2025 AMC 10B Problem 6

First locate the line inside the square. At $x = 0$ it is at $y = 1$; at $x = 2$ it is at $y = \tfrac53$. Both heights lie strictly between $0$ and $2$, so the line enters through the left edge and leaves through the right edge, and the lower region is the trapezoid with vertices $(0,0)$, $(2,0)$, $\left(2, \tfrac53\right)$, $(0,1)$. Its area is
$$\frac{1 + \tfrac53}{2}\cdot 2 = \frac{8}{3},$$
so each half must have area $\tfrac43$.

The part of the lower region with $0 \le x \le a$ is again a trapezoid: its parallel sides are the heights of the line at $x = 0$ and $x = a$, namely $1$ and $\tfrac{a}{3} + 1$, and its width is $a$. Hence
$$\frac{1 + \left(\tfrac{a}{3} + 1\right)}{2}\cdot a = a + \frac{a^2}{6}.$$

Set this equal to $\tfrac43$ and multiply through by $6$:
$$a^2 + 6a = 8 \quad\Longrightarrow\quad a^2 + 6a + 9 = 17 \quad\Longrightarrow\quad (a+3)^2 = 17.$$
Since $a > 0$, $a = \sqrt{17} - 3 \approx 1.12$, which indeed lies in $[0,2]$. So $s = 17$, $t = 3$, and $s + t = 20$.

The answer is $\boxed{\textbf{(C)}\ 20}$.

Checking the line's height at both edges of the square is what identifies the shape: had the line left through the top edge, the lower region would have been a pentagon or triangle and the trapezoid formula would not apply. Once the shape is a trapezoid, the area of the left piece is a quadratic in $a$, and the requested form $\sqrt{s} - t$ is a strong hint that completing the square is the intended finish. The same set-up handles any "line cuts a rectangle, find the equal-area vertical cut" problem.

• Halving the whole square rather than the lower region, that is setting a + a^2/6 = 2, which gives a = sqrt(21) - 3 and s + t = 24, not on the list.
• Assuming $a = 1$ by symmetry; the trapezoid is taller on the right, so the equal-area cut sits to the right of the midline.
• Keeping the negative root $a = -3 - \sqrt{17}$, or writing the positive root as $\sqrt{17} + 3$ while still reading off $t = 3$.