https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Shurong.ge&feedformat=atomAoPS Wiki - User contributions [en]2021-07-28T11:56:10ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Proof_that_2%3D1&diff=116709Proof that 2=12020-02-03T04:58:04Z<p>Shurong.ge: /* Error */</p>
<hr />
<div>==Proof==<br />
<br />
1) <math>a = b</math>. Given.<br />
<br />
2) <math>a^2 = ab</math>. Multiply both sides by a.<br />
<br />
3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides.<br />
<br />
4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides.<br />
<br />
5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math><br />
<br />
6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>.<br />
<br />
7) <math>2a = a</math>. Addition.<br />
<br />
8) <math>2 = 1</math>. Divide both sides by <math>a</math>.<br />
<br />
==Error==<br />
Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way.<br />
<br />
In this case, the quantity of <math>a-b</math> is <math>0</math> as <math>a = b</math>, since one cannot divide by zero, the proof is incorrect from that point on.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Proof_that_2%3D1&diff=116708Proof that 2=12020-02-03T04:57:34Z<p>Shurong.ge: /* Error */</p>
<hr />
<div>==Proof==<br />
<br />
1) <math>a = b</math>. Given.<br />
<br />
2) <math>a^2 = ab</math>. Multiply both sides by a.<br />
<br />
3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides.<br />
<br />
4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides.<br />
<br />
5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math><br />
<br />
6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>.<br />
<br />
7) <math>2a = a</math>. Addition.<br />
<br />
8) <math>2 = 1</math>. Divide both sides by <math>a</math>.<br />
<br />
==Error==<br />
The quantity of <math>a-b</math> is <math>0</math> as <math>a = b</math>, since one cannot divide by zero, the proof is incorrect from that point on.<br />
<br />
Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=1167072020 AMC 10A Problems/Problem 162020-02-03T04:53:22Z<p>Shurong.ge: </p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br />
<br />
== Problem ==<br />
<br />
A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math><br />
<br />
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math><br />
<br />
== Solution 1 ==<br />
=== Diagram ===<br />
<asy><br />
size(10cm);<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
filldraw((arc((0,0), 0.4, 0, 90))--(0,0)--cycle, gray);<br />
draw(arc((1,0), 0.4, 90, 180));<br />
filldraw((arc((1,0), 0.4, 90, 180))--(1,0)--cycle, gray);<br />
draw(arc((1,1), 0.4, 180, 270));<br />
filldraw((arc((1,1), 0.4, 180, 270))--(1,1)--cycle, gray);<br />
draw(arc((0,1), 0.4, 270, 360));<br />
filldraw(arc((0,1), 0.4, 270, 360)--(0,1)--cycle, gray);<br />
</asy><br />
<br />
Diagram by [[User:Shurong.ge|Shurong.ge]] Using Asymptote<br />
<br />
Note: The diagram only represents a small portion of the given <math>2020 * 2020</math> square. <br />
<br />
===Solution===<br />
<br />
We consider an individual one-by-one block.<br />
<br />
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write<br />
<br />
<cmath>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</cmath><br />
<br />
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes<br />
<br />
<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math><br />
<br />
<br />
== Solution 2 ==<br />
As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]]<br />
<br />
==Video Solution==<br />
https://youtu.be/RKlG6oZq9so<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=1167062020 AMC 10A Problems/Problem 162020-02-03T04:50:15Z<p>Shurong.ge: /* Solution 1 */</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br />
<br />
== Problem ==<br />
<br />
A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math><br />
<br />
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math><br />
<br />
== Solution 1 ==<br />
<asy><br />
size(10cm);<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
filldraw((arc((0,0), 0.4, 0, 90))--(0,0)--cycle, gray);<br />
draw(arc((1,0), 0.4, 90, 180));<br />
filldraw((arc((1,0), 0.4, 90, 180))--(1,0)--cycle, gray);<br />
draw(arc((1,1), 0.4, 180, 270));<br />
filldraw((arc((1,1), 0.4, 180, 270))--(1,1)--cycle, gray);<br />
draw(arc((0,1), 0.4, 270, 360));<br />
filldraw(arc((0,1), 0.4, 270, 360)--(0,1)--cycle, gray);<br />
</asy><br />
<br />
We consider an individual one-by-one block.<br />
<br />
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write<br />
<br />
<cmath>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</cmath><br />
<br />
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes<br />
<br />
<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math><br />
<br />
<br />
Diagram by [[User:Shurong.ge|Shurong.ge]] Using Asymptote<br />
<br />
== Solution 2 ==<br />
As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]]<br />
<br />
==Video Solution==<br />
https://youtu.be/RKlG6oZq9so<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=1167052020 AMC 10A Problems/Problem 162020-02-03T04:49:20Z<p>Shurong.ge: </p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br />
<br />
== Problem ==<br />
<br />
A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math><br />
<br />
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math><br />
<br />
== Solution 1 ==<br />
<asy><br />
size(10cm);<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
filldraw((arc((0,0), 0.4, 0, 90))--(0,0)--cycle, gray);<br />
draw(arc((1,0), 0.4, 90, 180));<br />
filldraw((arc((1,0), 0.4, 90, 180))--(1,0)--cycle, gray);<br />
draw(arc((1,1), 0.4, 180, 270));<br />
filldraw((arc((1,1), 0.4, 180, 270))--(1,1)--cycle, gray);<br />
draw(arc((0,1), 0.4, 270, 360));<br />
filldraw(arc((0,1), 0.4, 270, 360)--(0,1)--cycle, gray);<br />
</asy><br />
<br />
We consider an individual one-by-one block.<br />
<br />
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write<br />
<br />
<cmath>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</cmath><br />
<br />
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes<br />
<br />
<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math><br />
<br />
<br />
Diagram by [[Shurong.ge]] Using Asymptote<br />
<br />
== Solution 2 ==<br />
As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]]<br />
<br />
==Video Solution==<br />
https://youtu.be/RKlG6oZq9so<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=1167042020 AMC 10A Problems/Problem 162020-02-03T04:48:44Z<p>Shurong.ge: /* Solution 1 */</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br />
<br />
== Problem ==<br />
<br />
A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math><br />
<br />
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math><br />
<br />
== Solution 1 ==<br />
<asy><br />
size(10cm);<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
filldraw((arc((0,0), 0.4, 0, 90))--(0,0)--cycle, gray);<br />
draw(arc((1,0), 0.4, 90, 180));<br />
filldraw((arc((1,0), 0.4, 90, 180))--(1,0)--cycle, gray);<br />
draw(arc((1,1), 0.4, 180, 270));<br />
filldraw((arc((1,1), 0.4, 180, 270))--(1,1)--cycle, gray);<br />
draw(arc((0,1), 0.4, 270, 360));<br />
filldraw(arc((0,1), 0.4, 270, 360)--(0,1)--cycle, gray);<br />
</asy><br />
<br />
We consider an individual one-by-one block.<br />
<br />
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write<br />
<br />
<cmath>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</cmath><br />
<br />
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes<br />
<br />
<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math><br />
<math>\textbf{Diagram by [[Shurong.ge]] Using Asymptote}</math><br />
<br />
== Solution 2 ==<br />
As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]]<br />
<br />
==Video Solution==<br />
https://youtu.be/RKlG6oZq9so<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=1167032020 AMC 10A Problems/Problem 162020-02-03T04:47:24Z<p>Shurong.ge: </p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br />
<br />
== Problem ==<br />
<br />
A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math><br />
<br />
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math><br />
<br />
== Solution 1 ==<br />
<asy><br />
size(10cm);<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
filldraw((arc((0,0), 0.4, 0, 90))--(0,0)--cycle, gray);<br />
draw(arc((1,0), 0.4, 90, 180));<br />
filldraw((arc((1,0), 0.4, 90, 180))--(1,0)--cycle, gray);<br />
draw(arc((1,1), 0.4, 180, 270));<br />
filldraw((arc((1,1), 0.4, 180, 270))--(1,1)--cycle, gray);<br />
draw(arc((0,1), 0.4, 270, 360));<br />
filldraw(arc((0,1), 0.4, 270, 360)--(0,1)--cycle, gray);<br />
</asy><br />
<br />
We consider an individual one-by-one block.<br />
<br />
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write<br />
<br />
<cmath>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</cmath><br />
<br />
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes<br />
<br />
<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math><br />
<br />
== Solution 2 ==<br />
As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]]<br />
<br />
==Video Solution==<br />
https://youtu.be/RKlG6oZq9so<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_23&diff=1167022020 AMC 10A Problems/Problem 232020-02-03T04:25:37Z<p>Shurong.ge: </p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #20]] and [[2020 AMC 10A Problems|2020 AMC 10A #23]]}}<br />
<br />
== Problem ==<br />
Let <math>T</math> be the triangle in the coordinate plane with vertices <math>(0,0), (4,0),</math> and <math>(0,3).</math> Consider the following five isometries (rigid transformations) of the plane: rotations of <math>90^{\circ}, 180^{\circ},</math> and <math>270^{\circ}</math> counterclockwise around the origin, reflection across the <math>x</math>-axis, and reflection across the <math>y</math>-axis. How many of the <math>125</math> sequences of three of these transformations (not necessarily distinct) will return <math>T</math> to its original position? (For example, a <math>180^{\circ}</math> rotation, followed by a reflection across the <math>x</math>-axis, followed by a reflection across the <math>y</math>-axis will return <math>T</math> to its original position, but a <math>90^{\circ}</math> rotation, followed by a reflection across the <math>x</math>-axis, followed by another reflection across the <math>x</math>-axis will not return <math>T</math> to its original position.)<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25</math><br />
<br />
== Solution ==<br />
<asy><br />
size(10cm);<br />
Label f; <br />
f.p=fontsize(6); <br />
xaxis(-6,6,Ticks(f, 2.0)); <br />
yaxis(-6,6,Ticks(f, 2.0));<br />
<br />
filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1));<br />
</asy><br />
<br />
First, any combination of motions we can make must reflect <math>T</math> an even number of times. This is because every time we reflect <math>T</math>, it changes orientation. Once <math>T</math> has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed <math>3</math> transformations and an even number of them must be reflections, we either reflect <math>T</math> <math>0</math> times or <math>2</math> times.<br />
<br />
<br />
<br />
Case 1: 0 reflections on T<br />
<br />
<br />
<br />
In this case, we must use <math>3</math> rotations to return <math>T</math> to its original position. Notice that our set of rotations, <math>\{90^\circ,180^\circ,270^\circ\}</math>, contains every multiple of <math>90^\circ</math> except for <math>0^\circ</math>. We can start with any two rotations <math>a,b</math> in <math>\{90^\circ,180^\circ,270^\circ\}</math> and there must be exactly one <math>c \equiv -a - b \pmod{360^\circ}</math> such that we can use the three rotations <math>(a,b,c)</math> which ensures that <math>a + b + c \equiv 0^\circ \pmod{360^\circ}</math>. That way, the composition of rotations <math>a,b,c</math> yields a full rotation. For example, if <math>a = b = 90^\circ</math>, then <math>c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}</math>, so <math>c = 180^\circ</math> and the rotations <math>(90^\circ,90^\circ,180^\circ)</math> yields a full rotation.<br />
<br />
The only case in which this fails is when <math>c</math> would have to equal <math>0^\circ</math>. This happens when <math>(a,b)</math> is already a full rotation, namely, <math>(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),</math> or <math>(270^\circ,90^\circ)</math>. However, we can simply subtract these three cases from the total. Selecting <math>(a,b)</math> from <math>\{90^\circ,180^\circ,270^\circ\}</math> yields <math>3 \cdot 3 = 9</math> choices, and with <math>3</math> that fail, we are left with <math>6</math> combinations for case 1.<br />
<br />
<br />
<br />
Case 2: 2 reflections on T<br />
<br />
<br />
<br />
In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps <math>T</math> back to itself, inserting a rotation before, between, or after these two reflections would change <math>T</math>'s final location, meaning that any combination involving two reflections across the x-axis would not map <math>T</math> back to itself. The same applies to two reflections across the y-axis.<br />
<br />
Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a <math>180^\circ</math> rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us <math>3! = 6</math> combinations for case 2.<br />
<br />
Combining both cases we get <math>6 + 6 = \boxed{\textbf{(A) } 12}</math><br />
<br />
==Solution 2(Rewording solution 1)==<br />
<br />
As in the previous solution, note that we must have either 0 or 2 reflections because of orientation since reflection changes orientation that is impossible to fix by rotation. We also know we can't have the same reflection twice, since that would give a net of no change and would require an identity rotation.<br />
<br />
Suppose there are no reflections. Denote <math>90^{\circ}</math> as 1, <math>180^{\circ}</math> as 2, and <math>270^{\circ}</math> as 3, just for simplification purposes. We want a combination of 3 of these that will sum to either 4 or 8(0 and 12 is impossible since the minimum is 3 and the max is 9). 4 can be achieved with any permutation of <math>(1-1-2)</math> and 8 can be achieved with any permutation of <math>(2-3-3)</math>. This case can be done in <math>3+3=6</math> ways.<br />
<br />
Suppose there are two reflections. As noted already, they must be different, and as a result will take the triangle to the opposite side of the origin if we don't do any rotation. We have 1 rotation left that we can do though, and the only one that will return to the original position is 2, which is <math>180^{\circ}</math> AKA reflection across origin. Therefore, since all 3 transformations are distinct. The three transformations can be applied anywhere since they are commutative(think quadrants). This gives <math>6</math> ways.<br />
<br />
<math>6+6=\boxed{(A)12}</math><br />
<br />
== Video Solution ==<br />
<br />
https://youtu.be/5TjrDCxTm7Q - <math>Phineas1500</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2020|ab=A|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_19&diff=1166992020 AMC 10A Problems/Problem 192020-02-03T04:04:48Z<p>Shurong.ge: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
As shown in the figure below, a regular dodecahedron (the polyhedron consisting of <math>12</math> congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br />
<br />
<math>\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810</math><br />
<br />
== Diagram ==<br />
<asy><br />
import graph;<br />
unitsize(5cm);<br />
pair A = (0.082, 0.378);<br />
pair B = (0.091, 0.649);<br />
pair C = (0.249, 0.899);<br />
pair D = (0.479, 0.939);<br />
pair E = (0.758, 0.893);<br />
pair F = (0.862, 0.658);<br />
pair G = (0.924, 0.403);<br />
pair H = (0.747, 0.194);<br />
pair I = (0.526, 0.075);<br />
pair J = (0.251, 0.170);<br />
pair K = (0.568, 0.234);<br />
pair L = (0.262, 0.449);<br />
pair M = (0.373, 0.813);<br />
pair N = (0.731, 0.813);<br />
pair O = (0.851, 0.461);<br />
path[] f;<br />
f[0] = A--B--C--M--L--cycle;<br />
f[1] = C--D--E--N--M--cycle;<br />
f[2] = E--F--G--O--N--cycle;<br />
f[3] = G--H--I--K--O--cycle;<br />
f[4] = I--J--A--L--K--cycle;<br />
f[5] = K--L--M--N--O--cycle;<br />
draw(f[0]);<br />
axialshade(f[1], white, M, gray(0.5), (C+2*D)/3);<br />
draw(f[1]);<br />
filldraw(f[2], gray);<br />
filldraw(f[3], gray);<br />
axialshade(f[4], white, L, gray(0.7), J);<br />
draw(f[4]);<br />
draw(f[5]);<br />
</asy><br />
<br />
== Solution ==<br />
Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.<br />
<br />
We have <math>5</math> choices for which face we visit first on the top ring. From there, we have <math>9</math> choices for how far around the top ring we go before moving down: <math>1,2,3,</math> or <math>4</math> faces around clockwise, <math>1,2,3,</math> or <math>4</math> faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring. We then have <math>2</math> choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly <math>2</math> lower-ring faces) and then once again <math>9</math> choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.<br />
<br />
Multiplying together all the numbers of choices we have, we get <math>5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}</math>.<br />
<br />
== Solution 2 ==<br />
Swap the faces as vertices and the vertices as faces. Then, this problem is the same as<br />
[https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 2016 AIME I #3]<br />
which had an answer of <math>\boxed{\textbf{(E) } 810}</math>.<br />
<math>\textbf{- Emathmaster}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/RKlG6oZq9so<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3<br />
{{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=User:Shurong.ge&diff=115848User:Shurong.ge2020-01-29T23:13:44Z<p>Shurong.ge: /* Other Websites I Use */</p>
<hr />
<div>==About Me==<br />
Go onto my user page to learn about me.<br />
Here are some quick facts:<br />
*I am a human<br />
*I have my own AoPS wiki page<br />
*I like stating obvious facts<br />
<br />
==What I am learning==<br />
At AoPS, I am taking Geometry, Algebra 2, and 6th grade LA at the Bellevue campus.<br />
<br />
==Other Websites I Use==<br />
*Edabit<br />
*Repl<br />
*Scratch (ish)<br />
<br />
==Closest Friends on AoPS==<br />
===ThriftyPiano===<br />
Thriftypiano in real life earned 1st place in a state wide math competition, the following section is an excerpt from Thriftypiano's intro speech:<br />
<br />
"Thriftypiano is unknown. Nobody knows where he is, who he is, why he exists, and what he is. All we know is that he exists and is male.<br />
He created this song:<br />
When I was a little biddy boy<br />
My grandmother bought me a cute little toy<br />
Silver bells hanging on a string<br />
She told me it was my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
Then momma took me to Grammar school<br />
Always went by the best of rule *<br />
Everytime that bell would ring<br />
Catch me playing with my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while climbing the garden wall,<br />
I slipped and had a terrible fall<br />
I fell so hard I heard bells ring,<br />
But I held on to My ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while swimming cross turtle creek<br />
Man them snappers all 'round at my feet<br />
Was so hard swimming cross that thing<br />
with both hands holding my dingaling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
And this song:<br />
A duck walked up to a lemonade stand<br />
And he said to the man, running the stand<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said<br />
"No we just sell lemonade. But it's cold<br />
And it's fresh<br />
And it's all home-made. Can I get you<br />
Glass?"<br />
The duck said,<br />
"I'll pass".<br />
Then he waddled away.<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-bada-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?<br />
The man said,<br />
"No, like I said yesterday<br />
We just sell lemonade OK?<br />
Why not give it a try?"<br />
The duck said,<br />
"Goodbye."good day<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (bum bum bum) Got any grapes?<br />
