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 .
Diagram
Diagram by RedFireTruck (talk)
Solution 1
Let be the origin, so and Using equilateral triangle properties tells us that and as well. Therefore, and Applying the Shoelace Theorem to triangle gives
so
Solution 2
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.
Note: A much easier way to go about finding without having to use Stewart's Theorem is to simply drop the altitudes from M and E to AC, thus hitting AC at points X and Y. Then clearly AEY and AMX are similar with ratio 2. But we know that . Additionally, from similar triangles meaning we can now just do pythagorean theorem on right triangle to get - SuperJJ
Solution 3
Medians are equal, so
is equilateral triangle.
The height of is distance from to midpoint is
is the median of
The area of
vladimir.shelomovskii@gmail.com, vvsss
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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