by its members, is a forum founded on February 26th, 2018. It is still active as of February 7th, 2020.<br />
It was founded by natiecswim and sm24136. The first few members to join were em24136, IceWolf10, and alligator1234. The cofounders stayed for one full year, then passed on leadership to IceWolf10 and Awanglnc about a month after the anniversary.<br />
alligator1234, IceWolf10, and BZ1 were essential to the early development of the PP army. Later on, haha0201 gave the forum a huge boost in activity, creating games and contributing a lot.<br />
Currently, it is under the leadership of IceWolf10, anc3, and michaelwen. Join now!</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=117206IceWolf102020-02-08T02:40:18Z<p>Icewolf10: </p>
<hr />
<div>Hello!<br />
Please join the Pink Penguinaphants Army, if you haven't already.</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=IceWolf10&diff=117204IceWolf102020-02-08T02:38:08Z<p>Icewolf10: IceWolf10's Bio</p>
<hr />
<div>Hello!<br />
<br />
My name is IceWolf10.<br />
Please join the Pink Penguinaphants Army, if you haven't already.</div>Icewolf10https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Symbols&diff=96536LaTeX:Symbols2018-07-27T22:54:44Z<p>Icewolf10: /* Operators */</p>
<hr />
<div>{{Latex}}<br />
<br />
This article will provide a short list of commonly used LaTeX symbols. <br />
<br />
== Common Symbols ==<br />
<br />
=== Operators ===<br />
*<math>\div</math><br />
*<math>\frac{2}{1}</math><br />
*<math>+</math><br />
*<math>-</math><br />
*<math>\dfrac{1}{2}</math><br />
*<math>\cdot</math><br />
<br />
== Finding Other Symbols ==<br />
<br />
Here are some external resources for finding less commonly used symbols:<br />
<ul><br />
<li><br />
[http://detexify.kirelabs.org/classify.html Detexify] is an app which allows you to draw the symbol you'd like and shows you the <math>\LaTeX</math> code for it!<br />
<br/><br/></li><br />
<br />
<li><br />
MathJax (what allows us to use <math>\LaTeX</math> on the web) maintains a [http://docs.mathjax.org/en/latest/tex.html#supported-latex-commands list of supported commands].<br />
<br/><br/></li><br />
<br />
<li><br />
[http://mirrors.ctan.org/info/symbols/comprehensive/symbols-a4.pdf The Comprehensive LaTeX Symbol List].<br />
<br/><br/></li><br />
</ul><br />
<br />
-------------------------------------------------------------------------------------------------------------<br />
<br />
==Operators==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\pm</math>||\pm||<math>\mp</math>||\mp||<math>\times</math>||\times<br />
|-<br />
|<math>\div</math>||\div||<math>\cdot</math>||\cdot||<math>\ast</math>||\ast<br />
|-<br />
|<math>\star</math>||\star||<math>\dagger</math>||\dagger||<math>\ddagger</math>||\ddagger<br />
|-<br />
|<math>\amalg</math>||\amalg||<math>\cap</math>||\cap||<math>\cup</math>||\cup<br />
|-<br />
|<math>\uplus</math>||\uplus||<math>\sqcap</math>||\sqcap||<math>\sqcup</math>||\sqcup<br />
|-<br />
|<math>\vee</math>||\vee||<math>\wedge</math>||\wedge||<math>\oplus</math>||\oplus<br />
|-<br />
|<math>\ominus</math>||\ominus||<math>\otimes</math>||\otimes||<math>\circ</math>||\circ<br />
|-<br />
|<math>\bullet</math>||\bullet||<math>\diamond</math>||\diamond||<math>\lhd</math>||\lhd<br />
|-<br />
|<math>\rhd</math>||\rhd||<math>\unlhd</math>||\unlhd||[[Image:Unrhd.gif]]||\unrhd<br />
|-<br />
|<math>\oslash</math>||\oslash||<math>\odot</math>||\odot||<math>\bigcirc</math>||\bigcirc<br />
|-<br />
|<math>\triangleleft</math>||\triangleleft||<math>\Diamond</math>||\Diamond||<math>\bigtriangleup</math>||\bigtriangleup<br />
|-<br />
