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Difference between revisions of "1999 AHSME Problems/Problem 23"

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We see that the figure contains <math>43</math> unit triangles, and therefore its area is <math>\boxed{\frac{43\sqrt{3}}4}</math>.
 
We see that the figure contains <math>43</math> unit triangles, and therefore its area is <math>\boxed{\frac{43\sqrt{3}}4}</math>.
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==Solution 2==
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=======Solution 2=======
 
<center><asy>
 
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unitsize(0.5cm);
 
unitsize(0.5cm);

Revision as of 16:37, 19 March 2020

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

Solution 1

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 $\boxed{\frac{43\sqrt{3}}4}$.

=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)); draw((1,1.732)--(2,0)); 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("$E$",(2,0),S); label("$F$",(1,1.732),W); 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); label("$1$",(1,1.732)--(2,0),NE); label("$1$",(0,0)--(2,0),S); label("$1$",(0,0)--(1,1.732),NW); label("$1$",(6,10.392)--(7,12.124),NW); label("$1$",(10,0)--(14,0),S); [/asy]

An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\frac{1^2\sqrt{3}}{4}\implies\frac{1\sqrt{3}}{4}$, and $\frac{2^2\sqrt{3}}{4}\implies\frac{4\sqrt{3}}{4}\implies\sqrt{3}$ and the area of the large equilateral triangle is $\frac{7^2\sqrt{3}}{4}\implies\frac{49\sqrt{3}}{4}$ so the area of the hexagon would be $\frac{49\sqrt{3}}{4}-\frac{\sqrt {3}}{4}-\frac{\sqrt{3}}{4}-\sqrt{3}\implies\boxed{\frac{43\sqrt{3}}{4}}$

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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