Difference between revisions of "1996 AHSME Problems/Problem 19"
Talkinaway (talk | contribs) (Created page with "==See also== {{AHSME box|year=1996|num-b=18|num-a=20}}") |
Talkinaway (talk | contribs) |
||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
+ | The midpoints of the sides of a regular hexagon <math>ABCDEF</math> are joined to form a smaller hexagon. What fraction of the area of <math>ABCDEF</math> 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> | ||
+ | |||
+ | <math> \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} </math> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | <math>\triangle OCD</math> is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside <math>\triangle OCD</math> to the the area of <math>\triangle OCD</math> itself, it will be the same ratio. | ||
+ | |||
+ | Let <math>OC=CD=OD =2</math> so that the area of the triangle is <math>\frac{\sqrt{3}s^2}{4} = \sqrt{3}</math>. | ||
+ | |||
+ | Notice that <math>\triangle OCD</math> is made up of a kite and two <math>30-60-90</math> triangles. The two hypotenuses of these two triangles form <math>CD</math>, so the hypotenuse of each triangle must be <math>\frac{2}{2} = 1</math>. Thus, the legs of each triangle are <math>\frac{1}{2}</math> and <math>\frac{\sqrt{3}}{2}}</math>, and the area of two of these triangles is <math>\frac{1}{2}\cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}</math>. | ||
+ | |||
+ | Subtracting the area of the two triangles from the area of the equilateral triangle, we find that the area of the kite is <math>\sqrt{3} - \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}</math>. | ||
+ | |||
+ | Thus, the ratio of areas is <math>\frac{3}{4}</math>, which is option <math>\boxed{D}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We see six isosceles <math>120-30-30</math> triangles at the vertices of the large hexagon. If we let the side of the larger hexagon be <math>1</math>, we find that the two congruent sides of these triangles are both <math>\frac{1}{2}</math>. By the [[Law of Sines]], we find that the third side <math>x</math> is: | ||
+ | |||
+ | <math>\frac{\frac{1}{2}}{\sin 30^\circ} = \frac{x}{\sin 120^\circ}</math> | ||
+ | |||
+ | |||
+ | <math>1 = \frac{x}{\frac{\sqrt{3}}{2}}</math> | ||
+ | |||
+ | <math>x = \frac{\sqrt{3}}{2}</math> | ||
+ | |||
+ | 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 <math>\frac{\sqrt{3}}{2}</math>, since the large hexagon has a unit side. The ratio of the areas is the square of the ratio of the sides, which is <math>\frac{3}{4}</math>, or option <math>\boxed{D}</math>. | ||
+ | |||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=18|num-a=20}} | {{AHSME box|year=1996|num-b=18|num-a=20}} |
Revision as of 19:38, 19 August 2011
Contents
Problem
The midpoints of the sides of a regular hexagon are joined to form a smaller hexagon. What fraction of the area of is enclosed by the smaller hexagon?
Solution
is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside to the the area of itself, it will be the same ratio.
Let so that the area of the triangle is .
Notice that is made up of a kite and two triangles. The two hypotenuses of these two triangles form , so the hypotenuse of each triangle must be . Thus, the legs of each triangle are and $\frac{\sqrt{3}}{2}}$ (Error compiling LaTeX. ! Extra }, or forgotten $.), and the area of two of these triangles is .
Subtracting the area of the two triangles from the area of the equilateral triangle, we find that the area of the kite is .
Thus, the ratio of areas is , which is option .
Solution 2
We see six isosceles triangles at the vertices of the large hexagon. If we let the side of the larger hexagon be , we find that the two congruent sides of these triangles are both . By the Law of Sines, we find that the third side is:
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 , since the large hexagon has a unit side. The ratio of the areas is the square of the ratio of the sides, which is , or option .
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 |