Difference between revisions of "2008 AMC 12B Problems/Problem 15"
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− | The equilateral triangles form trapezoids with side lengths <math> 1, 1, 1, 2</math> (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths <math> 1</math> and an angle of <math> 30^{\circ}</math> in between them, so the total area of these triangles (which is the area of <math> S | + | The equilateral triangles form trapezoids with side lengths <math> 1, 1, 1, 2</math> (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths <math> 1</math> and an angle of <math> 30^{\circ}</math> in between them, so the total area of these triangles (which is the area of <math> S - R</math>) is, by the [[Law of Sines]], <math> 4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1</math> which makes the answer <math> \boxed{C}</math>. |
==See Also== | ==See Also== |
Revision as of 22:53, 13 March 2015
Problem
On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let be the region formed by the union of the square and all the triangles, and be the smallest convex polygon that contains . What is the area of the region that is inside but outside ?
Solution
The equilateral triangles form trapezoids with side lengths (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths and an angle of in between them, so the total area of these triangles (which is the area of ) is, by the Law of Sines, which makes the answer .
See Also
Region with Squares and Triangles
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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