Difference between revisions of "2016 AMC 12A Problems/Problem 17"
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Note that we can also use coordinates to solve this problem. WLOG, set the side length of square <math>ABCD</math> equal to <math>6</math>. | Note that we can also use coordinates to solve this problem. WLOG, set the side length of square <math>ABCD</math> equal to <math>6</math>. | ||
− | This makes the coordinates of the square <math>EFGH</math> equal to <math>(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),</math> and | + | This makes the coordinates of the square <math>EFGH</math> equal to <math>(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),</math> and <math>(3, -\sqrt{3})</math>. |
Using the first two points, this gives <math>EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12 | Using the first two points, this gives <math>EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12 |
Latest revision as of 17:47, 15 June 2020
Problem 17
Let be a square. Let and be the centers, respectively, of equilateral triangles with bases and each exterior to the square. What is the ratio of the area of square to the area of square ?
Solution
The center of an equilateral triangle is its centroid, where the three medians meet.
The distance along the median from the centroid to the base is one third the length of the median.
Let the side length of the square be . The height of is so the distance from to the midpoint of is
(from above) (side length of the square).
Since is the diagonal of square ,
Solution(Coordinates)
Note that we can also use coordinates to solve this problem. WLOG, set the side length of square equal to .
This makes the coordinates of the square equal to and .
Using the first two points, this gives .
Thus, .
Because the side length of is , .
Therefore,
- Pleaseletmewin
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.