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2010 AMC 8 Problems/Problem 17: Difference between revisions

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==Problem==
==Problem==
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?
<asy>
<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);
import graph; size(300);  
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)+linetype("2pt 2pt")); draw((1,1)--(1,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((2,2)--(2,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((3,2)--(3,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((4,2)--(4,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((5,1)--(5,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((0,0)--(5,1.5),linewidth(1.2pt));
real lsf = 0.5;  
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>
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);
//+linetype("2pt 2pt"));
draw((1,1)--(1,0),linewidth(1.2pt);
//+linetype("2pt 2pt"));  
draw((2,2)--(2,0),linewidth(1.2pt);
//+linetype("2pt 2pt"));  
draw((3,2)--(3,0),linewidth(1.2pt);
//+linetype("2pt 2pt"));  
draw((4,2)--(4,0),linewidth(1.2pt);
//+linetype("2pt 2pt"));  
draw((5,1)--(5,0),linewidth(1.2pt);
//+linetype("2pt 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>
 
<math> \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} </math>
<math> \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} </math>


Revision as of 12:37, 19 March 2015

Problem

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 $\dfrac{XQ}{QY}$? 

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);
//+linetype("2pt 2pt"));
 draw((1,1)--(1,0),linewidth(1.2pt);
//+linetype("2pt 2pt")); 
draw((2,2)--(2,0),linewidth(1.2pt);
//+linetype("2pt 2pt")); 
draw((3,2)--(3,0),linewidth(1.2pt);
//+linetype("2pt 2pt")); 
draw((4,2)--(4,0),linewidth(1.2pt);
//+linetype("2pt 2pt")); 
draw((5,1)--(5,0),linewidth(1.2pt);
//+linetype("2pt 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);
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$\textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$

Solution

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, $\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}$

See Also

2010 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

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