jasperE3 wrote: For each $a$, $1<a<2$, the graphs of functions $y=1-|x-1|$ and $y=|2x-a|$ determine a figure. Prove that the area of this figure is less than $\frac13$. Figure is quadrilateral $\left(\frac a3,\frac a3\right),(1,1),\left(\frac{a+2}3,\frac{4-a}3\right),\left(\frac a2,0\right)$ Whose area is easily computed as $\frac{-a^2+4a-2}6$ Whose supremum indeed is $\boxed{\frac 13}$, reached when $a\to 2^-$