The man said,<br />
Look, this is getting old.<br />
I mean, lemonade's all we've ever sold.<br />
Why not give it a go?"<br />
The duck said,<br />
"How 'bout, no."<br />
Then he waddled away<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said,<br />
"THAT'S IT!<br />
If you don't stay away, duck,<br />
I'll glue you to a tree and leave you there all day, stuck<br />
So don't get to close!"<br />
The duck said,<br />
"Adios."<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man that was running the stand,<br />
"Hey! (Bum bum bum) got any glue?"<br />
"What"<br />
"Got any glue?"<br />
"No, why would I– oh!"<br />
And one more question for you;<br />
"Got any grapes?"<br />
(Bum bum bum, bum bum bum)<br />
And the man just stopped.<br />
Then he started to smile.<br />
He started to laugh.<br />
He laughed for a while.<br />
He said,<br />
"Come on duck, let's walk to the store.<br />
I'll buy you some grapes<br />
So you won't have to ask anymore."<br />
So they walked to the store<br />
And the man bought some grapes.<br />
He gave one to the duck and the duck said,<br />
"Hmmm..No thanks. But you know what sounds good?<br />
It would make my day.<br />
Do you think this store<br />
Do you think this store<br />
Do you think this store has any lemonade?"<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)" -ThriftyPiano<br />
<br />
===Potato2017===<br />
He is a 5th grader, but has an IQ of 190. Thriftypiano AND the Stanford test says so.<br />
<br />
Edited by Thrifty Piano<br />
<br />
===Bluesky11===<br />
He is a very good friend, it's sad we got split into different middle schools...<br />
<br />
==Coding==<br />
Yes, I code in Python and Java<br />
<br />
==Makers of This Page==<br />
Page created by Shurong.ge; edited by ThriftyPiano</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=User:Shurong.ge&diff=115847User:Shurong.ge2020-01-29T23:13:15Z<p>Shurong.ge: /* About Me */</p>
<hr />
<div>==About Me==<br />
Go onto my user page to learn about me.<br />
Here are some quick facts:<br />
*I am a human<br />
*I have my own AoPS wiki page<br />
*I like stating obvious facts<br />
<br />
==What I am learning==<br />
At AoPS, I am taking Geometry, Algebra 2, and 6th grade LA at the Bellevue campus.<br />
<br />
==Other Websites I Use==<br />
Edabit<br />
Repl<br />
<br />
==Closest Friends on AoPS==<br />
===ThriftyPiano===<br />
Thriftypiano in real life earned 1st place in a state wide math competition, the following section is an excerpt from Thriftypiano's intro speech:<br />
<br />
"Thriftypiano is unknown. Nobody knows where he is, who he is, why he exists, and what he is. All we know is that he exists and is male.<br />
He created this song:<br />
When I was a little biddy boy<br />
My grandmother bought me a cute little toy<br />
Silver bells hanging on a string<br />
She told me it was my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
Then momma took me to Grammar school<br />
Always went by the best of rule *<br />
Everytime that bell would ring<br />
Catch me playing with my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while climbing the garden wall,<br />
I slipped and had a terrible fall<br />
I fell so hard I heard bells ring,<br />
But I held on to My ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while swimming cross turtle creek<br />
Man them snappers all 'round at my feet<br />
Was so hard swimming cross that thing<br />
with both hands holding my dingaling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
And this song:<br />
A duck walked up to a lemonade stand<br />
And he said to the man, running the stand<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said<br />
"No we just sell lemonade. But it's cold<br />
And it's fresh<br />
And it's all home-made. Can I get you<br />
Glass?"<br />
The duck said,<br />
"I'll pass".<br />
Then he waddled away.<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-bada-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?<br />
The man said,<br />
"No, like I said yesterday<br />
We just sell lemonade OK?<br />
Why not give it a try?"<br />
The duck said,<br />
"Goodbye."good day<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (bum bum bum) Got any grapes?<br />
The man said,<br />
Look, this is getting old.<br />
I mean, lemonade's all we've ever sold.<br />
Why not give it a go?"<br />
The duck said,<br />
"How 'bout, no."<br />
Then he waddled away<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said,<br />
"THAT'S IT!<br />
If you don't stay away, duck,<br />
I'll glue you to a tree and leave you there all day, stuck<br />
So don't get to close!"<br />
The duck said,<br />
"Adios."<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man that was running the stand,<br />
"Hey! (Bum bum bum) got any glue?"<br />
"What"<br />
"Got any glue?"<br />
"No, why would I– oh!"<br />
And one more question for you;<br />
"Got any grapes?"<br />
(Bum bum bum, bum bum bum)<br />
And the man just stopped.<br />
Then he started to smile.<br />
He started to laugh.<br />
He laughed for a while.<br />
He said,<br />
"Come on duck, let's walk to the store.<br />
I'll buy you some grapes<br />
So you won't have to ask anymore."<br />
So they walked to the store<br />
And the man bought some grapes.<br />
He gave one to the duck and the duck said,<br />
"Hmmm..No thanks. But you know what sounds good?<br />
It would make my day.<br />
Do you think this store<br />
Do you think this store<br />
Do you think this store has any lemonade?"<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)" -ThriftyPiano<br />
<br />
===Potato2017===<br />
He is a 5th grader, but has an IQ of 190. Thriftypiano AND the Stanford test says so.<br />
<br />
Edited by Thrifty Piano<br />
<br />
===Bluesky11===<br />
He is a very good friend, it's sad we got split into different middle schools...<br />
<br />
==Coding==<br />
Yes, I code in Python and Java<br />
<br />
==Makers of This Page==<br />
Page created by Shurong.ge; edited by ThriftyPiano</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_22&diff=1158462015 AMC 10A Problems/Problem 222020-01-29T23:12:16Z<p>Shurong.ge: </p>
<hr />
<div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #17]] and [[2015 AMC 10A Problems|2015 AMC 10A #22]]}}<br />
==Problem==<br />
<br />
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br />
<br />
<math>\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
We count how many valid arrangements there are and then divide by <math>2^8=256</math>.<br />
<br />
First, arbitrarily pick a person A and consider whether he is standing or sitting.<br />
If he is sitting, then the number of valid arrangements is the same as the number of valid arrangements of 7 people in a line.<br />
If he is standing, then the two people next to him must be sitting, so the number of valid arrangements is the same as the number of valid arrangements of 5 people in a line.<br />
<br />
Let <math>a_n</math> denote the number of ways to arrange <math>n</math> people in a line such that no two adjacent people are standing. We determine a recurrence relation for <math>a_n,\ n\geq 2</math> as follows. If the first person in the line is standing, then the next person must be sitting, and there are <math>a_{n-2}</math> ways to arrange the rest. If the first person in the line is sitting, then there are <math>a_{n-1}</math> ways to arrange the rest. Thus, <math>a_n=a_{n-1}+a_{n-2}</math> and using the fact that <math>a_0=1</math> and <math>a_1=2</math>, we get that <math>a_n</math> is the <math>(n+2)</math>nd Fibonacci number <math>F_{n+2}</math>. In particular, <math>a_5=F_7=13</math> and <math>a_7=F_9=34</math>.<br />
<br />
So our desired count is <math>a_7+a_5=34+13=47</math>, and the answer is <math>\boxed{\textbf{(A) } \frac{47}{256}}</math>.<br />
<br />
===Solution 2===<br />
We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by <math>2^8 = 256</math> at the end. We casework on how many people are standing.<br />
<br />
Case <math>1:</math> <math>0</math> people are standing. This yields <math>1</math> arrangement.<br />
<br />
Case <math>2:</math> <math>1</math> person is standing. This yields <math>8</math> arrangements.<br />
<br />
Case <math>3:</math> <math>2</math> people are standing. This yields <math>\dbinom{8}{2} - 8 = 20</math> arrangements, because the two people cannot be next to each other.<br />
<br />
Case <math>4:</math> <math>4</math> people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding <math>2</math> possible arrangements.<br />
<br />
More difficult is:<br />
<br />
Case <math>5:</math> <math>3</math> people are standing. First, choose the location of the first person standing (<math>8</math> choices). Next, choose <math>2</math> of the remaining people in the remaining <math>5</math> legal seats to stand, amounting to <math>6</math> arrangements considering that these two people cannot stand next to each other. However, we have to divide by <math>3,</math> because there are <math>3</math> ways to choose the first person given any three. This yields <math>\dfrac{8 \cdot 6}{3} = 16</math> arrangements for Case <math>5.</math><br />
<br />
Alternate Case <math>5:</math> Use complementary counting. Total number of ways to choose 3 people from 8 which is <math>\dbinom{8}{3}</math>. Sub-case <math>1:</math> three people are next to each other which is <math>\dbinom{8}{1}</math>. Sub-case <math>2:</math> two people are next to each other and the third person is not <math>\dbinom{8}{1}</math> <math>\dbinom{4}{1}</math>. This yields <math>\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16</math> <br />
<br />
Summing gives <math>1 + 8 + 20 + 2 + 16 = 47,</math> and so our probability is <math>\boxed{\textbf{(A) } \dfrac{47}{256}}</math>.<br />
<br />
===Solution 3===<br />
We will count how many valid standing arrangements there are counting rotations as distinct and divide by <math>256</math> at the end.<br />
Line up all <math>8</math> people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires <math>2</math> spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.<br />
<br />
If there are <math>4</math> standing, there are <math>{4 \choose 4}=1</math> ways to place them.<br />
For <math>3,</math> there are <math>{3+2 \choose 3}=10</math> ways.<br />
etc.<br />
Summing, we get <math>{4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34</math> ways.<br />
<br />
Now we consider that the far right person can be standing as well, so we have<br />
<math>{3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13</math> ways<br />
<br />
Together we have <math>34+13=47</math>, and so our probability is <math>\boxed{\textbf{(A) } \dfrac{47}{256}}</math>.<br />
<br />
===Solution 4===<br />
We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by <math>2^8 = 256</math> at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only <math>7</math> people. If they stand, we count the arrangements with <math>6</math> instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with <math>1</math> person there are two ways and with <math>2</math> people there are three ways. Carrying out the Fibonacci recursion until we get to <math>8</math> people, we find there are <math>55</math> standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for <math>4</math> people to stand in a line, which is <math>8</math> from our sequence. Therefore our probability is <math>\frac{55 - 8}{256} = \boxed{\textbf{(A) } \dfrac{47}{256}}</math><br />
<br />
===Solution 5===<br />
We will count the number of valid arrangements and then divide by <math>2^8</math> at the end. We proceed with casework on how many people are standing.<br />
<br />
Case <math>1:</math> <math>0</math> people are standing. This yields <math>1</math> arrangement.<br />
<br />
Case <math>2:</math> <math>1</math> person is standing. This yields <math>8</math> arrangements.<br />
<br />
Case <math>3:</math> <math>2</math> people are standing. To do this, we imagine having 6 people with tails in a line first. Notate "tails" with <math>T</math>. Thus, we have <math>TTTTTT</math>. Now, we look to distribute the 2 <math>H</math>'s into the 7 gaps made by the <math>T</math>'s. We can do this in <math>{7 \choose 2}</math> ways. However, note one way does not work, because we have two H's at the end, and the problem states we have a table, not a line. So, we have <math>{7 \choose 2}-1=20</math> arrangements.<br />
<br />
Case <math>4:</math> <math>3</math> people are standing. Similarly, we imagine 5 <math>T</math>'s. Thus, we have <math>TTTTT</math>. We distribute 3 <math>H</math>'s into the gaps, which can be done <math>{6 \choose 3}</math> ways. However, 4 arrangements will not work. (See this by putting the H's at the ends, and then choosing one of the remaining 4 gaps: <math>{4 \choose 1}=4</math>) Thus, we have <math>{6 \choose 3}-4=16</math> arrangements. <br />
<br />
Case <math>5:</math> <math>4</math> people are standing. This can clearly be done in 2 ways: <math>HTHTHTHT</math> or <math>THTHTHTH</math>. This yields <math>2</math> arrangements.<br />
<br />
Summing the cases, we get <math>1+8+20+16+2=47</math> arrangements.<br />
Thus, the probability is <math>\boxed{\textbf{(A) } \dfrac{47}{256}}</math><br />
<br />
==Video Solution (Alternative)==<br />
If you didn't quite follow this process, then you can look at this Ruzczyk video: https://www.youtube.com/watch?v=krlnSWWp0I0. I <br />
<br />
== See Also ==<br />
{{AMC10 box|year=2015|ab=A|num-b=21|num-a=23}}<br />
{{AMC12 box|year=2015|ab=A|num-b=16|num-a=18}}<br />
<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Combinatorics Problems]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Category:Integers&diff=115845Category:Integers2020-01-29T22:39:26Z<p>Shurong.ge: Created blank page</p>
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<div></div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Mu_Alpha_Theta&diff=115844Mu Alpha Theta2020-01-29T22:37:48Z<p>Shurong.ge: </p>
<hr />
<div>'''Mu Alpha Theta''' is a national honor society in the United States for high school and two year college students. Mu Alpha Theta chapters host numerous [[mathematics competitions]] and each year a national Mu Alpha Theta convention is held during the summer that includes a large competition.<br />
== The Competition ==<br />
Mu Alpha Theta competitions can be divided up into regionals, invitationals, states, and national competitions. There is no need to progress through the levels- you simply register and you can participate. There are many different tests with different topic themes. High School students are often organized into three groups- Mu, Alpha and Theta. Theta is the lowest level, Alpha is higher and Mu is the highest.<br />
<br />
== The Test ==<br />
=== Individual ===<br />
The test is a multiple choice test with 4 choices and an extra choice called NOTA, which means None Of The Above. Test topics usually follow school curriculum subjects; however they are far more advanced. The main groups in the regional and invitational levels are:<br />
*[[Algebra]] 1<br />
*[[Geometry]]<br />
*[[Algebra]] 2<br />
*[[Pre-Calculus]]<br />
*[[Calculus]]<br />
You may only take one type of test a year for every competition, depending on how many competitions your region has a year. <br />
At the states and nationals, there are many more different topic tests ranging from trigonometry to analytic geometry to complex numbers.<br />
(Full list coming soon.)<br />
The tests are often made by other students.<br />
They generally have 30 questions. However, at nationals and state, some tests have 40 questions. There is also a one hour time limit except for History of Math which has a 30 minute time limit.<br />
The scoring system usually is as follows:<br />
*4 points for each correct question<br />
*0 points for each unanswered question<br />
*-1 for each incorrect question<br />
However at Nationals the scoring system is as follows:<br />
*5 points for each correct question<br />
*1 point for each unanswered question<br />
*0 points for each incorrect question<br />
<br />
=== Team Rounds ===<br />
In team rounds, teams of 4 people work fast to answer a problem in 4 minutes. However, the faster a team turns in an answer, the more points recieved.<br />
*Correct answer in first minute<br />
**16 points<br />
*Correct answer in second minute<br />
**12 points<br />
*Correct answer in third minute<br />
**8 points<br />
*Correct answer in fourth minute<br />
**4 points<br />
This round is NOT multiple choice.<br />
Questions, just as in the individual round, often follow a certain subject.<br />
Questions usually follow the certain rubic:<br />
<br />
A=<insert some kind of math problem here><br />
<br />
B=<Insert some kind of math problem here><br />
<br />
C=<Insert some kind of math problem here><br />
<br />
D=<Insert some kind of math problem here><br />
<br />
Find AB-CD.<br />
<br />
However, this is not always the case.<br />
<br />
== Example Problems ==<br />
<br />
Probability/Permutations/Combinations- Theta 2003<br />
<br />
#10 The set S is {#, !, @, *, $, %}. How many different proper subsets are possible?<br />
A.6 B.63 C.64 D.127 E. NOTA <br />
<br />
#30 On a celebrity match of the Weakest Link, Big Bird and Barney make it to the final round. the final round consists of five questions per player. Play is stopped if it becomes evident that one of the contestants cannot possibly win. If Big Bird has answered his questions correctly 75% of the time and Barney has answered his questions correctly 60% of the time, what is the probability that the final round will end after each player has answered exactly four questions?<br />
<br />
A. 3753/160000 B.2943/40000 C.3753/40000 D.3753/32000 E. NOTA<br />
<br />
2001- Number theory Topic Test- Theta<br />
<br />
#The 4-digit number 6A6B is divisibly by 72. What is the sum of all possible values of A?<br />
A. 2 B. 7 C. 9 D. 11 E. NOTA<br />
<br />
== Science Fair ==<br />
At many regional and state science and engineering fairs, Mu Alpha Theta gives away prizes to high school students. These are prizes for investigating a modern mathematical problem.<br />
<br />
== Resources ==<br />
* [http://www.mualphatheta.org/ Mu Alpha Theta homepage] -- includes dozens of past tests.<br />
* [http://www.mualphatheta.org/index.php?mu-alpha-theta-scholarships Mu Alpha Theta scholarships]<br />
<br />
== See also ==<br />
* [[Mu Alpha Theta school chapters]]<br />
* [[Mathematics competition resources]]<br />
* [[Math books]]<br />
* [[Mathematics scholarships]]<br />
<br />
<br />
<br />
[[Category:Mathematics competitions]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Mean&diff=115843Mean2020-01-29T22:36:29Z<p>Shurong.ge: /* Types of Means */</p>
<hr />
<div>The '''mean''' of a set of real numbers usually refers to the [[arithmetic mean]] of the set (also known as the [[average]]). For example, the arithmetic mean of the members of the set {3, 5, 10} is<br />
<br />
<math> \frac{3 + 5 + 10}{3} = \frac{18}{3} = 6. </math><br />
<br />
<br />
However, there are numerous other kinds of various means used in [[mathematics]] and [[statistics]].<br />
<br />
<br />
== Types of Means ==<br />
* [[Arithmetic mean]]<br />
* [[Geometric mean]]<br />
* [[Harmonic mean]]<br />
* [[Power mean]]<br />
* [[quadratic mean]] (also known as the root mean square)<br />
<br />
The arithmetic mean, geometric mean, harmonic mean, and root mean square are all special cases of the power mean.<br />
<br />
== Inequalities and Optimization ==<br />
There are numerous [[inequality | inequalities]] that relate different types of means. The most common are part of the RMS-AM-GM-HM inequality chain. This inequality chain is a set of special cases of the [[Power mean inequality]].<br />
<br />
== See Also ==<br />
<br />
* [[Median of a set]]<br />
<br />
* [[Mode of a set]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Talk:Gmaas&diff=115712Talk:Gmaas2020-01-27T23:46:37Z<p>Shurong.ge: </p>
<hr />
<div>Follow the following steps to summon [[Gmaas]]:<br />
<br />
1. Draw a circle and circumscribe it with a regular hexagon and an equilateral triangle.<br />
<br />
