Revision as of 17:35, 23 February 2016 by RandomPieKevin (talk | contribs)


Just kidding. I am Kevin.

I started competitive math in the beginning of 8th grade and I took (950) Introduction to Geometry with SamE (Sam Elder) in the beginning of 2015 (still 8th grade). Then, I started (995) Algebra B with jonjoseph (Jon Joseph) in the middle of 2015 with bluespruce and Ridley-C. Then, I started (1020) Intermediate Algebra with djquarfoot (David Quarfoot) in the beginning of 9th grade.

I have improved from a 12 question guy on the AMC 8 to a 19-20 question guy on the AMC 10 in the past year and a half.

Also, I failed the 2016 AMC 10A... o.O

I did decently on the 2016 AMC 10B... 18 correct...


I'm pretty good at writing proofs...

Take a look at this one for 2012 AMC 10B Problem 16:

To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle with a side of length $4$. We must find the area of this triangle to include the figure formed in between the circles. Since the equilateral triangle has two 30-60-90 triangles inside, we can find the height and the base of each 30-60-90 triangle from the ratios: $1: \sqrt{3}: 2.$ The height is $2\sqrt{3}$ and the base is $2$. Multiplying the height and base together with $\dfrac{1}{2}$, we get $2\sqrt{3}$. Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the 30-60-90 triangle by $2$: \[2\cdot 2\sqrt{3} = 4\sqrt{3}.\]

To find the area of the remaining sectors, which are $\dfrac{5}{6}$ of the original circles once we remove the triangle, we know that the sectors have a central angle of $300^\circ$ since the equilateral triangle already covered that area. Since there are $3$ $\dfrac{1}{6}$ pieces gone from the equilateral triangle, we have, in total, $\dfrac{1}{2}$ of a circle (with radius $2$) gone. Each circle has an area of $\pi r^2 = 4pi$, so three circles gives a total area of $12\pi$. Subtracting the half circle, we have: \[12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.\]

Summing the areas from the equilateral triangle and the remaining circle sections gives us: $\boxed{\textbf{(A)} 10\pi + 4\sqrt3}$.

Yeah pretty good? (The first paragraph was already written by someone else, but I edited it. Also, there was some other stuff that I cleaned up and modified so that it looks as good as it is right now. I'll post some more proofs that I've written below if you want...

Here's another proof:

We are asked to calculate \[S = \sum_{k=1}^n \frac 1{k(k+1)(k+2)}.\]We can start by using partial fraction decomposition to help telescope this sum. We know that $\dfrac{1}{(k)(k+2)} = \dfrac{1}{2}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right)$ from partial fraction decomposition: \[\dfrac{1}{2}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right) = \dfrac{1}{2}\left(\dfrac{k+2}{k(k+2)}-\dfrac{k}{(k(k+2)}\right) = \dfrac{1}{2}\left(\dfrac{2}{(k)(k+2)}\right) = \dfrac{1}{(k)(k+2)}.\] Substituting it in, we get: \[S = \dfrac{1}{2}\sum_{k=1}^n \left(\dfrac{1}{2}-\dfrac{1}{k+2}\right)\dfrac{1}{k+1} = \dfrac{1}{2}\sum_{k = 1}^n \left(\dfrac{1}{k(k+1)} - \dfrac{1}{(k+1)(k+2)}\right) = \dfrac{1}{2}\left(\sum_{k = 1}^n \dfrac{1}{k(k+1)}- \sum_{k = 1}^n \dfrac{1}{(k+1)(k+2)} \right).\]Now, we have a telescoping series. All of the terms except for the first and last terms cancel out. We have: \[\frac{1}{2}\left(\frac{1}{2} - \frac{1}{(n+1)(n+2)}\right) = \boxed{\frac{1}{4} - \frac{1}{2(n+1)(n+2)}}.\]

And here's another one:

We have $\log_8 3 = P$ and $\log_3 5 = Q$, and we are asked to express $\log_{10} 5$ in terms of $P$ and $Q$.

We start by looking for any substitutions that we can make. It seems like the first two must be substituted into the third in order to express the third in terms of $P$ and $Q$.

