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Difference between revisions of "User:Idk12345678"

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[https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5#Solution_3.28similar_to_solution_2.29 2005 AIME II Problem 5]
 
[https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5#Solution_3.28similar_to_solution_2.29 2005 AIME II Problem 5]
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[https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_3 2008 AMC 10B Problem 3]
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[https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_6#Solution_2 2008 AMC 10B Problem 6]
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[https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_9#Solution_3 2008 AMC 10B Problem 9]
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[https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_1#Solution_4 2008 AIME II Problem 1]
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[https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_10#Solution_5 2009 AMC 10A Problem 10]
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[https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_2#Solution_3 2009 AIME II Problem 2]
  
 
[https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_16#Solution_3 2013 AMC 10A Problem 16]
 
[https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_16#Solution_3 2013 AMC 10A Problem 16]
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[https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_9#Solution_3 2014 AMC 10B Problem 9]
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[https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_10#Solution_2 2018 AMC 10A Problem 10]
  
 
[https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_14#Solution_11 2020 AMC 10A Problem 14]
 
[https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_14#Solution_11 2020 AMC 10A Problem 14]
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[https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_2#Solution_5 2024 AIME I Problem 2]
 
[https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_2#Solution_5 2024 AIME I Problem 2]
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[https://artofproblemsolving.com/wiki/index.php/2024_AMC_10B_Problems/Problem_10#Solution_4 2024 AMC 10B Problem 10]
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==Some Proofs I wrote==
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===<math>(x+y)^n \equiv x^n + y^n \pmod{n}</math> if <math>n</math> is prime. ===
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Proof: Expanding <math>(x+y)^n</math> out, all the coefficients are of the form <math>n \choose r</math> by the binomial theorem. To prove the original result we must show that if <math>r \neq 1</math> and <math>r \neq n</math>, then <cmath>{n \choose r} \equiv 0 \pmod{n}</cmath>. Because <cmath>{n \choose r} = \frac{n!}{r!(n-r)!}</cmath>, <cmath>{n \choose r} \times r!(n-r)! = n!</cmath>, which is divisible by <math>n</math>, so the original expression must be divisible by <math>n</math>. However if <math>n</math> is prime, <cmath>\gcd(n, r!(n-r)!) = 1</cmath>, since <math>r!</math> does not contain <math>n</math>(because <math>r<n</math>). Therefore, in order for <cmath>{n \choose r} \times r!(n-r)!</cmath> to be divisible by <math>n</math>, <math>n \choose r</math> is divisible by <math>n</math>. All the coefficients of the expansion(besides the coefficients of <math>x^n</math> and <math>y^n</math>)  are of the form <math>n \choose r</math>, and <cmath>{n \choose r} \equiv 0 \pmod{n}</cmath>, so they cancel out and <cmath>(x+y)^n \equiv x^n + y^n \pmod{n}</cmath> if <math>n</math> is prime. <math>\square</math>
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===Volume of Cylinder, Cone, and Sphere===
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If we have a function <math>f(x)</math>, that can be rotated to make a shape, the area underneath it will turn into the volume. However, since we are revolving it in a circular motion, the area will actually become the radius. Another way of seeing this is splitting it into infinite circles and adding up all of them. Therefore, for a function <math>f(x)</math>, we have the volume of the solid of revolution to be <cmath>\pi \int_{a}^{b} (f(x))^2 \,dx </cmath>.
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Cylinder: A cylinder can be expressed a solid of revolution by revolving the line <math>y = r</math> around the <math>x</math>-axis. To find the volume, we can find the area under the curve, and then when we revolve it, it becomes the volume. The radius is <math>r</math> and the height, <math>h</math>, is the upper bound of integration. We have <cmath>\pi \int_{0}^{h} r^2 \,dx </cmath>. Integrating, we get <cmath>\pi (r^2h - r^2(0)) = \pi r^2h</cmath>. This is the formula of a cylinder.
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Cone: If you are given the height and radius of the cone, and you have the point <math>(0,0)</math> on your line(since the vertex is 0), then <math>f(h) = r</math>, because the height is the x-coordinate and the radius is the y(for the same reason seen above in the cylinder). Now, since we have <math>(0,0)</math>, we know the y-intercept, and we can only have one slope. If <math>h=x</math>, and <math>m</math> is the slope, then we have <math>r = mh</math>, and therefore <math>m = \frac{r}{h}</math>, so the equation is <math>f(x) = \frac{rx}{h}</math>. For the integral, we get <cmath>\pi  \int_{0}^{h} \frac{r^2x^2}{h^2} \,dx  = \pi \frac{r^2h}{3}</cmath>.
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Sphere: The equation of a sphere should be a circle, but that is a relation and not a function. Therefore, we can use the top half of a circle, and the bottom half will get filled in when it rotates. Therefore, we get <math>f(x) = \sqrt{r^2 - x^2}</math>. The diameter is <math>-r</math> to <math>r</math>, so that is where we integrate. <cmath>\pi  \int_{-r}^{r} r^2 - x^2 \,dx  = \frac{4\pi r^3}{3}</cmath>.

