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1996 AHSME Problems/Problem 19

Revision as of 19:38, 19 August 2011 by Talkinaway (talk | contribs)

Problem

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon?

[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir(origin--A)); label("$B$", B, dir(origin--B)); label("$C$", C, dir(origin--C)); label("$D$", D, dir(origin--D)); label("$E$", E, dir(origin--E)); label("$F$", F, dir(origin--F)); [/asy]

$\text{(A)}\ \frac{1}{2}\qquad\text{(B)}\ \frac{\sqrt 3}{3}\qquad\text{(C)}\ \frac{2}{3}\qquad\text{(D)}\ \frac{3}{4}\qquad\text{(E)}\ \frac{\sqrt 3}{2}$

Solution

[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); pair O=(0,0); dot(A^^B^^C^^D^^E^^F^^O); label("$A$", A, dir(origin--A)); label("$B$", B, dir(origin--B)); label("$C$", C, dir(origin--C)); label("$D$", D, dir(origin--D)); label("$E$", E, dir(origin--E)); label("$F$", F, dir(origin--F)); label("$O$", O, N); draw(O--D--C--cycle); [/asy]

$\triangle OCD$ is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside $\triangle OCD$ to the the area of $\triangle OCD$ itself, it will be the same ratio.

Let $OC=CD=OD =2$ so that the area of the triangle is $\frac{\sqrt{3}s^2}{4} = \sqrt{3}$.

Notice that $\triangle OCD$ is made up of a kite and two $30-60-90$ triangles. The two hypotenuses of these two triangles form $CD$, so the hypotenuse of each triangle must be $\frac{2}{2} = 1$. Thus, the legs of each triangle are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}}$ (Error compiling LaTeX. Unknown error_msg), and the area of two of these triangles is $\frac{1}{2}\cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$.

Subtracting the area of the two triangles from the area of the equilateral triangle, we find that the area of the kite is $\sqrt{3} - \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$.

Thus, the ratio of areas is $\frac{3}{4}$, which is option $\boxed{D}$.

Solution 2

We see six isosceles $120-30-30$ triangles at the vertices of the large hexagon. If we let the side of the larger hexagon be $1$, we find that the two congruent sides of these triangles are both $\frac{1}{2}$. By the Law of Sines, we find that the third side $x$ is:

$\frac{\frac{1}{2}}{\sin 30^\circ} = \frac{x}{\sin 120^\circ}$


$1 = \frac{x}{\frac{\sqrt{3}}{2}}$

$x = \frac{\sqrt{3}}{2}$

This third side of the triangle is the length of the edge of the hexagon. The ratio of the sides of the small hexagon to the large hexagon is $\frac{\sqrt{3}}{2}$, since the large hexagon has a unit side. The ratio of the areas is the square of the ratio of the sides, which is $\frac{3}{4}$, or option $\boxed{D}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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