2. Write the numerical value of <math>17^{36}</math> along the edge of the circle.<br />
<br />
3. Write the numerical value of <math>33^{29}</math> along the edge of the hexagon.<br />
<br />
4. Write the numerical value of <math>\cot(0)</math> 5 centimeters above the hexagon.<br />
<br />
5. Write the numerical value of <math>\tan(\frac{\pi}{2} rad)</math> 5 centimeters below the hexagon.<br />
<br />
6. Write the numerical value of <math>\ln(0)</math> inside the circle.<br />
<br />
7. Write the numerical value of the melting point of water inside the circle. Include units and write 15 significant digits.<br />
<br />
8. Write the numerical value of the gravitational constant inside the circle. Include units and write 25 significant digits.<br />
<br />
9. Carry the paper with the circle and hexagon in your left hand.<br />
<br />
10. Recite the value of <math>\pi^{e^2}</math> and include <math>\pi^{e^2}</math> (rounded to the nearest septillionth) digits.<br />
<br />
11. Travel at the speed of light with that paper and Gmaas will be summoned.<br />
<div><br />
<br />
<div><br />
EDIT: The author congratulates the previous editor for his research on the cutting edge of Gmaasology. He/she has been awarded the Nobel Prize in Gmaasology for his discovery. However, the aforementioned summoning has several problems: it does not specify the dimensions of the hexagon or the circle; the fact that it has to be drawn on a paper; the size of the paper; the pressure of the water being melted; whether <math>\frac{\pi}{2}</math> radians, degrees, or gradians; the exact location if the value of <math>17^{36}</math> and <math>33^{29}</math> in the diagram and the base those numbers should be written in; the units of the melting point of water or the units of the gravitational constant; or the location where Gmaas will be summoned. He might be summoned on the other side of the universe.<br />
<div><br />
<br />
<div><br />
The person who wrote the steps to summon Gmaas says:<br />
<div><br />
<br />
<div><br />
When I followed this procedure to summon Gmaas, I used a paper that is 15 inches long and 11 inches wide. It is currently unknown whether paper of other dimensions can be used to summon Gmaas. For step 7, the pressure of the water being melted is at 611.66 Pascals, which matches the triple point of water. Also, the units of the melting points of water and the gravitational constant should be written in terms of SI base units. The circle used should have a radius of 5 inches. If you follow this procedure, Gmaas will be summoned for about 3.53 nanoseconds at a random position in the planet you are on. The exact time Gmaas will be summoned depends on how accurately you follow the procedure. For example, if the melting point of water and the gravitational constant are accurate to 200 significant digits, then Gmaas will be summoned for 77.8 microseconds. If they are accurate to 400 significant digits, Gmaas will be summoned for 1.7 seconds. The function that relates the number of significant digits and the time Gmaas will be summoned is still unknown.<br />
<br />
EDIT: Based on the two data points, the function that relates the number of significant digits and the time Gmaas will be summoned is <math>t=.0081111s+.5444</math>.<br />
<br />
Edit: Gmaas no longer appears with a hexagon. You must now draw a heptagon.<br />
<br />
Edit: Gmaas never appears for more than 10 seconds, and burns the paper with him, forcing you do do it all over again to summon him again<br />
<br />
Edit: If you are able to edit these steps, then you have been partially (<math>\frac{1}{1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000}</math>) enlightened by Gmaas<br />
<br />
Edit Names For Challenge:<br />
juliankuang<br />
<br />
== Gmaas ==<br />
<br />
NOOOOO. THEY DELETED THE PAGE!!! https://artofproblemsolving.com/wiki/index.php?title=Gmaas#Known_Facts_About_gmaas<br />
<br />
<br />
HOW DARE THEY! THIS IS HORRIBLE!!!!!! THIS IS NOT GOOD AT ALL!!!! - asdf334 ;( currently mourning this loss<br />
<br />
<br />
NOPE. IT'S THERE NOW LAST TIME I CHECKED. -MATHGUY49<br />
<br />
<br />
Collball: Talk: Gmaas is not the place to post your article.<br />
<br />
<br />
<br />
(Mistyketchum28: EDIT: ALL HAIL GMAAS<br />
btw ONLY THE PART IN PARENTHESIS IS WRITTEN BY ME <br />
IM NOT CoPYING)<br />
<br />
NOOOOOOOOOOOOOOOOOOOOOO GMAAS PaGE IS NOW PROTECTED NOOOOOOOOOOOOOOOOoo - juliankuang, crying in bed right now<br />
<br />
== No more sseraj ==<br />
<br />
By the way, Samer Seraj (GMAAS's slave) doesn't use the account sseraj anymore.<br />
<br />
== Locking the GMAAS page ==<br />
<br />
It's a sad thing the 5space locked the GMAAS page. But at least we still have this Talk page<br />
<br />
--[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 18:47, 26 December 2019 (EST)<br />
<br />
It might not be 5space. Let us just put hope, even if absurd the idea is, that the Gmaas page will be unprotected soon. juliankuang 14:15, 31 December 2019 (EST) juliankuang<br />
<br />
It is 5space, that's what the history page says. I think he locked it because there was a user that blanked the page like 3 times.<br />
<br />
--[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 18:47, 30 December 2019 (EST)<br />
<br />
--[[User:ARay10|ARay10]] ([[User talk:Aray10|talk]]) 02:08 2 January 2020 (EST)<br />
Even if it is locked, honor Gmass! He is really important.<br />
<br />
Respect Gmaas. Spell His name correctly. juliankuang 10:17, 2 January 2020 (EST)<br />
<br />
Actually it's spelled GMAAS, that's what sseraj called him.<br />
--[[User:Piphi|piphi]] ([[User talk:Piphi|talk]]) 14:15, 2 January 2020 (EST)<br />
<br />
But there is no more sseraj ARay 10 20:34, 3 January 2020 (EST)<br />
<br />
==Some of Gmass's facts==<br />
<br />
Notice, this is not all, as the real Gmass page has to be the biggest.<br />
<br />
<br />
<br />
<br />
<br />
gmaaas is our leader. To prove yourself worthy of GMAAS, you don't have to memorize facts: _________________________________________________<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. THE FACTS ABOUT THE GREAT EPIC AWESOME GMAAS: Words fail to describe the epic nature of GMAAS, for he is too almighty and powerful to be bound by the plebeian and trifling words. But even the Great Epic Awesome Plenipotentiary GMAAS cannot get rid of a language that's been around for a <math>999999999999999999999999999999999999999999^{9999999999999999999999999999999999999}</math> years and counting. That's older than he is. EDIT: It is not older than he is because Gmaas is infinity years old. EDIT EDIT: He has now<br />
<br />
1. GMAAS's theorem states that for any math problem, GMAAS knows the answer to it. This theorem was proved by GMAAS. But then GMAAS forgot about the theorem so later, the mathematician named ARay10 proved it again: [b]Proof of the MAAS theorem[/b]: The GMAAS theorem states that for every math problem, GMAAS knows the answer. Using the 3.141592653589793238462643383... GMAAS theorem, stating that "26. GMAAS's Theorem states that GMAAS knows the answer to any math problem. EDIT: That theorem was proved by GMAAS. EDIT: GMAAS's Theorem has real-world applications: because GMAAS knows the answer to any math problem, you can use GMAAS to solve math problems. GMAAS is busy, so he charges a fee of one dollar for 1,000,000 math problems. EDIT: The fee has gone up. It is now 1,000,000 dollars for one math problem. Gmaas's technician made a mistake and reciprocated the fee," you must pay <math>1,000,000</math> to solve a problem using the Games theorem. You wouldn't waste that much money to solve one problem. Therefore, I proved that you cannot use the Games theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Games can turn things into anything. Using that fact, you can use the Games theorem to solve any problem.<br />
<br />
2. The GMATS were supposed to be called the GMAAS, but the manager, games, wanted to eat some gnats. Games have stopped eating gnats because they taste dreadful. EDIT: He now eats gnats again. He thinks that they taste like pi(e). EDIT: He ate 3,141,592,653,589,793,238,462,643,383 so far.<br />
<br />
3. GMAAS's archenemy is John Wick. John Wick once gave Gmaas 31,415,926,535,897,932,384 Oren Berries, but Gmaas thought they were Oran Berries. That is why Games hates John Wick. Germans knew they were Oren Berries because Gmaas knows everything, but he decided to play along.<br />
<br />
4. Gmaas wants every fact to have a pi reference. Gmaas created pi. EDIT: He made it 3.1415926 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 games knows!!!) EDIT: His favorite is memedog EDIT EDIT: Gmaas does not like meme dog. Games was the one who told the AoPSSheriff to give warnings for posting meme dog.<br />
<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.EDIT: Gmass doesn't like music that much; it hurts his ears. EDIT EDIT: GMAAS is now confirmed to like al video game OSTs, especially Undertale. His favorite song is Megalovania. EDIT EDIT EDIT: The previous edit is false.<br />
<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: 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: But Thanos stole them after Gmaas died for 0.31415926535 seconds.<br />
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14. Gmaas killed Thanos with a swat of Gmaas's tail. Gmaas barfed and created a furball, which leads to a new dimension, where he put a newly revived Thanos to become a farmer when Thor aimed for the head. Gmaas got mad and made him fat. EDIT: The presence of Gmaas killed Thanos.<br />
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15. Gmaas won an infinite amount of games against AlphaGo and Gary Kasparov while eating dinner, chasing sseraj, and doing a handstand.<br />
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16. Gmaas owns a pet: sseraj. Gmaas likes to play chess with his pets. EDIT: Gmaas has long moved on from chess because he was too good for it. EDIT EDIT: Gmaas has now invented a new form of chess<br />
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17. Gmaas is so interesting that an entire science has been devoted to studying him: Gmaasology.<br />
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18. Gmaas is the only known living being who has a Ph.D. in Gmaasology.<br />
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19. Most universities, including Harvard, are beginning to offer MAs in Gmaasology.<br />
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20. Gmaasology is one of the most eminent fields of science, falling behind Physics, Biology, Chemistry, Economics, Geology, and Computer Programming.<br />
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21. Gmaas wants his AoPS Wiki page to be the longest ever. EDIT: Sadly, the Gmaas page is only the fourth-longest page on AoPS wiki. It has around 50,000 bytes. EDIT: The Gmaas page is getting closer and closer! Gmaas has beaten Primitive Pythagorean Triple and Proofs without words and is the second-longest AoPS Wiki page. It now has around 60,000 bytes. However, it will still be a challenger to overcome 2008 most iT Problems, which has around 73,000 bytes. EDIT: Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Gmaas's article has 89,119 bytes.<br />
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22. Gmaas has the longest lifespan of any cat. He has lived for 3,141,592,653,589,793,238,462,643,383,279 years. (He is older than the universe.)<br />
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23. Gmaas is the ancient Greek god of cats and catfish. EDIT: He is the god of everything<br />
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24. Gmaas can turn things into gold. EDIT: Gmaas can turn anything into anything.<br />
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25. Gmaas can eat lava. EDIT: Everyone can eat lava once. After you eat it once, you die. EDIT EDIT: Gmaas can eat anything.<br />
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26. Gmaas's Theorem states that Gmaas knows the answer to any math problem. EDIT: That theorem was proved by Gmaas EDIT: Gmaas's Theorem has real-world applications: because Gmaas knows the answer to any math problem, you can use Gmaas to solve math problems. Gmaas is busy, so he charges a fee of one dollar for 1,000,000 math problems. EDIT: The fee has gone up. It is now 1,000,000 dollars for one math problem. Gmaas's technician made a mistake and reciprocated the fee.<br />
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27. Gmaas can only be described as Gmaas. EDIT: Gmaas has an age, but his age changes all the time. Every second his age increases by one second.<br />
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28. Gmaas should be capitalized to show respect. EDIT: It is normal to capitalize on people's names. EDIT: Gmaas isn't a person, he is a divine entity that takes the form of a cat, and should, therefore, be worshipped.<br />
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29. The steps to summon Gmaas can be found at Talk: Gmaas.<br />
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30. Gmaas owns your AoPS account 31.415926% of the time. EDIT: There is an exception to this: Gmaas owns his own account 99.9999% of the time. The only time when Gmaas did not control his AoPS account was when he had slow internet. For Gmaas, "slow internet" happens when it takes more than a nanosecond to load a webpage. EDIT: Gmaas owns every AoPS account at some point. EDIT: You are controlled by Gmaas. So actually, Gmaas owns your AoPS account 100% of the time, and his AoPS account is owned by him 314% of the time.<br />
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31. Everyone, except for me, is Gmaas. EDIT: You are also Gmaas. Gmaas is not me nor you. Gmaas is in us all and also not in us all. Why? The power of Gmaas. Gmaas is an energy field created by all living things. It surrounds us and penetrates us. It binds the galaxy together.<br />
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32. Gmaas used to be a dog, but he didn't like to be a dog. So he became a cat.<br />
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33. Gmaas can be spotted in The Matrix at 31:41:59, metric time. EDIT: This fact is incorrect because there have only been 30 sightings, all of them inconsistent.<br />
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34. The Metropolitan Museum of Art and the Louvre are Gmaas's private art collections from 3:14 AM to 3:15 AM. EDIT: The only exception was on the 3141st day after its opening. EDIT: All the paintings in these museums are secretly portraits of Gmaas in his most glorious human/animal forms.<br />
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35. Gmaas is a Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is owned by Gmaas who is blah blah blah blah blah blah blah blah blah blah blah blah blah blah.<br />
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36. Gmaas says hello. EDIT: Gmaas does not speak; he only uses mind signals. Speaking is too primitive for Gmaas.<br />
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37. For one day, Gmaas was Richard Rusczyk, Gmaas, and David Patrick at the same time. It felt strange, so Gmaas stopped.<br />
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38. Gmaas, in addition to being the oldest cat in history, the most powerful cat in history, and the most confusing cat in history is also the largest cat in history.<br />
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39. Gmaas is dead--he was eaten for lunch. EDIT: The above sentence is incorrect. Gmaas has not sent a gamma-ray burst to destroy the planet, which has happened every time someone has eaten Gmaas.<br />
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40. Gmaas is the creator of the BCPI Ai project.<br />
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41. Gmaas exists in <math>2\pi^2</math> dimensions because he doesn't like string theory.<br />
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42. Christmas was supposed to be called Gmaasmas. Until it wasn't. EDIT: Christ is Gmaas. EDIT: Why? Because of the power of Gmaas.<br />
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43. <math>\pi</math> is a representation of how many pies Gmaas has eaten. EDIT: The number <math>\pi</math> was created by Gmaas. He took a ten-sided die and flipped it an infinite number of times. The numbers he rolled became the digits of <math>\pi</math>.<br />
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44. <math>e</math> is a representation of how many days Gmaas forgot to eat. EDIT: The number <math>e</math> was created by Gmaas. He took a book that has infinite pages and flipped to a random page. The digits of the page number became digits of <math>e</math>. EDIT: That is impossible. The previous fact would mean that <math>e</math> has the last digit, which it does not. EDIT: That is why Gmaas is still flipping to this day. EDIT: Why would he still be doing that? That would be boring. EDIT: Gmaas is not flipping the book now because he can flip an infinite amount of pages in <math>\frac{1}{\infty}</math> seconds. EDIT: Then why isn't he done? EDIT: The power of Gmaas. EDIT: The editors of the Gmaas article like to make long chains of edits, don't we? : Yes, we do. : I do too. So do I. EDIT: The number of edits only verifies how high-quality this holy bible of Gmaas is, because, with each edit, this bible becomes better. That is exactly why we are editing this. I will make one more edit, just to respect Gmaas EDIT: Gmass rules!<br />
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45. Gmaas is the creator of everything including nothing.<br />
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46. According to recent DNA tests, famous historical people, such as Euclid, Julius Caesar, Omar Khayyam, George Washington, and Ramanujan are Gmaas in disguise. Gmaas has been thousands of people; only 99.9% of them were important historical figures. EDIT: They aren't dead. They're still alive. They are all dead reincarnations of Gmaas. Gmaas only has one reincarnation at a time. He is right now a cat living in sseraj's house.<br />
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47. Gmaas is a cat. EDIT: How many times do we say this? EDIT: Very many times. EDIT: Gmaas can be anything, but he chooses to be a cat. EDIT: He has been a human dozen of times. Gmaas painted the Lascaux cave paintings.<br />
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48. Gmaas has the following powers: trout, carp, earthworm, and catfish. Gmaas never uses any of them because Gmaas has an infinite number of powers. EDIT: Gmaas has used his catfish power several times. EDIT: Gmaas once used all his powers 3,141,592,653,589,793,238,462,643,383,279 times in 1 second.<br />
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49. Gmaas was once spotted in Minecraft chewing a tree. EDIT: The tree broke. EDIT: Gmaas once broke a Nokia. EDIT: Gmaas can break anything except for Gmaas's logic. It is too strong. EDIT: Gmaas is strong.<br />
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50. Gmaas was spotted in Roblox eating a taco cat. EDIT: Tacocat is just what happens when Gmaas sheds. Shedding is annoying for Gmaas, so he sheds no more. Therefore, there are only 271,828,182 taco cats in the world. EDIT: However, Tacocats reproduce, so they are not in danger of extinction.<br />
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51. Gmaas would like to go to Taco Bell, but Gmaas goes to Wendy's instead. No one knows why. EDIT: Because Taco Bell doesn't serve tacocats.<br />
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52. Gmaas real name is Grayson Maas. He is the CEO of AoPS. EDIT: Gmaas's real name is Gmaas. He is not the CEO of AoPS. EDIT: Gmaas has always been the master of AoPS as he is AoPS.<br />
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53. Gmaas has a pet pufferfish named Pafferfash. EDIT: He also has a goldfish named Sylar.<br />
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54. Gmaas has colonized the universe. EDIT: Gmaas created the universe, so he is allowed to claim control over it. EDIT EDIT: Gmaas created every universe.<br />
