Area under envelope of a family of lines

Each line in the family of lines has an equation of $y = \cfrac{0 - (1 - t)}{t - 0}x + 1 - t$ or $y = x - \cfrac{x}{t} + 1 - t$ .

The envelope is made up of maximum heights of these lines at a given $x$ -coordinate, so $\cfrac{dy}{dt} = \cfrac{x}{t^2} - 1 = 0$ , which solves to $t = \pm \sqrt{x}$ .

Substituting $t = \pm \sqrt{x}$ into $y = x - \cfrac{x}{t} + 1 - t$ and simplifying gives $y = x - 2\sqrt{x} + 1$ for $y < 1$ .

The area under the curve is then $\displaystyle \int_{0}^{1} (x - 2\sqrt{x} + 1) \,dx = \cfrac{1}{6}$ , so $p = 1$ , $q = 6$ , and $p + q = \boxed{7}$ .

The straight line from $(t,0)$ to $(0,1-t)$ has equation $(1-t)x + ty \; = \; t(1-t)$ We obtain the envelope by differentiating this partially with respect to $t$ , so $y - x \; = \; 1 - 2t$ and eliminating $t$ from these two equations. This yields $2(x+y) \; = \; 1 + (x-y)^2$ Solving for $y$ gives $y = x + 1 - 2\sqrt{x}$ , and hence the desired area is $\tfrac16$ , making the answer $\boxed{7}$ .

It is interesting to rotate the coordinates by $45^\circ$ , using the coordinates $X = \tfrac{1}{\sqrt{2}}(x - y)$ and $Y = \tfrac{1}{\sqrt{2}}(x + y)$ , which makes the equation of the envelope $2\sqrt{2}Y \; = \; 1 + 2X^2$ and hence the curve is a parabola.