2025 AMC 10B Problem 6

Question

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?

18

19

20

21

22

The lower region left of

x = a

is a trapezoid with vertical sides

1

and

\tfrac{a}{3} + 1

and width

a

The total lower area is

\tfrac{8}{3},

so set

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

equal to half of it,

\tfrac{4}{3},

and solve the quadratic

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},

so

a^2 + 6a - 8 = 0

and

a = -3 + \sqrt{17}.

Then

s = 17,

t = 3,

and

s + t = 20.

Therefore, the answer is C.

Problem 6 in other years