|<math>\bigtriangledown</math>||\bigtriangledown||<math>\Box</math>||\Box||<math>\triangleright</math>||\triangleright<br />
|-<br />
|<math>\setminus</math>||\setminus||<math>\wr</math>||\wr||<math>\sqrt{x}</math>||\sqrt{x}<br />
|-<br />
|<math>x^{\circ}</math>||x^{\circ}||<math>\triangledown</math>||\triangledown||<math>\sqrt[n]{x}</math>||\sqrt[n]{x}<br />
|-<br />
|<math>a^x</math>||a^x||<math>a^{xyz}</math>||a^{xyz}<br />
|}<br />
<br />
==Relations==<br />
{| class="latextable"<br />
!Symbol !! Command !!Symbol !! Command!!Symbol !! Command<br />
|-<br />
| <math>\le</math>||\le||<math>\ge</math>||\ge||<math>\neq</math>||\neq<br />
|-<br />
| <math>\sim</math>||\sim||<math>\ll</math>||\ll||<math>\gg</math>||\gg<br />
|-<br />
| <math>\doteq</math>||\doteq||<math>\simeq</math>||\simeq||<math>\subset</math>||\subset<br />
|-<br />
| <math>\supset</math>||\supset||<math>\approx</math>||\approx||<math>\asymp</math>||\asymp<br />
|-<br />
| <math>\subseteq</math>||\subseteq||<math>\supseteq</math>||\supseteq||<math>\cong</math>||\cong<br />
|-<br />
| <math>\smile</math>||\smile||<math>\sqsubset</math>||\sqsubset||<math>\sqsupset</math>||\sqsupset<br />
|-<br />
| <math>\equiv</math>||\equiv||<math>\frown</math>||\frown||<math>\sqsubseteq</math>||\sqsubseteq<br />
|-<br />
| <math>\sqsupseteq</math>||\sqsupseteq||<math>\propto</math>||\propto||<math>\bowtie</math>||\bowtie<br />
|-<br />
| <math>\in</math>||\in||<math>\ni</math>||\ni||<math>\prec</math>||\prec<br />
|-<br />
| <math>\succ</math>||\succ||<math>\vdash</math>||\vdash||<math>\dashv</math>||\dashv<br />
|-<br />
| <math>\preceq</math>||\preceq||<math>\succeq</math>||\succeq||<math>\models</math>||\models<br />
|-<br />
| <math>\perp</math>||\perp||<math>\parallel</math>||\parallel||<br />
|-<br />
| <math>\mid</math>||\mid||<math>\bumpeq</math>||\bumpeq||<br />
|}<br />
Negations of many of these relations can be formed by just putting \not before the symbol, or by slipping an n between the \ and the word. Here are a few examples, plus a few other negations; it works for many of the others as well.<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\nmid</math>||\nmid||<math>\nleq</math>||\nleq||<math>\ngeq</math>||\ngeq<br />
|-<br />
| <math>\nsim</math>||\nsim||<math>\ncong</math>||\ncong||<math>\nparallel</math>||\nparallel<br />
|-<br />
| <math>\not<</math>||\not<||<math>\not></math>||\not>||<math>\not=</math>||\not=<br />
|-<br />
| <math>\not\le</math>||\not\le||<math>\not\ge</math>||\not\ge||<math>\not\sim</math>||\not\sim<br />
|-<br />
|<math>\not \approx</math>||\not\approx||<math>\not\cong</math>||\not\cong||<math>\not\equiv</math>||\not\equiv<br />
|-<br />
| <math>\not\parallel</math>||\not\parallel||<math>\nless</math>||\nless||<math>\ngtr</math>||\ngtr<br />
|-<br />
| <math>\lneq</math>||\lneq||<math>\gneq</math>||\gneq||<math>\lnsim</math>||\lnsim<br />
|-<br />
| <math>\lneqq</math>||\lneqq||<math>\gneqq</math>||\gneqq<br />
|}<br />
<br />
To use other relations not listed here, such as =, >, and <, in LaTeX, you may just use the symbols on your keyboard.<br />
<br />
==Greek Letters==<br />
{| class="latextable"<br />
|+ Lowercase Letters<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\alpha</math>||\alpha||<math>\beta</math>||\beta||<math>\gamma</math>||\gamma||<math>\delta</math>||\delta<br />
|-<br />
| <math>\epsilon</math>||\epsilon||<math>\varepsilon</math>||\varepsilon||<math>\zeta</math>||\zeta||<math>\eta</math>||\eta<br />
|-<br />
| <math>\theta</math>||\theta||<math>\vartheta</math>||\vartheta||<math>\iota</math>||\iota||<math>\kappa</math>||\kappa<br />
|-<br />
