2009 AMC 12B Problems/Problem 14

Revision as of 21:54, 17 August 2021 by Mewto (talk | contribs) (Solutions)
The following problem is from both the 2009 AMC 10B #17 and 2009 AMC 12B #14, so both problems redirect to this page.


Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$, divides the entire region into two regions of equal area. What is $c$?

[asy] unitsize(0.2cm); 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("$(c,0)$",(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 1

For $c\geq 1.5$ the shaded area is at most $1.5$, which is too little. Hence $c<1.5$, 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 $(c,0)$, $(3,0)$, and $(3,3)$. The area of the entire triangle is $\frac{3(3-c)}2$, therefore the area of the shaded part is $\frac{7-3c}{2}$.

The entire figure has area $5$, hence we want the shaded part to have area $\frac 52$. Solving for $c$, we get $c=\boxed{\frac 23}$. 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: $(3-c) * 3 = 7$.

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

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

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 $3-c$ and a height of $3$. The area of this triangle region is now (using the formula $A=bh/2$ for a triangle)$(9-3c)/2$. 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 top of the unshaded portion, the figure is now a right trapezoid. Using the formula for a trapezoid, you can derive that its area is $(9+3c)/2$. 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 $c$. $(9-3c)/2 -1 = (9+3c)/2 -3$. $c$ is now solved to be $\frac{2}{3}$. $\boxed{\mathrm{(C)}}$.

~Solution by athens2016

Solution 4

Test all solutions by solving for the triangle area. When you find a triangle minus 1 equals $\frac{5}{2}$, that should be your answer. $\boxed{\mathrm{(C)}}$

Solution 5

Use the shoelace theorem to get the area of the unshaded portion in terms of a. Divide this area by the whole area which is 5. This fraction should be equal to $1/2$. You can easily solve for the answer from there. -mewto

See Also

2009 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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