2018 AMC 10B Problems/Problem 24
Contents
[hide]Problem
Let be a regular hexagon with side length . Denote , , and the midpoints of sides , , and , respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of and ?
Answer:
Solution
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (AXFZ, XBCY, and ZYED), and 3 right triangles (With one vertice on each of X, Y, and Z). Now we know that one base of each trapezoid is just the side length of the hexagon which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal) with a height of (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of for a total area of (Alternatively, we could have calculated the area of hexagon ABCDEF and subtracted the area of triangle XYZ, which, as we showed before, had a side length of 3/2). Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertice on X, is similar to the triangle with a base of YC = 1/2. Using similar triangles we calculate the base to be 1/4 and the height to be giving us an area of per triangle, and a total area of . Adding the two areas together, we get . Finding the total area, we get . Taking the complement, we get , so the answer is C --- Arpitr20
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Problem
Let be a regular hexagon with side length . Denote , , and the midpoints of sides , , and , respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of and ?
Answer:
Solution
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (AXFZ, XBCY, and ZYED), and 3 right triangles (With one vertice on each of X, Y, and Z). Now we know that one base of each trapezoid is just the side length of the hexagon which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal) with a height of (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of for a total area of (Alternatively, we could have calculated the area of hexagon ABCDEF and subtracted the area of triangle XYZ, which, as we showed before, had a side length of 3/2). Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertice on X, is similar to the triangle with a base of YC = 1/2. Using similar triangles we calculate the base to be 1/4 and the height to be giving us an area of per triangle, and a total area of . Adding the two areas together, we get . Finding the total area, we get . Taking the complement, we get , so the answer is C --- Arpitr20
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.