We can use the change of base formula on the third expression: \[\log_{10} 5 = \dfrac{\log_3 5}{\log_3 10}.\]We can substitute $Q$ into the numerator: \[\dfrac{\log_3 5}{\log_3 10} = \dfrac{Q}{\log_3 10}.\]Next, we can try to substitute $P$ and $Q$ into the denominator by using another logarithmic property: \[\dfrac{Q}{\log_3 10} = \dfrac{Q}{\log_3 2+\log_3 5}.\]We can substitute $Q$ into a term in the denominator: \[\dfrac{Q}{\log_3 2 + \log_3 5} = \dfrac{Q}{\log_3 2 + Q}.\] Summing up our manipulations, we have: \[\log_{10} 5 = \dfrac{\log_3 5}{\log_3 10} = \dfrac{Q}{\log_3 10} = \dfrac{Q}{\log_3 2+\log_3 5} = \dfrac{Q}{\log_3 2 + Q}.\] We need to substitute something into the $\log_3 2$ in the denominator. We can try to manipulate the equation for $P$ to create something in terms of $P$ that is equal to $\log_3 2$: \[P=\log_8 3 = \dfrac{\log_3 3}{\log_3 8} = \dfrac{1}{3\log_3 2}.\]We see that there is a $\log_3 2$ in the denominator of the equation, so we can manipulate the equation to write it in terms of $P$: \begin{align*} P & = \dfrac{1}{3\log_3 2} \\ 3P & = \dfrac{1}{\log_3 2} \\ \dfrac{1}{3P} & = \log_3 2. \end{align*} We can substitute this into the equation:\[\log_{10} 5 = \dfrac{Q}{\log_3 2 + Q} = \dfrac{Q}{\dfrac{1}{3P} + Q}.\] Multiplying the numerator and denominator by $3P$, we get: \[\boxed{\log_{10} 5 = \dfrac{3PQ}{1+3PQ}}.\]

I created this question and the solution:

Simplify: \[\dfrac{(1000-998)+(998-996)+(996-994)+\dots+(202-200)}{\sqrt[4]{272-\sqrt{272-\sqrt{272-\dots}}}}\;\cdot\; 1+\cfrac{12}{1+\cfrac{12}{1+\cfrac{12}{1+\dots}}}.\] Solution:

We are given: \[\dfrac{(1000-998)+(998-996)+(996-994)+\dots+(202-200)}{\sqrt[4]{272-\sqrt{272-\sqrt{272-\dots}}}}\;\cdot\; 1+\cfrac{12}{1+\cfrac{12}{1+\cfrac{12}{1+\dots}}}.\]We will start by simplifying the numerator of the first fraction. We see that the middle terms cancel out, for example: \[(998-996)+(996-994) = 998-994.\]This pattern continues, and we are left with \[1000-200.\]Simplifying gives: \[\boxed{(1000-998)+(998-996)+(996-994)+\dots+(202-200)=800}.\]Next, we will simplify the denominator of the first fraction. Since we are solving for the value of the denominator, we can set it equal to $y$: \[x=\sqrt[4]{272-\sqrt{272-\sqrt{272-\dots}}}.\]There isn't anything that looks obvious, so we will get rid of the $4^{th}$ root: \[y=\sqrt{\sqrt{272-\sqrt{272-\sqrt{272-\dots}}}}.\]Now, we see that we can square both sides to get: \[y^2=\sqrt{272-\sqrt{272-\sqrt{272-\dots}}}.\]If we let $x$ equal the right side of the current equation, we can solve for $y$, which is our desired value: \[x=\sqrt{272-\sqrt{272-\sqrt{272-\dots}}}.\]We see that we can substitute an $x$ into the equation: \[x=\sqrt{272-x}.\] Simplifying: \begin{align*} x&=\sqrt{272-x} \\ x^2&=272-x \\ 0&=x^2+x-272 \\ 0&=(x-16)(x+17) \\ x&=16, -17 \end{align*} We know that $x$ cannot be negative, so we now have: \begin{align*} y^2 &= x \\ y^2 &= 16 \\ y &= \pm 4 \end{align*} We know that $y$ cannot be negative, so we have: \[\boxed{\sqrt[4]{272-\sqrt{272-\sqrt{272-\dots}}}=4}.\]Next, we have to simplify \[1+\cfrac{12}{1+\cfrac{12}{1+\cfrac{12}{1+\dots}}}.\]We want to solve for the value of the fraction, so we will set $z$ equal to it: \[z=1+\cfrac{12}{1+\cfrac{12}{1+\cfrac{12}{1+\dots}}}.\]Looking at the equation, we see that there is a pattern. Substituting $z$ in gives: \[z=1+\dfrac{12}{z}.\] Simplifying: \begin{align*} z^2&=z+12 \\ z^2-z-12 & = 0 \\ (z-4)(z+3) & =0 \\ z&=-3, 4 \end{align*} We know that the value of $z$ cannot be negative, so \[\boxed{1+\cfrac{12}{1+\cfrac{12}{1+\cfrac{12}{1+\dots}}} = 4}.\]Finally, we can put all of the information together: \[\dfrac{(1000-998)+(998-996)+(996-994)+\dots+(202-200)}{\sqrt[4]{272-\sqrt{272-\sqrt{272-\dots}}}}\;\cdot\; 1+\cfrac{12}{1+\cfrac{12}{1+\cfrac{12}{1+\dots}}} = \dfrac{800}{4} \; \cdot \; 4.\]Simplifying gives: \[\large\boxed{800}.\]

