Difference between revisions of "2015 AIME I Problems/Problem 4"
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Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 18:52, 15 February 2021 (EST) | Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 18:52, 15 February 2021 (EST) | ||
− | ==Solution== | + | ==Solution 1== |
− | Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>. | + | Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula <math>\textbf{OR}</math> by Shoelace Theorem, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>. |
+ | |||
==Solution 2== | ==Solution 2== | ||
Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS. | Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS. |
Revision as of 22:48, 31 August 2021
Contents
[hide]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) 18:52, 15 February 2021 (EST)
Solution 1
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 by Shoelace Theorem, , so is .
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.