https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Fath2012&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:25:55ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_4&diff=1402972011 AMC 12A Problems/Problem 42020-12-22T22:53:25Z<p>Fath2012: </p>
<hr />
<div>== Problem ==<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?<br />
<br />
<math><br />
\textbf{(A)}\ 12 \qquad<br />
\textbf{(B)}\ \frac{37}{3} \qquad<br />
\textbf{(C)}\ \frac{88}{7} \qquad<br />
\textbf{(D)}\ 13 \qquad<br />
\textbf{(E)}\ 14 </math><br />
<br />
== Solution ==<br />
Let us say that there are <math>f</math> fifth graders. According to the given information, there must be <math>2f</math> fourth graders and <math>4f</math> third graders. The average time run by each student is equal to the total amount of time run divided by the number of students. This gives us <math>\frac{12\cdot 4f + 15\cdot 2f + 10\cdot f}{4f + 2f + f} = \frac{88f}{7f} = \frac{88}{7} \Rightarrow \boxed{C}</math><br />
<br />
<br />
<br />
If you want to simplify the problem even more, just imagine/assume that only <math>1</math> fifth grader existed. Then you can simply get rid of the variables.<br />
<br />
==Video Solution==<br />
https://youtu.be/3LWBLXzcSKo<br />
<br />
~savannahsolver<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=3|num-a=5|ab=A}}<br />
{{AMC10 box|year=2011|ab=A|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_3&diff=1402962011 AMC 12A Problems/Problem 32020-12-22T22:51:10Z<p>Fath2012: </p>
<hr />
<div>== Problem ==<br />
A small bottle of shampoo can hold <math>35</math> milliliters of shampoo, whereas a large bottle can hold <math>500</math> milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math><br />
\textbf{(A)}\ 11 \qquad<br />
\textbf{(B)}\ 12 \qquad<br />
\textbf{(C)}\ 13 \qquad<br />
\textbf{(D)}\ 14 \qquad<br />
\textbf{(E)}\ 15 </math><br />
<br />
== Solution ==<br />
To find how many small bottles we need, we can simply divide <math>500</math> by <math>35</math>. This simplifies to <math>\frac{100}{7}=14 \frac{2}{7}. </math> Since the answer must be an integer greater than <math>14</math>, we have to round up to <math>15</math> bottles, or <math>\boxed{\textbf{E}}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/A6oQF25ayzo<br />
<br />
~savannahsolver<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=2|num-a=4|ab=A}}<br />
{{AMC10 box|year=2011|num-b=1|num-a=3|ab=A}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_1&diff=1402942011 AMC 12A Problems/Problem 12020-12-22T22:47:21Z<p>Fath2012: </p>
<hr />
<div>== Problem ==<br />
A cell phone plan costs <math>20</math> dollars each month, plus <math>5</math> cents per text message sent, plus <math>10</math> cents for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math><br />
\textbf{(A)}\ 24.00 \qquad<br />
\textbf{(B)}\ 24.50 \qquad<br />
\textbf{(C)}\ 25.50 \qquad<br />
\textbf{(D)}\ 28.00 \qquad<br />
\textbf{(E)}\ 30.00 </math><br />
<br />
<br />
== Solution ==<br />
The base price of Michelle's cell phone plan is <math>20</math> dollars. <br />
If she sent <math>100</math> text messages and it costs <math>5</math> cents per text, then she must have spent <math>500</math> cents for texting, or <math>5</math> dollars. She talked for <math>30.5</math> hours, but <math>30.5-30</math> will give us the amount of time that she has to pay an additional amount for. <br />
<math>30.5-30=.5</math> hours <math>=30</math> minutes.<br />
Since the price for phone calls is <math>10</math> cents per minute, the additional amount Michelle has to pay for phone calls is <math>30*10=300</math> cents, or <math>3</math> dollars.<br />
Adding <math>20+5+3</math> dollars <math>=28</math> dollars <math>=\boxed{\textbf{D}}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/tWz5T2Db9AY<br />
<br />
~savannahsolver<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|before=First Problem|num-a=2|ab=A}}<br />
{{AMC10 box|year=2011|before=First Problem|num-a=2|ab=A}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_19&diff=1401842010 AMC 12B Problems/Problem 192020-12-22T05:41:41Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #19]] and [[2010 AMC 10B Problems|2010 AMC 10B #24]]}}<br />
<br />
== Problem 19 ==<br />
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than <math>100</math> points. What was the total number of points scored by the two teams in the first half?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math><br />
<br />
<br />
== Solution 1 ==<br />
<br />
<br />
Let <math>a,ar,ar^{2},ar^{3}</math> be the quarterly scores for the Raiders. We know <math>r > 1</math> because the sequence is said to be increasing. We also know that each of <math>a, ar, ar^2, ar^3</math> is an integer. We start by showing that '''<math>r</math> must also be an integer.'''<br />
<br />
Suppose not, and say <math>r = m/n</math> where <math>m>n>1</math>, and <math>\gcd(m,n)=1</math>. Then <math>n, n^2, n^3</math> must all divide <math>a</math> so <math>a=n^3k</math> for some integer <math>k</math>. Then <math>S_R = n^3k + n^2mk + nm^2k + m^3k < 100</math> and we see that even if <math>k=1</math> and <math>n=2</math>, we get <math>m < 4</math>, which means that the only option for <math>r</math> is <math>r=3/2</math>. A quick check shows that even this doesn't work. Thus <math>r</math> must be an integer.<br />
<br />
Let <math>a, a+d, a+2d, a+3d</math> be the quarterly scores for the Wildcats. Let <math>S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d</math>. Let <math>S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)</math>. Then <math>S_R<100</math> implies that <math>r<5</math>, so <math>r\in \{2, 3, 4\}</math>. The Raiders win by one point, so<cmath>a(1+r)(1+r^2) = 4a+6d+1.</cmath><br />
*If <math>r=4</math> we get <math>85a = 4a+6d+1</math> which means <math>3(27a-2d) = 1</math>, which is absurd.<br />
*If <math>r=3</math> we get <math>40a = 4a+6d+1</math> which means <math>6(6a-d) = 1</math>, which is also absurd.<br />
*If <math>r=2</math> we get <math>15a = 4a+6d+1</math> which means <math>11a-6d = 1</math>. Reducing modulo 6 we get <math>a \equiv 5\pmod{6}</math>. Since <math>15a<100</math> we get <math>a<7</math>. Thus <math>a=5</math>. It then follows that <math>d=9</math>.<br />
Then the quarterly scores for the Raiders are <math>5, 10, 20, 40</math>, and those for the Wildcats are <math>5, 14, 23, 32</math>. Also <math>S_R = 75 = S_W + 1</math>. The total number of points scored by the two teams in the first half is <math>5+10+5+14=\boxed{\textbf{(E)}\ 34}</math>.<br />
<br />
<br />
== Solution 2 ==<br />
<br />
Let <math>a,ar,ar^{2},ar^{3}</math> be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let <math>a,a+d,a+2d,a+3d</math> be the quarterly scores for the Wildcats. The sum of the Raiders scores is <math>a(1+r+r^{2}+r^{3})</math> and the sum of the Wildcats scores is <math>4a+6d</math>. Now we can narrow our search for the values of <math>a,d</math>, and <math>r</math>. Because points are always measured in positive integers, we can conclude that <math>a</math> and <math>d</math> are positive integers. We can also conclude that <math>r</math> is a positive integer by writing down the equation:<br />
<br />
<cmath>a(1+r+r^{2}+r^{3})=4a+6d+1</cmath><br />
<br />
Now we can start trying out some values of <math>r</math>. We try <math>r=2</math>, which gives<br />
<br />
<cmath>15a=4a+6d+1</cmath><br />
<br />
<cmath>11a=6d+1</cmath><br />
<br />
We need the smallest multiple of <math>11</math> (to satisfy the <100 condition) that is <math>\equiv 1 \pmod{6}</math>. We see that this is <math>55</math>, and therefore <math>a=5</math> and <math>d=9</math>.<br />
<br />
So the Raiders' first two scores were <math>5</math> and <math>10</math> and the Wildcats' first two scores were <math>5</math> and <math>14</math>. <br />
<br />
<cmath>5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}</cmath><br />
<br />
==Video Solution==<br />
https://youtu.be/krRrPxRdgD0<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}}<br />
{{AMC10 box|year=2010|num-b=23|num-a=25|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_19&diff=1401832010 AMC 12B Problems/Problem 192020-12-22T05:41:26Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #19]] and [[2010 AMC 10B Problems|2010 AMC 10B #24]]}}<br />
<br />
== Problem 19 ==<br />
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than <math>100</math> points. What was the total number of points scored by the two teams in the first half?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math><br />
<br />
<br />
== Solution 1 ==<br />
<br />
<br />
Let <math>a,ar,ar^{2},ar^{3}</math> be the quarterly scores for the Raiders. We know <math>r > 1</math> because the sequence is said to be increasing. We also know that each of <math>a, ar, ar^2, ar^3</math> is an integer. We start by showing that '''<math>r</math> must also be an integer.'''<br />
<br />
Suppose not, and say <math>r = m/n</math> where <math>m>n>1</math>, and <math>\gcd(m,n)=1</math>. Then <math>n, n^2, n^3</math> must all divide <math>a</math> so <math>a=n^3k</math> for some integer <math>k</math>. Then <math>S_R = n^3k + n^2mk + nm^2k + m^3k < 100</math> and we see that even if <math>k=1</math> and <math>n=2</math>, we get <math>m < 4</math>, which means that the only option for <math>r</math> is <math>r=3/2</math>. A quick check shows that even this doesn't work. Thus <math>r</math> must be an integer.<br />
<br />
Let <math>a, a+d, a+2d, a+3d</math> be the quarterly scores for the Wildcats. Let <math>S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d</math>. Let <math>S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)</math>. Then <math>S_R<100</math> implies that <math>r<5</math>, so <math>r\in \{2, 3, 4\}</math>. The Raiders win by one point, so<cmath>a(1+r)(1+r^2) = 4a+6d+1.</cmath><br />
*If <math>r=4</math> we get <math>85a = 4a+6d+1</math> which means <math>3(27a-2d) = 1</math>, which is absurd.<br />
*If <math>r=3</math> we get <math>40a = 4a+6d+1</math> which means <math>6(6a-d) = 1</math>, which is also absurd.<br />
*If <math>r=2</math> we get <math>15a = 4a+6d+1</math> which means <math>11a-6d = 1</math>. Reducing modulo 6 we get <math>a \equiv 5\pmod{6}</math>. Since <math>15a<100</math> we get <math>a<7</math>. Thus <math>a=5</math>. It then follows that <math>d=9</math>.<br />
Then the quarterly scores for the Raiders are <math>5, 10, 20, 40</math>, and those for the Wildcats are <math>5, 14, 23, 32</math>. Also <math>S_R = 75 = S_W + 1</math>. The total number of points scored by the two teams in the first half is <math>5+10+5+14=\boxed{\textbf{(E)}\ 34}</math>.<br />
<br />
<br />
== Solution 2 ==<br />
<br />
Let <math>a,ar,ar^{2},ar^{3}</math> be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let <math>a,a+d,a+2d,a+3d</math> be the quarterly scores for the Wildcats. The sum of the Raiders scores is <math>a(1+r+r^{2}+r^{3})</math> and the sum of the Wildcats scores is <math>4a+6d</math>. Now we can narrow our search for the values of <math>a,d</math>, and <math>r</math>. Because points are always measured in positive integers, we can conclude that <math>a</math> and <math>d</math> are positive integers. We can also conclude that <math>r</math> is a positive integer by writing down the equation:<br />
<br />
<cmath>a(1+r+r^{2}+r^{3})=4a+6d+1</cmath><br />
<br />
Now we can start trying out some values of <math>r</math>. We try <math>r=2</math>, which gives<br />
<br />
<cmath>15a=4a+6d+1</cmath><br />
<br />
<cmath>11a=6d+1</cmath><br />
<br />
We need the smallest multiple of <math>11</math> (to satisfy the <100 condition) that is <math>\equiv 1 \pmod{6}</math>. We see that this is <math>55</math>, and therefore <math>a=5</math> and <math>d=9</math>.<br />
<br />
So the Raiders' first two scores were <math>5</math> and <math>10</math> and the Wildcats' first two scores were <math>5</math> and <math>14</math>. <br />
<br />
<cmath>5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}</cmath><br />
<br />
==Video Solution==<br />
https://youtu.be/krRrPxRdgD0<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}}<br />
{{AMC10 box|year=2010|num-b=17|num-a=19|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_17&diff=1401792010 AMC 12B Problems/Problem 172020-12-22T05:37:09Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #17]] and [[2010 AMC 10B Problems|2010 AMC 10B #23]]}}<br />
<br />
== Problem ==<br />
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math><br />
<br />
== Solution 1 ==<br />
Observe that all tableaus must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each tableau, there exists a valid tableau diagonally symmetrical across the diagonal extending from the top left to the bottom right. <br />
<br />
<br />
*'''Case 1: Center 4'''<br />
<cmath>\begin{tabular}{|c|c|c|} \hline 1&2&\\<br />
\hline 3&4&8\\<br />
\hline &&9\\<br />
\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\<br />
\hline 3&4&\\<br />
\hline &8&9\\<br />
\hline \end{tabular}</cmath><br />
<br />
3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. <math>2*6=12</math><br />
<br />
*'''Case 2: Center 5'''<br />
<cmath>\begin{tabular}{|c|c|c|} \hline 1&2&3\\<br />
\hline 4&5&\\<br />
\hline &8&9\\<br />
\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\<br />
\hline 3&5&\\<br />
\hline &8&9\\<br />
\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\<br />
\hline 3&5&8\\<br />
\hline &&9\\<br />
\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\<br />
\hline 4&5&8\\<br />
\hline &&9\\<br />
\hline \end{tabular}</cmath><br />
<br />
Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that <math>4<5</math>, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, <math>2*9=18</math><br />
<br />
*'''Case 3: Center 6'''<br />
By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured.<br />
<math>2*6=12</math><br />
<br />
<cmath>12+18+12=\boxed{\textbf{D)}42}</cmath><br />
<br />
<br />
~BJHHar<br />
<br />
== Solution 3==<br />
This solution is trivial by the hook length theorem. The hooks look like this:<br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\<br />
\hline 4 & 3 & 2\\<br />
\hline 3 & 2 & 1\\<br />
\hline \end{tabular}</math><br />
<br />
So, the answer is <math>\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}</math> = <math>\boxed{\text{(D) }42}</math><br />
<br />
P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.<br />
<br />
==Video Solution==<br />
https://youtu.be/ZfnxbpdFKjU?t=422<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}}<br />
{{AMC10 box|year=2010|num-b=22|num-a=24|ab=B}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_11&diff=1401772010 AMC 12B Problems/Problem 112020-12-22T05:36:22Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #11]] and [[2010 AMC 10B Problems|2010 AMC 10B #21]]}}<br />
<br />
== Problem 11 ==<br />
A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math><br />
<br />
== Solution ==<br />
View the palindrome as some number with form (decimal representation):<br />
<math>a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>. But because the number is a palindrome, <math>a_3 = a_0, a_2 = a_1</math>. Recombining this yields <math>1001a_3 + 110a_2</math>. 1001 is divisible by 7, which means that as long as <math>a_2 = 0</math>, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (<math>9 \cdot 10</math>) possibilities for palindromes. However, if <math>a_2 = 7</math>, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to <math>18/90 = \boxed {\frac{1}{5} } = \boxed {E}</math><br />
<br />
== Addendum (Alternate) == <br />
<math>7\mid 1001a^3+110b^2</math> and <math>1001 \equiv 0 (\textrm{mod} 7)</math>. Knowing that <math>a</math> does not factor (pun intended) into the problem, note 110's prime factorization and <math>7\mid b</math>. There are only 10 possible digits for b (0-9), but <math>7\mid b</math> only holds if <math>b=0, 7</math>. This is 2 of the 10 digits, so <math>\frac{2}{10}=\boxed{\textbf{E)}\frac{1}{5}}</math><br />
<br />
~BJHHar<br />
==Video Solution==<br />
https://youtu.be/ZfnxbpdFKjU<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=10|num-a=12|ab=B}}<br />
{{AMC10 box|year=2010|num-b=20|num-a=22|ab=B}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_16&diff=1401762010 AMC 12B Problems/Problem 162020-12-22T05:26:44Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}}<br />
<br />
== Problem 16 ==<br />
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}</math><br />
<br />
== Solution 1 ==<br />
<br />
We group this into groups of <math>3</math>, because <math>3|2010</math>. This means that every residue class mod 3 has an equal probability.<br />
<br />
If <math>3|a</math>, we are done. There is a probability of <math>\frac{1}{3}</math> that that happens.<br />
<br />
Otherwise, we have <math>3|bc+b+1</math>, which means that <math>b(c+1) \equiv 2\pmod{3}</math>. So either <cmath>b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}</cmath> or <cmath>b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3</cmath> which will lead to the property being true. There are a <math>\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}</math> chance for each bundle of cases to be true. Thus, the total for the cases is <math>\frac{2}{9}</math>. But we have to multiply by <math>\frac{2}{3}</math> because this only happens with a <math>\frac{2}{3}</math> chance. So the total is actually <math>\frac{4}{27}</math>.<br />
<br />
The grand total is <cmath>\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}</cmath><br />
<br />
== Solution 2 (Minor change from Solution 1) ==<br />
<br />
Just like solution 1, we see that there is a <math>\frac{1}{3}</math> chance of <math>3|a</math> and <math>\frac{2}{9}</math> chance of <math>3|1+b+bc</math> <br />
<br />
Now, we can just use PIE (Principles of Inclusion and Exclusion) to get our answer to be <math>\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{E) \frac{13}{27}}</math><br />
<br />
-Conantwiz2023<br />
<br />
==Solution 3 (Fancier version of Solution 1)==<br />
<br />
As with solution one, we conclude that if <math>a\equiv0\mod 3</math> then the requirements are satisfied. We then have:<br />
<cmath>a(bc+c)+a\equiv0 \mod 3</cmath><br />
<cmath>a(bc+c)\equiv-a \mod 3</cmath><br />
<cmath>c(b+1)\equiv-1 \mod 3</cmath><br />
<cmath>b+1\equiv \frac{-1}{c} \mod 3</cmath><br />
Which is true for one <math>b</math> when <math>c\not\equiv 0 \mod 3</math> because the integers<math>\mod 3</math> form a field under multiplication and addition with absorbing element <math>0</math>.<br />
<br />
This gives us <math>P=\frac{1}{3}+\frac{4}{27}=\boxed{\text{(E) }\frac{13}{27}}</math>.<br />
<br />
~Snacc<br />
<br />
==Video Solution==<br />
https://youtu.be/FQO-0E2zUVI?t=437<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}<br />
{{AMC10 box|year=2010|num-b=17|num-a=19|ab=B}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_8&diff=1401752010 AMC 12B Problems/Problem 82020-12-22T05:20:57Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #8]] and [[2010 AMC 10B Problems|2010 AMC 10B #17]]}}<br />
<br />
== Problem 8 ==<br />
Every high school in the city of Euclid sent a team of <math>3</math> students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?<br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math><br />
<br />
== Solution ==<br />
There are <math>x</math> schools. This means that there are <math>3x</math> people. Because no one's score was the same as another person's score, that means that there could only have been <math>1</math> median score. This implies that <math>x</math> is an odd number. <math>x</math> cannot be less than <math>23</math>, because there wouldn't be a <math>64</math>th place if there were. <math>x</math> cannot be greater than <math>23</math> either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is <math>23 \Rightarrow \boxed{B}</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>a</math> be Andrea's place. We know that she was the highest on our team, so <math>a < 37</math>. <br />
<br />
Since <math>a</math> is the median, there are <math>a-1</math> to the left and right of the median, so the total number of people is <math>2a-1</math> and the number of schools is <math>(2a-1)/3</math>. This implies that <math>2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}</math>. <br />
<br />
Also, since <math>2a-1</math> is the rank of the last-place person, and one of Andrea's teammates already got 64th place, <math>2a-1 > 64 \implies a \ge 33</math>.<br />
<br />
Putting it all together: <math>33 \le a < 37</math> and <math>a \equiv 2 \pmod{3}</math>, so clearly <math>a = 35</math>, and the number of schools as we got before is <math>(2a-1)/3 = 69/3 = \boxed{23}</math>.<br />
<br />
~adihaya<br />
<br />
==Video Solution==<br />
https://youtu.be/FQO-0E2zUVI?t=297<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=B}}<br />
{{AMC10 box|year=2010|num-b=16|num-a=18|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_10&diff=1401742010 AMC 12B Problems/Problem 102020-12-22T05:15:48Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #10]] and [[2010 AMC 10B Problems|2010 AMC 10B #14]]}}<br />
<br />
== Problem 10 ==<br />
The average of the numbers <math>1, 2, 3,\cdots, 98, 99,</math> and <math>x</math> is <math>100x</math>. What is <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}</math><br />
<br />
== Solution ==<br />
We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\cdot50</math>. Then, we know that the sum of the series is <math>99\cdot50+x</math>. There are <math>100</math> terms, so we can divide this sum by <math>100</math> and set it equal to <math>100x</math>:<br />
<cmath>\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x</cmath><br />
Using difference of squares:<br />
<cmath>x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}</cmath><br />
Thus, the answer is <math>\boxed{\text{B}}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/vYXz4wStBUU?t=413<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=9|num-a=11|ab=B}}<br />
{{AMC10 box|year=2010|num-b=13|num-a=15|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_6&diff=1401732010 AMC 12B Problems/Problem 62020-12-22T05:11:37Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #6]] and [[2010 AMC 10B Problems|2010 AMC 10B #12]]}}<br />
<br />
== Problem 6 ==<br />
At the beginning of the school year, <math>50\%</math> of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and <math>50\%</math> answered "No." At the end of the school year, <math>70\%</math> answered "Yes" and <math>30\%</math> answered "No." Altogether, <math>x\%</math> of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80</math><br />
<br />
== Solution ==<br />
Clearly, the minimum possible value would be <math>70 - 50 = 20\%</math>. The maximum possible value would be <math>30 + 50 = 80\%</math>. The difference is <math>80 - 20 = \boxed{60}</math> <math>(D)</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/vYXz4wStBUU?t=160<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=5|num-a=7|ab=B}}<br />
{{AMC10 box|year=2010|num-b=11|num-a=13|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_7&diff=1401722010 AMC 12B Problems/Problem 72020-12-22T05:04:14Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #7]] and [[2010 AMC 10B Problems|2010 AMC 10B #10]]}}<br />
