Difference between revisions of "2014 AMC 10B Problems/Problem 13"
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==Solution 3== | ==Solution 3== | ||
− | We know the area of a triangle can be found through the formula K (Area) = r (inradius) <math>\cdot</math> s (semiperimeter). Since the hexagon fully enveloped inside the triangle touches all <math>3</math> sides, we can visualize that hexagon as <math>6</math> congruent equilateral triangles, each with side lengths <math>1</math>. Draw a circle that circumscribes the hexagon. Using the equilateral triangles, we can see that the circle has a radius of <math>1</math>. Since the circle touches all 3 sides of the triangle, we can say that <math>1</math> is the inradius of <math>\triangle{ABC}</math>. We can find the semiperimeter of <math>\triangle{ABC}</math> by using the <math>30-60-90</math> rule of triangles on the <math>6</math> congruent triangles inside <math>\triangle{ABC}</math> to find that the perimeter is <math>6\sqrt{3}</math>. Thus, the semiperimeter is <math>\dfrac{6\sqrt{3}}{2} = 3\sqrt{3}</math>. Thus, the area of the triangle is <math>1 \cdot 3\sqrt{3} = \boxed{\textbf{(B) | + | We know the area of a triangle can be found through the formula K (Area) = r (inradius) <math>\cdot</math> s (semiperimeter). Since the hexagon fully enveloped inside the triangle touches all <math>3</math> sides, we can visualize that hexagon as <math>6</math> congruent equilateral triangles, each with side lengths <math>1</math>. Draw a circle that circumscribes the hexagon. Using the equilateral triangles, we can see that the circle has a radius of <math>1</math>. Since the circle touches all 3 sides of the triangle, we can say that <math>1</math> is the inradius of <math>\triangle{ABC}</math>. We can find the semiperimeter of <math>\triangle{ABC}</math> by using the <math>30-60-90</math> rule of triangles on the <math>6</math> congruent triangles inside <math>\triangle{ABC}</math> to find that the perimeter is <math>6\sqrt{3}</math>. Thus, the semiperimeter is <math>\dfrac{6\sqrt{3}}{2} = 3\sqrt{3}</math>. Thus, the area of the triangle is <math>1 \cdot 3\sqrt{3} = \boxed{\textbf{(B)} 3\sqrt{3}}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2014|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:18, 10 September 2023
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
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 | ||
All AMC 10 Problems and Solutions |
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