Difference between revisions of "2015 AIME I Problems/Problem 4"
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==Problem== | ==Problem== | ||
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>. | Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>. | ||
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+ | ==Solution== | ||
+ | Let point A be at (0,0). Then, B is at (16,0), and C is at (20,0). Due to symmetry, it is allowed to assume D and E are in quadrant 1. By equilateral triangle calculations, Point D is at (8,<math>8\sqrt{3}</math>), and Point E is at (18,<math>2\sqrt{3}</math>). By Midpoint Formula, M is at (9,<math>\sqrt{3}</math>), and N is at (14,<math>4\sqrt{3}</math>). Distance formula shows that BM=BN=MN=<math>2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, x=13<math>\sqrt{3}</math>, so <math>x^2</math> is 507. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=3|num-a=5}} | {{AIME box|year=2015|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:58, 20 March 2015
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 A be at (0,0). Then, B is at (16,0), and C is at (20,0). Due to symmetry, it is allowed to assume D and E are in quadrant 1. By equilateral triangle calculations, Point D is at (8,), and Point E is at (18,). By Midpoint Formula, M is at (9,), and N is at (14,). Distance formula shows that BM=BN=MN=. Therefore, by equilateral triangle area formula, x=13, so is 507.
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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