<br />
== Problem 7 ==<br />
Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math><br />
<br />
== Solution 1 ==<br />
<br />
Let <math>x</math> be the time it is not raining, and <math>y</math> be the time it is raining, in hours.<br />
<br />
We have the system: <math>30x+20y=16</math> and <math>x+y=2/3</math><br />
<br />
Solving gives <math>x=\frac{4}{15}</math> and <math>y=\frac{2}{5}</math><br />
<br />
We want <math>y</math> in minutes, <math>\frac{2}{5}*60=24 \Rightarrow C</math><br />
<br />
== Solution 2 ==<br />
Let <math>x</math> be the time it is raining. Thus, the number of minutes it is not raining is <math>40-x</math> . <br />
<br />
Since we are calculating the time in minutes, it is best to convert the speeds in minutes. Thus, the speed per minute when it is not raining is <math>\frac{1}{2}</math> per minute, and <math>\frac{1}{3}</math> per minute when it is not raining. <br />
Thus, we have the equation, <math>\frac{1}{2}</math> * x + <math>\frac{1}{3}</math> * (40-x) = 16 <br />
<br />
Solving, gives <math>x</math> = <math>16</math> , so the amount of time it is not raining is <math>40</math> - <math>16</math> = <math>24</math><br />
<br />
~coolmath2017<br />
<br />
==Video Solution==<br />
https://youtu.be/I3yihAO87CE?t=429<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=6|num-a=8|ab=B}}<br />
{{AMC10 box|year=2010|num-b=9|num-a=11|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_5&diff=1401712010 AMC 12B Problems/Problem 52020-12-22T05:02:16Z<p>Fath2012: </p>
<hr />
<div>== Problem 5 ==<br />
Lucky Larry's teacher asked him to substitute numbers for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> in the expression <math>a-(b-(c-(d+e)))</math> and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry substituted for <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> were <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. What number did Larry substitute for <math>e</math>?<br />
<br />
<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math><br />
<br />
== Solution ==<br />
We simply plug in the numbers<br />
<cmath>1 - 2 - 3 - 4 + e = 1 - (2 - (3 - (4 + e)))</cmath><br />
<cmath>-8 + e = -2 - e</cmath><br />
<cmath>2e = 6</cmath><br />
<cmath>e = 3 \;\;(D)</cmath><br />
<br />
==Video Solution==<br />
https://youtu.be/I3yihAO87CE?t=303<br />
<br />
~IceMatrix<br />
<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=4|num-a=6|ab=B}}<br />
{{AMC10 box|year=2010|num-b=8|num-a=10|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_3&diff=1401702010 AMC 12B Problems/Problem 32020-12-22T05:01:47Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #3]] and [[2010 AMC 10B Problems|2010 AMC 10B #8]]}}<br />
<br />
== Problem 3 ==<br />
A ticket to a school play cost <math>x</math> dollars, where <math>x</math> is a whole number. A group of 9<sub>th</sub> graders buys tickets costing a total of &#36;<math>48</math>, and a group of 10<sub>th</sub> graders buys tickets costing a total of &#36;<math>64</math>. How many values for <math>x</math> are possible?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
== Solution ==<br />
We find the greatest common factor of <math>48</math> and <math>64</math> to be <math>16</math>. The number of factors of <math>16</math> is <math>5</math> which is the answer <math>(E)</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/I3yihAO87CE?t=179<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=2|num-a=4|ab=B}}<br />
{{AMC10 box|year=2010|num-b=7|num-a=9|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_4&diff=1401692010 AMC 12B Problems/Problem 42020-12-22T04:55:56Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #4]] and [[2010 AMC 10B Problems|2010 AMC 10B #5]]}}<br />
<br />
== Problem 4 ==<br />
A month with <math>31</math> days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
== Solution ==<br />
<math>31 \equiv 3 \pmod {7}</math> so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is <math>\boxed{B}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/uAc9VHtRRPg?t=329<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=3|num-a=5|ab=B}}<br />
{{AMC10 box|year=2010|num-b=4|num-a=6|ab=B}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_1&diff=1401682010 AMC 12B Problems/Problem 12020-12-22T04:49:23Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #1]] and [[2010 AMC 10B Problems|2010 AMC 10B #2]]}}<br />
<br />
== Problem ==<br />
Makarla attended two meetings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?<br />
<br />
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35</math><br />
<br />
== Solution ==<br />
The total number of minutes in here <math>9</math>-hour work day is <math>9 \times 60 = 540.</math><br />
The total amount of time spend in meetings in minutes is <math>45 + 45 \times 2 = 135.</math><br />
The answer is then <math>\frac{135}{540}</math> <math>= \boxed{25\%}</math> or <math>\boxed{(C)}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/uAc9VHtRRPg?t=82<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|before=First Question|num-a=2|ab=B}}<br />
{{AMC10 box|year=2010|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_23&diff=1399502010 AMC 12A Problems/Problem 232020-12-19T01:01:28Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #23]] and [[2010 AMC 10A Problems|2010 AMC 10A #24]]}}<br />
<br />
== Problem ==<br />
<br />
The number obtained from the last two nonzero digits of <math>90!</math> is equal to <math>n</math>. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math><br />
<br />
== Hints and Method of Attack ==<br />
Let <math>P</math> be the result of dividing <math>90!</math> by tens such that <math>P</math> is not divisible by <math>10</math>. We want to consider <math>P \mod 100</math>. But because <math>100</math> is not prime, and because <math>P</math> is obviously divisible by <math>4</math> (if in doubt, look at the answer choices), we only need to consider <math>P \mod 25</math>.<br />
<br />
However, <math>25</math> is a very particular number. <math>1 * 2 * 3 * 4 \equiv -1 \mod 25</math>, and so is <math>6 * 7 * 8 * 9</math>. How can we group terms to take advantage of this fact?<br />
<br />
There might be a problem when you cancel out the <math>10</math>s from <math>90!</math>. One method is to cancel out a factor of <math>2</math> from an existing number along with a factor of <math>5</math>. But this might prove cumbersome, as the grouping method will not be as effective. Instead, take advantage of ''inverses'' in modular arithmetic. Just leave the negative powers of <math>2</math> in a "storage base," and take care of the other terms first. Then, use Fermat's Little Theorem to solve for the power of <math>2</math>.<br />
<br />
== Solution 1==<br />
<br />
We will use the fact that for any integer <math>n</math>,<br />
<cmath>\begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\<br />
&=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\<br />
&=24\pmod{25}\equiv -1\pmod{25}.\end{align*}</cmath><br />
<br />
First, we find that the number of factors of <math>10</math> in <math>90!</math> is equal to <math>\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21</math>. Let <math>N=\frac{90!}{10^{21}}</math>. The <math>n</math> we want is therefore the last two digits of <math>N</math>, or <math>N\pmod{100}</math>. If instead we find <math>N\pmod{25}</math>, we know that <math>N\pmod{100}</math>, what we are looking for, could be <math>N\pmod{25}</math>, <math>N\pmod{25}+25</math>, <math>N\pmod{25}+50</math>, or <math>N\pmod{25}+75</math>. Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because <math>N\pmod{100}</math> has to be a multiple of 4.<br />
<br />
If we divide <math>N</math> by <math>5^{21}</math> by taking out all the factors of <math>5</math> in <math>N</math>, we can write <math>N</math> as <math>\frac M{2^{21}}</math> where<br />
<cmath>M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,</cmath><br />
where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form <math>5n</math> is replaced by <math>n</math>, and every number in the form <math>25n</math> is replaced by <math>n</math>.<br />
<br />
The number <math>M</math> can be grouped as follows:<br />
<br />
<cmath>\begin{align*}M= &(1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots(86\cdot 87\cdot 88\cdot 89)\\<br />
&\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\<br />
&\cdot (1\cdot 2\cdot 3).\end{align*}</cmath><br />
<br />
Where the first line is composed of the numbers in <math>90!</math> that aren't multiples of five, the second line is the multiples of five '''and not 25''' after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.<br />
<br />
Using the identity at the beginning of the solution, we can reduce <math>M</math> to<br />
<br />
<cmath>\begin{align*}M&\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\<br />
&= 1\cdot -21\cdot 6\\<br />
&= -1\pmod{25} =24\pmod{25}.\end{align*}</cmath><br />
<br />
Using the fact that <math>2^{10}=1024\equiv -1\pmod{25}</math> (or simply the fact that <math>2^{21}=2097152</math> if you have your powers of 2 memorized), we can deduce that <math>2^{21}\equiv 2\pmod{25}</math>. Therefore <math>N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}</math>.<br />
<br />
Finally, combining with the fact that <math>N\equiv 0\pmod 4</math> yields <math>n=\boxed{\textbf{(A)}\ 12}</math>.<br />
<br />
==Solution 2==<br />
Let <math>P</math> be <math>90!</math> after we truncate its zeros. Notice that <math>90!</math> has exactly (floored) <math>\left\lfloor\frac{90}{5}\right\rfloor + \left\lfloor\frac{90}{25}\right\rfloor = 21</math> factors of 5; thus, <cmath>P = 2^{-21}*5^{-21}*90!.</cmath> We shall consider <math>P</math> modulo 4 and 25, to determine its residue modulo 100. It is easy to prove that <math>P</math> is divisible by 4 (consider the number of 2s dividing <math>90!</math> minus the number of 5s dividing <math>90!</math>), and so we only need to consider <math>P</math> modulo 25.<br />
<br />
Now, notice that for integers <math>a, n</math> we have<cmath>(5n + a)(5n - a) \equiv -a^2 \mod 25.</cmath><br />
<br />
Thus, for integral a: <cmath>(10a + 1)(10a + 2)(10a + 3)(10a + 4)(10a + 6)(10a + 7)(10a + 8)(10a + 9) \equiv (-1)(-4)(-9)(-16) \equiv 576 \equiv 1 \mod 25.</cmath> Using this process, we can essentially remove all the numbers which had not formerly been a multiple of 5 in <math>90!</math> from consideration.<br />
<br />
Now, we consider the remnants of the 5, 10, 15, 20, ..., 90 not yet eliminated. The 10, 20, 30, ..., 90 becomes 1, 2, 3, 4, 1, 6, 7, 8, 9, whose product is 1 mod 25. Also, the 5, 5, 15, 25, ..., 85 becomes 1, 1, 3, 1, 7, 9, 11, 13, 3, 17 and <math>2^{-12}</math>. We deduce that from multiplying out the 1, 1, 3, 1, 7, ..., 17 is equivalent to 2 modulo 25, and so we need to compute <math>2^{-11}</math>. But this is simply by Fermat's Little Theorem <math>2^9 = 512 \equiv 12 \mod 25</math>. Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show that <math>P = 12 \mod 100</math> and so the answer is <math>\boxed{12}</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2010|num-b=22|num-a=24|ab=A}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_19&diff=1399482010 AMC 12A Problems/Problem 192020-12-19T00:58:22Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #19]] and [[2010 AMC 10A Problems|2010 AMC 10A #23]]}}<br />
<br />
== Problem ==<br />
Each of <math>2010</math> boxes in a line contains a single red marble, and for <math>1 \le k \le 2010</math>, the box in the <math>k\text{th}</math> position also contains <math>k</math> white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let <math>P(n)</math> be the probability that Isabella stops after drawing exactly <math>n</math> marbles. What is the smallest value of <math>n</math> for which <math>P(n) < \frac{1}{2010}</math>?<br />
<br />
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math><br />
<br />
== Solution 1==<br />
The probability of drawing a white marble from box <math>k</math> is <math>\frac{k}{k + 1}</math>, and the probability of drawing a red marble from box <math>k</math> is <math>\frac{1}{k+1}</math>.<br />
<br />
To stop after drawing <math>n</math> marbles, we must draw a white marble from boxes <math>1, 2, \ldots, n-1,</math> and draw a red marble from box <math>n.</math> Thus, <cmath>P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.</cmath><br />
<br />
So, we must have <math>\frac{1}{n(n + 1)} < \frac{1}{2010}</math> or <math>n(n+1) > 2010.</math><br />
<br />
Since <math>n(n+1)</math> increases as <math>n</math> increases, we can simply test values of <math>n</math>; after some trial and error, we get that the minimum value of <math>n</math> is <math>\boxed{\textbf{(A) }45}</math>, since <math>45(46) = 2070</math> but <math>44(45) = 1980.</math><br />
<br />
== Solution 2(cheap) ==<br />
Do the same thing as Solution 1, but when we get to <math>n(n+1)>2010</math> just test all the answer choices in ascending order(from A to E), and stop when one of the answer choices is greater than <math>2010</math>. We get <math>45(46)=2070</math>, which is greater than <math>2010</math>, so we are done. The answer is <math>\textbf{(A)}</math><br />
<br />
-vsamc<br />
<br />
==Video Solution by the Beauty of Math==<br />
https://youtu.be/47XsxmQ5Ej4<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=A|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_17&diff=1399472010 AMC 12A Problems/Problem 172020-12-19T00:55:46Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #17]] and [[2010 AMC 10A Problems|2010 AMC 10A #19]]}}<br />
<br />
== Problem ==<br />
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math><br />
<br />
== Solution 1==<br />
It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on <math>\triangle ABC</math>, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>.<br />
<br />
If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore<br />
<br />
<math>\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)</math>.<br />
<br />
<br />
Based on the initial conditions,<br />
<br />
<cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath><br />
<br />
Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
==Solution 2==<br />
Step 1: Use [[Law of Cosines]] in the same manner as the previous solution to get <math>AC=\sqrt{r^2+r+1}</math>. <br />
<br />
Step 2: <math>\triangle{ABC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}</math>. The area of the hexagon is equal to <math>[ACE] + 3[ABC]</math>. We are given that <math>70\%</math> of this area is equal to <math>[ACE]</math>; solving for <math>AC</math> in terms of <math>r</math> gives <math>AC=\sqrt{7r}</math>.<br />
<br />
Step 3: <math>\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0</math> and by [[Vieta's Formulas]] , we get <math>\boxed{\textbf{E}}</math>.<br />
<br />
Note: Since <math>r</math> has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.<br />
<br />
==Solution 3==<br />
Find the area of the triangle <math>ACE</math> as how it was done in solution 1. Find the sum of the areas of the congruent triangles <math>ABC, CDE, EFA</math> as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles <math>ABC, CDE, EFA</math> is <math>30\%</math> of the area of the hexagon. Hence <math>\frac{7}{3}</math> times the latter is equal to the triangle <math>ACE</math>. Hence <math>\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)</math>. We can simplify this to <math>7r=r^2+r+1\implies r^2-6r+1=0</math>. By Vieta's, we get the sum of all possible values of <math>r</math> is <math>-\frac{-6}{1}=6\text{ or } \boxed{\textbf{E}}</math>.<br />
-vsamc<br />
(Edited by pinkbunny1228)<br />
<br />
===Proof Triangle ACE is Equilateral.===<br />
We know triangles ABC, CDE, and EFA are the same by SAS congruence, so the side opposite the 120 degree is also the same (since the triangles are congruent). Thus ACE is congruent.<br />
Q.E.D.<br />
~mathboy282<br />
<br />
==Video Solution by the Beauty of Math==<br />
https://youtu.be/rsURe5Xh-j0?t=961<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=A|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_16&diff=1399452010 AMC 12A Problems/Problem 162020-12-19T00:48:28Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #16]] and [[2010 AMC 10A Problems|2010 AMC 10A #18]]}}<br />
<br />
== Problem ==<br />
Bernardo randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8,9\}</math> and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8\}</math> and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br />
<br />
<math>\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}</math><br />
<br />
== Solution 1 ==<br />
We can solve this by breaking the problem down into <math>2</math> cases and adding up the probabilities.<br />
<br />
<br />
Case <math>1</math>: Bernardo picks <math>9</math>.<br />
If Bernardo picks a <math>9</math> then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a <math>9</math> is <math>\frac{1 \cdot \binom{8}{2}}{\binom{9}{3}} = \frac{\frac{8\cdot7}{2}}{\frac{9\cdot8\cdot7}{3\cdot2\cdot1}}=\frac{1}{3}</math>.<br />
<br />
<br />
Case <math>2</math>: Bernardo does not pick <math>9</math>.<br />
Since the chance of Bernardo picking <math>9</math> is <math>\frac{1}{3}</math>, the probability of not picking <math>9</math> is <math>\frac{2}{3}</math>.<br />
<br />
If Bernardo does not pick 9, then he can pick any number from <math>1</math> to <math>8</math>. Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.<br />
<br />
Ignoring the <math>9</math> for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.<br />
<br />
We get this probability to be <math>\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}</math><br />
<br />
Probability of Bernardo's number being greater is<br />
<cmath>\frac{1-\frac{1}{56}}{2} = \frac{55}{112}</cmath><br />
<br />
Factoring the fact that Bernardo could've picked a <math>9</math> but didn't:<br />
<br />
<cmath>\frac{2}{3}\cdot{\frac{55}{112}} = \frac{55}{168}</cmath><br />
<br />
Adding up the two cases we get <math>\frac{1}{3}+\frac{55}{168} = \boxed{\frac{37}{56}\ \textbf{(B)}}</math><br />
<br />
~Edits by mathboy282<br />
<br />
===Note===<br />
We have for case 1 <math>\frac{1 \cdot \binom{8}{2}}{\binom{9}{3}}</math> since <math>1</math> is the number of ways to pick 9 and <math>\binom{8}{2}</math> is the number of ways to pick the rest 2 numbers. <math>\binom{9}{3}</math> is just from 9 numbers pick 3.<br />
<br />
~mathboy282<br />
<br />
<br />
== Solution 2 ==<br />
<br />
From Bernardo's set, there are <math>\binom{9}{3} = 84</math> numbers that he can randomly choose. From Silvia's set, there are <br />
<math>\binom{8}{3} = 56</math> numbers that she can randomly choose. Since Bernardo and Silvia can choose their numbers independently, there are <math>84 \cdot 56</math> pairs of numbers that you can compare. For example, if Bernardo chooses 321 and Silvia chooses 543, that is one pair. We can sort Bernardo's numbers from the greatest to the smallest. We can do the same for Silvia's numbers. So, Bernardo's greatest <math>84 - 56 = 28</math> numbers are all bigger than Silvia's numbers. Here, we have <math>28 \cdot 56</math> pairs satisfying that Bernardo's number will be greater than Silvia's. If Bernardo chooses the 29th greatest number, which is the same as Silvia's greatest number, hence there will be <math>56 - 1 = 55</math> pairs satisfying that Bernardo's 29th greatest number will be greater than Silvia's. Similarly, if Bernardo chooses the 30th greatest number, there will be <math>55 - 1 = 54</math> pairs satisfying that Bernardo's 30th greatest number will be greater than Silvia's. This pattern continues. So if Bernardo chooses the 29th greatest number or below, he will have <math>55 + 54 + 53 + \cdots + 1</math> pairs where his number will be greater than Silvia's. <br />
<br />
In total, Bernardo's probability of having a greater number than Silvia's is <math>\frac{28 \cdot 56 + (55 + 54 + 53 + \cdots + 1)}{84 \cdot 56} = \frac{37}{56}</math>, which is <math>\textbf{(B)}</math>.<br />
<br />
<br />
- Leo M.<br />
<br />
==Video Solution by the Beauty of Math==<br />
https://youtu.be/rsURe5Xh-j0?t=590<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=A|num-b=17|num-a=19}}<br />
{{AMC12 box|year=2010|num-b=15|num-a=17|ab=A}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_9&diff=1399442010 AMC 12A Problems/Problem 92020-12-19T00:47:30Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #9]] and [[2010 AMC 10A Problems|2010 AMC 10A #17]]}}<br />
<br />
== Problem ==<br />
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?<br />
<br />
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
Imagine making the cuts one at a time. The first cut removes a box <math>2\times 2\times 3</math>. The second cut removes two boxes, each of dimensions <math>2\times 2\times 0.5</math>, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is <math>12 + 4 + 4 = 20</math>.<br />
<br />
Therefore the volume of the rest of the cube is <math>3^3 - 20 = 27 - 20 = \boxed{7\ \textbf{(A)}}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
We can use [[Principle of Inclusion-Exclusion]] (PIE) to find the final volume of the cube.<br />
<br />
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has <math>2 \times 2 \times 3=12</math> cubic inches. However, we can not just sum their volumes, as<br />
the central <math>2\times 2\times 2</math> cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.<br />
<br />
Hence the total volume of the cuts is <math>3(2 \times 2 \times 3) - 2(2\times 2\times 2) = 36 - 16 = 20</math>.<br />
<br />
Therefore the volume of the rest of the cube is <math>3^3 - 20 = 27 - 20 = \boxed{7\ \textbf{(A)}}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.<br />
<br />
Each edge can be seen as a <math>2\times 0.5\times 0.5</math> box, and each corner can be seen as a <math>0.5\times 0.5\times 0.5</math> box.<br />
<br />
<math>12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \boxed{7\ \textbf{(A)}}</math>.<br />
<br />
==Video Solution by the Beauty of Math==<br />
https://youtu.be/rsURe5Xh-j0?t=354<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=A|num-b=16|num-a=18}}<br />
{{AMC12 box|year=2010|num-b=8|num-a=10|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_14&diff=1399432010 AMC 12A Problems/Problem 142020-12-19T00:46:15Z<p>Fath2012: </p>
<hr />
<div>{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #14]] and [[2010 AMC 10A Problems|2010 AMC 10A #16]]}}<br />
<br />
== Problem ==<br />
Nondegenerate <math>\triangle ABC</math> has integer side lengths, <math>\overline{BD}</math> is an angle bisector, <math>AD = 3</math>, and <math>DC=8</math>. What is the smallest possible value of the perimeter?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37</math><br />
<br />
== Solution ==<br />
By the [[Angle Bisector Theorem]], we know that <math>\frac{AB}{BC} = \frac{3}{8}</math>. If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then <math>AB + BC = AD + DC = AC</math>, contradicting the [[Triangle Inequality]]. If we use the next lowest values (<math>AB = 6</math> and <math>BC = 16</math>), the Triangle Inequality is satisfied. Therefore, our answer is <math>6 + 16 + 3 + 8 = \boxed{33}</math>, or choice <math>\textbf{(B)}</math>.<br />
