2010 AMC 8 Problems/Problem 17

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The diagram shows an octagon consisting of $10$ unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base $5$. If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\dfrac{XQ}{QY}$?

[asy] import graph; size(300);  real lsf = 0.5;  pen dp = linewidth(0.7) + fontsize(10);  defaultpen(dp);  pen ds = black;  pen xdxdff = rgb(0.49,0.49,1); draw((0,0)--(6,0),linewidth(1.2pt));  draw((0,0)--(0,1),linewidth(1.2pt));  draw((0,1)--(1,1),linewidth(1.2pt));  draw((1,1)--(1,2),linewidth(1.2pt));  draw((1,2)--(5,2),linewidth(1.2pt));  draw((5,2)--(5,1),linewidth(1.2pt));  draw((5,1)--(6,1),linewidth(1.2pt));  draw((6,1)--(6,0),linewidth(1.2pt));  draw((1,1)--(5,1),linewidth(1.2pt)); draw((1,1)--(1,0),linewidth(1.2pt));  draw((2,2)--(2,0),linewidth(1.2pt));  draw((3,2)--(3,0),linewidth(1.2pt));  draw((4,2)--(4,0),linewidth(1.2pt));  draw((5,1)--(5,0),linewidth(1.2pt)); draw((0,0)--(5,1.5),linewidth(1.2pt)); dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf);  dot((0,1),ds);  dot((1,1),ds);  dot((1,2),ds);  dot((5,2),ds);  label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds);  label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds);  dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds);  dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds);  dot((4,2),ds); dot((5,1.5),ds);  label("$Q$", (5.14,1.51),NE*lsf);  clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle); [/asy]

$\textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$

Solution 1

We see that half the area of the octagon is $5$. We see that the triangle area is $5-1 = 4$. That means that $\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}$. \[\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}\] Meaning, $\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}$

Solution 2

Like stated in solution 1, we know that half the area of the octagon is $5$.

That means that the area of the trapezoid is $5+1=6$.

$5(XQ+2)/2=6$. Solving for $XQ$, we get $XQ=2/5$.

Subtracting $2/5$ from $1$, we get $QY=3/5$.

Therefore, the answer comes out to $\boxed{\textbf{(D) }\frac{2}{3}}$


Solution 3

We can move the cube on the bottom right to the top left, thus creating a rectangle. We thus know that now this has been separated to 2 triangle, with equal area, although, obviously, one is not a perfect triangle. From this we can subtract 1 from one of the 5's (the area of one of the triangle-shapes-polygons) and add it to the other, since we have just moved the block, original in the triangle-shaped-polygon on the bottom, and thus, we get 4/6, or $\boxed{\textbf{(D) }\frac{2}{3}}$ -RealityWrites

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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