I also wrote this short "article" to help a fellow AoPS user with some range questions:

Range is the set of values that a function can output.

Therefore, it is determined by the domain, since the values that you plug in will get you $1$ value out. However, you can determine the range without the domain in many cases.

Here is an example:

We have the function $f$ such that: \[f(x)=\sqrt{16-x^2}.\]Find the domain and range of this function.

I'm sure you can find the domain, any values of $x$ that do not make the value inside the square root negative. Therefore, the domain of $f(x)$ is: \[\boxed{[-4, 4]}.\] Now, for the tricky part (Jyzhang12), we need to find the range. Since this is a square root, we need to find the smallest possible value inside the square root. We see that the smallest value allowed in the square root is $0$, or else it will be negative. Therefore, the lower bound of the range is: \[\sqrt{0} = 0.\]Next, we will find the upper bound of the range. Since this is a square root, we need to find the maximum possible value inside the square root. We see that it is $16$, when $x=0$. Therefore, the upper bound of the range is: \[\boxed{\sqrt{16} = 4}.\] Our range is: \[\boxed{[0, 4]}.\] Next, another type of function has a special method (but still logical - very, very, logical!!! Don't memorize anything!).

We have the function $f$ such that \[f(x)=\dfrac{2x+4}{x+5}.\]Find the domain and range of this function.

We need to find the domain of this function. As a general rule, whenever you have $x$ in the denominator, you always exclude the value of $x$ that makes the denominator $0$, since you cannot divide by $0$. Therefore, we have: \begin{align*} x+5 & = 0 \\ x & = -5 \end{align*} Our domain is: \[\boxed{(-\infty, -5)\cup(-5, +\infty)}.\] Next, we need to find the range. We don't see anything obvious, so we will start by plugging in a few values. If $x=0$, then $f(x)=0.8$. If $x=100$, then $f(x)\approx 1.942$. If $x=100000$, then $f(x)\approx 1.999$.

We see that as $x$ approaches infinity, $f(x)$ approaches $2$. We must explain why in order to understand why is has that behavior.

In the numerator, we have $2x+4$ and we have $x+5$ in the denominator. The $2x$ and the $x$ can divide to form $2$. However, you will never get there, as there are the constant terms adding onto the variables. As $x$ approaches infinity, the constants don't really matter, as you have a super large number added to $4$. The only value that $y$ cannot be, since it only approaches it as $x$ approaches infinity, is $2$.

Therefore, our range is: \[\boxed{(-\infty, 2)\cup(2, +\infty)}.\] There are many different types of parent functions and it would take numerous pages to fit them all. The main point is to understand why, not memorize (which is what most public schools (and private schools) make you do).

Invalid username
Login to AoPS