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55. Some people think that Gmaas is human. However, this has never been proven. Many AoPSers believe Gmaas is a cat. EDIT: Gmaas is a cat. EDIT: Yeah, we've said that already. EDIT: It does not matter how many times we say it; it will always be true. EDIT: Gmaas is a cat. EDIT: Right now he is a cat, but many of his lives have been other species. EDIT: He has been a dog, as said before, but it was too boring.<br />
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56. Gmaas started Pastafarianism. But then converted to Catholicism because Gmaas knows all EDIT: In that religion, one can only eat pasta.<br />
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57. Gmaas could eat your hand, but he would not because hands taste bad.<br />
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58. According to the Interuniversal Gmaas Society, 17.548 percent of the universe's population thinks that Gmaas is spelled "Gmass". EDIT: In case you don't know already, the name Gmass is spelled Gmaas. Alternative spellings include GMAAS, gmaas, gmaas, Gmaas, Gmail, Maas, G. Maas, G. Mass, Gabriel Maas, Genius of Maas, General Maas, Greek Mass, and the big fluffy kitty who lives in sseraj's house. These are no longer accepted spellings, and Gmaas is the current acceptable spelling.<br />
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59. The Interuniversal Gmaas Society was founded in 1314 on May 9th at 2:06:53 PM. EDIT: Ever since then, Gmaas day has been celebrated on May 9th.<br />
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60. The Interuniversal Gmaas Society has just reconstructed a lost book about Gmaas from Lucretius's De Rerum Natura. They researched this for three years. EDIT: A few years ago the Society compiled a biography of Gmaas's last twenty lives. EDIT: The only copies of these biographies are locked in the Gmaasian Library beneath the Library of Congress. EDIT: See the Gmaasology page for more information on these projects and others made by the Interuniversal Gmaas Society.<br />
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61. The Interuniversal Gmaas society has found out all of the information and more. For details on the history of the Interuniversal Gmaas Society, see the Gmaathamatics page.<br />
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62. GMAAS likes to surprise unsuspecting people. People cannot surprise GMAAS because GMAAS knows what everyone is doing and will do.<br />
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63. Gmaas loves sparkly mechanical pencils. EDIT: Gmaas has eaten several mechanical pencils. EDIT: He absorbed them and became colorful for 0.271828182846 seconds. EDIT: Then, he came colorful for 3.1415926535897932384626433832 more seconds.<br />
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64. This page is Gmaas's holy book where people go to worship Gmaas in Maas. EDIT: This is an unusual holy where everyone edits it. EDIT: That is the point. Gmaas is too lazy to write or to hire someone to write his holy book, so he lets people write it for free. EDIT: Gmaas does not like being called lazy. Our language is simply too unimportant for him to waste time on.<br />
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65. Gmaas tastes like a furry meatball. EDIT: No one has ever tasted Gmaas's sacred body before. EDIT: Once, he was a catfish, and someone ate him for dinner. Gmaas was angry and the entire planet exploded. (This was on the planet, Demeter. It blew up into so many pieces that the Asteroid Belt was formed. The biggest asteroid in the belt is called Ceres, the Roman version of Demeter, in honor of the lost planet.)<br />
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66. Gmaas's favorite food is pepperoni pizza. EDIT: Gmaas's favorite food is turnips. EDIT: Gmaas hates catnip and turnips, and pepperoni. He only likes alien alienish H2So4. EDIT: He does love turnips. He has a secret turnip garden under sseraj's house.<br />
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67. Gmaas is Johnny Johnny's Papa. EDIT: We'll never know. EDIT: Gmaas caught Johnny Johnny eating sugar and lying. He is Johnny Johnny's Papa.<br />
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68. Gmaas is in you, and Gmaas is in you, and Gmaas is in me. EDIT: Gmaas is in all cats. EDIT: He is not in every cat. The Guinness Book of World Records has a cat that does not have any Gmaas. EDIT: Gmaas invented The Guinness Book of World Records. EDIT: Gmaas invented the world. EDIT: Gmaas will also destroy the world.<br />
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69. Gmaas always remembers not to destroy the universe. EDIT: Once he forgot and had to travel back in time to stop it.<br />
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70. Gmaas created many user's accounts. EDIT: All accounts on AoPS are Gmaas in disguise since you are Gmaas in disguise. Except for maybe Geoflex and CrazyEyeMoody.EDIT: Everyone is Gmass, even Geoflex and CrazyEyeMoody EDIT: How? EDIT: The power of Gmass.<br />
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71. Gmaas is both living and nonliving. EDIT: He is living 90% of the time. Every once in a while he takes a break and dies.<br />
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72. Gmaas cannot comprehend the stupidity of humans.<br />
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73. All hail the Gmaas cloud! EDIT: Gmaas is a cloud: a cloud of electrons and nuclei.<br />
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74. Some things are beyond possible human comprehension. Nothing is beyond Gmaas. EDIT: The only thing beyond Gmaas is his tail; he has never managed to eat it. EDIT: Once he ate it. It tasted bad, so he didn't ever eat it again.<br />
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75. Gmaas eats food and resides at King Arthur's throne when he feels like it. EDIT: King Arthur is dead. However, Welsh folk stories say that he will come back if the Welsh people are in trouble. EDIT: King Arthur lives on paper flour bags. He came back when the Welsh people didn't have enough bread. EDIT: Because Gmaas ate 31415926535897932984626433 loaves of bread.<br />
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76. Gmaas owns a rabbit. EDIT: Gmaas ate it on March 19, 2019. It reincarnated on the other side of the planet. EDIT: Gmaas ate it again on April 31, 2019.<br />
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77. Gmaas is both singular and plural. EDIT: It is usually singular. EDIT: The plural of Gmaas in English and Latin is Gmaases. (Gmaas is a third declension noun: Gmaas, Gmaasis, m.) EDIT: But there is no use for the plural, because there is only one Gmaas.<br />
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78. Gmaas eats disbelievers as if they were donuts for breakfast. (Yet I'm somehow still alive. Do you think Gmaas ate my soul?) EDIT: Yes, I do. Gmaas is just typing through your account. EDIT: Gmaas typed through everyone's account. EDIT: Gmaas owns most AoPSers, including me. So that's why I'm typing. EDIT: That is why everyone here is typing.<br />
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79. Gmaas knows Jon Snow's parents. EDIT: Gmaas was Jon Snow's parents for 0.3141592653589793238 seconds, but it felt weird. So he became Gmaas again.<br />
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80. All Gmaas article editors will be escorted to Gmaas heaven after they die. EDIT: I hope so because I edited this article. EDIT: Me too. EDIT EDIT: Me three. EDIT EDIT EDIT: That are good!<br />
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81. Gmaas won all the wars. EDIT: He did not win the Intergmaasian War, which was between Gmaas's head and his tail. His tail won. Two flees died in the war. EDIT: Stefán Karl Stefansson died in the war, which made Gmaas very sad. Because of this, Gmaas started the world peace movement.<br />
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82. Gmail was named after Gmaas. EDIT: Google was named after Gmaas. EDIT: Almost every word starting with G is named after Gmaas. EDIT: Every word starting with G is named after Gmaas.<br />
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83. Gmaas ate cat-food today. EDIT: Gmaas also ate it yesterday. EDIT: Only because there was no alien alienish H2So4.<br />
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84. Gmaas won Battle For Dream Island and Total Drama Island. EDIT: Gmaas made Dream Island. And he ate it. It tasted like dirt.<br />
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85. Somehow Gmaas exists at all places at the same time. EDIT: Gmaas does not exist in my kitchen. EDIT: I'll just go check. Aaaaaaaaah! He is in my kitchen. He is eating everything! EDIT: Seriously? Oh ya, the power of Gmass.<br />
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86. Gmaas can lift with Gmaas's will. EDIT: Gmaas can lift with no one's will while he is sleepwalking.<br />
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87. Gmaas is the rightful heir to the Iron Throne. EDIT: Gmaas made the Iron Throne.<br />
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88. Gmaas is Teemo in League of Legends because Gmaas made LoL and they made an honorary Gmaas character. EDIT: LoL must be used in this article more than once. EDIT EDIT: It is. In fact, twice in this message.<br />
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89. Gmaas started the Game of Thrones. EDIT: Gmaas will end the Game of Thrones.<br />
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90. Gmaas has killed himself hundreds of times. He was reincarnated as a different species each time. EDIT: Gmaas has only died two times. Other times he was putting on a magic show. The kids were very impressed. EDIT: Gmaas has died and reincarnated thousands of times.<br />
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91. Gmaas created everything after puking. EDIT: He could not have puked because he is a god. Gods do not do that. EDIT: Gmaas does that. EDIT: Why? EDIT: The power of Gmaas. EDIT: He almost never does it, only if he wants to.. Also he did not puke and made everything. EDIT: No, because, GMAAS.<br />
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92. Gordan's last name was named after Gmaas. EDIT: But Gmaas did not want people disrupting his beauty sleep. So he changed it.<br />
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93. Gmaas is more powerful than Gohan.<br />
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94. Gmaas is over 90000 years old. EDIT: Gmaas is trillions of years old.<br />
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95. Gmaas has 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 cat lives, maybe even more. Gmaas has 0 dog lives. EDIT: He has 314,159,265 fish lives. EDIT: Out of Gmaas's 314,159,265 fish lives, 271,828,182 are catfish lives. He has used up 141,421,356 of them. EDIT: Gmaas has infinitely lives of all types. EDIT: How? EDIT: The power of Gmaas.<br />
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96. Who wrote Harry Potter? None other than Gmaas himself. EDIT: That is incorrect. EDIT: The above post is irrelevant. Gmaas created J. K. Rowling. Therefore, he created Harry Potter. EDIT: Gmaas has strong logic. No one can break it. Not even Gmaas himself. He is still trying to this day. EDIT: Why? EDIT: The power of Gmass.<br />
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97. Gmaas created the catfish. EDIT: See a few posts above for more information about Gmaas and catfish.<br />
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98. Gmaas has proven that the universe is infinite by traveling to the edge of the universe in a second. EDIT: He has never repeated this experiment because Gmaas is busy and has better things to do.<br />
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99. Gmaas founded Target, but then Gmaas sued them for making the mascot look like a dog when it was supposed to look like Gmaas. EDIT: They went broke because Gmaas sued them but then Gmaas ate a fudge popsicle that made him super hyper and he made Target not broke anymore in his hyperness.<br />
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100. Everyone has a bit of Gmaas inside them. EDIT: I don't. EDIT: Yes, you do. EDIT: Gmass: LoL<br />
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101. Gmaas likes to eat popsicles. Especially the fudge ones that get him hyper. EDIT: Gmaas is a popsicle. EDIT: Then how come Gmaas hasn't melted? EDIT: The power of Gmaas.<br />
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102. When Gmaas is hyper, he runs across Washington D.C. and grabs unsuspecting pedestrians, steals their phones, hacks into them, and downloads PubG onto their phone.<br />
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103. Gmaas's favorite cereal is fruit loops. Gmaas thinks it tastes like unicorns jumping on rainbows. EDIT: Gmaas eats unicorns jumping on rainbows like a toddler eats Goldfish.<br />
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104. Gmaas thinks that the McChicken has too much mayonnaise. EDIT: Gmaas thinks McDonald's is not good enough for him. He like KFC better.<br />
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105. Gmaas is a champion pillow-fighter. EDIT: Gmaas invented pillows.<br />
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106. Gmaas colonized Mars. EDIT: Gmaas also colonized Jupiter, Pluto, and several other galaxies. Gmaas cloned little Gmaas robots (with Gmaas's amazingly robotic skill of coding) and put them all over a galaxy called Gmaasalaxy. EDIT: Gmaas has colonized the whole universe.<br />
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This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam ==== This article is spam This article is spam 107. Gmaas can make every device play "The Duck Song" at will. EDIT: "The Duck Song" was copied off of the "Gmaas song," but the animators though Gmaas wasn't catchy enough.<br />
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108. Gmaas once caught the red dot and ate it. EDIT: Gmaas is a red dot.<br />
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109. Gmaas's favorite color is Gmaasian blue, a very rare blue that shines with the brightness of lightning. EDIT: it's <math>1000^{1000}</math> times brighter than lightning. Gmaasian blue occurs when a meteor hits the earth and usually is only seen for a few seconds. Only Gmaas has good enough eyesight to see it for that short a time.<br />
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110. Gmaas can create wormholes and false vacuums. EDIT: He is made out of exotic matter.<br />
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111. Gmaas is a champion PVP Minecraft player.<br />
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112. Gmaas doesn't like tacos. The last time he tried one he turned into a mouse and then caught himself.<br />
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113. Gmaas is the coach of True Ninja Music and Myth.<br />
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114. Gmaas caught a CP 6000 Mewtwo with a normal Pokeball in Pokemon Go.<br />
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115. Gmaas founded Costco.<br />
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116. Gmaas does not need to attend the FIFA World Cup. If Gmaas did, he would start the game with a goal and break the ankles of everyone watching the World Cup, including you, at the same time in a fraction of a second, even if you are watching from a device. EDIT: Gmaas created your device. EDIT: Gmaas is your device.<br />
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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: 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: Gmaas has a winter house on Jupiter. EDIT: Gmaas has a spring house on Earth. EDIT: Gmaas also experiences a fifth season called wintrautumn. It happens during November and December and is the cross between fall and winter. Everything is dreary, but there is no snow. Gmaas spends wintrautumn in his house on Venus, where he is incinerated daily. He reincarnates every night through the power of Gmaas. EDIT: Gmaas invented a new season on Feb. 31, 2019, called winter-summer. It is spring but transcends spring.<br />
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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: 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: 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: How? 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 />
<br />
136. I am Gmaas. EDIT: No you are not. You only have part of Gmaas inside of you. EDIT EDIT: I am also Gmaas. EDIT EDIT EDIT: But it is I who am Gmaas. EDIT EDIT EDIT EDIT: Gmaas is in us all. EDIT EDIT EDIT EDIT EDIT: Gmaas is all of us yet none of us. EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas is a cat. EDIT EDIT EDIT EDIT EDIT EDIT EDIT: He is only in a cat form right now. He can be whatever he wants to be. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: This chain of edits has superseded another chain of edits on this page for the record of the longest chain of edits on AoPS Wiki. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: But the edits in this chain tend to be shorter than most. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: You're right. But we sure do love strings of edits, don't we? EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas makes the edits. He is in your head spinning edits in your brain. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas is made to be edited. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas pays people to edit. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Some people edit even without Gmaas's payment. They do it because they are believers. EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT EDIT: Gmaas is Gmaas. You are not Gmaas. Gmaas does not pay people to edit; we do it because we want to.<br />
<br />
137. In 2018 Gmaas once challenged Magnus Carlsen to a chess game. Gmaas won every round. EDIT: Gmaas ate the chess pieces afterward. They tasted funny.<br />
<br />
All hail Gmaas!<br />
<br />
P.S. Gmaas won USAMO 5 times in a row. That's because he invented USAMO. When Gmaas took it nobody knew about it and he was the only one who took it. So of course, he won. How dare you spell Gmaas with a lowercase G? He is the supreme and awesome ruler of the universe!<br />
<br />
P.P.S. Gmaas invented the word Yeet.<br />
<br />
P.P.P.S. Gmass is the first person who played the game YECK (Moth poth 2019 reference)<br />
<br />
P.P.P.P.S. Gmaas is a non-human who can snap anyone out of existence without anything.<br />
<br />
P.P.P.P.P.S. Gmaas was the first person who did math.<br />
<br />
P.P.P.P.P.P.S. Gmaas invented proof by dictatorship.<br />
<br />
P.P.P.P.P.P.P.S Gmaas invented proof by procrastination.<br />
<br />
P.P.P.P.P.P.P.P.S Gmaas invented proofs!<br />
<br />
P.P.P.P.P.P.P.P.P.S Gmaas invented everything.<br />
<br />
P.P.P.P.P.P.P.P.P.P.S Gmaas did not invent roblox. Gmaas invented Minecraft.<br />
<br />
==Notice(Gmass's page)==<br />
<br />
Most of the starting facts are false. Plz don't post these false facts, 1 of the reasons 5space locked it(along with blanking)</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=User:Shurong.ge&diff=115711User:Shurong.ge2020-01-27T23:45:24Z<p>Shurong.ge: /* ThriftyPiano */</p>
<hr />
<div>==About Me==<br />
Go onto my user page to learn about me.<br />
Here are some quick facts:<br />
I am a human<br />
I have my own AoPS wiki page<br />
<br />
==What I am learning==<br />
At AoPS, I am taking Geometry, Algebra 2, and 6th grade LA at the Bellevue campus.<br />
<br />
==Other Websites I Use==<br />
Edabit<br />
Repl<br />
<br />
==Closest Friends on AoPS==<br />
===ThriftyPiano===<br />
Thriftypiano in real life earned 1st place in a state wide math competition, the following section is an excerpt from Thriftypiano's intro speech:<br />
