|
|
| (20 intermediate revisions by 5 users not shown) |
| Line 1: |
Line 1: |
| − | 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...
| |
| − |
| |
| − | ________________________________________________________________________________________________________________________
| |
| − |
| |
| − | 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 <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 find the area of the remaining sectors, which are <math>\dfrac{5}{6}</math> of the original circles once we remove the triangle, we know that the sectors have a central angle of <math>300^\circ</math> since the equilateral triangle already covered that area. Since there are <math>3</math> <math>\dfrac{1}{6}</math> pieces gone from the equilateral triangle, we have, in total, <math>\dfrac{1}{2}</math> of a circle (with radius <math>2</math>) gone. Each circle has an area of <math>\pi r^2 = 4pi</math>, so three circles gives a total area of <math>12\pi</math>. Subtracting the half circle, we have: <cmath>12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.</cmath>
| |
| − |
| |
| − | 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. I'll post some more proofs that I've written below if you want...
| |
