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

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The entire figure has area <math>5</math>, hence we want the shaded part to have area <math>\frac 52</math>. Solving for <math>a</math>, we get <math>a=\boxed{\frac 23}</math>.  The answer is <math>\mathrm{(C)}</math>.
 
The entire figure has area <math>5</math>, hence we want the shaded part to have area <math>\frac 52</math>. Solving for <math>a</math>, we get <math>a=\boxed{\frac 23}</math>.  The answer is <math>\mathrm{(C)}</math>.
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 +
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== Solution 2 ==
 +
The unit square is of area 1, so the five unit squares have area 5.
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Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of are 3.5.
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It can then be inferred:
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(3-a) * 3 = 7.
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3-a = <math>\frac{7}{3}</math>.
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3 - <math>\frac{7}{3}</math> = <math>\frac{9-7}{3}</math> = <math>\frac{2}{3}</math>. <math>\mathrm{(C)}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 13:14, 13 October 2016

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

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 unit square is of area 1, so the five unit squares have area 5. Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of are 3.5. It can then be inferred: (3-a) * 3 = 7. 3-a = $\frac{7}{3}$. 3 - $\frac{7}{3}$ = $\frac{9-7}{3}$ = $\frac{2}{3}$. $\mathrm{(C)}$.

See Also

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

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