Difference between revisions of "2014 AMC 10B Problems/Problem 13"
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− | We note that the <math>6</math> triangular sections in <math>\triangle{ABC}</math> 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 <math>\dfrac{ | + | <math> \textbf {(A) } 2\sqrt{3} \qquad \textbf {(B) } 3\sqrt{3} \qquad \textbf {(C) } 1+3\sqrt{2} \qquad \textbf {(D) } 2+2\sqrt{3} \qquad \textbf {(E) } 3+2\sqrt{3} </math> |
+ | [[Category: Introductory Geometry Problems]] | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | We note that the <math>6</math> triangular sections in <math>\triangle{ABC}</math> 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 <math>\dfrac{3\sqrt{3}}{2}</math>. The area of <math>\triangle{ABC}</math>, which is equivalent to two of these hexagons together, is <math>\boxed{\textbf{(B)} 3\sqrt{3}}</math>. | ||
+ | |||
+ | ==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 <math>\sqrt{3}</math>, so the side length of triangle ABC is <math>2\sqrt{3}</math>. Using the equilateral triangle area formula, we figure out that the answer is <math>\boxed{\textbf{(B)} 3\sqrt{3}}</math>. | ||
+ | (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 <math>\text{Area = inradius} \cdot \text{semiperimeter}</math>. | ||
+ | |||
+ | As 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 applying 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> | ||
+ | ~NSAoPS | ||
==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}} |
Latest revision as of 18:08, 19 December 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 .
As 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 applying 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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.