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)
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,dir(45));
dot("
",B,dir(90));
dot("
",C,dir(135));
dot("
",D,dir(180));
dot("
",E,dir(-135));
dot("
",F,dir(-90));
dot("
",G,dir(-45));
dot("
",H,dir(0));
dot("",X,dir(135/2));
dot("",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]
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. 