<br />
"Thriftypiano is unknown. Nobody knows where he is, who he is, why he exists, and what he is. All we know is that he exists and is male.<br />
He created this song:<br />
When I was a little biddy boy<br />
My grandmother bought me a cute little toy<br />
Silver bells hanging on a string<br />
She told me it was my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
Then momma took me to Grammar school<br />
Always went by the best of rule *<br />
Everytime that bell would ring<br />
Catch me playing with my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while climbing the garden wall,<br />
I slipped and had a terrible fall<br />
I fell so hard I heard bells ring,<br />
But I held on to My ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while swimming cross turtle creek<br />
Man them snappers all 'round at my feet<br />
Was so hard swimming cross that thing<br />
with both hands holding my dingaling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
And this song:<br />
A duck walked up to a lemonade stand<br />
And he said to the man, running the stand<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said<br />
"No we just sell lemonade. But it's cold<br />
And it's fresh<br />
And it's all home-made. Can I get you<br />
Glass?"<br />
The duck said,<br />
"I'll pass".<br />
Then he waddled away.<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-bada-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?<br />
The man said,<br />
"No, like I said yesterday<br />
We just sell lemonade OK?<br />
Why not give it a try?"<br />
The duck said,<br />
"Goodbye."good day<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (bum bum bum) Got any grapes?<br />
The man said,<br />
Look, this is getting old.<br />
I mean, lemonade's all we've ever sold.<br />
Why not give it a go?"<br />
The duck said,<br />
"How 'bout, no."<br />
Then he waddled away<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said,<br />
"THAT'S IT!<br />
If you don't stay away, duck,<br />
I'll glue you to a tree and leave you there all day, stuck<br />
So don't get to close!"<br />
The duck said,<br />
"Adios."<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man that was running the stand,<br />
"Hey! (Bum bum bum) got any glue?"<br />
"What"<br />
"Got any glue?"<br />
"No, why would I– oh!"<br />
And one more question for you;<br />
"Got any grapes?"<br />
(Bum bum bum, bum bum bum)<br />
And the man just stopped.<br />
Then he started to smile.<br />
He started to laugh.<br />
He laughed for a while.<br />
He said,<br />
"Come on duck, let's walk to the store.<br />
I'll buy you some grapes<br />
So you won't have to ask anymore."<br />
So they walked to the store<br />
And the man bought some grapes.<br />
He gave one to the duck and the duck said,<br />
"Hmmm..No thanks. But you know what sounds good?<br />
It would make my day.<br />
Do you think this store<br />
Do you think this store<br />
Do you think this store has any lemonade?"<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)" -ThriftyPiano<br />
<br />
===Potato2017===<br />
He is a 5th grader, but has an IQ of 190. Thriftypiano AND the Stanford test says so.<br />
<br />
Edited by Thrifty Piano<br />
<br />
===Bluesky11===<br />
He is a very good friend, it's sad we got split into different middle schools...<br />
<br />
==Coding==<br />
Yes, I code in Python and Java<br />
<br />
==Makers of This Page==<br />
Page created by Shurong.ge; edited by ThriftyPiano</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=User:Shurong.ge&diff=115710User:Shurong.ge2020-01-27T23:44:54Z<p>Shurong.ge: </p>
<hr />
<div>==About Me==<br />
Go onto my user page to learn about me.<br />
Here are some quick facts:<br />
I am a human<br />
I have my own AoPS wiki page<br />
<br />
==What I am learning==<br />
At AoPS, I am taking Geometry, Algebra 2, and 6th grade LA at the Bellevue campus.<br />
<br />
==Other Websites I Use==<br />
Edabit<br />
Repl<br />
<br />
==Closest Friends on AoPS==<br />
===ThriftyPiano===<br />
Thriftypiano in real life did not earn 1st place in a state wide math competition, the following section is an excerpt from Thriftypiano's intro speech:<br />
<br />
"Thriftypiano is unknown. Nobody knows where he is, who he is, why he exists, and what he is. All we know is that he exists and is male.<br />
He created this song:<br />
When I was a little biddy boy<br />
My grandmother bought me a cute little toy<br />
Silver bells hanging on a string<br />
She told me it was my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
Then momma took me to Grammar school<br />
Always went by the best of rule *<br />
Everytime that bell would ring<br />
Catch me playing with my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while climbing the garden wall,<br />
I slipped and had a terrible fall<br />
I fell so hard I heard bells ring,<br />
But I held on to My ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while swimming cross turtle creek<br />
Man them snappers all 'round at my feet<br />
Was so hard swimming cross that thing<br />
with both hands holding my dingaling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
And this song:<br />
A duck walked up to a lemonade stand<br />
And he said to the man, running the stand<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said<br />
"No we just sell lemonade. But it's cold<br />
And it's fresh<br />
And it's all home-made. Can I get you<br />
Glass?"<br />
The duck said,<br />
"I'll pass".<br />
Then he waddled away.<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-bada-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?<br />
The man said,<br />
"No, like I said yesterday<br />
We just sell lemonade OK?<br />
Why not give it a try?"<br />
The duck said,<br />
"Goodbye."good day<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (bum bum bum) Got any grapes?<br />
The man said,<br />
Look, this is getting old.<br />
I mean, lemonade's all we've ever sold.<br />
Why not give it a go?"<br />
The duck said,<br />
"How 'bout, no."<br />
Then he waddled away<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said,<br />
"THAT'S IT!<br />
If you don't stay away, duck,<br />
I'll glue you to a tree and leave you there all day, stuck<br />
So don't get to close!"<br />
The duck said,<br />
"Adios."<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man that was running the stand,<br />
"Hey! (Bum bum bum) got any glue?"<br />
"What"<br />
"Got any glue?"<br />
"No, why would I– oh!"<br />
And one more question for you;<br />
"Got any grapes?"<br />
(Bum bum bum, bum bum bum)<br />
And the man just stopped.<br />
Then he started to smile.<br />
He started to laugh.<br />
He laughed for a while.<br />
He said,<br />
"Come on duck, let's walk to the store.<br />
I'll buy you some grapes<br />
So you won't have to ask anymore."<br />
So they walked to the store<br />
And the man bought some grapes.<br />
He gave one to the duck and the duck said,<br />
"Hmmm..No thanks. But you know what sounds good?<br />
It would make my day.<br />
Do you think this store<br />
Do you think this store<br />
Do you think this store has any lemonade?"<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)" -ThriftyPiano<br />
<br />
===Potato2017===<br />
He is a 5th grader, but has an IQ of 190. Thriftypiano AND the Stanford test says so.<br />
<br />
Edited by Thrifty Piano<br />
<br />
===Bluesky11===<br />
He is a very good friend, it's sad we got split into different middle schools...<br />
<br />
==Coding==<br />
Yes, I code in Python and Java<br />
<br />
==Makers of This Page==<br />
Page created by Shurong.ge; edited by ThriftyPiano</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=User:Shurong.ge&diff=115709User:Shurong.ge2020-01-27T23:42:55Z<p>Shurong.ge: /* ThriftyPiano */</p>
<hr />
<div>==About Me==<br />
Go onto my user page to learn about me.<br />
Here are some quick facts:<br />
*I am a human<br />
*I have my own AoPS wiki page<br />
<br />
==What I am learning==<br />
At AoPS, I am taking Geometry, Algebra 2, and 6th grade LA at the Bellevue campus.<br />
<br />
==Other Websites I Use==<br />
*Edabit<br />
*Repl<br />
<br />
==Closest Friends on AoPS==<br />
===ThriftyPiano===<br />
<br />
The following section is an excerpt from [[Thriftypiano]]'s (dumb) intro speech:<br />
<br />
"Thriftypiano is unknown. Nobody knows where he is, who he is, why he exists, and what he is. All we know is that he exists and is male.<br />
He created this song:<br />
When I was a little biddy boy<br />
My grandmother bought me a cute little toy<br />
Silver bells hanging on a string<br />
She told me it was my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
Then momma took me to Grammar school<br />
Always went by the best of rule *<br />
Everytime that bell would ring<br />
Catch me playing with my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while climbing the garden wall,<br />
I slipped and had a terrible fall<br />
I fell so hard I heard bells ring,<br />
But I held on to My ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while swimming cross turtle creek<br />
Man them snappers all 'round at my feet<br />
Was so hard swimming cross that thing<br />
with both hands holding my dingaling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
And this song:<br />
A duck walked up to a lemonade stand<br />
And he said to the man, running the stand<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said<br />
"No we just sell lemonade. But it's cold<br />
And it's fresh<br />
And it's all home-made. Can I get you<br />
Glass?"<br />
The duck said,<br />
"I'll pass".<br />
Then he waddled away.<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-bada-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?<br />
The man said,<br />
"No, like I said yesterday<br />
We just sell lemonade OK?<br />
Why not give it a try?"<br />
The duck said,<br />
"Goodbye."good day<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (bum bum bum) Got any grapes?<br />
The man said,<br />
Look, this is getting old.<br />
I mean, lemonade's all we've ever sold.<br />
Why not give it a go?"<br />
The duck said,<br />
"How 'bout, no."<br />
Then he waddled away<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said,<br />
"THAT'S IT!<br />
If you don't stay away, duck,<br />
I'll glue you to a tree and leave you there all day, stuck<br />
So don't get to close!"<br />
The duck said,<br />
"Adios."<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man that was running the stand,<br />
"Hey! (Bum bum bum) got any glue?"<br />
"What"<br />
"Got any glue?"<br />
"No, why would I– oh!"<br />
And one more question for you;<br />
"Got any grapes?"<br />
(Bum bum bum, bum bum bum)<br />
And the man just stopped.<br />
Then he started to smile.<br />
He started to laugh.<br />
He laughed for a while.<br />
He said,<br />
"Come on duck, let's walk to the store.<br />
I'll buy you some grapes<br />
So you won't have to ask anymore."<br />
So they walked to the store<br />
And the man bought some grapes.<br />
He gave one to the duck and the duck said,<br />
"Hmmm..No thanks. But you know what sounds good?<br />
It would make my day.<br />
Do you think this store<br />
Do you think this store<br />
Do you think this store has any lemonade?"<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)" -ThriftyPiano<br />
<br />
===Potato2017===<br />
He is a 5th grader, but has an IQ of 190. Thriftypiano AND the Stanford test says so.<br />
<br />
Edited by Thrifty Piano<br />
<br />
===Bluesky11===<br />
He is a very good friend, it's sad we got split into different middle schools...<br />
<br />
===RaymondZhu===<br />
He got first place in that 4th grade statewide math competition, him and Potato2017 had a tie-breaker for first, he came out victorious. He is currently in 2 schools at once because of his legendary math skills.<br />
<br />
==Coding==<br />
Yes, I code in Python and Java<br />
<br />
==Makers of This Page==<br />
Page created by Shurong.ge; edited by ThriftyPiano and RaymondZhu.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=User:Shurong.ge&diff=115708User:Shurong.ge2020-01-27T23:41:59Z<p>Shurong.ge: /* Closest Friends on AoPS */</p>
<hr />
<div>==About Me==<br />
Go onto my user page to learn about me.<br />
Here are some quick facts:<br />
*I am a human<br />
*I have my own AoPS wiki page<br />
<br />
==What I am learning==<br />
At AoPS, I am taking Geometry, Algebra 2, and 6th grade LA at the Bellevue campus.<br />
<br />
==Other Websites I Use==<br />
*Edabit<br />
*Repl<br />
<br />
==Closest Friends on AoPS==<br />
===ThriftyPiano===<br />
<br />
[[Thriftypiano]] in real life did not earn 1st place in a 4th grade statewide math competition, the following section is an excerpt from [[Thriftypiano]]'s (dumb) intro speech:<br />
<br />
"Thriftypiano is unknown. Nobody knows where he is, who he is, why he exists, and what he is. All we know is that he exists and is male.<br />
He created this song:<br />
When I was a little biddy boy<br />
My grandmother bought me a cute little toy<br />
Silver bells hanging on a string<br />
She told me it was my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
Then momma took me to Grammar school<br />
Always went by the best of rule *<br />
Everytime that bell would ring<br />
Catch me playing with my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while climbing the garden wall,<br />
I slipped and had a terrible fall<br />
I fell so hard I heard bells ring,<br />
But I held on to My ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while swimming cross turtle creek<br />
Man them snappers all 'round at my feet<br />
Was so hard swimming cross that thing<br />
with both hands holding my dingaling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
And this song:<br />
A duck walked up to a lemonade stand<br />
And he said to the man, running the stand<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said<br />
"No we just sell lemonade. But it's cold<br />
And it's fresh<br />
And it's all home-made. Can I get you<br />
Glass?"<br />
The duck said,<br />
"I'll pass".<br />
Then he waddled away.<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-bada-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?<br />
The man said,<br />
"No, like I said yesterday<br />
We just sell lemonade OK?<br />
Why not give it a try?"<br />
The duck said,<br />
"Goodbye."good day<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (bum bum bum) Got any grapes?<br />
The man said,<br />
Look, this is getting old.<br />
I mean, lemonade's all we've ever sold.<br />
Why not give it a go?"<br />
The duck said,<br />
"How 'bout, no."<br />
Then he waddled away<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said,<br />
"THAT'S IT!<br />
If you don't stay away, duck,<br />
I'll glue you to a tree and leave you there all day, stuck<br />
So don't get to close!"<br />
The duck said,<br />
"Adios."<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man that was running the stand,<br />
"Hey! (Bum bum bum) got any glue?"<br />
"What"<br />
"Got any glue?"<br />
"No, why would I– oh!"<br />
And one more question for you;<br />
"Got any grapes?"<br />
(Bum bum bum, bum bum bum)<br />
And the man just stopped.<br />
Then he started to smile.<br />
He started to laugh.<br />
He laughed for a while.<br />
He said,<br />
"Come on duck, let's walk to the store.<br />
I'll buy you some grapes<br />
So you won't have to ask anymore."<br />
So they walked to the store<br />
And the man bought some grapes.<br />
He gave one to the duck and the duck said,<br />
"Hmmm..No thanks. But you know what sounds good?<br />
It would make my day.<br />
Do you think this store<br />
Do you think this store<br />
Do you think this store has any lemonade?"<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)" -ThriftyPiano<br />
<br />
===Potato2017===<br />
He is a 5th grader, but has an IQ of 190. Thriftypiano AND the Stanford test says so.<br />
<br />
Edited by Thrifty Piano<br />
<br />
===Bluesky11===<br />
He is a very good friend, it's sad we got split into different middle schools...<br />
<br />
===RaymondZhu===<br />
He got first place in that 4th grade statewide math competition, him and Potato2017 had a tie-breaker for first, he came out victorious. He is currently in 2 schools at once because of his legendary math skills.<br />
<br />
==Coding==<br />
Yes, I code in Python and Java<br />
<br />
==Makers of This Page==<br />
Page created by Shurong.ge; edited by ThriftyPiano and RaymondZhu.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=1156872018 AMC 10B Problems/Problem 232020-01-27T20:35:30Z<p>Shurong.ge: Undo revision 115686 by Shurong.ge (talk)</p>
<hr />
<div>==Problem==<br />
<br />
How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation <br />
<cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath><br />
where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple?<br />
<br />
<math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math><br />
<br />
<br />
==Solution==<br />
Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes<br />
<br />
<cmath>x\cdot y + 63 = 20x + 12y</cmath><br />
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath><br />
<br />
Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as<br />
<br />
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath><br />
<cmath>(x - 12)(y - 20) = 177</cmath><br />
<br />
From here we can already see that this is a quadratic, and thus must have <math>2</math> solutions. But, let's continue, to see if one of the solutions is extraneous.<br />
<br />
Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the <math>\text{GCD}</math> must be a divisor of the <math>\text{LCM}</math>, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions.<br />
(awesomeag)<br />
<br />
Edited by IronicNinja, Firebolt360, and mprincess0229~<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=JWGHYUeOx-k<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=1156862018 AMC 10B Problems/Problem 232020-01-27T20:35:03Z<p>Shurong.ge: </p>
<hr />
<div>==Problem==<br />
<br />
How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation <br />
<cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath><br />
where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple?<br />
<br />
<math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math><br />
<br />
<br />
==Solution==<br />
Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes<br />
<br />
<cmath>x\cdot y + 63 = 20x + 12y</cmath><br />
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath><br />
<br />
Using [[Simon's Favorite Factoring Trick]]or SFFT, we rewrite this equation as<br />
<br />
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath><br />
<cmath>(x - 12)(y - 20) = 177</cmath><br />
<br />
From here we can already see that this is a quadratic, and thus must have <math>2</math> solutions. But, let's continue, to see if one of the solutions is extraneous.<br />
<br />
Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the <math>\text{GCD}</math> must be a divisor of the <math>\text{LCM}</math>, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions.<br />
(awesomeag)<br />
<br />
Edited by IronicNinja, Firebolt360, and mprincess0229~<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=JWGHYUeOx-k<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=1156852018 AMC 10B Problems/Problem 232020-01-27T20:32:41Z<p>Shurong.ge: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation <br />
<cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath><br />
where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple?<br />
<br />
<math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math><br />
<br />
<br />
==Solution==<br />
Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes<br />
<br />
<cmath>x\cdot y + 63 = 20x + 12y</cmath><br />
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath><br />
<br />
Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as<br />
<br />
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath><br />
<cmath>(x - 12)(y - 20) = 177</cmath><br />
<br />