<br />
== Solution 2(Trick) ==<br />
We find that <math>\frac{AB}{BC}=\frac{3}{8}</math> by the [[Angle Bisector Theorem]] so we let the lengths be <math>3n</math> and <math>8n</math>, respectively where <math>n</math> is a positive integer. Also since <math>AD=3</math> and <math>BC=8</math>, we notice that the perimeter of the triangle is the sum of these, namely <math>3n+8n+3+8=11n+11.</math> This can be factored into <math>11(n+1)</math> and so the sum must be a multiple of <math>11</math>. The only answer choice which is a multiple of <math>11</math> is <math>\boxed{\textbf{(B)} 33}</math>. ~mathboy282<br />
<br />
==Video Solution by the Beauty of Math==<br />
https://youtu.be/rsURe5Xh-j0<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_8&diff=1399422010 AMC 12A Problems/Problem 82020-12-19T00:43:20Z<p>Fath2012: </p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?<br />
<br />
<math>\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ</math><br />
<br />
== Solution ==<br />
<br />
<center>[[File:AMC 2010 12A Problem 8.png]]</center><br />
<br />
<br />
Let <math>\angle BAE = \angle ACD = x</math>.<br />
<br />
<cmath>\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\<br />
\angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\<br />
\angle EAC &= 60^\circ - x\\<br />
\angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath><br />
<br />
Since <math>\frac{AC}{AB} = \frac{1}{2}</math> and the angle between the hypotenuse and the shorter side is <math>60^\circ</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>.<br />
<br />
== Solution 2(Trig and Angle Chasing) ==<br />
Let <math>AB=2a, AC=a</math>. Let <math>\angle BAE=\angle ACD=x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle FCE=60</math>, so <math>\angle ACB=60+x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle CFE=60</math>. Angles <math>AFD</math> and <math>CFE</math> are vertical, so <math>\angle AFD=60</math>. By triangle <math>ADF</math>, we have <math>\angle ADF=120-x</math>, and because of line <math>AB</math>, we have <math>\angle BDC=60+x</math>. Because Of line <math>BC</math>, we have <math>\angle AEB=120</math>, and by line <math>CD</math>, we have <math>\angle DFE=120</math>. By quadrilateral <math>BDFE</math>, we have <math>\angle ABC=60-x</math>.<br />
<br />
By the Law of Sines, we have <math>\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies \sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)</math>. By the sine addition formula(which states <math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)</math> by the way), we have <math>2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)</math>. Because cosine is an even function, and sine is an odd function, we have <math>2\sin(60)\cos(x)-2\cos(60)\sin(x)=\sin(60)\cos(x)+\cos(60)\sin(x) \implies \sin(60)\cos(x)=3\cos(60)\sin(x)</math>. We know that <math>\sin(60)=\frac{\sqrt{3}}{2}</math>, and <math>\cos(60)=\frac{1}{2}</math>, hence <math>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</math>. The only value of <math>x</math> that satisfies <math>60+x<180</math>(because <math>60+x</math> is an angle of the triangle) is <math>x=30^{\circ}</math>. We seek to find <math>\angle ACB</math>, which as we found before is <math>60+x</math>, which is <math>90</math>. The answer is <math>90, \text{or} \textbf{(C)}</math><br />
<br />
-vsamc<br />
<br />
== Solution 3 (Similar Triangles) ==<br />
Notice that <math>\angle AEB=\angle AFC = 120^{\circ}</math> and <math>\angle ACF=\angle AEB</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2BC</math>, <math>AE=2CF=2FE</math>, as triangle CFE is equilateral. Therefore, <math>AF=FE=FC</math>, and since <math>\angle AFC=120^{\circ}</math>,<math>x=30</math>. Thus, the measure of <math>\angle ACE</math> equals to <math>\angle FCE+\angle ACF=90^{\circ}, \text{or} \textbf{(C)}</math> <br />
-HarryW<br />
<br />
==Video Solution by the Beauty of Math==<br />
https://youtu.be/kU70k1-ONgM?t=785<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=A|num-b=13|num-a=15}}<br />
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_19&diff=1128252019 AMC 8 Problems/Problem 192019-12-14T00:00:27Z<p>Fath2012: /* bruh 2 */</p>
<hr />
<div>==Problem 19==<br />
In a tournament there are six team that play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br />
<br />
<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math><br />
<br />
==Solution 1==<br />
<br />
After fully understanding the problem, we immediately know that the three top teams, say team <math>A</math>, team <math>B</math>, and team <math>C</math>, must beat the other three teams <math>D</math>, <math>E</math>, <math>F</math>. Therefore, <math>A</math>,<math>B</math>,<math>C</math> must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against <math>D</math>, <math>E</math>, and <math>F</math>. For games between <math>A</math>, <math>B</math>, <math>C</math>, we have 2 cases. In both cases, there is an equality of points between <math>A</math>, <math>B</math>, and <math>C</math>.<br />
<br />
Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have <math>(1+1)*2=4</math> points (they play twice). Therefore, this case brings a total of <math>4+18=22</math> points.<br />
<br />
Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have <math>3</math> points, so a team gets <math>3*2=6</math> points if they each win a game and lose a game. This case brings a total of <math>18+6=24</math> points. <br />
<br />
Therefore, we use Case 2 since it brings the greater amount of points, or <math>\boxed{24}</math>, so the answer is <math>\boxed{C}</math>.<br />
<br />
~A1337h4x0r<br />
<br />
--------------------------<br />
Note that case 2 can be easily seen to be better as follows. Let <math>x_A</math> be the number of points <math>A</math> gets, <math>x_B</math> be the number of points <math>B</math> gets, and <math>x_C</math> be the number of points <math>C</math> gets. Since <math>x_A = x_B = x_C</math>, to maximize <math>x_A</math>, we can just maximize <math>x_A + x_B + x_C</math>. But in each match, if one team wins then the total sum increases by <math>3</math> points, whereas if they tie, the total sum increases by <math>2</math> points. So it is best if there are the fewest ties possible.<br />
<br />
==Solution 2==<br />
<br />
bruh: (1st match(3) + 2nd match(1)) * number of teams(6) = 24, <math>\boxed{C}</math>.<br />
<br />
Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So bruh means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won & drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is <math>\boxed{C}</math>.<br />
<br />
bruh bruh<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&diff=1120432019 AMC 8 Problems/Problem 112019-11-22T17:45:53Z<p>Fath2012: /* Solution 2 */</p>
<hr />
<div>==Problem 11==<br />
The eighth grade class at Lincoln Middle School has <math>93</math> students. Each student takes a math class or a foreign language class or both. There are <math>70</math> eighth graders taking a math class, and there are <math>54</math> eight graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?<br />
<br />
<math>\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70</math><br />
<br />
==Solution 1==<br />
Let <math>x</math> be the number of students taking both a math and a foreign language class.<br />
<br />
By P-I-E, we get <math>70 + 54 - x</math> = <math>93</math>. <br />
<br />
Solving gives us <math>x = 31</math>.<br />
<br />
But we want the number of students taking only a math class.<br />
<br />
Which is <math>70 - 31 = 39</math>.<br />
<br />
<math>\boxed{\textbf{(D)}\ 39}</math><br />
<br />
~phoenixfire<br />
==Solution 2==<br />
We have <math>70 + 54 = 124</math> people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that <math>31</math> people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get <math>70 - 31 = \boxed{\textbf{C} \, 39}</math> -fath2012<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=10|num-a=12}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&diff=1120422019 AMC 8 Problems/Problem 112019-11-22T17:45:34Z<p>Fath2012: My solution added, very similar to variable method</p>
<hr />
<div>==Problem 11==<br />
The eighth grade class at Lincoln Middle School has <math>93</math> students. Each student takes a math class or a foreign language class or both. There are <math>70</math> eighth graders taking a math class, and there are <math>54</math> eight graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?<br />
<br />
<math>\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70</math><br />
<br />
==Solution 1==<br />
Let <math>x</math> be the number of students taking both a math and a foreign language class.<br />
<br />
By P-I-E, we get <math>70 + 54 - x</math> = <math>93</math>. <br />
<br />
Solving gives us <math>x = 31</math>.<br />
<br />
But we want the number of students taking only a math class.<br />
<br />
Which is <math>70 - 31 = 39</math>.<br />
<br />
<math>\boxed{\textbf{(D)}\ 39}</math><br />
<br />
~phoenixfire<br />
==Solution 2==<br />
We have <math>70 + 54 = 124</math> people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that <math>31</math> people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get <math>70 - 31 = \boxed{\textbf{C} \, 39}</math><br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=10|num-a=12}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1120412019 AMC 8 Problems/Problem 242019-11-22T17:38:19Z<p>Fath2012: Added Line break</p>
<hr />
<div><br />
<br />
<br />
==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?<br />
<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{(B) 30}</math> ~heeeeeeeheeeee<br />
<br />
==Solution 2 (Mass Points)==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(7cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br />
draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br />
draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br />
draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br />
draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((0.28,2.39),dotstyle); <br />
label("$A$", (0.36,2.59), NE * labelscalefactor); <br />
dot((-2.8,-1.17),dotstyle); <br />
label("$B$", (-2.72,-0.97), NE * labelscalefactor); <br />
dot((3.78,-1.05),dotstyle); <br />
label("$C$", (3.86,-0.85), NE * labelscalefactor); <br />
dot((1.2887445398528459,1.3985482236874887),dotstyle); <br />
label("$D$", (1.36,1.59), NE * labelscalefactor); <br />
dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br />
label("$F$", (-0.64,-0.93), NE * labelscalefactor); <br />
dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br />
label("$E$", (-0.2,0.57), NE * labelscalefactor); <br />
label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br />
label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); <br />
label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br />
label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); <br />
label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br />
<br />
First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of 3. By equality, point <math>B</math> has a mass of 3 also. <br />
<br />
Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>.<br />
<br />
Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases.<br />
<br />
So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)</math><br />
-Brudder<br />
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{B}</math><br />
-Brudder<br />
<br />
==Solution 3==<br />
<math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{(B) 30}</math>. ~~SmileKat32<br />
<br />
==Solution 4 (Similar Triangles)==<br />
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown:<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
</asy><br />
Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>.<br />
<br />
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>.<br />
<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$60$", (A+G+D)/3);<br />
label("$30$", (B+E+F)/3);<br />
</asy><br />
(Credit to MP8148 for the idea)<br />
<br />
==Solution 5 (Area Ratios)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. <br />
(Credit to scrabbler94 for the idea)<br />
<br />
==Solution 6(Coordbash)==<br />
Let <math>ADB</math> be a right triangle, and <math>BD=CD</math><br />
<br />
Let <math>A=(-2\sqrt{30}, 0)</math><br />
<br />
<math>B=(0, 4\sqrt{30})</math><br />
<br />
<math>C=(4\sqrt{30}, 0)</math><br />
<br />
<math>D=(0, 0)</math><br />
<br />
<math>E=(0, 2\sqrt{30})</math><br />
<br />
The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math><br />
<br />
The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math><br />
<br />
Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math><br />
<br />
Now we can find <math>EF=BF=2\sqrt{15}</math><br />
<br />
<math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{(B) 30}\blacksquare</math><br />
<br />
-Trex4days<br />
<br />
== Solution 7 ==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(15cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br />
<br />
/* draw figures */<br />
draw(circle((0,0), 5), linewidth(2)); <br />
draw((-4,-3)--(4,3), linewidth(2)); <br />
draw((-4,-3)--(0,5), linewidth(2)); <br />
draw((0,5)--(4,3), linewidth(2)); <br />
draw((12,-1)--(-4,-3), linewidth(2)); <br />
draw((0,5)--(0,-5), linewidth(2)); <br />
draw((-4,-3)--(0,-5), linewidth(2)); <br />
draw((4,3)--(0,2.48), linewidth(2)); <br />
draw((4,3)--(12,-1), linewidth(2)); <br />
draw((-4,-3)--(4,3), linewidth(2)); <br />
/* dots and labels */<br />
dot((0,0),dotstyle); <br />
label("E", (0.27,-0.24), NE * labelscalefactor); <br />
dot((-5,0),dotstyle); <br />
dot((-4,-3),dotstyle); <br />
label("B", (-4.45,-3.38), NE * labelscalefactor); <br />
dot((4,3),dotstyle); <br />
label("$D$", (4.15,3.2), NE * labelscalefactor); <br />
dot((0,5),dotstyle); <br />
label("A", (-0.09,5.26), NE * labelscalefactor); <br />
dot((12,-1),dotstyle); <br />
label("C", (12.23,-1.24), NE * labelscalefactor); <br />
dot((0,-5),dotstyle); <br />
label("$G$", (0.19,-4.82), NE * labelscalefactor); <br />
dot((0,2.48),dotstyle); <br />
label("I", (-0.33,2.2), NE * labelscalefactor); <br />
dot((0,0),dotstyle); <br />
label("E", (0.27,-0.24), NE * labelscalefactor); <br />
dot((0,-2.5),dotstyle); <br />
label("F", (0.23,-2.2), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
Let <math>A[\Delta XYZ]</math> = <math>Area</math> <math>of</math> <math>Triangle</math> <math>XYZ</math> <br />
<br />
<br />
<math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math><br />
<br />
<br />
<math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts)<br />
<br />
<br />
Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>.<br />
<br />
<br />
<math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic)<br />
<br />
<br />
But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>.<br />
<br />
\therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).<br />
<br />
In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>.<br />
<br />
Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math><br />
<br />
<br />
Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>.<br />
<br />
<math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>.<br />
<br />
Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math><br />
<br />
Thus <math>A[\Delta BEF] = 30</math><br />
<br />
<br />
~phoenixfire & flamewavelight<br />
<br />
== Solution 8 ==<br />
The diagram is very inaccurate. <br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF, M;<br />
B = (0,0); C = (3,0); M = (1.45,0);<br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
draw(EE--M);<br />
label("$M$",M,S);<br />
label("$1$",A--DD,N);<br />
label("$2$",DD--C,N);<br />
label("$2$",EE--M,N);<br />
</asy><br />
Note: All numbers above <math>\overline{AC}</math> and <math>\overline{EM}</math> are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated. <br /><br />
Using the ratio of <math>\overline{AD}</math> and <math>\overline{CD}</math>, we find the area of <math>\triangle ADB</math> is <math>120</math> and the area of <math>\triangle BDC</math> is <math>240</math>. Also using the fact that <math>E</math> is the midpoint of <math>\overline{BD}</math>, we know <math>\triangle ADE = \triangle ABE = 60</math>.<br />
Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. We immediatley know that <math>\triangle BEM \sim BDC</math> by <math>2</math>. Using that we can conclude <math>EM</math> has ratio <math>1</math>. Using <math>\triangle EFM \sim \triangle AFC</math>, we get <math>EF:AE = 1:2</math>. Therefore using the fact that <math>\triangle EBF</math> is in <math>\triangle ABF</math>, the area has ratio <math>\triangle BEF : \triangle ABE=1:2</math> and we know <math>\triangle ABE</math> has area <math>60</math> so <math>\triangle BEF</math> is <math>\boxed{\textbf{B} \, 30}</math>. - fath2012<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1120002019 AMC 8 Problems/Problem 242019-11-22T06:38:33Z<p>Fath2012: Gave credit to myself</p>
<hr />
<div><br />
<br />
<br />
==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?<br />
<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{(B) 30}</math> ~heeeeeeeheeeee<br />
<br />
==Solution 2 (Mass Points)==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(7cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br />
draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br />
draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br />
draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br />
draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((0.28,2.39),dotstyle); <br />
label("$A$", (0.36,2.59), NE * labelscalefactor); <br />
dot((-2.8,-1.17),dotstyle); <br />
label("$B$", (-2.72,-0.97), NE * labelscalefactor); <br />
dot((3.78,-1.05),dotstyle); <br />
label("$C$", (3.86,-0.85), NE * labelscalefactor); <br />
dot((1.2887445398528459,1.3985482236874887),dotstyle); <br />
label("$D$", (1.36,1.59), NE * labelscalefactor); <br />
dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br />
label("$F$", (-0.64,-0.93), NE * labelscalefactor); <br />
dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br />
label("$E$", (-0.2,0.57), NE * labelscalefactor); <br />
label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br />
label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); <br />
label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br />
label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); <br />
label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br />
<br />
First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of 3. By equality, point <math>B</math> has a mass of 3 also. <br />
<br />
Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>.<br />
<br />
Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases.<br />
<br />
So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)</math><br />
-Brudder<br />
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{B}</math><br />
-Brudder<br />
<br />
==Solution 3==<br />
<math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{(B) 30}</math>. ~~SmileKat32<br />
<br />
==Solution 4 (Similar Triangles)==<br />
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown:<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
</asy><br />
Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>.<br />
<br />
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>.<br />
<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$60$", (A+G+D)/3);<br />
label("$30$", (B+E+F)/3);<br />
</asy><br />
(Credit to MP8148 for the idea)<br />
<br />
==Solution 5 (Area Ratios)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. <br />
(Credit to scrabbler94 for the idea)<br />
<br />
==Solution 6(Coordbash)==<br />
Let <math>ADB</math> be a right triangle, and <math>BD=CD</math><br />
<br />
Let <math>A=(-2\sqrt{30}, 0)</math><br />
<br />
<math>B=(0, 4\sqrt{30})</math><br />
<br />
<math>C=(4\sqrt{30}, 0)</math><br />
<br />
<math>D=(0, 0)</math><br />
<br />
<math>E=(0, 2\sqrt{30})</math><br />
<br />
The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math><br />
<br />
The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math><br />
<br />
Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math><br />
<br />
Now we can find <math>EF=BF=2\sqrt{15}</math><br />
<br />
<math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{(B) 30}\blacksquare</math><br />
<br />
-Trex4days<br />
<br />
== Solution 7 ==<br />
<math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math><br />
<br />
<br />
<math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts)<br />
<br />
<br />
Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>.<br />
<br />
<br />
<math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic)<br />
<br />
<br />
But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>.<br />
<br />
\therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).<br />
<br />
In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>.<br />
<br />
Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math><br />
<br />
<br />
Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>.<br />
<br />
<math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>.<br />
<br />
Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math><br />
<br />
Thus <math>A[\Delta BEF] = 30</math><br />
<br />
<br />
~phoenixfire & flamewavelight<br />
<br />
== Solution 8 ==<br />
The diagram is very inaccurate. <br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF, M;<br />
B = (0,0); C = (3,0); M = (1.45,0);<br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
draw(EE--M);<br />
label("$M$",M,S);<br />
label("$1$",A--DD,N);<br />
label("$2$",DD--C,N);<br />
label("$2$",EE--M,N);<br />
</asy><br />
Note: All numbers above <math>\overline{AC}</math> and <math>\overline{EM}</math> are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated. <br />
Using the ratio of <math>\overline{AD}</math> and <math>\overline{CD}</math>, we find the area of <math>\triangle ADB</math> is <math>120</math> and the area of <math>\triangle BDC</math> is <math>240</math>. Also using the fact that <math>E</math> is the midpoint of <math>\overline{BD}</math>, we know <math>\triangle ADE = \triangle ABE = 60</math>.<br />
Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. We immediatley know that <math>\triangle BEM \sim BDC</math> by <math>2</math>. Using that we can conclude <math>EM</math> has ratio <math>1</math>. Using <math>\triangle EFM \sim \triangle AFC</math>, we get <math>EF:AE = 1:2</math>. Therefore using the fact that <math>\triangle EBF</math> is in <math>\triangle ABF</math>, the area has ratio <math>\triangle BEF : \triangle ABE=1:2</math> and we know <math>\triangle ABE</math> has area <math>60</math> so <math>\triangle BEF</math> is <math>\boxed{\textbf{B} \, 30}</math>. - fath2012<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1119992019 AMC 8 Problems/Problem 242019-11-22T06:38:06Z<p>Fath2012: Finsihed /* Solution 8 */</p>
