Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 2)
 
(7 intermediate revisions by one other user not shown)
Line 64: Line 64:
 
label("$2$",(14,0)--(12,3.464),NE);
 
label("$2$",(14,0)--(12,3.464),NE);
 
label("$1$",(1,1.732)--(2,0),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("$2$",(6,10.392)--(7,12.124),NW);
 
−
label("$1$",(10,0)--(14,0),S);
 
−
</asy></center>
 
−
An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners.
 
−
Labeling the triangle <math>X Y</math> and <math>Z</math> and drawing <math>AB</math> of length one will remove one equilateral triangle of side length <math>1</math>, and drawing <math>CD</math> will take out another equilateral triangle of side length <math>2</math>.Labeling the other sides of the smaller equilateral triangles, we can find that <math>XY</math>, or the side length of the equilateral triangle is <math>7</math>. Now, because we know what the side length of the triangle is, what <math>DY</math> is, and it is given that <math>DE</math> is <math>4</math>, we can find the length of <math>EZ</math>, <math>7-4-2=1</math>. 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 <math>\frac{1^2\sqrt{3}}{4}\implies\frac{1\sqrt{3}}{4}</math>, and <math>\frac{2^2\sqrt{3}}{4}\implies\frac{4\sqrt{3}}{4}\implies\sqrt{3}</math> and the area of the large equilateral triangle is <math>\frac{7^2\sqrt{3}}{4}\implies\frac{49\sqrt{3}}{4}</math> so the area of the hexagon would be <math>\frac{49\sqrt{3}}{4}-\frac{\sqrt
 
−
{3}}{4}-\frac{\sqrt{3}}{4}-\sqrt{3}\implies\boxed{\frac{43\sqrt{3}}{4}}</math>
 
−
 
−
==Solution 2==
 
−
<center><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("$2$",(1,1.732)--(2,0),NE);
 
 
label("$1$",(0,0)--(2,0),S);
 
label("$1$",(0,0)--(2,0),S);
 
label("$1$",(0,0)--(1,1.732),NW);
 
label("$1$",(0,0)--(1,1.732),NW);
 
label("$1$",(6,10.392)--(7,12.124),NW);
 
label("$1$",(6,10.392)--(7,12.124),NW);
−
label("$1$",(10,0)--(14,0),S);
+
label("$2$",(10,0)--(14,0),S);
 
</asy></center>
 
</asy></center>
 
An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners.  
 
An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners.  

Latest revision as of 21:49, 5 August 2024

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("$2$",(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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_23&oldid=224558"