From here we can already see that this is a quadratic, and thus must have <math>2</math> solutions. But, let's continue, to see if one of the solutions is extraneous.<br />
<br />
Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the <math>\text{GCD}</math> must be a divisor of the <math>\text{LCM}</math>, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions.<br />
(awesomeag)<br />
<br />
Edited by IronicNinja, Firebolt360, and mprincess0229~<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=JWGHYUeOx-k<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_16&diff=1156842018 AMC 10B Problems/Problem 162020-01-27T20:30:49Z<p>Shurong.ge: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>a_1,a_2,\dots,a_{2018}</math> be a strictly increasing sequence of positive integers such that <cmath>a_1+a_2+\cdots+a_{2018}=2018^{2018}.</cmath><br />
What is the remainder when <math>a_1^3+a_2^3+\cdots+a_{2018}^3</math> is divided by <math>6</math>?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
==Solution 1==<br />
<br />
One could simply list out all the residues to the third power <math>\mod 6</math>. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent <math>\mod 6</math>. This is due to the fact that <math>a_k</math> need not be relatively prime to <math>6</math>.)<br />
<br />
Therefore the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}</math><br />
<br />
Note from Williamgolly: we can wlog assume <math>a_1,a_2... a_2017=0(mod6)</math> and have <math>a_2018=2(mod6)</math> to make life easier<br />
<br />
==Solution 2==<br />
<br />
Note that <math>\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k</math><br />
<br />
Note that <math><br />
a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018})<br />
\equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6<br />
</math><br />
Therefore, <math>-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}</math>.<br />
<br />
Thus, <math>a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3</math>. However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is <math>\boxed{\text{(E) }4}</math><br />
<br />
==Solution 3 (Partial Proof)==<br />
First, we can assume that the problem will have a consistent answer for all possible values of <math>a_1</math>. For the purpose of this solution, we will assume that <math>a_1 = 1</math>.<br />
<br />
We first note that <math>1^3+2^3+...+n^3 = (1+2+...+n)^2</math>. So what we are trying to find is what <math>\left(2018^{2018}\right)^2=\left(2018^{4036}\right)</math> mod <math>6</math>. We start by noting that <math>2018</math> is congruent to <math>2 \pmod{6}</math>. So we are trying to find <math>\left(2^{4036}\right) \pmod{6}</math>. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of <math>2</math> and see that <math>2^1</math> is <math>2</math> mod <math>6</math>, <math>2^2</math> is <math>4</math> mod <math>6</math>, <math>2^3</math> is <math>2</math> mod <math>6</math>, <math>2^4</math> is <math>4</math> mod <math>6</math>, and so on... So we see that since <math>\left(2^{4036}\right)</math> has an even power, it must be congruent to <math>4 \pmod{6}</math>, thus giving our answer <math>\boxed{\text{(E) }4}</math>. You can prove this pattern using mods. But I thought this was easier.<br />
<br />
-TheMagician<br />
<br />
==Solution 4 (Lazy solution)==<br />
First, we can assume that the problem will have a consistent answer for all possible values of <math>a_1</math>. For the purpose of this solution, assume <math>a_1, a_2, ... a_{2017}</math> are multiples of 6 and find <math>2018^{2018} \pmod{6}</math> (which happens to be <math>4</math>). Then <math>{a_1}^3 + ... + {a_{2018}}^3</math> is congruent to <math>64 \pmod{6}</math> or just <math>4</math>. <br />
<br />
-Patrick4President<br />
<br />
==Solution 5 (Fermat's Little Theorem)==<br />
<br />
First note that each <math>a_{i}^3 \equiv a_{i} \pmod 3</math> by Fermat's Little Theorem. This implies that <math>a_{1}^3+...+a_{2018}^3 \equiv a_{1}+...+a_{2018} \equiv 2^{2018} \equiv 1 \pmod{3}</math>. Also, all <math>a_{i}^2 \equiv a_{i} \pmod{2}</math>, hence <math>a_{i}^3 \equiv (a_{i})(a_{i}^2) \equiv a_{i}^2 \equiv a_{i} \pmod{2}</math> by Fermat's Little Theorem.Thus, <math>a_{1}^3+...a_{2018}^3 \equiv 2^{2018} \equiv 0 \pmod{2}</math>. Now set <math>x=a_{1}^3+...+a_{2018}^3</math>. Then, we have the congruences <math>x \equiv 0 \pmod{2}</math> and <math>x \equiv 1 \pmod{3}</math>. By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that <math>x \equiv 4 \pmod{6}</math>. Thus, the answer is <math>\boxed{ (E)4}</math><br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_16&diff=1156832018 AMC 10B Problems/Problem 162020-01-27T20:29:18Z<p>Shurong.ge: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>a_1,a_2,\dots,a_{2018}</math> be a strictly increasing sequence of positive integers such that <cmath>a_1+a_2+\cdots+a_{2018}=2018^{2018}.</cmath><br />
What is the remainder when <math>a_1^3+a_2^3+\cdots+a_{2018}^3</math> is divided by <math>6</math>?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
==Solution 1==<br />
<br />
One could simply list out all the residues to the third power <math>\mod 6</math>. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent <math>\mod 6</math>. This is due to the fact that <math>a_k</math> need not be relatively prime to <math>6</math>.)<br />
<br />
Therefore the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}</math><br />
<br />
Note from Williamgolly: we can wlog assume <math>a_1,a_2... a_2017=0(mod6)</math> and have <math>a_2018=2(mod6)</math> to make life easier<br />
<br />
==Solution 2==<br />
<br />
Note that <math>\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k</math><br />
<br />
Note that <math><br />
a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018})<br />
\equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6<br />
</math><br />
Therefore, <math>-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}</math>.<br />
<br />
Thus, <math>a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3</math>. However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is <math>\boxed{\text{(E) }4}</math><br />
<br />
==Solution 3 (Partial Proof)==<br />
First, we can assume that the problem will have a consistent answer for all possible values of <math>a_1</math>. For the purpose of this problem, we will assume that <math>a_1 = 1</math><br />
<br />
We first note that <math>1^3+2^3+...+n^3 = (1+2+...+n)^2</math>. So what we are trying to find is what <math>\left(2018^{2018}\right)^2=\left(2018^{4036}\right)</math> mod <math>6</math>. We start by noting that <math>2018</math> is congruent to <math>2 \pmod{6}</math>. So we are trying to find <math>\left(2^{4036}\right) \pmod{6}</math>. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of <math>2</math> and see that <math>2^1</math> is <math>2</math> mod <math>6</math>, <math>2^2</math> is <math>4</math> mod <math>6</math>, <math>2^3</math> is <math>2</math> mod <math>6</math>, <math>2^4</math> is <math>4</math> mod <math>6</math>, and so on... So we see that since <math>\left(2^{4036}\right)</math> has an even power, it must be congruent to <math>4 \pmod{6}</math>, thus giving our answer <math>\boxed{\text{(E) }4}</math>. You can prove this pattern using mods. But I thought this was easier.<br />
<br />
-TheMagician<br />
<br />
==Solution 4 (Lazy solution)==<br />
Assume <math>a_1, a_2, ... a_{2017}</math> are multiples of 6 and find <math>2018^{2018} \pmod{6}</math> (which happens to be <math>4</math>). Then <math>{a_1}^3 + ... + {a_{2018}}^3</math> is congruent to <math>64 \pmod{6}</math> or just <math>4</math>. <br />
<br />
-Patrick4President<br />
<br />
==Solution 5 (Fermat's Little Theorem)==<br />
<br />
First note that each <math>a_{i}^3 \equiv a_{i} \pmod 3</math> by Fermat's Little Theorem. This implies that <math>a_{1}^3+...+a_{2018}^3 \equiv a_{1}+...+a_{2018} \equiv 2^{2018} \equiv 1 \pmod{3}</math>. Also, all <math>a_{i}^2 \equiv a_{i} \pmod{2}</math>, hence <math>a_{i}^3 \equiv (a_{i})(a_{i}^2) \equiv a_{i}^2 \equiv a_{i} \pmod{2}</math> by Fermat's Little Theorem.Thus, <math>a_{1}^3+...a_{2018}^3 \equiv 2^{2018} \equiv 0 \pmod{2}</math>. Now set <math>x=a_{1}^3+...+a_{2018}^3</math>. Then, we have the congruences <math>x \equiv 0 \pmod{2}</math> and <math>x \equiv 1 \pmod{3}</math>. By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that <math>x \equiv 4 \pmod{6}</math>. Thus, the answer is <math>\boxed{ (E)4}</math><br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_12&diff=1156822018 AMC 10A Problems/Problem 122020-01-27T19:33:42Z<p>Shurong.ge: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
How many ordered pairs of real numbers <math>(x,y)</math> satisfy the following system of equations?<br />
<cmath>x+3y=3</cmath> <br />
<cmath>\big||x|-|y|\big|=1</cmath><br />
<math>\textbf{(A) } 1 \qquad <br />
\textbf{(B) } 2 \qquad <br />
\textbf{(C) } 3 \qquad <br />
\textbf{(D) } 4 \qquad <br />
\textbf{(E) } 8 </math><br />
<br />
==Solutions==<br />
===Solution 1===<br />
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.<br />
<br />
The graph looks something like this:<br />
<asy><br />
draw((-3,0)--(3,0), Arrows);<br />
draw((0,-3)--(0,3), Arrows);<br />
draw((2,3)--(0,1)--(-2,3), blue);<br />
draw((-3,2)--(-1,0)--(-3,-2), blue);<br />
draw((-2,-3)--(0,-1)--(2,-3), blue);<br />
draw((3,-2)--(1,0)--(3,2), blue);<br />
draw((-3,2)--(3,0), red);<br />
dot((-3,2));<br />
dot((3/2,1/2));<br />
dot((0,1));<br />
</asy><br />
Now, it becomes clear that there are <math>\boxed{\textbf{(C) } 3}</math> intersection points.<br />
<br />
===Solution 2===<br />
<math>x+3y=3</math> can be rewritten to <math>x=3-3y</math>. Substituting <math>3-3y</math> for <math>x</math> in the second equation will give <math>||3-3y|-y|=1</math>. Splitting this question into casework for the ranges of <math>y</math> will give us the total number of solutions. <br />
<br />
<math>\textbf{Case 1:}</math> <math>y>1</math>:<br />
<math>3-3y</math> will be negative so <math>|3-3y| = 3y-3.</math><br />
<math>|3y-3-y| = |2y-3| = 1</math><br />
Subcase 1: <math>y>\frac{3}{2}</math><br />
<math>2y-3</math> is positive so <math>2y-3 = 1</math> and <math>y = 2</math> and <math>x = 3-3(2) = -3</math><br />
Subcase 2: <math>1<y<\frac{3}{2}</math><br />
<math>2y-3</math> is negative so <math>|2y-3| = 3-2y = 1</math>. <math>2y = 2</math> and so there are no solutions (<math>y</math> can't equal to <math>1</math>)<br />
<br />
<math>\textbf{Case 2:}</math> <math>y = 1</math>:<br />
It is fairly clear that <math>x = 0.</math><br />
<br />
<math>\textbf{Case 3:}</math> <math>y<1</math>:<br />
<math>3-3y</math> will be positive so <math>|3-3y-y| = |3-4y| = 1</math><br />
Subcase 1: <math>y>\frac{4}{3}</math><br />
<math>3-4y</math> will be negative so <math>4y-3 = 1</math> <math>\rightarrow</math> <math>4y = 4</math>. There are no solutions (again, <math>y</math> can't equal to <math>1</math>)<br />
Subcase 2: <math>y<\frac{4}{3}</math><br />
<math>3-4y</math> will be positive so <math>3-4y = 1</math> <math>\rightarrow</math> <math>4y = 2</math>. <math>y = \frac{1}{2}</math> and <math>x = \frac{3}{2}</math>.<br />
Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is <math>\boxed{\textbf{(C) } 3}</math>.<br />
<br />
===Solution 3===<br />
Note that <math>||x| - |y||</math> can take on either of four values: <math>x + y</math>, <math>x - y</math>, <math>-x + y</math>, <math>-x -y</math>. <br />
Solving the equations (by elimination, either adding the two equations or subtracting),<br />
we obtain the three solutions: <math>(0, 1)</math>, <math>(-3,2)</math>, <math>(1.5, 0.5)</math> so the answer is <math>\boxed{\textbf{(C) } 3}</math>. One of those equations overlap into <math>(0, 1)</math> so there's only 3 solutions.<br />
<br />
~trumpeter, ccx09<br />
~minor edit, XxHalo711<br />
<br />
===Solution 4===<br />
Just as in solution <math>2</math>, we derive the equation <math>||3-3y|-|y||=1</math>. If we remove the absolute values, the equation collapses into four different possible values. <math>3-2y</math>, <math>3-4y</math>, <math>2y-3</math>, and <math>4y-3</math>, each equal to either <math>1</math> or <math>-1</math>. Remember that if <math>P-Q=a</math>, then <math>Q-P=-a</math>. Because we have already taken <math>1</math> and <math>-1</math> into account, we can eliminate one of the conjugates of each pair, namely <math>3-2y</math> and <math>2y-3</math>, and <math>3-4y</math> and <math>4y-3</math>. Find the values of <math>y</math> when <math>3-2y=1</math>, <math>3-2y=-1</math>, <math>3-4y=1</math> and <math>3-4y=-1</math>. We see that <math>3-2y=1</math> and <math>3-4y=-1</math> give us the same value for <math>y</math>, so the answer is <math>\boxed{\textbf{(C) } 3}</math><br />
<br />
~Zeric Hang<br />
<br />
===Solution 5===<br />
Just as in solution <math>2</math>, we derive the equation <math>x=3-3y</math>. Squaring both sides in the second equation gives <math>x^2+y^2-2|xy|=1</math>. Putting <math>x=3-3y</math> and doing a little calculation gives <math>10y^2-18y+9-2|3y-3y^2|=1</math>. From here we know that <math>3y-3y^2</math> is either positive or negative.<br />
<br />
When positive, we get <math>2y^2-3y+1=0</math> and then, <math>y=1/2</math> or <math>y=1</math>.<br />
When negative, we get <math>y^2-3y+2=0</math> and then, <math>y=2</math> or <math>y=1</math>. Clearly, there are <math>3</math> different pairs of values and that gives us <math>\boxed{\textbf{(C) } 3}</math><br />
<br />
~OlutosinNGA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=11|num-a=13}}<br />
{{AMC12 box|year=2018|ab=A|num-b=9|num-a=11}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_12&diff=1156812018 AMC 10A Problems/Problem 122020-01-27T19:32:49Z<p>Shurong.ge: /* Solution 1 */ : Deletes unnecessary comment (vandalism)</p>
<hr />
<div>== Problem ==<br />
<br />
How many ordered pairs of real numbers <math>(x,y)</math> satisfy the following system of equations?<br />
<cmath>x+3y=3</cmath> <br />
<cmath>\big||x|-|y|\big|=1</cmath><br />
<math>\textbf{(A) } 1 \qquad <br />
\textbf{(B) } 2 \qquad <br />
\textbf{(C) } 3 \qquad <br />
\textbf{(D) } 4 \qquad <br />
\textbf{(E) } 8 </math><br />
<br />
==Solutions==<br />
===Solution 1===<br />
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.<br />
<br />
The graph looks something like this:<br />
<asy><br />
draw((-3,0)--(3,0), Arrows);<br />
draw((0,-3)--(0,3), Arrows);<br />
draw((2,3)--(0,1)--(-2,3), blue);<br />
draw((-3,2)--(-1,0)--(-3,-2), blue);<br />
draw((-2,-3)--(0,-1)--(2,-3), blue);<br />
draw((3,-2)--(1,0)--(3,2), blue);<br />
draw((-3,2)--(3,0), red);<br />
dot((-3,2));<br />
dot((3/2,1/2));<br />
dot((0,1));<br />
</asy><br />
Now, it becomes clear that there are <math>\boxed{\textbf{(C) } 3}</math> intersection points.<br />
<br />
===Solution 2===<br />
<math>x+3y=3</math> can be rewritten to <math>x=3-3y</math>. Substituting <math>3-3y</math> for <math>x</math> in the second equation will give <math>||3-3y|-y|=1</math>. Splitting this question into casework for the ranges of <math>y</math> will give us the total number of solutions. <br />
<br />
<math>\textbf{Case 1:}</math> <math>y>1</math>:<br />
<math>3-3y</math> will be negative so <math>|3-3y| = 3y-3.</math><br />
<math>|3y-3-y| = |2y-3| = 1</math><br />
Subcase 1: <math>y>\frac{3}{2}</math><br />
<math>2y-3</math> is positive so <math>2y-3 = 1</math> and <math>y = 2</math> and <math>x = 3-3(2) = -3</math><br />
Subcase 2: <math>1<y<\frac{3}{2}</math><br />
<math>2y-3</math> is negative so <math>|2y-3| = 3-2y = 1</math>. <math>2y = 2</math> and so there are no solutions (<math>y</math> can't equal to <math>1</math>)<br />
<br />
<math>\textbf{Case 2:}</math> <math>y = 1</math>:<br />
It is fairly clear that <math>x = 0.</math><br />
<br />
<math>\textbf{Case 3:}</math> <math>y<1</math>:<br />
<math>3-3y</math> will be positive so <math>|3-3y-y| = |3-4y| = 1</math><br />
Subcase 1: <math>y>\frac{4}{3}</math><br />
<math>3-4y</math> will be negative so <math>4y-3 = 1</math> <math>\rightarrow</math> <math>4y = 4</math>. There are no solutions (again, <math>y</math> can't equal to <math>1</math>)<br />
Subcase 2: <math>y<\frac{4}{3}</math><br />
<math>3-4y</math> will be positive so <math>3-4y = 1</math> <math>\rightarrow</math> <math>4y = 2</math>. <math>y = \frac{1}{2}</math> and <math>x = \frac{3}{2}</math>.<br />
Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is <math>\boxed{\textbf{(C) } 3}</math>.<br />
<math>\text{\LaTeX}</math> edit by pretzel, very minor <math>\text{\LaTeX}</math> edits by Bryanli, very very minor <math>\text{\LaTeX}</math> edit by ssb02<br />
<br />
===Solution 3===<br />
Note that <math>||x| - |y||</math> can take on either of four values: <math>x + y</math>, <math>x - y</math>, <math>-x + y</math>, <math>-x -y</math>. <br />
Solving the equations (by elimination, either adding the two equations or subtracting),<br />
we obtain the three solutions: <math>(0, 1)</math>, <math>(-3,2)</math>, <math>(1.5, 0.5)</math> so the answer is <math>\boxed{\textbf{(C) } 3}</math>. One of those equations overlap into <math>(0, 1)</math> so there's only 3 solutions.<br />
<br />
~trumpeter, ccx09<br />
~minor edit, XxHalo711<br />
<br />
===Solution 4===<br />
Just as in solution <math>2</math>, we derive the equation <math>||3-3y|-|y||=1</math>. If we remove the absolute values, the equation collapses into four different possible values. <math>3-2y</math>, <math>3-4y</math>, <math>2y-3</math>, and <math>4y-3</math>, each equal to either <math>1</math> or <math>-1</math>. Remember that if <math>P-Q=a</math>, then <math>Q-P=-a</math>. Because we have already taken <math>1</math> and <math>-1</math> into account, we can eliminate one of the conjugates of each pair, namely <math>3-2y</math> and <math>2y-3</math>, and <math>3-4y</math> and <math>4y-3</math>. Find the values of <math>y</math> when <math>3-2y=1</math>, <math>3-2y=-1</math>, <math>3-4y=1</math> and <math>3-4y=-1</math>. We see that <math>3-2y=1</math> and <math>3-4y=-1</math> give us the same value for <math>y</math>, so the answer is <math>\boxed{\textbf{(C) } 3}</math><br />
<br />
~Zeric Hang<br />
<br />
===Solution 5===<br />
Just as in solution <math>2</math>, we derive the equation <math>x=3-3y</math>. Squaring both sides in the second equation gives <math>x^2+y^2-2|xy|=1</math>. Putting <math>x=3-3y</math> and doing a little calculation gives <math>10y^2-18y+9-2|3y-3y^2|=1</math>. From here we know that <math>3y-3y^2</math> is either positive or negative.<br />
<br />
When positive, we get <math>2y^2-3y+1=0</math> and then, <math>y=1/2</math> or <math>y=1</math>.<br />
When negative, we get <math>y^2-3y+2=0</math> and then, <math>y=2</math> or <math>y=1</math>. Clearly, there are <math>3</math> different pairs of values and that gives us <math>\boxed{\textbf{(C) } 3}</math><br />
<br />
~OlutosinNGA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=11|num-a=13}}<br />
{{AMC12 box|year=2018|ab=A|num-b=9|num-a=11}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_24&diff=1156802017 AMC 10B Problems/Problem 242020-01-27T19:31:02Z<p>Shurong.ge: /* Diagram */</p>
<hr />
<div>==Problem 24==<br />
The vertices of an equilateral triangle lie on the hyperbola <math>xy=1</math>, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169</math><br />
<br />
==Diagram==<br />
<asy><br />
size(15cm);<br />
Label f; <br />
f.p=fontsize(6); <br />
xaxis(-8,8,Ticks(f, 2.0)); <br />
yaxis(-8,8,Ticks(f, 2.0)); <br />
real f(real x) <br />
{ <br />
return 1/x; <br />
} <br />
draw(graph(f,-8,-0.125));<br />
