2025 AMC 10B Problem 6

The line

y = \tfrac{1}{3}x + 1

divides the square region defined by

0 \le x \le 2

and

0 \le y \le 2

into an upper region and a lower region. The line

x = a

divides the lower region into two regions of equal area. Then

a

can be written as

\sqrt{s} - t,

where

s

and

t

are positive integers. What is

s + t?

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Solution

The lower region has area

\int_0^2\left(\tfrac{x}{3} + 1\right)dx = \tfrac{2}{3} + 2 = \tfrac{8}{3}.

The slice with

0 \le x \le a

is a trapezoid of area

a + \tfrac{a^2}{6}.

We want that to be half the total, namely

\tfrac{4}{3},

a^2 + 6a - 8 = 0

and

a = -3 + \sqrt{17}.

Then

s = 17,

t = 3,

and

s + t = 20.

Therefore, the answer is C.