Difference between revisions of "2015 AIME I Problems/Problem 4"
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I noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral: | I noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral: | ||
− | One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle </ | + | One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>. |
− | The other way is to use the Mean Geometry Theorem. Note that < | + | The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. (The weights are actually <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so they are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral. |
==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 21:59, 25 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 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 B 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 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 actually
and
, so they 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.