3 ms·

While we're all solving puzzles, can anyone help me with this one? No doubt you've doodled this shape in your graph paper before: http://i.imgur.com/oY29sBc.pn

• anonymoushn 14y ago
  This is a quadratic Bézier curve http://en.wikipedia.org/wiki/B%C3%A9zier_curve http://en.wikipedia.org/wiki/B%C3%A9zier_curve
  • jacobolus 14y ago
    Or in other words, it’s a section of a parabola. (As it happens, it’s a parabola tilted 45°, but still.)
• fahadkhan 14y ago
  sqrt(x)+sqrt(y)=C
• chnuschper 14y ago
  It's y=(sqrt(x) - 1)^2 Consider the function l(t) which gives the appropriate line function f(x) for the line going through (t,0). The function we are looking for can then be defined as c(x) := max(l(t)(x), t in (0;1]). Just solve for t and pop it into l(t), et voila.
  • jules 14y ago
    Sorry, I accidentally down-voted you when I tried to up-vote you. This answer is absolutely correct. Here is my very similar derivation: All lines go through (t,0) and (0,1-t). Now take two such lines, one with constant t and another with constant s and find their intersection. We have: y = 1-t - (1-t)/t * x y = 1-s - (1-s)/s * x So: 1-t - (1-t)/t * x = 1-s - (1-s)/s s-t = (s-t)/(st) * x So: x = st The points on the curve will be generated by the intersection of two almost adjacent lines (this is easy to see geometrically), so we take s =~ t. Then we have: x = t^2 y = (1-t)^2 So we have: y = (1 - sqrt(x))^2 Or, my favorite form: sqrt(x) + sqrt(y) = 1