2015 AIME I Problems/Problem 4
Problem
Point lies on line segment with and . Points and lie on the same side of line forming equilateral triangles and . Let be the midpoint of , and be the midpoint of . The area of is . Find .
Solution
Let point be at . Then, is at , and is at . Due to symmetry, it is allowed to assume and are in quadrant 1. By equilateral triangle calculations, Point is at , and Point is at . By Midpoint Formula, is at , and is at . The distance formula shows that . Therefore, by equilateral triangle area formula, , so is .
Solution 2
Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle to solve for its area.
Solution 3
Note that and . Also, . Thus, by SAS.
From this, it is clear that a rotation about will map to . This rotation also maps to . Thus, and . Thus, is equilateral.
Using the Law of Cosines on , Thus, .
Using Stewart's Theorem on ,
Calculating the area of , Thus, , so . Our final answer is .
Admittedly, this is much more tedious than the coordinate solutions.
I also noticed that there are two more ways of showing that is equilateral:
One way is to show that , , and are related by a spiral similarity centered at .
The other way is to use the Mean Geometry Theorem. Note that and are similar and have the same orientation. Note that is the weighted average of and , is the weighted average of and , and is the weighted average of and . The weights are the same for all three averages. (The weights are actually just and , so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, is similar to both and , which means that is equilateral.
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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