Difference between revisions of "2018 AMC 10B Problems/Problem 24"
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(→Solution) |
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| Line 21: | Line 21: | ||
Y=(5/4, sqrt(3)/4); | Y=(5/4, sqrt(3)/4); | ||
Z=(-1/4, sqrt(3)/4); | Z=(-1/4, sqrt(3)/4); | ||
| + | M=(0,sqrt(3)/2); | ||
| + | N=(,); | ||
| + | O=(,); | ||
draw(A--B--C--D--E--F--cycle); | draw(A--B--C--D--E--F--cycle); | ||
| Line 26: | Line 29: | ||
draw(A--C--E--cycle); | draw(A--C--E--cycle); | ||
draw(X--Y--Z--cycle); | draw(X--Y--Z--cycle); | ||
| + | draw(M—N—O—cycle); | ||
label("$A$",A,NW); | label("$A$",A,NW); | ||
| Line 36: | Line 40: | ||
label("$Y$", Y, ESE); | label("$Y$", Y, ESE); | ||
label("$Z$", Z, WSW); | label("$Z$", Z, WSW); | ||
| + | label("$M$",M, W); | ||
| + | label("$N$", N, NE); | ||
| + | label("$O$", O, SE); | ||
</asy> | </asy> | ||
Revision as of 16:53, 16 February 2018
Problem
Let 
be a regular hexagon with side length 
. Denote 
, 
, and 
the midpoints of sides 
, 
, and 
, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of 
and 
?
Answer: 
Solution
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,sqrt(3));
B=(1,sqrt(3));
C=(3/2,sqrt(3)/2);
D=(1,0);
E=(0,0);
F=(-1/2,sqrt(3)/2);
X=(1/2, sqrt(3));
Y=(5/4, sqrt(3)/4);
Z=(-1/4, sqrt(3)/4);
M=(0,sqrt(3)/2);
N=(,);
O=(,);
draw(A--B--C--D--E--F--cycle);
draw(A--C--E--cycle);
draw(X--Y--Z--cycle);
draw(M—N—O—cycle);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,ESE);
label("$D$",D,SE);
label("$E$",E,SW);
label("$F$",F,WSW);
label("$X$", X, N);
label("$Y$", Y, ESE);
label("$Z$", Z, WSW);
label("$M$",M, W);
label("$N$", N, NE);
label("$O$", O, SE);
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See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 AMC 10 Problems and Solutions | ||
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
