1999 AHSME Problems/Problem 23

Revision as of 16:54, 17 April 2015 by TBNR (talk | contribs) (there was an extra "}" in the LaTeX)

Problem

The equiangular convex hexagon $ABCDEF$ has $AB = 1, BC = 4, CD = 2,$ and $DE = 4.$ The area of the hexagon is $\mathrm{(A) \ } \frac {15}2\sqrt{3} \qquad \mathrm{(B) \ }9\sqrt{3} \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }\frac{39}4\sqrt{3} \qquad \mathrm{(E) \ } \frac{43}4\sqrt{3}$

Solution

Equiangularity means that each internal angle must be exactly $120^\circ$. 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]

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]

We see that the figure contains $43$ unit triangles, and therefore its area is $\boxed{\frac{43\sqrt{3}}4}$.

See also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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