2009 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2009 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

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Concepts:triangle areaarea decompositioncoordinate geometry

Difficulty rating: 1610

14.

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\dfrac{1}{2}

35\dfrac{3}{5}

23\dfrac{2}{3}

34\dfrac{3}{4}

45\dfrac{4}{5}

Solution:

The five squares have total area 5,5, so each region must have area 52.\dfrac{5}{2}.

The line from (a,0)(a, 0) to (3,3)(3, 3) together with the axes bounds a triangle of base 3a3 - a and height 33; the region on the lower-right side of the line is this triangle with one unit square removed. Setting 3(3a)21=52 \dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2} gives 3(3a)=7,3(3 - a) = 7, so a=23.a = \dfrac{2}{3}.

Thus, the correct answer is C.

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