Difference between revisions of "User:RandomPieKevin"
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Take a look at this one for 2012 AMC 10B Problem 16: | Take a look at this one for 2012 AMC 10B Problem 16: | ||
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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 <math>4</math>. 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: <math>1: \sqrt{3}: 2.</math> The height is <math>2\sqrt{3}</math> and the base is <math>2</math>. Multiplying the height and base together with <math>\dfrac{1}{2}</math>, we get <math>2\sqrt{3}</math>. Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the 30-60-90 triangle by <math>2</math>: <cmath>2\cdot 2\sqrt{3} = 4\sqrt{3}.</cmath> | 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 <math>4</math>. 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: <math>1: \sqrt{3}: 2.</math> The height is <math>2\sqrt{3}</math> and the base is <math>2</math>. Multiplying the height and base together with <math>\dfrac{1}{2}</math>, we get <math>2\sqrt{3}</math>. Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the 30-60-90 triangle by <math>2</math>: <cmath>2\cdot 2\sqrt{3} = 4\sqrt{3}.</cmath> | ||
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Summing the areas from the equilateral triangle and the remaining circle sections gives us: <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math>. | Summing the areas from the equilateral triangle and the remaining circle sections gives us: <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math>. | ||
| − | 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. | + | 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... |
Revision as of 17:23, 22 February 2016
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 
. 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: 
The height is 
and the base is 
. Multiplying the height and base together with 
, we get . Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the 30-60-90 triangle by : ![\[2\cdot 2\sqrt{3} = 4\sqrt{3}.\]](http://latex.artofproblemsolving.com/0/7/2/0724eb2af08977be3554453187958c0bf167d5ad.png)
To find the area of the remaining sectors, which are 
of the original circles once we remove the triangle, we know that the sectors have a central angle of 
since the equilateral triangle already covered that area. Since there are 
pieces gone from the equilateral triangle, we have, in total, of a circle (with radius ) gone. Each circle has an area of 
, so three circles gives a total area of 
. Subtracting the half circle, we have: ![\[12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.\]](http://latex.artofproblemsolving.com/2/8/2/282489b5df7de97c913d0ddc6ad6f7c1c4b4c59a.png)
Summing the areas from the equilateral triangle and the remaining circle sections gives us: 
.
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...