| <math>\lambda</math>||\lambda||<math>\mu</math>||\mu||<math>\nu</math>||\nu||<math>\xi</math>||\xi<br />
|-<br />
|<math>\pi</math>||\pi||<math>\varpi</math>||\varpi||<math>\rho</math>||\rho||<math>\varrho</math>||\varrho<br />
|-<br />
| <math>\sigma</math>||\sigma||<math>\varsigma</math>||\varsigma||<math>\tau</math>||\tau||<math>\upsilon</math>||\upsilon<br />
|-<br />
| <math>\phi</math>||\phi||<math>\varphi</math>||\varphi||<math>\chi</math>||\chi||<math>\psi</math>||\psi<br />
|-<br />
| <math>\omega</math>||\omega<br />
|}<br />
<br />
<br />
{| class="latextable"<br />
|+ Capital Letters<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\Gamma</math>||\Gamma||<math>\Delta</math>||\Delta||<math>\Theta</math>||\Theta||<math>\Lambda</math>||\Lambda<br />
|-<br />
| <math>\Xi</math>||\Xi||<math>\Pi</math>||\Pi||<math>\Sigma</math>||\Sigma||<math>\Upsilon</math>||\Upsilon<br />
|-<br />
| <math>\Phi</math>||\Phi||<math>\Psi</math>||\Psi||<math>\Omega</math>||\Omega<br />
|}<br />
<br />
<br />
== Arrows ==<br />
<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\gets</math>||\gets||<math>\to</math>||\to<br />
|-<br />
|<math>\leftarrow</math>||\leftarrow||<math>\Leftarrow</math>||\Leftarrow<br />
|-<br />
|<math>\rightarrow</math>||\rightarrow||<math>\Rightarrow</math>||\Rightarrow<br />
|-<br />
|<math>\leftrightarrow</math>||\leftrightarrow||<math>\Leftrightarrow</math>||\Leftrightarrow<br />
|-<br />
|<math>\mapsto</math>||\mapsto||<math>\hookleftarrow</math>||\hookleftarrow<br />
|-<br />
|<math>\leftharpoonup</math>||\leftharpoonup||<math>\leftharpoondown</math>||\leftharpoondown<br />
|-<br />
|<math>\rightleftharpoons</math>||\rightleftharpoons||<math>\longleftarrow</math>||\longleftarrow<br />
|-<br />
|<math>\Longleftarrow</math>||\Longleftarrow||<math>\longrightarrow</math>||\longrightarrow<br />
|-<br />
|<math>\Longrightarrow</math>||\Longrightarrow||<math>\longleftrightarrow</math>||\longleftrightarrow<br />
|-<br />
|<math>\Longleftrightarrow</math>||\Longleftrightarrow||<math>\longmapsto</math>||\longmapsto<br />
|-<br />
|<math>\hookrightarrow</math>||\hookrightarrow||<math>\rightharpoonup</math>||\rightharpoonup<br />
|-<br />
|<math>\rightharpoondown</math>||\rightharpoondown||<math>\leadsto</math>||\leadsto<br />
|-<br />
|<math>\uparrow</math>||\uparrow||<math>\Uparrow</math>||\Uparrow<br />
|-<br />
|<math>\downarrow</math>||\downarrow||<math>\Downarrow</math>||\Downarrow<br />
|-<br />
|<math>\updownarrow</math>||\updownarrow||<math>\Updownarrow</math>||\Updownarrow<br />
|-<br />
|<math>\nearrow</math>||\nearrow||<math>\searrow</math>||\searrow<br />
|-<br />
|<math>\swarrow</math>||\swarrow||<math>\nwarrow</math>||\nwarrow<br />
|}<br />
(For those of you who hate typing long strings of letters, \iff and \implies can be used in place of \Longleftrightarrow and \Longrightarrow respectively.)<br />
<br />
==Dots==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!<br />
|- <br />
|<math>\cdot</math>||\cdot|| |<math>\vdots</math>||\vdots|| <br />
|- <br />
|<math>\dots</math>||\dots|| |<math>\ddots</math>||\ddots||<br />
|-<br />
|<math>\cdots</math>||\cdots|| |<math>\iddots</math>||\iddots||<br />
<br />
|}<br />
<br />
==Accents==<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\hat{x}</math>||\hat{x}||<math>\check{x}</math>||\check{x}||<math>\dot{x}</math>||\dot{x}<br />
|-<br />
|<math>\breve{x}</math>||\breve{x}||<math>\acute{x}</math>||\acute{x}||<math>\ddot{x}</math>||\ddot{x}<br />
|-<br />
|<math>\grave{x}</math>||\grave{x}||<math>\tilde{x}</math>||\tilde{x}||<math>\mathring{x}</math>||\mathring{x}<br />
|-<br />
|<math>\bar{x}</math>||\bar{x}||<math>\vec{x}</math>||\vec{x}<br />
