2009 AMC 12B Problems/Problem 14

Problem

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 $a$ ? $[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(a,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]$ $\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$

Solution

For $a\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $a<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.

Then the area of the shaded part is one less than the area of the triangle with vertices $(a,0)$ , $(3,0)$ , and $(3,3)$ . Its area is obviously $\frac{3(3-a)}2$ , therefore the area of the shaded part is $\frac{7-3a}{2}$ .

The entire figure has area $5$ , hence we want the shaded part to have area $\frac 52$ . Solving for $a$ , we get $a=\boxed{\frac 23}$ .

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2009 AMC 12B Problems/Problem 14

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