Difference between revisions of "2015 AMC 8 Problems/Problem 2"
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− | ==Solution== | + | ==Solution 1== |
Since octagon <math>ABCDEFGH</math> is a regular octagon, it is split into 8 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 3 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{\text{(D) }\dfrac{7}{16}}.</math> | Since octagon <math>ABCDEFGH</math> is a regular octagon, it is split into 8 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 3 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{\text{(D) }\dfrac{7}{16}}.</math> | ||
+ | |||
+ | ==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("<math>A</math>",A,dir(45)); | ||
+ | dot("<math>B</math>",B,dir(90)); | ||
+ | dot("<math>C</math>",C,dir(135)); | ||
+ | dot("<math>D</math>",D,dir(180)); | ||
+ | dot("<math>E</math>",E,dir(-135)); | ||
+ | dot("<math>F</math>",F,dir(-90)); | ||
+ | dot("<math>G</math>",G,dir(-45)); | ||
+ | dot("<math>H</math>",H,dir(0)); | ||
+ | dot("<math>X</math>",X,dir(135/2)); | ||
+ | dot("<math>O</math>",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 <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 18:50, 25 November 2015
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?
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.