|}<br />
When applying accents to i and j, you can use \imath and \jmath to keep the dots from interfering with the accents:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\vec{\jmath}</math>||\vec{\jmath}||<math>\tilde{\imath}</math>||\tilde{\imath}<br />
|}<br />
\tilde and \hat have wide versions that allow you to accent an expression:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\widehat{7+x}</math>||\widehat{7+x}||<math>\widetilde{abc}</math>||\widetilde{abc}<br />
|}<br />
<br />
==Others==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command <br />
|-<br />
|<math>\infty</math>||\infty||<math>\triangle</math>||\triangle||<math>\angle</math>||\angle<br />
|-<br />
|<math>\aleph</math>||\aleph||<math>\hbar</math>||\hbar||<math>\imath</math>||\imath<br />
|-<br />
|<math>\jmath</math>||\jmath||<math>\ell</math>||\ell||<math>\wp</math>||\wp<br />
|-<br />
|<math>\Re</math>||\Re||<math>\Im</math>||\Im||<math>\mho</math>||\mho<br />
|-<br />
|<math>\prime</math>||\prime||<math>\emptyset</math>||\emptyset||<math>\nabla</math>||\nabla<br />
|-<br />
|<math>\surd</math>||\surd||<math>\partial</math>||\partial||<math>\top</math>||\top<br />
|-<br />
|<math>\bot</math>||\bot||<math>\vdash</math>||\vdash||<math>\dashv</math>||\dashv<br />
|-<br />
|<math>\forall</math>||\forall||<math>\exists</math>||\exists||<math>\neg</math>||\neg<br />
|-<br />
|<math>\flat</math>||\flat||<math>\natural</math>||\natural||<math>\sharp</math>||\sharp<br />
|-<br />
|<math>\backslash</math>||\backslash||<math>\Box</math>||\Box||<math>\Diamond</math>||\Diamond<br />
|-<br />
|<math>\clubsuit</math>||\clubsuit||<math>\diamondsuit</math>||\diamondsuit||<math>\heartsuit</math>||\heartsuit<br />
|-<br />
|[[Image:Spadesuit.gif]]||\spadesuit||<math>\Join</math>||\Join||<math>\blacksquare</math>||\blacksquare<br />
|-<br />
|<math>\S</math>||\S||<math>\P</math>||\P||<math>\copyright</math>||\copyright<br />
|-<br />
|<math>\pounds</math>||\pounds||<math>\overarc{ABC}</math>||\overarc{ABC}||<math>\underarc{XYZ}</math>||\underarc{XYZ}<br />
|-<br />
|<math>\bigstar</math>||\bigstar||<math>\in</math>||\in||<math>\cup</math>||\cup<br />
|-<br />
|<math>\square</math>||\square||<br />
|-<br />
|<math>\smiley</math>||\smiley||<br />
|-<br />
|<math>\mathbb{R}</math>||\mathbb{R} (represents all real numbers)||<br />
|-<br />
|<math>\checkmark</math>||\checkmark||<br />
|-<br />
|<math>\cancer</math>||\cancer||<br />
|}<br />
<br />
==Command Symbols==<br />
Some symbols are used in commands so they need to be treated in a special way.<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\textdollar</math>||\textdollar or \$||<math>\&</math>||\&||<math>\%</math>||\%||<math>\#</math>||\#<br />
|-<br />
|<math>\_</math>||\_||<math>\{</math>||\{||<math>\}</math>||\}||<math>\backslash</math>||\backslash<br />
|}<br />
<br />
(Warning: Using <nowiki>$</nowiki> for <math>\textdollar</math> will result in <math>\$</math>. This is a bug as far as we know. Depending on the version of <math>\LaTeX</math> this is not always a problem.)<br />
<br />
==European Language Symbols==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>{\oe}</math>||{\oe}||<math>{\ae}</math>||{\ae}||<math>{\o}</math>||{\o}<br />
|-<br />
|<math>{\OE}</math>||{\OE}||<math>{\AE}</math>||{\AE}||<math>{\AA}</math>||{\AA}||<math>{\O}</math>||{\O}<br />
|-<br />
|<math>{\l}</math>||{\l}||<math>{\ss}</math>||{\ss}||<math>\text{!`}</math>||!`<br />
|-<br />
|<math>{\L}</math>||{\L}||<math>{\SS}</math>||{\SS}||<br />
|}<br />
<br />
==Bracketing Symbols==<br />
