2014 AMC 10B Problems/Problem 13
Problem
Six regular hexagons surround a regular hexagon of side length as shown. What is the area of ?
Solution 1
We note that the triangular sections in can be put together to form a hexagon congruent to each of the seven other hexagons. By the formula for the area of the hexagon, we get the area for each hexagon as . The area of , which is equivalent to two of these hexagons together, is .
Solution 2
The measure of an interior angle in a hexagon is 120 degrees. Each side of the 6 triangles that make up the remainder of triangle ABC are isosceles with 2 side lengths of 1 and an angle of 120 degrees. Therefore, by the Law of Cosines, we calculate that the longest side of this triangle is , so the side length of triangle ABC is . Using the equilateral triangle area formula, we figure out that the answer is . (note that it may not be so nice to use trigonometry in AMC10 contest, however, it is a more efficient way of solving those geometry question. ~Kai Gao)
Solution 3
We know the area of a triangle can be found through the formula K (Area) = r (inradius) s (semiperimeter). Since the hexagon fully enveloped inside the triangle touches all sides, we can visualize that hexagon as congruent equilateral triangles, each with side lengths . Draw a circle that circumscribes the hexagon. Using the equilateral triangles, we can see that the circle has a radius of . Since the circle touches all 3 sides of the triangle, we can say that is the inradius of . We can find the semiperimeter of by using the rule of triangles on the congruent triangles inside to find that the perimeter is . Thus, the semiperimeter is . Thus, the area of the triangle is ~NSAoPS
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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