2025 AMC 10B Problem 6

Attempt Problem 6 of the 2025 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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6.

The line y=13x+1y = \tfrac{1}{3}x + 1 divides the square region defined by 0x20 \le x \le 2 and 0y20 \le y \le 2 into an upper region and a lower region. The line x=ax = a divides the lower region into two regions of equal area. Then aa can be written as st,\sqrt{s} - t, where ss and tt are positive integers. What is s+t?s + t?

1818

1919

2020

2121

2222

Answer: C
Concepts:trapezoidareaquadratic
Difficulty rating: 1410
Solution:

The lower region has area 02(x3+1)dx=23+2=83.\int_0^2\left(\tfrac{x}{3} + 1\right)dx = \tfrac{2}{3} + 2 = \tfrac{8}{3}. The slice with 0xa0 \le x \le a is a trapezoid of area a+a26.a + \tfrac{a^2}{6}. We want that to be half the total, namely 43,\tfrac{4}{3}, so a2+6a8=0a^2 + 6a - 8 = 0 and a=3+17.a = -3 + \sqrt{17}. Then s=17,s = 17, t=3,t = 3, and s+t=20.s + t = 20. Therefore, the answer is C.

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