<hr />
<div><br />
<br />
<br />
==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?<br />
<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{(B) 30}</math> ~heeeeeeeheeeee<br />
<br />
==Solution 2 (Mass Points)==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(7cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br />
draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br />
draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br />
draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br />
draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((0.28,2.39),dotstyle); <br />
label("$A$", (0.36,2.59), NE * labelscalefactor); <br />
dot((-2.8,-1.17),dotstyle); <br />
label("$B$", (-2.72,-0.97), NE * labelscalefactor); <br />
dot((3.78,-1.05),dotstyle); <br />
label("$C$", (3.86,-0.85), NE * labelscalefactor); <br />
dot((1.2887445398528459,1.3985482236874887),dotstyle); <br />
label("$D$", (1.36,1.59), NE * labelscalefactor); <br />
dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br />
label("$F$", (-0.64,-0.93), NE * labelscalefactor); <br />
dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br />
label("$E$", (-0.2,0.57), NE * labelscalefactor); <br />
label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br />
label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); <br />
label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br />
label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); <br />
label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br />
<br />
First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of 3. By equality, point <math>B</math> has a mass of 3 also. <br />
<br />
Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>.<br />
<br />
Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases.<br />
<br />
So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)</math><br />
-Brudder<br />
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{B}</math><br />
-Brudder<br />
<br />
==Solution 3==<br />
<math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{(B) 30}</math>. ~~SmileKat32<br />
<br />
==Solution 4 (Similar Triangles)==<br />
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown:<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
</asy><br />
Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>.<br />
<br />
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>.<br />
<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$60$", (A+G+D)/3);<br />
label("$30$", (B+E+F)/3);<br />
</asy><br />
(Credit to MP8148 for the idea)<br />
<br />
==Solution 5 (Area Ratios)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. <br />
(Credit to scrabbler94 for the idea)<br />
<br />
==Solution 6(Coordbash)==<br />
Let <math>ADB</math> be a right triangle, and <math>BD=CD</math><br />
<br />
Let <math>A=(-2\sqrt{30}, 0)</math><br />
<br />
<math>B=(0, 4\sqrt{30})</math><br />
<br />
<math>C=(4\sqrt{30}, 0)</math><br />
<br />
<math>D=(0, 0)</math><br />
<br />
<math>E=(0, 2\sqrt{30})</math><br />
<br />
The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math><br />
<br />
The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math><br />
<br />
Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math><br />
<br />
Now we can find <math>EF=BF=2\sqrt{15}</math><br />
<br />
<math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{(B) 30}\blacksquare</math><br />
<br />
-Trex4days<br />
<br />
== Solution 7 ==<br />
<math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math><br />
<br />
<br />
<math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts)<br />
<br />
<br />
Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>.<br />
<br />
<br />
<math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic)<br />
<br />
<br />
But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>.<br />
<br />
\therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).<br />
<br />
In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>.<br />
<br />
Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math><br />
<br />
<br />
Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>.<br />
<br />
<math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>.<br />
<br />
Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math><br />
<br />
Thus <math>A[\Delta BEF] = 30</math><br />
<br />
<br />
~phoenixfire & flamewavelight<br />
<br />
== Solution 8 ==<br />
The diagram is very inaccurate. <br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF, M;<br />
B = (0,0); C = (3,0); M = (1.45,0);<br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
draw(EE--M);<br />
label("$M$",M,S);<br />
label("$1$",A--DD,N);<br />
label("$2$",DD--C,N);<br />
label("$2$",EE--M,N);<br />
</asy><br />
Note: All numbers above <math>\overline{AC}</math> and <math>\overline{EM}</math> are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated. <br />
Using the ratio of <math>\overline{AD}</math> and <math>\overline{CD}</math>, we find the area of <math>\triangle ADB</math> is <math>120</math> and the area of <math>\triangle BDC</math> is <math>240</math>. Also using the fact that <math>E</math> is the midpoint of <math>\overline{BD}</math>, we know <math>\triangle ADE = \triangle ABE = 60</math>.<br />
Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. We immediatley know that <math>\triangle BEM \sim BDC</math> by <math>2</math>. Using that we can conclude <math>EM</math> has ratio <math>1</math>. Using <math>\triangle EFM \sim \triangle AFC</math>, we get <math>EF:AE = 1:2</math>. Therefore using the fact that <math>\triangle EBF</math> is in <math>\triangle ABF</math>, the area has ratio <math>\triangle BEF : \triangle ABE=1:2</math> and we know <math>\triangle ABE</math> has area <math>60</math> so <math>\triangle BEF</math> is <math>\boxed{\textbf{B} \, 30}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1119962019 AMC 8 Problems/Problem 242019-11-22T05:12:38Z<p>Fath2012: </p>
<hr />
<div><br />
<br />
<br />
==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?<br />
<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{(B) 30}</math> ~heeeeeeeheeeee<br />
<br />
==Solution 2 (Mass Points)==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(7cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br />
draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br />
draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br />
draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br />
draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((0.28,2.39),dotstyle); <br />
label("$A$", (0.36,2.59), NE * labelscalefactor); <br />
dot((-2.8,-1.17),dotstyle); <br />
label("$B$", (-2.72,-0.97), NE * labelscalefactor); <br />
dot((3.78,-1.05),dotstyle); <br />
label("$C$", (3.86,-0.85), NE * labelscalefactor); <br />
dot((1.2887445398528459,1.3985482236874887),dotstyle); <br />
label("$D$", (1.36,1.59), NE * labelscalefactor); <br />
dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br />
label("$F$", (-0.64,-0.93), NE * labelscalefactor); <br />
dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br />
label("$E$", (-0.2,0.57), NE * labelscalefactor); <br />
label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br />
label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); <br />
label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br />
label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); <br />
label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br />
<br />
First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of 3. By equality, point <math>B</math> has a mass of 3 also. <br />
<br />
Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>.<br />
<br />
Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases.<br />
<br />
So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)</math><br />
-Brudder<br />
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{B}</math><br />
-Brudder<br />
<br />
==Solution 3==<br />
<math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{(B) 30}</math>. ~~SmileKat32<br />
<br />
==Solution 4 (Similar Triangles)==<br />
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown:<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
</asy><br />
Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>.<br />
<br />
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>.<br />
<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$60$", (A+G+D)/3);<br />
label("$30$", (B+E+F)/3);<br />
</asy><br />
(Credit to MP8148 for the idea)<br />
<br />
==Solution 5 (Area Ratios)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. <br />
(Credit to scrabbler94 for the idea)<br />
<br />
==Solution 6(Coordbash)==<br />
Let <math>ADB</math> be a right triangle, and <math>BD=CD</math><br />
<br />
Let <math>A=(-2\sqrt{30}, 0)</math><br />
<br />
<math>B=(0, 4\sqrt{30})</math><br />
<br />
<math>C=(4\sqrt{30}, 0)</math><br />
<br />
<math>D=(0, 0)</math><br />
<br />
<math>E=(0, 2\sqrt{30})</math><br />
<br />
The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math><br />
<br />
The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math><br />
<br />
Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math><br />
<br />
Now we can find <math>EF=BF=2\sqrt{15}</math><br />
<br />
<math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{(B) 30}\blacksquare</math><br />
<br />
-Trex4days<br />
<br />
== Solution 7 ==<br />
<math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math><br />
<br />
<br />
<math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts)<br />
<br />
<br />
Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>.<br />
<br />
<br />
<math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic)<br />
<br />
<br />
But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>.<br />
<br />
\therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).<br />
<br />
In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>.<br />
<br />
Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math><br />
<br />
<br />
Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>.<br />
<br />
<math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>.<br />
<br />
Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math><br />
<br />
Thus <math>A[\Delta BEF] = 30</math><br />
<br />
<br />
~phoenixfire & flamewavelight<br />
<br />
== Solution 8 ==<br />
The diagram is very inaccurate. <br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF, M;<br />
B = (0,0); C = (3,0); M = (1.45,0);<br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
draw(EE--M);<br />
label("$M$",M,S);<br />
label("$1$",A--DD,N);<br />
label("$2$",DD--C,N);<br />
</asy><br />
Note: All numbers above <math>\overline{AC}</math> are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated. <br />
Using the ratio of <math>\overline{AD}</math> and <math>\overline{CD}</math>, we find the area of <math>\triangle ADB</math> is <math>120</math> and the area of <math>\triangle BDC</math> is <math>240</math>. Also using the fact that <math>E</math> is the midpoint of <math>\overline{BD}</math>, we know <math>\triangle ADE = \triangle ABE = 60</math>.<br />
Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. We immediatley know that <math>\triangle BEM \sim BDC</math> by <math>2</math>. <br />
<br />
Sorry. The solution is still incomplete and the author is working to finish it. <br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1119912019 AMC 8 Problems/Problem 242019-11-22T04:15:28Z<p>Fath2012: </p>
<hr />
<div><br />
<br />
<br />
==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?<br />
<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{(B) 30}</math> ~heeeeeeeheeeee<br />
<br />
==Solution 2 (Mass Points)==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(7cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br />
draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br />
draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br />
draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br />
draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((0.28,2.39),dotstyle); <br />
label("$A$", (0.36,2.59), NE * labelscalefactor); <br />
dot((-2.8,-1.17),dotstyle); <br />
label("$B$", (-2.72,-0.97), NE * labelscalefactor); <br />
dot((3.78,-1.05),dotstyle); <br />
label("$C$", (3.86,-0.85), NE * labelscalefactor); <br />
dot((1.2887445398528459,1.3985482236874887),dotstyle); <br />
label("$D$", (1.36,1.59), NE * labelscalefactor); <br />
dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br />
label("$F$", (-0.64,-0.93), NE * labelscalefactor); <br />
dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br />
label("$E$", (-0.2,0.57), NE * labelscalefactor); <br />
label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br />
label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); <br />
label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br />
label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); <br />
label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br />
<br />
First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of 3. By equality, point <math>B</math> has a mass of 3 also. <br />
<br />
Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>.<br />
<br />
Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases.<br />
<br />
So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)</math><br />
-Brudder<br />
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{B}</math><br />
-Brudder<br />
<br />
==Solution 3==<br />
<math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{(B) 30}</math>. ~~SmileKat32<br />
<br />
==Solution 4 (Similar Triangles)==<br />
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown:<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
</asy><br />
Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>.<br />
<br />
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>.<br />
<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$60$", (A+G+D)/3);<br />
label("$30$", (B+E+F)/3);<br />
</asy><br />
(Credit to MP8148 for the idea)<br />
<br />
==Solution 5 (Area Ratios)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. <br />
(Credit to scrabbler94 for the idea)<br />
<br />
==Solution 6(Coordbash)==<br />
Let <math>ADB</math> be a right triangle, and <math>BD=CD</math><br />
<br />
Let <math>A=(-2\sqrt{30}, 0)</math><br />
<br />
<math>B=(0, 4\sqrt{30})</math><br />
<br />
<math>C=(4\sqrt{30}, 0)</math><br />
<br />
<math>D=(0, 0)</math><br />
<br />
<math>E=(0, 2\sqrt{30})</math><br />
<br />
The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math><br />
<br />
The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math><br />
<br />
Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math><br />
<br />
Now we can find <math>EF=BF=2\sqrt{15}</math><br />
<br />
<math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{(B) 30}\blacksquare</math><br />
<br />
-Trex4days<br />
<br />
== Solution 7 ==<br />
<math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math><br />
<br />
<br />
<math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts)<br />
<br />
<br />
Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>.<br />
<br />
<br />
<math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic)<br />
<br />
<br />
But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>.<br />
<br />
\therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).<br />
<br />
In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>.<br />
<br />
Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math><br />
<br />
<br />
Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>.<br />
<br />
<math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>.<br />
<br />
Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math><br />
<br />
Thus <math>A[\Delta BEF] = 30</math><br />
<br />
<br />
~phoenixfire & flamewavelight<br />
<br />
== Solution 8 ==<br />
The diagram is very inaccurate. <br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF, M;<br />
B = (0,0); C = (3,0); M = (1.45,0);<br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
draw(EE--M);<br />
label("$M$",M,S);<br />
label("$1$",A--DD,N);<br />
label("$2$",DD--C,N);<br />
</asy><br />
Note: All numbers above <math>\overline{AC}</math> are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated. <br />
Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. <br />
<br />
Sorry. The solution is still incomplete and the author is working to finish it. <br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Changelog&diff=111990Changelog2019-11-22T04:09:22Z<p>Fath2012: </p>
<hr />
<div><br />
<h2>Tip BBCode tag added</h2><br />
<h3> Date: Unknown </h3><br />
[tip = spam]foo[/tip] shows "spam" when you hover on foo. <br />
There is a bug where the HTML generated by the tip command in the AoPS classroom gets escaped.<br />
<br />
Update: Tips just show as plaintext in classroom. <br />
<p><br />
<br />
<h2>Contribution Guidelines</h2><br />
You are welcome to add anything to this changelog because this publicly available on the AoPS wiki. Please do take your time to get spelling and grammar correct. Please make sure your contribution sounds informational.</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_2&diff=1119892019 AMC 8 Problems/Problem 22019-11-22T04:01:24Z<p>Fath2012: </p>
<hr />
<div>=Problem 2=<br />
Three identical rectangles are put together to form rectangle, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle?<br />
<br />
<asy><br />
draw((0,0)--(3,0));<br />
draw((0,0)--(0,2));<br />
draw((0,2)--(3,2));<br />
draw((3,2)--(3,0));<br />
dot((0,0));<br />
dot((0,2));<br />
dot((3,0));<br />
dot((3,2));<br />
draw((2,0)--(2,2));<br />
draw((0,1)--(2,1));<br />
label("A",(0,0),S);<br />
label("B",(3,0),S);<br />
label("C",(3,2),N);<br />
label("D",(0,2),N);<br />
</asy><br />
<br />
<math>\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150</math><br />
<br />
<br />
<br />
==Solution 1==<br />
<br />
We know that the length of the shorter side of the 3 identical rectangles are all 5 so we can use that by seeing that the longer side of the right rectangle is the same as 2 of the shorter sides of the other 2 left rectangles. This means that <math>2\cdot{5}\ = 10</math> which is the longer side of the right rectangle, and because all the rectangles are congruent, we see that each of the rectangles have a longer side of 10 and a shorter side of 5. Now the bigger rectangle has a shorter length of 10(because the shorter side of the bigger rectangle is the bigger side of the shorter rectangle, which is 10) and so the bigger side of the bigger rectangle is the bigger side of the smaller rectangle + the smaller side of the smaller rectangle, which is <math>10 + 5 = 15</math> . Thus, the area is <math>15\cdot{10}\ = 150</math> for choice <math>\boxed{\textbf{(E)}\ 150}</math> ~~Saksham27<br />
<br />
==Solution 2==<br />
Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is <math>5 \cdot 2 = 10</math>. So the area of the identical rectangles is <math>5 \cdot 10 = 50</math>. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is <math>50 \cdot 3 = \boxed{\textbf{(E)}\ 150}</math>. -fath2012<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=110735Gmaas2019-10-30T23:32:59Z<p>Fath2012: </p>
<hr />
<div>GMAAS is our god. We shall worship him. To prove yourself worthy of the great GMAAS, you must memorize facts about him:<br />
<br />
-10. Thriftypiano will rule the world. All facts about GMAAS are fake. Worship your real leader EDIT: Stop it! This is not true.<br />
<br />
-3.14159265358979. GMAAS is our almighty leader. He will help us be more like him in all ways. If you wish to become more like GMAAS, make sure everytime you speak his name, it is in all capitals. Otherwise he will shoot a flaming ball of hair at you. YOU HAVE BEEN WARNED!<br />
<br />
-1.5. Gmaas wants you to worship him. And you will. Or else.<br />
<br />
-1. Gmaas looks like this when he is mad: https://cdn.artofproblemsolving.com/images/7/4/b/74b21fec95225e1621c71ec8f17f343c48a726e0.jpg Gmaas permits to post this picture.<br />
<br />
0. THEM 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.<br />
<br />
1. Gmaas' 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 usernameyourself proved it again: [b]Proof of the Gmaas theorem[/b]: The Gmaas theorem states that for every math problem, gmaas knows the answer. Using the 1.26 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 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 EDIT 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 Gmaas theorem. You wouldn't waste that much money to solve one problem. Therefore, I proved that you cannot use the Gmaas theorem to solve problems. But using theorem 1.24 "24. Gmaas can turn things into gold. EDIT: Gmaas can turn anything into anything," Gmaas may turn anything into anything. This means he can turn grass into millions of dollars. I take one million and solve the problem. Therefore, you can use the Gmaas theorem to solve any problem.<br />
<br />
2. The GMATS were supposed to be called the GMAAS, but the manager, gmaas, wanted to eat some gnats. Gmaas has stopped eating gnats because they taste dreadful. EDIT: He now eats gnats again. He thinks that they taste like pie. EDIT EDIT: He ate 3,141,592,653,589,793,238,462,643,383 so far.<br />
<br />
3. Gmaas' 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 Gmaas hates John Wick. Gmaas knew they were Oren Berries because Gmaas knows everything, but he decided to play along.<br />
<br />
4. Gmaas wants every fact to have a pi reference. Gmaas created pi. EDIT: He made it 314 times.<br />
<br />
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 gmaas knows!!!)<br />
<br />
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 />
<br />
7. Thanos wishes he could be like Gmaas. It's why he got the Infinity Stones and snapped.<br />
<br />
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 />
<br />
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.<br />
<br />
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 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 />
<br />
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 />
<br />
12. Gmaas farted and created a false vacuum, but then he burped, destroying the false vacuum.<br />
<br />
13. Gmaas was the first person to use the Infinity Gauntlet. EDIT: Gmaas created the Infinity Gauntlet and the Infinity Stones. EDIT EDIT: But Thanos stole them after Gmaas died for 0.31415926535 seconds.<br />
<br />
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 />
<br />
15. Gmaas won an infinite amount of games against AlphaGo and Gary Kasparov while eating dinner, chasing sseraj, and doing a handstand.<br />
<br />
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.<br />
<br />
17. Gmaas is so interesting that an entire science has been devoted to studying him: Gmaasology.<br />
<br />
18. Gmaas is the only known living being who has a Ph.D. in Gmaasology.<br />
<br />
19. Most universities, including Harvard, are beginning to offer MAs in Gmaasology.<br />
<br />
20. Gmaasology is one of the most eminent fields of science, falling behind Physics, Biology, Chemistry, Economics, Geology, and Computer Programming.<br />