In mathematics, sometimes we need to enclose expressions in brackets or braces or parentheses. Some of these work just as you'd imagine in LaTeX; type ( and ) for parentheses, [ and ] for brackets, and | and | for absolute value. However, other symbols have special commands:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\{</math>||\{||<math>\}</math>||\}||<math>\|</math>||<nowiki>\|</nowiki><br />
|-<br />
| <math>\backslash</math>||\backslash||<math>\lfloor</math>||\lfloor||<math>\rfloor</math>||\rfloor<br />
|-<br />
| <math>\lceil</math>||\lceil||<math>\rceil</math>||\rceil||<math>\langle</math>||\langle<br />
|-<br />
| <math>\rangle</math>||\rangle<br />
|}<br />
You might notice that if you use any of these to typeset an expression that is vertically large, like<br />
<br />
:<tt>(\frac{a}{x} )^2</tt><br />
<br />
the parentheses don't come out the right size:<br />
<br />
:<math>(\frac{a}{x})^2</math><br />
<br />
If we put \left and \right before the relevant parentheses, we get a prettier expression:<br />
<br />
:<tt>\left(\frac{a}{x} \right)^2</tt><br />
<br />
gives<br />
<br />
:<math>\left(\frac{a}{x} \right)^2</math><br />
<br />
And with system of equations:<br />
<br />
<tt>\left\{\begin{array}{l}x+y=3\\2x+y=5\end{array}\right.</tt><br />
<br />
Gives<br />
<br />
<math>\left\{\begin{array}{l}x+y=3\\2x+y=5\end{array}\right.</math><br />
<br />
See that there's a dot after <tt>\right</tt>. You must put that dot or the code won't work.<br />
<br />
<br />
And, if you type this<br />
<br />
<tt>\underbrace{a_0+a_1+a_2+\cdots+a_n}_{x}</tt><br />
<br />
Gives<br />
<br />
<math>\underbrace{a_0+a_1+a_2+\cdots+a_n}_{x}</math><br />
<br />
Or<br />
<br />
<tt>\overbrace{a_0+a_1+a_2+\cdots+a_n}^{x}</tt><br />
<br />
Gives<br />
<br />
<math>\overbrace{a_0+a_1+a_2+\cdots+a_n}^{x}</math><br />
<br />
<br />
\left and \right can also be used to resize the following symbols:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\uparrow</math>||\uparrow||<math>\downarrow</math>||\downarrow||<math>\updownarrow</math>||\updownarrow<br />
|-<br />
| <math>\Uparrow</math>||\Uparrow||<math>\Downarrow</math>||\Downarrow||<math>\Updownarrow</math>||\Updownarrow<br />
|}<br />
<br />
==Multi-Size Symbols==<br />
Some symbols render differently in inline math mode and in display mode. Display mode occurs when you use <nowiki>\[...\]</nowiki> or <nowiki>$$...$$</nowiki>, or environments like \begin{equation}...\end{equation}, \begin{align}...\end{align}. Read more in the [[LaTeX:Commands|commands]] section of the guide about how symbols which take arguments above and below the symbols, such as a summation symbol, behave in the two modes.<br />
<br />
In each of the following, the two images show the symbol in display mode, then in inline mode.<br />
<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\sum \textstyle\sum</math>||\sum||<math>\int \textstyle\int</math>||\int||<math>\oint \textstyle\oint</math>||\oint<br />
|-<br />
|<math>\prod \textstyle\prod</math>||\prod||<math>\coprod \textstyle\coprod</math>||\coprod||<math>\bigcap \textstyle\bigcap</math>||\bigcap<br />
|-<br />
|<math>\bigcup \textstyle\bigcup</math>||\bigcup||<math>\bigsqcup \textstyle\bigsqcup</math>||\bigsqcup||<math>\bigvee \textstyle\bigvee</math>||\bigvee<br />
|-<br />
|<math>\bigwedge \textstyle\bigwedge</math>||\bigwedge||<math>\bigodot \textstyle\bigodot</math>||\bigodot||<math>\bigotimes \textstyle\bigotimes</math>||\bigotimes<br />
|-<br />
|<math>\bigoplus \textstyle\bigoplus</math>||\bigoplus||<math>\biguplus \textstyle\biguplus</math>||\biguplus<br />
|}<br />
<br />
==See Also==<br />
*[[LaTeX:Commands | Next: Commands]]<br />
*[[LaTeX:Layout | Previous: Layout]]</div>Icewolf10