1992 AHSME Problems/Problem 9
Problem
![[asy] draw((-7,0)--(7,0),black+linewidth(.75)); draw((-3*sqrt(3),0)--(-2*sqrt(3),3)--(-sqrt(3),0)--(0,3)--(sqrt(3),0)--(2*sqrt(3),3)--(3*sqrt(3),0),black+linewidth(.75)); draw((-2*sqrt(3),0)--(-1*sqrt(3),3)--(0,0)--(sqrt(3),3)--(2*sqrt(3),0),black+linewidth(.75)); [/asy]](http://latex.artofproblemsolving.com/f/1/2/f124b9722fb0d38fc417e782945e0f8ef8523a10.png)
Five equilateral triangles, each with side 
, are arranged so they are all on the same side of a line containing one side of each vertex. Along this line, the midpoint of the base of one triangle is a vertex of the next. The area of the region of the plane that is covered by the union of the five triangular regions is
Solution
First, we calculate the area of 1 triangle. For an equilateral triangle with side s, its area is $\frac\{\sqrt{3}s^{2}}{4}$ (Error compiling LaTeX. Unknown error_msg). If the side of the equilateral triangle is $\2\sqrt{3}$ (Error compiling LaTeX. Unknown error_msg), the area of one such triangle is $\3\sqrt{3}$ (Error compiling LaTeX. Unknown error_msg).
There are 5 equilateral triangles in total, overlapping by an area of 4 smaller equilateral triangles. Each smaller triangle is one fourth as big as the big equilateral triangles. Therefore, we subtract the overlapping area, which is equivalent to the area of 1 big equilateral triangle.
Hence, the total area is equal to the area of 4 equilateral triangles, which is $\3\sqrt{3} * 4 = 12\sqrt{3}$ (Error compiling LaTeX. Unknown error_msg).
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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| All AHSME Problems and Solutions | ||
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