<br />
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 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 EDIT EDIT: Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Gmaas's article has 89,119 bytes.<br />
<br />
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 />
<br />
23. Gmaas is the ancient Greek god of cats and catfish.<br />
<br />
24. Gmaas can turn things into gold. EDIT: Gmaas can turn anything into anything.<br />
<br />
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 />
<br />
26. Gmaas's Theorem states that Gmaas knows the answer to any math problem. EDIT: That theorem was proved by Gmaas. EDIT 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 EDIT 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 />
<br />
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 />
<br />
28. Gmaas should be capitalized to show respect. EDIT: It is normal to capitalize on people's names. EDIT EDIT: Gmaas isn't a person, he is a divine entity that takes the form of a cat, and should, therefore, be worshipped.<br />
<br />
29. The steps to summon Gmaas can be found at Talk: Gmaas.<br />
<br />
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 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 />
<br />
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 />
<br />
32. Gmaas used to be a dog, but he didn't like to be a dog. So he became a cat.<br />
<br />
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 />
<br />
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 EDIT: All the paintings in these museums are secretly portraits of Gmaas in his most glorious human/animal forms.<br />
<br />
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 />
<br />
36. Gmaas says hello. EDIT: Gmaas does not speak; he only uses mind signals. Speaking is too primitive for Gmaas.<br />
<br />
37. For one day, Gmaas was Richard Rusczyk, Gmaas, and David Patrick at the same time. It felt strange, so Gmaas stopped.<br />
<br />
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 />
<br />
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 />
<br />
40. Gmaas is the creator of the BCPI Ai project.<br />
<br />
41. Gmaas exists in <math>2\pi^2</math> dimensions because he doesn't like string theory.<br />
<br />
42. Christmas was supposed to be called Gmaasmas. Until it wasn't. EDIT: Christ is Gmaas. EDIT EDIT: Why? Because of the power of Gmaas.<br />
<br />
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 />
<br />
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 EDIT: That is impossible. The previous fact would mean that <math>e</math> has the last digit, which it does not. EDIT EDIT EDIT: That is why Gmaas is still flipping to this day. EDIT EDIT EDIT EDIT: Why would he still be doing that? That would be boring. EDIT EDIT EDIT EDIT 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 EDIT EDIT EDIT EDIT EDIT: Then why isn't he done? EDIT EDIT EDIT EDIT EDIT EDIT EDIT: The power of Gmaas. EDIT EDIT EDIT EDIT EDIT EDIT 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 EDIT EDIT EDIT EDIT EDIT EDIT 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 EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmass rules! <br />
<br />
45. Gmaas is the creator of everything including nothing.<br />
<br />
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 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 />
<br />
47. Gmaas is a cat. EDIT: How many times do we say this? EDIT EDIT: Very many times. EDIT EDIT EDIT: Gmaas can be anything, but he chooses to be a cat. EDIT EDIT EDIT EDIT: He has been a human dozen of times. Gmaas painted the Lascaux cave paintings.<br />
<br />
<br />
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 EDIT: Gmaas once used all his powers 3,141,592,653,589,793,238,462,643,383,279 times in 1 second.<br />
<br />
49. Gmaas was once spotted in Minecraft chewing a tree. EDIT: The tree broke. EDIT EDIT: Gmaas once broke a Nokia. EDIT EDIT EDIT: Gmaas can break anything except for Gmaas's logic. It is too strong. EDIT EDIT EDIT EDIT: Gmaas is strong.<br />
<br />
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 EDIT: However, Tacocats reproduce, so they are not in danger of extinction.<br />
<br />
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 />
<br />
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 EDIT: Gmaas has always been the master of AoPS as he is AoPS.<br />
<br />
53. Gmaas has a pet pufferfish named Pafferfash. EDIT: He also has a goldfish named Sylar.<br />
<br />
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 />
<br />
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 EDIT: Yeah, we've said that already. EDIT EDIT EDIT: It does not matter how many times we say it; it will always be true. EDIT EDIT EDIT EDIT: Gmaas is a cat. EDIT EDIT EDIT EDIT EDIT: Right now he is a cat, but many of his lives have been other species. EDIT EDIT EDIT EDIT EDIT EDIT: He has been a dog, as said before, but it was too boring.<br />
<br />
56. Gmaas started Pastafarianism. But then converted to Catholicism because Gmaas knows all EDIT: In that religion, one can only eat pasta.<br />
<br />
57. Gmaas could eat your hand, but he would not because hands taste bad.<br />
<br />
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 />
<br />
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 EDIT: The only copies of these biographies are locked in the Gmaasian Library beneath the Library of Congress. EDIT EDIT 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 EDIT: He absorbed them and became colorful for 0.271828182846 seconds. EDIT EDIT 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 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 EDIT 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 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 EDIT: Gmaas hates catnip and turnips, and pepperoni. He only likes alien alienish H2So4. EDIT EDIT 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 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 EDIT: He is not in every cat. The Guinness Book of World Records has a cat that does not have any Gmaas. EDIT EDIT EDIT: Gmaas invented The Guinness Book of World Records. EDIT EDIT EDIT EDIT: Gmaas invented the world. EDIT EDIT EDIT EDIT 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 EDIT: Everyone is Gmass, even Geoflex and CrazyEyeMoody EDIT EDIT EDIT: How? EDIT EDIT EDIT 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 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 EDIT: King Arthur lives on paper flour bags. He came back when the Welsh people didn't have enough 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 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 EDIT: The plural of Gmaas in English and Latin is Gmaases. (Gmaas is a third declension noun: Gmaas, Gmaasis, m.)<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 EDIT: Gmaas typed through everyone's account. EDIT EDIT EDIT: Gmaas owns most AoPSers, including me. So that's why I'm typing. EDIT EDIT EDIT 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 EDIT: Me too. EDIT EDIT EDIT: Me three.<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 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 EDIT: Almost 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 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 EDIT: I'll just go check. Aaaaaaaaah! He is in my kitchen. He is eating everything! EDIT EDIT 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 EDIT: LoL must be used in this article more than once. EDIT 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 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 EDIT: Gmaas does that. EDIT EDIT EDIT: Why? EDIT EDIT EDIT EDIT: The power of Gmaas. EDIT EDIT EDIT EDIT EDIT: He almost never does it, only if he wants to.. Also he did not puke and made everything.<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 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 EDIT EDIT: Gmaas has infinitely lives of all types. EDIT EDIT EDIT EDIT: How? EDIT EDIT EDIT EDIT 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 EDIT: The above post is irrelevant. Gmaas created J. K. Rowling. Therefore, he created Harry Potter. EDIT EDIT EDIT: Gmaas has strong logic. No one can break it. Not even Gmaas himself. He is still trying to this day. EDIT EDIT EDIT EDIT: Why? EDIT EDIT EDIT EDIT 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 EDIT: Yes, you do. EDIT EDIT 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 EDIT: Then how come Gmaas hasn't melted? EDIT EDIT 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 EDIT: Gmaas has colonized the whole universe.<br />
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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 EDIT: Gmaas is your device.<br />
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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 />
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118. Gmaas caught a CP 20,000 Mewtwo with a normal Pokeball and no berries blindfolded first try in Pokemon Go.<br />
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119. When Gmaas flips coins, they always land tails, except when Gmaas makes bets with Zeus.<br />
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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 />
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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 />
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122. Gmaas's burps created all gaseous planets.<br />
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123. Gmaas beat Luke Robatille in an epic showdown of catnip consumption.<br />
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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 EDIT: More accurate estimates predict that Gmaas's wealth is infinite.<br />
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125. Gmaas has a summer house on Mars. EDIT: Gmaas has a fall house on Venus. EDIT EDIT: Gmaas has a winter house on Jupiter. EDIT EDIT EDIT: Gmaas has a spring house on Earth. EDIT EDIT EDIT 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 EDIT EDIT EDIT EDIT: Gmaas invented a new season on Feb. 31, 2019, called winter-summer. It is spring but transcends spring.<br />
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126. The Earth and all known planets are Gmaas's hairballs.<br />
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127. Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time using a time-turner.<br />
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128. Gmaas attended Hogwarts and was a perfect. EDIT: Gmaas is headmaster. EDIT EDIT: Hogwarts was supposed to be called Hogmaas.<br />
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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 EDIT: Gmaas hates having a haircuts, so he reincarnated Voldemort to punish Harry.Edit: Not true at all.<br />
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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 />
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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 EDIT: How? EDIT EDIT EDIT: The power of Gmaas.<br />
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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 />
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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 />
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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 />
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135. Gmaas attended all the Ivy Leagues.<br />
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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 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.<br />
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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 />
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138. Gmaas was captured in 2017 but was released due to sympathy. EDIT: Gmaas was only captured in his concrete form; his abstract form cannot be processed by a feeble human brain.<br />
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139. Gmaas's fur is white, black, grey, yellow, red, blue, green, brown, pink, orange, turquoise, and purple at the same time. EDIT: Gmaas can make his fur any color he wants. EDIT EDIT: Gmaas's fur color can be ultraviolet.<br />
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140. Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br />
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141. Gmaas crossed the Delaware River with Washington. EDIT: Gmaas also crossed the Atlantic with the pilgrims.<br />
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142. If you can capture a Gmaas hair, Gmaas will give you some of his Gmaas power. EDIT: Gmaas does not shed and will eat anyone who has Gmaas hair. EDIT EDIT: Gmaas used to shed. The sheddings became tacocats.<br />
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143. Chuck Norris makes Gmaas jokes. EDIT: Those jokes all praise Gmaas.<br />
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144. Gmaas is also the ruler of Oceania, Eastasia, and Eurasia. EDIT: Gmaas wrote the book "1984."<br />
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145. Gmaas killed Big Brother by farting on him. Though Gmaas was caught by the Ministry of Love, Gmaas escaped easily. EDIT: Gmaas used to be Big Brother. EDIT EDIT: Gmaas destroyed the Ministry of Love.<br />
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146. Gmaas was not affected by Thanos's snap; in fact Gmaas is the creator of the Infinity Stones. EDIT: Gmaas is Thanos.<br />
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147. Everyone knows that Gmaas is a god. EDIT: That is because he is a god.<br />
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148. Gmaas also owns Animal Farm. Napoleon was Gmaas's servant. EDIT: Napoleon was Gmaas's slave.<br />
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149. Gmaas is the only person who knows where Amelia Earhart is. EDIT: Gmaas! Please tell us! It would be a great contribution to historical knowledge! EDIT EDIT: Gmaas responds, "No."<br />
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150. Gmaas is the only cat that has been proven transcendental. EDIT: Gmaas has been proved to be normal as well.<br />
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151. Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmaas isn't very happy about that, either. EDIT EDIT: That is why he looks grumpy in the memes.<br />
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152. The reason why AIME cutoffs aren't out yet is that Gmaas refused to grade them due to too much problem misplacement. EDIT: Gmaas controls the MAA. EDIT EDIT: Gmaas founded the MAA. EDIT EDIT EDIT: Gmaas is the MAA.<br />
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153. Gmaas dueled Grumpy Cat and won. Gmaas wasn't trying. EDIT: Gmaas killed Grumpy cat. EDIT EDIT: Gmaas revived Grumpy cat because he has not other rival.<br />
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154. Gmaas sits on the statue of Pallas and says forevermore. EDIT: When Gmaas was a statue, people hid him in a basement and forgot about him. Then civilization collapsed and the Middle Ages began. People finally discovered and melted down the statue of Gmaas around 1350 to make a church bell. Gmaas was free and reincarnated again, bringing about the Renaissance. EDIT EDIT: Despite the efforts of the recently founded National Gmatthematical Society of Florence, the statue was melted down. For more information about the National Gmatthematical Society of Florence, see the Gmaasology page.<br />
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155. Gmaas is a big fan of Edgar Allan Poe because he is actually Poe. EDIT: He became Poe during his thirty-year depression in the 19th century.<br />
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156. Gmaas does merely not use USD; he owns it.<br />
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157. In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br />
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158. "Actually, my name is spelled "GMAAS"." (Citation needed)<br />
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159. Gmaas is Link in the Legend of Zelda. He despises Calamity Ganon in Breath of the Wild. "Too much of a hog..."<br />
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160. Gmaas is the smartest living being in the universe. EDIT: Notice how this sentences says "living being," so Gmaas might be dummer than some dead person. EDIT EDIT: Gmaas is the smartest being that has lived, lives, and will live. EDIT EDIT EDIT: How? EDIT EDIT EDIT EDIT: The power of Gmaas. EDIT EDIT EDIT EDIT EDIT: HOW DARE YOU QUESTION THE POWER OF GMAAS!?!?!?<br />
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161. Gmaas helped Sun Wukong on the Journey to the West.<br />
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162. Gmaas was the creator of Wikipedia.<br />
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163. Gmaas can hack any website he desires. EDIT: He can edit any website with the words Gmaas in it. EDIT EDIT: AoPS includes the word Gmaas right here, so Gmaas can hack it. EDIT EDIT EDIT: Gmaas can edit any website he desires.<br />
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164. Gmaas is the basis of Greek and Egyptian mythology. EDIT: He is also the basis for Aztec mythology. He once had a craving for human flesh, so he created human sacrifice.<br />
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165. Gmaas once sold Google to a man for around 12 dollars! EDIT: Gmaas has not spent those 12 dollars and is waiting for the economy to crash. EDIT EDIT: Why would he do that? EDIT EDIT EDIT: The power of Gmaas. EDIT EDIT EDIT EDIT: Gmaas is smarter than you think he is. Do you think you know how smart Gmaas is now? If so, then you are still wrong.<br />
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166. Gmaas uses a HP printer. It is specifically a HP 21414144124124142141414412412414214141441241241421414144124124142141414412412414 printer.<br />
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167. Gmaas owns all AoPS staff including Richard Rusczyk. EDIT: Richard Rusczyk is one of Gmaas's many code names.<br />
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168. Gmaas saw Yoda's birth. EDIT: Gmaas was Yoda's father. EDIT EDIT: Was Gmaas green like Yoda back then? EDIT EDIT EDIT: Yes. EDIT EDIT EDIT EDIT: Why? EDIT EDIT EDIT EDIT EDIT: The power of Gmaas.<br />
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169. Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas. EDIT: This is all not true. EDIT EDIT: This is all true. EDIT EDIT EDIT: Gmaas has an infinite number of lives. EDIT EDIT EDIT EDIT: Gmaas does not have an infinite number of lives. He just has an unlimited number of lives. EDIT EDIT EDIT EDIT EDIT Gmass has an infinite and unlimited number of lives.<br />
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170. sseraj once spelled Gmaas as gmASS by accident in Introduction to Geometry (1532). sseraj’s life was never the same afterwards. EDIT: Gmaas created sseraj. He is a Gmaas, but without the face, body, legs, and tail.<br />
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171. Gmaas has beaten Chuck Norris, The Rock, and John Cena all together in a fight.<br />
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172. Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., Soviet, Russian, and Chinese citizen at the same time. EDIT: Gmaas is a citizen of every country in the world. Gmaas seems to enjoy the country of AoPS best, however.<br />
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173. "I am sand" destroyed Gmaas in FTW. EDIT: Because Gmaas accidentally died, he defeated "I am sand" 3,141,592,653 times after.<br />
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174. sseraj posted a picture of Gmaas with a game controller in Introduction to Geometry (1532).<br />
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175. Gmaas plays Roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also loves Catch that fish. EDIT Gmass likes Minecraft better than Roblox<br />
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176. Gmaas is Roy Moore's horse in the shape of a cat.<br />
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177. Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over <math>289547987693</math> roblox and <math>190348</math> in CPR.<br />
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178. This is all hypothetical. EDIT: This is all factual. For reference, see an earlier post.<br />
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179. Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar. EDIT: That is incorrect. Gmaas's real name is Gmaas. EDIT EDIT: Gmass named himself. That's why his name is amazing.<br />
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180. Gmaas is capable of salmon powers. EDIT: Gmaas is capable of everything. Nothing is beyond Gmaas.<br />
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181. Gmaas told Richard Rusczyk to make AoPS.<br />
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182. The Gmaas is everything. Yes, you are part of the Gmaas-Dw789. EDIT: Gmaas is older than the universe. He is more than everything.<br />
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183. The Gmaas knows every dimension up to 99999999999999999999999999999999999999999999999999999999999999999999999999999999999th dimension. EDIT: He is working on the higher dimensions presently. It takes up 1% of his time. He spends the remaining 99% of his time eating, sleeping, and being. EDIT EDIT: Does that mean he stops being while he learns higher dimensions? EDIT EDIT EDIT: Yes. EDIT EDIT EDIT EDIT: Why? EDIT EDIT EDIT EDIT EDIT: The power of Gmaas.<br />
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184. Gmaas went into a black hole and exited the white hole. He ended up in the 15th dimension where people drink tea every day. He stole 31,415,926,535,897,932,384 buckets of tea.<br />
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185. Gmaas is "TIRED OF PEOPLE ADDING TO HIS PAGE!!" (Maas 45). EDIT: This is Gmaas's holy, so he wants to have a longer version. For reference, see Maas 3,141,592,653,589,793,238,462. Gmaas is too lazy to write this book himself or to hire someone to do this for him, so he has assembled a group of unpaid freelancers on AoPS to write his holy book for him. He is watching it grow. However, Gmaas sometimes sees a comment that he does not like on the Gmaas article. He possesses an editor of Gmaas and makes him/her delete it. EDIT EDIT: AoPS itself is just an elaborately concealed piece of Gmaas propaganda. EDIT EDIT EDIT: How dare you reveal the truth! *Zaps previous editor out of existence* Now, where were we...<br />
