1999 AHSME Problems/Problem 23
Problem
The equiangular convex hexagon 
has 
and 
The area of the hexagon is
Solution
Solution 1
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 
.
Solution 2
![[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); label("$X$",(7,13),S); label("$Y$",(14,0),S); label("$Z$",(0,0),S); label("$A$",(6,10.392),W); label("$B$",(8,10.392),E); label("$C$",(12,3.464),E); label("$D$",(10,0),S); label("$1$",(7,12.124)--(8,10.392),NE); label("$1$",(6,10.392)--(8,10.392),S); label("$4$",(8,10.392)--(12,3.464),NE); label("$2$",(10,0)--(12,3.464),NW); label("$2$",(14,0)--(12,3.464),NE); [/asy]](http://latex.artofproblemsolving.com/e/1/2/e129c7fa1a0e8df1f8a8090963aaddf5d55b3d2b.png)
An equiangular (but not necessarily regular) hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners.
Labeling the triangle 
and 
and drawing 
of length one will remove one equilateral triangle of side length 
, and drawing 
will take out another equilateral triangle of side length 
.Labeling the other sides of the triangle, we can find that 
, or the side length of the equilateral triangle is 
.
See also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
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