Difference between revisions of "1983 AHSME Problems/Problem 4"
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F = (1, 1.732); | F = (1, 1.732); | ||
draw(A--B--C--D--E--F--A); | draw(A--B--C--D--E--F--A); | ||
| − | label("A", A, | + | label("$A$", A, NW); |
| − | label("B", | + | label("$B$", B, 3W); |
| − | label("C", | + | label("$C$", C, SW); |
| − | label("D", D, | + | label("$D$", D, SE); |
| − | label("E", E, E); | + | label("$E$", E, E); |
| − | label("F", F, | + | label("$F$", F, NE); |
draw(B--D, dashed+linewidth(0.5)); | draw(B--D, dashed+linewidth(0.5)); | ||
draw(B--E, dashed+linewidth(0.5)); | draw(B--E, dashed+linewidth(0.5)); | ||
Revision as of 08:10, 2 September 2017
Problem 4
Position 
such that 
and 
are parallel, as are sides 
and 
,
and sides 
and 
. Each side has length of 
and it is given that 
.
The area of the figure is
Solution
![[asy] pair A, B, C, D, E, F; A = (0, 1.732); B = (0.5, 0.866); C = (0,0); D = (1, 0); E = (1.5, 0.866); F = (1, 1.732); draw(A--B--C--D--E--F--A); label("$A$", A, NW); label("$B$", B, 3W); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); label("$F$", F, NE); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]](http://latex.artofproblemsolving.com/b/d/d/bddb9e658c2a3f38841b9348a1aa73ffaeadbc5e.png)
Drawing the diagram as described, we create a convex hexagon with all side lengths equal to 1. In this case, it is a natural approach to divide the figure up into four equilateral triangles. The area, A, of one such equilateral triangle is 
, which gives a total of 
, or 
.
See Also
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
