2009 AMC 10B Problem 17

Attempt Problem 17 of the 2009 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 2009 AMC 10B solutions, or check the answer key.

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17.

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)(a,0) to (3,3),(3,3), divides the entire region into two regions of equal area. What is a?a?

12\dfrac12

35\dfrac35

23\dfrac23

34\dfrac34

45\dfrac45

Answer: C
Concepts:triangle areaarea decompositioncoordinate geometry
Difficulty rating: 1540
Solution:

The five unit squares have total area 5,5, so each region must have area 52.\dfrac52.

The region to the lower right of the line is a right triangle with legs 3a3-a and 3,3, minus the one unit square it does not cover. Setting its area to 52\dfrac52 gives 3(3a)21=52, \dfrac{3(3-a)}{2}-1=\dfrac52, so 3(3a)=73(3-a)=7 and a=23.a=\dfrac23.

Thus, the correct answer is C.

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