Revision as of 17:23, 22 February 2016 by RandomPieKevin (talk | contribs) (Solution)


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...

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