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 | + | 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>. | ||
+ | |||
+ | That means that the area of the 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 | ||
+ | |||
+ | ==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 <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
+ | -RealityWrites | ||
+ | |||
+ | ==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== | ==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 21:04, 23 December 2023
Contents
Problem
The diagram shows an octagon consisting of unit squares. The portion below
is a unit square and a triangle with base
. If
bisects the area of the octagon, what is the ratio
?
Solution 1
We see that half the area of the octagon is . We see that the triangle area is
. That means that
.
Meaning,
Solution 2
Like stated in solution 1, we know that half the area of the octagon is .
That means that the area of the trapezoid is .
. Solving for
, we get
.
Subtracting from
, we get
.
Therefore, the answer comes out to
~kempwood
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
-RealityWrites
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
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.