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

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==Problem==
 +
 
Point <math>O</math> is the center of the regular octagon <math>ABCDEFGH</math>, and <math>X</math> is the midpoint of the side <math>\overline{AB}.</math> What fraction of the area of the octagon is shaded?
 
Point <math>O</math> is the center of the regular octagon <math>ABCDEFGH</math>, and <math>X</math> is the midpoint of the side <math>\overline{AB}.</math> What fraction of the area of the octagon is shaded?
  
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draw(E--O--X);
 
draw(E--O--X);
 
</asy>
 
</asy>
−
 
+
==Solutions==
−
==Solution 1==
+
===Solution 1===
  
 
Since octagon <math>ABCDEFGH</math> is a regular octagon, it is split into <math>8</math> equal parts, such as triangles <math>\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO</math>, etc.  These parts, since they are all equal, are <math>\frac{1}{8}</math> of the octagon each.  The shaded region consists of <math>3</math> of these equal parts plus half of another, so the fraction of the octagon that is shaded is <math>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\textbf{(D) }\dfrac{7}{16}}.</math>
 
Since octagon <math>ABCDEFGH</math> is a regular octagon, it is split into <math>8</math> equal parts, such as triangles <math>\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO</math>, etc.  These parts, since they are all equal, are <math>\frac{1}{8}</math> of the octagon each.  The shaded region consists of <math>3</math> of these equal parts plus half of another, so the fraction of the octagon that is shaded is <math>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\textbf{(D) }\dfrac{7}{16}}.</math>
  
−
==Solution 2==
+
===Solution 2===
 
<asy>
 
<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;
Line 83: Line 85:
 
The octagon has been divided up into <math>16</math> identical triangles (and thus they each have equal area). Since the shaded region occupies <math>7</math> out of the <math>16</math> total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>.
 
The octagon has been divided up into <math>16</math> identical triangles (and thus they each have equal area). Since the shaded region occupies <math>7</math> out of the <math>16</math> total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>.
  
−
==Solution 3==
+
-Flare
  
−
For starters what I find helpful is to divide the whole octagon up into triangles as shown here:
+
===Solution 3===
 +
 
 +
For starters, what I find helpful is to divide the whole octagon up into triangles as shown here:
 
<asy>
 
<asy>
 
pair A,B,C,D,E,F,G,H,O,X;
 
pair A,B,C,D,E,F,G,H,O,X;
Line 123: Line 127:
 
</asy>
 
</asy>
  
−
Now it is just a matter of counting the larger triangles remember that <math>\triangle BOX</math> and <math>\triangle XOA</math> are not full triangles and are only half for these purposes. We count it up and we get a total of <math>\frac{3.5}{8}</math> of the shape shaded. We then simplify it to get our answer of <math>\boxed{\textbf{(D)}~\frac{7}{16}}</math>.
+
Now, it is just a matter of counting the larger triangles. Remember that <math>\triangle BOX</math> and <math>\triangle XOA</math> are not full triangles and are only half for these purposes. We count it up and we get a total of <math>\frac{3.5}{8}</math> of the shape shaded. We then simplify it to get our answer of <math>\boxed{\textbf{(D)}~\frac{7}{16}}</math>.
 +
 
 +
== Solution 4 ==
 +
<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));
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dot("$O$",O,dir(0));
 +
draw(E--O--X);
 +
draw(C--O--B);
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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>
 +
 
 +
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. After that, you turn it into a fraction, 28/64, and simplify to get  <math>\boxed{\textbf{(D)}~\frac{7}{16}}</math>.
 +
 
 +
==Solution 5 (Quick and Efficient)==
 +
<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));
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dot("$H$",H,dir(0));
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dot("$X$",X,dir(135/2));
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dot("$O$",O,dir(0));
 +
draw(E--O--X);
 +
</asy>
 +
We see that the midpoint of <math>\overline{AB}</math> is <math>X</math>. We can also notice that point <math>X</math> to point <math>E</math> *ALMOST* makes the octagon half shaded. Since we know an octagon has 8 sides, <math>X</math> is one-half of <math>\overline{AB}</math> which means instead of counting in eights, we need to count in sixteenths. Now back to the problem. Let's say that the midpoint of <math>\overline{EF}</math> is <math>Y</math>, then <math>\overline{XY}</math> cuts the octagon perfectly in half, but since there is no <math>Y</math>, we are now <math>\frac1{16}</math> away from one-half. We know that <math>\frac1{2}=\frac8{16}</math>, so we can just do <math>\frac8{16}-\frac1{16}</math> which gets us, for the final answer,  <math>\boxed{\textbf{(D)}~\frac{7}{16}}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Am24 AM24]
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!)==
 +
https://youtu.be/azFKEreETAw
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/NbIav9YlPEY
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/51K3uCzntWs?t=2314
 +
 
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
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{{AMC8 box|year=2015|num-b=1|num-a=3}}
 
{{AMC8 box|year=2015|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 19:32, 26 June 2025

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

[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]

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. After that, you turn it into a fraction, 28/64, and simplify to get $\boxed{\textbf{(D)}~\frac{7}{16}}$.

Solution 5 (Quick and Efficient)

[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] We see that the midpoint of $\overline{AB}$ is $X$. We can also notice that point $X$ to point $E$ *ALMOST* makes the octagon half shaded. Since we know an octagon has 8 sides, $X$ is one-half of $\overline{AB}$ which means instead of counting in eights, we need to count in sixteenths. Now back to the problem. Let's say that the midpoint of $\overline{EF}$ is $Y$, then $\overline{XY}$ cuts the octagon perfectly in half, but since there is no $Y$, we are now $\frac1{16}$ away from one-half. We know that $\frac1{2}=\frac8{16}$, so we can just do $\frac8{16}-\frac1{16}$ which gets us, for the final answer, $\boxed{\textbf{(D)}~\frac{7}{16}}$.

~AM24

Video Solution (HOW TO THINK CRITICALLY!!)

https://youtu.be/azFKEreETAw

~Education, the Study of Everything

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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