User:Idk12345678

Revision as of 12:38, 15 November 2024 by Idk12345678 (talk | contribs) (My Solutions)
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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. \[\pi  \int_{-r}^{r} r^2 - x^2 \,dx  = \frac{4\pi r^3}{3}\].