Latest revision as of 12:38, 15 November 2024

My Solutions

2001 AMC 10 Problem 10

2001 AMC 10 Problem 12

2002 AIME II Problem 3

2005 AIME II Problem 5

2008 AMC 10B Problem 3

2008 AMC 10B Problem 6

2008 AMC 10B Problem 9

2008 AIME II Problem 1

2009 AMC 10A Problem 10

2009 AIME II Problem 2

2013 AMC 10A Problem 16

2014 AMC 10B Problem 9

2018 AMC 10A Problem 10

2020 AMC 10A Problem 14

2023 AIME I Problem 2

2024 AIME I Problem 2

2024 AMC 10B Problem 10

Some Proofs I wrote

$(x+y)^n \equiv x^n + y^n \pmod{n}$ if $n$ is prime.

Proof: Expanding $(x+y)^n$ out, all the coefficients are of the form $n \choose r$ by the binomial theorem. To prove the original result we must show that if $r \neq 1$ and $r \neq n$, then \[{n \choose r} \equiv 0 \pmod{n}\]. Because \[{n \choose r} = \frac{n!}{r!(n-r)!}\], \[{n \choose r} \times r!(n-r)! = n!\], which is divisible by $n$, so the original expression must be divisible by $n$. However if $n$ is prime, \[\gcd(n, r!(n-r)!) = 1\], since $r!$ does not contain $n$(because $r<n$). Therefore, in order for \[{n \choose r} \times r!(n-r)!\] to be divisible by $n$, $n \choose r$ is divisible by $n$. All the coefficients of the expansion(besides the coefficients of $x^n$ and $y^n$) are of the form $n \choose r$, and \[{n \choose r} \equiv 0 \pmod{n}\], so they cancel out and \[(x+y)^n \equiv x^n + y^n \pmod{n}\] if $n$ is prime. $\square$

Volume of Cylinder, Cone, and Sphere

If we have a function $f(x)$, that can be rotated to make a shape, the area underneath it will turn into the volume. However, since we are revolving it in a circular motion, the area will actually become the radius. Another way of seeing this is splitting it into infinite circles and adding up all of them. Therefore, for a function $f(x)$, we have the volume of the solid of revolution to be \[\pi \int_{a}^{b} (f(x))^2 \,dx\].

Cylinder: A cylinder can be expressed a solid of revolution by revolving the line $y = r$ around the $x$-axis. To find the volume, we can find the area under the curve, and then when we revolve it, it becomes the volume. The radius is $r$ and the height, $h$, is the upper bound of integration. We have \[\pi \int_{0}^{h} r^2 \,dx\]. Integrating, we get \[\pi (r^2h - r^2(0)) = \pi r^2h\]. This is the formula of a cylinder.

Cone: If you are given the height and radius of the cone, and you have the point $(0,0)$ on your line(since the vertex is 0), then $f(h) = r$, because the height is the x-coordinate and the radius is the y(for the same reason seen above in the cylinder). Now, since we have $(0,0)$, we know the y-intercept, and we can only have one slope. If $h=x$, and $m$ is the slope, then we have $r = mh$, and therefore $m = \frac{r}{h}$, so the equation is $f(x) = \frac{rx}{h}$. For the integral, we get \[\pi \int_{0}^{h} \frac{r^2x^2}{h^2} \,dx = \pi \frac{r^2h}{3}\].

Sphere: The equation of a sphere should be a circle, but that is a relation and not a function. Therefore, we can use the top half of a circle, and the bottom half will get filled in when it rotates. Therefore, we get $f(x) = \sqrt{r^2 - x^2}$. The diameter is $-r$ to $r$, so that is where we integrate. \[\pi \int_{-r}^{r} r^2 - x^2 \,dx = \frac{4\pi r^3}{3}\].

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