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Difference between revisions of "2015 AMC 8 Problems/Problem 2"
Line 128: Line 128:
  
 
== Solution 4 ==
 
== Solution 4 ==
 +
<asy>
 
A=dir(45);
 
A=dir(45);
 
B=dir(90);
 
B=dir(90);
Line 142: Line 143:
 
draw(A--B--C--D--E--F--G--H--cycle);
 
draw(A--B--C--D--E--F--G--H--cycle);
  
−
dot("<math>A</math>",A,dir(45));
+
dot("$A$",A,dir(45));
−
dot("<math>B</math>",B,dir(90));
+
dot("$B$",B,dir(90));
−
dot("<math>C</math>",C,dir(135));
+
dot("$C$",C,dir(135));
−
dot("<math>D</math>",D,dir(180));
+
dot("$D$",D,dir(180));
−
dot("<math>E</math>",E,dir(-135));
+
dot("$E$",E,dir(-135));
−
dot("<math>F</math>",F,dir(-90));
+
dot("$F$",F,dir(-90));
−
dot("<math>G</math>",G,dir(-45));
+
dot("$G$",G,dir(-45));
−
dot("<math>H</math>",H,dir(0));
+
dot("$H$",H,dir(0));
−
dot("<math>X</math>",X,dir(135/2));
+
dot("$X$",X,dir(135/2));
−
dot("<math>O</math>",O,dir(0));
+
dot("$O$",O,dir(0));
 
draw(E--O--X);
 
draw(E--O--X);
 
draw(C--O--B);
 
draw(C--O--B);

Revision as of 17:46, 17 January 2023

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

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

Solution 4

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);
 (Error making remote request. Unknown error_msg)

We can divide the octagon into 8 parts and pretend that the area is 64. We know that X is the midpoint of BA and that each space between two points is 8 because 64/8=8. This means that BX=4 because 8/2=4. Then we add that to 3*8 because there are 3 spaces between points that are each 8 ft. After that, you turn it into a fraction, 28/64 and simplify to get $\boxed{\textbf{(D)}~\frac{7}{16}}$.

Video Solution

https://youtu.be/NbIav9YlPEY

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=2314

~ pi_is_3.14

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

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