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186. Gmaas is a Gmaas who is actually Gmaas. EDIT: Gmaas is omnipotent and omniscient. However, he is not always benevolent.<br />
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187. Gmaas has a penguin servant who runs GMAASINC. The penguin may or may not be dead. He had another penguin, who is mostly alive.<br />
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188. Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown. EDIT: Gmass is the doctor<br />
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189. Gmaas knows how to hack into top secret AoPS community pages. EDIT: AoPS is just Gmaas's elaborately-concealed blog.<br />
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190. Gmaas used to be river clan cat who crossed the event horizon of a black hole and came out the other end. EDIT: He eventually created sseraj and has lived with sseraj ever since.<br />
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191. Gmaas is king of the first men, the anduls.<br />
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192. Gmaas is a well known professor at Meowston Academy. EDIT: He is professor of Gmaasology there. EDIT EDIT: Gmaas founded Meowston Academy. EDIT EDIT EDIT: It is actually called Gmaaston Academy<br />
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193. Gmaas comes from Gmaas Land, a magical place where pie is infinite and everything starts with a g<br />
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194. Gmaas is the CEO of Caterpillar.<br />
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195. Gmaas drinks at Starbucks everyday.<br />
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196. Gmaas is addicted to tuna, along with other more potent fish, such as salmon and trout.<br />
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197. Gmaas likes turning into fish and catching himself. EDIT: So far, Gmaas has spent 2,718,281 of his many lives as a catfish. However, sometimes, he has been catched as a catfish by other people. Luckily, Gmaas has always escaped.<br />
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198. Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat. (He also out-grumped grumpy cat!!!) EDIT: He is in the Guinness Book of World Records for fluffiest cat ever.<br />
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199. He was last sighted at 1665 Alligator Swamp-A 4/1/18 at 3:14:15.92653589793238462 PM.<br />
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200. Gmaas is the owner of sseraj, not his pet. EDIT: Gmaas also created sseraj.<br />
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201. Gmaas is the embodiment of life, the universe, and beyond.<br />
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202. Gmaas watches memes about Gmaas. EDIT: All memes originate with Gmaas, but they are often posted by others. EDIT EDIT: Gmaas is getting rather tired of memes.<br />
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203. After death Gmaas became the god of hyperdeath and obtained over 9000 souls.<br />
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204. One of Gmaas's many real names is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso.<br />
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205. Gmaas is a certified Slytherin. EDIT: Gmaas is actually a certified Ravenclaw.<br />
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206. Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom. EDIT: Gmaas has superpowers that allowed him to overcome the horrors of Mr. Toilet while locked in the bathroom.<br />
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207. Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br />
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208. Gmaas conquered the moon and imprinted his face on it until asteroids came and erased it.<br />
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209. Gmaas is a supreme overlord who must be given<cmath>1000^{1,000,000,000,000,000,000,000^{1,000,000,000,000,000,000,000}}</cmath>minecraft diamonds. EDIT: Gmaas owns Mojang, but Gmaas sued them when it was supposed to be called Gmaascraft. EDIT EDIT: A previous fact said that Gmaas only lost one lawsuit. Why is this Minecraft not called Gmaascraft if he won the lawsuit?<br />
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210. Gmaas is the Doctor Who lord, sports Dalek-painted cars, and eats human finger cheese, custard, and black holes. EDIT: Gmaas once ate a white hole. Gmaas then exploded and made a new universe.<br />
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211. Gmaas is everyone's favorite animal.<br />
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212. Gmaas is a Persian Smoke.<br />
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213. Gmaas lives with sseraj.<br />
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214. Gmaas dislikes geometry but enjoys number theory. EDIT: sseraj once posted a picture of Gmaas being grumpy after seeing an olympiad geometry problem.<br />
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215. Gmaas is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj. EDIT: Gmaas does not need food. He just eats food because it tastes good. EDIT EDIT: Gmaas does not like vegetables.<br />
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216. Gmaas has<cmath>\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAAAAS</cmath>supercars, excluding the Purrari and the 138838383 Teslas.<br />
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217. Gmaas employs AoPS. EDIT: Gmaas is the reason why AoPS exists. EDIT EDIT: Gmaas is the reason why everything exists.<br />
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218. Gmaas is a Gmaas with yellow fur and white hypnotizing eyes. EDIT: His fur is not always yellow. His fur can be wha== This article is spam == 219. Gmaas has the ability to divide by zero. EDIT: Gmaas knows what 1/0 is. What is it? We'll never know. EDIT EDIT: 1/0 = 0. How? The power of Gmaas. EDIT EDIT EDIT: The previous editor was wrong. 1/0 = Gmaas. EDIT EDIT EDIT EDIT: The previous editor was wrong. <math>\frac{1}{\infty}=</math> GMAAS<br />
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220. Gmaas was born with a tail that is a completely different color from the rest of his fur. EDIT: Gmaas can change the color of his fur.<br />
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221. Gmaas's stare is very hypnotizing and effective at getting table scraps. EDIT: Gmaas's stare turned Medusa into rock, King Midas into gold, and sseraj into sseraj.<br />
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222. Gmaas always appears several minutes before certain classes start as an administrator. EDIT: This is becoming more and more infrequent. Sadly, sseraj rarely posts pictures of Gmaas before classes anymore. EDIT EDIT: Why is this happening? EDIT EDIT EDIT: The power of Gmaas.<br />
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223. Gmaas is an AoPS administrator under the alias Grayson Maas. EDIT: Gmaas always posts class surveys. He has a fake photo on his bio and has a real one on his community page. However, Gmaas himself has not yet posted on the Gmaas forum. EDIT EDIT: He also has not edited the Gmaas article, the Talk: Gmaas article, the Gmaasology article, or the sseraj article. EDIT EDIT EDIT: The final one is surprising because Gmaas knows more about sseraj than even sseraj does.<br />
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224. Gmaas died from too many Rubik's cubes in an Introduction to Algebra A class, but he was revived by the Dark Lord at 12:13:37 AM the next day. EDIT: This was thanks in return for Gmaas reincarnating him during the Triumverate Tournament.<br />
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225. It is uncertain whether or not Gmaas is a cat or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke). EDIT: He is both. EDIT EDIT: Actually, Gmaas is a cat. Gmaas said so, and science says so. EDIT EDIT EDIT: The profile for Gmaas on the list of AoPS teachers does not mention him being a cat. EDIT EDIT EDIT EDIT: The above post is irrelevant. Gmaas hides his identity under the alias Grayson Maas. EDIT EDIT EDIT EDIT EDIT: Gmaas us everything yet nothing. EDIT EDIT EDIT EDIT EDIT EDIT: How is that? The power of Gmaas.<br />
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226. Gmaas is distant relative of the chair of the department of Gmaasology at Princeton. EDIT: In the last few years, many universities have been opening Gmaasology departments. For more information, see the Gmaasology page.<br />
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227. Gmaas cannot be Force choked. Darth Vader learned that the hard way...<br />
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228. Gmaas is famous, and mods always talk about him before class starts.<br />
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229. Gmaas's favorite food is AoPS textbooks because they help him digest problems. EDIT: Gmaas wrote all AoPS textbooks but is never listed in the acknowledgments section. EDIT EDIT: That is because Gmaas is very humble except for when he isn't.<br />
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230. Gmaas tends to reside in sseraj's fridge. EDIT: Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer. EDIT EDIT: Gmaas's fur can protect him from the harsh conditions of a freezer. EDIT EDIT EDIT: Then Gmaas ate all the food in sseraj's freezer. He enjoyed the ice cream in the freezer the most. EDIT EDIT EDIT EDIT: Gmaas also ate sseraj's freezer.<br />
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231. Gmaas once demanded Epic Games to give him 5,000,000 V-bucks for his 569823rd birthday. EDIT: This is why Gmaas no longer has an Epic Games account. EDIT EDIT: Gmaas created Epic Games, though.<br />
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232. Gmaas sightings are not very common. There have only been 30 confirmed sightings of Gmaas in the wild.<br />
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233. Gmaas is a sage omniscient cat.<br />
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235. Gmaas is looking for suitable places other than sseraj's fridge to live in.<br />
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236. List of places where Gmaas sightings have happened:<br />
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- 1. The Royal Scoop ice cream store in Bonita Beach Florida<br />
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- 2. Inside the abandoned hospital on Rosevelt Island in New York City, which is also a feral cat sanctuary.<br />
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- 3. MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
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- 4. Gmaasology 2 (1440)<br />
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- 5. Alligator Swamp A 1072<br />
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- 6. Alligator Swamp B 1073<br />
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- 7. Gmaasology A (1488)<br />
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- 8. Introduction to Gmaasology A (1170)<br />
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- 9. Introduction to Gmaasology B (1529)<br />
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- 10. Welcome to Panda Town Gate 1076<br />
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- 11. Welcome to Gmaas Town Gate 1221<br />
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- 12. Welcome to Gmaas Town Gate 1125<br />
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- 13. 33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
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- 14. The other side of the ice in Antarctica<br />
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- 15. Feisty Alligator Swamp 1115<br />
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- 16. Intermediate Gmaas and Gmaasology 1207<br />
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- 17. Posting student surveys<br />
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- 18. USF Castle Walls - Elven Tribe 1203<br />
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- 19. The Dark Lord's Hut 1210<br />
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- 20. Gmathamatical Problem Series 1200<br />
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- 21. Intermediate Gmaasology 1138<br />
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- 22. Intermediate Number Theory 1476<br />
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- 23. Introduction to Gmaasology 1204 on July 27, 2016<br />
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- 24. Gmaasology B 1112<br />
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- 25. Intermediate Algebra 1561 7:17 PM 12/11/16<br />
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- 26. Nowhere Else, Tasmania<br />
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- 27. The Gmaathamatics School of Gmaasology<br />
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237. Gmaas can communicate with, and sometimes control, any other cats; however, this is very rare, as cats normally have a very strong will.<br />
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238. A picture of Gmaas: http://i.imgur.com/PP9xi.png<br />
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239. Gmaas met Mike Miller. EDIT: Gmaas ate Mike Miller.<br />
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240. Gmaas got mad at sseraj once, so Gmaas locked sseraj in sseraj's freezer.<br />
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241. Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment. EDIT: sseraj could not eat all of them so, he put the remaining turnips in the fire.<br />
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242. Gmaas once ate a pie. EDIT: He only ate the first 31415926535897932384626433832 digits.<br />
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153 A Gmaas bite is 314159265358979323846264 psi.<br />
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154 Many people have met Gmaas, although many of these sightings are dubious. EDIT: It is just like alien sightings.<br />
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155 Gmaas can talk but normally hates talking.<br />
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156 Gmaas likes to eat his own fur. That's why he doesn't have any as of late<br />
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157 Gmaas is bigger than an ant. EDIT: so is a regular cat<br />
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158 Gmaas once ate the king of the ants. EDIT: Ants do not have kings; they have queens.<br />
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159 Gmaas lives somewhere over the rainbow. EDIT: He lives in sseraj's house. EDIT EDIT: sseraj lives somewhere over the rainbow.<br />
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160 Gmaas is an obviously omnipotent cat.<br />
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- sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop.'<br />
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171 Gmaas EDIT: This is a waste of a fact.<br />
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172 sseraj has posted pictures of Gmaas in Introduction to Algebra before class started with the title "caption contest." Anyone who posted a caption mysteriously vanished in the middle of the night. EDIT: This has happened many times, including in Introduction to Geometric Gmaasology 1533, among other active classes. The person writing this did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br />
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173 Gmaas once slept in your bed and made it gray. EDIT: My bed is not gray. EDIT EDIT: Nor is mine.<br />
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174 Richard Rusczyk actually Gmaas in disguise. EDIT: That is false. Gmaas is only in one form at a time. EDIT EDIT: This is also false, you are Gmaas and Gmaas owns you. EDIT EDIT EDIT: This is false too. Gmaas can only be in one form.<br />
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175 Gmaas is suspected to be a cat in disguise. EDIT: He is a cat in disguise. See the results on the Gmaas poll in the Gmaas forum.<br />
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176 Gmaas is a cat but has characteristics of every other animal on Earth. EDIT: Gmaas does not have the characteristic of being able to eat plankton. EDIT EDIT: Yes he does. He just has not decided to reincarnate as a blue whale yet.<br />
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177 Pegasus was modeled off Gmaas. EDIT: Pegasus used to be Gmaas's pet, but Pegasus escaped and joined Bellerophon.<br />
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178 Gmaas is the ruler of the universe and has been known to be the creator of the species "Gmaasians". EDIT: He is the universe<br />
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179 There is a rumor that Gmaas is starting a poll. EDIT: He has. It is on the Gmaas forum, made by his minions. EDIT: We are all his minions.<br />
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180 Gmaas is a Persian smoke who ran away, founded GmaasClan, and became a kitty-pet.<br />
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181 There is a sport called "Gmaas Hunting" where people try to successfully capture Gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. Many people have tried Gmaas Hunting, but they never have been successful.<br />
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182 Gmaas burped and caused an earthquake.<br />
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183 Gmaas once drank from a teacup.<br />
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184 GMAAS IS HERE.... PURRRRRRRRRRRRRRRRRRRR EDIT: Gmaas is everywhere. EDIT EDIT: He is behind you.<br />
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185 Gmaas made and currently owns the Matrix. EDIT: The above fact is true. Therefore, this is an illusion. EDIT EDIT: Gmaas is not an illusion.<br />
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186 Gmaas is the reason Salah will become better than Ronaldo.<br />
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187 Who is Gmaas, really? EDIT: Gmaas is a cat. EDIT EDIT: Who cares how much we mention it? Gmass is a cat.<br />
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188 Gmaas is a heavenly being. EDIT: He is immortal. EDIT EDIT: He is not immortal. He just tricked you to think he is immortal. Gmaas is good at tricking.<br />
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189 Gmaas is a cat with no fur or tail. In fact he's a human wearing a cat costume. EDIT: This fact is false. 31415 times false.<br />
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190 Illuminati was a manifestation of Gmaas, but Gmaas decided Illuminati was not great enough for his godly self.<br />
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191 sseraj has met Gmaas, and Gmaas is his best friend. EDIT: sseraj is Gmaas's great great great great great grandson.<br />
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192 Gmaas read Twilight. EDIT: ...and SURVIVED.<br />
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193 There is a secret code when put into super smash where Gmaas would be a playable character. Too bad he didn't say it.<br />
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194 Gmaas was a tribute to one of the Hunger Games, came out a Victor, and now lives in District 4. EDIT: This fact is true and not true. He is everywhere.<br />
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195 Gmaas is the only known creature that will survive the destruction of Earth in 99,999,999 years. EDIT: The Earth will probably die in around 5,000,000,000 years.<br />
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196 5space (side admin) is one of Gmaas's slaves. EDIT: Gmaas owns all AoPS site administrators. EDIT EDIT: He owns Himself.<br />
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197 Gmaas photos − https://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br />
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http://disneycreate.wikia.com/wiki/File:Troll_cat_gif_(1).gif<br />
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He was also sighted here.<br />
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198 Gmaas in Popular Culture:<br />
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BREAKING NEWS: A Gmaasologist has found a possible cousin to Gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Another eminent Gmaasologist is now looking into it.<br />
199 Someone is writing a book about Gmaas.<br />
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200 Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br />
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201 Oryx the mad god is actually Gmaas wearing a suit of armor. This explains why he is never truly killed.<br />
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202 Potential sighting of Gmaas [1]<br />
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203 Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, The Name of the Doctor, Silence in the Library, The Idiots Lantern and many more.<br />
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204 Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles.<br />
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205 Gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br />
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206 An eminent Gmaasologist is also writing a story about him. He is continuing the book that was started by the professor of Gmaasology at Harvard. When he is done he will post it here.<br />
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207 Gmaas is a time traveler from 0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br />
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208 No one knows if Gmaas is a Mr. Mime in a cat skin, the other way around, or just a downright combination of both. EDIT: Gmaas is an immortal in a cat body.<br />
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209 In it, it mentions these four links as things Gmaas is having trouble (specifically technical difficulties). What could it mean? Links:<br />
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https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br />
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https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br />
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https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br />
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https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br />
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210 Another possible Gmaas sighting?<br />
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211 <math>Another</math> sighting? [3]<br />
