Difference between revisions of "1999 AHSME Problems/Problem 23"
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<math> \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}</math> | <math> \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}</math> | ||
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Equiangularity means that each internal angle must be exactly <math>120^\circ</math>. | Equiangularity means that each internal angle must be exactly <math>120^\circ</math>. | ||
The information given by the problem statement looks as follows: | The information given by the problem statement looks as follows: |
Revision as of 16:41, 10 February 2019
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.