2010 AMC 8 Problems/Problem 17
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
?
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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Solution
We see that half the area of the octagon is
. We see that the triangle area is
. That means that
.
Meaning,
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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