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212 Yet Another Gmaas sighting? [4]<br />
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213 Gmaas has been sighted several times on the Global Announcements forum.<br />
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214 Gmaas uses the following transportation: <img> http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg </img> EDIT: He also sometimes uses a TARDIS<br />
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215 When Gmaas was angry, he started world wars 1 & 2. It is only because of Gmaas that we have not had World War 3. EDIT: He was starting to have to get angry in the Cold War, especially around the Cuban Missile Crisis and the Ronald Regan era, but he decided to eat food, which calmed him down.<br />
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216 Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br />
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217 Gmaas belongs to a secret club of transcendental cats. EDIT: That club is located underneath the Eiffel Tower. EDIT EDIT: Gustave Eiffel was Gmaas in disguise and designed the club.<br />
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218 Gmaas plays Geometry Dash. His username is D3m0nG4m1n9. EDIT: Gmaas only does this to punish himself. Olympiad Geometry is his least favorite thing to do. EDIT EDIT: None of these are true.<br />
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219 Gmaas likes to whiz on the wilzo.<br />
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220 Gmaas has been spotted in many AoPS classes, such as AMC 8 Basics. EDIT: Many of the AoPS classes Gmaas has visited can be found in the Gmaas sightings section of this article.<br />
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221 Gmaas is cool. EDIT: He was not cool in summer, so he invented air conditioning. EDIT EDIT: He is air conditioning, in fact he is the entire universe.<br />
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222 Gmaas hemoon card that does over 9000000 dmg.<br />
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223 Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br />
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224 Kirby once swallowed Gmaas. Gmaas had to spit him out.<br />
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225 Gmaas was the creator of pokemon, and his pokemon card can OHKO anyone in one turn. He is invisible and he will always move first.<br />
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226 Gmaas beat Dongmin in The Genius Game Seasons 1, 2, 3, 4, 5, 6, and 7.<br />
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227 Gmaas has five letters. Pizza also has five letters. Pizzas are round. Eyes are round. There is an eye in the illuminati symbol. EDIT: Gmaas is a fluffy cat.<br />
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228 Gmaas knows both 'table' and 'tabular' in LaTeX, and can do them in his sleep. EDIT: Gmaas invented LaTeX in his sleep.<br />
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229 Gmaas does not hate cheddar cheese, but he doesn't love it either.<br />
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230 Gmaas is a cat and not a cat. EDIT: He is a cat.<br />
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231 Gmaas was born on the sun. EDIT: Not the sun, the suns. He was born on all the suns at once.<br />
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232 Gmaas eats tape. EDIT: Gmaas invented tape. EDIT EDIT: Gmaas eats everything. Gmaas eats computers.<br />
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233 Gmaas likes Bubble Gum.<br />
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234 Thomas Edison did not invent the lightbulb; Gmaas did.<br />
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235 Gmaas invented the alphabet. EDIT: Gmaas thought that hieroglyphs were too complicated, so he invented the alphabet.<br />
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236 Gmaas eats metal. EDIT: Gmaas once ate so much metal that he turned into a bronze statue.<br />
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237 Gmaas is over 9000 years old! EDIT: This is just a DBZ reference, and bears no reality to his true age. EDIT EDIT: Gmaas is trillions of years old.<br />
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238 Gmaas is a cat and not a cat. EDIT: He is a cat. EDIT EDIT: This fact has been stated 31,415 times in this article.<br />
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239 Gmaas started the Iron Age. EDIT: He thought civilization was getting too advanced. So he sent out the Sea People to destroy civilization.<br />
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240 Gmaas made the dinosaurs go extinct. EDIT: That happened when he got angry that a dinosaur stepped on his toe.<br />
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241 Gmaas created life. Then he destroyed it, and in doing so, destroyed himself. Until his son, Gmaas II took over and saved the planet.<br />
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242 Gmaas created AoPS. EDIT: AoPS was actually born out of a small fraction of Gmaas's abstract reality, and only the sheer amount math can keep it here. (It is also rumored that when he reclaims it, the USF will be deleted, as that is where 83% of the factions of his abstract reality lives, and when people leave USF, more and more escapes.)<br />
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243 Gmaas created mathematics.<br />
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244 Gmaas does not like Roblox.<br />
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245 Gmaas told Steve Jobs to start a company.<br />
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246 Gmaas invented Geometry Dash. EDIT: Gmaas hates Olympiad Geometry. [Source: sseraj]<br />
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247 Gmaas got to<math>\infty</math>in Flappy Bird.<br />
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248 Gmaas invented Helix Jump.<br />
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249 Gmaas can play Happy Birthday on the Violin.<br />
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250 Gmaas has mastered Paganini. EDIT: Paganini was Gmaas in disguise. EDIT EDIT: Gmaas was Paganini in one of his 100,000 human lives. He does not have a human life now because he has a cat life.<br />
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251 Gmaas discovered Atlantis after one dive underwater.<br />
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252 Gmaas made a piano with 89 keys. EDIT: The piano was made out of chocolate. EDIT: He also made a piano bench made out of chocolate. EDIT EDIT EDIT: Gmaas ate them<br />
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253 Gmaas can see the future and how to change it.<br />
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254 Gmaas has every super power you can imagine. EDIT: Gmaas has more superpowers than you can imagine.<br />
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255 Gmaas made a Violin with 9 strings.<br />
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256 Gmaas can read 5 books at once. EDIT: It is slightly difficult for Gmaas to do this because he has no fingers. But somehow, Gmaas finds a way. EDIT: He uses his mind to hold open the books.<br />
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257 Gmaas eats rubber bands. EDIT: Gmaas once reincarnated as a rubber band. He thought it was the worst period in his life. Eventually he was snapped and reincarnated as a bacterium, which was also boring.<br />
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258 Gmaas married Mrs. Norris. EDIT: Mrs. Norris is his second worst enemy. They say keep your enemies close. EDIT: Gmaas eventually divorced Mrs. Norris for her astounding lack of inteligence and short life span.<br />
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259 Gmaas can fly faster than anything. EDIT: No one can travel faster than light. EDIT EDIT: Gmaas can.<br />
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260 Grumpy cat is his son. EDIT: Grumpy cat is his worst enemy and his son.<br />
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261 Gmaas eats paper. EDIT: According to Gmaas, recycled paper tastes the very good. Paper with steak juice on it tastes even better.<br />
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262 Gmaas likes lollipops. EDIT: Gmaas was reincarnated as a lollypop. He did not appreciate having spit all over him while he dissolved.<br />
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263 Gmaas nibbles on pencils.<br />
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- Gmaas is alive. EDIT: He is both alive and dead. (For reference, see later post.) EDIT EDIT: He can make people alive and dead.<br />
<br />
- Gmaas is Ninja in Fortnite. EDIT: Gmaas possesses Ninja in Fortnite.<br />
<br />
- Gmaas is love; Gmaas is life. EDIT: Scientist do not know how life was formed. Gmaas does.<br />
<br />
This article is spam<br />
- Gmaas taught Richard Rusczyk everything he knows about mathematics. EDIT: Many famous mathematicians have been Gmaas in disguise.== This article is spam ==<br />
<br />
- Gmaas can control matter by looking at it. EDIT: He once looked at a galaxy while he was angry. Everything in the galaxy exploded.<br />
<br />
- Gmaas created AoPS. EDIT: He created 99% of all websites. 98% of those websites are blank and unreachable, but the ones that are not are interesting.<br />
<br />
- Gmaas is a quantum particle. EDIT: Gmaas is multiple quantum particles. He is a macroscopic collection of them.<br />
<br />
This article is spam<br />
- Gmaas is alive and dead at the same time. EDIT: He is usually alive.<br />
<br />
This article is spam<br />
- Gmaas likes snow. EDIT: Once Gmaas was reincarnated as a polar bear around 1,000,000 years ago. He had the best time of his lives. EDIT EDIT: While he was a polar bear, he ate many fish.<br />
<br />
- GMAAS LIKES THE NEW AOPS UPDATE! EDIT: Are you sure he does? Did you ask him? EDIT EDIT: Yes, I did.<math>\textrm{Gmaas is right behind you}</math>EDIT: Gmaas is everywhere. He is in a quantum superposition. He is still in the superposition until someone sees him. The chance of him being behind you is 0.0000000000000000000000000000000000000000000000000000001%. But, you never know. Is he behind you? Is he behind me right now? He is!!! Aaaaaaaaaah... THUMP<br />
<br />
- Gmaas likes his eggs hard boiled.<br />
<br />
- Gmaas doesn't take showers; he only takes bloodbaths.<br />
<br />
- When the bogeyman goes to sleep, he checks his closet for Gmaas.<br />
<br />
- When Gmaas crosses the street, cars have to look both ways.<br />
<br />
This article is spam<br />
- Gmaas has counted to infinity an infinite amount of times.<br />
<br />
This article is spam<br />
- Gmaas pulled the pin in a grenade. 1 billion people died. Then he threw it.<br />
<br />
This article is spam<br />
- Gmaas controls the Illuminati and controls all the world's resources. He also is the ruler of a communist society on Mars.<br />
<br />
- Gmaas has taken the AHSME/AMC 10, AIME, USAJMO, USAMO, and IMO all the years it has come out. It is rumored that for 2018, Gmaas got a 163.5 on the AMC 12A, a 156 on the AMC 12B, and 17 problems right on the AIME I. EDIT: He also got 51 on USAMO.<br />
<br />
- Gmaas has designed most of the AMC tests. EDIT: He didn't design them in 2015 because he was hibernating too long.<br />
<br />
- Gmaas is a koala. EDIT: Gmaas was actually the first koala. He wanted to reincarnate himself into something new, so he created a whole other species. [Source Wikipedia: Koala] EDIT: Gmaas is a Koala when they are not alive, which is 10% of the time. When they are alive, Gmaas is a cat.<br />
<br />
- Time is Gmaas's vacation home. EDIT: Gmaas created time as a science project.<br />
<br />
- Gmaas was once reincarnated as Chuck Norris, but he missed being a cat. Then he became a cat.<br />
<br />
- Gmaas once resided with a certain physicist named Schrödinger, but he left because Schrödinger was too confusing. Granted, nothing is too confusing for Gmaas but Schrödinger also made Gmaas alive and dead, which felt weird.<br />
<br />
- Gmaas once decided to eat a carrot. He decided turnips were better.<br />
<br />
- Gmaasology used to be called Gmaathamatics. It was renamed a few years ago. For more information, see the Gmaasology page. That page needs expanding.<br />
<br />
- \gmaas should be made into a LaTeX command. Its function should be to summon Gmaas.<br />
<br />
-<math>\gmaas</math> GMAAS has entered the room<br />
<br />
- Grumpy cat pretends to be a descendent of Gmaas because he is jealous. He is not a descendant of Gmaas.<br />
<br />
- Every IMO Gold Medalist is two of Gmaas, standing on top of each other in a trenchcoat.<br />
<br />
- Gmaas has been everywhere in the world. EDIT: Gmaas has been everywhere in the universe and multiverse.<br />
<br />
- The degree used to be called the Gmaas Arc, but the name was too long for most people, so they switched the name.<br />
<br />
- Gmaas is famous on AoPS. EDIT: Gmaas is AoPS.<br />
<br />
- Gmaas invented every math theorem and in fact invented math.<br />
<br />
- Gmaas made 31.41592653589793238462643383279502884% of memes. EDIT: This is incorrect. This is only an estimate.<br />
<br />
- Gmaas is a Gmaas since Gmaas likes to Gmaas and Gmaas is cool because he is named Gmaas.<br />
<br />
- Gmaas once ate Big Chungus. EDIT: Gmaas also once ate cat food. EDIT EDIT: Gmaas hates cat food. EDIT EDIT EDIT: Gmaas also hates dog food. EDIT EDIT EDIT EDIT: Gmaas does not need to eat, he is an all-conquering god who never/rarely dies. EDIT EDIT EDIT EDIT EDIT: Gmaas is Big Chungus EDIT EDIT EDIT EDIT EDIT EDIT: He was Big Chungus for 3.14159265358979323846327656180937734 seconds. EDIT EDIT EDIT EDIT EDIT EDIT EDIT: This is only an rough estimate<br />
<br />
- Garfield is Gmaas in disguise. EDIT: But Gmaas also created Garfield. He took a Gar and a Field, and added them. Boom. He made Garfield.<br />
<br />
-Gmaas didn't create the world. He didn't create the universe. The didn't create multiverses. He is the world. He is the universe. He is the multiverses<br />
<br />
-Gmaas has a cult with many followers. This is the page for his followers. They pray to him and remember him for his deeds.<br />
<br />
- Garfield ate Gmaas but then Gmaas ate garfield. Then they ate each other.<br />
<br />
- Gmaas has an existential crisis every 99.0000000000000000000000000000000000009123659812374 seconds.<br />
<br />
- Gmaas married and then divorced a picture of himself EDIT: He then destroyed all evidence of this by swallowing it and barfing it out as a hairball<br />
<br />
- Gmaas needed light to read and he turned into a lamp called the sun and created 51.59265358979% of life. EDIT: Gmaas uses a new type of tetradecimal number system where the digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, G, M, A, and S. The number 416877 is Gmaas's favorite integer. (Convert it to this number system)<br />
<br />
-None of this is true. Except for maybe that last one. EDIT: This fact is therefore not true. EDIT EDIT This fact is a paradox. Gmaas does not like paradoxes. Therefore Gmaas does not like this fact.<br />
<br />
-Gmaas has controlled everyone's mind and made them eat moldy cheeseburgers.<br />
<br />
-You can't PM or email Gmaas. He's a pretty tough guy to talk to. He makes the great hulk look like a mouse. To contact him you have to summon Gmaas' royal guards at his palace. Or you can contact him using a sophisticated and secretive instant messaging system. Which is really just pretending to read minds.<br />
<br />
- remember, EVERY SINGLE thing you do, Gmaas is watching, and when you do bad, or insult him, you will feel regret very soon. Also, if u find cat hair everywhere, but you are not a cat owner, he is probably visiting your home to see how much math u we doing. So START DOING YOUR MATH KIDS! EDIT Gmaas is displeased if you use bad grammar like "u" and "r". Use full words.<br />
<br />
-Gmaas once woke up one day, farted, and accidentally created the known universe.<br />
<br />
-Gmaas can destroy every living creature and every imaginary creature, including himself, using only 1% of his power. EDIT: That means he can destroy himself, but he defeats everyone, so this is a paradox. However, this is still possible because of the power of GMAAS.<br />
<br />
-Spelling Gmaas with a lowercase G is disrespectful. Such a insult shall never happen, or the world may be imbalanced to the point of collapsing.<br />
<br />
-Out of all the life that ever existed, 31.415926535897% of it are reincarnations of Gmaas.<br />
<br />
-Gmaas can make so many edits to this page, that it would destroy the known universe.<br />
<br />
-Gmaas can jump higher than a house (since a house cannot jump)<br />
<br />
-Gmaas is leader of StarClan. EDIT: Gmaas created starclan. EDIT EDIT: Gmaas is starclan. EDIT EDIT EDIT: Gmaas wrote these edits. EDIT EDIT EDIT EDIT: Gmaas wrote everything.<br />
<br />
-Gmaas went to Area 51. He said that there were no aliens.<br />
<br />
-Gmass was once yoda.<br />
<br />
-Gmass is the best cat ever; ThriftyPiano is super bad.<br />
<br />
(Why is this a thing?) Because we are all worshippers of Gmaas.<br />
<br />
This article is NOT spam otherwise it would have been deleted by now. Edit: This article is spam.<br />
<br />
Gmaas is actually an AoPS site admin... (And a human) not a cat... His username is Gmaasrocks32 EDIT: His username is all the accounts.<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 />
He <br />
P.P.P.P.P.P.P.P.P.P.P.S Gmaas is superior to all other site admins, including RRusczyk himself.Edit: He is not superior. EDIT EDIT: Whoever is saying this article is spam and he is not superior his mean. None of his facts are true. EDIT EDIT EDIT: This is TriphtyPiano or whatever you call him he does not deserve the respect.<br />
<br />
P.P.P.P.P.P.P.P.P.P.P.P.S. ThriftyPiano is telling all the false facts. Do not listen to him. Or whatever u call him, we don't respect him, but we do with Gmass. Whenever you see This article is spam, it is his doing<br />
<br />
P.P.P.P.P.P.P.P.P.P.P.P.P.S. fath2012 would like you to print this page out</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Changelog&diff=110734Changelog2019-10-30T23:31:19Z<p>Fath2012: </p>
<hr />
<div><br />
<h2>Tip BBCode tag added</h2><br />
<h3> Date: Unknown </h3><br />
[tip = spam]foo[/tip] shows "spam" when you hover on foo. <br />
There is a bug where the HTML generated by the tip command in the AoPS classroom gets escaped.<br />
<br />
<p><br />
<br />
<h2>Contribution Guidelines</h2><br />
You are welcome to add anything to this changelog because this publicly available on the AoPS wiki. Please do take your time to get spelling and grammar correct. Please make sure your contribution sounds informational.</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Changelog&diff=110728Changelog2019-10-30T04:07:40Z<p>Fath2012: </p>
<hr />
<div><br />
<h2>Tip BBCode tag added</h2><br />
<h3> Date: Unknown </h3><br />
[tip = spam]foo[/tip] shows "spam" when you hover on foo. <br />
There is a bug where the HTML generated by the tip command in the AoPS classroom gets escaped.<br />
<br />
<p><br />
You are welcome to add anything to this changelog because this publicly available on the AoPS wiki. Please do take your time to get spelling and grammar correct.</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Changelog&diff=110724Changelog2019-10-30T01:11:58Z<p>Fath2012: </p>
<hr />
<div><br /><br />
<br /><br />
<h1>Public Community Changelog</h1><br />
<h2>Tip BBCode tag added</h2><br />
<h3> Date: Unknown </h3><br />
[tip = spam]foo[/tip] shows "spam" when you hover on foo. <br />
There is a bug where the html generated by the tip command in the AoPS classroom gets escaped.</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Changelog&diff=110723Changelog2019-10-30T01:11:42Z<p>Fath2012: </p>
<hr />
<div><br /><br />
<h3>Public Community Changelog</h1><br />
<h2>Tip BBCode tag added</h2><br />
<h3> Date: Unknown </h3><br />
[tip = spam]foo[/tip] shows "spam" when you hover on foo. <br />
There is a bug where the html generated by the tip command in the AoPS classroom gets escaped.</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Changelog&diff=110722Changelog2019-10-30T01:08:15Z<p>Fath2012: Created page with "<h1>Public Community Changelog</h1> <h2>Tip BBCode tag added</h2> <h3> Date: Unknown </h3> [tip = spam]foo[/tip] shows "spam" when you hover on foo. There is a bug where the..."</p>
<hr />
<div><h1>Public Community Changelog</h1><br />
<h2>Tip BBCode tag added</h2><br />
<h3> Date: Unknown </h3><br />
[tip = spam]foo[/tip] shows "spam" when you hover on foo. <br />
There is a bug where the html generated by the tip command in the AoPS classroom gets escaped.</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User:Fath2012&diff=110063User:Fath20122019-09-30T02:43:11Z<p>Fath2012: </p>
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<div><div class = "404"><br />
<h1> 404 not found</h1><br />
<p><br />
Page not found. Maybe the user has deleted it?<br />
</p><br />
HTML test has succeeded!<br />
</div></div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User:Fath2012&diff=105914User:Fath20122019-05-22T01:11:06Z<p>Fath2012: Latex</p>
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<div><div class = "404"><br />
<h1> 404 not found</h1><br />
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Page not found. Maybe the user has deleted it?<br />
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HTML test has succeeded!<br />
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<br />
\begin{align*} &\cos A = \frac{\text{adjacent leg}}{\text{hypotenuse}}, \\ &\sin A = \frac{\text{opposite leg}}{\text{hypotenuse}}, \, \\ &\tan A = \frac{\text{opposite leg}}{\text{adjacent leg}}, \\ &\sec A = \frac{\text{hypotenuse}}{\text{adjacent leg}}, \, \\ &\csc A = \frac{\text{hypotenuse}}{\text{opposite leg}}, \, \\ &\text{ and } \\ &\cot A = \frac{\text{adjacent leg}}{\text{opposite leg}}. \end{align*}<br />
<br />
\begin{align*} \sin 30^\circ & = \frac{1}{2}, \\ \cos 30^\circ & = \frac{\sqrt{3}}{2}, \\ \tan 30^\circ & = \frac{\sqrt{3}}{3},\\ \\ \sin 45^\circ & = \frac{\sqrt{2}}{2}, \\ \cos 45^\circ & = \frac{\sqrt{2}}{2}, \\ \tan 45^\circ & = 1, \\ \\ \sin 60^\circ & = \frac{\sqrt{3}}{2}, \\ \cos 60^\circ & = \frac{1}{2}, \\ \text{ and } \\ \tan 60^\circ & = \sqrt{3}. \end{align*}<br />
<br />
<img src="https://latex.artofproblemsolving.com/8/6/5/865494988ab38b5d8adf1df23b29902b8831d48f.png"></img><br />
\begin{align*} &\sin^2\theta + \cos^2\theta = 1, \\ &\cos\theta = \cos(-\theta), \\ &\sin\theta = -\sin(-\theta), \\ &\cos\theta = -\cos(180^\circ - \theta), \\ &\text{ and } \\ &\sin\theta = \sin(180^\circ - \theta). \end{align*}<br />
Credits to heatherfinotti for these charts</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User:Fath2012&diff=105913User:Fath20122019-05-22T01:10:34Z<p>Fath2012: </p>
