User:RandomPieKevin

Revision as of 18:26, 22 February 2016 by RandomPieKevin (talk | contribs)

HELLO!!! I AM RANDOMPIEKEVIN!!!

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...

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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: \[\boxed{\log_{10} 5 = \dfrac{3PQ}{1+3PQ}}.\]

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