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| − | + | ==My Solutions== | |
[https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_10#Solution_3.28strategic_guess_and_check.29 2001 AMC 10 Problem 10] | [https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_10#Solution_3.28strategic_guess_and_check.29 2001 AMC 10 Problem 10] | ||
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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] | ||
| − | + | [https://artofproblemsolving.com/wiki/index.php/2024_AMC_10B_Problems/Problem_10#Solution_4 2024 AMC 10B Problem 10] | |
| − | ==<math>(x+y)^n \equiv x^n + y^n \pmod{n}</math> if <math>n</math> is prime. == | + | ==Some Proofs I wrote== |
| + | |||
| + | ===<math>(x+y)^n \equiv x^n + y^n \pmod{n}</math> if <math>n</math> is prime. === | ||
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> | 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> | ||
| − | ==== | + | ===Volume of Cylinder, Cone, and Sphere=== |
| + | 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>. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | 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
Contents
My Solutions
Some Proofs I wrote
if 
is prime.
if Proof: Expanding 
out, all the coefficients are of the form 
by the binomial theorem. To prove the original result we must show that if 
and 
, then ![\[{n \choose r} \equiv 0 \pmod{n}\]](http://latex.artofproblemsolving.com/c/6/3/c63848a8ef72c7440c7022bd218dad3ced5ab858.png)
. Because ![\[{n \choose r} = \frac{n!}{r!(n-r)!}\]](http://latex.artofproblemsolving.com/3/c/a/3ca41bab24e0ced5a3bd2665f4e96651799fe7cc.png)
, ![\[{n \choose r} \times r!(n-r)! = n!\]](http://latex.artofproblemsolving.com/7/b/e/7bef983037ba8643bc0efb980a61e51f9e25e8b3.png)
, which is divisible by 
, so the original expression must be divisible by . However if is prime, ![\[\gcd(n, r!(n-r)!) = 1\]](http://latex.artofproblemsolving.com/0/d/0/0d095da7ac98ad36f4c2dd61cc45e45e38906992.png)
, since 
does not contain (because 
). Therefore, in order for ![\[{n \choose r} \times r!(n-r)!\]](http://latex.artofproblemsolving.com/6/a/b/6abc4a965f1ab5a8620d40b4d393876dded1647b.png)
to be divisible by , is divisible by . All the coefficients of the expansion(besides the coefficients of 
and 
) are of the form , and , so they cancel out and ![\[(x+y)^n \equiv x^n + y^n \pmod{n}\]](http://latex.artofproblemsolving.com/3/c/1/3c1d9f3704248e7d2b3af8c5b48ebf8acbcd48b6.png)
if is prime. 
Volume of Cylinder, Cone, and Sphere
If we have a function 
, 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 , we have the volume of the solid of revolution to be ![\[\pi \int_{a}^{b} (f(x))^2 \,dx\]](http://latex.artofproblemsolving.com/0/1/f/01f54f46d45f8eb302ed841e4ae8a37def5746cb.png)
.
Cylinder: A cylinder can be expressed a solid of revolution by revolving the line 
around the 
-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 
and the height, 
, is the upper bound of integration. We have ![\[\pi \int_{0}^{h} r^2 \,dx\]](http://latex.artofproblemsolving.com/d/4/0/d40f57835771a4ce23aa2a1e70fa537cd527f409.png)
. Integrating, we get ![\[\pi (r^2h - r^2(0)) = \pi r^2h\]](http://latex.artofproblemsolving.com/6/f/8/6f8ecf414ce5bd07fd11ba5b3e1e1e9c7502670e.png)
. This is the formula of a cylinder.
Cone: If you are given the height and radius of the cone, and you have the point 
on your line(since the vertex is 0), then 
, 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 , we know the y-intercept, and we can only have one slope. If 
, and 
is the slope, then we have 
, and therefore 
, so the equation is 
. For the integral, we get ![\[\pi \int_{0}^{h} \frac{r^2x^2}{h^2} \,dx = \pi \frac{r^2h}{3}\]](http://latex.artofproblemsolving.com/3/8/0/38004c058583c7e32616cdc7ff663869079d1cce.png)
.
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 
. The diameter is 
to , so that is where we integrate. ![\[\pi \int_{-r}^{r} r^2 - x^2 \,dx = \frac{4\pi r^3}{3}\]](http://latex.artofproblemsolving.com/8/9/6/89682a177f64199845cf77e35a4b09ae2930b55e.png)
.
