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? | ||
Line 31: | Line 33: | ||
draw(E--O--X); | draw(E--O--X); | ||
</asy> | </asy> | ||
+ | ==Solutions== | ||
+ | ===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> | |
− | + | ===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 81: | Line 83: | ||
</asy> | </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>. | + | 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=== | ||
+ | |||
+ | 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 <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>. | ||
==See Also== | ==See Also== |
Latest revision as of 15:54, 16 January 2021
Problem
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?
Solutions
Solution 1
Since octagon is a regular octagon, it is split into equal parts, such as triangles , etc. These parts, since they are all equal, are of the octagon each. The shaded region consists of of these equal parts plus half of another, so the fraction of the octagon that is shaded is
Solution 2
The octagon has been divided up into identical triangles (and thus they each have equal area). Since the shaded region occupies out of the total triangles, the answer is .
Solution 3
For starters what I find helpful is to divide the whole octagon up into triangles as shown here:
Now it is just a matter of counting the larger triangles remember that and are not full triangles and are only half for these purposes. We count it up and we get a total of of the shape shaded. We then simplify it to get our answer of .
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.