Difference between revisions of "2009 AMC 12B Problems/Problem 14"

Revision as of 09:56, 4 July 2013

The following problem is from both the 2009 AMC 10B #17 and 2009 AMC 12B #14, so both problems redirect to this page.

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

Solution 1

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 $a=\boxed{\frac 23}$ . The answer is $\mathrm{(C)}$ .

Solution 2

The total area is 5, so the area of the shaded area is $2.5$ . If we add a unit square in the lower right corner, the area is $3.5$ . Therefore $\frac{(3-a)\cdot 3}{2}=3.5$ , or $3-a=\frac{7}{3}$ . Therefore $a=\frac{2}{3}$ .

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC10 box|year=2009|ab=B|num-b=16|num-a=18}}
 {{AMC12 box|year=2009|ab=B|num-b=13|num-a=15}}
+{{MAA Notice}}

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