# Difference between revisions of "2015 AMC 8 Problems/Problem 2"

## Problem

Point $O$ is the center of the regular octagon $ABCDEFGH$, and $X$ is the midpoint of the side $\overline{AB}.$ What fraction of the area of the octagon is shaded?

$\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}$

$[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("A",A,dir(45)); dot("B",B,dir(90)); dot("C",C,dir(135)); dot("D",D,dir(180)); dot("E",E,dir(-135)); dot("F",F,dir(-90)); dot("G",G,dir(-45)); dot("H",H,dir(0)); dot("X",X,dir(135/2)); dot("O",O,dir(0)); draw(E--O--X); [/asy]$

## Solutions

### Solution 1

Since octagon $ABCDEFGH$ is a regular octagon, it is split into $8$ equal parts, such as triangles $\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO$, etc. These parts, since they are all equal, are $\frac{1}{8}$ 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 $\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\textbf{(D) }\dfrac{7}{16}}.$

### Solution 2

$[asy] pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); a=midpoint(B--C); b=midpoint(C--D); c=midpoint(D--E); d=midpoint(E--F); e=midpoint(F--G); f=midpoint(G--H); g=midpoint(H--A); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("A",A,dir(45)); dot("B",B,dir(90)); dot("C",C,dir(135)); dot("D",D,dir(180)); dot("E",E,dir(-135)); dot("F",F,dir(-90)); dot("G",G,dir(-45)); dot("H",H,dir(0)); dot("X",X,dir(135/2)); dot("O",O,dir(0)); draw(E--O--X); draw(B--F); draw(A--O); draw(D--H); draw(C--G); draw(a--e); draw(b--f); draw(c--g); draw(d--O); [/asy]$

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 $\boxed{\textbf{(D)}~\dfrac{7}{16}}$.

### Solution 3

For starters what I find helpful is to divide the whole octagon up into triangles as shown here: $[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("A",A,dir(45)); dot("B",B,dir(90)); dot("C",C,dir(135)); dot("D",D,dir(180)); dot("E",E,dir(-135)); dot("F",F,dir(-90)); dot("G",G,dir(-45)); dot("H",H,dir(0)); dot("X",X,dir(135/2)); dot("O",O,dir(0)); draw(E--O--X); draw(C--O--B); draw(B--O--A); draw(A--O--H); draw(H--O--G); draw(G--O--F); draw(F--O--E); draw(E--O--D); draw(D--O--C); [/asy]$

Now it is just a matter of counting the larger triangles remember that $\triangle BOX$ and $\triangle XOA$ are not full triangles and are only half for these purposes. We count it up and we get a total of $\frac{3.5}{8}$ of the shape shaded. We then simplify it to get our answer of $\boxed{\textbf{(D)}~\frac{7}{16}}$.