1994 AIME Problems/Problem 6

Revision as of 18:02, 16 October 2008 by Azjps (talk | contribs) (sol, asymptote needed)

Problem

The graphs of the equations

$y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,$

are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side $2/\sqrt{3}.\,$ How many such triangles are formed?

Solution

We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.


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Solving the above equations for $k=\pm 10$, we see that the hexagon in question is regular, with side length $\frac{20}{\sqrt{3}}$. Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just $\left(\frac{20/\sqrt{3}}{2\sqrt{3}}\right)^2 = 100$. Thus, the total number of unit triangles is $6 \times 100 = 600$.

There are $6 \cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = \boxed{660}$.

See also

1994 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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