Difference between revisions of "2015 AMC 8 Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
| − | + | <asy> | |
pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; | pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; | ||
A=dir(45); | A=dir(45); | ||
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draw(A--B--C--D--E--F--G--H--cycle); | draw(A--B--C--D--E--F--G--H--cycle); | ||
| − | dot(" | + | dot("$A$",A,dir(45)); |
| − | dot(" | + | dot("$B$",B,dir(90)); |
| − | dot(" | + | dot("$C$",C,dir(135)); |
| − | dot(" | + | dot("$D$",D,dir(180)); |
| − | dot(" | + | dot("$E$",E,dir(-135)); |
| − | dot(" | + | dot("$F$",F,dir(-90)); |
| − | dot(" | + | dot("$G$",G,dir(-45)); |
| − | dot(" | + | dot("$H$",H,dir(0)); |
| − | dot(" | + | dot("$X$",X,dir(135/2)); |
| − | dot(" | + | dot("$O$",O,dir(0)); |
draw(E--O--X); | draw(E--O--X); | ||
draw(B--F); | draw(B--F); | ||
| Line 79: | Line 79: | ||
draw(c--g); | draw(c--g); | ||
draw(d--O); | draw(d--O); | ||
| − | + | </asy> | |
The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>. | The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>. | ||
Revision as of 17:51, 25 November 2015
Point 
is the center of the regular octagon 
, and 
is the midpoint of the side 
What fraction of the area of the octagon is shaded?
![[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("$A$",A,dir(45)); dot("$B$",B,dir(90)); dot("$C$",C,dir(135)); dot("$D$",D,dir(180)); dot("$E$",E,dir(-135)); dot("$F$",F,dir(-90)); dot("$G$",G,dir(-45)); dot("$H$",H,dir(0)); dot("$X$",X,dir(135/2)); dot("$O$",O,dir(0)); draw(E--O--X); [/asy]](http://latex.artofproblemsolving.com/0/6/6/06635650c87a3cbd84380b9105fc49cc48e3792e.png)
Solution 1
Since octagon is a regular octagon, it is split into 8 equal parts, such as triangles 
, etc. These parts, since they are all equal, are 
of the octagon each. The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is 
Solution 2
![[asy] pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); a=midpoint(B--C); b=midpoint(C--D); c=midpoint(D--E); d=midpoint(E--F); e=midpoint(F--G); f=midpoint(G--H); g=midpoint(H--A); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("$A$",A,dir(45)); dot("$B$",B,dir(90)); dot("$C$",C,dir(135)); dot("$D$",D,dir(180)); dot("$E$",E,dir(-135)); dot("$F$",F,dir(-90)); dot("$G$",G,dir(-45)); dot("$H$",H,dir(0)); dot("$X$",X,dir(135/2)); dot("$O$",O,dir(0)); draw(E--O--X); draw(B--F); draw(A--O); draw(D--H); draw(C--G); draw(a--e); draw(b--f); draw(c--g); draw(d--O); [/asy]](http://latex.artofproblemsolving.com/0/6/5/0650daad76500b862bb75dbf6f0a81620b3471c5.png)
The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is 
.
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