draw(graph(f,0.125,8));<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, <math>A = (1,1)</math>, so <math>AI = BI = CI = 2\sqrt{2}</math>, so since <math>\triangle AIB</math> is isosceles and <math>\angle AIB = 120^{\circ}</math>, then by Law of Cosines, <math>AB = 2\sqrt{6}</math>. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to <math>\frac {s}{\sqrt{3}}</math>. Therefore, the area of the triangle is <math>\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}</math>, so the square of the area of the triangle is <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 2==<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = \left(a,\frac{1}{a}\right)</math> and <math>C=\left(\frac{1}{a},a\right)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 3==<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (1, 1)</math> and let point <math>A</math> be <math>(-1, -1)</math>. It is known that the centroid is equidistant from the three vertices of <math>\triangle ABC</math>. Because we have the coordinates of both <math>A</math> and <math>G</math>, we know that the distance from <math>G</math> to any vertice of <math>\triangle ABC</math> is <math>\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}</math>. Therefore, <math>AG=BG=CG=2\sqrt{2}</math>. It follows that from <math>\triangle ABG</math>, where <math>AG=BG=2\sqrt{2}</math> and <math>\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}</math>, <math>[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}</math> using the formula for the area of a triangle with sine <math>\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)</math>. Because <math>\triangle ACG</math> and <math>\triangle BCG</math> are congruent to <math>\triangle ABG</math>, they also have an area of <math>2\sqrt{3}</math>. Therefore, <math>[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}</math>. Squaring that gives us the answer of <math>\boxed{\textbf{(C) }108}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_24&diff=1156792017 AMC 10B Problems/Problem 242020-01-27T19:28:01Z<p>Shurong.ge: </p>
<hr />
<div>==Problem 24==<br />
The vertices of an equilateral triangle lie on the hyperbola <math>xy=1</math>, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169</math><br />
<br />
==Diagram==<br />
<asy><br />
size(5cm);<br />
Label f; <br />
f.p=fontsize(6); <br />
xaxis(-8,8,Ticks(f, 2.0)); <br />
yaxis(-8,8,Ticks(f, 2.0)); <br />
real f(real x) <br />
{ <br />
return 1/x; <br />
} <br />
draw(graph(f,-8,-0.125));<br />
draw(graph(f,0.125,8));<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, <math>A = (1,1)</math>, so <math>AI = BI = CI = 2\sqrt{2}</math>, so since <math>\triangle AIB</math> is isosceles and <math>\angle AIB = 120^{\circ}</math>, then by Law of Cosines, <math>AB = 2\sqrt{6}</math>. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to <math>\frac {s}{\sqrt{3}}</math>. Therefore, the area of the triangle is <math>\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}</math>, so the square of the area of the triangle is <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 2==<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = \left(a,\frac{1}{a}\right)</math> and <math>C=\left(\frac{1}{a},a\right)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 3==<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (1, 1)</math> and let point <math>A</math> be <math>(-1, -1)</math>. It is known that the centroid is equidistant from the three vertices of <math>\triangle ABC</math>. Because we have the coordinates of both <math>A</math> and <math>G</math>, we know that the distance from <math>G</math> to any vertice of <math>\triangle ABC</math> is <math>\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}</math>. Therefore, <math>AG=BG=CG=2\sqrt{2}</math>. It follows that from <math>\triangle ABG</math>, where <math>AG=BG=2\sqrt{2}</math> and <math>\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}</math>, <math>[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}</math> using the formula for the area of a triangle with sine <math>\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)</math>. Because <math>\triangle ACG</math> and <math>\triangle BCG</math> are congruent to <math>\triangle ABG</math>, they also have an area of <math>2\sqrt{3}</math>. Therefore, <math>[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}</math>. Squaring that gives us the answer of <math>\boxed{\textbf{(C) }108}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_24&diff=1156782017 AMC 10B Problems/Problem 242020-01-27T19:23:11Z<p>Shurong.ge: /* Solution 1 */</p>
<hr />
<div>==Problem 24==<br />
The vertices of an equilateral triangle lie on the hyperbola <math>xy=1</math>, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169</math><br />
<br />
==Solution 1==<br />
<asy><br />
size(5cm);<br />
Label f; <br />
f.p=fontsize(6); <br />
xaxis(-8,8,Ticks(f, 2.0)); <br />
yaxis(-8,8,Ticks(f, 2.0)); <br />
real f(real x) <br />
{ <br />
return 1/x; <br />
} <br />
draw(graph(f,-8,-0.125));<br />
draw(graph(f,0.125,8));<br />
</asy><br />
<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, <math>A = (1,1)</math>, so <math>AI = BI = CI = 2\sqrt{2}</math>, so since <math>\triangle AIB</math> is isosceles and <math>\angle AIB = 120^{\circ}</math>, then by Law of Cosines, <math>AB = 2\sqrt{6}</math>. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to <math>\frac {s}{\sqrt{3}}</math>. Therefore, the area of the triangle is <math>\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}</math>, so the square of the area of the triangle is <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 2==<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = \left(a,\frac{1}{a}\right)</math> and <math>C=\left(\frac{1}{a},a\right)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 3==<br />
Without loss of generality, let the centroid of <math>\triangle ABC</math> be <math>G = (1, 1)</math> and let point <math>A</math> be <math>(-1, -1)</math>. It is known that the centroid is equidistant from the three vertices of <math>\triangle ABC</math>. Because we have the coordinates of both <math>A</math> and <math>G</math>, we know that the distance from <math>G</math> to any vertice of <math>\triangle ABC</math> is <math>\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}</math>. Therefore, <math>AG=BG=CG=2\sqrt{2}</math>. It follows that from <math>\triangle ABG</math>, where <math>AG=BG=2\sqrt{2}</math> and <math>\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}</math>, <math>[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}</math> using the formula for the area of a triangle with sine <math>\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)</math>. Because <math>\triangle ACG</math> and <math>\triangle BCG</math> are congruent to <math>\triangle ABG</math>, they also have an area of <math>2\sqrt{3}</math>. Therefore, <math>[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}</math>. Squaring that gives us the answer of <math>\boxed{\textbf{(C) }108}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=LaTeX&diff=115677LaTeX2020-01-27T19:14:32Z<p>Shurong.ge: </p>
<hr />
<div>{{Latex}}<br />
The <math>\LaTeX</math> typesetting system (pronounced "Lay-Tek" by most, or "Lah-Tek" by some, or even "Lay-Tex) is widely used to produce well-formatted [[math|mathematical]] and scientific writing. With <math>\LaTeX</math>, it is very easy to produce expressions like <br />
<cmath><br />
\sqrt{\frac {a^2+b^2+c^2}3} \geq \frac {a+b+c}3 \geq \sqrt[3]{abc} \geq \frac 3 { \frac 1a + \frac 1b + \frac 1c } .<br />
</cmath> Nearly every serious student of math and science will use <math>\LaTeX</math> frequently. Through these web pages, you will learn much of what you'll need to express math and science like a professional.<br />
<br />
* [http://www.artofproblemsolving.com/wiki/index.php/LaTeX:LaTeX_on_AoPS Click here] to start learning how to use <math>\LaTeX</math> on AoPS<br />
{{Asymptote}}<br />
{{main|Asymptote}}<br />
'''Asymptote''' is a powerful vector graphics language designed for creating mathematical diagrams and figures. It can output images in either eps or pdf format and is compatible with the standard mathematics typesetting language, [[LaTeX]]. It is also a complete programming language and has cleaner syntax than its predecessor, [http://netlib.bell-labs.com/who/hobby/MetaPost.html MetaPost], which was a language used only for two-dimensional graphics.<br />
<br />
Here is an example of an image that can be produced using Asymptote:<br />
<br />
<center>[[Image:Figure1.jpg]]</center><br />
<br />
In a sense, Asymptote is the ruler and compass of typesetting.<br />
<br />
<br />
You can use Asymptote on the AoPSWiki right now, by enclosing the Asymptote code within <tt><nowiki><asy>...</asy></nowiki></tt> tags. For example, the following code<br />
<pre><nowiki><asy><br />
draw((0,0)--(3,7),red);<br />
dot((0,0));<br />
dot((3,7));<br />
label("Produced with Asymptote "+version.VERSION,point(S),2S);<br />
</asy></nowiki></pre><br />
created the picture <br />
<center><asy><br />
draw((0,0)--(3,7),red);<br />
dot((0,0));<br />
dot((3,7));<br />
label("Produced with Asymptote "+version.VERSION,point(S),2S);<br />
</asy></center><br />
And on the AoPS forums you can use <tt><nowiki>[asy]..[/asy]</nowiki></tt><br />
<br />
Another example:<br />
<br />
<pre><nowiki>[asy]<br />
pair A,B,C,X,Y,Z; <br />
A = (0,0);<br />
B = (1,0);<br />
C = (0.3,0.8);<br />
draw(A--B--C--A);<br />
X = (B+C)/2;<br />
Y = (A+C)/2;<br />
Z = (A+B)/2;<br />
draw(A--X, red);<br />
draw(B--Y,red);<br />
draw(C--Z,red);<br />
[/asy]</nowiki></pre><br />
<br />
<asy><br />
pair A,B,C,X,Y,Z;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (0.3,0.8);<br />
draw(A--B--C--A);<br />
X = (B+C)/2;<br />
Y = (A+C)/2;<br />
Z = (A+B)/2;<br />
draw(A--X, red);<br />
draw(B--Y,red);<br />
draw(C--Z,red);<br />
</asy></div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_15&diff=1156751953 AHSME Problems/Problem 152020-01-27T18:41:54Z<p>Shurong.ge: /* Solution */</p>
<hr />
<div>==Problem 15==<br />
A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is: <br />
<br />
<math>\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\ <br />
\textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\\ \textbf{(C)}\ \frac{1}{2}\text{ the area of the circular piece}\\ \textbf{(D)}\ \frac{1}{4}\text{ the area of the circular piece}\\ \textbf{(E)}\ \text{none of these} </math><br />
<br />
==Solution==<br />
The maximum diameter of the circular piece is the same as the side length of the square piece, so the circle is tangent to the square on all four sides. The maximum size a square piece that you can cut from the circle now has 4 edges that are the same as the square created by connecting the four midpoints of the original square. If the side length of the original square is <math>s</math>, then the sides of the final square have length <math>\sqrt{\frac{s^2}{2}}</math>, so its area is <math>\frac{s^2}{2}</math>, <math>\frac{1}{2}</math> the area of the original square. <br />
<math>\textbf{(B)}</math><br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1953|num-b=14|num-a=16}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Absolute_zero&diff=115674Absolute zero2020-01-27T18:35:51Z<p>Shurong.ge: </p>
<hr />
<div>Absolute zero is the theoretical point in [[temperature]] where all kinetic motion of an atom at an atomic scale stops. (This is [[thermal energy]])<br />
<br />
<br />
== At absolute zero: ==<br />
<br />
All gases have the same pressure at absolute zero.<br />
<br />
-273.15 degrees [[Celsius]]<br />
<br />
0 degrees [[Kelvin]] (Celsius increments)<br />
<br />
-459.67 degrees [[Fahrenheit]]<br />
<br />
0 degrees [[Rankine]] (Fahrenheit increment)<br />
<br />
<br />
[[Category:Physics]]<br />
<br />
{{Stub}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Prealgebra&diff=115673Prealgebra2020-01-27T18:28:41Z<p>Shurong.ge: Page was blanked</p>
<hr />
<div>==Definition==<br />
Pre (before) - algebra<br />
<br />
A course which the level is lower than algebra.<br />
<br />
==AoPS Course==<br />
Prealgebra is a subject in mathematics that is generally offered for students as a preparation for algebra. It is a mix of different middle-school topics to get the student ready for algebra.<br />
<br />
===Example Problem===<br />
If Bob is <math>65</math> years old and is <math>5</math> times <math>3</math> more years than hold old John is, how old is John.<br />
<br />
Answer: 10<br />
<br />
{{stub}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Dividend&diff=115672Dividend2020-01-27T18:25:03Z<p>Shurong.ge: </p>
<hr />
<div>==Definition==<br />
===Easy-to-Understand===<br />
The number you want to divide.<br />
===Professional===<br />
A number to be divided by another number.<br />
==Examples==<br />
===Specific===<br />
In <math>20/5=4</math>, <math>20</math> is the dividend.<br />
===General===<br />
In <math>x/y = z</math>, <math>x</math> is the dividend.<br />
==See Also==<br />
*[[Divisor]]<br />
*[[Quotient]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Dividend&diff=115671Dividend2020-01-27T18:24:35Z<p>Shurong.ge: </p>
<hr />
<div>==Definition==<br />
===Easy-to-Understand===<br />
The number you want to divide.<br />
===Professional===<br />
A number to be divided by another number.<br />
==Examples==<br />
===Specific===<br />
In <math>20/5=4</math>, <math>20</math> is the dividend.<br />
===General===<br />
In <math>x/y = z</math>, <math>x</math> is the dividend.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Proof_that_2%3D1&diff=115670Proof that 2=12020-01-27T18:18:18Z<p>Shurong.ge: </p>
<hr />
<div>==Proof==<br />
<br />
1) <math>a = b</math>. Given.<br />
<br />
2) <math>a^2 = ab</math>. Multiply both sides by a.<br />
<br />
3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides.<br />
<br />
4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides.<br />
<br />
5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math><br />
<br />
6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>.<br />
<br />
7) <math>2a = a</math>. Addition.<br />
<br />
8) <math>2 = 1</math>. Divide both sides by <math>a</math>.<br />
<br />
==Error==<br />
The quantity of <math>a-b</math> is <math>0</math> as <math>a = b</math>, since one cannot divide by zero, the proof is incorrect from that point on.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Proof_that_2%3D1&diff=115669Proof that 2=12020-01-27T18:16:56Z<p>Shurong.ge: </p>
<hr />
<div>==Proof==<br />
<br />
1) <math>a = b</math>. Given.<br />
<br />
2) <math>a^2 = ab</math>. Multiply both sides by a.<br />
<br />
3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides.<br />
<br />
4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides.<br />
<br />
5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math><br />
<br />
6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>.<br />
<br />
7) <math>2a = a</math>. Addition.<br />
<br />
8) <math>2 = 1</math>. Divide both sides by <math>a</math>.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Proof_that_2%3D1&diff=115668Proof that 2=12020-01-27T18:11:51Z<p>Shurong.ge: /* Proof */</p>
<hr />
<div>==Proof==<br />
<br />
1) <math>a = b</math>. Given.<br />
<br />
2) <math>a^2 = ab</math>. Multiply both sides by a.<br />
<br />
3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides.<br />
<br />
4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides.<br />
<br />
5) <math>(a+b) = b</math>. Add both sides by <math>(a-b)</math><br />
<br />
6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>.<br />
<br />
7) <math>2a = a</math>. Addition.<br />
<br />
8) <math>2 = 1</math>. Divide both sides by <math>a</math>.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=Proof_that_2%3D1&diff=115667Proof that 2=12020-01-27T18:09:18Z<p>Shurong.ge: </p>
<hr />
<div>==Proof==<br />
<br />
1) <math>a = b</math>. Given.<br />
<br />
2) <math>a^2 = ab</math>. Multiply both sides by a.<br />
<br />
3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides.<br />
<br />
4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides.<br />
<br />
5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math><br />
<br />
6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>.<br />
<br />
7) <math>2a = a</math>. Addition.<br />
<br />
8) <math>2 = 1</math>. Divide both sides by <math>a</math>.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_24&diff=1156642016 AMC 10B Problems/Problem 242020-01-27T17:54:42Z<p>Shurong.ge: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
How many four-digit integers <math>abcd</math>, with <math>a \neq 0</math>, have the property that the three two-digit integers <math>ab<bc<cd</math> form an increasing arithmetic sequence? One such number is <math>4692</math>, where <math>a=4</math>, <math>b=6</math>, <math>c=9</math>, and <math>d=2</math>.<br />
<br />
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20</math><br />
<br />
==Solution==<br />
The numbers are <math>10a+b, 10b+c,</math> and <math>10c+d</math>. Note that only <math>d</math> can be zero for the numbers ab, bc, and cd cannot start with a zero and that <math>a\le b\le c</math>.<br />
<br />
To form the sequence, we need <math>(10c+d)-(10b+c)=(10b+c)-(10a+b)</math>. This can be rearranged as <math>10(c-2b+a)=2c-b-d</math>. Notice that since the left-hand side is a multiple of <math>10</math>, the right-hand side can only be <math>0</math> or <math>10</math>. (A value of <math>-10</math> would contradict <math>a\le b\le c</math>.) Therefore we have two cases: <math>a+c-2b=1</math> and <math>a+c-2b=0</math>.<br />
<br />
=== Case 1===<br />
If <math>c=9</math>, then <math>b+d=8,\ 2b-a=8</math>, so <math>5\le b\le 8</math>. This gives <math>2593, 4692, 6791, 8890</math>.<br />
If <math>c=8</math>, then <math>b+d=6,\ 2b-a=7</math>, so <math>4\le b\le 6</math>. This gives <math>1482, 3581, 5680</math>.<br />
If <math>c=7</math>, then <math>b+d=4,\ 2b-a=6</math>, so <math>b=4</math>, giving <math>2470</math>.<br />
There is no solution for <math>c=6</math>.<br />
Added together, this gives us <math>8</math> answers for Case 1.<br />
<br />
===Case 2===<br />
This means that the digits themselves are in arithmetic sequence. This gives us <math>9</math> answers, <math>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789</math>.<br />
<cmath></cmath><br />
Adding the two cases together, we find the answer to be <math>8+9=</math> <math>\boxed{\textbf{(D) }17}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=User:Shurong.ge&diff=115496User:Shurong.ge2020-01-25T01:50:18Z<p>Shurong.ge: /* Closest Friends on AoPS */</p>
<hr />
<div>==About Me==<br />
Go onto my user page to learn about me.<br />
Here are some quick facts:<br />
*I am a human<br />
*I have my own AoPS wiki page<br />
<br />
==What I am learning==<br />
At AoPS, I am taking Geometry, Algebra 2, and 6th grade LA at the Bellevue campus.<br />
<br />
==Other Websites I Use==<br />
*Edabit<br />
*Repl<br />
<br />
==Closest Friends on AoPS==<br />
===ThriftyPiano===<br />
[[Thriftypiano]] in real life did not earn 1st place in a 4th grade statewide math competition, the following section is an excerpt from [[Thriftypiano]]'s (dumb) intro speech:<br />
<br />
"Thriftypiano is unknown. Nobody knows where he is, who he is, why he exists, and what he is. All we know is that he exists and is male.<br />
He created this song:<br />
When I was a little biddy boy<br />
My grandmother bought me a cute little toy<br />
Silver bells hanging on a string<br />
She told me it was my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
Then momma took me to Grammar school<br />
Always went by the best of rule *<br />
Everytime that bell would ring<br />
Catch me playing with my ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while climbing the garden wall,<br />
I slipped and had a terrible fall<br />
I fell so hard I heard bells ring,<br />
But I held on to My ding-a-ling-a-ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
One day while swimming cross turtle creek<br />
Man them snappers all 'round at my feet<br />
Was so hard swimming cross that thing<br />
with both hands holding my dingaling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
My Ding-A-Ling My Ding-A-Ling I want you to play with My Ding-A-Ling<br />
And this song:<br />
A duck walked up to a lemonade stand<br />
And he said to the man, running the stand<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said<br />
"No we just sell lemonade. But it's cold<br />
And it's fresh<br />
And it's all home-made. Can I get you<br />
Glass?"<br />
The duck said,<br />
"I'll pass".<br />
Then he waddled away.<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-bada-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?<br />
The man said,<br />
"No, like I said yesterday<br />
We just sell lemonade OK?<br />
