2015 AMC 8 Problems/Problem 2
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)
Contents
Solution 1
Since octagon is a regular octagon, it is split into 
equal parts, such as triangles 
, etc. These parts, since they are all equal, are 
of the octagon each. The shaded region consists of 
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 
identical triangles (and thus they each have equal area). Since the shaded region occupies 
out of the total triangles, the answer is 
.
Solution 3
For starters what I find helpful is to divide the whole octagon up into triangles as shown here:
![[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); draw(C--O--B); draw(B--O--A); draw(A--O--H); draw(H--O--G); draw(G--O--F); draw(F--O--E); draw(E--O--D); draw(D--O--C); [/asy]](http://latex.artofproblemsolving.com/b/a/8/ba8f1ed7f1c93637ee903d0935139cee9be6f540.png)
Now it is just a matter of counting the larger triangles remember that 
and 
are not full triangles and are only half for these purposes. We count it up and we get a total of 
of the shape shaded. We then simplify it to get our answer: 
or 
.
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. 
