2009 AMC 12B Problem 14

Question

Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from

(a, 0)

to

(3, 3),

divides the entire region into two regions of equal area. What is

a?

\dfrac{1}{2}

\dfrac{3}{5}

\dfrac{2}{3}

\dfrac{3}{4}

\dfrac{4}{5}

The total area is

5,

so each region must be

\dfrac{5}{2}

The region on the lower-right side of the line is a triangle of base

3 - a

and height

3,

minus one unit square

The five squares have total area

5,

so each region must have area

\dfrac{5}{2}.

The line from

(a, 0)

to

(3, 3)

together with the axes bounds a triangle of base

3 - a

and height

3

; the region on the lower-right side of the line is this triangle with one unit square removed. Setting

\dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2}

gives

3(3 - a) = 7,

so

a = \dfrac{2}{3}.

Thus, the correct answer is C.

Problem 14 in other years