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<div><div class = "404"><br />
<h1> 404 not found</h1><br />
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Page not found. Maybe the user has deleted it?<br />
</p><br />
HTML test has succeeded!<br />
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<math><br />
\begin{align*} &\cos A = \frac{\text{adjacent leg}}{\text{hypotenuse}}, \\ &\sin A = \frac{\text{opposite leg}}{\text{hypotenuse}}, \, \\ &\tan A = \frac{\text{opposite leg}}{\text{adjacent leg}}, \\ &\sec A = \frac{\text{hypotenuse}}{\text{adjacent leg}}, \, \\ &\csc A = \frac{\text{hypotenuse}}{\text{opposite leg}}, \, \\ &\text{ and } \\ &\cot A = \frac{\text{adjacent leg}}{\text{opposite leg}}. \end{align*}</math><br />
<br />
<math>\begin{align*} \sin 30^\circ & = \frac{1}{2}, \\ \cos 30^\circ & = \frac{\sqrt{3}}{2}, \\ \tan 30^\circ & = \frac{\sqrt{3}}{3},\\ \\ \sin 45^\circ & = \frac{\sqrt{2}}{2}, \\ \cos 45^\circ & = \frac{\sqrt{2}}{2}, \\ \tan 45^\circ & = 1, \\ \\ \sin 60^\circ & = \frac{\sqrt{3}}{2}, \\ \cos 60^\circ & = \frac{1}{2}, \\ \text{ and } \\ \tan 60^\circ & = \sqrt{3}. \end{align*}<br />
</math><br />
<img src="https://latex.artofproblemsolving.com/8/6/5/865494988ab38b5d8adf1df23b29902b8831d48f.png"></img><br />
<math>\begin{align*} &\sin^2\theta + \cos^2\theta = 1, \\ &\cos\theta = \cos(-\theta), \\ &\sin\theta = -\sin(-\theta), \\ &\cos\theta = -\cos(180^\circ - \theta), \\ &\text{ and } \\ &\sin\theta = \sin(180^\circ - \theta). \end{align*}</math><br />
Credits to heatherfinotti for these charts</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User:Fath2012&diff=105912User:Fath20122019-05-22T01:09:42Z<p>Fath2012: </p>
<hr />
<div><div class = "404"><br />
<h1> 404 not found</h1><br />
<p><br />
Page not found. Maybe the user has deleted it?<br />
</p><br />
HTML test has succeeded!<br />
</div><br />
<latex><br />
\begin{align*} &\cos A = \frac{\text{adjacent leg}}{\text{hypotenuse}}, \\ &\sin A = \frac{\text{opposite leg}}{\text{hypotenuse}}, \, \\ &\tan A = \frac{\text{opposite leg}}{\text{adjacent leg}}, \\ &\sec A = \frac{\text{hypotenuse}}{\text{adjacent leg}}, \, \\ &\csc A = \frac{\text{hypotenuse}}{\text{opposite leg}}, \, \\ &\text{ and } \\ &\cot A = \frac{\text{adjacent leg}}{\text{opposite leg}}. \end{align*}<br />
<br />
\begin{align*} \sin 30^\circ & = \frac{1}{2}, \\ \cos 30^\circ & = \frac{\sqrt{3}}{2}, \\ \tan 30^\circ & = \frac{\sqrt{3}}{3},\\ \\ \sin 45^\circ & = \frac{\sqrt{2}}{2}, \\ \cos 45^\circ & = \frac{\sqrt{2}}{2}, \\ \tan 45^\circ & = 1, \\ \\ \sin 60^\circ & = \frac{\sqrt{3}}{2}, \\ \cos 60^\circ & = \frac{1}{2}, \\ \text{ and } \\ \tan 60^\circ & = \sqrt{3}. \end{align*}<br />
</latex><br />
<img src="https://latex.artofproblemsolving.com/8/6/5/865494988ab38b5d8adf1df23b29902b8831d48f.png"></img><br />
\begin{align*} &\sin^2\theta + \cos^2\theta = 1, \\ &\cos\theta = \cos(-\theta), \\ &\sin\theta = -\sin(-\theta), \\ &\cos\theta = -\cos(180^\circ - \theta), \\ &\text{ and } \\ &\sin\theta = \sin(180^\circ - \theta). \end{align*}</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Getting_Started_With_Python_Programming&diff=105911Getting Started With Python Programming2019-05-22T01:06:13Z<p>Fath2012: Added dark theme info</p>
<hr />
<div>This guide takes you through the process of getting started with programming using the Python programming language. The only language that AoPS teaches (as of May 2019) in a class is Python. <br />
<br />
The sections flow from one to the next so it's recommended to read through this document in order from top to bottom. <br />
<br />
If you find that this is too easy, '''make sure you've read everything through (at least roughly)''', and check out [[Basic Programming With Python]].<br />
<br />
==Installing Python==<br />
<br />
Python is a useful and popular computer programming language. Confusingly, Python has two major versions (2 and 3) and they are not fully compatible. We recommend using the most recent release of version 3. (This is the version that our [http://www.artofproblemsolving.com/School/courseinfo.php?course_id=python1 Introduction to Programming with Python course] uses -- if you are enrolled in that class, you '''must''' have Python 3.) There is absolutely nothing wrong with Python 2, as it is what most of today's technology supports and uses, but Python 3 is well on the way of replacing Python 2, so it will be more useful in a few years.<br />
<br />
Python is open-source software and it is '''free''' to install and use. Here are installation instructions:<br />
<br />
#Go to the Python download page at http://www.python.org/downloads. Near the top of the page, there will be a list of download links for the Python 3.7.x installer. (The x will be replaced by a number -- as of April 2019 the version is 3.7.3.) If you are given multiple options, click on the link that corresponds to your computer type (Windows or Mac, 32-bit or 64-bit -- if you're not sure, use the 32-bit version.) Some browsers will save the file automatically, others may pop up a box asking you if you want to save the file, in which case you should click the "save file" option. Depending on how your browser is configured, you may be asked where to save the file. If this is the case, keep track of where you save the installer.<br />
#Find where the installer was downloaded and double click on it to run it. On most browsers, you should simply be able to double-click the installer from the browser's "Downloads" window or menu. You may also have to click "Run" or "Yes" to a security window -- do this if necessary.<br />
#The setup wizard should launch. You should just click "Next" for every option in the setup wizard (i.e. use the defaults), unless you have some specific reason not to.<br />
#Familiarize yourself with the Python shell and IDLE text editor by running through the two sections below.<br />
#If you like everything dark, you may go to Options->Configure IDLE->Highlights and change it to the built in dark theme<br />
<br />
==Programming==<br />
<br />
Yay, it's time to program! The next few sections will talk about some very basic programming. We will program a few programs as a demonstration.<br />
<br />
===Using the Python Shell===<br />
The program that you'll use to run Python is called IDLE. It may be listed on your computer as "IDLE (Python GUI)". <br />
* On a Mac, IDLE should be in the Applications folder. <br />
* On Windows, IDLE should be accessible from the Start menu in a folder named "Python 3.7" (or something similar).<br />
The icon for IDLE looks something like this [[File:Idleicon.png]] or this [[File:Idleiconmac.png]].<br />
<br />
When you first open IDLE, you'll see the Python Shell (the numbers on your shell might be different than those shown below): <br />
<br />
[[File:Idle2-1.png]]<br />
<br />
(The screenshots in this article are taken using IDLE on a Mac with the font increased. Thus IDLE may look a little bit different for you but should still function similarly.)<br />
<br />
Note that the first line is the version of Python, which is 3.1.2 in the screenshot but should be 3.7.something if you installed it as directed above. Another thing to note is that in the lower right hand corner of the Python Shell you can see that it says "Ln: 4 Col: 4". This is just telling you where in the document your cursor is. In this case it's on line 4 and over in column 4. (The line and column number may be slightly different for your installation.)<br />
<br />
When you first start up Python on a Mac, you might get the following warning:<br />
: >>> WARNING: The version of Tcl/Tk (8.5.9) in use may be unstable.<br />
: Visit http://www.python.org/download/mac/tcltk/ for current information.<br />
If you get this warning, you'll need to update a graphics driver on your computer. Follow the link shown above and download and install the ActiveTcl driver that's recommended for the version of OS X that your Mac is running. This most likely will be 8.5.15.0, which you can also download directly from http://www.activestate.com/activetcl/downloads (IMPORTANT: you only need to do this step if you get the warning printed above when you start IDLE for the first time. If you don't get the warning, then everything is good to go.)<br />
<br />
<br />
The Python Shell is very useful for quick one-liners and short sequences of commands:<br />
<br />
[[File:Idle2-2.2.png]]<br />
<br />
Here we see a number of familiar operations: + for addition, - for subtraction, * for multiplication, and / for division. The last operation shown in the example, denoted by **, happens to be exponentiation. One neat feature to note about Python is that it can store arbitrarily large numbers (limited by the amount of memory your computer has). Trying some hefty exponentiation, we can see that we can compute with some pretty big numbers such as <math>2^{1000}</math> as illustrated below.<br />
<br />
[[File:Idle2-3.png]]<br />
<br />
While Python can make for a pretty good calculator, it can do a whole lot more. One example is when dealing with strings as follows:<br />
<br />
[[File:Idle2-4.png]]<br />
<br />
Here we are concatenating the three strings "python", "is", and "cool" by using the + operator. Notice that previously we used + to add numbers but now with strings, Python concatenates them! You may also note that the output of the operation gives us a string with single quotes around it. In Python, you are able to use single quotes or double quotes to denote a string. You can use them interchangeably.<br />
<br />
As a final example, we can even write code in the Python Shell that extends beyond a single line as shown below. We also see our first example of a <math>\verb=for=</math> loop.<br />
<br />
[[File:Idle2-5.png]]<br />
<br />
As you type the above, the Python Shell will automatically indent the second line for you. To let the Python Shell know that you're done and are ready for it to run your code, you'll need to put in an extra blank line by hitting the Enter key again. At that point it should run your code and print your output.<br />
<br />
Take some time to play around with the Python Shell. You'll want to go through a more extensive introduction to programming to learn the full extent of what you can do with Python, but you can still do some pretty nifty stuff by just playing around. The Python Shell also has an extensive built-in help system -- just type '''help()''' at the ">>>" prompt to get started and then follow the instructions it gives you.<br />
<br />
===The IDLE Text Editor===<br />
For most programming needs, you'll want to edit your program in a separate document and then run it. Luckily, IDLE comes with its own built-in text editor.<br />
<br />
To get started, go to the File menu of the Python Shell and click on "New Window". This should give you a blank document with the title "Untitled" as shown below:<br />
<br />
[[File:Idle2-6.png]]<br />
<br />
You'll need to save your file before running it, so you might as well save it now. Make sure that you name your file with a file name with a file extension of .py (so it ends with .py), so your computer knows that it is a Python program. Here, we save ours as test.py:<br />
<br />
[[File:Idle2-7.png]]<br />
<br />
To get acquainted with the text editor, let's write our first Python program! Let's write a program that will do the following task:<br />
<br />
'''Print all the integers from 1 to 50 inclusive.'''<br />
<br />
We can achieve this by using a loop that can loop through all the integers. Luckily, Python has a function just for doing that! We use a for loop with the following code:<br />
<br />
[[File:Test_.png]]<br />
<br />
Note that as you type, the keywords like "for", "in", "range" and "print" get colored in orange or purple!<br />
Also, note that you must copy the exact same indentation. Even though the editor automatically indents for you when you type <code>for i in range(1, 51):</code>, proper indentation in Python is super important! If you don't do it correctly, the program will not compile correctly.<br />
<br />
You can indent by pressing Tab on your keyboard.<br />
<br />
This for loop means to iterate from 1 to 51 excluding the 51 and including the 1. Every iteration, Python will print out the number that it is iterating through.<br />
<br />
Now that you've written this code, you probably want to run it and test it out. You can do so by going to the Run menu and hitting "Run Module" (or by pressing F5 on your keyboard). The ===RESTART=== line means that Python is clearing all the work you've previously done before it starts running your program. The program should execute and print all the integers to the Python Shell. If it didn't, then make sure your code exactly matches the code above.<br />
<br />
[[File:Test__results.png]]<br />
<br />
If your code worked, congratulations! You have written your very first program! Now, let's try another more useful one.<br />
<br />
'''Find the sum of all the positive multiples of 3 below 1000.'''<br />
<br />
We first need to create a new file. Go into the Python Shell and click on New Window again. Remember, we must save our file first. We can save it as test2.py. Now, on to the coding!<br />
We can solve this by keeping a running total: we'll start with the smallest positive multiple of 3 and go up one multiple at a time keeping track of the sum, stopping once we hit 1000. We can do this with the following code:<br />
<br />
[[File:Idle2-8.png]]<br />
<br />
This is called a while loop. While loops keep iterating until a statement becomes false.<br />
Notice that as you type the above code, the keywords ("while" and "print") will automatically get colored -- this makes the code easier to read. Also, after typing the line "while i < 1000:", the editor will automatically start indenting for you. When you get to the line "print(total)", you'll need to use the backspace key to remove the indentation. It is important that the code look exactly like it does in the screenshot above. Again, in Python, proper indentation is very important!<br />
<br />
This program basically works by incrementing the variable <math>\verb=i=</math> by 3 every time and adding it to the variable <math>\verb=total=</math>. The <math>\verb%+=%</math> operation might be intimidating at first. However, the statement <math>\verb%i += 3%</math> is just a shorthand for <math>\verb%i = i + 3%</math>. (So <math>\verb%a += b%</math> means <math>\verb&a = a + b&</math>.)<br />
<br />
Run your program, and you should get this:<br />
<br />
[[File:Idle2-9.png]]<br />
<br />
Again, the ===RESTART=== line just means that Python is clearing all the work you've previously done before it starts running your program. Then, the program runs and we get our answer, 166833. If you instead get an error message or a different answer, check that your program exactly matches the screenshot above, and try it again.<br />
<br />
Congrats! You have written your first two Python programs!<br />
<br />
==What's Next?==<br />
<br />
Now that you've learned the very basics of getting Python going, there's a bunch of tutorials you can look at which are [http://wiki.python.org/moin/BeginnersGuide/NonProgrammers listed] on the Python website. Go check them out! Another great resource is "Stack Overflow," a forums website built for people who would like to talk about, and get help with programming. It is also recommended that you check out a wiki article discussing more advanced python, namely [[Basic Programming With Python]].<br />
<br />
Or, you can take our Introduction to Programming with Python online course!</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User:Fath2012&diff=105910User:Fath20122019-05-21T20:59:36Z<p>Fath2012: </p>
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<div><div class = "404"><br />
<h1> 404 not found</h1><br />
<p><br />
Page not found. Maybe the user has deleted it?<br />
</p><br />
HTML test has succeeded!<br />
</div></div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User_talk:Fath2012&diff=105909User talk:Fath20122019-05-21T19:55:50Z<p>Fath2012: /* Talk Page Help */</p>
<hr />
<div>===Talk Page Help===<br />
You said you didn't know how to use talk pages so I'm here to help.<br/><br />
<br/><br />
Step One: Post something<br/><br />
<br/><br />
Step Two: Type <code>~ ~ ~ ~</code> Without spaces<br/><br />
<br/><br />
Step Three: There is no step three<br />
<br/><br />
Banana_cream_Pi 15:28, 21 May 2019 (EDT)<br />
15:55, 21 May 2019 (EDT)<br />
<br />
===Adopt a page===<br />
[[Adopt a Page]]<br />
Banana_cream_Pi 15:53, 21 May 2019 (EDT)</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:FAQ&diff=105902AoPS Wiki talk:FAQ2019-05-21T19:14:13Z<p>Fath2012: added me, <-- I don't actually know how to use talk pages</p>
<hr />
<div>== The Important Question ==<br />
Someone should add "can I be a mod?" -AwesomeToad<br />
:Yes, even you can add it. -levans<br />
:Heh, added. --[[User:Dojo|Dojo]] 18:44, 18 June 2011 (EDT)<br />
<br />
== Q&A Template ==<br />
Can someone make a template for questions and answers? I'm not thrilled with the indented, green diamond questions. -levans<br />
: I changed it to Wikipedia-style FAQ. That hopefully looks better. <span style="color:#0000BF">theoneforce, SCotFF <sup>([[User:theoneforce|read the file]]) ([[User talk:theoneforce|submit an after-action report]])</sup></span> 10:00, 20 June 2011 (EDT)<br />
::I formatted it to make it more attractive. [[User:Btilm305|<span style="border-radius:1em;-webkit-border-radius:1em;-moz-border-radius:1em;border:1px solid black;font-size:11px;background-color:green;color:white;padding:1px 4px 1px 5px">'''Btilm305'''</span>]] 21:23, 23 June 2011 (EDT)<br />
<br />
== More complete FAQ? ==<br />
I think we should have a more complete FAQ that even covers the basics. For example, the question "Who can see my post rating?" is good, but people might ask, "What is post rating?". [[User:Chaotic iak|Chaotic iak]] 23:31, 23 June 2011 (EDT)<br />
:This is a good idea! I am probably going to lock this page from editing because it is now on the forum sidebar. But, I will have a page where people can make new question and answers. -- Unsigned comment by [[User:Btilm305|Btilm305]] at 00:33, 24 June 2011 (EDT)<br />
::I've added the subpage [[AoPSWiki:FAQ/Additions]] to do exactly that. Hopefully that should be good enough. <span style="color:#0000BF">theoneforce, SCotFF <sup>([[User:theoneforce|read the file]]) ([[User talk:theoneforce|submit an after-action report]])</sup></span> 20:21, 24 June 2011 (EDT)<br />
== <br />
"The user does not use the forums to cheat on Alcumus, homework, contests, or similar activities." - wiki<br />
We should also add the wiki, and the blogs. -fath2012</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=Asymptote:_Macros_and_Packages&diff=105480Asymptote: Macros and Packages2019-04-24T04:43:42Z<p>Fath2012: Added information about making a function with no output</p>
<hr />
<div>{{Asymptote}}<br />
<br />
==Definitions==<br />
You can define your own functions in Asymptote. As an example, let's say you wanted to make a function called <tt>newfunction</tt> that takes a pair <math>(a,b)</math> and a real value <math>r</math> as input, and returns the pair <math>(a+r,b+r)</math>. In addition, you want it to simply return the pair <math>(a,b)</math> if no value of <math>r</math> is specified, so you want <math>r</math> to default to <math>0</math>. The code would be as follows:<br />
pair newfunction(pair z, real r=0)<br />
{<br />
real a,b;<br />
a=z.x;<br />
b=z.y;<br />
return (a+r,b+r);<br />
}<br />
<br />
Put this definition in an asymptote document and then test it using some command like <br />
draw(newfunction((20,30))--newfunction((20,30),30)--(0,0)--cycle); <br />
See if it works!<br />
<br />
Notice that the function must be declared a pair since it returns a pair, and each of the variables must be declared some data type too. The default value of <math>r</math> was set to <math>0</math> by <math>r=0</math>, and the actual function procedure goes in between <tt>{}</tt>. To define a function with no output simply put <tt>void</tt> before the function name. This is the general format for a function definition.<br />
<br />
==Packages==<br />
Asymptote comes with several packages that contain useful functions for various purposes. For example, the package <tt>graph.asy</tt> contains the function <br />
Circle(pair p, real r, int n=400);<br />
which is a more accurate circle (having 400 nodes by default) than the built-in <tt>circle</tt> command. To use this function and others in graph.asy, simply put the command<br />
import graph;<br />
at the top of your Asymptote document. Graph also has more advanced functions such as the ability to [http://artofproblemsolving.com/wiki/index.php?title=Asymptote:_Graphing Graph a function]<br />
<br />
You can create your own package by simply creating a new .asy file (say <tt>MyMacros.asy</tt>) with your own definitions in it, and saving it in the directory in which Asymptote is installed (<tt>C:\Program Files\Asymptote</tt> by default). Then <tt>import MyMacros;</tt> in your document, and you'll be set!<br />
<br />
===The Olympiad Package===<br />
We have created an Olympiad package for Asymptote which includes macros for all the constructions that come up most often in Olympiad geometry problems! You can obtain the package olympiad.asy by clicking [http://math.berkeley.edu/~monks/images/olympiad.asy here] or [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&t=165767 here] (the latter link has a few usage examples). <br />
<br />
This package includes the following definitions:<br />
<br />
[[Image:Olympiad1.gif]]<br />
<br />
[[Image:Olympiad2.gif]]<br />
<br />
[[Image:Olympiad3.gif]]<br />
<br />
[[Image:Olympiad4.gif]]<br />
<br />
[[Image:Olympiad5.gif]]<br />
<br />
'''Note:''' A sequence of variables without type declarations indicates that they are the same type as the variable preceding it. For example, the notation <tt>concurrent(pair A, B, C, D, E, F)</tt> indicates that all of the variables should have type pair.<br />
<br />
<tt>*</tt> These boolean functions test for equality within <math>10^{-5}</math> ps points in order to avoid approximation errors.<br />
<br />
[[Asymptote: Advanced|Next: Advanced]]</div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User:Fath2012&diff=105155User:Fath20122019-04-07T04:41:15Z<p>Fath2012: </p>
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<div><div class = "404"><br />
<h1> 404 not found</h1><br />
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Page not found. Maybe the user has deleted it?<br />
</p><br />
<br />
</div></div>Fath2012https://artofproblemsolving.com/wiki/index.php?title=User:Fath2012&diff=105154User:Fath20122019-04-07T04:37:55Z<p>Fath2012: Created page with "<div class = "404"> <h1> 404 not found</h1> <p> Page not found. Maybe the user has deleted it? </p> </div>"</p>
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<div><div class = "404"><br />
<h1> 404 not found</h1><br />
<p><br />
Page not found. Maybe the user has deleted it?<br />
</p><br />
</div></div>Fath2012