2009 AMC 10B Problem 17

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?

\dfrac12

\dfrac35

\dfrac23

\dfrac34

\dfrac45

The whole region has area

5,

so each part must be

\dfrac52

The lower-right part is a triangle of base

3-a

and height

3,

with one unit square removed

The five unit squares have total area

5,

so each region must have area

\dfrac52.

The region to the lower right of the line is a right triangle with legs

3-a

and

3,

minus the one unit square it does not cover. Setting its area to

\dfrac52

gives

\dfrac{3(3-a)}{2}-1=\dfrac52,

so

3(3-a)=7

and

a=\dfrac23.

Thus, the correct answer is C.

Problem 17 in other years