Why not give it a try?"<br />
The duck said,<br />
"Goodbye."good day<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (bum bum bum) Got any grapes?<br />
The man said,<br />
Look, this is getting old.<br />
I mean, lemonade's all we've ever sold.<br />
Why not give it a go?"<br />
The duck said,<br />
"How 'bout, no."<br />
Then he waddled away<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man running the stand,<br />
"Hey! (Bum bum bum) Got any grapes?"<br />
The man said,<br />
"THAT'S IT!<br />
If you don't stay away, duck,<br />
I'll glue you to a tree and leave you there all day, stuck<br />
So don't get to close!"<br />
The duck said,<br />
"Adios."<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)<br />
'Til the very next day.<br />
(Bum bum bum bum ba-ba-dum)<br />
When the duck walked up to the lemonade stand<br />
And he said to the man that was running the stand,<br />
"Hey! (Bum bum bum) got any glue?"<br />
"What"<br />
"Got any glue?"<br />
"No, why would I– oh!"<br />
And one more question for you;<br />
"Got any grapes?"<br />
(Bum bum bum, bum bum bum)<br />
And the man just stopped.<br />
Then he started to smile.<br />
He started to laugh.<br />
He laughed for a while.<br />
He said,<br />
"Come on duck, let's walk to the store.<br />
I'll buy you some grapes<br />
So you won't have to ask anymore."<br />
So they walked to the store<br />
And the man bought some grapes.<br />
He gave one to the duck and the duck said,<br />
"Hmmm..No thanks. But you know what sounds good?<br />
It would make my day.<br />
Do you think this store<br />
Do you think this store<br />
Do you think this store has any lemonade?"<br />
Then he waddled away.<br />
(Waddle waddle)<br />
Then he waddled away.<br />
(Waddle waddle waddle)<br />
Then he waddled away<br />
(Waddle waddle)" -ThriftyPiano<br />
<br />
===Potato2017===<br />
He is a 5th grader, but has an IQ of 190. Thriftypiano AND the Stanford test says so.<br />
<br />
Edited by Thrifty Piano<br />
<br />
===Bluesky11===<br />
He is a very good friend, it's sad we got split into different middle schools...<br />
<br />
===RaymondZhu===<br />
He got first place in that 4th grade statewide math competition, him and Potato2017 had a tie-breaker for first, he came out victorious. He is currently in 2 schools at once because of his legendary math skills.<br />
<br />
==Coding==<br />
Yes, I code in Python and Java<br />
<br />
==Makers of This Page==<br />
Page created by Shurong.ge; edited by ThriftyPiano and RaymondZhu.</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=SAS_Similarity&diff=115495SAS Similarity2020-01-25T01:48:58Z<p>Shurong.ge: </p>
<hr />
<div>==Definition==<br />
===AoPS===<br />
SAS stands for Side-Angle-Side, for two [[triangle]]s to be [[similar|similar triangles]] by SAS similarity, they must have a pair of congruent angle and the two sides next to the angle must be proportional.<br />
===Mathwords Definition===<br />
Side-angle-side similarity. When two triangles have corresponding angles that are congruent and corresponding sides with identical ratios as shown below, the triangles are similar.<br />
<br />
==Diagram==<br />
<asy><br />
dot((0,0));<br />
label("A",(0,0),SW);<br />
dot((5,0));<br />
label("B",(5,0),SE);<br />
dot((3,4));<br />
label("C",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.1;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
<asy><br />
size((8cm));<br />
dot((0,0));<br />
label("D",(0,0),SW);<br />
dot((5,0));<br />
label("E",(5,0),SE);<br />
dot((3,4));<br />
label("F",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.0675;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
If <math>m\angle CAB = m\angle FDE</math> and <math>\dfrac{CA}{FD} = \dfrac{AB}{DE}</math>, then the triangles are similar by SAS similarity.<br />
<br />
==See Also==<br />
*[[Similarity]]<br />
*[[Congruence]]<br />
<br />
==Categories==<br />
<br />
[[Category:Geometry]] [[Category:Mathematics]][[Category:Stubs]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=SAS_Similarity&diff=115494SAS Similarity2020-01-25T01:44:57Z<p>Shurong.ge: /* Categories */</p>
<hr />
<div>==Definition==<br />
SAS stands for Side-Angle-Side, for two [[triangle]]s to be [[similar|similar triangles]] by SAS similarity, they must have a pair of congruent angle and the two sides next to the angle must be proportional.<br />
==Diagram==<br />
<asy><br />
dot((0,0));<br />
label("A",(0,0),SW);<br />
dot((5,0));<br />
label("B",(5,0),SE);<br />
dot((3,4));<br />
label("C",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.1;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
<asy><br />
size((8cm));<br />
dot((0,0));<br />
label("D",(0,0),SW);<br />
dot((5,0));<br />
label("E",(5,0),SE);<br />
dot((3,4));<br />
label("F",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.0675;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
If <math>m\angle CAB = m\angle FDE</math> and <math>\dfrac{CA}{FD} = \dfrac{AB}{DE}</math>, then the triangles are similar by SAS similarity.<br />
<br />
==Categories==<br />
<br />
[[Category:Geometry]] [[Category:Mathematics]][[Category:Stubs]]</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=SAS_Similarity&diff=115493SAS Similarity2020-01-25T01:43:24Z<p>Shurong.ge: /* Diagram */</p>
<hr />
<div>==Definition==<br />
SAS stands for Side-Angle-Side, for two [[triangle]]s to be [[similar|similar triangles]] by SAS similarity, they must have a pair of congruent angle and the two sides next to the angle must be proportional.<br />
==Diagram==<br />
<asy><br />
dot((0,0));<br />
label("A",(0,0),SW);<br />
dot((5,0));<br />
label("B",(5,0),SE);<br />
dot((3,4));<br />
label("C",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.1;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
<asy><br />
size((8cm));<br />
dot((0,0));<br />
label("D",(0,0),SW);<br />
dot((5,0));<br />
label("E",(5,0),SE);<br />
dot((3,4));<br />
label("F",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.0675;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
If <math>m\angle CAB = m\angle FDE</math> and <math>\dfrac{CA}{FD} = \dfrac{AB}{DE}</math>, then the triangles are similar by SAS similarity.<br />
<br />
==Categories==<br />
{{Category: Stubs}}<br />
{{Category: Geometry}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=SAS_Similarity&diff=115491SAS Similarity2020-01-25T01:41:48Z<p>Shurong.ge: </p>
<hr />
<div>==Definition==<br />
SAS stands for Side-Angle-Side, for two [[triangle]]s to be [[similar|similar triangles]] by SAS similarity, they must have a pair of congruent angle and the two sides next to the angle must be proportional.<br />
==Diagram==<br />
<asy><br />
dot((0,0));<br />
label("A",(0,0),SW);<br />
dot((5,0));<br />
label("B",(5,0),SE);<br />
dot((3,4));<br />
label("C",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.1;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
<asy><br />
size((8cm));<br />
dot((0,0));<br />
label("D",(0,0),SW);<br />
dot((5,0));<br />
label("E",(5,0),SE);<br />
dot((3,4));<br />
label("F",(3,4),N);<br />
draw((0,0)--(5,0)--(3,4)--cycle);<br />
markscalefactor = 0.0675;<br />
draw(anglemark((5,0),(0,0),(3,4)));<br />
</asy><br />
==Categories==<br />
{{Category: Stubs}}<br />
{{Category: Geometry}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=SAS_Similarity&diff=115486SAS Similarity2020-01-25T01:33:19Z<p>Shurong.ge: Created page with "==Definition== SAS stands for Side-Angle-Side, for two triangles to be similar triangles by SAS similarity, they must have a pair of congruent angle and the tw..."</p>
<hr />
<div>==Definition==<br />
SAS stands for Side-Angle-Side, for two [[triangle]]s to be [[similar|similar triangles]] by SAS similarity, they must have a pair of congruent angle and the two sides next to the angle must be proportional.<br />
==Categories==<br />
{{Category: Stubs}}<br />
{{Category: Geometry}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&diff=1152762016 AMC 8 Problems/Problem 222020-01-23T20:11:54Z<p>Shurong.ge: /* Solution 3 (Coordinate Geometry) */</p>
<hr />
<div>Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br />
label("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
label("$1$", (0.5, 4), N);<br />
label("$1$", (1.5, 4), N);<br />
label("$1$", (2.5, 4), N);<br />
label("$4$", (3.2, 2), E);<br />
</asy><br />
<br />
<math>\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5</math><br />
<br />
==Solution 1==<br />
Draw G in between B and C<br />
Draw H, J, K beneath C, G, B respectively .<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, grey);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, grey);<br />
draw((1,0)--(1,4));<br />
draw((1.5,0)--(1.5,4));<br />
draw((2,0)--(2,4));<br />
label("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
label("$G$", (1.5, 4.2));<br />
label("$H$", (1, -0.2));<br />
label("$J$", (1.5, -0.2));<br />
label("$K$", (2, -0.2));<br />
label("$1$", (0.5, 4), N);<br />
label("$1$", (2.5, 4), N);<br />
label("$4$", (3.2, 2), E);<br />
</asy><br />
Let us take a look at rectangle CDEH. I have labeled E' for convenience. <br />
<asy><br />
fill((0,0)--(1,4)--(1,2)--cycle, grey);<br />
draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0));<br />
draw((0,0)--(1,4)--(1,2)--(0,0));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$H$", (1, -0.2));<br />
label("$E'$", (1.2, 2));<br />
</asy><br />
We can clearly see that CEE' has <math>\frac{1}{4}</math> the area of CDEH because it has half the base and the same height. Similarly, the shaded regions in CGJH, BGJK, and ABKF all have <math>\frac{1}{4}</math> the area of their rectangle. So, the total shaded region is just <math>\frac{1}{4}</math> the area of the total region, or <math>\frac{1}{4} \times 3 \times 4</math>, or <math>\boxed{\textbf{(C) }3}</math><br />
==Solution 2==<br />
The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math><br />
<br />
==Solution 3 (Coordinate Geometry)==<br />
<br />
Set coordinates to the points:<br />
<br />
Let <math>E=(0,0)</math>, <math>F=(3,0)</math><br />
<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); <br />
draw((3,0)--(1,4)--(0,0)); <br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black); <br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black); <br />
label(scale(0.7)*"$A(3,4)$",(3.25,4.2)); <br />
label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); <br />
label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); <br />
label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); <br />
label(scale(0.7)*"$E(0,0)$", (0,-0.2)); <br />
label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); <br />
label(scale(0.7)*"$F(3,0)$", (3,-0.2)); <br />
label(scale(0.7)*"$1$", (0.3, 4), N); <br />
label(scale(0.7)*"$1$", (1.5, 4), N); <br />
label(scale(0.7)*"$1$", (2.7, 4), N); <br />
label(scale(0.7)*"$4$", (3.2, 2), E); <br />
</asy><br />
<br />
Now, we easily discover that line <math>CF</math> has lattice coordinates at <math>(1,4)</math> and <math>(3,0)</math>. Hence, the slope of line <math>CF=-2</math> <br />
<br />
Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</math><br />
<br />
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.<br />
<br />
Hence, setting them equal to find the intersection point...<br />
<br />
<math>y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3</math>.<br />
<br />
Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.<br />
<br />
Now, we can see that <br />
<br />
<math>E=(0,0)</math> <br />
<br />
<math>Z=(\dfrac{3}{2},3)</math> <br />
<br />
<math>C=(1,4)</math>.<br />
<br />
Now use the [[Shoelace Theorem|shoelace theorem]].<br />
<br />
<math>\frac{(0*3 + \dfrac{3}{2}*4 + 1*0)-(\dfrac{3}{2}*0 + 1*3 + 4*0)}{2} = \frac{6-3}{2} = \frac{3}{2}</math><br />
<br />
Using the well known [[Shoelace Theorem|shoelace theorem]], we find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>.<br />
<br />
Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math><br />
{{AMC8 box|year=2016|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&diff=1152752016 AMC 8 Problems/Problem 222020-01-23T20:11:35Z<p>Shurong.ge: </p>
<hr />
<div>Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br />
label("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
label("$1$", (0.5, 4), N);<br />
label("$1$", (1.5, 4), N);<br />
label("$1$", (2.5, 4), N);<br />
label("$4$", (3.2, 2), E);<br />
</asy><br />
<br />
<math>\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5</math><br />
<br />
==Solution 1==<br />
Draw G in between B and C<br />
Draw H, J, K beneath C, G, B respectively .<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, grey);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, grey);<br />
draw((1,0)--(1,4));<br />
draw((1.5,0)--(1.5,4));<br />
draw((2,0)--(2,4));<br />
label("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
label("$G$", (1.5, 4.2));<br />
label("$H$", (1, -0.2));<br />
label("$J$", (1.5, -0.2));<br />
label("$K$", (2, -0.2));<br />
label("$1$", (0.5, 4), N);<br />
label("$1$", (2.5, 4), N);<br />
label("$4$", (3.2, 2), E);<br />
</asy><br />
Let us take a look at rectangle CDEH. I have labeled E' for convenience. <br />
<asy><br />
fill((0,0)--(1,4)--(1,2)--cycle, grey);<br />
draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0));<br />
draw((0,0)--(1,4)--(1,2)--(0,0));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$H$", (1, -0.2));<br />
label("$E'$", (1.2, 2));<br />
</asy><br />
We can clearly see that CEE' has <math>\frac{1}{4}</math> the area of CDEH because it has half the base and the same height. Similarly, the shaded regions in CGJH, BGJK, and ABKF all have <math>\frac{1}{4}</math> the area of their rectangle. So, the total shaded region is just <math>\frac{1}{4}</math> the area of the total region, or <math>\frac{1}{4} \times 3 \times 4</math>, or <math>\boxed{\textbf{(C) }3}</math><br />
==Solution 2==<br />
The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math><br />
<br />
==Solution 3 (Coordinate Geometry)==<br />
<br />
Set coordinates to the points:<br />
<br />
Let <math>E=(0,0)</math>, <math>F=(3,0)</math><br />
<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); <br />
draw((3,0)--(1,4)--(0,0)); <br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black); <br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black); <br />
label(scale(0.7)*"$A(3,4)$",(3.25,4.2)); <br />
label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); <br />
label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); <br />
label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); <br />
label(scale(0.7)*"$E(0,0)$", (0,-0.2)); <br />
label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); <br />
label(scale(0.7)*"$F(3,0)$", (3,-0.2)); <br />
label(scale(0.7)*"$1$", (0.3, 4), N); <br />
label(scale(0.7)*"$1$", (1.5, 4), N); <br />
label(scale(0.7)*"$1$", (2.7, 4), N); <br />
label(scale(0.7)*"$4$", (3.2, 2), E); <br />
</asy><br />
<br />
Now, we easily discover that line <math>CF</math> has lattice coordinates at <math>(1,4)</math> and <math>(3,0)</math>. Hence, the slope of line <math>CF=-2</math> <br />
<br />
Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</math><br />
<br />
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.<br />
<br />
Hence, setting them equal to find the intersection point...<br />
<br />
<math>y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3</math>.<br />
<br />
Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.<br />
<br />
Now, we can see that <br />
<br />
<math>E=(0,0)</math> <br />
<br />
<math>Z=(\dfrac{3}{2},3)</math> <br />
<br />
<math>C=(1,4)</math>.<br />
<br />
Now use the [[Shoelace Theorem|shoelace theorem]].<br />
<math>\frac{(0*3 + \dfrac{3}{2}*4 + 1*0)-(\dfrac{3}{2}*0 + 1*3 + 4*0)}{2} = \frac{6-3}{2} = \frac{3}{2}</math><br />
<br />
Using the well known [[Shoelace Theorem|shoelace theorem]], we find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>.<br />
<br />
Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math><br />
{{AMC8 box|year=2016|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shurong.gehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&diff=1152742016 AMC 8 Problems/Problem 222020-01-23T20:10:30Z<p>Shurong.ge: /* Solution 3 */</p>
<hr />
<div>Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br />
label("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
label("$1$", (0.5, 4), N);<br />
label("$1$", (1.5, 4), N);<br />
label("$1$", (2.5, 4), N);<br />
label("$4$", (3.2, 2), E);<br />
</asy><br />
<br />
<math>\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5</math><br />
<br />
==Solution 1==<br />
Draw G in between B and C<br />
Draw H, J, K beneath C, G, B respectively .<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, grey);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, grey);<br />
draw((1,0)--(1,4));<br />
draw((1.5,0)--(1.5,4));<br />
draw((2,0)--(2,4));<br />
label("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
label("$G$", (1.5, 4.2));<br />
label("$H$", (1, -0.2));<br />
label("$J$", (1.5, -0.2));<br />
label("$K$", (2, -0.2));<br />
label("$1$", (0.5, 4), N);<br />
label("$1$", (2.5, 4), N);<br />
label("$4$", (3.2, 2), E);<br />
</asy><br />
Let us take a look at rectangle CDEH. I have labeled E' for convenience. <br />
<asy><br />
fill((0,0)--(1,4)--(1,2)--cycle, grey);<br />
draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0));<br />
draw((0,0)--(1,4)--(1,2)--(0,0));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$H$", (1, -0.2));<br />
label("$E'$", (1.2, 2));<br />
</asy><br />
We can clearly see that CEE' has <math>\frac{1}{4}</math> the area of CDEH because it has half the base and the same height. Similarly, the shaded regions in CGJH, BGJK, and ABKF all have <math>\frac{1}{4}</math> the area of their rectangle. So, the total shaded region is just <math>\frac{1}{4}</math> the area of the total region, or <math>\frac{1}{4} \times 3 \times 4</math>, or <math>\boxed{\textbf{(C) }3}</math><br />
==Solution 2==<br />
The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math><br />
<br />
==Solution 3 (Coordinate Geometry)==<br />
<br />
Set coordinates to the points:<br />
<br />
Let <math>E=(0,0)</math>, <math>F=(3,0)</math><br />
<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); <br />
draw((3,0)--(1,4)--(0,0)); <br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black); <br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black); <br />
label(scale(0.7)*"$A(3,4)$",(3.25,4.2)); <br />
label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); <br />
label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); <br />
label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); <br />
label(scale(0.7)*"$E(0,0)$", (0,-0.2)); <br />
label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); <br />
label(scale(0.7)*"$F(3,0)$", (3,-0.2)); <br />
label(scale(0.7)*"$1$", (0.3, 4), N); <br />
label(scale(0.7)*"$1$", (1.5, 4), N); <br />
label(scale(0.7)*"$1$", (2.7, 4), N); <br />
label(scale(0.7)*"$4$", (3.2, 2), E); <br />
</asy><br />
<br />
Now, we easily discover that line <math>CF</math> has lattice coordinates at <math>(1,4)</math> and <math>(3,0)</math>. Hence, the slope of line <math>CF=-2</math> <br />
<br />
Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</math><br />
<br />
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.<br />
<br />
Hence, setting them equal to find the intersection point...<br />
<br />
<math>y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3</math>.<br />
<br />
Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.<br />
<br />
Now, we can see that <br />
<br />
<math>E=(0,0)</math> <br />
<br />
<math>Z=(\dfrac{3}{2},3)</math> <br />
<br />
<math>C=(1,4)</math>.<br />
<br />
Now use the shoelace theorem.<br />
<math>\frac{(0*3 + \dfrac{3}{2}*4 + 1*0)-(\dfrac{3}{2}*0 + 1*3 + 4*0)}{2} = \frac{6-3}{2} = \frac{3}{2}</math><br />
<br />
Using the well known [[shoelace theorem]], we find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>.<br />
<br />
Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math><br />
{{AMC8 box|year=2016|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shurong.ge