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Difference between revisions of "2010 AMC 8 Problems/Problem 17"

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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));
 +
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>
 +
 
 
<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>
  
−
==Solution==
+
==Solution 1==
 +
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.
 +
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath>
 +
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math>
 +
 
 +
==Solution 2==
 +
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>.
 +
 
 +
After moving the square from the bottom right to the top left, the area of the resulting trapezoid is <math>5+1=6</math>.
 +
 
 +
<math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>.
 +
 
 +
Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>.
 +
 
 +
Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math>
 +
 
 +
~kempwood
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/j3QSD5eDpzU?t=937
 +
 
 +
 
 +
==Video by MathTalks==
 +
 
 +
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/N7Yu9-bLqls
  
−
(D)<math>\frac{2}{3}</math>
+
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=16|num-a=18}}
 
{{AMC8 box|year=2010|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 20:00, 23 October 2024

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 $\dfrac{XQ}{QY}$?

[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 $5$.

After moving the square from the bottom right to the top left, the area of the resulting 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 $\boxed{\textbf{(D) }\frac{2}{3}}$

~kempwood

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=937


Video by MathTalks

https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be

Video Solution by WhyMath

https://youtu.be/N7Yu9-bLqls

~savannahsolver

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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