1999 AHSME Problems/Problem 23
Problem
The equiangular convex hexagon 
has 
and 
The area of the hexagon is
Solution
Equiangularity means that each internal angle must be exactly 
.
The information given by the problem statement looks as follows:
![[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW)); dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label("$A$",O,SW); label("$B$",O+E,SE); label("$C$",O+E+4*NE,E); label("$D$",O+E+4*NE+2*NW,NE); label("$E$",O-3*E+4*NE+2*NW,NW); [/asy]](http://latex.artofproblemsolving.com/2/f/c/2fcaeaff120ed09924376ada47cd9c2603a51efc.png)
We can now place this incomplete polygon onto a triangular grid, finish it, compute its area in unit triangles, and multiply the result by the area of the unit triangle.
![[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW) -- (O-3*E+3*NE+2*NW) -- cycle, 0.8red+4bp); for (int i=-5; i<=1; ++i) { draw( (O+i*E-1.5*NE)--(O+i*E+6.5*NE), dashed ); } for (int i=-2; i<=5; ++i) { draw( (O+i*E-1.5*NW)--(O+i*E+6.5*NW), dashed ); } for (int i=-1; i<=6; ++i) { draw( (O-2.5*E+i*NW)--(O+5.5*E+i*NW), dashed ); } dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label("$A$",O,SW); label("$B$",O+E,SE); label("$C$",O+E+4*NE,E); label("$D$",O+E+4*NE+2*NW,NE); label("$E$",O-3*E+4*NE+2*NW,NW); label("$F$",(O-3*E+3*NE+2*NW),W); [/asy]](http://latex.artofproblemsolving.com/4/9/c/49c7f5471656c5757d32fd94b3554f8d4a6eb3b1.png)
We see that the figure contains 
unit triangles, and therefore its area is $\boxed{\frac{43\sqrt{3}}4}}$ (Error compiling LaTeX. Unknown error_msg).
See also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
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| All AHSME Problems and Solutions | ||
