Difference between revisions of "2015 AIME I Problems/Problem 4"
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==Solution== | ==Solution== | ||
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>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>. | ||
+ | ==Solution 2== | ||
+ | Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle <math>BMN</math> to solve for its area. | ||
==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 10:54, 23 March 2015
Contents
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.
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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