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Geometry Difficulty 3.6 AMC 10/12 Find the answer China

On a coordinate plane there are two regions, MM and NN:
MM is confined by
{y≥0,y≤x,y≤2−x \begin{cases} y \ge 0, \\ y \le x, \\ y \le 2-x \end{cases}
and NN is determined by the inequalities t≤x≤t+1t \le x \le t+1, 0≤t≤10 \le t \le 1. Then the size of the common area of MM and NN is given by f(t)=‾f(t) = \underline{\hspace{2cm}}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

As shown in the figure, we have
f(t)=Sshaded area=S△AOB−S△OCD−S△BEF=1−12t2−12(1−t)2=−t2+t+12,0≤t≤1. \begin{aligned} f(t) &= S_{\text{shaded area}} \\ &= S_{\triangle AOB} - S_{\triangle OCD} - S_{\triangle BEF} \\ &= 1 - \frac{1}{2}t^2 - \frac{1}{2}(1-t)^2 \\ &= -t^2 + t + \frac{1}{2}, \quad 0 \le t \le 1. \end{aligned}
Figure 1

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