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

Revision as of 12:33, 10 August 2019

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 area 3.5. It can then be inferred: $(3-a) * 3 = 7$ .

$3-a=\frac{7}{3}$ , so $3-\frac{7}{3}=a$ .

$3-\frac{7}{3} = \frac{9-7}{3} = \frac{2}{3}$ . $\boxed{\mathrm{(C)}}$ .

Solution 3

The shaded space of the object can become a triangle by adding a unit square to its bottom. This triangle has a base length of $3-a$ and a height of $3$ . The area of this triangle region is now (using the formula $A=bh/2$ for a triangle) $(9-3a)/2$ . But, remember that we have to subtract the area of the extra unit square from this area to get the area of the shaded region. If you add 3 unit squares to the unshaded portion, the figure is now a right trapezoid. Using the formula for a trapezoid, you can derive that its area is $(9+3a)/2$ . After you subtract the area of the 3 extra unit squares, we have derived the area of the unshaded portion. Since these areas are both equal, we set them equal to each other and solve for $a$ . $(9-3a)/2 -1 = (9+3a)/2 -3$ $a$ is now solved to be $\frac{2}{3}$ . $\boxed{\mathrm{(C)}}$ .

~Solution by athens2016

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.

@@ Line 49: / Line 49: @@
 <math>a</math> is now solved to be <math>\frac{2}{3}</math>. <math>\boxed{\mathrm{(C)}}</math>.
+~Solution by athens2016
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