2025 AMC 10B Problem 6

1. See where the line goes. At $x = 0$ the line is at height $y = 1$. At $x = 2$ it is at height $y = \frac{2}{3} + 1 = \frac53$. Both heights are between $0$ and $2$, so the line goes in through the left side of the square and out through the right side.

2. Find the area of the lower region. Below the line is a trapezoid with vertical sides of lengths $1$ and $\frac53$ and width $2$:
$$\text{area} = \frac{1 + \frac53}{2} \times 2 = \frac83.$$
Each half must have area $\frac83 \div 2 = \frac43$.

3. Find the area to the left of $x = a$. This piece is also a trapezoid. Its left side has height $1$, its right side has height $\frac{a}{3} + 1$, and its width is $a$:
$$\text{area} = \frac{1 + \left(\frac a3 + 1\right)}{2} \times a = \left(1 + \frac{a}{6}\right) a = a + \frac{a^2}{6}.$$

4. Set it equal to $\frac43$ and solve. Multiply both sides of $a + \frac{a^2}{6} = \frac43$ by $6$:
$$a^2 + 6a = 8.$$
The answer is supposed to look like $\sqrt{s} - t$, which hints at completing the square. Add $9$ to both sides:
$$a^2 + 6a + 9 = 17, \qquad (a + 3)^2 = 17.$$
Since $a$ is positive, $a + 3 = \sqrt{17}$, so $a = \sqrt{17} - 3 \approx 1.12$. That is between $0$ and $2$, as it should be.

1. So $s = 17$, $t = 3$, and $s + t = 20$.

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

Checking the line's height at both edges tells you the shape of the region, and a trapezoid has an easy area formula. Cutting it with a vertical line gives a smaller trapezoid whose area is a quadratic in $a$. When an answer is supposed to be of the form $\sqrt{s} - t$, completing the square is almost always the intended finish.

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 $\boxed{\textbf{(C)}\ 20}$.