AMC10 2025 B

We solve this by finding the areas of trapezoids. The line is $y = \frac{1}{3}x + 1$. Total Area ($A_{\text{total}}$) of Lower Region: The region from $x=0$ to $x=2$ is a trapezoid. \[A_{\text{total}} = \frac{1}{2}(y(0) + y(2)) \times (2 - 0) = \frac{1}{2}\left(1 + \frac{5}{3}\right)(2) = \frac{8}{3}\] Half Area ($A_{\text{half}}$): The area of the region from $x=0$ to $x=a$ must be $\frac{1}{2}A_{\text{total}}$. \[A_{\text{half}} = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3}\] Solve for $a$: This half-region is also a trapezoid. \[A_{\text{half}} = \frac{1}{2}(y(0) + y(a)) \times (a - 0) = \frac{4}{3}\] \[\frac{1}{2}\left(1 + \left(\frac{1}{3}a + 1\right)\right)a = \frac{4}{3}\] \[\frac{1}{2}\left(2 + \frac{a}{3}\right)a = \frac{4}{3}\] \[a + \frac{a^2}{6} = \frac{4}{3}\] Multiplying by 6 gives the quadratic equation: \[a^2 + 6a - 8 = 0\] \[a = \frac{-6 \pm \sqrt{6^2 - 4(1)(-8)}}{2} = \frac{-6 \pm \sqrt{68}}{2} = \frac{-6 \pm 2\sqrt{17}}{2} = -3 \pm \sqrt{17}\] Since $a$ must be positive, $a = \sqrt{17} - 3$. \[s + t = 17 + 3 = 20\] The correct option is $C$. Note: We can find the area of the trapezoid by integrating. We have the area is $\int_{0}^{2} \frac{1}{3}x + 1 \ dx = \frac{1}{6}x^2 + x \bigg|_{x=0}^{2} = \frac{8}{3}$. But this sort of destroys the purpose of this simple solution without calculus. With calculus, the solution below would be the most efficient.