Last Post: 1:11:33am, 05/01/2018
I know this topic is dead but I was thinking about it last night so hard while I was trying to go to sleep that I was imagining calculus in my head... Anyway, here is the calculus method:
(forget what I said earlier about considering the plot turned 90 degrees)
Let's call the upward line f'(xA) and the downward line f'(xB). The solution is area C. The point where they meet is coordinate (x1,y1). Assume the square is a unit size, 1 x 1. What we want to find is the integral (area underneath) the first line up to the point where they meet (x1), and from that point the integral of the second line until it meets the x-axis at x=1.
C = f(xA) from 0 to x1 + f(xB) from x1 to 1.
f'(xA) = (1/(1/2))x + 0 = 2x
f'(xB) = -(1/1)x + 1 = 1 - x
f(xA) = integrate f'(xA) = x^2
f(xB) = integrate f'(xB) = x - (x^2)/2
Find x1: This is the point where f'(xA) = f'(xB).
2*x1 = 1 - x1
2*x1 + x1 = 1
x1 = 1/3
Solve for C = [f(xA) 0_x1] + [f(xB) x1_1].
C = [(1/3)^2 - (0)^2] + [(1 - (1^2)/2) - (1/3 - (1/3)^2/2)]
C = [1/9] + [1/2 - 5/18]
C = [2/18] + [4/18]
C = 1/3
This is the most similar to K3nnyan's geometric solution which was to solve for the point where they meet and find the area of the two triangles, it's different math but the same concept. To me, robrobd's algebraic solution was the most elegant. My solution here is the most rigorous proof, it doesn't require any geometric knowledge (like how to split a rectangular area into two triangles) or trigonometry like Magus's calculator method. The reason mine is more robust is because the other methods would fall apart if you replaced one of the lines with, say, a circular arc, or some other arbitrary curve.
Edit: This was fun :)
I don't mean to say my method was "better" than anyone and I'm not hating on anyone's math. I agree it's really fun finding all the different ways it can be solved.
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- ThanatosMace